Manual Solutions to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
MODERN CONTROL SYSTEMS SOLUTION MANUAL
Richard C. Dorf
University of California, Davis
Robert H. Bishop
The University of Texas at Austin
A companion to MODERN CONTROL SYSTEMS ELEVENTH EDITION Richard C. Dorf Robert H. Bishop
Prentice Hall Upper Saddle River, NJ 07458
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
c 2008 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. The author and publisher of this book have used their best eorts in preparing this book. These eorts include the development, research, and testing of the theories and programs to determine their eectiveness. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of the furnishing, performance, or use of these programs.
A Camera-ready copy for this manual was prepared by the author using L TEX 2 .
MATLAB is a registered trademark of The MathWorks, Inc. 24 Prime Park Way, Natick, MA 01760-1520. Phone. [phone], Fax: [phone] Email: [e-mail] LabVIEW MathScript is a registered trademark of the National Instruments Corporation 11500 N Mopac Expwy, Austin, TX 78759-3504 Phone. [phone], Fax: [phone] E-mail: [e-mail] Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN 0-13-227029-3 Pearson Education Limited (UK) Pearson Education Australia Pty Ltd Prentice Hall Canada Ltd Pearson Educacion de Mexico, S.A. de C.V. Pearson Education Japan KK Pearson Education China Ltd
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
PREFACE
In each chapter, there are ve problem types: Exercises Problems Advanced Problems Design Problems/Continuous Design Problem Computer Problems In total, there are over 850 problems. The abundance of problems of increasing complexity gives students condence in their problem-solving ability as they work their way from the exercises to the design and computer-based problems. It is assumed that instructors (and students) have access to MATLAB, the Control System Toolbox or the LabVIEW and MathScript. All of the comptuer solutions in this Solution Manual were developed and tested on a Window XP platform using MATLAB 7.3 Release 2006b and the Control System Toolbox Version 7.1 and LabVIEW 8.2. It is not possible to verify each solution on all the available computer platforms that are compatible with MATLAB and LabVIEW MathScript. Please forward any incompatibilities you encounter with the scripts to Prof. Bishop at the email address given below. The authors and the sta at Prentice Hall would like to establish an open line of communication with the instructors using Modern Control Systems. We encourage you to contact Prentice Hall with comments and suggestions for this and future editions. Robert H. Bishop [e-mail]
iii
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
TABLE-OF-CONTENTS
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
Introduction to Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Mathematical Models of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 State Variable Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Feedback Control System Characteristics . . . . . . . . . . . . . . . . . . . . . . . 126 The Performance of Feedback Control Systems . . . . . . . . . . . . . . . . . 166 The Stability of Linear Feedback Systems . . . . . . . . . . . . . . . . . . . . . . 216 The Root Locus Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Frequency Response Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Stability in the Frequency Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 The Design of Feedback Control Systems . . . . . . . . . . . . . . . . . . . . . . . 492 The Design of State Variable Feedback Systems . . . . . . . . . . . . . . . . 574 Robust Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633 Digital Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691
iv
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
1
Introduction to Control Systems
There are, in general, no unique solutions to the following exercises and problems. Other equally valid block diagrams may be submitted by the student.
Exercises
E1.1 A microprocessor controlled laser system:
Controller Process
Desired power output
Error
power
Microprocessor
Current i(t)
Laser
Measurement
Power out
Measured
Power Sensor
E1.2
A driver controlled cruise control system:
Controller Process
Foot pedal Desired speed
-
Driver
Measurement
Car and Engine
Actual auto speed
Visual indication of speed
Speedometer
E1.3
Although the principle of conservation of momentum explains much of the process of y-casting, there does not exist a comprehensive scientic explanation of how a y-sher uses the small backward and forward motion of the y rod to cast an almost weightless y lure long distances (the 1
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
2
CHAPTER 1
Introduction to Control Systems
current world-record is 236 ft). The y lure is attached to a short invisible leader about 15-ft long, which is in turn attached to a longer and thicker Dacron line. The objective is cast the y lure to a distant spot with deadeye accuracy so that the thicker part of the line touches the water rst and then the y gently settles on the water just as an insect might.
Fly- sher
Desired position of the y Wind disturbance
Controller
Process
-
Mind and body of the y- sher
Rod, line, and cast
Actual
position of the y
Measurement
Visual indication of the position of the y
Vision of the y- sher
E1.4
An autofocus camera control system:
One-way trip time for the beam
K1
Beam Emitter/ Receiver Beam return Subject Lens focusing motor
Conversion factor (speed of light or sound) Distance to subject
Lens
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
3
E1.5
Tacking a sailboat as the wind shifts:
Controller Actuators Wind Process
Desired sailboat direction
Error
-
Sailor
Rudder and sail adjustment
Sailboat
Actual sailboat direction
Measurement Measured sailboat direction
Gyro compass
E1.6
An automated highway control system merging two lanes of trac:
Controller Actuators Process
Desired gap
Error
-
Embedded computer
Brakes, gas or steering
Active vehicle
Actual gap
Measurement Measured gap
Radar
E1.7
Using the speedometer, the driver calculates the dierence between the measured speed and the desired speed. The driver throotle knob or the brakes as necessary to adjust the speed. If the current speed is not too much over the desired speed, the driver may let friction and gravity slow the motorcycle down.
Controller Actuators Process
Desired speed
Error
-
Driver
Throttle or brakes
Motorcycle
Actual motorcycle speed
Measurement Visual indication of speed
Speedometer
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
4 E1.8
CHAPTER 1
Introduction to Control Systems
Human biofeedback control system:
Controller Process Message to blood vessels
Desired body temp
-
Hypothalumus
Human body
Actual body temp
Measurement
Visual indication of body temperature
TV display
Body sensor
E1.9
E-enabled aircraft with ground-based ight path control:
Corrections to the ight path Controller Aircraft
Desired Flight Path
Optimal ight path
Gc(s)
Meteorological data
G(s)
Health Parameters Location and speed
Flight Path
Ground-Based Computer Network
Optimal ight path Meteorological data Health Parameters Location and speed
Desired Flight Path
E1.10
Unmanned aerial vehicle used for crop monitoring in an autonomous mode:
Trajectory error Controller UAV
Specified Flight Trajectory
Location with respect to the ground
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Corrections to the ight path
Gc(s)
Controller
G(s)
Aircraft
Flight Path
Gc(s)
G(s)
Flight Trajectory
Map Correlation Algorithm
Ground photo
Sensor
Camera
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
5
E1.11
An inverted pendulum control system using an optical encoder to measure the angle of the pendulum and a motor producing a control torque:
Actuator Process
Desired angle
Error
Voltage
Torque
-
Controller
Motor
Pendulum
Angle
Measurement
Measured angle
Optical encoder
E1.12
In the video game, the player can serve as both the controller and the sensor. The objective of the game might be to drive a car along a prescribed path. The player controls the car trajectory using the joystick using the visual queues from the game displayed on the computer monitor.
Controller Actuator Process
Desired game objective
Error
-
Player
Joystick
Video game
Game objective
Measurement
Player
(eyesight, tactile, etc.)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
6
CHAPTER 1
Introduction to Control Systems
Problems
P1.1 An automobile interior cabin temperature control system block diagram:
Controller
Process
Desired temperature set by the driver
Error
-
Thermostat and air conditioning unit
Measurement
Automobile cabin
Automobile cabin temperature
Measured temperature
Temperature sensor
P1.2
A human operator controlled valve system:
Controller Process
Desired uid output *
Error *
Visual indication of uid output *
Valve
Tank
Fluid output
Measurement
Meter
* = operator functions
P1.3
A chemical composition control block diagram:
Controller Process
Error Desired chemical composition
-
Valve
Mixer tube
Chemical composition
Measurement Measured chemical composition
Infrared analyzer
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
7
P1.4
A nuclear reactor control block diagram:
Controller Process
Error Desired power level
-
Motor and ampli er
Reactor and rods
Output power level
Measurement Measured chemical composition
Ionization chamber
P1.5
Measurement
A light seeking control system to track the sun:
Controller
Light source
Dual Photocells
Ligh intensity
Trajectory Planner
Desired carriage position
Controller
Error
Motor inputs
Process
-
K
Motor, carriage, and gears
Photocell carriage position
P1.6
If you assume that increasing workers wages results in increased prices, then by delaying or falsifying cost-of-living data you could reduce or eliminate the pressure to increase workers wages, thus stabilizing prices. This would work only if there were no other factors forcing the cost-of-living up. Government price and wage economic guidelines would take the place of additional controllers in the block diagram, as shown in the block diagram.
Process Market-based prices Controller
Initial wages
-
Industry
Government price guidelines
Prices
Controller
Wage increases
Government wage guidelines
Cost-of-living
K1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
8 P1.7
CHAPTER 1
Introduction to Control Systems
Assume that the cannon res initially at exactly 5:00 p.m.. We have a positive feedback system. Denote by t the time lost per day, and the net time error by ET . Then the follwoing relationships hold: t = 4/3 min. + 3 min. = 13/3 min. and ET = 12 days 13/3 min./day . Therefore, the net time error after 15 days is ET = 52 min.
P1.8
The student-teacher learning process:
Controller Process
Error Desired knowledge
Lectures
-
Teacher
Knowledge
Student
Measurement
Measured knowledge
Exams
P1.9
A human arm control system:
Controller u e Process s
Desired arm location
y
Brain
z
Nerve signals
Arm & muscles
d
Arm location
Measurement Pressure Visual indication of arm location
Eyes and pressure receptors
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
9
P1.10
An aircraft ight path control system using GPS:
Controller Actuators Process
Desired ight path from air tra c controllers
Error
-
Computer Auto-pilot
Ailerons, elevators, rudder, and engine power
Measurement
Aircraft
Flight path
Measured ight path
Global Positioning System
P1.11
The accuracy of the clock is dependent upon a constant ow from the orice; the ow is dependent upon the height of the water in the oat tank. The height of the water is controlled by the oat. The control system controls only the height of the water. Any errors due to enlargement of the orice or evaporation of the water in the lower tank is not accounted for. The control system can be seen as:
Controller Process
Desired height of the water in oat tank
-
Float level
Flow from upper tank to oat tank
Actual
height
P1.12
Assume that the turret and fantail are at 90 , if w = F -90 . The fantail operates on the error signal w - T , and as the fantail turns, it drives the turret to turn.
y
Wind
qW = Wind angle qF = Fantail angle qT = Turret angle
Error
Controller Torque
Process
Fantail
*
qF
qW qT
qW
*
Turret
-
Fantail
Gears & turret
qT
x
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
10 P1.13
CHAPTER 1
Introduction to Control Systems
This scheme assumes the person adjusts the hot water for temperature control, and then adjusts the cold water for ow rate control.
Controller Process
Desired water temperature
Error
-
Valve adjust
Hot water system
Hot water
Actual water temperature and ow rate Desired water ow rate
-
Valve adjust
Cold water system
Cold water
Measurement
Measured water ow Measured water temperature
Human: visual and touch
P1.14
If the rewards in a specic trade is greater than the average reward, there is a positive inux of workers, since q (t) = f1 (c(t) r (t)). If an inux of workers occurs, then reward in specic trade decreases, since c(t) = f2 (q (t)).
Controller Process
Average rewards r(t)
Error
-
f1(c(t)-r(t))
q(t)
-
f2(q(t))
Total of rewards c(t)
P1.15
A computer controlled fuel injection system:
Controller Process
Desired Fuel Pressure
-
Electronic Control Unit
Measurement Measured fuel pressure
High Pressure Fuel Supply Pump and Electronic Fuel Injectors
Fuel Pressure
Fuel Pressure Sensor
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
11
P1.16
With the onset of a fever, the body thermostat is turned up. The body adjusts by shivering and less blood ows to the skin surface. Aspirin acts to lowers the thermal set-point in the brain.
Controller Process
Desired temperature or set-point from body thermostat in the brain
-
Adjustments within the body
Body
Body temperature
Measurement Measured body temperature
Internal sensor
P1.17
Hitting a baseball is arguably one of the most dicult feats in all of sports. Given that pitchers may throw the ball at speeds of 90 mph (or higher!), batters have only about 0.1 second to make the decision to swingwith bat speeds aproaching 90 mph. The key to hitting a baseball a long distance is to make contact with the ball with a high bat velocity. This is more important than the bats weight, which is usually around 33 ounces (compared to Ty Cobbs bat which was 41 ounces!). Since the pitcher can throw a variety of pitches (fast ball, curve ball, slider, etc.), a batter must decide if the ball is going to enter the strike zone and if possible, decide the type of pitch. The batter uses his/her vision as the sensor in the feedback loop. A high degree of eye-hand coordination is key to successthat is, an accurate feedback control system. Dene the following variables: p = output pressure, fs = spring force = Kx, fd = diaphragm force = Ap, and fv = valve force = fs - fd . The motion of the valve is described by y = fv /m where m is the valve mass. The output pressure is proportional to the valve displacement, thus p = cy , where c is the constant of proportionality.
P1.18
Spring
Constant of proportionality
Screw displacement x(t)
K
fs
-
fv
Valve position
Valve
y
c
Output pressure p(t)
Diaphragm area
fd
A
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
12 P1.19
CHAPTER 1
Introduction to Control Systems
A control system to keep a car at a given relative position oset from a lead car:
Follower car
Position of follower
Throttle
Fuel throttle (fuel)
Lead car Video camera & processing algorithms
Position of lead
-
Actuator
u
Controller
-
Relative position
Reference photo
Desired relative position
P1.20
A control system for a high-performance car with an adjustable wing:
Process Road conditions
Controller
Actuator
Desired road adhesion
-
Computer
Adjustable wing
Race Car
Road adhesion
Measurement
Measured road adhesion
K
Tire internal strain gauges
P1.21
A control system for a twin-lift helicopter system:
Measurement Measured separation distance
Radar
Desired separation distance Desired altitude
Controller
Process Separation distance
Pilot
Helicopter
Altitude
Measurement Measured altitude
Altimeter
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
13
P1.22
The desired building deection would not necessarily be zero. Rather it would be prescribed so that the building is allowed moderate movement up to a point, and then active control is applied if the movement is larger than some predetermined amount.
Process Controller
Desired de ection
-
Hydraulic sti eners
Building
De ection
Measurement
Measured de ection
K
Strain gauges on truss structure
P1.23
The human-like face of the robot might have micro-actuators placed at strategic points on the interior of the malleable facial structure. Cooperative control of the micro-actuators would then enable the robot to achieve various facial expressions.
Controller Process Voltage
Error Desired actuator position
-
Ampli er
Electromechanical actuator
Actuator position
Measurement
Measured position
Position sensor
P1.24
We might envision a sensor embedded in a gutter at the base of the windshield which measures water levelshigher water levels corresponds to higher intensity rain. This information would be used to modulate the wiper blade speed.
Process
Controller
Desired wiper speed
-
Electronic Control Unit
Measurement
Wiper blade and motor
Wiper blade speed
K
Measured water level
Water depth sensor
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
14 P1.25
CHAPTER 1
Introduction to Control Systems
A feedback control system for the space trac control:
Controller Actuator Jet commands Process Applied forces
Desired orbit position
Error
-
Control law
Reaction control jets
Satellite
Actual
orbit position
Measurement Measured orbit position
Radar or GPS
P1.26
Earth-based control of a microrover to point the camera:
Microrover
Receiver/ Transmitter
Camera position command
Controller
Gc(s)
G(s)
Rover position
Camera
Camera Position
m Ca ap er
Measured camera position
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
M ea d re su
Sensor
iti os on m m co ap er d an iti os on
P1.27
Desired Charge Level
m ca
Control of a methanol fuel cell:
Controller Recharging System Fuel Cell
Methanol water solution
-
Gc(s)
GR(s)
Sensor
G(s)
Charge Level
Measured charge level
H(s)
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
15
Advanced Problems
AP1.1 Control of a robotic microsurgical device:
Controller Desired End-e ector Position
Microsurgical robotic manipulator
-
Gc(s)
Sensor
G(s)
End-e ector Position
H(s)
AP1.2
An advanced wind energy system viewed as a mechatronic system:
AERODYNAMIC DESIGN STRUCTURAL DESIGN OF THE TOWER ELECTRICAL AND POWER SYSTEMS
SENSORS Rotor rotational sensor Wind speed and direction sensor ACTUATORS Motors for manipulatiing the propeller pitch
Physical System Modeling
CONTROL SYSTEM DESIGN AND ANALYSIS ELECTRICAL SYSTEM DESIGN AND ANALYSIS POWER GENERATION AND STORAGE
Sensors and Actuators WIND ENERGY SYSTEM Signals and Systems
Software and Data Acquisition
Computers and Logic Systems
CONTROLLER ALGORITHMS DATA ACQUISTION: WIND SPEED AND DIRECTION ROTOR ANGULAR SPEED PROPELLOR PITCH ANGLE
COMPUTER EQUIPMENT FOR CONTROLLING THE SYSTEM SAFETY MONITORING SYSTEMS
AP1.3
The automatic parallel parking system might use multiple ultrasound sensors to measure distances to the parked automobiles and the curb. The sensor measurements would be processed by an on-board computer to determine the steering wheel, accelerator, and brake inputs to avoid collision and to properly align the vehicle in the desired space.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
16
CHAPTER 1
Introduction to Control Systems
Even though the sensors may accurately measure the distance between the two parked vehicles, there will be a problem if the available space is not big enough to accommodate the parking car.
Controller Actuators Process
Desired automobile position
Error
-
On-board computer
Steering wheel, accelerator, and brake
Measurement
Automobile
Actual automobile position
Position of automobile relative to parked cars and curb
Ultrasound
AP1.4
There are various control methods that can be considered, including placing the controller in the feedforward loop (as in Figure 1.3). The adaptive optics block diagram below shows the controller in the feedback loop, as an alternative control system architecture.
Process
Astronomical object Uncompensated image
Astronomical telescope mirror
Measurement
Compensated image
Wavefront corrector
Actuator & controller
Wavefront reconstructor
Wavefront sensor
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
17
Design Problems
CDP1.1
The machine tool with the movable table in a feedback control conguration:
Controller Actuator Process
Desired position x
Error
-
Ampli er
Positioning motor
Machine tool with table
Actual position x
Measurement
Measured position
Position sensor
DP1.1
Use the stereo system and ampliers to cancel out the noise by emitting signals 180 out of phase with the noise.
Controller Noise signal Process
Desired noise = 0
-
Shift phase by 180 deg
Positioning motor
Machine tool with table
Noise in cabin
Measurement
Microphone
DP1.2
An automobile cruise control system:
Desired shaft speed Controller Process
Desired speed of auto set by driver
1/K
-
Electric motor
Valve
Automobile and engine
K
Actual speed of auto
Measurement
Measured shaft speed
Shaft speed sensor
Drive shaf t speed
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
18 DP1.3
CHAPTER 1
Introduction to Control Systems
An automoted cow milking system:
Measurement Cow location
Vision system
Controller Actuator Process
Desired cup location
-
Motor and gears
Robot arm and cup gripper
Location of cup
Cow and milker
Milk
Measurement
Measured cup location
Vision system
DP1.4
A feedback control system for a robot welder:
Controller Process Voltage
Desired position
Error
-
Computer and ampli er
Motor and arm
Weld top position
Measurement
Measured position
Vision camera
DP1.5
A control system for one wheel of a traction control system:
Engine torque
Antislip controller Wheel dynamics
Wheel speed
+
+
-
-
Sensor
+ +
Actual slip
1/Rw
Rw = Radius of wheel
Sensor
Measured slip
Vehicle dynamics
Vehicle speed
Brake torque
Antiskid controller
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
19
DP1.6
A vibration damping system for the Hubble Space Telescope:
Controller Actuators Process Signal to cancel the jitter
Desired jitter = 0
Error
-
Computer
Gyro and reaction wheels
Spacecraft dynamics
Jitter of vibration
Measurement
Measurement of 0.05 Hz jitter
Rate gyro sensor
DP1.7
A control system for a nanorobot:
Controller Actuators Process
Desired nanorobot position
Error
-
Biocomputer
Plane surfaces and propellers
Nanorobot
Actual nanorobot position
Measurement
External beacons
Many concepts from underwater robotics can be applied to nanorobotics within the bloodstream. For example, plane surfaces and propellers can provide the required actuation with screw drives providing the propulsion. The nanorobots can use signals from beacons located outside the skin as sensors to determine their position. The nanorobots use energy from the chemical reaction of oxygen and glucose available in the human body. The control system requires a bio-computeran innovation that is not yet available. For further reading, see A. Cavalcanti, L. Rosen, L. C. Kretly, M. Rosenfeld, and S. Einav, Nanorobotic Challenges n Biomedical Application, Design, and Control, IEEE ICECS Intl Conf. on Electronics, Circuits and Systems, Tel-Aviv, Israel, December 2004.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
2
Mathematical Models of Systems
Exercises
E2.1 We have for the open-loop y = r2 and for the closed-loop e = r y and y = e2 . So, e = r e2 and e2 + e r = 0 .
16
14
12
10
8
y
6
open-loop
4
2
closed-loop
0
0
0.5
1
1.5
2 r
2.5
3
3.5
4
FIGURE E2.1 Plot of open-loop versus closed-loop.
For example, if r = 1, then e2 + e 1 = 0 implies that e = 0.618. Thus, y = 0.382. A plot y versus r is shown in Figure E2.1. 20
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
21
E2.2
Dene f (T ) = R = R0 e0.1T and R = f (T ) f (T0 ) , T = T T0 . Then, R = f (T ) f (T0 ) = where f T
T =T0 =20
f T
T =T0 =20
T +
= 0.1R0 e0.1T0 = 135,
when R0 = 10, 000. Thus, the linear approximation is computed by considering only the rst-order terms in the Taylor series expansion, and is given by R = 135T . E2.3 The spring constant for the equilibrium point is found graphically by estimating the slope of a line tangent to the force versus displacement curve at the point y = 0.5cm, see Figure E2.3. The slope of the line is K 1.
2 1.5 1 0.5 0 Spring breaks
Force (n)
-0.5 -1 -1.5 -2 -2.5 -3 -2 -1.5 Spring compresses -1 -0.5 0 0.5 1 1.5 2 2.5 3
y=Displacement (cm)
FIGURE E2.3 Spring force as a function of displacement.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
22 E2.4
CHAPTER 2
Mathematical Models of Systems
Since R(s) = we have Y (s) = 4(s + 50) . s(s + 20)(s + 10) 1 s
The partial fraction expansion of Y (s) is given by Y (s) = where A1 = 1 , A2 = 0.6 and A3 = 1.6 . Using the Laplace transform table, we nd that y (t) = 1 + 0.6e20t 1.6e10t . The nal value is computed using the nal value theorem: lim y (t) = lim s
s0
A1 A2 A3 + + s s + 20 s + 10
t
s(s2
4(s + 50) =1. + 30s + 200)
E2.5
The circuit diagram is shown in Figure E2.5.
R2
v+
A +
vin -
+ v0 -
R1
FIGURE E2.5 Noninverting op-amp circuit.
With an ideal op-amp, we have vo = A(vin v ),
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
23
where A is very large. We have the relationship v = Therefore, vo = A(vin and solving for vo yields vo = A 1+
AR1 R1 +R2
R1 vo . R1 + R2
R1 vo ), R1 + R2
vin . Then the expression for
1 Since A 1, it follows that 1 + RARR2 1+ vo simplies to
AR1 R1 +R2 .
vo = E2.6 Given
R1 + R2 vin . R1
y = f (x) = x1/2 and the operating point xo = 1/2 , we have the linear approximation y = K x where K= E2.7 df dx =
xo =1/2
1 1/2 x 2
xo =1/2
1 = . 2
The block diagram is shown in Figure E2.7.
+ Ea(s)
R(s)
G1(s)
G2(s)
I(s)
H(s)
FIGURE E2.7 Block diagram model.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
24
CHAPTER 2
Mathematical Models of Systems
Starting at the output we obtain I (s) = G1 (s)G2 (s)E (s). But E (s) = R(s) H (s)I (s), so I (s) = G1 (s)G2 (s) [R(s) H (s)I (s)] . Solving for I (s) yields the closed-loop transfer function I (s) G1 (s)G2 (s) = . R(s) 1 + G1 (s)G2 (s)H (s) E2.8 The block diagram is shown in Figure E2.8.
H2(s)
-
R(s)
-
K E(s)
-
W(s) -
G1(s)
A(s)
G2(s)
Z(s)
1 s
Y(s)
H3(s)
H1(s)
FIGURE E2.8 Block diagram model.
Starting at the output we obtain Y (s) = 1 1 Z (s) = G2 (s)A(s). s s
But A(s) = G1 (s) [H2 (s)Z (s) H3 (s)A(s) + W (s)] and Z (s) = sY (s), so 1 Y (s) = G1 (s)G2 (s)H2 (s)Y (s) G1 (s)H3 (s)Y (s) + G1 (s)G2 (s)W (s). s Substituting W (s) = KE (s) H1 (s)Z (s) into the above equation yields Y (s) = G1 (s)G2 (s)H2 (s)Y (s) G1 (s)H3 (s)Y (s) 1 + G1 (s)G2 (s) [KE (s) H1 (s)Z (s)] s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
25
and with E (s) = R(s) Y (s) and Z (s) = sY (s) this reduces to Y (s) = [G1 (s)G2 (s) (H2 (s) + H1 (s)) G1 (s)H3 (s) 1 1 G1 (s)G2 (s)K ]Y (s) + G1 (s)G2 (s)KR(s). s s Solving for Y (s) yields the transfer function Y (s) = T (s)R(s), where T (s) = E2.9 KG1 (s)G2 (s)/s . 1 + G1 (s)G2 (s) [(H2 (s) + H1 (s)] + G1 (s)H3 (s) + KG1 (s)G2 (s)/s
From Figure E2.9, we observe that Ff (s) = G2 (s)U (s) and FR (s) = G3 (s)U (s) . Then, solving for U (s) yields U (s) = and it follows that FR (s) = G3 (s) U (s) . G2 (s) 1 Ff (s) G2 (s)
Again, considering the block diagram in Figure E2.9 we determine Ff (s) = G1 (s)G2 (s)[R(s) H2 (s)Ff (s) H2 (s)FR (s)] . But, from the previous result, we substitute for FR (s) resulting in Ff (s) = G1 (s)G2 (s)R(s)G1 (s)G2 (s)H2 (s)Ff (s)G1 (s)H2 (s)G3 (s)Ff (s) . Solving for Ff (s) yields Ff (s) = G1 (s)G2 (s) R(s) . 1 + G1 (s)G2 (s)H2 (s) + G1 (s)G3 (s)H2 (s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
26
CHAPTER 2
Mathematical Models of Systems
H2(s)
U (s )
G2(s)
Ff (s)
R(s)
+-
G1(s)
U (s )
G3(s)
FR(s)
H2(s)
FIGURE E2.9 Block diagram model.
E2.10
The shock absorber block diagram is shown in Figure E2.10. The closedloop transfer function model is T (s) = Gc (s)Gp (s)G(s) . 1 + H (s)Gc (s)Gp (s)G(s)
Controller + R(s) Desired piston travel Gc(s) -
Gear Motor Gp(s)
Plunger and Piston System G(s) Y(s) Piston travel
Sensor
Piston travel measurement
H(s)
FIGURE E2.10 Shock absorber block diagram.
E2.11
Let f denote the spring force (n) and x denote the deection (m). Then K= f . x
Computing the slope from the graph yields: (a) xo = 0.14m K = f /x = 10 n / 0.04 m = 250 n/m (b) xo = 0m K = f /x = 10 n / 0.05 m = 200 n/m (c) xo = 0.35m K = f /x = 3n / 0.05 m = 60 n/m
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
27
E2.12
The signal ow graph is shown in Fig. E2.12. Find Y (s) when R(s) = 0.
Td(s)
1
-K
1 K2
G(s) Y (s)
-1
FIGURE E2.12 Signal ow graph.
The transfer function from Td (s) to Y (s) is Y (s) = If we set K1 K2 = 1 , then Y (s) = 0 for any Td (s). E2.13 Since we want to compute the transfer function from R2 (s) to Y1 (s), we can assume that R1 = 0 (application of the principle of superposition). Then, starting at the output Y1 (s) we obtain
H1(s)
-
G(s)Td (s) K1 K2 G(s)Td (s) G(s)(1 K1 K2 )Td (s) = . 1 (K2 G(s)) 1 + K2 G(s)
R1(s)
G1(s) G7(s)
+ +
+ +
G2(s) G9(s)
+ +
G3(s)
Y1(s)
G8(s) G5(s)
W(s)
R2(s)
G4(s)
G6(s)
Y2(s)
H2(s)
FIGURE E2.13 Block diagram model.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
28
CHAPTER 2
Mathematical Models of Systems
Y1 (s) = G3 (s) [H1 (s)Y1 (s) + G2 (s)G8 (s)W (s) + G9 (s)W (s)] , or [1 + G3 (s)H1 (s)] Y1 (s) = [G3 (s)G2 (s)G8 (s)W (s) + G3 (s)G9 (s)] W (s). Considering the signal W (s) (see Figure E2.13), we determine that W (s) = G5 (s) [G4 (s)R2 (s) H2 (s)W (s)] , or [1 + G5 (s)H2 (s)] W (s) = G5 (s)G4 (s)R2 (s). Substituting the expression for W (s) into the above equation for Y1 (s) yields Y1 (s) G2 (s)G3 (s)G4 (s)G5 (s)G8 (s) + G3 (s)G4 (s)G5 (s)G9 (s) = . R2 (s) 1 + G3 (s)H1 (s) + G5 (s)H2 (s) + G3 (s)G5 (s)H1 (s)H2 (s) E2.14 For loop 1, we have R1 i1 + L1 di1 1 + dt C1 (i1 i2 )dt + R2 (i1 i2 ) = v (t) .
And for loop 2, we have 1 C2 E2.15 i2 dt + L2 di2 1 + R2 (i2 i1 ) + dt C1 (i2 i1 )dt = 0 .
The transfer function from R(s) to P (s) is 4.2 P (s) =3 . 2 + 4s + 4.2 R(s) s + 2s The block diagram is shown in Figure E2.15a. The corresponding signal ow graph is shown in Figure E2.15b for P (s)/R(s) = s3 + 2s2 4.2 . + 4s + 4.2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
29
v1(s) v2(s) q(s) 1 s2+2s+4
R(s)
-
7
0.6 s
P(s)
(a)
1 V1 7 V2 0.6 s 1 s2 + 2 s + 4 P (s)
R(s )
-1
(b)
FIGURE E2.15 (a) Block diagram, (b) Signal ow graph.
E2.16
A linear approximation for f is given by f = f x x = 4kx3 x = 4kx o
x= xo
where xo = 1, f = f (x) f (xo ), and x = x xo . E2.17 The linear approximation is given by y = m x where m= y x .
x= xo
(a) When xo = 1, we nd that yo = 2.4, and yo = 13.2 when xo = 2. (b) The slope m is computed as follows: m= y x = 1 + 4.2x2 . o
x= xo
Therefore, m = 5.2 at xo = 1, and m = 18.8 at xo = 2. E2.18 The output (with a step input) is Y (s) = 10(s + 2) . s(s + 3)(s + 5)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
30
CHAPTER 2
Mathematical Models of Systems
The partial fraction expansion is Y (s) = 4 51 1 + 3 . 3s 3 s + 3 s+5
Taking the inverse Laplace transform yields y (t) = E2.19 4 5 3t +e 3e5t . 33
The input-output relationship is Vo A(K 1) = V 1 + AK where K= Assume A 1. Then, Vo K1 Z2 = = V K Z1 where Z1 = Therefore, Vo (s) R2 (R1 C1 s + 1) 2(s + 1) = = . V (s) R1 (R2 C2 s + 1) s+2 R1 R1 C1 s + 1 and Z2 = R2 . R2 C2 s + 1 Z1 . Z1 + Z2
E2.20
The equation of motion of the mass mc is mc xp + (bd + bs )xp + kd xp = bd xin + kd xin . Taking the Laplace transform with zero initial conditions yields [mc s2 + (bd + bs )s + kd ]Xp (s) = [bd s + kd ]Xin (s) . So, the transfer function is Xp (s) bd s + kd 0.7s + 2 = =2 . Xin (s) mc s2 + (bd + bs )s + kd s + 2.8s + 2
E2.21
The rotational velocity is (s) = 2.5(s + 2) 1 . 2 4s (s + 5)(s + 1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
31
Expanding in a partial fraction expansion yields (s) = 0.25 0.0234 0.1562 0.2734 + . 2 s s+5 (s + 1) s+1
Taking the inverse Laplace transform yields (t) = 0.25 + 0.0234e5t 0.1562tet 0.2734et . E2.22 The closed-loop transfer function is K1 K2 Y (s) = T (s) = 2 . R(s) s + (K1 + K2 K3 + K1 K2 )s + K1 K2 K3 E2.23 The closed-loop tranfser function is Y (s) 10 = T (s) = 2 . R(s) s + 21s + 10 E2.24 Let x = 0.6 and y = 0.8. Then, with y = ax3 , we have 0.8 = a(0.6)3 . Solving for a yields a = 3.704. A linear approximation is y yo = 3ax2 (x xo ) o or y = 4x 1.6, where yo = 0.8 and xo = 0.6. E2.25 The equations of motion are m1 x1 + k(x1 x2 ) = F m2 x2 + k(x2 x1 ) = 0 . Taking the Laplace transform (with zero initial conditions) and solving for X2 (s) yields X2 (s) = (m2 s2 k F (s) . + k)(m1 s2 + k) k2 1 s2 (s2 + 2)
Then, with m1 = m2 = k = 1, we have X2 (s)/F (s) = E2.26 .
The transfer function from Td (s) to Y (s) is Y (s)/Td (s) = G2 (s) . 1 + G1 G2 H (s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
32 E2.27
CHAPTER 2
Mathematical Models of Systems
The transfer function is R2 R4 C R2 R4 Vo (s) = s+ = 24s + 144 . V (s) R3 R1 R3
E2.28
(a) If G(s) = s2 1 + 15s + 50 and H (s) = 2s + 15 ,
then the closed-loop transfer function of Figure E2.28(a) and (b) (in Dorf & Bishop) are equivalent. (b) The closed-loop transfer function is T (s) = E2.29 s2 1 . + 17s + 65
(a) The closed-loop transfer function is T (s) = G(s) 10 =2 1 + G(s) s + 2s + 20 where G(s) = 10 . s2 + 2s + 10
(b) The output Y (s) (when R(s) = 1/s) is Y (s) = 0.12 0.24 0.12 + . s s + 10 s + 20
Step Response 0.8 0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 Time ( sec ) 4 5 6
FIGURE E2.29 Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
33
(c) The plot of y (t) is shown in Figure E2.29. The output is given by y (t) = 0.12 0.24e10t + 0.12e20t . E2.30 The partial fraction expansion is V (s) = a b + s + p1 s + p2
where p1 = 4 19.6j and p2 = 4 + 19.6j . Then, the residues are a = 10.2j The inverse Laplace transform is v (t) = 10.2je(4+19.6j )t + 10.2je(419.6j )t = 20.4e4t sin 19.6t . b = 10.2j .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
34
CHAPTER 2
Mathematical Models of Systems
Problems
P2.1 The integrodierential equations, obtained by Kirchos voltage law to each loop, are as follows: R1 i1 + and R3 i2 + P2.2 1 C2 i2 dt + R2 (i2 i1 ) + L1 d(i2 i1 ) =0 dt (loop 2) . 1 C1 i1 dt + L1 d(i1 i2 ) + R2 (i1 i2 ) = v (t) dt (loop 1)
The dierential equations describing the system can be obtained by using a free-body diagram analysis of each mass. For mass 1 and 2 we have M1 y1 + k12 (y1 y2 ) + by1 + k1 y1 = F (t) M2 y2 + k12 (y2 y1 ) = 0 . Using a force-current analogy, the analagous electric circuit is shown in Figure P2.2, where Ci Mi , L1 1/k1 , L12 1/k12 , and R 1/b .
FIGURE P2.2 Analagous electric circuit.
P2.3
The dierential equations describing the system can be obtained by using a free-body diagram analysis of each mass. For mass 1 and 2 we have M x1 + kx1 + k(x1 x2 ) = F (t) M x2 + k(x2 x1 ) + bx2 = 0 . Using a force-current analogy, the analagous electric circuit is shown in Figure P2.3, where CM L 1/k R 1/b .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
35
FIGURE P2.3 Analagous electric circuit.
P2.4
(a) The linear approximation around vin = 0 is vo = 0vin , see Figure P2.4(a). (b) The linear approximation around vin = 1 is vo = 2vin 1, see Figure P2.4(b).
(a) 0.4 4
(b)
0.3
3.5
3 0.2 2.5 0.1
2 vo linear approximation
vo
0
1.5
-0.1
1
0.5 -0.2 0 -0.3
-0.5
linear approximation
-0.4 -1
-0.5
0 vin
0.5
1
-1 -1
0 vin
1
2
FIGURE P2.4 Nonlinear functions and approximations.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
36 P2.5
CHAPTER 2
Mathematical Models of Systems
Given Q = K (P1 P2 )1/2 . Let P = P1 P2 and Po = operating point. Using a Taylor series expansion of Q, we have Q = Qo + where
1 Qo = KPo /2
Q P
P =Po
(P Po ) +
and
Q P
=
P =Po
K 1/2 P . 2o
Dene Q = Q Qo and P = P Po . Then, dropping higher-order terms in the Taylor series expansion yields Q = m P where m= P2.6 From P2.1 we have R1 i1 + and R3 i2 + 1 C2 i2 dt + R2 (i2 i1 ) + L1 d(i2 i1 ) =0. dt 1 C1 i1 dt + L1 d(i1 i2 ) + R2 (i1 i2 ) = v (t) dt K 2Po
1/2
.
Taking the Laplace transform and using the fact that the initial voltage across C2 is 10v yields [R1 + and [R2 L1 s]I1 (s) + [L1 s + R3 + Rewriting in matrix form we have
1 + L1 s + R2 ]I1 (s) + [R2 L1 s]I2 (s) = 0 C1 s 1 10 + R2 ]I2 (s) = . C2 s s
R1 +
1 C1 s
+ L 1 s + R2
R2 L 1 s L 1 s + R3 +
1 C2 s
R2 L 1 s
+ R2
I1 (s) I2 (s)
=
0 10/s
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
37
Solving for I2 yields
1 R2 + L 1 s 0 1 L1 s + R3 + C2 s + R2 = . 1 I2 (s) 10/s R2 + L 1 s R1 + C1 s + L1 s + R2
I1 (s)
or I2 (s) = where = (R1 + P2.7 1 1 + L1 s + R2 )(L1 s + R3 + + R2 ) (R2 + L1 s)2 . C1 s C2 s 10(R1 + 1/C1 s + L1 s + R2 ) s
Consider the dierentiating op-amp circuit in Figure P2.7. For an ideal op-amp, the voltage gain (as a function of frequency) is V2 (s) = where Z1 = R1 1 + R1 Cs Z2 (s) V1 (s), Z1 (s)
and Z2 = R2 are the respective circuit impedances. Therefore, we obtain V2 (s) = R2 (1 + R1 Cs) V1 (s). R1
Z
1
Z
R2
C
2
+ R1 V1(s) -
+
+ V2(s) -
FIGURE P2.7 Dierentiating op-amp circuit.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
38 P2.8
CHAPTER 2
Mathematical Models of Systems
Let G2 + Cs = Cs G2 Then, Vj = ij I1 or or V3 13 I1 / = . V1 11 I1 / Cs G1 + 2Cs Cs G2 Cs Cs + G2
.
Therefore, the transfer function is Cs 2Cs + G1 T (s) = 13 V3 = = V1 11 G2 2Cs + G1 Cs Cs Cs Cs + G2
Pole-zero map (x:poles and o:zeros) 3
2
o
1
Imag Axis
0
x
x
-1
-2
o
-3 -8
-7
-6
-5
-4 Real Axis
-3
-2
-1
0
FIGURE P2.8 Pole-zero map.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
39 = C 2 R1 R2 s2 + 2CR1 s + 1 . C 2 R1 R2 s2 + (2R1 + R2 )Cs + 1
Using R1 = 0.5, R2 = 1, and C = 0.5, we have T (s) = s2 + 4s + 8 (s + 2 + 2j )(s + 2 2j ) . = 2 + 8s + 8 s (s + 4 + 8)(s + 4 8)
The pole-zero map is shown in Figure P2.8. P2.9 From P2.3 we have M x1 + kx1 + k(x1 x2 ) = F (t) M x2 + k(x2 x1 ) + bx2 = 0 . Taking the Laplace transform of both equations and writing the result in matrix form, it follows that
0.4
M s2 + 2k k
M s2 + bs + k
k
X1 (s) X2 (s)
=
F (s) 0
,
Pole zero map
0.3
0.2
0.1 Imag Axis
0
- 0.1
-0.2
-0.3
-0.4 -0.03
-0.025
-0.02
-0.015 Real Axis
-0.01
-0.005
0
FIGURE P2.9 Pole-zero map.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
40
CHAPTER 2
Mathematical Models of Systems
or
where = (M s2 + bs + k)(M s2 + 2k) k2 . So, G(s) =
k F (s) 1 M s2 + bs + k = 2 + 2k X2 (s) k Ms 0 X1 (s) X1 (s) M s2 + bs + k = . F (s)
When b/k = 1, M = 1 , b2 /M k = 0.04, we have G(s) = s2 + 0.04s + 0.04 . s4 + 0.04s3 + 0.12s2 + 0.0032s + 0.0016
The pole-zero map is shown in Figure P2.9. P2.10 From P2.2 we have M1 y1 + k12 (y1 y2 ) + by1 + k1 y1 = F (t) M2 y2 + k12 (y2 y1 ) = 0 . Taking the Laplace transform of both equations and writing the result in matrix form, it follows that
M1 s2 + bs + k1 + k12 k12 Y1 (s)
or
M2 s2 + k12
k12
Y1 (s) Y2 (s)
=
F (s) 0
where
k12 F (s) 1 M2 s2 + k12 = Y2 (s) k12 M1 s2 + bs + k1 + k12 0
2 = (M2 s2 + k12 )(M1 s2 + bs + k1 + k12 ) k12 .
So, when f (t) = a sin o t, we have that Y1 (s) is given by Y1 (s) = aM2 o (s2 + k12 /M2 ) . 2 (s2 + o )(s)
For motionless response (in the steady-state), set the zero of the transfer function so that (s2 + k12 2 ) = s2 + o M2 or
2 o =
k12 . M2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
41
P2.11
The transfer functions from Vc (s) to Vd (s) and from Vd (s) to (s) are: Vd (s)/Vc (s) = K1 K2 , and (Lq s + Rq )(Lc s + Rc ) Km . (s)/Vd (s) = 2 + f s)((L + L )s + R + R ) + K K s (Js a a 3m d d
The block diagram for (s)/Vc (s) is shown in Figure P2.11, where (s)/Vc (s) = where (s) = s(Lc s + Rc )(Lq s + Rq )((Js + b)((Ld + La )s + Rd + Ra ) + Km K3 ) . K1 K2 Km (s) Vd (s) = , Vd (s) Vc (s) (s)
Vc
1 L cs+R c
Ic
K1
Vq
1 L qs+R q
Iq K2
Vd + Vb
1 (L d+L a)s+R d+R a
Id Km
Tm
1 Js+f
w
1 s
q
K3
FIGURE P2.11 Block diagram.
P2.12
The open-loop transfer function is Y (s) K = . R(s) s + 10 With R(s) = 1/s, we have Y (s) = The partial fraction expansion is Y (s) = K 10 1 1 , s s + 10 K . s(s + 10)
and the inverse Laplace transform is y (t) = K 1 e10t , 10
As t , it follows that y (t) K/10. So we choose K = 100 so that
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
42
CHAPTER 2
Mathematical Models of Systems
y (t) approaches 10. Alternatively we can use the nal value theorem to obtain y (t)t = lim sY (s) =
s0
K = 10 . 10
It follows that choosing K = 100 leads to y (t) 10 as t . P2.13 The motor torque is given by Tm (s) = (Jm s2 + bm s)m (s) + (JL s2 + bL s)nL (s) = n((Jm s2 + bm s)/n2 + JL s2 + bL s)L (s) where n = L (s)/m (s) = gear ratio . But Tm (s) = Km Ig (s) and Ig (s) = and Vg (s) = Kg If (s) = Kg Vf (s) . Rf + L f s 1 Vg (s) , (Lg + Lf )s + Rg + Rf
Combining the above expressions yields L (s) Kg Km = . Vf (s) n1 (s)2 (s) where 1 (s) = JL s2 + bL s + and 2 (s) = (Lg s + Lf s + Rg + Rf )(Rf + Lf s) . P2.14 For a eld-controlled dc electric motor we have (s)/Vf (s) = Km /Rf . Js + b Jm s2 + bm s n2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
43
With a step input of Vf (s) = 80/s, the nal value of (t) is (t)t = lim s (s) =
s0
80Km = 2.4 Rf b
or
Km = 0.03 . Rf b
Solving for (t) yields (t) = 80Km 1 1 L Rf J s(s + b/J ) = 80Km (1e(b/J )t ) = 2.4(1e(b/J )t ) . Rf b
At t = 1/2, (t) = 1, so (1/2) = 2.4(1 e(b/J )t ) = 1 implies Therefore, (s)/Vf (s) = P2.15 0.0324 . s + 1.08 b/J = 1.08 sec .
Summing the forces in the vertical direction and using Newtons Second Law we obtain x+ k x=0. m
The system has no damping and no external inputs. Taking the Laplace transform yields X (s) = s2 x0 s , + k/m
where we used the fact that x(0) = x0 and x(0) = 0. Then taking the inverse Laplace transform yields x(t) = x0 cos P2.16 Using Cramers rule, we have
k t. m
1 1.5 2 4
or
x1 x2
=
6 11
1 4 1.5 6 = 2 x2 1 11
x1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
44
CHAPTER 2
Mathematical Models of Systems
where = 4(1) 2(1.5) = 1 . Therefore, x1 = 4(6) 1.5(11) = 7.5 1 and x2 = 2(6) + 1(11) = 1 . 1
The signal ow graph is shown in Figure P2.16.
11 1/4 1 X1 -1.5
6
-1/2
X2
FIGURE P2.16 Signal ow graph.
So, x1 = P2.17 6(1) 1.5( 11 ) 4 = 7.5 3 1 4 and x2 =
1 11( 4 ) + 1 1 2 (6) 3 4
= 1 .
(a) For mass 1 and 2, we have M1 x1 + K1 (x1 x2 ) + b1 (x3 x1 ) = 0 M2 x2 + K2 (x2 x3 ) + b2 (x3 x2 ) + K1 (x2 x1 ) = 0 . (b) Taking the Laplace transform yields (M1 s2 + b1 s + K1 )X1 (s) K1 X2 (s) = b1 sX3 (s) K1 X1 (s) + (M2 s2 + b2 s + K1 + K2 )X2 (s) = (b2 s + K2 )X3 (s) . (c) Let G1 (s) = K2 + b2 s G2 (s) = 1/p(s) G3 (s) = 1/q (s) G4 (s) = sb1 , where p(s) = s2 M2 + sf2 + K1 + K2 and q (s) = s2 M1 + sf1 + K1 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
45
The signal ow graph is shown in Figure P2.17.
G4 G3 G1 G2 K1 K1 X1
X3
FIGURE P2.17 Signal ow graph.
(d) The transfer function from X3 (s) to X1 (s) is X1 (s) K1 G1 (s)G2 (s)G3 (s) + G4 (s)G3 (s) = . 2 X3 (s) 1 K1 G2 (s)G3 (s) P2.18 The signal ow graph is shown in Figure P2.18.
I1 V1 Y1 -Y 1 -Z 2 -Y 3 Va Ia
Z2
Y3
Z4 V2
FIGURE P2.18 Signal ow graph.
The transfer function is V2 (s) Y1 Z2 Y3 Z4 = . V1 (s) 1 + Y1 Z2 + Y3 Z2 + Y3 Z4 + Y1 Z2 Z4 Y3 P2.19 For a noninerting op-amp circuit, depicted in Figure P2.19a, the voltage gain (as a function of frequency) is Vo (s) = Z1 (s) + Z2 (s) Vin (s), Z1 (s)
where Z1 (s) and Z2 (s) are the impedances of the respective circuits. In the case of the voltage follower circuit, shown in Figure P2.19b, we have
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
46
CHAPTER 2
Mathematical Models of Systems
Z2 Z1
vin
(a)
+
v0
vin
(b)
+
v0
FIGURE P2.19 (a) Noninverting op-amp circuit. (b) Voltage follower circuit.
Z1 = (open circuit) and Z2 = 0. Therefore, the transfer function is Vo (s) Z1 = = 1. Vin (s) Z1 P2.20 (a) Assume Rg Rs and Rs R1 . Then Rs = R1 + R2 R2 , and vgs = vin vo , where we neglect iin , since Rg Rs . At node S, we have vo = gm vgs = gm (vin vo ) or Rs (b) With gm Rs = 20, we have vo 20 = = 0.95 . vin 21 (c) The block diagram is shown in Figure P2.20. vo gm Rs = . vin 1 + gm Rs
vin(s)
-
gmRs
vo(s)
FIGURE P2.20 Block diagram model.
P2.21
From the geometry we nd that z = k l1 l2 l2 (x y ) y . l1 l1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
47
The ow rate balance yields A dy = pz dt which implies Y (s) = pZ (s) . As
By combining the above results it follows that Y (s) = p l1 l2 l2 k (X (s) Y (s)) Y (s) . As l1 l1
Therefore, the signal ow graph is shown in Figure P2.21. Using Masons
-1 k X 1 -l 2 / l 1 (l 1 - l 2)/l 1
DZ
p/As Y
FIGURE P2.21 Signal ow graph.
gain formula we nd that the transfer function is given by Y (s) = X (s) 1+ where K1 = P2.22 k(l1 l2 )p p l1 A and K2 = l2 p . l1 A
k (l1 l2 )p l1 As k (l1 l2 )p l2 p l1 As + l1 As
=
K1 , s + K2 + K1
(a) The equations of motion for the two masses are M L2 1 + M gL1 + k L 2 L M L2 2 + M gL2 + k 2
2
(1 2 ) =
2
L f (t) 2
(2 1 ) = 0 .
With 1 = 1 and 2 = 2 , we have g k k f (t) + 1 + 2 + L 4M 4M 2M L g k k 1 + 2 . 2 = 4M L 4M 1 =
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
48
CHAPTER 2
Mathematical Models of Systems
a
F (t) 1/2ML
1/s
w1 1/s
q1
(a)
b
1/s
w2
1/s
q2
a
Imag(s)
+j g k L + 4M g k L + 2M
X O
+j g L
(b)
+j
X
Re(s)
FIGURE P2.22 (a) Block diagram. (b) Pole-zero map.
(b) Dene a = g/L + k/4M and b = k/4M . Then 1 (s) 1 s2 + a = . F (s) 2M L (s2 + a)2 b2 (c) The block diagram and pole-zero map are shown in Figure P2.22. P2.23 The input-output ratio, Vce /Vin , is found to be (R 1) + hie Rf Vce = . Vin hre + hie (hoe + Rf ) P2.24 (a) The voltage gain is given by RL 1 2 (R1 + R2 ) vo = . vin (R1 + R2 )(Rg + hie1 ) + R1 (R1 + R2 )(1 + 1 ) + R1 RL 1 2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
49
(b) The current gain is found to be ic2 = 1 2 . ib1 (c) The input impedance is vin (R1 + R2 )(Rg + hie1 ) + R1 (R1 + R2 )(1 + 1 ) + R1 RL 1 2 = , ib1 R1 + R2 and when 1 2 is very large, we have the approximation vin RL R1 1 2 . ib1 R1 + R2 P2.25 The transfer function from R(s) and Td (s) to Y (s) is given by Y (s) = G(s) R(s) = G(s)R(s) . Thus, Y (s)/R(s) = G(s) . Also, we have that Y (s) = 0 . when R(s) = 0. Therefore, the eect of the disturbance, Td (s), is eliminated. P2.26 The equations of motion for the two mass model of the robot are M x + b(x y ) + k(x y ) = F (t) my + b(y x) + k(y x) = 0 . Taking the Laplace transform and writing the result in matrix form yields
1 (G(s)R(s) + Td (s)) + Td (s) + G(s)R(s) G(s)
M s2 + bs + k (bs + k)
ms2
(bs + k) + bs + k
X (s) Y (s)
=
F (s) 0
.
Solving for Y (s) we nd that
1 Y (s) mM (bs + k ) = m b F (s) s2 [s2 + 1 + M m s +
k m
]
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
50 P2.27
CHAPTER 2
Mathematical Models of Systems
The describing equation of motion is mz = mg k Dening f (z, i) = g leads to z = f (z, i) . The equilibrium condition for io and zo , found by solving the equation of motion when z=z=0, is ki2 o 2 = zo . mg We linearize the equation of motion using a Taylor series approximation. With the denitions z = z zo and i = i io , f i ki2 mz 2 i2 . z2
we have z = z and z = z . Therefore, f z = f (z, i) = f (zo , io ) + z
z = zo i=io
z +
z = zo i=io
i +
But f (zo , io ) = 0, and neglecting higher-order terms in the expansion yields 2ki2 2kio o z = z i . 3 2 mzo mzo Using the equilibrium condition which relates zo to io , we determine that g 2g z = z i . zo io Taking the Laplace transform yields the transfer function (valid around the equilibrium point) Z (s) g/io =2 . I (s) s 2g/zo
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
51
P2.28
The signal ow graph is shown in Figure P2.28.
-d G P +b +a -m +e +g +f M +h -k B +c D
S
C
FIGURE P2.28 Signal ow graph.
(a) The PGBDP loop gain is equal to -abcd. This is a negative transmission since the population produces garbage which increases bacteria and leads to diseases, thus reducing the population. (b) The PMCP loop gain is equal to +efg. This is a positive transmission since the population leads to modernization which encourages immigration, thus increasing the population. (c) The PMSDP loop gain is equal to +ehkd. This is a positive transmission since the population leads to modernization and an increase in sanitation facilities which reduces diseases, thus reducing the rate of decreasing population. (d) The PMSBDP loop gain is equal to +ehmcd. This is a positive transmission by similar argument as in (3). P2.29 Assume the motor torque is proportional to the input current Tm = ki . Then, the equation of motion of the beam is J = ki , where J is the moment of inertia of the beam and shaft (neglecting the inertia of the ball). We assume that forces acting on the ball are due to gravity and friction. Hence, the motion of the ball is described by mx = mg bx
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
52
CHAPTER 2
Mathematical Models of Systems
where m is the mass of the ball, b is the coecient of friction, and we have assumed small angles, so that sin . Taking the Laplace transfor of both equations of motion and solving for X (s) yields X (s)/I (s) = P2.30 Given H (s) = k s + 1 gk/J . s2 (s2 + b/m)
where = 4s = 4 106 seconds and 0.999 k < 1.001. The step response is Y (s) = k 1 k k = . s + 1 s s s + 1/
Taking the inverse Laplace transform yields y (t) = k ket/ = k(1 et/ ) . The nal value is k. The time it takes to reach 98% of the nal value is t = 15.6s independent of k. P2.31 The closed-loop transfer function is T (s) = 500K (0.5s + 1) (0.5s + 1)(s + 1)(s + 10) + 500K
The nal value due to a step input of R(s) = A/s is v (t) A 500K . 500K + 10
If we want the nal value of the output to be as close to 50 m/s as possible, we will need to select K as large so that v (t) A = 50. However, to keep the percent overshoot to less than 9%, we need to limit the magnitude of K . As a trade-o we can let K = 0.0178 and select the magnitude of the input to be A = 106. The inverse Laplace transform of the closed-loop response with R(s) = 106/s is v (t) = 50 + 9.47e10.23t 59.38e1.38t cos 1.33t + 11.1e1.38t sin 1.33t The result is P.O. = 9% and the steady-state value of the output is approximately 50 m/s, as desired.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
53
P2.32
From the block diagram we have Y1 (s) = G2 (s)[G1 (s)E1 (s) + G3 (s)E2 (s)] = G2 (s)G1 (s)[R1 (s) H1 (s)Y1 (s)] + G2 (s)G3 (s)E2 (s) . Therefore, Y1 (s) = G1 (s)G2 (s) G2 (s)G3 (s) R1 (s) + E2 (s) . 1 + G1 (s)G2 (s)H1 (s) 1 + G1 (s)G2 (s)H1 (s)
And, computing E2 (s) (with R2 (s) = 0) we nd E2 (s) = H2 (s)Y2 (s) = H2 (s)G6 (s) or E2 (s) = G4 (s)G6 (s)H2 (s) Y1 (s) . G2 (s)(1 G5 (s)G6 (s)H2 (s)) G4 (s) Y1 (s) + G5 (s)E2 (s) G2 (s)
Substituting E2 (s) into equation for Y1 (s) yields Y1 (s) = G1 (s)G2 (s) R1 (s) 1 + G1 (s)G2 (s)H1 (s) G3 (s)G4 (s)G6 (s)H2 (s) + Y1 (s) . (1 + G1 (s)G2 (s)H1 (s))(1 G5 (s)G6 (s)H2 (s))
Finally, solving for Y1 (s) yields Y1 (s) = T1 (s)R1 (s) where T1 (s) = G1 (s)G2 (s)(1 G5 (s)G6 (s)H2 (s)) (1 + G1 (s)G2 (s)H1 (s))(1 G5 (s)G6 (s)H2 (s)) G3 (s)G4 (s)G6 (s)H2 (s) Similarly, for Y2 (s) we obtain Y2 (s) = T2 (s)R1 (s) . where T2 (s) = G1 (s)G4 (s)G6 (s) (1 + G1 (s)G2 (s)H1 (s))(1 G5 (s)G6 (s)H2 (s)) G3 (s)G4 (s)G6 (s)H2 (s) . .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
54 P2.33
CHAPTER 2
Mathematical Models of Systems
The signal ow graph shows three loops: L1 = G1 G3 G4 H2 L2 = G2 G5 G6 H1 L3 = H1 G8 G6 G2 G7 G4 H2 G1 . The transfer function Y2 /R1 is found to be Y2 (s) G1 G8 G6 1 G2 G5 G6 2 = , R1 (s) 1 (L1 + L2 + L3 ) + (L1 L2 ) where for path 1 1 = 1 and for path 2 2 = 1 L1 . Since we want Y2 to be independent of R1 , we need Y2 /R1 = 0. Therefore, we require G1 G8 G6 G2 G5 G6 (1 + G1 G3 G4 H2 ) = 0 .
P2.34
The closed-loop transfer function is G3 (s)G1 (s)(G2 (s) + K5 K6 ) Y (s) = . R(s) 1 G3 (s)(H1 (s) + K6 ) + G3 (s)G1 (s)(G2 (s) + K5 K6 )(H2 (s) + K4 )
P2.35
The equations of motion are m1 y1 + b(y1 y2 ) + k1 (y1 y2 ) = 0 m2 y2 + b(y2 y1 ) + k1 (y2 y1 ) + k2 y2 = k2 x Taking the Laplace transform yields (m1 s2 + bs + k1 )Y1 (s) (bs + k1 )Y2 (s) = 0 (m2 s2 + bs + k1 + k2 )Y2 (s) (bs + k1 )Y1 (s) = k2 X (s) Therefore, after solving for Y1 (s)/X (s), we have k2 (bs + k1 ) Y2 (s) = . 2 + bs + k )(m s2 + bs + k + k ) (bs + k )2 X (s) (m1 s 1 2 1 2 1
P2.36
(a) We can redraw the block diagram as shown in Figure P2.36. Then, T (s) = K1 /s(s + 1) K1 =2 . 1 + K1 (1 + K2 s)/s(s + 1) s + (1 + K2 K1 )s + K2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
55
(b) The signal ow graph reveals two loops (both touching): L1 = Therefore, T (s) = K1 /s(s + 1) K1 =2 . 1 + K1 /s(s + 1) + K1 K2 /(s + 1) s + (1 + K2 K1 )s + K1 K1 s(s + 1) and L2 = K1 K2 . s+1
(c) We want to choose K1 and K2 such that s2 + (1 + K2 K1 )s + K1 = s2 + 20s + 100 = (s + 10)2 . Therefore, K1 = 100 and 1 + K2 K1 = 20 or K2 = 0.19. (d) The step response is shown in Figure P2.36.
K1 s (s+1) 1 +K 2s
1 0.9 0.8 0.7 0.6 <---- time to 90% = 0.39 sec
R(s )
Y (s)
+
y(t)
0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 time(sec) 1.2 1.4 1.6 1.8 2
FIGURE P2.36 The equivalent block diagram and the system step response.
P2.37
(a) Given R(s) = 1/s2 , the partial fraction expansion is Y (s) = s2 (s 12 2 2/3 1/4 1 19/12 = + + 2 . + 1)(s + 3)(s + 4) s+1 s+3 s+4 s s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
56
CHAPTER 2
Mathematical Models of Systems
Therefore, using the Laplace transform table, we determine that the ramp response is 2 1 19 y (t) = 2et e3t + e4t + t , 3 4 12 t0.
(b) For the ramp input, y (t) 0.36 at t = 1.5 seconds (see Figure P2.37a). (c) Given R(s) = 1, the partial fraction expansion is Y (s) = 12 2 6 4 = + . (s + 1)(s + 3)(s + 4) s+1 s+3 s+4
Therefore, using the Laplace transform table, we determine that the impulse response is y (t) = 2et 6e3t + 4e4t , t0.
(d) For the impulse input, y (t) 0.39 at t = 1.5 seconds (see Figure P2.37b).
(a) Ramp input 1.6 0.7 (b) Impulse input
1.4
0.6
1.2 0.5 1 0.4 y(t) 0.8 y(t) 0.3 0.6 0.2 0.4 0.1 0 1 Time (sec) 2 3 0 0
0.2
0
1 Time (sec)
2
3
FIGURE P2.37 (a) Ramp input response. (b) Impulse input response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
57
P2.38
The equations of motion are m1 d2 x = (k1 + k2 )x + k2 y dt2 and m2 d2 y = k2 (x y ) + u . dt2
When m1 = m2 = 1 and k1 = k2 = 1, we have d2 x = 2x + y dt2 P2.39 and d2 y =xy+u . dt2
The equation of motion for the system is J d d2 + b + k = 0 , 2 dt dt
where k is the rotational spring constant and b is the viscous friction coecient. The initial conditions are (0) = o and (0) = 0. Taking the Laplace transform yields J (s2 (s) so ) + b(s (s) o ) + k (s) = 0 . Therefore, (s) =
b (s + J o ) (s + 2n )o =2 . 2 2 + bs + K) s + 2n s + n (s J J
Neglecting the mass of the rod, the moment of inertia is detemined to be J = 2M r 2 = 0.5 kg m2 . Also, n = k = 0.02 rad/s J and = b = 0.01 . 2Jn
Solving for (t), we nd that (t) = where tan = o en t sin(n 1 2 t + ) , 1 2
1 2 / ). Therefore, the envelope decay is e = o en t . 1 2
So, with n = 2 104 , o = 4000o and f = 10o , the elapsed time is
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
58
CHAPTER 2
Mathematical Models of Systems
computed as t= P2.40 1 ln n o = 8.32 hours . 1 2 f and vc (0) = 5V ,
When t < 0, we have the steady-state conditions i1 (0) = 1A , va (0) = 2V
where vc (0) is associated with the 1F capacitor. After t 0, we have 2 and i2 dt + 10i2 + 4(i2 i1 ) i1 = 0 . Taking the Laplace transform (using the initial conditions) yields 2(sI1 i1 (0)) + 2I1 + 4I1 4I2 = and 1 [ I2 vc (0)]+10I2 +4(I2 I1 ) = I1 (s) or s Solving for I2 (s) yields I2 = where (s) = Then, Vo (s) = 10I2 (s) . P2.41 The equations of motion are J1 1 = K (2 1 ) b(1 2 ) + T Taking the Laplace transform yields (J1 s2 + bs + K )1 (s) bs2 (s) = K2 (s) + T (s) and J2 2 = b(1 2 ) . s+3 5s 2 14s + 1 = 14s2 + 33s + 3 . 5s(s2 + 6s + 13) , 14(s + 2)(s) 5sI1 (s)+(14s+1)I2 (s) = 5s . 10 s+2 or (s + 3)I1 (s) 2I2 (s) = s+7 s+2 di1 + 2i1 + 4(i1 i2 ) = 10e2t dt
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
59
and (J2 s2 + bs)2 (s) bs1 (s) = 0 . Solving for 1 (s) and 2 (s), we nd that 1 (s) = where (s) = J1 J2 s3 + b(J1 + J2 )s2 + J2 Ks + bK . P2.42 Assume that the only external torques acting on the rocket are control torques, Tc and disturbance torques, Td , and assume small angles, (t). Using the small angle approximation, we have h=V J = Tc + Td , where J is the moment of inertia of the rocket and V is the rocket velocity (assumed constant). Now, suppose that the control torque is proportional to the lateral displacement, as Tc (s) = KH (s) , where the negative sign denotes a negative feedback system. The corresponding block diagram is shown in Figure P2.42.
Td Tc
+ +
(K2 (s) + T (s))(J2 s + b) (s)
and 2 (s) =
b(K2 (s) + T (s)) , (s)
H desired=0
K
+ -
1 Js 2
V s
H( s)
FIGURE P2.42 Block diagram.
P2.43
(a) The equation of motion of the motor is J d = Tm b , dt
where J = 0.1, b = 0.06, and Tm is the motor input torque.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
60
CHAPTER 2
Mathematical Models of Systems
(b) Given Tm (s) = 1/s, and (0) = 0.7, we take the Laplace transform of the equation of motion yielding s (s) (0) + 0.6 (s) = 10Tm or (s) = 0.7s + 10 . s(s + 0.6)
Then, computing the partial fraction expansion, we nd that (s) = B 16.67 15.97 A + = . s s + 0.6 s s + 0.6
The step response, determined by taking the inverse Laplace transform, is (t) = 16.67 15.97e0.6t , P2.44 Tm m = TL L . Also, the travel distance is the same for each gear, so r1 m = r2 L . The number of teeth on each gear is proportional to the radius, or r1 N 2 = r2 N 1 . So, m r2 N2 = = , L r1 N1 and N1 m = N2 L N1 L = m = nm , N2 where n = N1 /N2 . Finally, L N1 Tm = = =n. TL m N2 t0.
The work done by each gear is equal to that of the other, therefore
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
61
P2.45
The inertia of the load is JL = Also, from the dynamics we have T2 = JL 2 + bL 2 and T1 = nT2 = n(JL 2 + bL 2 ) . So, T1 = n2 (JL 1 + bL 1 ) , since 2 = n1 . Therefore, the torque at the motor shaft is T = T1 + Tm = n2 (JL 1 + bL 1 ) + Jm 1 + bm 1 . Lr 4 . 2
P2.46
Let U (s) denote the human input and F (s) the load input. The transfer function is P (s) = where = 1 + GH (s) + G1 KBH (s) + Gc E (s) + G1 KE (s) . G(s) + KG1 (s) Gc (s) + KG1 (s) U (s) + F (s) , (s) (s)
P2.47
Consider the application of Newtons law ( mv we obtain
F = mx). From the mass
mv x1 = F k1 (x1 x2 ) b1 (x1 x2 ). Taking the Laplace transform, and solving for X1 (s) yields X1 (s) = where 1 := mv s2 + b1 s + k1 . 1 b1 s + k1 F (s) + X2 (s), 1 (s) 1 (s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
62
CHAPTER 2
Mathematical Models of Systems
From the mass mt we obtain mt x2 = k2 x2 b2 x2 + k1 (x1 x2 ) + b1 (x1 x2 ). Taking the Laplace transform, and solving for X2 (s) yields X2 (s) = where 2 := mt s2 + (b1 + b2 )s + k1 + k2 . Substituting X2 (s) above into the relationship fpr X1 (s) yields the transfer function 2 (s) X1 (s) = . F (s) 1 (s)2 (s) (b1 s + k1 )2 P2.48 Using the following relationships h(t) = (1.6 (t) h(t))dt b1 s + k1 X1 (s), 2 (s)
(t) = (t) J (t) = Km ia (t) va (t) = 50vi (t) = 10ia (t) + vb (t) = Kvb we nd the dierential equation is d3 h Km + 1+ dt3 10JK P2.49 (a) The transfer function is V2 (s) (1 + sR1 C1 )(1 + sR2 C2 ) = . V1 (s) R1 C2 s (b) When R1 = 100 k, R2 = 200 k, C1 = 1 F and C2 = 0.1 F , we have V2 (s) 0.2(s + 10)(s + 50) = . V1 (s) s P2.50 (a) The closed-loop transfer function is T (s) = G(s) 6205 =3 . 2 + 1281s + 6205 1 + G(s) s + 13s d2 h Km dh 8Km + = vi . dt2 10JK dt J
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
63
1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 Time (secs) 1.2 1.4 1.6 1.8 2
FIGURE P2.50 Step response.
(b) The poles of T (s) are s1 = 5 and s2,3 = 4 j 35. Y (s) = 1
(c) The partial fraction expansion (with a step input) is 1.0122 0.0061 + 0.0716j 0.0061 0.0716j + + . s+5 s + 4 + j 35 s + 4 j 35
(d) The step response is shown in Figure P2.50. The real and complex roots are close together and by looking at the poles in the s-plane we have diculty deciding which is dominant. However, the residue at the real pole is much larger and thus dominates the response. P2.51 (a) The closed-loop transfer function is T (s) = s3 + 45s2 14000 . + 3100s + 14500
(b) The poles of T (s) are s1 = 5 and s2,3 = 20 j 50. (c) The partial fraction expansion (with a step input) is Y (s) = 0.9655 0.0310 0.0390j 1.0275 0.0310 + 0.0390j + + . s+5 s + 20 + j 50 s + 20 j 50
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
64
CHAPTER 2
Mathematical Models of Systems
1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 Time (secs) 1.2 1.4 1.6 1.8 2
FIGURE P2.51 Step response.
(d) The step response is shown in Figure P2.51. The real root dominates the response. (e) The nal value of y (t) is yss = lim Y (s) = lim T (s) = 0.97 .
s0 s0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
65
Advanced Problems
AP2.1 The transfer function from V (s) to (s) has the form (s) Km = . V (s) m s + 1 In the steady-state, ss = lim s
s0
Km 5 = 5Km . m s + 1 s
So, Km = 70/5 = 14 . Also, (t) = Vm Km (1 et/m ) where V (s) = Vm /s. Solving for m yields m = When t = 2, we have m = 2 = 3.57 . ln(1 30/70) t . ln(1 (t)/ss )
Therefore, the transfer function is (s) 14 = . V (s) 3.57s + 1 AP2.2 The closed-loop transfer function form R1 (s) to Y2 (s) is Y2 (s) G1 G4 G5 (s) + G1 G2 G3 G4 G6 (s) = R1 (s) where = [1 + G3 G4 H2 (s)][1 + G1 G2 H3 (s)] . If we select G5 (s) = G2 G3 G6 (s) then the numerator is zero, and Y2 (s)/R1 (s) = 0. The system is now decoupled.
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66 AP2.3
CHAPTER 2
Mathematical Models of Systems
(a) Computing the closed-loop transfer function: Y (s) = G(s)Gc (s) R(s) . 1 + Gc (s)G(s)H (s)
Then, with E (s) = R(s) Y (s) we obtain E (s) = 1 + Gc (s)G(s)(H (s) 1) R(s) . 1 + Gc (s)G(s)H (s)
If we require that E (s) 0 for any input, we need 1 + Gc (s)G(s)(H (s) 1) = 0 or H (s) = Gc (s)G(s) 1 n(s) = . Gc (s)G(s) d(s)
Since we require H (s) to be a causal system, the order of the numerator polynomial, n(s), must be less than or equal to the order of the denominator polynomial, d(s). This will be true, in general, only if both Gc (s) and G(s) are proper rational functions (that is, the numerator and denominator polynomials have the same order). Therefore, making E 0 for any input R(s) is possible only in certain circumstances. (b) The transfer function from Td (s) to Y (s) is Y (s) = Gd (s)G(s) Td (s) . 1 + Gc (s)G(s)H (s)
With H (s) as in part (a) we have Y (s) = (c) No. Since Y (s) = Gd (s)G(s) Td (s) = T (s)Td (s) , 1 + Gc (s)G(s)H (s) Gd (s) Td (s) . Gc (s)
the only way to have Y (s) 0 for any Td (s) is for the transfer function T (s) 0 which is not possible in general (since G(s) = 0). AP2.4 (a) With q (s) = 1/s we obtain (s) = Dene := QS + 1/R Ct and := 1/Ct . 1/Ct s+
QS +1/R Ct
1 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
67
Then, it follows that (s) = 1 / / = + . s+ s s+ s t e + = [1 et ] .
Taking the inverse Laplace transform yields (t) =
1 (b) As t , (t) = Qs+1/R . (c) To increase the speed of response, you want to choose Ct , Q, S and R such that
:= is large. AP2.5
Qs + 1/R Ct
Considering the motion of each mass, we have M3 x3 + b3 x3 + k3 x3 = u3 + b3 x2 + k3 x2 M2 x2 + (b2 + b3 )x2 + (k2 + k3 )x2 = u2 + b3 x3 + k3 x3 + b2 x1 + k2 x1 M1 x1 + (b1 + b2 )x1 + (k1 + k2 )x1 = u1 + b2 x2 + k2 x2 In matrix form the three equations can be written as
M1 0
0 M2 0
0
M3
0 x 1 b1 + b2 b2 0 0 x 2 + b2 b2 + b3 b3
x3
0
b3
b3
k1 + k2 + k2
k2 k2 + k3 k3
0
x1 u1 k3 x2 = u2 .
0
x1 x 2 x3
k3
x3
u3
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68
CHAPTER 2
Mathematical Models of Systems
Design Problems
CDP2.1
The model of the traction drive, capstan roller, and linear slide follows closely the armature-controlled dc motor model depicted in Figure 2.18 in Dorf and Bishop. The transfer function is T (s) = where JT = Jm + r 2 (Ms + Mb ) . rKm , s [(Lm s + Rm )(JT s + bm ) + Kb Km ]
Va(s) -
Km Lms+Rm
1 JTs+bm
w
1 s
q
r
X(s)
Back EMF
Kb
DP2.1
The closed-loop transfer function is Y (s) G1 (s)G2 (s) = . R(s) 1 + G1 (s)H1 (s) G2 (s)H2 (s) When G1 H1 = G2 H2 and G1 G2 = 1, then Y (s)/R(s) = 1. Therefore, select G1 (s) = 1 G2 (s) and H1 (s) = G2 (s)H2 (s) = G2 (s)H2 (s) . 2 G1 (s)
DP2.2
At the lower node we have v 11 + + G + 2i2 20 = 0 . 43
Also, we have v = 24 and i2 = Gv . So v 11 + + G + 2Gv 20 = 0 43
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
69
and G= DP2.3 20 v
1 4
+
1 3
3v
=
1 S. 12
Taking the Laplace transform of 1 31 y (t) = et e2t + t 4 42 yields Y (s) = 1 1 3 1 + 2. s + 1 4(s + 2) 4s 2s
Similarly, taking the Laplace transform of the ramp input yields R(s) = Therefore G(s) = DP2.4 Y (s) 1 = . R(s) (s + 1)(s + 2) 1 . s2
For an ideal op-amp, at node a we have vin va vo va + =0, R1 R1 and at node b vin vb = C vb , R2 from it follows that 1 1 + Cs Vb = Vin . R2 R2 Also, for an ideal op-amp, Vb Va = 0. Then solving for Vb in the above equation and substituting the result into the node a equation for Va yields Vo = Vin or Vo (s) R2 Cs 1 = . Vin (s) R2 Cs + 1
1 R2
2 1 R2 + Cs
1 R2
+ Cs 2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
70
CHAPTER 2
Mathematical Models of Systems
For vin (t) = At, we have Vin (s) = A/s2 , therefore vo (t) = A where = 1/R2 C . 2 t 2 e +t
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
71
Computer Problems
CP2.1 The m-le script is shown in Figure CP2.1.
pq = 1 15 P= -4 -1 Z= -10 value = 0
p=[1 5 4]; q=[1 10]; % Part (a) pq=conv(p,q) % Part (b) P=roots(p), Z=roots(q) % Part (c) value=polyval(p,-1)
54
40
FIGURE CP2.1 Script for various polynomial evaluations.
CP2.2
The m-le script and step response is shown in Figure CP2.2.
numc = [1]; denc = [1 1]; sysc = tf(numc,denc) numg = [1 2]; deng = [1 3]; sysg = tf(numg,deng) % part (a) sys_s = series(sysc,sysg); sys_cl = feedback(sys_s,[1]) % part (b) step(sys_cl); grid on
Transfer function: s+2 ------------s^2 + 5 s + 5
Step Response
From: U(1) 0.4
0.35
0.3
0.25
Amplitude
To: Y(1)
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time (sec.)
FIGURE CP2.2 Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
72 CP2.3
CHAPTER 2
Mathematical Models of Systems
Given y + 4y + 4y = u with y (0) = y = 0 and U (s) = 1/s, we obtain (via Laplace transform) Y (s) = s(s2 1 1 = . + 4s + 4) s(s + 2)(s + 2)
Expanding in a partial fraction expansion yields Y (s) = 0.25 0.25 0.5 . s s + 2 (s + 2)2
Taking the inverse Laplace transform we obtain the solution y (t) = 0.25 0.25e2t 0.5te2t . The m-le script and step response is shown in Figure CP2.3.
Step Response 0.25
0.2
Amplitude
0.15
0.1
0.05
n=[1]; d=[1 4 4]; sys = tf(n,d); t=[0:0.01:5]; y = step(sys,t); ya=0.25-0.25*exp(-2*t)-0.5*t.*exp(-2*t); plot(t,y,t,ya); grid; title('Step Response'); xlabel('Time (sec)'); ylabel('Amplitude');
0
0
1
2 Time (sec)
3
4
5
FIGURE CP2.3 Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
73
CP2.4
The mass-spring-damper system is represented by mx + bx + kx = f . Taking the Laplace transform (with zero initial conditions) yields the transfer function X (s)/F (s) = s2 1/m . + bs/m + k/m
The m-le script and step response is shown in Figure CP2.4.
m=10; k=1; b=0.5; num=[1/m]; den=[1 b/m k/m]; sys = tf(num,den); t=[0:0.1:150]; step(sys,t)
Step Response
From: U(1) 1.8
1.6
1.4
1.2
Amplitude
1 To: Y(1) 0.8 0.6 0.4 0.2 0 0
50
100
150
Time (sec.)
FIGURE CP2.4 Step response.
CP2.5
The spacecraft simulations are shown in Figure CP2.5. We see that as J is decreased, the time to settle down decreases. Also, the overhoot from 10o decreases as J decreases. Thus, the performance seems to get better (in some sense) as J decreases.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
74
CHAPTER 2
Mathematical Models of Systems
Nominal (solid); O -nominal 80% (dashed); O -nominal 50% (dotted) 18 16 14
Spacecraft attitude (deg)
12 10 8 6 4 2 0 0
10
20
30
40
50 Time (sec)
60
70
80
90
100
%Part (a) a=1; b=8; k=10.8e+08; J=10.8e+08; num=k*[1 a]; den=J*[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[1]); % % Part (b) and (c) t=[0:0.1:100]; % % Nominal case f=10*pi/180; sysf=sys_cl*f ; y=step(sysf,t); % % O -nominal case 80% J=10.8e+08*0.8; den=J*[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[1]); sysf=sys_cl*f ; y1=step(sysf,t); % % O -nominal case 50% J=10.8e+08*0.5; den=J*[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[1]); sysf=sys_cl*f ; y2=step(sysf,t); % plot(t,y*180/pi,t,y1*180/pi,'--',t,y2*180/pi,':'),grid xlabel('Time (sec)') ylabel('Spacecraft attitude (deg)') title('Nominal (solid); O -nominal 80% (dashed); O -nominal 50% (dotted)')
FIGURE CP2.5 Step responses for the nominal and o-nominal spacecraft parameters.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
75
CP2.6
The closed-loop transfer function is T (s) = 4s6 + 8s5 + 4s4 + 56s3 + 112s2 + 56s , (s)
p=
num1=[4]; den1=[1]; sys1 = tf(num1,den1); num2=[1]; den2=[1 1]; sys2 = tf(num2,den2); num3=[1 0]; den3=[1 0 2]; sys3 = tf(num3,den3); num4=[1]; den4=[1 0 0]; sys4 = tf(num4,den4); num5=[4 2]; den5=[1 2 1]; sys5 = tf(num5,den5); num6=[50]; den6=[1]; sys6 = tf(num6,den6); num7=[1 0 2]; den7=[1 0 0 14]; sys7 = tf(num7,den7); sysa = feedback(sys4,sys6,+1); sysb = series(sys2,sys3); sysc = feedback(sysb,sys5); sysd = series(sysc,sysa); syse = feedback(sysd,sys7); sys = series(sys1,syse) poles % pzmap(sys) % p=pole(sys) z=zero(sys)
Polezero map
2.5
7.0709 -7.0713 1.2051 + 2.0863i 1.2051 - 2.0863i 0.1219 + 1.8374i 0.1219 - 1.8374i -2.3933 -2.3333 -0.4635 + 0.1997i -0.4635 - 0.1997i z= 0 1.2051 + 2.0872i 1.2051 - 2.0872i -2.4101 -1.0000 + 0.0000i -1.0000 - 0.0000i
2
1.5
1
0.5
Imag Axis
0
-0.5
-1
-1.5
-2
-2.5 -8
-6
-4
-2
0
2
4
6
8
Real Axis
FIGURE CP2.6 Pole-zero map.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
76
CHAPTER 2
Mathematical Models of Systems
where (s) = s10 + 3s9 45s8 125s7 200s6 1177s5 2344s4 3485s3 7668s2 5598s 1400 . CP2.7 The m-le script and plot of the pendulum angle is shown in Figure CP2.7. With the initial conditions, the Laplace transform of the linear system is (s) = s2 0 s . + g/L
To use the step function with the m-le, we can multiply the transfer function as follows: (s) = s2 0 , 2 + g/L s s
which is equivalent to the original transfer function except that we can use the step function input with magnitude 0 . The nonlinear response is shown as the solid line and the linear response is shown as the dashed line. The dierence between the two responses is not great since the initial condition of 0 = 30 is not that large.
30
20
10 (deg)
0
L=0.5; m=1; g=9.8; theta0=30; % Linear simulation sys=tf([1 0 0],[1 0 g/L]); [y,t]=step(theta0*sys,[0:0.01:10]); % Nonlinear simulation [t,ynl]=ode45(@pend,t,[theta0*pi/180 0]); plot(t,ynl(:,1)*180/pi,t,y,'--'); xlabel('Time (s)') ylabel('\theta (deg)') function [yd]=pend(t,y) L=0.5; g=9.8; yd(1)=y(2); yd(2)=-(g/L)*sin(y(1)); yd=yd';
0 2 4 Time (s) 6 8 10
-10
-20
-30
FIGURE CP2.7 Plot of versus xt when 0 = 30 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
77
CP2.8
The system step responses for z = 3, 6, and 12 are shown in Figure CP2.8.
z=3 (solid), z=6 (dashed), z=12 (dotted) 1.8
1.6
1.4
1.2
1 x(t) 0.8 0.6 0.4 0.2 0 0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE CP2.8 The system response.
CP2.9
(a,b) Computing the closed-loop transfer function yields T (s) = G(s) s2 + 2s + 1 =2 . 1 + G(s)H (s) s + 4s + 3
The poles are s = 3, 1 and the zeros are s = 1, 1. (c) Yes, there is one pole-zero cancellation. The transfer function (after pole-zero cancellation) is T (s) = s+1 . s+3
(d) Only after all pole-zero cancellations have occurred is the transfer function of minimal complexity obtained.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
78
CHAPTER 2
Mathematical Models of Systems
Pole?Zero Map 1
0.8
0.6
0.4
0.2 Imaginary Axi s
0
?-0.2
?-0.4
?-0.6
?-0.8
?-1 ?-3
?-2.5
?-2
?-1.5 Real Axi s
?-1
?-0.5
0
ng=[1 1]; dg=[1 2]; sysg = tf(ng,dg); nh=[1]; dh=[1 1]; sysh = tf(nh,dh); sys=feedback(sysg,sysh) % pzmap(sys) % pole(sys) zero(sys)
>> Transfer function: s^2 + 2 s + 1 ------------s^2 + 4 s + 3
poles
p= -3 -1
zeros
z= -1 -1
FIGURE CP2.9 Pole-zero map.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
3
State Variable Models
Exercises
E3.1 One possible set of state variables is (a) the current iL2 through L2 , (b) the voltage vC2 across C2 , and (c) the current iL1 through L1 . We can also choose vC1 , the voltage across C1 as the third state variable, in place of the current through L1 . We know that the velocity is the derivative of the position, therefore we have dy =v , dt and from the problem statement dv = k1 v (t) k2 y (t) + k3 i(t) . dt This can be written in matrix form as 0 1 y 0 dy + i . = dt v k2 k1 v k3 x = Ax + Bu where A=
E3.2
Dene u = i, and let k1 = k2 = 1. Then,
0
1
1 1
,
B=
0 k3
, and
x=
y v
.
79
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
80 E3.3
CHAPTER 3
State Variable Models
The charactersitic roots, denoted by , are the solutions of det(I A) = 0. For this problem we have det(I A) = det
1
1 ( + 1)
Therefore, the characteristic roots are 1 3 1 3 1 = + j and 2 = j . 2 2 2 2 E3.4 The system in phase variable form is x = Ax + Bu y = Cx where
A=
= ( + 1) + 1 = 2 + + 1 = 0 .
0 0
1 0
8 6 4
0 1,
0 B= 0 ,
C=
100
.
20
E3.5
From the block diagram we determine that the state equations are x2 = (f k + d)x1 + ax1 + f u x1 = kx2 + u and the output equation is y = bx2 . Therefore, x = Ax + Bu y = Cx + Du , where A=
0
k
a (f k + d)
,
B=
1 f
,
C=
0b
and D = [0] .
E3.6
(a) The state transition matrix is (t) = eAt = I + At + 1 22 A t + 2!
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
81
But A2 = 0, thus A3 = A4 = = 0. So, (t) = eAt = I + At =
10 01
+
01 00
t =
1t 01
.
(b) The state at any time t 0 is given by x(t) = (t)x(0) and since x1 (0) = x2 (0) = 1, we determine that x1 (t) = x1 (0) + tx2 (0) = 1 + t x2 (t) = x2 (0) = 1 . E3.7 The state equations are x1 = x2 x2 = 100x1 20x2 + u or, in matrix form x=
0
1
100 20
x +
0 1
u .
So, the characteristic equation is determined to be det(I A) = det 1
= 2 + 20 + 100 = ( + 10)2 = 0 .
100 + 20
Thus, the roots of the characteristic equation are 1 = 2 = 10 . E3.8 The characteristic equation is
1 det(I A) = det 0
0 1 +2
0
5
Thus, the roots of the characteristic equation are 1 = 0 ,
= (2 + 2 + 5) = 0 .
2 = 1 + j 2 and 3 = 1 j 2 .
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82 E3.9
CHAPTER 3
State Variable Models
Analyzing the block diagram yields 1 x1 = x1 + x2 + r 2 3 x2 = x1 x2 r 2 3 y = x1 x2 r. 2 In state-variable form we have x=
1 1
1 2 3 2
The characteristic equation is
x+
1 1
r ,
y= 1
3 x+ 2
1
r.
5 1 s2 + s + 1 = (s + 2)(s + ) = 0 . 2 2 E3.10 (a) The characteristic equation is det[I A] = det
6
1 ( + 5)
So, the roots are 1 = 2 and 2 = 3. (b) We note that (s) = [sI A]
1
= ( + 5)+ 6 = ( + 2)( + 3) = 0 .
Taking the inverse Laplace transform yields the transition matrix (t) =
=
s
6
1 s+5
1
s+5 6 1 . = (s + 2)(s + 3) 1 s 6e2t 6e3t +
3e2t 2e3t e2t + e3t
2e2t
3e3t
E3.11
A state variable representation is x = Ax + Br y = Cx where A=
.
0
1
12 8
,
B=
0 1
,
C=
12 4
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
83
E3.12
The equation of motion is L where vc = 1 C i dt . di + Ri + vc = vin dt
Unit step response 1.6 1.4 1.2 x1: capacitor voltage 1
State response
0.8 0.6 0.4 0.2 x2: inductor current 0 -0.2
0
0.05
0.1
0.15 Time(sec)
0.2
0.25
0.3
FIGURE E3.12 State variable time history for a unit step input.
Selecting the state variables x1 = vc and x2 = i, we have 1 x2 C R 1 1 x2 = x2 x1 + vin . L L L x1 = This can be written in matrix form as x=
0
1/C
1/L R/L
x +
0 1/L
vin .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
84
CHAPTER 3
State Variable Models
When C = 0.001F , R = 4, and L = 0.1H , we have
x= E3.13
0 10
1000 40
x+
0 10
vin .
The step response is shown in Figure E3.12. (a) Select the state variables as x1 = y and x2 = . (b) The corresponding state equation is x1 = x1 ax2 + 2u x2 = bx1 4u or, in matrix form
x=
1 a b 0
x +
2 4
u
and x =
x1 x2
.
(c) The characteristic equation is
det[I A] = det So, the roots are
+1 a b
= 2 + + ab = 0 .
1 1 1 4ab . = 22 E3.14 Assume that the mass decay is proportional to the mass present, so that M = qM + Ku where q is the constant of proportionality. Select the state variable, x, to be the mass, M . Then, the state equation is x = qx + Ku . E3.15 The equations of motion are mx + kx + k1 (x q ) + bx = 0 mq + kq + bq + k1 (q x) = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
85
In state variable form we have
(k+k1 ) m x= 0
k1 m
0
1
b m
0
k1 m
0 0 1
b m
0 0
0 (k+k1 ) m
where x1 = x, x2 = x, x3 = q and x4 = q . E3.16 The governing equations of motion are
x
m1 x + k1 (x q ) + b1 (x q ) = u(t) m2 q + k2 q + b2 q + b1 (q x) + k1 (q x) = 0 . Let x1 = x, x2 = x, x3 = q and x4 = q . Then,
0
1
b m11
0
k1 m1
0
b1 m1
k1 x = m1 0
k1 m2
0
b1 m2
0
+ (k1m2k2 )
1
+ (b1m2b2 )
1 m x + 1 u(t) . 0
0
0
Since the output is y (t) = q (t), then y= 0010 x.
E3.17
At node 1 we have C1 v1 = and at node 2 we have C2 v2 = Let x1 = v1 and x2 = v2 . vb v2 v1 v2 + . R3 R2 va v1 v2 v1 + R1 R2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
86
CHAPTER 3
State Variable Models
Then, in matrix form we have x=
1 R1 C1
+
1 R2 C1
1 R2 C1
R21C2
1 R3 C2
+
1 R2 C2
E3.18
The governing equations of motion are Ri1 + L1
x+
1 R1 C1
0
1 R3 C2
0
va vb
.
di1 + v = va dt di2 L2 + v = vb dt dv . iL = i1 + i2 = C dt
Let x1 = i1 , x2 = i2 , x3 = v, u1 = va and u2 = vb . Then,
x=
R L1
0 0
1 C
0
1 C
1 L1 1 L2
0
y=
001
x + [0] u .
x+ 0
1 L1
1 L2
0
0
0
u
E3.19
First, compute the matrix sI A = Then, (s) is (s) = (sI A)1 where (s) = s2 + 4s + 3, and G(s) = 10 0
s+4 (s) 3 (s)
s
1
3 s+4
.
1 s+4 1 = (s) 3 s
1 (s) s (s)
0 1
=
s2
10 . + 4s + 3
E3.20
The linearized equation can be derived from the observation that sin when 0. In this case, the linearized equations are k g + + =0 . L m
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
87
Let x1 = and x2 = . Then in state variable form we have x = Ax y = Cx where A=
0
1
g/L k/m
E3.21
The transfer function is
,
C=
10
,
and
x(0) =
(0) (0)
.
G(s) = C [sI A]1 B + D = The unit step response is
s2
1 . + 2s + 1
y (t) = 1 + et + tet .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
88
CHAPTER 3
State Variable Models
Problems
P3.1 The loop equation, derived from Kirchos voltage law, is di 1 R 1 = v i vc dt L L L where vc = 1 C i dt .
(a) Select the state variables as x1 = i and x2 = vc . (b) The corresponding state equations are x1 = R 1 1 v x1 x2 L L L 1 x1 . x2 = C
(c) Let the input u = v . Then, in matrix form, we have x=
-R/L v 1/L 1/s x1 -1/L 1/C 1/s x2
R/L 1/L 1/C 0
x+
1/L 0
u .
FIGURE P3.1 Signal ow graph.
P3.2
Let a11 = b11 2 2R1 R2 , a22 = , (R1 + R2 )C (R1 + R2 )L 1 R2 = b12 = , b21 = b22 = . (R1 + R2 )C (R1 + R2 )L
The corresponding block diagram is shown in Figure P3.2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
89
2/(R1+R2)C
v1 1/(R1+R2)C 1/s x1
(a) v2 1/(R1+R2)C
R2 1/s x2
2R1R2/(R1+R2)C
a 11 v1 (b) b21 v2 b22 a 22 1/s x2 b11 b12 1/s x1
FIGURE P3.2 (a) Block diagram. (b) Signal ow graph.
P3.3
Using Kirchos voltage law around the outer loop, we have L diL vc + v2 v1 = 0 . dt
Then, using Kirchos current law at the node, we determine that C dvc = iL + iR , dt
where iR is the current through the resistor R. Considering the right loop we have iR R v2 + vc = 0 or iR = vc v2 + . R R
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
90
CHAPTER 3
State Variable Models
Thus, dvc vc iL v2 = + dt RC C RC and diL vc v1 v2 = + . dt dt L L
In matrix form, the state equations are
x1 x2
where x1 = iL and x2 = vc . The signal ow graph is shown in Figure P3.3.
=
0
1/L
1/C 1/RC
x1 x2
+
1/L 1/L 0 1/RC
v1 v2
,
v1 1/L -1/L -1/C v2 1/L 1/s 1/RC x1 -1/RC 1/s x2
FIGURE P3.3 Signal ow graph.
P3.4
(a) The block diagram model for phase variable form is shown in Figure P3.4a. The phase variable form is given by
x=
0 0
1 0
10 3 2 521 x.
0 0 1 x + 0 r
1
y=
(b) The block diagram in input feedforward form is shown in Figure P3.4b. The input feedforward form is given by
x=
10 0 0 100
2 1 0 1 3 0 1 x + 2 r (t) x.
5
y=
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
91
1 2 x3 x2 1 s 1 s
R(s)
+
--
1 s 2 3
x1
5
+
++
Y(s)
10 (a) 1 2
+
R(s)
5
. . x3 1 + + x2 s-
1+ s-
+
. x1
1 s 2
Y(s)
3 10 (b)
FIGURE P3.4 (a)Block diagram model for phase variable form. (b) Block diagram model for input feedforward form.
P3.5
(a) The closed-loop transfer function is T (s) = s3 s+1 . + 5s2 5s + 1
(b) A matrix dierential equation is x = Ax + Bu y = Cx where
A=
01 00
1 5 5
0 1,
0 B= 0 ,
C=
110
.
1
The block diagram is shown in Figure P3.5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
92
CHAPTER 3
State Variable Models
1 R(s)
+
--
1 s 5
x3
1 s
x2
1 s
x1
1
+
+
Y(s)
-5 1
FIGURE P3.5 Block diagram model.
P3.6
The node equations are dv1 vi v1 + iL =0 dt 4000 dv2 v2 0.0005 iL + i3 = 0 dt 1000 diL 0.002 + v2 v1 = 0 . dt 0.00025 Dene the state variables x1 = v1 Then, x = Ax + Bu where
1 A= 0
x2 = v2
x3 = iL .
500 500
0 4000 2 2000 , 0
P3.7
Given K = 1, we have KG(s)
1 B= 0
0
0
2000
0
(s + 1)2 1 = . s s(s2 + 1)
We then compute the closed-loop transfer function as T (s) = s1 + 2s2 + s3 s2 + 2s + 1 = . 3s3 + 5s2 + 5s + 1 3 + 5s1 + 5s2 + s3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
93
(a) The state variable model is
0 0
1 0
x=
1/3 5/3 5/3 121 x.
0 0 1 x + 0 r
1/3
y=
(b) We check the roots of the characteristic equation 5 5 1 det[sI A] = s3 + s2 + s + = 0 , 3 3 3 to nd s1 = 0.2551 and s2,3 = 0.7058 0.8991j .
All roots lie in the left hand-plane, therefore, the system is stable. P3.8 The state-space equations are x1 = x2 ku x2 = g x3 x3 = u . This is a set of nonlinear equations. P3.9 (a) The closed-loop transfer function is T (s) = 10 10s3 = , Js3 + (b + 10J )s2 + 10bs + 10K1 1 + 10.1s1 + s2 + 5s3
where K1 = 0.5, J = 1, and b = 0.1. (b) A state-space model is
0 0
1 0
=
x=
5 1 10.1 100 x.
0 0 1 x + 0 r
10
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
94
CHAPTER 3
State Variable Models
(c) The characteristic equation is
s det[sI A] = det 0
1 s 1
0 1 s + 10.1
5
The roots of the characteristic equation are s1 = 10.05 P3.10
= s3 + 10.1s2 + s + 5 = 0 .
and s2,3 = 0.0250 0.7049j .
All roots lie in the left hand-plane, therefore, the system is stable. (a) From the signal ow diagram, we determine that a state-space model is given by x= y=
K1 y1 y2
K2
K1 K2
=
10 01
x + x .
K1 K2 K1 K2
r1 r2
(b) The characteristic equation is det[sI A] = s2 + (K2 + K1 )s + 2K1 K2 = 0 . (c) When K1 = K2 = 1, then A=
1
1
1 1
The state transition matrix associated with A is =L P3.11
1
.
[sI A]
1
=e
The state transition matrix is (t) =
t
cos t
sin t
sin t cos t
.
(2t 1)et 2tet
(2t + 1)et
2tet
.
So, when x1 (0) = x2 (0) = 10, we have x(t) = (t)x(0)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
95
or x1 (t) = 10et x2 (t) = 10et P3.12 (a) The phase variable representation is
0 0
1 0
x=
48 44 12
0 0 1 x + 0 r
1
y = [40 8 0]x . (b) The canonical representation is 0 2 z = 0 4
0
0 6
0 0.5728 0 z + 4.1307
4.5638
r
y = [5.2372 0.4842 0.2191]z (c) The state transition matrix is (t) = where
. . . . 1 (t). 2 (t). 3 (t)
,
1 (t) = 6e6t + 12e4t 6e2t
e6t 3e4t + 3e2t
36e6t 48e4t + 12e2t
2 (t) =
5 3 6t 2e4t + 4 e2t 4e 9 e6t + 8e4t 5 e2t 2 2
27e6t 32e4t + 5e2t
3 (t) =
1 6t 1 e4t + 1 e2t 8e 4 8 3 6t 4t 1 e2t 4e +e 4 9 6t e 4e4t + 1 e2t 2 2
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
96 P3.13
CHAPTER 3
State Variable Models
(a) The RLC circuit state variable representation is x=
10 4 6 0
x+
4 0
u .
The characteristic equation is s2 + 10s + 24 = 0 . All roots of the characteristic equation (that is, s1 = 4 and s2 = 6) are in the left half-plane; therefore the system is stable. (b) The state transition matrix is (t) = (c) Given x1 (0) = 0.1 , we have i(t) = x1 (t) = 0.3e6t 0.2e4t vc (t) = x2 (t) = 0.3e6t + 0.3e4t . (d) When x(0) = 0 and u(t) = E , we have
t
3e6t + 3e4t 2e6t + 3e4t
3e6t 2e4t
2e6t + 2e4t
.
x2 (0) = 0 and e(t) = 0 ,
x(t) =
0
(t )Bu( )d ,
where Bu(t) = Integrating yields x1 (t) = (2e6t + 2e4t )E x2 (t) = (1 + 2e6t 3e4t )E . P3.14 A state space representation is x = Ax + Br , y = Cx
4E 0
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
97
where
0 0 0
1 0 0
0 1 0
A=
4 34 23 12
0 0 1
,
B=
0 0
, 0
C = [10 1 0 0] .
1
P3.15
A state variable representation is
0 0
1 0
x=
20 31 10
0 0 1 x+ 0 r
1
y = [20 5 0]x . The block diagram is shown in Figure P3.15.
5 R(s)
+
--
1 s 10
x3
1 s
x2
1 s
x1
20
+
+
Y(s)
31 20
FIGURE P3.15 Block diagram model.
P3.16
(a) The state and output equations are x1 x2 x3 y = x2 = x3 = 500x1 50x2 15x3 + 500r = x1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
98
CHAPTER 3
State Variable Models
or, in matrix form
0 0
1 0
x=
500 50 15 100 x
0 0 1 x + 0 r
500
y=
(b) The unit step response (shown in Figure P3.16) is stable and settles out in under 10 seconds.
1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0 0
1
2
3
4
5 Time (secs)
6
7
8
9
10
FIGURE P3.16 Unit step response.
(c) The characteristic equation is s3 + 15s2 + 50s + 500 = 0 and the roots are s1 = 13.9816 s2,3 = 0.5092 5.9584j . All the poles lie in the left half-planethe system is stable.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
99
P3.17
(a) The characteristic equation is s det(sI A) = det 0.0071 0
3 2
1 s + 0.111 0.07
0
s + 0.3
= s + 0.411s + 0.032s + 0.00213 = 0 .
0.12
The roots are s1 = 0.3343 and s2,3 = 0.0383 0.0700j .
All the poles lie in the left half-plane, therefore, the system is stable. (b) The solution of the system to a step of magnitude 0.285 is given by x1 (t) = 2.66 0.11e0.33t + e0.038t (2.77 cos 0.07t + 0.99 sin 0.07t) x2 (t) = 0.037e0.33t e0.038t (0.037 cos 0.07t + 0.23 sin 0.07t) x3 (t) = 0.069 0.075e0.33t + e0.038t (0.006 cos 0.07t 0.06 sin 0.07t)
x1 - solid; x2 - dotted; x3 - dashed 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3 -3.5
Step response)
0
20
40 Time (s)
60
80
100
FIGURE P3.17 Step response of magnitude 0.285 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
100 P3.18
CHAPTER 3
State Variable Models
The transfer function is G(s) = C(sI A)1 B = s3 4s 44 . 14s2 + 37s + 20
P3.19
Dene the state variables as x1 = 1 2 1 x2 = o 2 x3 = . o Then, the state equations of the robot are x1 = o x2 o x3 J2 o x2 = x1 J1 + J2 J1 o x3 = x2 + J1 + J2 or, in matrix form
0 x = o a 1
b x2 + J1 b x2 J2
b Km x3 + i J1 J1 o b x3 J2
1 b1
a
b 2 b2
1 0 b1 x + d i
0
where a= P3.20
J1 , (J1 + J2 )
b1 =
b , J1 o
b2 =
b and J2 o
d=
Km . J1 o
The state equation is given by x=
0
1
2 3
x
where x1 (0) = 1 and x2 (0) = 1. The state transition matrix is (t) = e2t + 2et e2t + et 2e2t 2et 2e2t et
.
The system response is
x1 (t) = e2t + 2et x1 (0) + e2t + et x2 (0)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
101 x2 (t) = 2e2t 2et x1 (0) + 2e2t et x2 (0) .
The state response is shown in Figure P3.20.
1 0.8 0.6 0.4 System response 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 0 1 2 3 Time (s) 4 5 6
x1
x2
FIGURE P3.20 Response with x1 (0) = 1 and x2 (0) = 1.
P3.21
The state equation is given by x=
0.693 6.7 1
0 0.693 9.2
The state transition matrix is (t) =
x
where x(0) =
0.3 1016 7 1016
.
e0.103433t 35.5786(e0.103433t e0.0753261t )
0 e0.075326t
The system response is x1 (t) = e0.103433t x1 (0)
.
x2 (t) = 35.5786 e0.103433t e0.0753261t x1 (0) + e0.075326t x2 (0) . The state response is shown in Figure P3.21.
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102
CHAPTER 3
7
State Variable Models
6
Nucleide densities in atoms per unit volume
5
X=Xenon 135 I=Iodine 135
4
3
2
1
0
-1
0
10
20 30 Time (hours)
40
50
FIGURE P3.21 Nuclear reactor state response to initial conditions.
P3.22
From the ow graph we obtain x1 = x2 + h1 u x2 = ho u a1 x2 ao x1 y = x1 . In matrix form we have x=
0
1
ao a1
x +
h1 ho
u
y = [1 0]x . The transfer function is G(s) = C(sI A)1 B = [1 0] where (s) = s2 + a1 s + ao .
s + a1 1 ao s
h1 ho
1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
103
Therefore, G(s) = (h1 (s + a1 ) + ho )/(s) = b1 s + bo . + a1 s + ao
s2
So, the vector dierential equation does represent the ow graph model. P3.23 The governing equations are L C1 di = v2 dt
dv1 1 1 + (v1 v ) + (v1 v2 ) = 0 dt R1 R2 dv2 1 v2 C2 + (v2 v1 ) + i + =0. dt R2 R3
Let u = v, x1 = i, x2 = v1 and x3 = v2 . Then,
0 0
1 C2 1 a
0
1 R1
x=
+
1 R2
1 L 1 C1 R2
1 R3 C2
1 R2 C2
1 R2 C2
+
x+
0
1 R1 C1
0
u
y = [0 0 1]x . P3.24 (a) The phase variable representation is
0 0
1 0
x=
y = [1 0 0]x .
30 31 10
0 0 x + 0 r 1
1
(b) The input feedforward representation is
10 x = 31
y = [1 0 0]x .
30 0 0
1 0 0 0 1 x + 0 r
1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
104
CHAPTER 3
State Variable Models
(c) The physical variable representation is 1 3 x = 0 2 0
y = [1 0 0]x .
0 5
0 0 1 x + 0 r
1
(d) The decoupled representation is 0 3 x = 0 2 y= P3.25
0
1 11 x. 63 2
0 5
0 1 0 x + 1 r
1
The matrix representation of the state equations is x=
30 02
When u1 = 0 and u2 = d = 1, we have
x +
11 01
u1 u2
+
0 1
d .
x1 = 3x1 + u2 x2 = 2x2 + 2u2 So we see that we have two independent equations for x1 and x2 . With U2 (s) = 1/s and zero initial conditions, the solution for x1 is found to be x1 (t) = L1 {X1 (s)} = L1 = L 1 and the solution for x2 is x2 (t) = L1 {X2 (s)} = L1 P3.26 2 s(s 2)
1
1 s(s 3) 1 11 + 3s 3 s 3
=
1 1 e3t 3
1 1 = L 1 + s s2
= 1+e2t .
Since (s) = (sI A)1 , we have (s) =
s+1 2
0 s+3
=
s+3 2
0 s+1
1 (s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
105
where (s) = (s + 1)(s + 3). The state transition matrix is (t) = L1 {(s)} = P3.27
et et e3t
0 e3t
.
The state variable dierential equation is x= and (s) = (sI A)1 =
0
1
y = [1 0]x .
25 6
x +
0 25
r
s+6 1 25 s
1 (s)
where (s) = s2 + 6s + 25. P3.28
Equating the change in angular momentum to the sum of the external torques yields J H cos = b k where b is the damping coecient, k is the spring constant, and J is the wheel moment of inertia. Dening the state variables x1 = and x2 = x and the input u = , we can write the equations of motion as x1 = x2 k b H x2 = x1 x2 + u cos x1 J J J With a small angle assumption (that is, cos x1 1) we have x=
0 k/J
1 b/J x.
y==
10
x +
0 H/J
u
P3.29
The governing equations of motion are m1 y1 + k(y1 y2 ) + by1 = u m2 y2 + k(y2 y1 ) + by2 = 0 y = y2 .
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106
CHAPTER 3
State Variable Models
Let x1 = y1 , x2 = y1 , x3 = y2 and x4 = y2 . Then
0 k x = m1 0
k m2
1
b m1
0
k m1
0 0 1
b m2
0 0
0
k m2
y=
0010
x.
0 1 m x + 1 u 0
0
P3.30
The equations of motion are I q1 + M gL sin q1 + k(q1 q2 ) = 0 J q2 k(q1 q2 ) = u . Let x1 = q1 , x2 = q1 , x3 = q2 , and x4 = q2 and linearize the equations using small angle assumptions (i.e. sin q1 q1 ). Then, we have x1 = x2 k M gL x1 (x1 x3 ) x2 = I I x3 = x4 k 1 x4 = (x1 x3 ) + u . J J
P3.31
Using Kirchos current law, we nd that C dv c = i2 + i3 dt
where i3 = current in R3 . Let i1 = current in R1 . Using Kirchos voltage law, we have L and R1 i1 + R2 i2 + vc = v1 . But i2 = i1 iL , so (R1 + R2 )i1 = v1 vc + R2 iL . diL = v1 R1 i1 dt
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
107
Using Kirchos voltage law once again, we calculate i3 as i3 = v2 vc . R3
Utilizing the above equations, we can solve for diL /dt and dvc /dt, as follows: diL R2 R1 R1 R2 = v1 + vc iL dt L(R1 + R2 ) L(R1 + R2 ) L(R1 + R2 ) vc v1 vc vc R1 iL v2 = + dt C (R1 + R2 ) C (R1 + R2 ) CR3 C (R1 + R2 ) CR3 Dene the state variables x1 = vc and x2 = iL . Then, in matrix form we have x=
+ (R13 (R2 +R3 ) CR R1 +R2 ) R1 L(R1 +R2 )
1 C (RR+R2 ) 1
R1 R2 L(R1 +R2 )
x+
1 C (R1 +R2 ) R2 L(R1 +R2 )
1 CR3
0
v1 v2
y = i2 = P3.32
1 (R1 +R2 )
(R1R1R2 ) +
x+
1 (R1 +R2 )
0
v1 v2
A state variable representation is x=
0
1
3 4
x +
0 30
u . s
The state transition matrix can be computed as follows: = L1 [sI A]1
1 s+4 = L 1 (s) 3
1
=
3 t 1 2 e3t 2e 3 et + 3 e3t 2 2
1 t 1 2 e3t 2e 1 et + 3 e3t 2 2
where (s) = s2 + 4s + 3 = (s + 1)(s + 3) . P3.33 A state variable representation is m1 = k1 m1 + r m2 = k1 m1 k2 m2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
108
CHAPTER 3
State Variable Models
where k1 and k2 are constants of proportionality. In matrix form, we have x = Ax + Br =
k1
0
k1 k2
x+
1 0
r
where x1 = m1 and x2 = m2 . Let k1 = k2 = 1 and assume that r (t) = 0 and x1 = 1 and x2 = 0. Then x(t) = (t)x(0) =
et
0
tet et
x(0) =
et tet
.
The simulation is shown in Figure P3.33.
1 0.9 0.8
0.4 0.35 0.3
0.7
state history, x(t)
0.6 0.5 0.4 0.3
x1
0.25
x2
x2
0.2 0.15 0.1
0.2 0.1 0 0 0.05 0 0 t=0 0.5 x1 1
5 time (sec)
10
FIGURE P3.33 Actual versus approximate state response.
P3.34
The system (including the feedback) is described by x = Ax =
0
1
1/2 1
x .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
109
The charactersitic equation is
det[I A] = det
1
1/2 + 1
= 2 + +
1 =0. 2
The roots of the characteristic equation are 1 1 1,2 = j . 2 2 The system response is
t t et/2 cos 2 + et/2 sin 2 t 2et/2 sin 2
x(t) = eAt x(0) = =e
et/2
sin
t 2
et/2
cos
t 2
et/2
sin
t 2
t/2
t 2 sin 2 t t cos 2 sin 2
x(0)
where x1 (0) = 0 and x2 (0) = 1. P3.35 (a) The state space representation is
0 x= 0
1 0
y = [8 0 0] x .
8 14 7
0 0 1 x + 0 r
1
(b) The element 11 (t) of the state transition matrix is 8 11 (t) = 3e4t 2e2t + et . 3 P3.36 The state equations are 1 8 h = x1 = [80 50h] = x1 + x2 50 5 = x2 = = x3 Km Km Kb Km Ka 353 25000 = x3 = ia = + vi = x3 + vi . J JRa JRa 30 3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
110
CHAPTER 3
State Variable Models
In state variable form, we have
8 1 5 x= 0 0
0 1
0
0 353 30
x +
0 0
25000 3
vi .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
111
Advanced Problems
AP3.1 With the state variables are dened as
x z= x ,
i
the nonlinear equations of motion are
z1 z = g 2 z3
z2
K (Io +z3 )2 m (Xo +z1 )2
1 L (v
Rz3 )
where the control is the voltage v . We assume that z1 = x is measurable. The linearized equations of motion are z = Az + Bv y = Cz
,
where
A=
0
2 2K Io 3 m Xo
1 0 0
0
KI 2m Xo2 o R L
0
The transfer function is
,
0 B = 0 , and C =
1 L
100
.
G(s) = C(sI A)1 B . With the constants R = 23.2 L = 0.508 m = 1.75 K = 2.9 104 Io = 1.06 Xo = 4.36 103
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
112
CHAPTER 3
State Variable Models
the transfer function is G(s) = AP3.2 s3 + 45.67s2 36.38 . + 4493s + 205195
The dierential equation describing the motion of y is my + by + ky = bu + ku . Taking Laplace tranforms (with zero initial conditions) yields the transfer function Y (s) (b/m)s + (k/m) =2 . U (s) s + (b/m)s + (k/m) In state space form, we have
x= y=
0
1
k/m b/m k/m b/m
x.
x +
0 1
u
AP3.3
The transfer function is Y (s) 2s2 + 6s + 5 =3 . R(s) s + 4s2 + 5s + 2 In (nearly) diagonal form, we have 1 1 A = 0 1 0
0 2
0 0,
0 B = 1 , and
C=
111
.
1
The matrix A is not exactly diagonal due to the repeated roots in the denominator of the transfer function. AP3.4 The dierential equations describing the motion of y and q are my + k2 y + k1 (y q ) = f bq + k1 (y q ) = f where k1 = 2 and k2 = 1. Assume the mass m = 1. Then with the state
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
T
113 yyq 01 , we have the state variable model
variables dened as z =
y=
z = 3 0
2/b 0 2/b z
0 z + 2
0 1 1/b
100
f
If we model a large bump at high speeds as an impulse and a small bump at low speeds as a step, then b = 0.8 provides good performance. In both cases, the ride settles out completely in about 10 seconds. AP3.5 The dierential equations describing the motion of x and are (M + m) + M L cos M L sin 2 = kx x g sin + cos x + L = 0 Assuming and are small, it follows that (M + m) + M L = kx x x + L = g Dene the state variables as z = able model is
k/m z= 0
xx 0 gM/m 0
T
. Then, the state vari z
0
1 0 0
k/(Lm) 0 g(M + m)/(Lm) 0
0 0 1
AP3.6 AP3.7
Computing the closed-loop system yields A BK =
1
1
K1 K2
The characteristic polynomial is
,
B=
0 1
, and
C=
21
.
|sI (A BK)| = s2 + (K2 + 1)s + K1 + K2 = 0. The roots are in the left-half plane whenever K2 +1 > 0 and K1 + K2 > 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
114
CHAPTER 3
State Variable Models
Design Problems
CDP3.1
The transfer model of the traction drive, capstan roller, and linear slide was given in CDP2.1 as X (s) rKm = , Va (s) s [(Lm s + Rm )(JT s + bm ) + Kb Km ] where JT = Jm + r 2 (Ms + Mb ) . Dene x1 = x, x2 = x, and x3 = x. Then, a state variable representation is x = Ax + Bva y = Cx where
0 A= 0
1 0
m +K Rm bLm JTb Km
0 1
m +R Lm bLm JTm JT
0
C=
100
.
,
B=
0 0
rKm Lm JT
DP3.1
(a) The equation of motion of the spring-mass-damper is my + by + ky = u or y= Select the state variables x1 = y Then, we have x = Ax + Bu y = Cx and x2 = y . b k 1 y y+ u. m m m
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
115
where A=
0
1
20 9
,
B= s 1
0 1
,
C=
10
.
A is the system matrix. The characteristic equation is det[I A] = det
= s2 + 9s + 20 = 0 .
20 s + 9
The roots of the characteristic equation are s1 = 4 and s2 = 5 , and the transistion matrix is (t) =
20e4t + 20e5t 4e4t + 5e5t
5e4t 4e5t
e4t e5t
.
(b) Assume the initial conditions are x1 (0) = 1 and x2 (0) = 2. The zeroinput response is shown in Figure DP3.1. (c) Suppose we redesign the system by choosing b and k to quickly damp out x2 and x1 . We can select b and k to achieve critical damping.
b/m=9, k/m=20 2 2 Critical damping: b/m=20, k/m=100
1.5 1 1 0 State response, x x1
0.5 State response, x
x1
0
-1
x2
-0.5 -2 -1 x2 -3 -1.5
-2
0
0.5
1 Time(sec)
1.5
2
-4
0
0.5
1 Time(sec)
1.5
2
FIGURE DP3.1 Zero input state response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
116
CHAPTER 3
State Variable Models
If we desire the characteristic polynomial to be pd (s) = (s + 10)2 = s2 + 20s + 100, then we need b = 20 and k = 100. DP3.2 The desired transfer function is Y (s) 10 =2 . U (s) s + 4s + 3 The transfer function derived from the phase variable representation is Y (s) d =2 . U (s) s + bs + a Therefore, we select d = 10, a = 3 and b = 4. DP3.3 Assume the aircraft lands precisely on the centerline. The linearized equations of motion are m3 x3 + KD x3 + K2 (x3 x2 ) = 0 m2 x2 + K2 (x2 x3 ) + K1 (x2 x1 ) = 0 2 m1 x1 = K2 (x1 x2 ) 2 where x1 (0) = x2 (0) = x2 (0) = x3 = 0 and x1 (0) = 60. The system response is shown in Figure DP3.3 where KD = 215. The aircraft settles out at 30 m, although initially it overshoots by about 10 m at 1 second.
45 40 35 30
Amplitude
25 20 15 10 5 0 0
1
2
3
4
5 Time (secs)
6
7
8
9
10
FIGURE DP3.3 Aircraft arresting gear response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
117
DP3.4
We can model the bungi cord system as a mass-spring-damper. This is actually an over-simplication because the bungi cord cannot push the jumper down as a spring wouldit can only exert a restoring force when the cord is stretched (that is, when the jumper exceeds the length, L, of the cord). The problem is nonlinear! When the distance of the jumper from the platform is less than L we should model the cord spring constant and damping as K = 0 and b = 0, respectively. Only gravity acts on the jumper. Also, when x (the jumper velocity) is negative (where we dene positive towards the ground), then we should model b = 0. A reasonable set of equations of motion are x1 = x2 K b x2 = x1 x2 + g m m where x1 is the distance measured from the top of the platform and x2 is the jumper velocity. For the initial conditions we have x1 (0) = 10 and x2 (0) = 0. A reasonable set of parameters for the bungi cord are L = 40 m, K = 40 N/m and b = 20 kg/m. The system response is shown in Figure DP3.4 for a person with m = 100 kg. The accelerations experienced by the jumper never exceed 1.5 g.
global MASS GRAVITY LENGTH K b MASS=100; HEIGHT=100; GRAVITY=9.806; LENGTH=40; SPRINGCONSTANT=40; SPRINGDAMPING=20; x0=[10;0]; t=0; dt=0.1; n=round(120/dt); for i=1:n; if x0(1)<LENGTH K=0; b=0; elseif x0(2)<0 b=0; else K=SPRINGCONSTANT; b=SPRINGDAMPING; end function [xdot] = bungi(t,x) tf=t+dt; global MASS GRAVITY LENGTH K b [ T,x] = ode45('bungi',[t tf ],x0); xdot(1)=x(2); xs(i,:)=x(length(x),:); t=tf; xdot(2)=-(K/MASS)*(x(1)LENGTH)-(b/MASS)*x(2)+GRAVITY; x0=x(length(x),:); ts(i)=tf; xdot=xdot'; end plot(ts,HEIGHT-xs(:,1)), grid
FIGURE DP3.4 (a) Bungi cord system response m-le script.
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118
CHAPTER 3
90
State Variable Models
80
70
60
Distance (m)
50
40
30
20
10
0
0
20
40
60 Time (sec)
80
100
120
FIGURE DP3.4 CONTINUED: Bungi cord system time history response.
DP3.5
Computing the closed-loop system yields A BK =
0
1
2 K1 3 K2
The characteristic polynomial is
,
B=
0 1
, and
C=
10
.
|sI (A BK)| = s2 + (K2 3)s + K1 + 2 = 0. Suppose that the desired poles are in the left-half plane and are denoted by p1 and p2 . Then the desired characteristic polynomial is (s + p1 )(s + p2 ) = s2 + (p1 + p2 )s + p1 p2 = 0. Equating coecients and solving for K = [K1 K2 ] yields K1 = p1 p2 2 K2 = p1 + p2 + 3.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
119
Computer Problems
CP3.1 The m-le script to compute the state-space models using the ss function is shown in Figure CP3.1.
a= x1 x2 x1 -8 -2.5 x2 2 0 b= u1 x1 4 x2 0 c= x1 x2 y1 -3.5 -1.5 d= u1 y1 3
% Part(a) num = [1]; den = [1 25]; sys = tf(num,den); sys_ss = ss(sys) % Part(b) num = [3 10 3]; den = [1 8 5]; sys = tf(num,den); sys_ss = ss(sys) % Part(c) num = [1 10]; den = [1 3 3 1]; sys = tf(num,den); sys_ss = ss(sys)
a= x1 x1 -25 b= u1 x1 1 c= x1 y1 1 d= u1 y1 0
a= x1 x2 x3 b= u1 x1 2 x2 0 x3 0 c= y1 d= u1 y1 0 x1 x2 x3 0 0.25 1.25 x1 x2 x3 -3 -1.5 -0.25 200 020
FIGURE CP3.1 Script to compute state-space models from transfer functions.
For example, in part (c) the state-space model is x = Ax + Bu y = Cx + Du ,
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120
CHAPTER 3
State Variable Models
where D = [0] and
3 1.5 A= 2 0
0
2
0.25 0, 0
CP3.2
The m-le script to compute the transfer function models using the tf function is shown in Figure CP3.2.
% Part (a) Transfer function: A=[0 1;2 4]; B=[0;1]; C=[1 0]; D=[0]; 1 sys_ss=ss(A,B,C,D); ------------s^2 - 4 s - 2 sys_tf = tf(sys_ss) % Part (b) A=[1 1 0;-2 0 4; 6 2 10]; B=[-1;0;1]; C=[0 1 0]; D=[0]; sys_ss=ss(A,B,C,D); sys_tf = tf(sys_ss) % Part (c) Transfer function: A=[0 1;-1 -2]; B=[0;1]; C=[-2 1]; D=[0]; s-2 sys_ss=ss(A,B,C,D); ------------s^2 + 2 s + 1 sys_tf = tf(sys_ss)
2 B= 0 ,
C=
0 0.25 1.25
.
0
Transfer function: 6s - 48 ----------------------s^3 - 11 s^2 + 4 s - 36
FIGURE CP3.2 Script to compute transfer function models from the state-space models.
CP3.3
For an ideal op-amp, the voltage gain (as a function of frequency) is Vo (s) = where Z1 = R1 + 1 C1 s Z2 (s) Vin (s), Z1 (s)
Z2 =
R2 1 + R2 C2 s
are the respective circuit impedances. Therefore, we obtain Vo (s) = R2 C1 s Vin (s). (1 + R1 C1 s)(1 + R2 C2 s)
The m-le script and step response is shown in Figure CP3.3.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
121
a= x1 x2 b= x1 x2 c= y1 d= y1 u1 0 x1 x2 2.50000 0 u1 4.00000 0 x1 x2 -3.00000 -1.00000 2.00000 0
R1=1000; R2=10000; C1=0.0005; C2=0.0001; numg=[R2*C1 0]; deng=conv([R1*C1 1],[R2*C2 1]); sys_tf=tf(numg,deng) % Part (a) % sys_ss=ss(sys_tf ) % Part (b) % step(sys_ss)
Continuous-time system.
Step Response 2.5
2
1.5 Amplitude
1
0.5
0
0
1
2
3 Time (sec.)
4
5
6
FIGURE CP3.3 The m-le script using the step function to determine the step response.
CP3.4
The m-le script and state history is shown in Figure CP3.4. The transfer function equivalent is G(s) = s3 + 5s2 1 . + 2s + 3
The computed state vector at t = 10 is the same using the simulation and the state transition matrix.
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122
CHAPTER 3
State Variable Models
x1 solid; x2 dotted; x3 dashed
1
0.8
0.6
0.4
0.2 x(t)
0
- 0.2
- 0.4
- 0.6
- 0.8
-1
0
1
2
3
4
5 time (sec)
6
7
8
9
10
a=[0 1 0; 0 0 1; -3 -2 -5]; b=[0;0;1]; c=[1 0 0]; d=[0]; % % Part (a) % sys_ss = ss(a,b,c,d) sys_tf = tf(sys_ss) % % Part (b) % x0 = [0 -1 1]; t = [0:0.1:10]; u = 0*t; [y,t,x] = lsim(sys_ss,u,t,x0); plot(t,x(:,1),t,x(:,2),':',t,x(:,3),'--'); xlabel('time (sec)'), ylabel('x(t)'), grid title('x1 - solid; x2 - dotted; x3 - dashed') xf_sim = x(length(t),:)' % % Part (c) % dt = 10; Phi = expm(a*dt); xf_phi = Phi*x0'
Transfer function: 1 --------------------s^3 + 5 s^2 + 2 s + 3
xf_sim = -0.2545 0.0418 0.1500 xf_phi = -0.2545 0.0418 0.1500
FIGURE CP3.4 The m-le script using the lsim function to determine the step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
123
CP3.5
The two state-space models represent the same transfer function, as shown in Figure CP3.5. The transfer function in both cases is G(s) = s3 + 8s2 4 . + 5s + 4
We see that a state-space representation of a transfer function is not unique.
a1=[0 1 0; 0 0 1; -4 -5 -8]; b1=[0;0;4]; c1=[1 0 0]; d1=[0]; % % Part (a) % sys_ss = ss(a1,b1,c1,d1); sys_tf = tf(sys_ss) % % Part (b) % a2=[ 0.5000 0.5000 0.7071; -0.5000 -0.5000 0.7071; -6.3640 -0.7071 -8.0000]; b2=[0;0;4]; c2=[0.7071 -0.7071 0]; d2=[0]; sys_ss = ss(a2,b2,c2,d2); sys_tf = tf(sys_ss)
Transfer function: 4 --------------------s^3 + 8 s^2 + 5 s + 4
Transfer function: 4 --------------------s^3 + 8 s^2 + 5 s + 4
FIGURE CP3.5 Comparison of the transfer functions of two state-space models.
CP3.6
The m-le script and impulse response are shown in Figure CP3.6. The controller state-space representation is x = 2x + u y=x and the plant state-space representation is x= y=
2 2 2 01 0 x
x+
0.5 0
u
The closed-loop system state variable representation is x = Ax + Bu y = Cx + Du ,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
124
CHAPTER 3
State Variable Models
where D = [0] and
2 2 A= 2 0
0 1 2
0.5 0,
0 B= 0 ,
C=
010
.
1
numc=[1]; denc=[1 2]; sys_tfc = tf(numc,denc)
numg=[1]; deng=[1 2 4]; sys_tfg = tf(numg,deng) % % Part (a) % sys_ssc = ss(sys_tfc) % % Part (b) % sys_ssg = ss(sys_tfg) % % Part (c) % sys_s = series(sys_ssc,sys_ssg); sys_cl = feedback(sys_s,[1]); impulse(sys_cl)
Impulse Response
From: U(1) 0.1
0.08
0.06
Amplitude
To: Y(1)
0.04
0.02
0
-0.02
0
1
2
3
4
5
6
Time (sec.)
FIGURE CP3.6 Computing the state-space representations and the impulse response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
125
CP3.7
The m-le script and system response is shown in Figure CP3.7.
a=[0 1;-2 -3]; b=[0;1]; c=[1 0]; d=[0]; sys = ss(a,b,c,d); x0=[1;0]; t=[0:0.1:10]; u=0*t; [y,t,x]=lsim(sys,u,t,x0); plot(t,x(:,1),t,x(:,2),'--') xlabel('Time (sec)') ylabel('State Response') legend('x1','x2',-1) grid
1
x1 x2
0.5 State Response 0 -0.5 0
1
2
3
4
5 Time (sec)
6
7
8
9
10
FIGURE CP3.7 Using the lsim function to compute the zero input response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
4
Feedback Control System Characteristics
Exercises
E4.1 (a) The system sensitivity to is given by
T TG S = SG S .
In this case, we have
T SG =
1 1 3s + 1 = = 1 + GH (s) 3s + 101 1 + 3100 s+1
and
G S =
s 3s = , s + 1 3s + 1
where = 3. Therefore,
T S =
3s . 3s + 101
(b) The closed-loop transfer function is T (s) = G(s) 100 100/101 0.99 = =3 = , 1 + GH (s) 3s + 101 c s + 1 101 s + 1
where the time-constant c = 3/101 = 0.0297 second. E4.2 (a) The system sensitivity to K2 is
T SK 2 =
T K2 1 = . K2 T 1 + K1 K2
126
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Exercises
127
(b) The transfer function from Td (s) to Vo (s) is Vo (s) = K2 Td (s) . 1 + K1 K2
(c) We would select K1 1, so that the transfer function from Td (s) to Vo (s) is small. E4.3 (a) The tracking error, E (s) = R(s) Y (s), is given by E (s) = R(s) A/s = . 1 + G(s) 1 + K/(s + 4)2
The steady-state error (computed using the nal value theorem) is ess = lim sE (s) = lim
s0 s0
A 1+
K (s+3)2
=
A . 1 + K/16
(b) A disturbance would be the wind shaking the robot arm. E4.4 (a) The tracking error, E (s) = R(s) Y (s), is given by E (s) = R(s) . 1 + KG(s)
The steady-state position error is computed (using the nal value theorem) to be ess = lim s
s0
A/s A =0. = lim s0 1 + 10K 1 + KG(s) s( s+1) 0.1 . s2
(b) The ramp input of 0.1 m/sec is given by R(s) =
Then, using the nal value theorem, we have ess or 0.1/s2 0.1 = lim s = lim 10K s0 s0 s + 10K 1 + s( s+1) s+1 ess = 0.01 0.1 = . 10K K
,
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128
CHAPTER 4
Feedback Control System Characteristics
We desire ess 0.0001 m, so K E4.5 0.01 = 100 . 0.0001
The light bounces o the surface of the slide and into a detector. If the light fails to hit the detector dead center, the unbalanced electric signal causes the motor to adjust the position of the light source, and simultaneously the lens. The closed-loop transfer function is T (s) = s2 5(s + 2) . + 15s + 10
E4.6
The step response is shown in Figure E4.6.
Step Response 1 0. 9 0. 8 0. 7 0. 6 Amplitude y(t)/A 0. 5 0. 4 0. 3 0. 2 0. 1 0
0
1
2
3
4 Time (sec. )
5
6
7
8
FIGURE E4.6 Step response.
E4.7
(a) The closed-loop transfer function is T (s) = KK1 . s + K1 (K + K2 )
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
129
(b) The sensitivities are
T SK =
T /T s + K1 K2 = K/K s + K1 (K + K2 ) s . s + K1 (K + K2 )
and
T SK 1 =
(c) The transfer function from Td (s) to Y (s) is 1 Y (s) = . Td (s) s + K1 (K2 + K ) Therefore, since E (s) = KY (s) (when R(s) = 0), we have E (s) = and ess = lim sE (s) =
s0
K Td (s) s + K1 (K2 + K ) K . K1 (K + K2 )
(d) With K = K2 = 1, we have T (s) = Then, Y (s) = and y (t) = 1 1 e2K1 t u(t) , 2 K1 1 s + 2K1 s K1 . s + 2K1
where u(t) is the unit step function. Therefore, select K1 = 10 for the fastest response. E4.8 The closed-loop transfer function is T (s) = 46.24K (s + 50)(s + 425) . (s + 200)(s + 425)(s2 + 16.7s + 72.9) + 19652K (s + 50)
The steady-state error is determined to be 1 6.3 ess = lim sE (s) = lim s(1 T (s)) = 1 lim T (s) = 1 T (0) = . s0 s0 s0 s 6.3 + K
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130
CHAPTER 4
Feedback Control System Characteristics
The plots of the steady-state error versus K and the percent overshoot P.O. versus K are shown in Figure E4.8 for 40 K 400.
0.14 45
0.12
40
0.1 percent overshoot (%) 0 100 200 K 300 400 35 steady-state error
0.08
30
0.06
25
0.04 20
0.02
0
15
0
100
200 K
300
400
FIGURE E4.8 (a) Steady-state error. (b) Percent overshoot.
E4.9
(a) The closed-loop transfer function is T (s) = K (s2 + 5s + 6) G(s) =3 1 + G(s)H (s) s + 15s2 + 56s + 60 + 14K
(b) With E (s) = R(s) Y (s) we obtain E (s) = 1 G(s) 1 G(s)(1 H (s)) R(s) = R(s) 1 + G(s)H (s) 1 + G(s)H (s) s3 + (15 K )s2 + (56 5K )s + (60 + 8K ) 1 = . s3 + 15s2 + 56s + 60 + 14K s
Then, using the nal value theorem we nd
s0
lim sE (s) =
(60 + 8K ) . 60 + 14K
(c) The transfer function from the disturbance Td (s) to the output is Y (s) = 1 s3 + 15s2 + 56s + 60 Td (s) = 3 Td (s) . 1 + G(s)H (s) s + 15s2 + 56s + 60 + 14K
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
131
The steady-state error to a unit step disturbance is lim sY (s) = lim s
s0
s0
s3 + 15s2 + 56s + 60 1 60 = . s3 + 15s2 + 56s + 60 + 14K s 60 + 14K
(d) The sensitivity is
T SK =
T K T G K = K T G K T 1 K = (1 + G(s)H (s))2 s + 10
1 + G(s)H (s) 1 = . G(s) 1 + G(s)H (s)
E4.10
(a) The closed-loop transfer function is T (s) = 100K1 (s + 5) Gc (s)G(s) =2 . 1 + Gc (s)G(s)H (s) s + 105s + (500 + 100K1 K2 )
The steady-state tracking error is E (s) = R(s) Y (s) = 1 Gc (s)G(s)(1 H (s)) R(s) 1 + Gc (s)G(s)H (s) s2 + (105 100K1 )s + 500 100K1 (5 K2 ) 1 = s2 + 105s + 500 + 100K1 K2 s
and
s0
lim sE (s) =
5 K1 (5 K2 ) . 5 + K1 K2
(b) The transfer function from the noise disturbance N (s) to the output Y (s) is Y (s) = Gc (s)G(s)H (s) 100K1 K2 N (s) = 2 N (s) . 1 + Gc (s)G(s)H (s) s + 105s + (500 + 100K1 K2 )
The steady-state error to a unit step N (s) = 1/s is
s0
lim sY (s) = lim s
s0
s2
100K1 K2 1 K1 K2 = . + 105s + (500 + 100K1 K2 ) s 5 + K1 K2
(c) The design trade-o would be to make K1 K2 as large as possible to improve tracking performance while keeping K1 K2 as small as possible to reject the noise. E4.11 The closed-loop transfer function is T (s) = s2 K . + 25s + K
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
132
CHAPTER 4
Feedback Control System Characteristics
The sensitivity is
T SK =
T /T s2 + 25s =2 . K/K s + 25s + K
E4.12
(a) The closed-loop transfer function is T (s) = The sensitivity is
T SK 1 =
Gc (s)G(s) K =2 . 1 + Gc (s)G(s)H (s) s + K1 s + K
T /T sK1 = 2 . K1 /K1 s + K1 s + K
(b) You would make K as large as possible to reduce the sensitivity to changes in K1 . But the design trade-o would be to keep K as small as possible to reject measurement noise. E4.13 (a) The closed-loop transfer function is T (s) = 120 Gc (s)G(s) =2 . 1 + Gc (s)G(s)H (s) s + 10s + 120
The steady-state tracking error is E (s) = R(s) Y (s) = 1 R(s) 1 + Gc (s)G(s) s2 + 10s 1 =2 s + 10s + 120 s
and
s0
lim sE (s) = 0 .
(b) The transfer function from the disturbance Td (s) to the output Y (s) is Y (s) = s2 1 Td (s) . + 10s + 120
The steady-state error to a unit step Td (s) = 1/s is
s0
lim sY (s) = lim s
s0
1 1 1 = . s2 + 10s + 120 s 120
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
133
Problems
P4.1 The tank level control block diagram is shown in Figure P4.1.
dH DQ1
+ -
G 1(s)
+
DH
+
K
FIGURE P4.1 Tank level control block diagram.
(a) For the open-loop system the transfer function is G1 (s) = Thus,
G SR 1 =
R . RCs + 1
G1 R 1 = . R G1 RCs + 1
For the closed-loop system, the transfer function is T (s) = Thus,
T SR =
G1 R = . 1 + KG1 RCs + 1 + KR
T R 1 = , R T RCs + 1 + KR T K KR = . K T RCs + 1 + KR
and
T SK =
(b) For the open-loop system H (s) =1. H (s) All disturbances show up directly in the output, thus the open-loop system has no capability to reject disturbances. On the other hand,
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134
CHAPTER 4
Feedback Control System Characteristics
for the closed-loop system we have H (s) 1 RCs + 1 = = . H (s) 1 + KG1 (s) RCs + 1 + KR By selecting K large, we reduce the eects of any disturbances. For example, consider a step disturbance. The steady-state error due to the disturbance is ess = lim s
s0
(RCs + 1) RCs + 1 + KR
A A = . s 1 + KR
As K gets larger, the steady-state error magnitude gets smaller, as desired. (c) Consider the step input Q1 (s) = A . s
Then, for the open-loop system we have ess = lim s (1 G1 )
s0
A = (1 R)A . s
The steady-state error is zero when R = 1, but is sensitive to changes in R. For the closed-loop system we have ess = lim s
s0
1 1 + KG1
A A = . s 1 + KR
By selecting K large, the eect of the disturbance is reduced and is relatively insensitive to changes in R. P4.2 (a) The open-loop transfer function is T (s) = Ka G(s) .
T Therefore, SK1 is undened and T SK a = 1 .
The closed-loop transfer function is T (s) = Therefore,
T SK 1 =
Ka G(s) . 1 + Ka K1 G(s)
T K1 Ka K1 G(s) = K1 T 1 + Ka K1 G(s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
135
and
T SK a =
1 . 1 + K1 Ka G(s)
(b) The tracking error, E (s) = d (s) (s) = (s), since d (s) = 0. The transfer function from the wave disturbance to the output (s) is (s) = G(s) Td (s) . 1 + K1 Ka G(s)
Consider a step disturbance input for the open- and closed-loop systems. For the open-loop system, we have ess = lim sG(s)
s0
A = A . s
Thus, we see that the open-loop system does not have the capability to reduce the eect of disturbances. For the closed-loop system, we have ess = lim s
s0
G(s) 1 + K1 Ka G(s)
2 A An = . 2 s 1 + K1 Ka n
We see that the larger we make K1 Ka , that smaller the eect of the wave disturbance on the output in steady-state. P4.3 (a) The open-loop transfer function is G(s) = K s + 1
where K = k1 ka Eb . Then, computing the sensitivity yields
G SK = 1 .
The closed-loop system transfer function is T (s) = K . s + KKth + 1
Similarly, computing the sensitivity yields
T SK =
1 s + 1 = . 1 + Kth G(s) s + 1 + KKth
(b) For the closed-loop system T (s) = 1/( s + 1) Te (s) Te (s) 1 + KKth /( s + 1) KKth
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136
CHAPTER 4
Feedback Control System Characteristics
when KKth 1. So, by choosing KKth large, we can reduce the eect of the disturbance. This cannot be done with the open-loop system. (c) Consider the step input Edes (s) = A . s
The tracking error for the open-loop system is E (s) = Edes (s) T (s) . Thus, ess = lim s 1
s0
K s + 1
A s
= (1 K )A .
So, ess = 0 when K = 1, but is sensitive to changes in K . The tracking error for the closed-loop system is E (s) = and ess = lim sE (s) = lim s
s0 s0
s + 1 + K (Kth 1) Edes (s) s + 1 + KKth
s + 1 + K (Kth 1) s + 1 + KKth
A A(1 + K (Kth 1)) = . s 1 + KKth
Selecting Kth = 1 and K 1 reduces the steady-state error. P4.4 (a) The overall transfer function is T (s) = Y (s) M G(s) + U QG(s) = . R(s) 1 + QG(s)
(b) From Eq. (4.16) in Dorf & Bishop, we have
T N D SG = SG SG .
In our case, we nd that
N SG = 1 ,
and
D SG =
QG(s) . 1 + QG(s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
137
Thus,
T SG = 1
QG(s) 1 = . 1 + QG(s) 1 + QG(s)
(c) The sensitivity does not depend upon U (s) or M (s). P4.5 The closed-loop transfer function is T (s) = G1 G(s) . 1 + G1 G(s)
(a) The sensitivity of T (s) to changes in ka is
T Ska =
1 . 1 + G1 G(s)
(b) The transfer function from Td (s) to (s) is (s) = G(s) Td (s) . 1 + G1 G(s)
Since we want (s) due to a disturbance, E (s) = (s) and ess = lim sE (s) = lim s
s0 s0
G(s) 1 + G1 G(s)
10 10 = . s ka
Since our maximum desired error magnitude is ess = we select ka 5730 . (c) The open-loop transfer function is (s) = G(s)Td (s) . So, ess = lim sG(s)
s0
0.10o = 0.001745 rad , 180
10 s
.
P4.6
The closed-loop transfer function is T (s) = G1 G(s) . 1 + G1 G(s)
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138
CHAPTER 4
Feedback Control System Characteristics
(a) The sensitivity is
T SK e =
1 (1 s + 1)(e s + 1) = . 1 + G1 G(s) (1 s + 1)(e s + 1) + K1 Ke
(b) The speed is aected by the load torque through the transfer function V (s) = Kg G(s) Td (s) . 1 + GG1 (s)
(c) Let R(s) = 30/s , and Ke K1 1 . When the car stalls, V (s) = 0. Using the nal value theorem, we nd
s0
lim s
Kg G(s) 1 + GG1 (s)
d G1 G(s) 30 + lim s s0 s 1 + GG1 (s) s +Kg Ke K1 Ke = d + 30 1 + Ke K1 1 + K1 Ke
.
Since Ke K1 1, we have Vss = d When Vss = 0, we have d = Thus, if Kg =2, K1 then d = 15 percent grade ( i.e. d = 15 ft rise per 100 ft horizontally) will stall the car. P4.7 (a) Let G1 (s) = k1 , G2 (s) = k2 , and s( s + 1) H (s) = k3 + k4 s . 30K1 . Kg Kg K1 + 30 .
Then the transfer function from TL (s) to Y (s) is Y (s) = G2 (s) k2 TL (s) = TL (s) . 1 + G1 G2 H (s) s( s + 1) + k1 k2 (k3 + k4 s)
(b) The sensitivity of the closed-loop system to k2 is
T Sk2 =
1 , 1 + G1 G2 H (s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
139
where T (s) is the closed loop transfer function T (s) = G1 G2 (s) k1 k2 = . 1 + G1 G2 H (s) s( s + 1) + k1 k2 (k3 + k4 s)
(c) The error is given by E (s) = R(s) T (s)R(s) . With R(s) = we have 1 1 ess = lim s(1 T (s)) = 1 T (0) = 1 . s0 s k3 P4.8 (a) The sensitivity is
T SK =
1 , s
(0.1s + 1)(s2 + 20s + 200) 1 = . 1 + G1 G(s) (0.1s + 1)(s2 + 20s + 200) + 200K
(b) The transfer function from Td (s) to Y (s) is Y (s) = P4.9 G(s) 200(0.1s + 1) Td (s) = Td (s) . 1 + G1 G(s) (0.1s + 1)(s2 + 20s + 200) + 200K
(a) Computing the derivative of R with respect to i yields dR 0.201R = . di (i 0.005)3/2 When vout = 35 volts, we have i= 35 = 7ma . 5000
At the operating point i = 7 ma, we nd from Figure P4.9(b) in Dorf & Bishop that R 20K (note: If we use the given formula, we nd that R 8.2K when i = 7 ma, thus we see that the formula is just an approximation to the plot). Using R = 20K , we have dR 0.402 104 = = 45 kohms/ma . di 0.896 104
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140
CHAPTER 4
Feedback Control System Characteristics
The transfer function (valid around the operating point) is T (s) = = Vout (s) K = Vin (s) (s + 1) + K K . s + 1 + 9KI
45I 5
The photosensor block diagram is shown in Figure P4.9.
vin
+
-
K ts + 1
vout
45 I i (ma)
FIGURE P4.9 Photosensor block diagram.
1 5
(b) The sensitivity of the system to changes in K is
T SK =
s + 1 . s + 1 + 9KI
P4.10
(a) and (b) The paper tension control block diagram is shown in Figure P4.10.
D T (s) V 1 + DV 1(s) k3 E o( s) + Km ts + 1 2 k1 w ( s)
o
2 +
R(s )
+
1 s
T( s)
-
k2
Y (s)
FIGURE P4.10 Paper tension control block diagram.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
141
(c) The closed-loop transfer function, Tc , is given by T (s) 2Km s( s+1) Tc (s) = = = . 4Km k2 2 + s + 4Km k2 R(s) 1 + k1 s( s+1) s k1 The sensitivity of Tc to changes in Km is
T SKcm = 2Km
1 1+
4Km k2 k1 s( s+1)
=
s2
s( s + 1) . + s + 4Km k2 k1
(d) The transfer function from V1 (s) to T (s) is T (s) = 1+
1 s 4Km k2 k1 s( s+1)
V1 (s) =
s+
4Km k2 k1 ( s+1)
1
V1 (s) .
When V1 (s) = A/s, we have T (s) = and
t
s2
( s + 1) A . 4Km k2 s + s + k1 Ak1 . 4Km k2
lim T (t) = lim sT (s) =
s0
P4.11
(a) The closed-loop transfer function is T (s) = Gc G(s) K K = = . 2 + 12s + 1 + K 1 + Gc G(s) (10s + 1)(2s + 1) + K 20s
T (b) The sensitivity SK is T SK =
T K 1 = = K T 1 + Gc G(s) 1+
1
K (10s+1)(2s+1)
.
(c) Dene E (s) = R(s) Y (s). Then E (s) = With R(s) = we have ess = lim sE (s) =
s0
R(s) 20s2 + 12s + 1 = R(s) . 1 + GGc (s) 20s2 + 12s + K + 1
A , s A . 1+K
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
142
CHAPTER 4
Feedback Control System Characteristics
(d) We want |e(t)| 0.02A as t . So, 0.02A implies K 49 . P4.12 (a) The two transfer functions are T1 (s) = and T2 (s) = Y (s) K1 K2 = . R(s) (1 + 0.09K1 )(1 + 0.09K2 ) Y (s) K1 K2 = R(s) 1 + 0.0099K1 K2 A K +1
(b) When K1 = K2 = 100 , then T1 (s) = and Ts (s) =
T (c) The sensitivity SK11 is T SK11 =
(100)2 = 100 1 + 0.0099(100)2 (100)2 = 100 . (1 + 0.09(100))2
T1 K1 1 = = 0.01 , K1 T1 1 + 0.0099K1 K2
T when K1 = K2 = 100. The sensitivity SK21 is T SK21 =
T2 K1 1 = = 0.1 , K1 T2 1 + 0.09K1
when K1 = 100. Thus,
T SK11 = T SK21 . 10
P4.13
(a) Let N (s) = G1 (s) + kG2 (s) and Td (s) = G3 (s) + kG4 (s). Then
T Sk =
N k D k G2 k G4 k = k N k D G1 + kG2 G3 + kG4 k(G2 G3 G1 G4 ) = . (G1 + kG2 )(G3 + kG4 )
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
143
(b) The closed-loop transfer function is T (s) = M G(s) + kU G(s) G1 (s) + kG2 (s) = . 1 + kGH (s) G3 (s) + kG4 (s)
Then using result from (a), we have
T Sk =
k(U G(s) M G2 H (s)) . (M G(s) + kU G(s))(1 + kGH (s))
P4.14
The closed-loop transfer function is T (s) = Then
T N D Sa = Sa Sa ,
N (s) 12(s + 5) G(s) = = . 1 + G(s) Td (s) s(s + a)(s + 1) + 10(s + 6)
but
N Sa = 0 .
Let G(s) = p(s) , q (s)(s + a)
where p(s) = 12(s + 5) and q (s) = s(s + 1). Then T (s) = and
T D Sa = Sa =
p(s) G(s) = , 1 + G(s) q (s)(s + a) + p(s)
a
dD a aq (s) s +a = = . da D q (s)(s + a) + p(s) 1 + G(s)
P4.15
(a) The closed-loop transfer function for the disturbance to the output is Y (s) G(s) = , Td (s) 1 + KG(s) with R = 0. The steady-state deviation is yss = lim s
s0
G(s) 1 + KG(s)
1 G(0) 1 = = . s 1 + KG(0) 1+K
So, with K = 5 we have yss = 1/6, and with K = 25 we have yss = 1/26.
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144
CHAPTER 4
Feedback Control System Characteristics
(b) Considering the rudder input, we have Y (s) = G(s)Td (s) + KG(s)R(s) G(s)(Td (s) + KR(s)) = . 1 + KG(s) 1 + KG(s)
Setting R(s) = Td (s) yields Y (s) = 0. K P4.16 (a) Let G1 (s) = Then T2 (s) = G1 (s) G2 Gc (s) To (s) + T2d (s) . 1 + G2 Gc (s) 1 + G2 Gc (s) 1 (1 s + 1)(2 s + 1) and G2 (s) = G1 (s) . 100
(b) We can equivalently consider the case of a step input, T2d = A/s, To = 0, and zero initial conditions. Thus, T2 (s) = Gc G2 5 A T2d = , 2 + 60s + 6 s 1 + Gc G2 500s
where Gc (s) = 500. The transient response is shown in the Figure P4.16 for a unit step input (A = 1).
Unit step response, A=1 1 0.9 0.8 0.7 0.6 Tp=34.3 sec p.o. = 12.8% Ts=66.7 sec
T2
0.5 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 time (sec) 60 70 80 90 100
FIGURE P4.16 Two tank temperature control system response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
145
(c) With E (s) = T2d (s) T2 (s) , we have E (s) = T2d (s) M (s)T2d (s) where M (s) = Then ess = lim s(1 M (s))
s0
G2 Gc (s) . 1 + G2 Gc (s)
5 A A = (1 M (0))A = (1 )A = . s 6 6
P4.17
(a) The closed-loop transfer function is (s) 600 6000 = =2 . 2 + s + 600 d (s) 0.1s s + 10s + 6000 The solution for a step input is (t) = 1 1.0021e5.0349t sin(77.2962t + 1.5058). (b) The transfer function from the disturbance to the output is (s) 1 = . 2 + s + 600 Td (s) 0.1s Thus, ss = lim s (s) =
s0
A . 600
Therefore, the disturbance input magnitude reduced by 600 at the output. (c) Using the nal value theorem we have (for d (s) = 1/s2 ) ess = lim sE (s) = lim s(1 T (s))d (s)
s0 s0 s0
= lim s
0.1s2 + s 0.1s2 + s + 600
1 1 = . s2 600
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146
CHAPTER 4
Feedback Control System Characteristics
Advanced Problems
AP4.1 The plant transfer function is Gp (s) = The closed-loop output is given by H (s) = 1 GGp (s) Q3 (s) + Hd (s) . 1 + GGp (s) 1 + GGp (s) R . RCs + 1
Therefore, with E (s) = Hd (s) H (s), we have E (s) = since Hd (s) = 0. (a) When G(s) = K , we have ess = lim sE (s) =
s0
1 Q3 (s) , 1 + GGp (s)
1 . 1 + KR
(b) When G(s) = K/s, we have ess = lim sE (s) = 0 .
s0
AP4.2
Dene G(s) = Then, (s)/d (s) = and E (s) = So, ess = lim sE (s) = lim s
s0 s0
Km Gc (s) . s(La s + Ra )(Js + f ) + Km Kb Gc (s)
nG(s) 1 + nG(s)
1 d (s) . 1 + nG(s)
1 A A AKb = = . 1 + nG(s) s 1 + nG(0) Kb + n
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
147
When d (s) = 0 and Td (s) = M/s, we have (s)/Td (s) = If Gc (s) = K , then ess = nM Ra Km (Kb + nK ) n(La s + Ra ) . s(La s + Ra )(Js + f ) + Km Kb + Km Gc n
and if Gc (s) = K/s, we determine that ess = 0. AP4.3 (a) The input R(s) is R(s) = 1 1 s s2
and the disturbance is Td (s) = 0. So, ess = lim s
s0
1 1 1 s = 0.8750 . R(s) = lim s0 1 + 10(2s+4) 1 + G(s)
7s(s+5)
(b) The error plot is shown in Figure AP4.3a.
2 1. 5 1 e(t) 0. 5 0 ?-0.5 ?-1 0
FIGURE AP4.3 (a) Error plot with d(t)=0.
2
4 6 Time (sec)
8
10
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
148
CHAPTER 4
Feedback Control System Characteristics
0
?-0.5
e(t)
?-1
?-1.5
-?2 0
2
4 6 Time (sec)
8
10
FIGURE AP4.3 CONTINUED: (b) Error plot with r(t)=0.
(c) The transfer function from Td (s) to Y (s) (with R(s) = 0 ) is Y (s)/Td (s) = 7s2 70 . + 55s + 40
The steady-state error due to a disturbance Td (s) = 1/s is ess = lim s
s0
7s2
70 1 = 1.75 . + 55s + 40 s
(d) The error e(t) is shown in Figure AP4.3b. AP4.4 (a) The closed-loop transfer function is (s)/V (s) = Ra Js2 Km . + Kb Km s + Km KKt
With v (t) = t, we have V (s) = 1/s2 , and Td (s) = 0. Using the nal value theorem yields ess = lim sE (s) = lim
s0
1 s+
KKm Ra Js+Km Kb
s0
=
Kb 0.1 = . K K
We desire that ess = 0.1 < 0.1 . K
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
149
Therefore, we should select K > 1. For example, we can take K = 8. (b) The transfer function from Td (s) to (s) is given by (s) 10s =2 . Td (s) s + 10s + 100 The error plot is shown in Figure AP4.4, where e(s) = (s) (V (s) = 0.)
0.2 0.18 0.16 0.14 0.12
e(t)
0.1 0.08 0.06 0.04 0.02 0 0 1 2 3 4 5 Time (sec) 6 7 8 9 10
FIGURE AP4.4 Error plot with a ramp disturbance input.
AP4.5
(a) The transfer function from the disturbance Td (s) to the output Y (s) is Y (s) s =3 . 2 + 4s + K Td (s) s + 4s The steady-state error (when Td (s) = 1/s) is ess = lim s
s0
s3
+
4s2
1 s =0. + 4s + K s
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150
CHAPTER 4
Feedback Control System Characteristics
(b) The closed-loop transfer function is Y (s) K =3 . R(s) s + 4s2 + 4s + K The steady-state error (when R(s) = 1/s2 ) is ess = lim s(1 T (s))
s0
s3 + 4s2 + 4s 4 1 = lim = . 2 s0 s(s3 + 4s2 + 4s + K ) s K
(c) Let K = 8. Then, Y (s) s =3 . Td (s) s + 4s2 + 4s + 8 The error plot is shown in Figure AP4.5, for r (t) = 0.
0.15
0.1
0.05
e(t)
0 -0.05 -0.1
0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE AP4.5 Error plot with a step disturbance input and K =8.
AP4.6
(a) The transfer function is Vo (s) 1 + RCs = . V (s) 2 + RCs
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
151
(b) The system sensitivity is dened as
G SC =
G/G . C/C
Therefore, the sensitivity is determined to be
G SC =
RCs = (2 + RCs)(1 + RCs) 1+
1
2 RCs
1+
1 RCs
.
(c) Let V (s) = 1/s. Then Vo (s) = 1 + RCs 1 0.5 0.5RC = + . 2 + RCs s s RCs + 2
Taking the inverse Laplace transform yields vo (t) = 0.5(1 + e2t/RC )u(t) where u(t) is the unit step function. A plot of vo (t) versus t/RC is shown in Figure AP4.6.
1 0.95 0.9 0.85 0.8
Vo
0.75 0.7 0.65 0.6 0.55 0.5 0 0.5 1 1.5 2 2.5 t / RC 3 3.5 4 4.5 5
FIGURE AP4.6 Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
152 AP4.7
CHAPTER 4
Feedback Control System Characteristics
(a) The transfer function from Td (s) to Y (s) is Y (s) 2s = . Td (s) s(s + 2) + 2K (b) The transfer function from N (s) to Y (s) is Y (s) 2K = . N (s) s(s + 2) + 2K (c) Let Td (s) = A/s and N (s) = B/s. Then, ess = yss = lim s
s0
A 2K B 2s + lim s = B . s(s + 2) + 2K s s0 s(s + 2) + 2K s
So, K has no eect on the steady-state errors. However, choosing K = 100 will minimize the eects of the disturbance Td (s) during the transient period. AP4.8 (a) The closed-loop transfer function is T (s) = Kb . s + Kb + 1
(b) The sensitivity is determined to be
T Sb =
T /T s+1 = . b/b s + Kb + 1
(c) The transfer function from Td (s) to Y (s) is Y (s) b = . Td (s) s + Kb + 1 So, choose K as large as possible, to make Y (s)/Td (s) as small as possible. Thus, select K = 50 .
T This also minimizes Sb at low frequencies.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
153
Design Problems
CDP4.1
The model of the traction drive, capstan roller, and linear slide was developed in CDP2.1: (s) Km = . Va (s) s [(Lm s + Rm )(JT s + bm ) + Kb Km ] The step response for the closed-loop system (with the tachometer not in the loop) and various values of the controller gain Ka is shown below.
% System parameters Ms=5.693; Mb=6.96; Jm=10.91e-03; r=31.75e-03; bm=0.268; Km=0.8379; Kb=0.838; Rm=1.36; Lm=3.6e-03; Lm=0; % Controller gain Ka=100; % Motor and slide model Jt=Jm+r^2*(Ms+Mb); num=[Km]; den=[Lm*Jt Rm*Jt+Lm*bm Kb*Km+Rm*bm 0]; sys=tf(num,den); %Closed-loop tf and step response sys_cl=feedback(Ka*sys,[1]); step(sys_cl)
1.5
1 Theta step response
Ka=2 Ka=5 Ka=10 Ka=100
0.5
0
0
0.2
0.4 Time (sec)
0.6
0.8
1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
154 DP4.1
CHAPTER 4
Feedback Control System Characteristics
(a) The transfer function from the load disturbance to the output speed is (s) G(s) s = =2 . Td (s) 1 + Gc G(s) s + 4s + K Thus, the eect on (s) (of a unit step disturbance) at steady-state is
t
lim (t) = lim s
s0
s2
s + 4s + K
1 =0. s
We see that the load disturbance has no eect on the output at steadystate. (b) The system response for 10 K 25 is shown in Figure DP4.1.
K=10,12,16,18,20,23,25 100.04 100.02 100 99.98 99.96 K=25
w(t)
99.94 99.92 99.9 99.88 99.86 99.84 0 0.5
K=10
1
1.5 Time(sec)
2
2.5
3
FIGURE DP4.1 Speed control system response.
For example , if we select K = 16, then n = 4, = response due to a unit step disturbance is (s) = s2 s + 4s + 16 1 s = 1 . (s + 2)2 + 12
1 2,
and the
Hence, if we are originally at (t) = 100 for t < , we have 1 (t) = 100 e2t sin 12t 12 t .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
155
DP4.2
With d = 0, we have (s) = G(s) s T (s) = 3 Td (s) . KK1 d 2 + 9s + KK s + 4s 1 + G(s) s 1
For Td = A/s, we have (s) = A . s3 + 4s2 + 9s + KK1
The system response to a unit step disturbance for various values of KK1 are shown in Figure DP4.2. From the plot we see that when KK1 is small the response is slow but not oscillatory. On the other hand, when KK1 is large the response is fast but highly oscillatory. In fact, if KK1 > 35, the system is unstable. Thus, we might select KK1 = 10 as a reasonable trade-o between fast performance and stability.
Unit step response for KK1=1,5,10,15,20,25
0.12 0.1 0.08 0.06 0.04
KK1=1
KK1=5
q
0.02 0 -0.02 -0.04 -0.06 KK1=25
0
1
2
3
4
5 time(sec)
6
7
8
9
10
FIGURE DP4.2 Aircraft roll angle control system response to a disturbance.
DP4.3
(a) The closed-loop transfer function is T (s) = (s) K =2 . d (s) s + 5s + KK1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
156
CHAPTER 4
Feedback Control System Characteristics
Then, E (s) = (1 T (s))d (s) = So, if 0.99 < K1 < 1.01 , then |ess | < 0.01 . (b) The transfer function from Td (s) to (s) is (s) = s Td (s) . s2 + 5s + KK1 s2 + 5s + K (K1 1) 1 . s2 + 5s + KK1 s
So, with E (s) = (s) and Td (s) = 2/s, we have
s0
lim sE (s) =
2 . KK1
Therefore, we select KK1 > 20 to obtain ess < 0.1. DP4.4 The steady-state error for a step input command is zero for any K1 . The transfer function from Td (s) to Y (s) is Y (s) G(s) 2 = =3 . 2 + 4s + 2K Td (s) 1 + KG(s) s + 5s Thus, the output at steady-state due to a step disturbance Td (s) = A/s is
s0
lim sY (s) =
A . K
We want to maximize K to reduce the eect of the disturbance. As we will see in Chapter 6, we cannot select K too high or the system will become unstable. That is why the problem statement suggests a maximum gain of K = 10. For the design we choose K = 10 . DP4.5 The transfer function from V (s) to Vo (s) is Vo (s)/V (s) = ks s+a
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
157
where k= R2 + R3 R2 and a = 1 . R1 C
Computing the step response, we nd that vo (t) = keat = 5e100t . Solving for R1 , R2 , R3 and C yields R1 C = 0.01 DP4.6 and R2 =4. R3
(a) The closed-loop transfer function is (s) = K/J d (s) . s2 + K/J
Since J > 0, the system is unstable when K < 0 and marginally stable when K > 0. (b) Since the system is marginally stable, the system response does not have a steady-state valueit oscillates indenitely. (c) The closed-loop transfer function is (s) = Js2 KD s + KP d (s) . + KD s + KP
The system is stable for all KD > 0 and KP > 0, given that J > 0. (d) The tracking error E (s) = d (s) (s) is E (s) = Js2 . Js2 + KD s + KP
Therefore, using the nal value theorem we obtain the steady-state value
s0
lim sE (s) = lim s
s0
Js2 1 =0. Js2 + KD s + KP s
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158
CHAPTER 4
Feedback Control System Characteristics
Computer Problems
CP4.1 The transfer function is G(s) = 5 . s2 + 2s + 20
An m-le script which generates the step response is shown in Figure CP4.1. The step response is also shown in Figure CP4.1.
Step Response 0.4
Amplitude
num=[5]; den=[1 2 20]; sys = tf(num,den); axis([0 6 0 1]); t=[0:0.01:6]; step(sys,t) y = step(sys,t); yss = y(length(t))
0.35
0.3
0.25
0.2
0.15
0.1
yss = 0.2496
0.05
0
0
1
2
3 Time (sec)
4
5
6
FIGURE CP4.1 Step response.
The step response is generated using the step function. In the script, the transfer function numerator is represented by num and the denominator is represented by den. The steady-state value is yss = 0.2 and the desired value is 1.0. Therefore, the steady-state error is ess = 0.8 . CP4.2 The step response and an m-le script which generates the step response is shown in Figure CP4.2. The closed-loop transfer function is determined to be T (s) = s2 75 . + 2s + 85
The percent overshoot is P.O. = 71% and the steady-state error is ess = 0.12.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
159
Step Response 1.6 System: sys_cl Peak amplitude: 1.51 Overshoot (%): 71 At time (sec): 0.343
1.4
1.2
num = [75]; den = [1 2 10]; sys = tf(num,den); sys_cl = feedback(sys,[1]) step(sys_cl) y = step(sys_cl);
1 Amplitude
0.8
0.6
0.4
0.2
0
0
1
2
3 Time (sec)
4
5
6
FIGURE CP4.2 Step response.
CP4.3
The step responses and the m-le script which generates the step responses is shown in Figure CP4.3.
7
6
K=10 K=200 K=500
K=[10,200,500]; t=[0:0.01:7]; for i=1:3 num=5*K(i); den=[1 15 K(i)]; sys = tf(num,den) y(:,i)= step(sys,t); end plot(t,y(:,1),t,y(:,2),'--',t,y(:,3),':') legend('K=10','K=200','K=500',-1)
5
4
3
2
1
0
0
1
2
3
4
5
6
7
FIGURE CP4.3 Step responses for K = 10, 100, 500.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
160 CP4.4
CHAPTER 4
Feedback Control System Characteristics
The step response and the m-le script which generates the step response is shown in Figure CP4.4. The closed-loop transfer function is determined to be T (s) = s2 10 . + 3.7s + 10
Using the m-le script, a trial-and-error search on k yields k = 3.7 . The percent overshoot P.O. = 10% and the steady-state value is 1, as expected.
1.4
k = 3.7; % Final value of k=3.7 numcg = [10]; dencg = [1 k 0]; sys_o = tf(numcg,dencg); sys_cl = feedback(sys_o,[1]) t = [0:0.1:5]; [y,t] = step(sys_cl,t); plot(t,y,[0 5],[1.1 1.1],'--'); grid xlabel('Time (sec)'); ylabel('y(t)');
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
Transfer function: 10 ---------------s^2 + 3.7 s + 10
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE CP4.4 Step response.
CP4.5
The closed-loop transfer function is T (s) = K sa+K K =2. K a
where K = 2. When a = 1 and R(s) = 1/s, the nal value is
s0
lim sT (s)R(s) = lim T (s) =
s0
The output is within 2% of the nal value at around t = 4.6 seconds. The plot of the step responses for a = 1, 0.5, 2, 5 is shown in Figure CP4.5. The output is unstable for a > 2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
161
K=2; t=[0:0.1:5]; num=K*[1]; a=[1 0.5 2 5]; for i=1:4 den=[1 -a(i)]; sys = tf(num,den); sys_cl = feedback(sys,[1]); y(:,i)=step(sys_cl,t); end plot(t,y(:,1),t,y(:,2),':',t,y(:,3),'--',t,y(:,4),'-.') axis([0 5 0 5]); xlabel('Time (sec)'), ylabel('y(t)') title('a=1 (solid); a=0.5 (dotted); a=2 (dashed); a=5 (dashdot)')
a=1 (solid); a=0.5 (dotted); a=2 (dashed); a=5 (dashdot) 5
4.5
4
3.5
3 y(t)
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE CP4.5 Step response for a=1, 0.5, 2, and 5.
CP4.6
The transfer function from the disturbance to the output is T (s) = G(s) 1 = . 2 + bs + k + K 1 + K0 G(s) Js 0
The disturbance response is shown in Figure CP4.6. The compensated system response is signicantly reduced from the uncompensated system response. The compensated system output is about 11 times less than the uncompensated system output. So, closed-loop feedback has the advantage of reducing the eect of unwanted disturbances on the output.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
162
CHAPTER 4
Feedback Control System Characteristics
J=1; k=5; c=0.9; num=[1/J]; den=[1 c/J k/J]; sys = tf(num,den); t=[0:0.1:10]; % yu=step(sys,t); % Part (a) K0=50; numk=[K0]; denk=[1]; sysk = tf(numk,denk); sys_cl = feedback(sys,sysk); yc=step(sys_cl,t); % Part (b) plot(t,yu,t,yc,'--') xlabel('Time (sec)'), ylabel('\theta') title('Uncompensated response (solid) & Compensated response (dashed)')
Uncompensated response (solid) & Compensated response (dashed) 0.35
0.3
0.25
0.2
q
0.15 0.1 0.05 0 0
1
2
3
4
5 Time (sec)
6
7
8
9
10
FIGURE CP4.6 Disturbance responses for both the uncompensated and compensated systems.
CP4.7
The step responses for the proportional and PI controller are shown in Figure CP4.7. The steady-state tracking error for the proportional controller is ess = 0.33 . Increasing the complexity of the controller from a proportional controller to a proportional plus integral (PI) controller allows the closed-loop system to track the unit step response with zero steady-state error. The cost is controller complexity, which translates into higher costs ($).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
163
numg=[10]; deng=[1 10]; sysg = tf(numg,deng); t=[0:0.001:0.5]; % Part (a) numc=[2]; denc=[1]; sysc = tf(numc,denc); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); yk=step(sys_cl,t); % Part (b) numc=[2 20]; denc=[1 0]; sysc = tf(numc,denc); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); yp=step(sys_cl,t); % plot(t,yk,t,yp,'--') xlabel('Time (sec)'),ylabel('y(t)') title('Proportional controller (solid) & PI controller (dashed)')
Proportional controller (solid) & PI controller (dashed) 1
0.9
0.8
0.7
0.6 y(t)
0.5
0.4
0.3
0.2
0.1
0
0
0.05
0.1
0.15
0.2
0.25 Time (sec)
0.3
0.35
0.4
0.45
0.5
FIGURE CP4.7 Step response for proportional controller and PI controller.
CP4.8
(a) The closed-loop transfer function is T (s) = G(s) 10s2 + 500s R(s) = 2 R(s) . 1 + G(s)H (s) s + 200s + 5000
The step response is shown in Figure CP4.8a.
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164
CHAPTER 4
Feedback Control System Characteristics
(b) The response of the system to the sinusoidal disturbance N (s) = s2 10 + 100
is shown in Figure CP4.8b. (c) In the steady-state, the magnitude of the peak response is 0.0095 and the frequency is 10 rad/sec (see Figure CP4.8b).
% Part (a) ng=10*[1 0]; dg=[1 100]; sysg=tf(ng,dg); nh=[5]; dh=[1 50]; sysh=tf(nh,dh); sys=feedback(sysg,sysh) gure(1) step(sys) % Part (b) sysn=-feedback(sysg*sysh,1) % This is the sinusoidal input syss=tf([10],[1 0 100]); gure(2) t=[0:0.001:7]; step(syss*sysn,t)
10
>> Transfer function: 10 s^2 + 500 s -----------------s^2 + 200 s + 5000
Step Response
8
Amplitude
6
4
2 Step Response 0.01 0 0 0.02 0.04 0.06 Time ( sec ) 0.005 Amplitude 0.08 0.1 0.12
(a)
0
-0.005
-0.01
0
1
2
3
4
5
6
7
Time ( sec )
(b)
FIGURE CP4.8 (a) Unit step response. (b) Response to sinusoidal noise input at = 10 rad/sec.
CP4.9
(a) The closed-loop transfer function is T (s) = Gc (s)G(s) K (s + 1) R(s) = R(s) . 2 + s + 6.5) + K (s + 1) 1 + G(s)Gc (s) (s + 15)(s
(b) The step responses are shown in Figure CP4.9a.
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Computer Problems
165
(c) The unit disturbance response of the system is shown in Figure CP4.9b. The steady-state value is 0.14.
Step Response
0.7
0.6
K=5 K=10 K=50
0.2 0.18 0.16 System: syscl Final Value: 0.14
0.5
0.14
Step response
Amplitude
0.4
0.12 0.1 0.08
0.3
0.2
0.06 0.04
0.1
0.02
0
0
0
1
2
3 4 Time (s)
5
6
7
0
1
2
3 Time (sec)
4
5
6
7
(a)
(b)
FIGURE CP4.9 (a) Unit step responses for K = [5, 10, 50]. (b) Disturbance unit step response.
CP4.10
The m-le is shown in Figure CP4.10a and the step responses in Figure CP4.10b.
5 4
K=10 K=12 K=15
K=[10, 12, 15]; t=[0:0.1:20]; ng=[20]; dg=[1 4.5 64]; sysg=tf(ng,dg); nh=[1]; dh=[1 1]; sysh=tf(nh,dh); for i=1:length(K) sys=K(i)*sysg; syscl=feedback(sys,sysh) y(:,i)= step(syscl,t); end plot(t,y(:,1),t,y(:,2),'--',t,y(:,3),':') xlabel('Time (s)') ylabel('Step response') legend('K=10','K=12','K=15',-1)
3 2 1 0 1 2 3
Step response
0
5
10 Time (s)
15
20
(a)
(b)
FIGURE CP4.10 (a) M-le script. (b) Unit step responses for K = [10, 12, 15].
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
5
The Performance of Feedback Control Systems
Exercises
E5.1 For a zero steady-state error, when the input is a step we need one integration, or a type 1 system. A type 2 system is required for ess = 0 for a ramp input. (a) The closed-loop transfer function is T (s) = Y (s) G(s) 100 100 = = =2 . 2 R(s) 1 + G(s) (s + 2)(s + 5) + 100 s + 2n s + n
E5.2
The steady-state error is given by ess = where R(s) = A/s and Kp = lim G(s) =
s0
A , 1 + Kp
100 = 10 . 10
Therefore, ess = A . 11
(b) The closed-loop system is a second-order system with natural frequency n = 110 , 166
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Exercises
167
and damping ratio 7 = = 0.334 . 2 110 Since the steady-state value of the output is 0.909, we must modify the percent overshoot formula which implicitly assumes that the steadystate value is 1. This requires that we scale the formula by 0.909. The percent overshoot is thus computed to be / 1 2 ) = 29% . P.O. = 0.909(100e E5.3 The closed-loop transfer function is Y (s) G(s) K K = = =2 . I (s) 1 + G(s) s(s + 14) + K s + 14s + K Utilizing Table 5.6 in Dorf & Bishop, we nd that the optimum coecients are given by
2 s2 + 1.4n s + n .
We have s2 + 14s + K ,
2 so equating coecients yields n = 10 and K = n = 100 . We can also compute the damping ratio as
=
14 = 0.7 . 2n
Then, using Figure 5.8 in Dorf & Bishop, we nd that P.O. 5%. E5.4 (a) The closed-loop transfer function is T (s) = G(s) 2(s + 8) =2 . 1 + G(s) s + 6s + 16
(b) We can expand Y (s) in a partial fraction expansion as Y (s) = (s2 2(s + 8) A 1 s+4 =A 2 + 6s + 16) s s s + 6s + 16 .
Taking the inverse Laplace transform (using the Laplace transform tables), we nd y (t) = A[1 1.07e3t sin( 7t + 1.21)] .
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168
CHAPTER 5
The Performance of Feedback Control Systems
(c) Using the closed-loop transfer function, we compute = 0.75 and n = 4. Thus, a 8 = = 2.67 , n 3 where a = 8. From Figure 5.13(a) in Dorf & Bishop, we nd (approximately) that P.O. = 4% . (d) This is a type 1 system, thus the steady-state error is zero and y (t) A as t . E5.5 (a) The closed-loop transfer function is T (s) = Y (s) G(s) 100 = =2 , R(s) 1 + GH (s) s + 100Ks + 100
where H (s) = 1 + Ks and G(s) = 100/s2 . The steady-state error is computed as follows: ess = lim s[R(s) Y (s)] = lim s[1 T (s)]
s0
= lim 1
s0
1+
100 s2 100 s2 (1 +
A = KA . Ks) s
s0
A s2
(b) From the closed-loop transfer function, T (s), we determine that n = 10 and = 100K = 5K . 2(10)
We want to choose K so that the system is critically damped, or = 1.0. Thus, K= 1 = 0.20 . 5
The closed-loop system has no zeros and the poles are at s1,2 = 50K 10 25K 2 1 . The percent overshoot to a step input is P.O. = 100e and P.O. = 0 for K > 0.2.
5K
125K 2
for
0 < K < 0.2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
169
E5.6
The closed-loop transfer function is Y (s) KG(s) K (s + 2) K (s + 2) = = =2 . R(s) 1 + KG(s) s(s + 1) + K (s + 2) s + s(K + 1) + 2K Therefore, n = 2K and = 2K +1 . So, 2K T (s) = a 4 = . n K+1 From Figure 5.13a in Dorf & Bishop, we determine that a 1.5 n when = 0.707. Thus, solving for K yields 4 = 1.5 K +1 or K = 1.67 .
E5.7
The pole-zero map is shown in Figure E5.7. Since the dominant poles
Polezero map
1
0.8
0.6
0.4
0.2
Imag Axis
0
- 0.2
- 0.4
- 0.6
- 0.8
-1 - 2.5
-2
- 1.5
-1
- 0.5
0
Real Axis
FIGURE E5.7 (a) Pole-zero map.
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170
CHAPTER 5
The Performance of Feedback Control Systems
Step Response
From: U(1) 0.35
0.3
0.25
Amplitude
0.2 To: Y(1) 0.15 0.1 0.05 0 0
1.6
3.2
4.8
6.4
8
Time (sec.)
FIGURE E5.7 CONTINUED: (b) Unit step response.
are real, you do not expect to have a large overshoot, as shown in Figure E5.7b. E5.8 (a) The closed-loop transfer function is T (s) = The damping ratio is 2 2 and the natural frequency is n = K . Therefore, we compute the percent overshoot to be 2 P.O. = 100e/ 1 = 4.3% = for = 0.707. The settling time is estimated via Ts = 4 8 = . n 2K s2 K . + 2Ks + K
(b) The settling time is less than 1 second whenever K > 32.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
171
E5.9
The second-order closed-loop transfer function is given by T (s) =
2 n . 2 s2 + 2n s + n
From the percent overshoot specication, we determine that P.O. 5% implies 0.69 . n > 1 .
From the settling time specication, we nd that Ts < 4 implies
And nally, from the peak time specication we have Tp < 1 implies n 1 2 > .
The constraints imposed on and n by the performance specications dene the permissible area for the poles of T (s), as shown in Figure E5.9.
Im(s) wn z = 0.69
1- z 2 = P
desired region for poles
46o Re(s)
wn
1-z 2 = - P
z wn = -1
FIGURE E5.9 Permissible area for poles of T (s).
E5.10
The system is a type 1. The error constants are Kp = and Kv = 1.0 .
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172
CHAPTER 5
The Performance of Feedback Control Systems
Therefore, the steady-state error to a step input is 0; the steady-state error to a ramp input is 1.0A0 , where A0 is the magnitude (slope) of the ramp input. E5.11 (a) The tracking error is given by E (s) = R(s) (s + 9)(s + 2)(s + 4) = R(s) . 1 + Gc G(s) (s + 9)(s + 2)(s + 4) + K (s + 6)
The steady-state tracking error (with R(s) = 1/s) is
s0
lim sE (s) =
72 . 72 + 6K
We require ess < 0.05, so solving for K yields K > 228. (b) The tracking error due to the disturbance is E (s) = G(s) (s + 9)(s + 6) Td (s) = Td (s) . 1 + Gc G(s) (s + 9)(s + 2)(s + 4) + K (s + 6)
The tracking error is shown in Figure E5.11.
0 -0.01 -0.02 -0.03
Amplitude
-0.04 -0.05 -0.06 -0.07 -0.08
0
0.1
0.2
0.3
0.4
0.5 Time (secs)
0.6
0.7
0.8
0.9
1
FIGURE E5.11 Tracking error due a step disturbance.
E5.12
The system is a type 0. The error constants are Kp = 0.2 and Kv = 0. The steady-state error to a ramp input is . The steady-state error to a
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
173
step input is ess = E5.13 1 = 0.833. 1 + Kp
(a) The tracking error is given by E (s) = [1 T (s)] R(s) . The steady-state tracking error (with R(s) = 1/s) is ess = lim s [1 T (s)] R(s) = lim [1 T (s)] = 1 T (0) .
s0 s0
The closed-loop transfer function is T (s) = K (s + 0.1) , s(s + 0.1)(s + 2) + K (s + 3)
(b) Use Gp (s) = 30. Then,
s0
and T (0) = 0.033. Therefore, ess = 1 T (0) = 0.967.
ess = lim s [1 T (s)Gp (s)] R(s) = 1 lim T (s)Gp (s) = 130 T (0) = 0 .
s0
E5.14
The plot of y (t) is shown in Figure E5.14.
1.4
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
0.2
0.4
0.6
0.8
1 Time (sec)
1.2
1.4
1.6
1.8
2
FIGURE E5.14 Plot of y (t) with T (s) (solid line) and approximate Ta (s) (dashed line).
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174
CHAPTER 5
The Performance of Feedback Control Systems
Using the dominant poles, the approximate closed-loop transfer function is Ta (s) = The actual transfer function is T (s) = E5.15 500 . (s + 10)(s2 + 10s + 50) s2 50 . + 10s + 50
The partial fraction expansion is y (t) = 10(z 1) t 10(z 8) 8t e+ e + 1.25 . 7z 56z
The plot of y (t) for z = 2, 4, 6 is shown in Figure E5.15.
z=2 (solid) & z=4 (dashed) & z=6 (dotted) 1.4
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
1
2
3 Time (sec)
4
5
6
FIGURE E5.15 Plot of y (t) for z =2, 4, 6.
E5.16
The desired pole locations for the 5 dierent cases are shown in Figure E5.16.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
175
Im(s)
desired region for poles
wn = 10
o 53
37
o
Re(s)
(a) 0.6 < z < 0 .8 and wn <10
desired region for poles
Im(s)
wn = 10
o
45
o 60
Re(s)
(b) 0.5 < z < 0 .707 and
wn > 1 0 Im(s)
desired region for poles
wn = 10
o
60
Re(s) wn = 5
(c) 0.5 < z a nd 5 < wn <10
FIGURE E5.16 Desired pole locations.
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176
CHAPTER 5
The Performance of Feedback Control Systems
Im(s)
desired region for poles
wn = 10
o
45
Re(s) wn = 5
(d) 0.707 >
z a nd 5 < wn <10 Im(s)
o
wn = 6
53 desired region for poles
Re(s)
(e) 0.6 < z and wn < 6
FIGURE E5.16 CONTINUED: Desired pole locations.
E5.17
The output is given by Y (s) = T (s)R(s) = K When K = 1, the steady-state error is ess = 0.2 which implies that
s0
G(s) R(s) . 1 + G(s)
lim sY (s) = 0.8 .
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Exercises
177
Since we want ess = 0, it follows that
s0
lim sY (s) = 1 ,
or 0.8K = 1 . Therefore, K = 1.25. E5.18 (a) The characteristic equation is
2 s2 = 2n s + n = s2 + 3.17s + 7 = 0 ,
from which it follows that n = 7 = 2.65,
=
3.17 = 0.6 . 2n
Therefore, we compute the percent overshoot and the estimated settling time to be 4 2 P.O. = 100e/ 1 = 9.53% and Ts = = 2.5 s . n (b) The unit step response is shown in Figure E5.18.
Step Response 1.4 System: sys Peak amplitude: 1.1 Overshoot (%): 9.53 At time (sec): 1.47
1.2
System: sys Settling Time (sec): 2.25
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5 Time (sec)
2
2.5
3
FIGURE E5.18 Unit step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
178 E5.19
CHAPTER 5
The Performance of Feedback Control Systems
(a) The closed-loop transfer function is T (s) = The damping ratio is 2 2 and the natural frequency is n = K . Therefore, we compute the percent overshoot to be 2 P.O. = 100e/ 1 = 4.3% = for = 0.707. The settling time is estimated via Ts = 4 8 = . n 2K s2 K . + 2Ks + K
(b) The settling time is less than 1 second whenever K > 32.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
179
Problems
P5.1 (a) The system error is E (s) = where R(s) = So,
t0
1 1+
Ka Km sm +1
R(s)
25o /sec . s
lim e(t) = lim sE (s) =
s0
25 . 1 + Ka Km
(b) If we desire ess 1o /sec, then 25o /s 1o /sec , 1 + Ka Km and solving for Ka Km yields Ka Km 24 . (c) The closed-loop transfer function is T (s) = Vb (s) Ka Km = . Vc (s) sm + 1 + Ka Km
The step response of the system (i.e. vc (t) = A) is vb (t) =
(Ka Km +1) AKa Km t m 1e 1 + Ka Km
.
So, at settling time, we have 1e
(1+Ka Km ) t m
0.98 ,
where m = 0.4. Setting t = 0.03 and solving for Ka Km yields Ka Km 52 . P5.2 (a) The settling time specication Ts = 4 < 0.6 n
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
180
CHAPTER 5
The Performance of Feedback Control Systems
is used to determine that n > 6.67. The P.O. < 20% requirement is used to determine < 0.45 which implies < 63o
and the P.O. > 10% requirement is used to determine > 0.60 which implies > 53o ,
since cos = . The desired region for the poles is shown in Figure P5.2.
Im(s)
desired region for poles
53
o 63
o
Re(s)
s = -6.67
FIGURE P5.2 Desired region for pole placement.
(b) The third root should be at least 10 times farther in the left halfplane, so |r3 | 10|n | = 66.7 . (c) We select the third pole such that r3 = 66.7. Then, with = 0.45 and n = 6.67, we determine that n = 14.8. So, the closed-loop transfer function is T (s) = 66.7(219.7) , (s + 66.7)(s2 + 13.3s + 219.7)
where the gain K = (66.7)(219.7) is chosen so that the steady-state
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
181
tracking error due to a step input is zero. Then, T (s) = or G(s) = P5.3 Given the input R(s) = we compute the steady-state error as ess 1 = lim s s0 1 + G(s) 1 1 = lim 2 3 s0 s G(s) s = lim
s0
G(s) , 1 + G(s)
T (s) . 1 T (s) 1 , s3
1 s2
K s2
=
1 . K
Since we require that ess 0.5 cm, we determine K2. P5.4 (a) The closed-loop transfer function is T (s) = Thus, n = K and = 1/n = 1/ K .
2 G(s) K n =2 =2 . 2 1 + G(s) s + 2s + K s + 2n s + n
Our percent overshoot requirement 5% implies that = 1/ 2 , of which in turn implies that n = 2. However, the corresponding time to peak would be 4.4 Tp = = 3.15 . 2 Our desired Tp = 1.1we cannot meet both specication simultaneously. (b) Let Tp = 1.1 and P.O. = 0.05, where is the relaxation factor 2 to be determined. We have that K = n and n = 1, so 1 = . K
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182
CHAPTER 5
The Performance of Feedback Control Systems
Thus, 2 P.O. = e/ 1 = e/ K 1 . Also, Tp = = 1.1 . K 1 Therefore, from the proceeding two equations we determine that P.O. = 0.05 = e1.1 . Solving for yields f () = ln + ln(0.05) + 1.1 = 0 . The plot of f () versus is shown in Figure P5.4. From the plot we
2 1 0 -1 -2
*
D=2.07
f(D)
-3 -4 -5 -6 -7 -8 0 0.5 1 1.5 2 2.5 3
D
FIGURE P5.4 Solving for the zeros of f.
see that = 2.07 results in f () = 0. Thus, P.O. = 0.05 = 10% Tp = 1.1 = 2.3 sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
183
So, we can meet the specications if they are relaxed by a factor of about 2 (i.e. = 2.07). P5.5 (a) The closed-loop transfer function is T (s) = s2 K1 K2 (s + 1) . + K1 K2 s + K1 K2
A percent overshoot less than 5% implies 0.69. So, choose = 2 0.69. Then set 2n = K1 K2 and n = K1 K2 . Then
2 2(0.69)n = n ;
and solving for n yields n = 1.38 .
2 Therefore K1 K2 = n = 1.9. When K1 K2 1.9 it follows that 0.69. (b) We have a type 2 system, so the steady-state tracking error to both a step and ramp input is zero. (c) For a step input, the optimum ITAE characteristic equation is 2 s2 + 1.4n s + n = 0 .
For a ramp input, the optimum ITAE characteristic equation is
2 s2 + 3.2n s + n = 0 . 2 Thus, K1 K2 = n = 3.2n . So, n = 3.2 and K1 K2 = 10.24.
P5.6
We have a ramp input, r (t) = t. So Kv = lim sG(s) = lim s
s0 s0
75 75(s + 1) = = 0.75 , s(s + 5)(s + 20) 100
and ess = P5.7 |R| 1 = = 1.33 . Kv 0.75
(a) The closed-loop transfer function is T (s) = Is2 K1 K2 . + K1 K2 K3 s + K1 K2
The steady-state tracking error for a ramp input is ess = lim sE (s) = lim s(1 T (s))R(s) = lim s(1 T (s))
s0 s0 s0
1 s2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
184
CHAPTER 5
The Performance of Feedback Control Systems
= lim
s0
Is + K1 K2 K3 = K3 . Is2 + K1 K2 K3 s + K1 K2
But we desire ess = 0.01 m, so K3 = 0.01. (b) For P.O. = 10%, we have = 0.6. Also, 2n = and
2 n =
0.01K1 K2 25
K1 K2 . 25
Thus, solving for K1 K2 yields K1 K2 = 36 104 . P5.8 (a) The closed-loop transfer function is T (s) = P (s) G(s)/s 20 = = . R(s) 1 + G(s)H (s)/s s(s + 30)
Therefore, the closed-loop system time constant is = 1/30 sec. (b) The transfer function from Td (s) to the output P (s) is P (s) G(s) 20 = = . Td (s) 1 + G(s)H (s)/s s + 30 The response to a unit step disturbance is 2 p(t) = (1 e30t ) . 3 At settling time, p(t) = 0.98pss = 0.65. Thus, solving for t(= Ts ) we determine that Ts = 0.13 sec. P5.9 We need to track at the rate = v 16000 = = 1.78 103 radians/sec . r 2500
The desired steady-state tracking error is ess Therefore, with ess = | | , Kv 1 degree = 0.1754 102 rad . 10
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
185
we compute Kv as Kv = 1.78 103 = 1.02 . 0.1754 102
This assumes that the system is type 1. P5.10 (a) The armature controlled DC motorblock diagram is shown in Figure P5.10.
ampli er
R(s )
K + -
Km R a+ L as
1 J s+b
w(s)
Kb
back emf
FIGURE P5.10 Armature controlled DC motor block diagram.
(b) The closed-loop transfer function is T (s) = where G(s) = Thus, T (s) = K , s2 + 2s + 1 + K Km . (Ra + La s)(Js + b) (s) KG(s) = , R(s) 1 + KKb G(s)
where Ra = La = J = b = Kb = Km = 1. The steady-state tracking error is ess = lim s(R(s) Y (s)) = lim s
s0 s0
A (1 T (s)) s K A = A(1 T (0)) = 1 = . 1+K 1+K
(c) For a percent overshoot of 15%, we determine that = 0.5. From our characteristic polynomial we have 2n = 2 and n = 1 + K . Solving for n yields n = 2, thus K = 3.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
186 P5.11
CHAPTER 5
The Performance of Feedback Control Systems
(a) The closed-loop transfer function is T (s) = Y (s) K = . R(s) s+K
To include the initial condition, we obtain the dierential equation: y (t) + Ky (t) = Kr (t) . Taking the Laplace transform yields: sY (s) y (to ) + KY (s) = K A s ,
where y (to ) = Q. Computing the inverse Laplace transform, L1 {Y (s)} yields y (t) = A(1 eKt ) + QeKt . Also, the tracking error is given by e(t) = A y (t) = eKt (A Q) . Thus, the performance index, I is determined to be (for K > 0) I=
0
(A Q)2 e2Kt dt = (A Q)2 = (A 2K Q)2 .
1 2K
e2Kt
0
(b) The minimum I is obtained when K = , which is not practical. (c) Set K at the maximum value allowable such that the process does not saturate. For example, if K = 50, then I= P5.12 (A Q)2 . 100
The optimum ITAE transfer function for a ramp input is T (s) =
2 3 3.25n s + n . 2 3 s3 + 1.75n s + 3.25n s + n
The steady-state tracking error, ess = 0, for a ramp input. The step response is shown in Figure P5.12 for n = 10. The percent overshoot is P.O. = 39%, and the settling time is Ts = 0.72 s .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
1.4 PO=39% 1.2 Ts=0.72s 1
187
0.8 y(t) 0.6 0.4 0.2 0 0
0.2
0.4
0.6
0.8
1 time (sec)
1.2
1.4
1.6
1.8
2
FIGURE P5.12 Step input system response.
P5.13
The step responses for the actual system and the approximate system are shown in Figure P5.13. It can be seen that the responses are nearly identical.
1
0.9
0.8
0.7
0.6 y(t)
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE P5.13 Closed-loop system step response: Actual T(s) (solid line) and second-order approximation (dashed line).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
188 P5.14
CHAPTER 5
The Performance of Feedback Control Systems
Consider L(s) = 2(c1 s + 1) . (s + 1)(s + 2)
After cancellation of like factors, we compute H (s)/L(s), H (s) s3 + 7s2 + 24s + 24 = . L(s) (s + 3)(s + 4)2(c1 s + 1) Therefore, M (s) = s3 + 7s2 + 24s + 24 , and (s) = 2[c1 s3 + (7c1 + 1)s2 + (12c1 + 7)s + 12] . Then, following the procedure outlined in Section 5.10, we have M o (0) = 24, M 1 (0) = 24, M 2 (0) = 14, M 3 (0) = 6, and
0 (0) = 24, 1 (0) = (12c1 + 7)2, 2 (0) = 2(2 (7c1 + 1)), 3 (0) = 12c1 . For q = 1: M2 = 240, and 2 = 4[144c2 + 25] . 1 Then, equating 2 and M2 , we nd c1 , c1 = 0.493 . So, L(s) = P5.15 2(0.493s + 1) 0.986s + 2 0.986(s + 2.028) =2 = . (s + 1)(s + 2) s + 3s + 2 (s + 1)(s + 2)
The open-loop transfer function is G(s) = 10 . (s + 1)(50Cs + 1)
Dene = 50C . Then, the closed-loop transfer function is Vo (s) 10 10/ = = +1 Vin (s) (s + 1)( s + 1) + 10 s2 + s + With
2 n = 11
.
11
1 +1 and = = , 2 n 2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
189
we can solve for , yielding 2 20 + 1 = 0 . Therefore, = 19.95 and 0.05. For each value of we determine C as follows: = 19.95 = 50C , implies C = 0.399F , and = 0.05 = 50C , implies C = 1mF . P5.16 (a) The closed-loop transfer function is T (s) = Y (s) 12K =2 . R(s) s + 12s + 12K
The percent overshoot specication P.O. 10% implies 0.59. From the characteristic equation we nd that
2 n = 12K
and n = 6 .
Solving for K yields 2(0.59) 12K = 12
which implies that K = 8.6 .
So, any gain in the interval 0 < K < 8.6 is valid. The settling time is Ts = 4/n = 4/6 seconds and satises the requirement. Notice that Ts is not a function of K . (b) The sensitivity is
T SK (s) =
1 s(s + 12) =2 1 + G(s) s + 12s + 120
when K = 10. (c) The sensitivity at DC (s = 0) is
T SK (0) = 0 .
(d) In this case, s = j 2 1 beat/sec = j 2 . So, the sensitivity at s = 2j is
T |SK (j 2 )| =
85.1084 = 0.77 . 110.31
P5.17
We select L(s) as L(s) = 1 , s + 1
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190
CHAPTER 5
The Performance of Feedback Control Systems
then H (s) 6s + 6 =3 . L(s) s + 6s2 + 11s + 6 Therefore, M (s) = 6s + 6 , and M o (0) = 6, M 1 (0) = 6, M 2 (0) = 0. Also, (s) = s3 + 6s2 + 11s + 6 , and o (0) = 6 , 1 (0) = 11 , 2 (0) = 12. So, computing M2 and 2 yields M2 = 362 , 2 = 49 . Finally, equating M2 = 2 yields 362 = 49 , or = 1.167 . Thus, L(s) = P5.18 0.857 1 = . 1.167s + 1 s + 0.857 and
(a) The closed-loop transfer function is T (s) = s3 + 6s2 8 . + 12s + 8
(b) The second-order approximation is L(s) = d2 s2 1 , + d1 s + 1
where d1 and d2 are to be determined using the methods of Section 5.10 in Dorf & Bishop. Given M (s) = 8d2 s2 + 8d1 s + 8 (s) = s3 + 6s2 + 12s + 8 we determine that M2 = 128d2 + 64d2 1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
1
191
0.9
0.8
0.7
0.6 y(t)
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE P5.18 Closed-loop system step response: Actual T (s) (solid line) and second-order approximation (dashed line).
M4 = 64d2 2 2 = 48 4 = 18 . Equating M2 = 2 and M4 = 4 , and solving for d1 and d2 yields d1 = 1.35 and d2 = 0.53 .
Thus, the second-order approximation is L(s) = 1 . + 1.35s + 1
0.53s2
(c) The plot of the step response for the actual system and the approximate system is shown in Figure P5.18. P5.19 The steady-state error is ess = lim (s + 10)(s + 12) + K (1 K1 ) 120 + K (1 K1 ) = . (s + 10)(s + 12) + K 120 + K
s0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
192
CHAPTER 5
The Performance of Feedback Control Systems
To achieve a zero steady-state tracking error, select K1 as follows: K1 = 1 + P5.20 The closed-loop transfer function is T (s) = s2 s+a . + (2k + a)s + 2ak + 1 120 . K
(a) If R(s) = 1/s, we have the tracking error E (s) = R(s) Y (s) = [1 T (s)]R(s) or E (s) = s2 + (2k + a 1)s + 2ak + 1 a 1 . s2 + (2k + a)s + 2ak + 1 s
From the nal value theorem we obtain ess = lim sE (s) =
s0
2ak + 1 a . 2ak + 1
Selecting k = (a 1)/(2a) leads to a zero steady-state error due to a unit step input. (b) To meet the percent overshoot specication we desire 0.69. From 2 T (s) we nd n = 2ak + 1 and 2n = 2k + a. Therefore, solving for a and k yields a = 1.5978 and k = 0.1871
when we select = 0.78. We select > 0.69 to account for the zero in the closed-loop transfer function which will impact the percent overshoot. With a and k, as chosen, we have T (s) = s + 1.598 s2 + 1.972s + 1.598
and the step response yields P.O. 4%. P5.21 The closed-loop transfer function is T (s) = 2(2s + ) . (s + 0.2K )(2s + ) + 4
(a) If R(s) = 1/s, we have the unit step response Y (s) = 2(2s + ) 1 . (s + 0.2K )(2s + ) + 4 s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
193
From the nal value theorem we obtain yss = lim sY (s) =
s0
2 . 0.2K + 4
Selecting K = 10 20/ leads to yss = 1 and a zero steady-state error due to a unit step input. (b) The characteristic equation is (s + 0.2K )(2s + ) + 4 = 2s2 + (0.4K + )s + 0.2K + 4 = 0 . So, with K = 10 20/ , the natural frequency and damping ratio are: n = 2+ and = 2 + 4 8 . 4 2 +
The settling time and percent overshoot are found using the standard design formulas 4 2 Ts = and P.O. = 100e 1 n with n and given above (as a function of ). Since the closed-loop system has a zero at s = /2, the formulas for Ts and P.O. will only be approximate. Also, note that for the closed-loop system poles to be in the left half-plane (that is, all the poles have negative real parts), we require that > 2 3 2 1.4642. As seen in the next chapter, this is the condition for stability. Having > 2 3 2 insures that the damping ratio is positive.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
194
CHAPTER 5
The Performance of Feedback Control Systems
Advanced Problems
AP5.1 (a) The steady-state error is ess = lim s(1 T (s))R(s) = 1 T (0) = 1
s0
96(3) =0. 8(36)
(b) Assume the complex poles are dominant. Then, we compute a = 0.75 , n since a = 3, = 0.67 and n = 6. Using Figure 5.13 in Dorf & Bishop, we estimate the settling time and percent overshoot to be P.O. = 45% and Ts = 4 = 1 second . n
(c) The step response is shown in Figure AP5.1. The actual settling time and percent overshoot are P.O. = 33% and Ts = 0.94 second .
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.2
0.4
0.6
0.8
1 Time (secs)
1.2
1.4
1.6
1.8
2
FIGURE AP5.1 Closed-loop system step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
195
AP5.2
The closed-loop transfer function is T (s) = s3 + 28s2 5440(z s + 1) . + (432 + 5440z )s + 5440
The closed-loop step responses are shown in Figure AP5.2. The performance results are summarized in Table AP5.2.
tau=0 (solid) & tau=0.05 (dashed) & tau=0.1 (dotted) & tau=0.5 (dot-dash) 1.4
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
0.2
0.4
0.6
0.8 Time (sec)
1
1.2
1.4
1.6
FIGURE AP5.2 Closed-loop system step responses.
z 0 0.05 0.1 0.5
TABLE AP5.2
Tr 0.16 0.14 0.10 0.04
Ts 0.89 0.39 0.49 1.05
P.O. 32.7% 4.5% 0% 29.2%
closed-loop poles p = 20, 4 16j p = 10.4, 8.77 21.06j p = 6.5, 10.74 26.84j p = 1.75, 13.12 54.16j
Performance summary.
As z increases from 0 to 0.1, the P.O. decreases and the response is faster
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
196
CHAPTER 5
The Performance of Feedback Control Systems
and more stable. However, as z is increased beyond 0.1, the P.O. and Ts increase, although Tr continues to decrease. AP5.3 The closed-loop transfer function is T (s) = p s3 1 . + (1 + 2p )s2 + 2s + 1
The closed-loop step responses for p = 0, 0.5, 2, 5 are shown in Figure AP5.3. The performance results are summarized in Table AP5.3.
tau=5 (solid) & tau=2 (dotted) & tau=0.5 (dashed) & tau=0 (dot-dash) 1.5
1
y(t) 0.5 0 0
5
10
15
20
25 Time (sec)
30
35
40
45
50
FIGURE AP5.3 Closed-loop system step responses.
p 0 0.5 2 5
TABLE AP5.3
Tr 4 3.6 4.6 6
Ts 5.8 7.4 22.4 45.8
P.O. 0% 4.75% 27.7% 46%
closed-loop poles p = 1, 1 p = 2.84, 0.58 0.6j p = 2.14, 0.18 0.45j p = 2.05, 0.07 0.3j
Performance summary.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
197
As p increases, the P.O., Tr and Ts also increase; adding the pole makes the system less stable with more overshoot. AP5.4 The system transfer function is Y (s) = 15 15K R(s) + Td (s) . (s + 5)(s + 7) + 15K (s + 5)(s + 7) + 15K
When considering the input response, we set Td (s) = 0, and similarly, when considering the disturbance response, we set R(s) = 0. The closedloop step input and disturbance responses for K = 1, 10, 100 are shown in Figure AP5.4. The performance results are summarized in Table AP5.4.
Unit step input response 1.6 0.35 Unit step distrubance response
1.4
0.3
1.2 0.25 1 0.2 y(t) 0.8 y(t) 0.15 0.6 0.1 0.4 0.05 0 0.2 0.4 0.6 Time (sec) 0.8 1 0 0
0.2
0
0.2
0.4 0.6 Time (sec)
0.8
1
FIGURE AP5.4 Closed-loop system input and disturbance responses (K =1: solid line, K =10: dotted line, and K =100:dashed line).
K 1 10 100
TABLE AP5.4
ess 0.7 0.19 0.023
Ts 0.45 0.6 0.59
P.O. 0% 17.3% 60.0%
|y/d|max 0.3 0.1 0.01
Performance summary.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
198
CHAPTER 5
The Performance of Feedback Control Systems
The best value of the gain is K = 10, which is compromise between (i) percent overshoot, and (ii) disturbance rejection and tracking error. AP5.5 The system transfer function is Y (s) = 50(s + )(s + 2) R(s) s(s + 3)(s + 4) + 50(s + )(s + 2) 50s(s + 2) Td (s) . + s(s + 3)(s + 4) + 50(s + )(s + 2)
Disturbance response: alpha=0 (solid) & alpha=10 (dashed) & alpha=100 (dotted) 10 9 8 7 6 5 y(t) 4 3 2 1 0 -1 0
0.05
0.1 Time (sec)
0.15
0.2
0.25
FIGURE AP5.5 Closed-loop system disturbance response.
When considering the input response, we set Td (s) = 0, and similarly, when considering the disturbance response, we set R(s) = 0. The steadystate tracking error is ess = lim s(1 T (s))R(s) = lim 1
s0 s0
50(s + )(s + 2) . s(s + 3)(s + 4) + 50(s + )(s + 2)
When = 0, we have ess = 1 100 = 0.11 , 100 + 12
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
199
and, for = 0 ess = 0 . The closed-loop step input and disturbance responses for = 0, 10, 100 are shown in Figure AP5.5. For disturbance rejection and steady-state tracking error the best value of the parameter is = 100 . However, when considering both the disturbance and input response we would select the parameter = 10 , since it oers a good compromise between input response overshoot (about 5% for = 10) and disturbance rejection/tracking error. AP5.6 (a) The closed-loop transfer function is T (s) = KKm . KKm + s(s + Km Kb + 0.01)
The steady-state tracking error for a ramp input R(s) = 1/s2 is ess = lim s(1 T (s))R(s) = lim s + Km Kb + 0.01 s0 KKm + s(s + Km Kb + 0.01) Km Kb + 0.01 = . KKm
s0
(b) With Km = 10 and Kb = 0.05 , we have Km Kb + 0.01 10(0.05) + 0.01 = =1. KKm 10K Solving for K yields K = 0.051 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
200
CHAPTER 5
The Performance of Feedback Control Systems
(c) The plot of the step and ramp responses are shown in Figure AP5.6. The responses are acceptable.
Step input response 1.4 1.2 1 y(t) 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 Time (sec) 12 14 16 18 20
Ramp input response 20
15 y(t)
10
5
0
0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE AP5.6 Closed-loop system step and ramp responses.
AP5.7
The performance is summarized in Table AP5.7 and shown in graphical form in Fig. AP5.7.
K 1000 2000 3000 4000 5000
TABLE AP5.7
Estimated Percent Overshoot 8.8 % 32.1 % 50.0 % 64.4 % 76.4 %
Performance summary.
Actual Percent Overshoot 8.5 % 30.2 % 46.6 % 59.4 % 69.9 %
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
201
80 70 Percent Overshoot (% ) 60 50 40 30 20 10 0 1000
FIGURE AP5.7 Percent overshoot versus K.
Actual P.O . Estimated P.O.
2000
3000 K
4000
5000
The closed-loop transfer function is T (s) = 100K . s(s + 50)(s + 100) + 100K
The impact of the third pole is more evident as K gets larger as the estimated and actual percent overshoot deviate in the range 0.3% at K = 1000 to 6.5% at K = 5000. AP5.8 The closed-loop transfer function is T (s) = K s2 + 120s + 110 1 + K s2 + as + b
where a = (5 + 120K )/(1 + K ) and b = (6 + 110K )/(1 + K ). Setting 2 b = n and a = 2n yields = 2
5+120K 1+K 6+110K 1+K
.
For the closed-loop transfer function to have complex roots, we require a2 4b < 0. This occurs when 0.0513 K 0.0014. When K = 0.0417, we have = 0, hence minimized. At this point, the system has roots on the imaginary axis and is marginally stable. When 0.0513 K 0.0417, the system is unstable.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
202
CHAPTER 5
The Performance of Feedback Control Systems
Design Problems
CDP5.1
The plant model with parameters given in Table CDP2.1 in Dorf and Bishop is given by: (s) 26.035 = , Va (s) s(s + 33.142) where we neglect the motor inductance Lm . The closed-loop transfer function from the disturbance to the output is (s) 26.035 =2 . Td (s) s + 33.142s + 26.035Ka For a unit step disturbance input the steady-state response is ss = 1 . Ka
Therefore, we want to use the maximum Ka while keeping the percent overshoot less than 5%. The step response for the closed-loop system (with the tachometer not in the loop) and Ka = 22 is shown below. Values of Ka greater than 22 lead to overshoots greater than 5%.
Step response 1.4 1.2 1 q(t)/A 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 Time (sec) 0.6 0.7 0.8 0.9 1
Unit disturbance response 0.05 0.04 0.03 q(t) 0.02 0.01 0 0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
203
DP5.1
(a) The closed-loop transfer function is 12.2K 12.2K (s) = =3 . 2 + 15.4s + 12.2K d (s) s(s + 2.2)(s + 7) + 12.2K s + 9.2s (b) For K = 0.7, we have the characteristic equation s3 + 9.2s2 + 15.4s + 8.54 = 0 , with roots s1 = 7.23 and s2,3 = 0.98 0.46j . For K = 3, we have the characteristic equation s3 + 9.2s2 + 15.4s + 36.6 = 0 , with roots s1 = 7.83 and s2,3 = 0.68 2.05j . And for K = 6, we have the characteristic equation s3 + 9.2s2 + 15.4s + 73.2 = 0 , with roots s1 = 8.4 and s2,3 = 0.4 2.9j .
(c) Assuming the complex conjugate pair are the dominant roots, we expect the following: (i) for K = 0.7: P.O.=0.13% and Tp = 6.8 sec (ii) for K = 3: P.O.=35.0% and Tp = 1.5 sec (iii) for K = 6: P.O.=65.2% and Tp = 1.1 sec (d),(e) We select K = 1.71 to have a P.O. = 16% and Tp = 2.18sec. All four cases (K = 0.7, 3, 6, 1.71) are shown in Figure DP5.1. In each case, the approximate transfer function is derived by neglecting the non-dominant real pole and adjusting the gain for zero steady-state error. The approximate transfer functions are 1.18 0.7908 = + 1.965s + 1.18 (s + 0.98 + 0.46j )(s + 0.98 0.46j ) 4.67 3.299 TK =3 (s) = 2 = s + 1.37s + 4.67 (s + 0.68 + 2.05j )(s + 0.68 2.05j ) 8.71 6.399 TK =6 (s) = 2 = s + 0.796s + 8.71 (s + 0.4 + 2.9j )(s + 0.4 2.9j ) 2.77 1.458 TK =1.71 (s) = 2 = s + 1.679s + 2.77 (s + 0.83 + 1.43j )(s + 0.83 1.43j ) TK =0.7 (s) = s2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
204
CHAPTER 5
The Performance of Feedback Control Systems
K=0.7 1.5 1.5
K=3
1 phi phi 0.5
1
0.5
0
0
5 time (sec) K=6
10
0
0
5 time (sec) K=1.71
10
2 1.5 phi 1 0.5 0 0 5 time (sec) 10 phi
1.5
1
0.5
0
0
5 time (sec)
10
FIGURE DP5.1 Step responses (actual response:solid lines; approximate response: dotted lines).
DP5.2
The closed-loop transfer function is T (s) =
2 Kn , 2 2 s3 + 2n s2 + n s + Kn
where = 0.2. From the second-order system approximation, we have Tp = n . 1 2
So, with = 0.2 given, we should select n large to make Tp small. Also, from the problem hint, let 0.1 < K/n < 0.3 . As a rst attempt, we can select n = 20. See Figure DP5.2 for various values of K/n . Our nal selection is K = 4 and n = 20 .
This results in P.O. = 2% and Tp = 0.9 second.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
1.4
205
1.2 K/wn=0.3 1
0.8 y(t)
0.6
K/wn=0.2
K/wn=0.1
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1 time (sec)
1.2
1.4
1.6
1.8
2
FIGURE DP5.2 Closed-loop system response.
DP5.3
The closed-loop transfer function is T (s) = s2 K . + qs + K
From the ITAE specication, we desire T (s) = But 2n = 1.4n which implies = 0.7 . s2
2 n . 2 + 1.4n s + n
Since we want Ts 0.5, we require n 8. So, n We can select n = 12. Then, T (s) = s2 144 . + 16.8s + 144 8 = 11.4 . 0.7
Therefore, K = 144 and q = 16.8. The predicted percent overshoot is P.O. = 4.5%.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
206 DP5.4
CHAPTER 5
The Performance of Feedback Control Systems
The open-loop transfer function is GGc (s) = 10K 10K/70 = . (s + 70)(s + 3)(s + 7) (s/70 + 1)(s + 3)(s + 7)
The second-order approximation is obtained by neglecting the fastest rstorder pole. Thus, GGc (s) K/7 . (s + 3)(s + 7)
The closed-loop transfer function is T (s) = K/7 . s2 + 10s + 21 + K/7
When 0.52, we have less than 15% overshoot. So, we have 2n = 10 and n = 21 + K/7.
Eliminating n and solving for K (with P.O. 15%) yields K 500.19 . Also, Kp = lim GGc (s) =
s0
K 7(21)
and ess = implies K 1078 . Therefore, we have an inconsistency. We require 1078 K to meet the steady-state requirement and K 500.18 to meet the percent overshoot requirement. It is not possible to meet both specications. DP5.5 The closed-loop characteristic equation is 1 + K1 G1 (s) + K2 G1 G2 (s) = 1 + K1 2K2 =0 s(s + 1) s(s + 1)(s + 2) 1 1 = < 0.12 K 1 + Kp 1 + 147
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
207
or s3 + 3s2 + (2 + K1 )s + 2(K1 K2 ) = 0 . Assuming that K1 > 0 and K2 > 0, the range of the gains for stability is 0 < K2 < K1 . DP5.6 The closed-loop transfer function is T (s) = s2 K1 . + (K1 K2 + 1)s + K1
The percent overshoot specication (P.O. < 4%) is satised when > 0.715. The peak time specication ( Tp = 0.2 sec) is satised when n = 22.47 and = 0.715. So, given
2 K1 = n
and
K1 K2 + 1 = 2n ,
we determine that the specications are satised when K1 = 504.81 and K2 = 0.0617 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
208
CHAPTER 5
The Performance of Feedback Control Systems
Computer Problems
CP5.1 With the impulse input we have R(s) = 1. The transfer function is Y (s) = 6 6 R(s) = . (s + 3)(s + 4) (s + 3)(s + 4)
Therefore, taking the inverse Laplace transforms yields the output response: y (t) = 6e3t 6e4t . The impulse response and the analytic response is shown in Figure CP5.1.
n=6; d=[1 7 12]; t=[0:0.1:6]; ya=6*exp(-3.*t)-6*exp(-4.*t); sys = tf(n,d) y=impulse(sys,t); plot(t,y,t,ya,'o') xlabel('Time (sec)'), ylabel('y(t)'), legend('Computer','Analytic',-1)
0.7 Computer Analytic
0.6
0.5
0.4 y(t) 0.3 0.2 0.1 0 0
1
2
3 Time (sec)
4
5
6
FIGURE CP5.1 Impulse responses.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
209
CP5.2
The ramp response is shown in Figure CP5.2. The error as t is ess .
Linear Simulation Results
2000 1800
1600
1400
n=[1 7]; d=[1 10 0 0]; t=[0:0.1:25]; sys= tf(n,d); sys_cl = feedback(sys,[1]); u=t; lsim(sys,u,t);
1200
Amplitude
To: Y(1)
1000
800
600
400
200
0
0
5
10
15
20
25
Time (sec.)
FIGURE CP5.2 Ramp responses.
CP5.3
The m-le script and the four plots are shown in Figure CP5.3. The plots can be compared to Figure 5.17 in Dorf & Bishop.
2 1 0 -1 -2 0 5 10 15 20 wn=2, zeta=0 2 1 0 -1 -2 0 5 10 15 20 wn=2, zeta=0.1
1 0.5 0 -0.5 -1 0 5
wn=1, zeta=0
1
wn=1, zeta=0.2
0.5
0
10
15
20
-0.5
0
5
10
15
20
FIGURE CP5.3 Impulse responses.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
210
CHAPTER 5
The Performance of Feedback Control Systems
w1=2; z1=0; w2=2; z2=0.1; w3=1; z3=0; w4=1; z4=0.2; t=[0:0.1:20]; % num1=[w1^2]; den1=[1 2*z1*w1 w1^2]; sys1 = tf(num1,den1); [y1,x1]=impulse(sys1,t); % num2=[w2^2]; den2=[1 2*z2*w2 w2^2]; sys2 = tf(num2,den2); [y2,x2]=impulse(sys2,t); % num3=[w3^2]; den3=[1 2*z3*w3 w3^2]; sys3 = tf(num3,den3); [y3,x3]=impulse(sys3,t); % num4=[w4^2]; den4=[1 2*z4*w4 w4^2]; sys4 = tf(num4,den4); [y4,x4]=impulse(sys4,t); % clf subplot(221),plot(t,y1),title('wn=2, zeta=0') subplot(222),plot(t,y2),title('wn=2, zeta=0.1') subplot(223),plot(t,y3),title('wn=1, zeta=0') subplot(224),plot(t,y4),title('wn=1, zeta=0.2')
FIGURE CP5.3 CONTINUED: Impulse response m-le script.
CP5.4
The closed-loop system is T (s) = s2 21 . + 2s + 21
Therefore, the natural frequency is m = 21 = 4.58 and the damping ratio is computed as 2n = 2 , which implies = 0.218 . The percent overshoot is estimated to be 2 P.O. = 100e/ 1 = 50% , since = 0.218. The actual overshoot is shown in Figure CP5.4.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
Step Response
From: U(1) 1.5
211
numc=[21]; denc=[1 0]; sysc = tf(numc,denc); numg=[1]; deng=[1 2]; sysg = tf(numg,deng); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]) step(sys_cl)
1
Amplitude
To: Y(1) 0.5 0 0
1
2
3
4
5
6
Time (sec.)
FIGURE CP5.4 Impulse responses.
CP5.5
The unit step response is shown in Figure CP5.5. The performance numbers are as follows: Mp = 1.04, Tp = 0.63, and Ts = 0.84.
Step Response
From: U(1) 1.4
1.2
1
numg=[50]; deng=[1 10 0]; sys = tf(numg,deng); sys_cl = feedback(sys,[1]); t=[0:0.01:2]; step(sys_cl,t);
Amplitude
0.8 To: Y(1) 0.6 0.4 0.2 0 0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (sec.)
FIGURE CP5.5 Closed-loop system step response m-le script.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
212 CP5.6
CHAPTER 5
The Performance of Feedback Control Systems
The m-le script and the simulations are shown in Figure CP5.6.
% Part (a) numc=[2]; denc=[1]; sys_c = tf(numc,denc); nums=[-10]; dens=[1 10]; sys_s = tf(nums,dens); numg=[-1 -5]; deng=[1 3.5 6 0]; sys_g = tf(numg,deng); sysa = series(sys_c,sys_s); sysb = series(sysa,sys_g); sys = feedback(sysb,[1]); f=0.5*pi/180; % Convert to rad/sec t=[0:0.1:10]; u=f*t; [y,x]=lsim(sys,u,t);(y(length(t),1)-u(1,length(t)))*180/pi subplot(211) plot(t,y*180/pi,t,u*180/pi,'--'), grid xlabel('Time (sec)'),ylabel('theta') title('Constant gain C(s) = 2: theta (solid) & input (dashed)') % Part (b) numc=[2 1]; denc=[1 0]; sys_c = tf(numc,denc); [numa,dena]=series(numc,denc,nums,dens); sysa = series(sys_c,sys_s); sysb = series(sysa,sys_g); sys = feedback(sysb,[1]); [y,x]=lsim(sys,u,t);(y(length(t),1)-u(1,length(t)))*180/pi subplot(212), plot(t,y*180/pi,t,u*180/pi,'--'), grid xlabel('Time (sec)'),ylabel('theta') title('PI controller C(s) = 2 + 1/s: theta (solid) & input (dashed)')
Constant gain C(s) = 2: theta (solid) & input (dashed) 5 4 3 2 1 0
theta
0
1
2
3
4
5 Time (sec)
6
7
8
9
10
PI controller C(s) = 2 + 1/s: theta (solid) & input (dashed) 6 5 4 theta 3 2 1 0 0 1 2 3 4 5 Time (sec) 6 7 8 9 10
FIGURE CP5.6 Closed-loop system response to a ramp input for two controllers.
For the constant gain controller, the attitude error after 10 seconds is ess = 0.3 deg. On the other hand, the PI controller has a zero steadystate error ess = 0 deg. So, we can decrease the steady-state error by
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
213
using a more sophisticated controller, in this case a PI controller versus a constant gain controller. CP5.7 The closed-loop characteristic equation is s3 + 12s2 + 610s + 500 = (s + 0.8324)(s2 + 11.1676s + 600.7027) = 0 . The natural frequency and damping ratio of the complex roots are n = 24.5 and = 0.23. From this we predict Mp = 1.48, Ts = 0.72, and Tp = 0.13. The actual response is shown in Figure CP5.7. The dierences
Step Response
From: U(1) 1.4
1.2
numg=[100 100]; deng=[1 2 100]; sysg = tf(numg,deng); numc=[0.1 5]; denc=[1 0]; sysc = tf(numc,denc); sys_o = series(sysg,sysc); sys_cl = feedback(sys_o,[1]) t=[0:0.01:3]; step(sys_cl,t); ylabel('theta dot')
1
theta dot
0.8 To: Y(1) 0.6 0.4 0.2 0 0
0.5
1
1.5
2
2.5
3
Time (sec.)
FIGURE CP5.7 Missile rate loop autopilot simulation.
can be explained by realizing that the system is not a second-order system. The closed-loop system actually has two zeros, one real pole, and two complex-conjugate poles: T (s) = (s + 50)(s + 1) . (s + 0.8324)(s2 + 11.1676s + 600.7027)
The eect of the pole at s = 0.8324 is diminished by the zero at s = 1. The third pole and the zeros aect the overall response such that the analytic formulas for second-order systems are not exact predictors of the transient response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
214 CP5.8
CHAPTER 5
The Performance of Feedback Control Systems
Figure CP5.8 shows an m-le to compute the closed-loop transfer function and to simulate and plot the step response.
Step Response 1 0.9 0.8 0.7 Amplitude 0.6 0.5 0.4 0.3 0.2 0.1 0 System: sys Settling Time (sec): 39.1 System: sys Peak amplitude: 0.979 Overshoot (%): 95.7 At time (sec): 0.533
numg=[10]; deng=[1 10]; sysg = tf(numg,deng); numh=[0.5]; denh=[10 0.5]; sysh = tf(numh,denh); sys = feedback(sysg,sysh) step(sys); Transfer function: 100 s + 5 --------------------10 s^2 + 100.5 s + 10
0
10
20
30 Time (sec)
40
50
60
FIGURE CP5.8 M-le to compute the transfer function and to simulate the step response.
CP5.9
Figure CP5.9 shows an m-le to compute the closed-loop transfer function and to simulate and plot the ramp response. The steady-state error is 7.5.
Linear Simulation Results 100 90 80 70 Amplitude 60 50 40 30 20 10 0
numg=[10]; deng=[1 20 75 0]; sysg = tf(numg,deng); sys = feedback(sysg,1) t=[0:0.1:100]; u=t; % Unit ramp input lsim(sys,u,t);
0
10
20
30
40
50 Time (sec)
60
70
80
90
100
FIGURE CP5.9 M-le to compute the transfer function and to simulate the ramp response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
215
CP5.10
Figure CP5.10 shows an m-le to compute the closed-loop transfer function and to simulate and plot the impulse, step, and ramp responses. Notice that the closed-loop system is unstable.
Impulse Response Amplitude 10 0 -10
Amplitude
numg=[1]; deng=[1 2 0]; sysg = tf(numg,deng); numc=[0.5 2]; denc=[1 0]; sysc = tf(numc,denc); syss=series(sysg,sysc); sys = feedback(syss,1) t=[0:0.1:20]; subplot(311) impulse(sys,t); subplot(312) step(sys,t); subplot(313) u=t; % Unit ramp input lsim(sys,u,t);
0
2
4
6
8
10 Time (sec) Step Response
12
14
16
18
20
Amplitude
10 0 -10
0
2
4
6
8 10 12 Time (sec) Linear Simulation Results
14
16
18
20
40 20 0
0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE CP5.10 M-le to compute the transfer function and to simulate the ramp response.
CP5.11
For the original system, we nd Ts = 2.28 and P.O. = 80.6%. For the 2nd-order approximation we nd Ts = 2.16 and P.O. = 101%
2.5
2nd order approximation
2
num=77*[1 2]; den=conv([1 7],[1 4 22]); sys = tf(num,den) na=(77/7)*[1 2]; da=[1 4 22]; sysa=tf(na,da); t=[0:0.01:5]; y=step(sys,t); ya=step(sysa,t); plot(t,y,t,ya,'--') xlabel('Time (s)'), ylabel('Step response')
3rd order system response
Step response 1.5
1
0.5
0
0
1
2 Time (s)
3
4
5
FIGURE CP5.11 Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
6
The Stability of Linear Feedback Systems
Exercises
E6.1 The Routh array is s3 s2 s1 so where b= 3K (K + 2) 5 . 3K 1 3K b 4 K +2 5 0
For stability, we require K > 0 and b > 0. Therefore, using the condition that b > 0, we obtain 3K 2 + 6K 5 > 0 , and solving for K yields K > 0.63 and K < 2.63. We select K > 0.53, since we also have the condition that K > 0. E6.2 The Routh array is s3 s2 s1 so 1 9 -2/3 24 2 24 0
The system is unstable since the rst column shows two sign changes. 216
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
217
E6.3
The Routh array is s4 s3 s2 s1 s0 1 9.5 43.59 17.82 10 45.7 20 10 10
By the Routh-Hurwitz criterion, the system is stable (i.e., all the numbers in the rst column are positive). E6.4 The closed-loop transfer function is T (s) = s3 K (s 1) . + 3s2 + (2 K )s + K
Therefore, the characteristic equation is s3 + 3s2 + (2 K )s + K = 0 . The corresponding Routh array is given by s3 s2 s1 so where b= 3(2 K ) K 6 4K = . 3 3 1 3 b K (2 K ) K 0
For stability we require K > 0 and b > 0 . Thus, the range of K for stability is 0 < K < 1.5. E6.5 The closed-loop transfer function is T (s) = s3 + 10s2 K . + 27s + 18 + K
When K = 20, the roots of the characteristic polynomial are s1,2 = 1.56 j 1.76
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218
CHAPTER 6
The Stability of Linear Feedback Systems
and s3 = 6.88 . E6.6 When K = 252, the roots of the characteristic equation are on the imaginary axis. The roots are s1,2 = j 5.2 E6.7 and s3 = 10 . K (s + 2) =0, s(s 1)
(a) The closed-loop system characteristic equation is 1 + GH (s) = 1 + or
s2 + (K 1)s + 2K = 0 . We have the relationships n = 2K and 2n = K 1, where = 0.707. Thus, 1 2 2K = K 1 , 2 or 2 2 and K 2 6K + 1 = 0 . Solving for K yields K = 5.83 and K = 0.17. However, for stability we require K > 1 (from the Routh array), so we select K = 5.83. (b) The two roots on the imaginary axis when K = 1 are s1,2 = j 2. E6.8 The closed-loop system characteristic equation is
3 2
=
K 1 2K
2
,
+ 20s2 + (100 + K )s + 20K = 0 .
The corresponding Routh array is s3 s2 s1 so 1 20 b 20K (100 + K ) 20K 0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
219
where b= 20(100 + K ) 20K 20(100) = = 100 . 20 20
Therefore, the system is stable for all K > 0. E6.9 The characteristic equation is s3 + 2s2 + (K + 1)s + 6 = 0 , and the Routh array is given by s3 s2 s1 so where b= Setting b = 0, yields K 2 = 0 or E6.10 E6.11 E6.12 K>2. 2(K + 1) 6 =K 2 . 2 1 2 b 46 K +1 6 0
Stable with your eyes open and (generally) unstable with your eyes closed. The system is unstable. The poles are s1 = 5.66, s2 = 0.90 and s3,4 = 0.28 j 0.714. The characteristic equation associated with the system matrix is s3 + 3s2 + 5s + 6 = 0 . The roots of the characteristic equation are s1 = 2 and s2,3 = 5 j 1.66. The system is stable.
E6.13
The roots of q (s) are s1 = 4, s2 = 3, s3,4 = 1 j 2 and s5,6 = j 0.5. The system is marginally stable. The Routh array is
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220
CHAPTER 6
The Stability of Linear Feedback Systems
s6 s5 s4 s3 s2 s1 so The auxillary equation is
1 9 24.44 31.909 60 0
31.25 61.25 66.11 9.2273 15 0
67.75 14.75 15 0
15
60s2 + 15 = 0 . Solving the auxillary equation yields two roots at s1,2 = j 0.5. After accounting for the row of zeros, the completed Routh array veries that the system has no poles in the right half-plane. E6.14 The Routh array is s4 s3 s2 s1 so 1 9 35.33 74.26 50 45 87 50 0 50
The system is stable. The roots of q (s) are s1,2 = 3 j 4, s3 = 2 and s4 = 1. E6.15 The characteristic equation is s3 + 6s2 + 11s + 6 = 0 . The system is stable. The roots of the characteristic equation are s1 = 1, s2 = 2 and s3 = 3. E6.16 The roots of q (s) are s1 = 20 and s2,3 = j 2.24. The system is marginally stable. The Routh array is
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
221 s3 s2 s1 so 1 20 0 5 100 0
The auxillary equation is 20s2 + 100 = 0 . The roots are s = j 2.24. So, the system has roots at s = j 2.24. Completing the Routh array (after accounting for the row of zeros) veries that no poles lie in the right half-plane. E6.17 (a) Unstable. (b) Unstable. (c) Stable. E6.18 E6.19 (a) The roots are s1,2 = 2 and s3 = 1. (b) The roots are s1,2,3 = 3. The characteristic equation is (sn 2)3 + 10(sn 2)2 + 29(sn 2) + K = 0 or s3 + 4s2 + sn 26 + K = 0 . n n The Routh array is s3 s2 s1 so 1 4
30K 4
1 K 26 0
K 26
If K = 30, then the auxillary equation is 4s2 + 4 = 0 or sn = j . n Therefore, s = sn 2 implies s = 2 j . E6.20 E6.21 This system is not stable. The output response to a step input is a ramp y (t) = kt. The characteristic polynomial is s3 + 3s2 + ks + 6 = 0 .
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222
CHAPTER 6
The Stability of Linear Feedback Systems
The Routh array is s3 s2 s1 so So, k > 2 for stability. E6.22 The transfer function is G(s) = C(sI A)1 B + D =[ 1
s 0 0 ] 0
1 3
3k 6 3
k 6
6
1 s k
0 1 s+k
k
=[ 1
where (s) = s3 + ks2 + ks + k. Thus, the transfer function is G(s) = The Routh array is s3 s2 s1 so For stability k > 1. E6.23 The closed-loop transfer function is T (s) = s2 (s Ks + 1 . + p) + Ks + 1 1 k k1 k k k s3 + ks2 1 . + ks + k
] 00
s2
+ ks + k k ks
0 0
1
s+k
s2 + ks
ks k s2
1 0 s 0
1
1 (s)
Therefore, the characteristic equation is s3 + ps2 + Ks + 1 = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
223
The Routh array is s3 s2 s1 so 1 p (pK 1)/p 1 K 1
We see that the system is stable for any value of p > 0 and pK 1 > 0. E6.24 The closed-loop transfer function is T (s) = 2s2 10 . + (K 20)s + 10
Therefore, the characteristic equation is 2s2 + (K 20)s + 10 = 0 . The Routh array is s2 s1 so 2 K 20 10 10
We see that the system is stable for any value of K > 20.
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224
CHAPTER 6
The Stability of Linear Feedback Systems
Problems
P6.1 (a) Given s2 + 5s + 2 , we have the Routh array s2 s1 so 1 5 2 2 0
Each element in the rst column is positive, thus the system is stable. (b) Given s3 + 4s2 + 8s + 4 , we have the Routh array s3 s2 s1 so 1 4 7 4 8 4 0
Each element in the rst column is positive, thus the system is stable. (c) Given s3 + 2s2 4s + 20 , we determine by inspection that the system is unstable, since it is necessary that all coecients have the same sign. There are two roots in the right half-plane. (d) Given s4 + s3 + 2s2 + 10s + 8 , we have the Routh array s4 s3 s2 s1 so 1 1 -8 11 8 2 10 8 0 8 0 0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
225
There are two sign changes in the rst column, thus the system is unstable with two roots in the right half-plane. (e) Given s4 + s3 + 3s2 + 2s + K , we have the Routh array s4 s3 s2 s1 so 1 1 1 2K K 3 2 K 0 K 0
Examining the rst column, we determine that the system is stable for 0 < K < 2. (f) Given s5 + s4 + 2s3 + s + 6 , we know the system is unstable since the coecient of the s2 term is missing. There are two roots in the right half-plane. (g) Given s5 + s4 + 2s3 + s2 + s + K , we have the Routh array s5 s4 s3 s2 s1 so 1 1 1 K K K 2 1 1K K 0 1 K
Examining the rst column, we determine that for stability we need K > 0 and K < 0. Therefore the system is unstable for all K . P6.2 (a) The closed-loop characteristic polynomial is s4 + 27.88s3 + 366.4s2 + 1500s + 1500ka = 0 . The Routh array is
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226
CHAPTER 6
The Stability of Linear Feedback Systems
s4 s3 s2 s1 so where
1 27.88 312.6 b 1500ka
366.4 1500 1500ka
1500ka
b = 1500 133.78ka . Examining the rst column of the Routh array, we nd that b > 0 and 1500ka > 0 for stability. Thus, 0 < ka < 11.21 . (b) With Ts = 1.5 = we determine that n = 2.67 . So, shift the axis by s = so 2.67, and (so 2.67)4 + 27.88(so 2.67)3 + 366.4(so 2.67)2 + 1500(so 2.67) + 1500ka = s4 + 17.2s3 + 185.85s2 + 63.55so 1872.8 + 1500ka . o o o The Routh array is s4 s3 s2 s1 so where b = 240.38 141.63ka . Examining the rst column of the Routh array, we nd that b > 0 and 1500ka 1872.8 > 0. Thus, 1.25 < ka < 1.69. 1 17.2 182.16 b 1500ka -1872.8 185.85 63.55 1500ka -1872.8 1500ka -1872.8 4 , n
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
227
P6.3
(a) Given G(s) = and H (s) = 1 , 0.005s + 1 K , (s + 1)(s + 2)(0.5s + 1)
the closed-loop transfer function is T (s) = 0.0025s4 K (0.005s + 1) . + 0.5125s3 + 2.52s2 + 4.01s + 2 + K
Therefore, the characteristic equation is 0.0025s4 + 0.5125s3 + 2.52s2 + 4.01s + (2 + K ) = 0 . The Routh array is given by s4 s3 s2 s1 so 0.0025 0.5125 2.50 3.6 0.205K 2+K 2.52 4.01 2+K 0 2+K 0
Examining the rst column, we determine that for stability we require 2 < K < 17.6 . (b) Using K = 9, the roots of the characteristic equation are s1 = 200 , s2,3 = 0.33 2.23j , and s4 = 4.35 .
Assuming the complex roots are dominant, we compute the damping ratio = 0.15. Therefore, we estimate the percent overshoot as 2 P.O. = 100e/ 1 = 62% . The actual overshoot is 27%, so we see that assuming that the complex poles are dominant does not lead to accurate predictions of the system response. P6.4 (a) The closed-loop characteristic equation is 1 + GH (s) = 1 + K (s + 40) =0, s(s + 10)(s + 20)
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228
CHAPTER 6
The Stability of Linear Feedback Systems
or s3 + 30s2 + 200s + Ks + 40K = 0 . The Routh array is s3 s2 s1 so 1 30 200 40K
K 3
200 + K 40K 0
Therefore, for stability we require 200 K/3 > 0 and 40K > 0. So, the range of K for stability is 0 < K < 600 . (b) At K = 600, the auxilary equation is 30s2 + 40(600) = 0 or s2 + 800 = 0 .
The roots of the auxiliary equation are s = j 28.3 . (c) Let K = 600/2 = 300. Then, to the shift the axis, rst dene so = s + 1. Substituting s = so 1 into the characteristic equation yields (so 1)3 +30(so 1)2 +500(so 1)+12000 = s3 +27s2 +443so +11529 . o o The Routh array is s3 s2 s1 so 1 27 16 11529 443 11529 0
All the elements of the rst column are positive, therefore all the roots lie to left of s = 1. We repeat the procedure for s = so 2 and obtain s3 + 24s2 + 392so + 10992 = 0 . o o The Routh array is
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
229 s3 s2 s1 so 1 24 -66 10992 392 10992 0
There are two sign changes in the rst column indicating two roots to right of s = 2. Combining the results, we determine that there are two roots located between s = 1 and s = 2. The roots of the characteristic equation are s1 = 27.6250 and s2,3 = 1.1875 20.8082j .
We see that indeed the two roots s2,3 = 1.187520.8082j lie between -1 and -2. P6.5 (a) Given the characteristic equation, s3 + 3s2 + 4s + 2 = 0 , we compute the roots s1 = 1, and s2,3 = 1 j . s4 + 9s3 + 30s2 + 42s + 20 = 0 are s1 = 1, s2 = 2, and s3,4 = 3 j 1. s3 + 19s2 + 110s + 200 = 0 are s1 = 4, s2 = 5, and s3 = 10. P6.6 (a) The characteristic equation is 1 + G(s) = 0 , or s3 + s2 + 10s + 2 = 0 . The roots are: s1 = 0.2033, and s2,3 = 0.3984 j 3.1112. s4 + 10s3 + 35s2 + 50s + 24 = 0 . The roots are s1 = 1, s2 = 2, s3 = 3, and s4 = 4.
(b) The roots of the characteristic equation
(c) The roots of the characteristic equation
(b) The characteristic equation is
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
230
CHAPTER 6
The Stability of Linear Feedback Systems
(c) The characteristic equation is s3 + 11s2 + 29s + 6 = 0 . The roots are s1 = 0.2258, s2 = 3.8206 and s3 = 6.9536. P6.7 (a) The closed-loop characteristic equation is s3 + 101s2 + (100 + 10KKa )s + 100KKa = 0 . The Routh array is s3 s2 s1 so where b = 100 + 910 KKa > 0 . 101 1 101 b 100KKa 100 + 10KKa 100KKa
Thus, examing the rst column, we determine that KKa > 0 stabilizes the system. (b) The tracking error is e(s) = lim s(1 T (s))
s0
100 100 = . 2 s KKa
We require E (s) < 1o = 0.01745. So, KKa > 100 = 5729 . 0.01745
When KKa = 5729, the roots of the characteristic polynomial are s1 = 10.15 P6.8 and s2,3 = 45.43 j 233.25 .
(a) The closed-loop characteristic equation is 1+ or s3 + 7s2 + 14s + 8(1 + K ) = 0 . The Routh array is K =0, (0.5s + 1)(s + 1)( 1 s + 1) 4
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
231 s3 s2 s1 so 1 7 b 8(1 + K ) 7(14) 8(1 + K ) . 7 14 8(1 + K )
where b=
For stability, we require b > 0 and 8(1 + K ) > 0. Therefore, the range of K for stability is 1 < K < 11.25 . (b) Let K = 11.25/3 = 3.75. Then, the closed-loop transfer function is T (s) = s3 + 7s2 3.37 . + 14s + 38
The settling time to a step input is Ts 6 seconds. (c) We want Ts = 4 sec, so Ts = 4 = 4 n implies n = 1 .
Our desired characteristic polynomial is
2 2 2 (s + b)(s2 + 2n s + n ) = s3 + (2 + b)s2 + (n + 2b)s + bn
where we have used the fact that n = 1 and n and b are to be determined. Our actual characteristic polynomial is s3 + 7s2 + 14s + 8(1 + K ) = 0 . Comparing the coecients of the actual and desired characteristic polynomials, we nd the following relationships:
2 n
2+b=7 + 2b = 14 2 bn = 8(1 + K ) .
Solving these three equations yields b=5, n = 2 and K = 1.5 .
The actual settling time is Ts = 4.17 sec. This is not exactly our
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
232
CHAPTER 6
The Stability of Linear Feedback Systems
desired Ts since we have the contribution of the additional pole at s = 5. The closed-loop poles are s1 = 5 and s2,3 = 1 1.73j . P6.9 (a) The closed-loop characteristic equation is 1 + GH (s) = 1 + or s3 + 140s2 + 4400s + 40000 + 10K = 0 . The Routh array is s3 s2 s1 so where b= 140(4400) (40000 + 10K ) . 140 1 140 b 40000 + 10K 4400 40000 + 10K 10K , (s + 100)(s + 20)2
Examining the rst column and requiring all the terms to be positive, we determine that the system is stable if 4000 < K < 57600 . (b) The desired characteristic polynomial is
2 2 2 (s + b)(s2 +1.38n s + n) = s3 +(1.38n + b)s2 +(n +1.38n b)s + bn
where we have used the fact that = 0.69 to achieve a 5% overshoot, and n and b are to be determined. The actual characteristic polynomial is s3 + 140s2 + 4400s + 40000 + 10K = 0 . Equating the coecients of the actual and desired characteristic polynomials, and solving for K , b, and n yields b = 104.2 , n = 25.9 and K = 3003 .
So, a suitable gain is K = 3003.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
233
P6.10
(a) The closed-loop characteristic equation is s4 + 7s3 + 20s2 + (24 + K )s + 10K = 0 . The Routh array is s4 s3 s2 s1 so where b= Setting b > 0 yields 2784 398K K 2 > 0 , which holds when 404.88 < K < 6.876 . Examining the rst column, we also nd that K < 116 and K > 0 for stability. Combining all the stability regions, we determine that for stability 0 < K < 6.876 . (b) When K = 6.876, the roots are s1,2 = 3.5 1.63j , and s3,4 = 2.1j .
116K 7
1 7
116K 7
20 24 + K 10K
10K 0
b 10K
(24 + K ) 70K
116K 7
.
P6.11
Given s3 + (1 + K )s2 + 10s + (5 + 15K ) = 0 , the Routh array is s3 s2 s1 so 1 1+K b 5 + 15K 10 5 + 15K
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234
CHAPTER 6
The Stability of Linear Feedback Systems
where b= (1 + K )10 (5 + 15K ) 5 5K = . 1+K 1+K
Given that K > 0, we determine that the system is stable when 55K > 0 or 0<K <1. When K = 1, the s2 row yields the auxilary equation 2s2 + 20 = 0 . The roots are s = j 10. So, the system frequency of oscillation is 10 rads/sec. P6.12 The system has the roots s1,2 = j 2 , s3,4 = j 2 , s 5 = 3 , and s6 = 2 .
Therefore, the system is not stable since there are repeated roots on the j -axis. P6.13 (a) Neglecting the zeros and poles, we have the characteristic equation s4 + 30s3 + 325s2 + 2500s + K = 0 . The Routh array is s4 s3 s2 s1 so where b= 604166.67 30K . 241.67 1 30 241.67 b K 325 2500 K K 0
Therefore, the system is stable for 0 < K < 20139. (b) Without neglecting the zeros and poles, the closed-loop characteristic equation is s6 + 90s5 + 5525s4 + 12400s3 + (1255000 + K )s2 + (8500000 + 30K )s + 1125K = 0 .
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Problems
235
This is stable for 0 < K < 61818 . We see that the additional poles and zero makes the system stable for a much larger gain K . P6.14 (a) The Routh array is s3 s2 s1 so 1 5 3.8 6 5 6
Examining the rst column of the Routh array, we see no sign changes. So, the system is stable. (b) The roots of the system are s1 = 0.3246 and s2,3 = 2.3377 3.6080j . (c) The step response is shown in Figure P6.14.
Step Response 0.18 0.16 0.14 0.12 Amplitude 0.1 0.08 0.06 0.04 0.02 0 0 5 10 Time ( sec ) 15
FIGURE P6.14 Unit step response.
P6.15
The closed-loop transfer function is T (s) = s3 + 3s2 K +1 . + 3s + K + 1
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236
CHAPTER 6
The Stability of Linear Feedback Systems
The Routh array is s3 s2 s1 so 1 3
8K 3
3 K+1
K +1
So, for stability we require 1 < K < 8. P6.16 The system characteristic equation is s2 + (h k)s + ab kh = 0 . For stability we require h > k and ab > kh. If k > h, the system is unstable. P6.17 (a) The characteristic equation is s3 + 9s2 + (K 10)s + 2K = 0 . The Routh array is s3 s2 s1 so For stability K > 90/7 . (b) When K = 90/7, the system is marginally stable. The roots are s1,2 = j 20/7 , at the j -axis crossing. P6.18 The closed-loop characteristic equation is q (s) = s5 + s4 + 4s3 + 4Ks2 + 2Ks + K . The range of stability for the vertical-lifto vehicle is 0.5361 < K < 0.9326 . Therefore, for K = 1, the system is unstable. 1 9
7K 90 9
K 10 2K
2K
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Problems
237
P6.19
The state transition matrix is (k p1 )ep1 t (k2 p2 )ep2 t ep1 t ep2 t 1 2 (t, 0) = p1 t + k ep2 t p1 t + p ep2 t p2 p1 k1 e p1 e 1 2
where p1 p2 = k1 and p1 + p2 = k2 . We assume that p1 = p2 . In the case when p1 = p2 , the state transition matrix will change, but the factors ep1 t and ep2 t will remain. The eigenvalues of A are given by the solution to det |I A| = 2 + k2 + k1 = 0 .
2 Therefore, the eigenvalues are 1,2 = k2 /2 k2 4k1 . If k2 > 0 and k1 > 0, then the eigenvalues are in the left half-plane, and the system is stable. The transfer function is given by
G(s) = C (sI A)1 B =
s1 . s2 + k2 s + k1
Therefore the characteristic equation is s2 + k2 s + k1 = 0 and the poles 2 are s1,2 = k2 /2 k2 4k1 . If k2 > 0 and k1 > 0, then the poles are in the left half-plane, and the system is stable. Notice that the values of 1,2 and s1,2 are the same. Also, the eigenvalues are the same as the values of p1 and p2 . So, if the eigenvalues are negative, then the elements of the state transition matrix will decay exponentially.
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238
CHAPTER 6
The Stability of Linear Feedback Systems
Advanced Problems
AP6.1 The Routh array is s4 s3 s2 s1 so where b= 20K1 4 100K2 . 5K1 1 1 20
20K1 4 20
K1 4 K2 0
K2
b K2
For stability, we require K2 > 0, K1 > 0.2, and b > 0. Therefore, using the condition that b > 0, we obtain K2 < 0.2K1 0.04 . The stability region is shown in Figure AP6.1.
0. 4
0.35
0. 3
0.25
K2
0. 2
0.15
0. 1
0.05 STABLE REGION 0 0. 2 0. 4 0. 6 0. 8 1 K1 1. 2 1. 4 1. 6 1. 8 2
FIGURE AP6.1 Stability region.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
239
AP6.2
The Routh array is s4 s3 s2 s1 so where b= (320 K )(K 41) 40.5K . 320 K 1 9
320K 9
31 K 41 0.5K 0
0.5K
b 0.5K
Therefore, using the condition that b > 0, we obtain the stability range for K : 48.178 < K < 272.32 . AP6.3 (a) The steady-state tracking error to a step input is ess = lim s(1 T (s))R(s) = 1 T (0) = 1 .
s0
We want |1 | < 0.05 . This yields the bounds for 0.95 < < 1.05 . (b) The Routh array is s3 s2 s1 so where b= 2 + 1 . 1+ 1 1+ b 1 1 0
Therefore, using the condition that b > 0, we obtain the stability range for : > 0.618 .
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240
CHAPTER 6
The Stability of Linear Feedback Systems
(c) Choosing = 1 satises both the steady-state tracking requirement and the stability requirement. AP6.4 The closed-loop transfer function is T (s) = The Routh array is s3 s2 s1 so where b= p2 + p K . 1+p 1 1+p b K p K 0 s3 K . + (p + 1)s2 + ps + K
Therefore, using the condition that b > 0, we obtain the the relationship K < p2 + p . The plot of K as a function of p is shown in Figure AP6.4.
120
100
80
60
K
40 STABLE REGION 20
0 0
1
2
3
4
5 p
6
7
8
9
10
FIGURE AP6.4 Stability region.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
241
AP6.5
The closed-loop transfer function is T (s) = 30K1 K2 . (s + 1 + K1 K3 )(s 10)(2s + K2 K3 4) + 30K1 K2 K4 s3 s2 s1 so
The Routh array is 2 b d c a c 0
2 where a = 9K2 K3 + 16 + K1 K2 K3 24K1 K3 , b = 2K1 K3 + K2 K3 22, 2 and c = 10K2 K3 + 40 10K1 K2 K3 + 40K1 K3 and d = (ab 2c)/b . The conditions for stability are
2K1 K3 + K2 K3 22 > 0 2 10K2 K3 + 40 10K1 K2 K3 + 40K1 K3 > 0 2 2(10K2 K3 + 40 10K1 K2 K3 + 40K1 K3 ) + (9K2 K3 2 +16 + K1 K2 K3 24K1 K3 )(2K1 K3 + K2 K3 22) > 0 Valid values for the various gains are: K1 = 50, K2 = 30, K3 = 1, and K4 = 0.3. The step response is shown in Figure AP6.5.
Step Response 350
300
250
Amplitude
200
150
100
50
0
0
5
10
15 Time (sec)
20
25
30
FIGURE AP6.5 Stability region.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
242
CHAPTER 6
The Stability of Linear Feedback Systems
Design Problems
CDP6.1
The plant model with parameters given in Table CDP2.1 in Dorf and Bishop is given by: (s) 26.035 = . Va (s) s(s + 33.142) In the above transfer function we have neglected the motor inductance Lm . The closed-loop transfer function from the input to the output is 26.035Ka (s) =2 . R(s) s + 33.142s + 26.035Ka The Routh array is s2 s1 s0 1 33.142 26.035Ka 26.035Ka 0
Stability is achieved for any 0 < Ka < . DP6.1 The closed-loop characteristic polynomial is 1 1 1 s3 + s2 (5 + p + K ) + s( Kp + K + 5p) + K = 0 . 5 5 5 (i) When p = 2, we have 1 3 s3 + s2 (7 + K ) + s(10 + K ) + K = 0 . 5 5 The Routh array is s3 s2 s1 so where b= (7 + K/5)(10 + 3K/5) K . 1 7 + 5K 1 7+ b K
K 5
10 + 3 K 5 K
When 32.98 < K < 17.69, we nd that b > 0. Examining the other terms in the rst column of the array, we nd that the system
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
243
is stable for any K > 0. (ii) When p = 0, we have 1 1 s3 + s2 (5 + K ) + s( K ) + K = 0 . 5 5 The Routh array is s3 s2 s1 so where b=
1 (5 + 5 K ) 1 K K K 2 /25 5 = . (5 + K/5) (5 + K/5)
1 5 + 1K 5 b K
1 5K
K
Again, examination of the rst column reveals that any K > 0 results in a stable system. So, we just need to select any K > 0; e.g. K = 10. DP6.2 (a) The closed-loop characteristic equation is 1+ or s3 + 10s2 + 10Ks + 10 = 0 . The Routh array is s3 s2 s1 so where b= 10K 1 . 1 1 10 b 1 10K 10 10(Ks + 1) =0, s2 (s + 10)
For stability, we require K > 0.1. (b) The desired characteristic polynomial is (s2 + as + b)(s + 5) = s3 + s2 (a + 5) + s(5a + b) + 5b = 0 . Equating coecients with the actual characteristic equation we can
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244
CHAPTER 6
The Stability of Linear Feedback Systems
solve for a, b and K , yielding b = 2, a = 5, and K= 5a + b 27 = . 10 10
(c) The remaining two poles are s1 = 4.56 and s2 = 0.438. (d) The step response is shown in Figure DP6.2.
1
0.9
0.8
0.7
0.6 y(t)
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10 time (sec)
12
14
16
18
20
FIGURE DP6.2 Mars guided vehicle step response.
DP6.3
(a) The closed-loop characteristic equation is 2 s3 + ( + 2)s2 + (K + 1)s + 2K = 0 . The Routh array is s3 s2 s1 so where b= ( + 2)(K + 1) 4K . ( + 2) 2 +2 b 2K K+1 2K
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Design Problems
245
Examining the rst column of the Routh array, we determine that for stability > 0, K > 0 and setting b > 0 yields the relationships: (1) K < +2 2 when > 3 2 3 (2) K > 0 when 0 < 2 . 3
The plot of versus K is shown in Figure DP6.3a.
5 4.5 4 3.5 3
tau
2.5 2 1.5 1 0.5
STABLE REGION 1 2 3 K 4 5 6 7
0
FIGURE DP6.3 (a) The plot of versus K .
(b) The steady-state error is ess = So, ess 1 = . A 2K We require that ess 0.25A, therefore K2. One solution is to select = 0.5, then we remain within the stable region. (c) The step response is shown in Figure DP6.3b. The percent overshoot is P.O. = 57%. A , Kv where Kv = 2K .
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246
CHAPTER 6
The Stability of Linear Feedback Systems
1.6 P.O. = 56.77 % 1.4 1.2 1
y(t)
0.8 0.6 0.4 0.2 0 0
2
4
6
8
10 time (sec)
12
14
16
18
20
FIGURE DP6.3 CONTINUED: (b) Closed-loop system step response.
DP6.4
(a) The closed-loop characteristic polynomial is s3 + Ks2 + [(2 + m)K 1]s + 2mK = 0 . The Routh array is s3 s2 s1 so 1 K b 2mK 2K + mK 1 2mK
Examining the rst column of the Routh array, we see that for stability we require m > 0, K > 0, and b > 0, where b= or K> 1 + 2m . 2+m (2K + mK 1)K 2mK = (2 + m)K (1 + 2m) > 0 , K
The plot of K vs m is shown in Figure DP6.4a.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
247
1.6
1.4 STABLE REGION 1.2
1
K
0.8
0.6
0.4
0
0.5
1
1.5
2
2.5 m
3
3.5
4
4.5
5
FIGURE DP6.4 (a) The plot of K versus m.
1.8 P.O. = 64.3208 % 1.6
1.4
1.2
1 y(t) 0.8 0.6 0.4 0.2 0 0
1
2
3
4
5 time (sec)
6
7
8
9
10
FIGURE DP6.4 CONTINUED: (b) Shuttle attitude control step response.
(b) The steady-state error is 1 1 ess = = < 0.10 , A Kv 2mK
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248
CHAPTER 6
The Stability of Linear Feedback Systems
or mK > 5. For example, we can select m = 0.5 and K = 2. (c) See Figure DP6.4b for the step response where P.O. = 64.3%. DP6.5 The closed-loop transfer function is T (s) = s3 10s2 K . + 20s + K
+
The range of K for stability is 0 < K < 200. If we let K = Km /N where Km = 200, then N = 6.25 results in a step response with P.O. = 15.7% and Ts = 1.96 seconds. DP6.6 The closed-loop system is given by
x=
0
1
2 K1 2 K2
x+
0 1
r
The characteristic polynomial is s2 +(2+ K2 )s + K1 2 = 0. So the system is stable for K1 > 2 and K2 > 2. Selecting K = 10 2 results in closed-loop eigenvalues at s = 2 2j . The closed-loop step response has a settling time of 2.11 s and a percent overshoot of 4.32%.
Im(s)
sin-1 = sin-1 0.69=43.63
Re(s)
desired region for eigenvalues
n = -1
FIGURE DP6.6 Performance region.
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Design Problems
249
DP6.7
(a) The inner loop closed-loop transfer function is Y (s) 20s =3 . 2 + 20s + 20K U (s) s + 10s 1 The Routh array is s3 s2 s1 so 1
20020K1 10
20 20K1
20K1
For stability 0 < K1 < 10. (b) The fastest response (that is, the quickest settling time) occurs when K1 = 2.2 (c) With K1 = 2.2, the closed-loop transfer function is Y (s) 20K2 s =3 . R(s) s + 10s2 + (20 + 20K2 )s + 44 The Routh array is s3 s2 s1 so For stability, we require 200K2 + 156 > 0 . Therefore, K2 > 0.78. DP6.8 The closed-loop characteristic equation is s2 + 4KD s + 4(KP + 1) = 0. So, it is possible to nd KP and KD to stabilize the system. For example, any KP > 0 and KD > 0 leads to stability. Choosing KP 9 results in a steady-state tracking error less than 0.1 due to a unit step input. Then, the damping ratio = 2/2 is achieved by selecting 2 KP + 1 KD = . 2 1 10
200K2 +156 10
20(K2 + 1) 44
44
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250
CHAPTER 6
The Stability of Linear Feedback Systems
Computer Problems
CP6.1 The m-le script is shown in Figure CP6.1.
ans = -2.1795 -0.4102 + 1.7445i -0.4102 - 1.7445i ans = pa=[1 3 5 7]; roots(pa) pb=[1 3 4 4 10]; roots(pb) pc=[1 0 2 1]; roots(pc) -1.8868 + 1.1412i -1.8868 - 1.1412i 0.3868 + 1.3810i 0.3868 - 1.3810i ans = 0.2267 + 1.4677i 0.2267 - 1.4677i -0.4534
FIGURE CP6.1 Computing the polynomial roots with the rootsfunction.
CP6.2
The m-le script is shown in Figure CP6.2.
K1=1;K2=2;K3=5; den=[1 2 1]; num1=K1*[1 -1 2];num2=K2*[1 -1 2];num3=K3*[1 -1 2]; sys1 = tf(num1,den); sys2 = tf(num2,den); sys3 = tf(num3,den); sys1_cl=feedback(sys1,[1]); sys2_cl=feedback(sys2,[1]); sys3_cl=feedback(sys3,[1]); p1 = pole(sys1_cl), p2 = pole(sys2_cl), p3 = pole(sys3_cl)
ans = -2.5000e -01 + 1.1990e+00i -2.5000e -01 - 1.1990e+00i
ans =
ans = 2.5000e -01 + 1.3307e+00i 2.5000e -01 - 1.3307e+00i 0 + 1.2910e+00i 0 - 1.2910e+00i
FIGURE CP6.2 K = 1 is stable;K = 2 is marginally stable; and K = 5 is unstable.
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Computer Problems
251
CP6.3
The closed-loop transfer function and the roots of the characteristic equation are shown in Figure CP6.3.
Transfer function: s+1 ---------------------s^3 + 4 s^2 + 7 s + 11 r= -2.8946 -0.5527 + 1.8694i -0.5527 - 1.8694i
numg=[1 1]; deng=[1 4 6 10]; sysg = tf(numg,deng); sys = feedback(sysg,[1]) r=pole(sys)
FIGURE CP6.3 Closed-loop transfer function and roots.
CP6.4
There are no poles in the right half-plane, but the system is unstable since there are multiple poles on the j -axis at s = j and s = j (see Figure CP6.4).
Step Response
From: U(1) 25
20
Amplitude
num=[1]; den=[1 2 2 4 1 2]; sys = tf(num,den); pole(sys) t = 0:0.1:100; step(sys,t) ans = -2.0000 0.0000 + 1.0000i 0.0000 - 1.0000i -0.0000 + 1.0000i -0.0000 - 1.0000i
15
10
5 To: Y(1)
0
-5
-10
-15
-20
-25
0
10
20
30
40
50
60
70
80
90
100
Time (sec.)
FIGURE CP6.4 Unstable system step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
252 CP6.5
CHAPTER 6
The Stability of Linear Feedback Systems
The closed-loop system poles for the slow/fast pilots are shown in Figure CP6.5. The maximum allowable time delay is 0.4776 seconds. At the maximum allowable time delay, the system has roots on the j -axis at s = 2.1j . The slow pilot destabilizes the aircraft.
nume=[-10]; dene=[1 10]; syse = tf(nume,dene); numg=[-1 -5]; deng=[1 3.5 6 0]; sysg = tf(numg,deng); % % Fast pilot % tau=0.25; tau1=2; K=1; tau2=0.5; nump=-K*[tau1*tau tau-2*tau1 -2]; denp=[tau2*tau tau+2*tau2 2]; sysp = tf(nump,denp); sysa = series(sysp,syse); sysb = series(sysa, sysg); sys = feedback(sysb,[1]); fast_pilot=pole(sys) % % Slow pilot % tau=0.50; tau1=2; K=1; tau2=0.5; nump=-K*[tau1*tau tau-2*tau1 -2]; denp=[tau2*tau tau+2*tau2 2]; sysp = tf(nump,denp); sysa = series(sysp,syse); sysb = series(sysa, sysg); sys = feedback(sysb,[1]); slow_pilot = pole(sys) % % Maximum pilot time delay, tau = 0.4776 sec % tau=0.4776; tau1=2; K=1; tau2=0.5; nump=-K*[tau1*tau tau-2*tau1 -2]; denp=[tau2*tau tau+2*tau2 2]; sysp = tf(nump,denp); sysa = series(sysp,syse); sysb = series(sysa, sysg); sys = feedback(sysb,[1]); max_pilot_delay=pole(sys)
closed-loop system poles fast_pilot = -9.3293 -9.3293 -4.0580 -0.2102 -0.2102 -0.3629 + 2.3290i - 2.3290i + 2.4146i - 2.4146i
slow_pilot = -8.9844 -5.0848 + 1.3632i -5.0848 - 1.3632i 0.0138 + 2.0742i 0.0138 - 2.0742i -0.3734
max_pilot_delay = -8.9054 -5.2049 + 1.2269i -5.2049 - 1.2269i 0.0000 + 2.1012i 0.0000 - 2.1012i -0.3725
FIGURE CP6.5 Closed-loop system poles for an aircraft with a pilot in-the-loop.
CP6.6
The closed-loop transfer function is T (s) = 1 . + (K 3)s + K + 1
s3
+
5s2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
253
Utilizing the Routh-Hurwitz approach, for stability we determine that K>4. When K = 4, the roots of the characteristic equation are s1 = 5 and s2,3 = j . The m-le script which generates a plot of the roots of the characteristic equation as a function of K is shown in Figure CP6.6.
K=[0:0.1:5]; n=length(K); for i=1:n numg=[1]; deng=[1 5 K(i)-3 K(i)]; sys_o = tf(numg,deng); sys_cl = feedback(sys_o,[1]); p(:,i)=pole(sys_cl); end plot(real(p),imag(p),'x'), grid text(-0.9,0.95,'K=4 -->'); text(-0.2,1.3,'K=5'); text(0,0.2,'K=0') % From a Routh-Hurwitz analysis we nd that % minimum K for stability is K=4 Kmax=4; numg=[1]; deng=[1 5 Kmax-3 Kmax]; sysg = tf(numg,deng); sys_cl = feedback(sysg,[1]); pole(sys_cl)
1.5
K=5 1 K=4 -->
0.5 K=0 0
-0.5
-1
-1.5 -6
-5
-4
-3
-2
-1
0
1
FIGURE CP6.6 Roots of the characteristic equation as a function of K , where 0 < K < 5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
254 CP6.7
CHAPTER 6
The Stability of Linear Feedback Systems
The characteristic equation is p(s) = s3 + 10s2 + 15s + 10 .
A=[0 1 0;0 0 1;-10 -15 -10]; b=[0;0;10];c=[1 1 0]; d=[0]; sys = ss(A,b,c,d); % % Part (a) % p=poly(A) % % Part (b) % roots(p) % % Part (c) % step(sys)
p= 1.0000 10.0000 15.0000 10.0000 r= -8.3464 -0.8268 + 0.7173i -0.8268 - 0.7173i
Step Response
From: U(1) 1.4
1.2
1
Amplitude
0.8 To: Y(1) 0.6 0.4 0.2 0 0
1.4
2.8
4.2
5.6
7
Time (sec.)
FIGURE CP6.7 Characteristic equation from the state-space representation using the poly function.
The roots of the characteristic equation are s1 = 8.3464 and s2,3 = 0.8268 0.7173j .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
255
The system is stable since all roots of the characteristic equation are in the left half-plane. The unit step response and associated m-le script are shown in Figure CP6.7. CP6.8 The characteristic equation is s3 + 10s2 + 10s + 5K1 = 0 . (a) The Routh array is s3 s2 s1 so 1 10
1005K1 10
10 5K1
5K1
From the Routh-Hurwitz criterion, we obtain the limits 0 < K1 < 20 for stability. (b) The plot of the pole locations is 0 < K1 < 30 is shown in Figure CP6.8. As seen in Figure CP6.8, when K1 > 20, the pole locations move into the right half-plane.
Root Locus 4 3 2 Imaginary Axi s 1 0 ?-1 ?-2 ?-3 ?-4 ?-12 ?-10 ?-8 ?-6 ?-4 ?-2 0 2
k=20
Real Axi s
FIGURE CP6.8 Pole locations for 0 < K1 < 30.
CP6.9
(a) The characteristic equation is s3 + 2s2 + s + k 4 = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
256
CHAPTER 6
The Stability of Linear Feedback Systems
The Routh array is s3 s2 s1 so 1 2
6k 2
1 k4
For stability, we obtain 4 < k < 6. (b) The pole locations for 0 < k < 10 are shown in Figure CP6.9. We see that for 0 < k < 4 the system is unstable. Similarly, for 6 < k < 10, the system is unstable.
Root Locus 2
k4
pole locations when k=0
1.5 1 Imaginary Axi s 0.5 0
k=10
pole location when k=0 k=10 increasing k
gk
?-0.5 ?-1 ?-1.5 ?-2 ?-3 ?-2
k=4 k=6
inc
?-1 Real Axi s
rea
sin
0
1
FIGURE CP6.9 Pole locations for 0 < k < 10.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
7
The Root Locus Method
Exercises
E7.1 (a) For the characteristic equation 1+K s2 s(s + 4) =0, + 2s + 2
the root locus is shown in Figure E7.1.
4 3 2 1
x
Imag Axis
0 -1 -2 -3 -4
o
o
x
-4
-3
-2
-1
0 Real Axis
1
2
3
4
FIGURE E7.1 s(s+4) Root locus for 1 + K s2 +2s+2 = 0.
257
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258
CHAPTER 7
The Root Locus Method
(b) The system characteristic equation can be written as (1 + K )s2 + (2 + 4K )s + 2 = 0 . Solving for s yields s= When (2 + 4K )2 8(1 + K ) = 0 ,
K then we have two roots at s1,2 = (1+2K ) . Solving for K yields K = 1+ 0.31. (c) When K = 0.31, the roots are
(1 + 2K ) (1 + K )
(2 + 4K )2 8(1 + K ) . 2(1 + K )
s1,2 =
(1 + 0.62) = 1.24 . (1.31)
(d) When K = 0.31, the characterisitc equation is s2 + 2.472s + 1.528 = (s + 1.24)2 = 0 . Thus, n = 1.24 and = 1, the system is critically damped. The settling time is Ts 4 sec. E7.2 (a) The root locus is shown in Figure E7.2. When K = 6.5, the roots of the characteristic equation are s1,2 = 2.65 j 1.23 and s3,4 = 0.35 j 0.8 .
The real part of the dominant root is 8 times smaller than the other two roots. (b) The dominant roots are (s + 0.35 + j 0.8)(s + 0.35 j 0.8) = s2 + 0.7s + 0.7625 . From this we determine that n = 0.873 Thus, the settling time is 4 4 = = 11.43 sec . n 0.35 2 The percent overshoot is P.O. = e/ 1 = 25.4%. Ts = and = 0.7 = 0.40 . 2(0.873)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
259
4 3 2 1
* x * x x
* K=6.5
Imag Axis
0 -1
*
x
*
-2 -3 -4 -4
-3
-2
-1
0 Real Axis
1
2
3
4
FIGURE E7.2 1 Root locus for 1 + K s(s+2)(s2 +4s+5) = 0.
E7.3
The root locus is shown in Figure E7.3. The roots are s1 = 8.7, s2,3 = 1.3 j 2.2 when K = 7.35 and = 0.5.
4 3 2 1 zeta=0.5
*
o
<----- K=7.35
Imag Axis
0 -1 -2 -3 -4 -10
*
x
x
o *
-8
-6
-4 Real Axis
-2
0
2
4
FIGURE E7.3 2 s+8 Root locus for 1 + K s 2+4+4) = 0. s (s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
260 E7.4
CHAPTER 7
The Root Locus Method
The root locus is shown in Figure E7.4.
2 1.5 1 0.5
x
Imag Axis
0 -0.5 -1 -1.5 -2 -4
x
o
-3.5
-3
-2.5
-2
-1.5 Real Axis
-1
-0.5
0
0.5
1
FIGURE E7.4 s+1 Root locus for 1 + K s2 +4s+5 = 0.
The departure angles and entry points are d = 225o , 225o and b = 2.4 . E7.5 (a) The root locus is in Figure E7.5. The breakaway points are b1 = 13.0 , b2 = 5.89 . (b) The asymptote centroid is cent = 18 , and asym = 90o . (c) The gains are K1 = 1.57 and K2 = 2.14 at the breakaway points.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
40
261
30
20
10 Imag Axis
0
-10
< asymptote
-20
-30
-40
-15
-10 Real Axis
-5
0
5
FIGURE E7.5 s2 +2s Root locus for 1 + K (s4 +38s3 +515s2 +10 s+6000) = 0. +2950
E7.6
The system is unstable for K > 100.
Root Locus 30
20
10 Imaginary Axis
0
10
20
30 50
40
30
20 10 Real Axis
0
10
20
FIGURE E7.6 20 Root locus for 1 + K s(s2 +20K+100) = 0. s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
262 E7.7
CHAPTER 7
The Root Locus Method
The root locus is shown in Figure E7.7. The characteristic equation has
Root Locus 20
15
10
asymptote >
5 Imaginary Axis
System: sys Gain: 25.2 Pole: 1.59 + 1.19i Damping: 0.8 Overshoot (%): 1.53 Frequency (rad/sec): 1.98
0
5
10
15
20 25
20
15
10 Real Axis
5
0
5
10
FIGURE E7.7 s+10 Root locus for 1 + K s(s+5)(s+6)(s+8) = 0.
4 poles and 1 zero. The asymptote angles are = +60o , 60o , 180o centered at cent = 3. When K = 25.2 then = 0.8 for the complex roots. E7.8 The characteristic equation is 1+K or s3 + 9s2 + Ks + K = 0 . For all the roots to be equal and real, we require (s + r )3 = s3 + 3rs2 + 3r 2 s + r 3 = 0 . Equating terms and solving for K yields K = 27. All three roots are equal at s = 3, when K = 27. The root locus is shown in Figure E7.8. (s + 1) =0, s2 (s + 9)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
263
8 6 4 2
Imag Axis
3 roots at s=-3 0 -2 -4 -6 -8 -15
x o x
-10
-5 Real Axis
0
5
FIGURE E7.8 Root locus for 1 + K s2s+1 = 0. (s+9)
E7.9
The characteristic equation is 1+K or s3 + 2s2 + 5s + K = 0 . (a) The system has three poles at s = 0 and 1 j 2. The number of asymptotes is np nz = 3 centered at cent = 2/3, and the angles are asymp at 60o , 180o . (c) The Routh array is 1 =0 s(s2 + 2s + 5)
(b) The angle of departure, d , is 90o + d +116.6o = 180o , so d = 26.6o . s3 s2 s1 so 1 2 b K 5 K
where b = 5 K/2. So, when K = 10 the roots lie on the imaginary
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
264
CHAPTER 7
The Root Locus Method
axis. The auxilary equation is 2s2 + 10 = 0 which implies s1,2 = j 5 .
(d) The root locus is shown in Figure E7.9.
4 3 2 1
x
Imag Axis
asymptote ---> 0 -1 -2 -3 -4 -4
x x
-3
-2
-1
0 Real Axis
1
2
3
4
FIGURE E7.9 1 Root locus for 1 + K s(s2 +2s+5) = 0.
E7.10
(a) The characteristic equation is 1+ Therefore, K= and dK s2 + 4s + 2 = =0. ds (s + 2)2 Solving s2 +4s+2 = 0 yields s = 0.586 and 3.414. Thus, the system (s2 + s) , (s + 2) K (s + 2) =0. s(s + 1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
265
(b) The desired characteristic polynomial is
breakaway and entry points are at s = 0.586 and s = 3.414. (s + 2 + aj )(s + 2 aj ) = s2 + 4s + 4 + a2 = 0 , where a is not specied. The actual characteristic polynomial is s2 + (1 + K )s + 2K = 0 . Equating coecients and solving for K yields K = 3 and a = Thus, when K = 3, the roots are s1,2 = 2 2j . 2.
(c) The root locus is shown in Figure E7.10.
2 1.5 1 0.5 K=3, s=-2+1.414j
*
Imag Axis
0 -0.5 -1 -1.5 -2 -4
s=-3.41
o
x
s=-0.58
x
*
-3.5
-3
-2.5
-2
-1.5 Real Axis
-1
-0.5
0
0.5
1
FIGURE E7.10 Root locus for 1 + K s(s+2 = 0. s+1)
E7.11
The root locus is shown in Figure E7.11 for the characteristic equation 1+ K (s + 2.5) =0. (s2 + 2s + 2)(s2 + 4s + 5)
From the root locus we see that we can only achieve = 0.707 when K = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
266
CHAPTER 7
The Root Locus Method
5 4 3 2 1
x x
Imag Axis
<---- zeta=0.707 & K=0
0 -1 -2 -3 -4 -5 -5 -4 -3
o
x
x
-2
-1
0 Real Axis
1
2
3
4
5
FIGURE E7.11 Root locus for 1 +
K (s+2.5) (s2 +2s+2)(s2 +4s+5)
= 0.
E7.12
(a) The root locus is shown in Figure E7.12 for the characteristic equation 1+ K (s + 1) =0. s(s2 + 6s + 18)
(b) The roots of the characteristic equation are (i) K = 10: s1,2 = 2.8064 4.2368j and s3 = 0.3872 (ii) K = 20: s1,2 = 2.7134 5.2466j and s3 = 0.5732
(c) The step response performance of the system is summarized in Table E7.12.
K Ts (sec) P.O. Tr (sec)
TABLE E7.12
10 9.0 0 4.8
20 5.5 0 2.6
System performance when K = 10 and K = 20.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
267
Root Locus 15
10
5 Imaginary Axis
0
5
10
15 3.5
3
2.5
2
1.5 Real Axis
1
0.5
0
0.5
FIGURE E7.12 Root locus for 1 +
K (s+1) s(s2 +6s+18)
= 0.
E7.13
(a) The characteristic equation is s(s + 1)(s + 3) + 4s + 4z = 0 . Rewriting with z as the parameter of interest yields 1+z 4 =0. s(s + 1)(s + 3) + 4s
The root locus is shown in Figure E7.13a. (b) The root locations for z = 0.6 , 2.0 , and 4.0 are shown in Figure E7.13a. When z = 0.6, we have = 0.76 and n = 2.33. Therefore, the predicted step response is P.O. = 2.4% and Ts = 2.3 sec ( = 0.6) .
When z = 2.0, we have = 0.42 and n = 1.79. Therefore, the predicted step response is P.O. = 23% and Ts = 5.3 sec ( = 2.0) .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
268
CHAPTER 7
The Root Locus Method
4 3 2 1
+ x * o
* z=0.6 o z=2.0 + z=4.0
Imag Axis
0 -1
+
o
*
x
x
*
o +
-2 -3 -4 -4
-3
-2
-1
0 Real Axis
1
2
3
4
FIGURE E7.13 (a) Root locus for 1 + z s(s+1)(4+3)+4s = 0. s
Finally, when z = 4.0, we have = 0.15 and n = 2.19. Therefore, the predicted step response is P.O. = 62% and Ts = 12 sec. (c) The actual step responses are shown in Figure E7.13b.
1.6
1.4
___ z=0.6 - - - z=2.0
1.2
..... z=4.0
1 y(t)
0.8
0.6
0.4
0.2
0
0
2
4
6
8 time (sec)
10
12
14
16
FIGURE E7.13 CONTINUED: (b) Step Responses for z = 0.6, 2.0, and 4.0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
269
E7.14
(a) The root locus is shown in Figure E7.14 for the characteristic equation 1+ K (s + 10) =0. s(s + 5)
The breakaway point is sb = 2.93; the entry point is se = 17.1.
10 8 6 4 2 K=5, s=-5+5j
*
Imag Axis
0 -2 -4 -6 -8 -10 -20
s=-17.1
o
x
s=-2.93
x
-15
-10 Real Axis
-5
0
5
FIGURE E7.14 Root locus for 1 +
K (s+10) s(s+5)
= 0.
(b) We desire = 1/ 2 = 0.707. So, the desired characteristic polynomial is 1 2 s2 + 2 n s + n = 0 . 2 Comparing the desired characteristic polynomial to the actual we nd the relationships 2 2n = 5 + K . n = 10K and Solving for K and n yields K = 5 and n = 7.07. The roots are s1,2 = 5 j 5 when K = 5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
270 E7.15
CHAPTER 7
The Root Locus Method
(a) The characteristic equation 1+K (s + 10)(s + 2) =0 s3
has the root locus in Figure E7.15.
Root Locus 15 10 5 0 ?-5 ?-10 ?-15 ?-25
Imaginary Axi s
K=1.67
?-20
?-15
?-10 Real Axi s
-5
0
5
FIGURE E7.15 Root locus for 1 +
K (s+10)(s+2) s3
= 0.
(b) The Routh array is s3 s2 s1 so 1 K b 20K 12K 20K
when b = 12K 20. For stability, we require all elements in the rst column to be positive. Therefore, K > 1.67 . (c) When K > 3/4, we have ess = lim sE (s) = lim s
s0 s0
1 1 s2 2 = lim 3 =0. s0 s + K (s + 1)(s + 3) 1 + GH (s) s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
271
E7.16
The expansion for eT s is eT s = 1 T s + If (T s) << 1, then eT s 1 T s = a + bs , c + ds (T s)2 ... 2!
where a, b, c and d are constants to be determined. Using long division,
40 30 20 10 K=21
*
Imag Axis
0 -10 -20 -30 -40 -40
x
x
o
-30
-20
-10
0
10 Real Axis
20
30
40
50
60
FIGURE E7.16 Root locus for 1 +
K (20s) (s+1)(20+s)
= 0.
we expand (a + bs)/(c + ds) and match as many coecients as possible. In this case, we determine that a = c = (2/T ) and also that b = d = 1. In this case, with T = 0.1, we have eT s = 20 s (s 20) = . 20 + s (s + 20)
So, the characteristic equation is 1+ K (s 20) , (s + 1)(s + 20)
and the root locus is shown in Figure E7.16. Using a Routh-Hurwitz
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272
CHAPTER 7
The Root Locus Method
analysis with the characteristic polynomial s2 + (21 K )s + 20 + 20K = 0 , we determine that the system is stable for 1 < K < 21. E7.17 (a) The root locus is in Figure E7.17a.
2 1.5 1 0.5
Imag Axis
0 -0.5 -1 -1.5 -2 -2
x
x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE E7.17 (a) Root locus for 1 +
K s(s1)
= 0.
The root locus is always in the right half-plane; the system is unstable for K > 0. (b) The characteristic equation is 1+ K (s + 2) =0, s(s 1)(s + 20)
and the root locus is shown in Figure E7.17b. The system is stable for K > 22.3 and when K = 22.3, the roots are s1,2 = j 1.53 and s3 = 19 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
273
10 8 6 4 2
Imag Axis
* x o
K=22.3
0 -2 -4 -6 -8 -10 -30 -25
xx
-20
-15
-10 Real Axis
-5
0
5
10
FIGURE E7.17 CONTINUED: (b) Root locus for 1 +
K (s+2) s(s+20)(s1)
= 0.
E7.18
The root locus is shown in Figure E7.18.
6
4
2
Imag Axis
+
x
+ x
K=8.15
0
x + x
+
-2
-4
-6 -6
-4
-2
0 Real Axis
2
4
6
FIGURE E7.18 Root locus for 1 +
K s(s+3)(s2 +2s+2)
= 0.
When K = 8.15, the roots are s1,2 = j 1.095 and s3,4 = 2.5 j 0.74.
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274 E7.19
CHAPTER 7
The Root Locus Method
The characteristic equation is 1+ s(s + 3)(s2 K =0, + 6s + 64)
and the root locus is shown in Figure E7.19. When K = 1292.5, the roots are s1,2 = j 4.62 and s3,4 = 4.49 j 6.36 .
15
10
x +
5
+
K=1292.5
Imag Axis
0
x
x
-5
+ x
+
-10
-15 -15
-10
-5
0 Real Axis
5
10
15
FIGURE E7.19 Root locus for 1 +
K s(s+3)(s2 +6s+64)
= 0.
E7.20
The characteristic equation is 1+ K (s + 1) =0, s(s 1)(s + 4) K>6. The maximum damping ratio of the stable complex roots is = 0.2 .
and the root locus is shown in Figure E7.20. The system is stable for
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
275
8 6 4 2 max zeta=0.2
Imag Axis
0 -2 -4 -6 -8 -8
x
o
x
x
-6
-4
-2
0 Real Axis
2
4
6
8
FIGURE E7.20 Root locus for 1 +
K (s+1) s(s1)(s+4)
= 0.
E7.21
The gain is K = 10.8 when the complex roots have = 0.66.
10
5 K=10.8
Imag Axis
+
x o
0
x
+ +
x
-5
-10 -10 -5 0 Real Axis 5 10
FIGURE E7.21 Root locus for 1 +
Ks s3 +5s2 +10
= 0.
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276 E7.22
CHAPTER 7
The Root Locus Method
The root locus is shown in Figure E7.22. The characteristic equation is 1+ K (s2 + 20)(s + 1) =0. (s2 2)(s + 10)
5
4
3
2
1 Imag Axis
0
-1
-2
-3
-4
-5
-10
-8
-6
-4 Real Axis
-2
0
2
FIGURE E7.22 Root locus for 1 +
K (s2 +20)(s+1) (s2 2)(s+10)
= 0.
E7.23
The characteristic equation is 5s2 + as + 4 = 0 , which can rewritten as 1+ as =0. 5s2 + 4
The roots locus (with a as the parameter) is shown in Figure E7.23.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
277
1.5 1 0.5
x
Imag Axis
0 -0.5 -1 -1.5 -1.5 -1 -0.5
o
x
0 Real Axis
0.5
1
1.5
FIGURE E7.23 Root locus for 1 +
as 5s2 +4
= 0.
E7.24
The transfer function is G(s) = C(sI A)1 B + D = [ 1 0 ] = s2
1
s
1
0 1
2 s+k
1 . + ks + 2
Therefore, the characteristic equation is s2 + ks + 2 = 0 , or 1+k s2 s =0. +2
The root locus for 0 < k < is shown in Figure E7.24. The closed-loop system is stable for all 0 < k < .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
278
CHAPTER 7
The Root Locus Method
Root Locus 1.5 1 0.5 0 -0.5 -1 -1.5 -2.5
Imaginary Axi s
-2
-1.5
-1 Real Axi s
-0.5
0
FIGURE E7.24 Root locus for 1 + k s2s = 0. +2
E7.25
The characteristic equation is 1+K 10 =0. s(s + 25)
The root locus shown in Figure E7.25 is stable for all 0 < K < .
Root Locus 15
10
5 Imaginary Axis
0
5
10
15 30
25
20
15 10 Real Axis
5
0
5
FIGURE E7.25 = 0. Root locus for 1 + K s(s10 +25)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
279
E7.26
The characteristic polynomial is det
s
1
s+K 3 s+K +2 1+K s+1 =0. s2 + 2s 3
or
=0
The root locus shown in Figure E7.26 is stable for all 0 < K < 3.
Root Locus 0.8
0.6
0.4
Imaginary Axis
0.2
0
0.2
0.4
0.6
0.8 12
10
8
6 4 Real Axis
2
0
2
FIGURE E7.26 s+1 Root locus for 1 + K s2 +2s3 = 0.
E7.27
The characteristic equation is 1+p s2 s =0. + 4s + 40
The root locus shown in Figure E7.25 is stable for all 0 < p < .
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280
CHAPTER 7
The Root Locus Method
Root Locus 8
6
4
Imaginary Axis
2
0
2
4
6
8 12
10
8
6 4 Real Axis
2
0
2
FIGURE E7.27 s Root locus for 1 + p s2 +4s+40 = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
281
Problems
P7.1
Root Locus 30
20
10 Imaginary Axis
0
10
20
30 50
40
30
20 10 Real Axis
0
10
20
Root Locus 5 4 3 2 Imaginary Axis 1 0 1 2 3 4 5 7
6
5
4
3 2 Real Axis
1
0
1
2
FIGURE P7.1 (a) Root locus for 1 +
K s(s+10)(s+8)
= 0, and (b) 1 +
K (s2 +2s+2)(s+1)
= 0.
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282
CHAPTER 7
The Root Locus Method
Root Locus 20
15
10
Imaginary Axis
5
0
5
10
15
20 8
7
6
5
4 3 Real Axis
2
1
0
1
Root Locus 2.5 2 1.5 1 Imaginary Axis 0.5 0 0.5 1 1.5 2 2.5 14
12
10
8
6 Real Axis
4
2
0
2
FIGURE P7.1 CONTINUED: (c) Root locus for 1 +
K (s+5) s(s+2)(s+7)
= 0, and (d)1 +
K (s2 +4s+8) s2 (s+7)
= 0.
P7.2
The root locus is shown in Figure P7.2 for the characteristic equation 1+ 10Kv (s + 10) =0. s(s + 1)(s + 100)
The damping ratio is = 0.6 when Kv = 0.8, 135 and 648. The roots of the characteristic equation are: (a) Kv = 0.8 : s1 = 99.9, s2,3 = 0.54 j 0.71
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
283
(b) Kv = 135 : s1 = 85.9, s2,3 = 7.5 j 10
(c) Kv = 648 : s1 = 11.7, s2,3 = 44.6 j 59.5
30
20
10
Imag Axis
0
x
o
xx
-10
-20
-30
-100
-80
-60 Real Axis
-40
-20
0
FIGURE P7.2 Root locus for 1 +
10Kv (s+10) s(s+1)(s+100)
= 0.
P7.3
(a) The breakaway point is s = 0.88 at K = 4.06. (b) The characteristic equation can be written as s(s + 2)(s + 5) + K = 0 . The Routh array is s3 s2 s1 so where b= 70 K . 7 1 7 b K 10 K 0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
284
CHAPTER 7
The Root Locus Method
When K = 70, the system has roots on j -axis at s = j 10. (c) When K = 6, the roots are s1,2 = 0.83 j 0.66, s3 = 5.34. (d) The characteristic equation 1+ K =0 s(s + 2)(s + 5)
has the root locus shown in Figure P7.3.
10 8 6 4 2 Imag Axis 0 -2 -4 -6 -8 -10 -10
-8
-6
-4
-2
0 Real Axis
2
4
6
8
10
FIGURE P7.3 Root locus for 1 +
K s(s+2)(s+5)
= 0.
P7.4
The characteristic equation for the large antenna is 1 + G1 G(s) = 1 + or 1+ 1000ka =0. (s + 10)(s2 + 14.4s + 100) 100ka =0, (0.1s + 1)(s2 + 14.4s + 100)
The root locus is shown in Figure P7.4. Using Rouths criteria, we nd that the system is stable for 1 < ka < 4.83 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
285
20 15 10
x *
<-- K=4.827
5
Imag Axis
0 -5
x
x
-10 -15 -20 -20
*
-15
-10
-5
0 Real Axis
5
10
15
20
FIGURE P7.4 Root locus for 1 +
1000ka (s2 +14.14s+100)(s+10)
= 0.
When ka = 4.83, we have s1,2 = j 15.53. P7.5 (a) The characteristic equation for hands-o control is 1+ 25K2 (s + 0.03)(s + 1) =0. (s + 0.4)(s2 0.36s + 0.16)(s + 9)
The root locus is shown in Figure P7.5a. The damping ratio is = 0.707 when K2 = 1.6 or K2 = 0.74. (b) The transfer function from Td (s) to Y (s) is Y (s) = where Gf (s) = K2 (s + 1) . s+9 G2 (s)Td (s) , 1 + G2 (s)Gf (s)
Using the nal value theorem, we determine that yss = lim s
s0
G2 (s) 1 11.7 = 1 + G2 (s)Gf (s) s 1 + 11.7
K2 9
= 3.8 ,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
286
CHAPTER 7
The Root Locus Method
5 4 3 2 1 <-- K=0.74
x x o xo x
<-- K=1.6
Imag Axis
0 -1 -2 -3 -4 -5 -10
-8
-6
-4
-2 Real Axis
0
2
4
FIGURE P7.5 (a) Root locus for 1 +
25K2 (s+0.03)(s+1) (s+9)(s2 0.36s+0.16)(s+0.4)
= 0.
20 15 10 5
x x o x x x o
Imag Axis
0 -5 -10 -15 -20 -20
-15
-10
-5
0 Real Axis
5
10
15
20
FIGURE P7.5 CONTINUED: (b) Root locus for 1 +
25K1 (s+0.03)(s+9) (s+0.045)(s2 +12s+1)(s+1.33)(s2 +7.66s+29.78)
= 0.
where we have selected K2 = 1.6. For K2 = 0.74, we nd that yss = 5.96.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
287
(c) The closed-loop characteristic equation with the pilot loop added is 1+ 25K1 (s + 0.03)(s + 9) =0. (s + 0.045)(s + 1.33)(s2 + 7.66s + 29.78)(s2 + 12s + 1)
The root locus is shown in Figure P7.5b. (d) Using K1 = 2, we determine that ess = 0.44 . P7.6 (a) The characteristic equation is 1+ K (s + 0.20)(s2 + 4s + 6.25) =0. (s + 0.9)(s 0.6)(s 0.1)(s + 4)
The root locus is shown in Figure P7.6.
4 zeta=0.5 3 2
o
zeta*wn=-1/3
K=4 -->
1
Imag Axis
0 -1
x
x
o
x
x
o
-2 -3 -4 -6
-5
-4
-3
-2 Real Axis
-1
0
1
2
FIGURE P7.6 Root locus for 1 +
K (s+0.2)(s2 +4s+6.25) (s+0.9)(s0.6)(s0.1)(s+4)
= 0.
(b) For Ts < 12 sec, we require n > 1/3. Also, we want > 0.5. So, we seek roots for a stable system with n > 1/3 and > 0.5. This occurs when K > 4.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
288 P7.7
CHAPTER 7
The Root Locus Method
(a) The characteristic equation for the speed control system is 1+ where K= 0.004 R and = 0.75 = 0.0001875 . 4000 K =0, (s + 4)2 (s + )
The root locus is shown in Figure P7.7. At = 0.6, we have K = 19.1,
6
4
2 <-- K=19.1
Imag Axis
0
x
x
-2
-4
-6 -6
-5
-4
-3
-2
-1 Real Axis
0
1
2
3
4
FIGURE P7.7 Root locus for 1 +
K (s+4)2 (s+1.875e04)
= 0.
therefore R = 0.00021 . When K = 19.1 the roots are s1,2 = 1.1 j 1.43 (b) The steady-state error is lim s (s) = lim s
s0
and
s3 = 5.80 .
s0
(0.25s + 1)2 L(s) (0.25s + 1)2 (Js + b) + 1/R
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
289 = 1 L LR , b + 1/R
when R < 0.1. P7.8 (a) The characteristic equation for the speed control system with the hydroturbine is 1+ where K= 0.002 R and = 0.75 = 0.0001875 . 4000 K (s + 1) =0, (s + 4)(s + 2)(s + )
The root locus is shown in Figure P7.8. At = 0.6, we have K = 2.85,
2 1.5 1 0.5 K=2.85 -->
Imag Axis
0 -0.5 -1 -1.5 -2 -6
x
x
x
o
-5
-4
-3
-2
-1 Real Axis
0
1
2
3
4
FIGURE P7.8 Root locus for 1 +
K (s+1) (s+4)(s+2)(s+ )
= 0.
therefore R = 0.0007 . When K = 2.85 the roots are 0.45 j 0.60, and -5.1.
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290
CHAPTER 7
The Root Locus Method
(b) The steady-state error is
s0
lim s (s) = lim s
s0
= when R < 0.1. P7.9
1 L LR , f + 1/R
(0.25s + 1)(0.5s + 1) L(s) (0.25s + 1)(0.5s + 1)(Js + f ) + (s + 1)/R
The characteristic equation is 1+K (s + 0.5)(s + 0.1)(s2 + 2s + 289) =0 s(s + 30)2 (s 0.4)(s + 0.8)(s2 + 1.45s + 361) K = 4000 , the roots are s1,2 = 0.82 j 19.4
50
where K = K1 K2 . The root locus is shown in Figure P7.9. When
40
30
20
10 Imag Axis
0
-10
-20
-30
-40
-50
-35
-30
-25
-20
-15 Real Axis
-10
-5
0
5
FIGURE P7.9 Root locus for 1 +
K (s+0.5)(s+0.1)(s2 +2s+289) s(s+30)2 (s0.4)(s+0.8)(s2 +1.45s+361)
= 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
291 s3 s4 s5 s6 s7 = 39.8 = 14.9 = 5.0 = 0.38 = 0.14 .
P7.10
(a) The characteristic equation is 1+ K1 K2 (s + 2)2 =0. (s + 10)(s + 100)(s2 + 1.5s + 6.25)
The root locus is shown in Figure P7.10.
10 8 6 4 2
x
Imag Axis
0 -2 -4 -6 -8 -10 -120
x
x
o
x
-100
-80
-60 Real Axis
-40
-20
0
FIGURE P7.10 Root locus for 1 +
K1 K2 (s+2)2 (s+10)(s+100)(s2 +1.5s+6.25)
= 0.
(b) The gain K1 K2 = 1620 when = 0.707. Therefore, K2 = 81000 ,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
292
CHAPTER 7
The Root Locus Method
since K1 = 0.02 at medium weight cruise condition. (c) At lightweight cruise condition K1 = 0.2 . Using K2 = 81000, we nd the roots are s1,2 = 54 j 119 s3,4 = 2 j 0.6 . The roots s3,4 become negligible and the roots at s1,2 become highly oscillatory. Hence, in this case = 0.41 . P7.11 (a) The closed-loop characteristic equation is 1+ where K2 = 10 . Then, the root locus is shown in Figure P7.11a. 20Ka (s2 + s + 0.02) =0, s(s + 1)2 (s2 + 2s + 0.8)
3
2
1 Ka=0.035 -->
Imag Axis
0
x
x o
x
o x
-1
-2
-3 -3
-2
-1
0 Real Axis
1
2
3
FIGURE P7.11 20s2 (a) Root locus for 1 + Ka s(s+1)2+20s+0.4 .8) = 0, where K2 = 10. (s2 +2s+0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
293
(b) When Ka < 0.035 , all the roots have a damping greater than or equal to 0.60. (c) Select Ka = 0.035 . Then, the characteristic equation with K2 as the parameter is 1 + K2 0.07(s2 + s) =0. s5 + 4s4 + 5.8s3 + 3.6s2 + 0.8s + 0.014
The root locus is shown in Figure P7.11b.
3 Ka=0.035 2
1
Imag Axis
0
x
x o x
x
x o
-1
-2
-3 -3
-2
-1
0 Real Axis
1
2
3
FIGURE P7.11 0.07s(s+1) CONTINUED: (b) Root locus for 1+ K2 s(s+1)2 (s2 +2s+0.8)+0.014 = 0, where Ka = 0.035.
P7.12
(a) The closed-loop transfer function is T (s) = 1.8s2 (s Ka Km (s + 25)(s + 15) . + 2) + Ka Km (s + 25)(s + 15) + 1.6Km s(s + 2)
So, with E (s) = R(s) Y (s), we have E (s) = (1 T (s))R(s) and ess = lim sE (s) = 1 T (0) = 0 .
s0
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294
CHAPTER 7
The Root Locus Method
Therefore, when the system is stable, it has zero steady-state error. (b) The characteristic equation is s3 + (3.6 + Ka )s2 + (3.2 + 40Ka )s + 375Ka . The Routh array is s3 s2 s1 so 1 3.6 + Ka b 375K 3.2 + 40Ka 375Ka
Solving for b > 0 leads to 0 < Ka < 0.05 or Ka > 5.64 for stability. (c) The characteristic equation can be written as 1+ Ka (s + 25)(s + 15) =0. s(s + 2)(s + 1.6)
The root locus is shown in Figure P7.12. (d) When K > 40 ,
40 30 20 10
Imag Axis
0 -10 -20 -30 -40 -70
o
o
xx x
-60
-50
-40
-30 Real Axis
-20
-10
0
10
FIGURE P7.12 (s+25)(s+15) Root locus for 1 + Ka s(s+2)(s+1.6) = 0, where Km = 1.8.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
295
the roots are s1 = 123 and s2,3 = 15.6 j 31.2 .
From the step response we nd P.O. = 5% Tp = 0.67 sec Ts = 0.25 sec . P7.13 (a) The characteristic equation is 1+ s(s + 3)(s2 K =0. + 4s + 7.84)
(b) When K = 13.5, the roots are
The root locus is shown in Figure P7.13. The breakaway point is s = 1.09 at K = 9.72. s1,2 = 0.84 j 0.84 s3,4 = 2.66 j 1.55 .
6
4
2
+
x +
Imag Axis
0
x + +
x
-2
x
-4
-6 -6
-4
-2
0 Real Axis
2
4
6
FIGURE P7.13 Root locus for 1 +
K s(s+3)(s2 +4s+7.84)
= 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
296
CHAPTER 7
The Root Locus Method
(c) The roots s = 0.84 j 0.84 are dominant roots. (d) For the dominant roots, we determine that = 0.7 and n = 1.19. Therefore, the settling time is Ts = P7.14 The characteristic equation is 1+ K (s + 2)(s + 3) =0. s2 (s + 1)(s + 10)(s + 50) 4 = 4.8 n sec .
The root locus is shown in Figure P7.14. When K = 15609, the roots are s1 = 55.7 s2 = 3.39 s3 = 1.92 s4,5 = j 16.07 .
When K = 370, the roots are s1 = 50.17
100 80 60 40 20 Imag Axis 0 -20 -40 -60 -80 -100 -100
s2 = 9.42
s3 = 1.41
s4,5 = j 1.82 .
-80
-60
-40
-20
0 Real Axis
20
40
60
80
100
FIGURE P7.14 (s+2)(s+3) Root locus for 1 + K s2 (s+1)(s+10)(s+50) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
297
The crossover points are s = j 16.07 Therefore, the system is stable for 370 < K < 15609 . P7.15 The characteristic equation is 1+ K (s2 + 30s + 625) . s(s + 20)(s2 + 20s + 200)(s2 + 60s + 3400) and s = j 1.82 .
The root locus is shown in Figure P7.15. When K = 30000, the roots are s1 = 18.5, s2 = 1.69, s3,4 = 9.8 j 8.9, and s5,6 = 30.1 j 49.9. The real root near the origin dominates, and the step response is overdamped.
100 80 60
x
40 20
o x
Imag Axis
0 -20 -40
x
x x o
x
-60 -80 -100 -100 -80 -60 -40 -20 0 Real Axis 20 40 60 80 100
FIGURE P7.15 s2 Root locus for 1 + K s(s+20)(s2 +20+30s+625 +60s+3400) = 0. s+200)(s2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
298 P7.16
CHAPTER 7
The Root Locus Method
(a) Let = 0. Then, rst reduce the motor and rolls to an equivalent G(s) as follows: G(s) =
0.25 s(s+1) 25 + s(0.+1) s
1
=
0.25 0.25 = . s(s + 1) + 0.25 (s + 0.5)2
The loop transfer function is then L(s) = 2(s + 0.5)Ka (0.25) 0.5Ka = . 2 (s + 0.5)2 s(s + 1) s(s + 1)2 (s + 0.5)
The characteristic equation is 1 + Ka 0.5 =0. s(s + 1)2 (s + 0.5)
The root locus is shown in Figure P7.16.
2 1.5 1 0.5
Imag Axis
+ +
0
+
x
x +
x
-0.5 -1 -1.5 -2 -2
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE P7.16 Root locus for 1 +
0.5Ka s(s+1)2 (s+0.5)
= 0.
(b) When K = 0.123, the roots of the characteristic equation are s1,2 = 1.1 j 0.27 s3,4 = 0.15 j 0.15 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
299
(c) When becomes nonnegligible, the root locus will have an additional pole, and the root locus will change accordingly. P7.17 The characteristic equation is
2 (M1 s2 + bs + k1 + k12 )(M2 s2 + k12 ) k12 = 0 . 2 If we let M1 = k1 = b = 1, and assume k12 < 1 so that k12 is negligible and k1 + k12 k1 , then the characteristic equation is
The roots at s = 0.15 j 0.15 have a damping ratio of = 0.707.
(s2 + s + 1)(M2 s2 + k12 ) = 0 where k= k12 . M2
or
1+
k =0, s2
The root locus is shown in Figure P7.17. All the roots lie on the j axis. If we select k12 = o , M2 then we cancel the vibration.
3
2
1
root locus -->
Imag Axis
0
x
-1
-2
-3 -3
-2.5
-2
-1.5
-1 Real Axis
-0.5
0
0.5
1
FIGURE P7.17 Root locus for 1 +
k s2
= 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
300 P7.18
CHAPTER 7
The Root Locus Method
The characteristic equation is s3 + (1 + 2 )s2 + (2 + 4)s + 4 = 0 . When = 0 we have 1+ s2 4s =0. + 2s + 4
The root locus for = 0 is shown in Figure P7.18.
3 <-- zeta=0.6 2
x *
beta=0
1
Imag Axis
0
o
-1
* x
-2
-3 -3
-2
-1
0 Real Axis
1
2
3
FIGURE P7.18 4s Root locus for 1 + s2 +2s+4 = 0, where = 0.
For = 0.3, the poles are s = 1.6 j 1.2 . Then, we have 1+ When = 0.121 s1,2 = 1.51 j 1.51 s3 = 7.24 . (s + 2)s2 =0. s2 + (2 + 4)s + 4
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
301
Thus, = 0.707 and n = 1.5 .
So, the performance specs are met. Also, Gc (s) = P7.19 0.3s + 1 2.48(s + 3.33) = . 0.121s + 1 (s + 8.26)
The characteristic equation is 1+ Ka (s2 + 3.6s + 81) =0. s(s + 1)(s + 5)
The root locus is shown in Figure P7.19.
10
o
8 6 4 2
Imag Axis
0 -2 -4 -6 -8
*x
x
* *
x
o
-10 -10
-8
-6
-4
-2 Real Axis
0
2
4
FIGURE P7.19 s2 +3.6 Root locus for 1 + Ka s(s+1)(s+81 = 0. s+5)
When Ka = 0.0265, the roots are s1,2 = 0.45 j 0.45 s3 = 5.11 . Thus, the complex roots have a damping ratio of = 0.707.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
302 P7.20
CHAPTER 7
The Root Locus Method
The characteristic equation is s3 + (2 + where K= 4 = 0.3 = 0.121 . 12 2 4 )s + ( + K )s + = 0 ,
The root sensitivity to changes in K is found to be
r1 SK =
r1 = 1.18 149.75o . K/K
The root sensitivity to changes in the pole at s = 2 is found to be r1 r1 S = 1.65 137o , = /2 P7.21 where the pole is s + 2 + .
(a) Let the pole be (s + 4 + ) and neglect 2 terms. Then, the characteristic equation is 1+ 2s2 + (8 + 2)s + 8 =0 s3 + (8 + )s2 + (16 + 8)s + 16 + K
where = 0.000788 and K = 19.1.
3
2
x
1
Imag Axis
0
x
o
o
-1
x
-2
-3 -10
-8
-6
-4 Real Axis
-2
0
2
FIGURE P7.21 2s2 +(8+2 )s+8 Root locus for 1 + s3 +(8+)s2 +(16+8)s+16+K = 0, ( = 0.000788 and K = 19.1).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
303
The root sensitivity is determined to be r1 r1 S = 3.314 132o . = /4 (b) Let R = Ro + R, where R = 0.00021. Then, r1 r SR1 = 1.31 107o . = R/R P7.22 The characteristic equation is s3 + 2s2 + s + K , where K = 0.24 for = 0.707. The root sensitivity to changes in the pole at s = 1 is found to be r1 r1 S = 0.95 126o , = where the pole is s + 1 + . P7.23 The characteristic equation is s3 + 5s2 + (6 + K )s + K , where K = 6.3 for = 0.707. The root sensitivity to changes in the pole at s = 2 is found to be r1 r1 S = 1.25 169.4o , = /2
where the pole is s + 2 + . The root sensitivity to changes in the zero at s = 1 is found to be r1 r1 S = 0.55 34.4o , = where the zero is s + 1 + . P7.24 The root locus for each of the four cases shown is shown in Figure P7.24. The four open-loop transfer functions are (a) KF (s) = (b) KF (s) = (c) s2 + 7s + 8.25 s3 + 6s2 + 5s s5 30s4
s+8 + + 296s3 + 1170s2 + 1575s 1 s2 + 6s + 6.75 KF (s) = 6 (d) KF (s) = 3 s + 2s5 + s4 s + 5s2 + 4s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
304
CHAPTER 7
The Root Locus Method
5
(a)
10 5
(b)
Imag Axis
0
Imag Axis
ox
ox x
0 -5
x
ox x x
x
-5 -10
-5 Real Axis
0
-10 -20
-10 Real Axis
0
2 1
(c)
5
(d)
Imag Axis
0 -1 -2 -2
Imag Axis
x
x
0
ox
ox x
0 Real Axis
2
-5 -10
-5 Real Axis
0
FIGURE P7.24 Root locus for the four cases.
P7.25
The characteristic equation is 1 + KGc (s)G(s) = 0 , therefore, KGc (s)G(s) = 1 . Squaring both sides yields K 2 G2 (s)G2 (s) = 1 and c 1 K 2 G2 (s)G2 (s) = 0 . c The root locus with 0 < K 2 < is shown in Figure P7.25. The value of K 2 for which the locus crosses the imaginary axis is K 2 = 2/3 , therefore K = 2/3 = 0.8165 corresponds to the j -axis crossing (at s = 0). You can check that 1 + KGc (s)G(s) = 0 for K = 0.8165 and s = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
305
3
2
1 Imag Axis
0
-1
-2
-3 -3 -2 -1 0 Real Axis 1 2 3
FIGURE P7.25 Root locus for the equation 1 K 2 G2 (s)G2 (s) = 0. c
P7.26
(a) The characteristic equation is 1+ K (s + 2)2 =0. s(s2 + 1)(s + 8)
The root locus is shown in Figure P7.26. (b) Using Rouths criteria, we determine that K > 14 for stability. (c) From the Routh array, we determine that for K = 14, we have two purely imaginary poles at s = j 8 . (d) When K > 50, the real part of the complex roots is approximately equal to the real part of the two real roots and therefore the complex roots are not dominant roots.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
306
CHAPTER 7
The Root Locus Method
15
10
5 Imag Axis
0
-5
-10
-15 -15 -10 -5 0 Real Axis 5 10 15
FIGURE P7.26 (s+2)2 Root locus for 1 + K s(s2 +1)(s+8) = 0.
P7.27
The characteristic equation is 1+ K (s2 + 0.105625) =0. s(s2 + 1)
The root locus is shown in Figure P7.27a. The locus enters the axis at s = 0.67 and leaves the axis at s = 0.48 . Dene p(s) = K = (s3 + s) . s2 + 0.105625
Then, a plot of p(s) vs s is shown in Figure P7.27b, where it can be seen that p(s) has two inection points at s = 0.67 and s = 0.48 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
307
2 1.5 1 0.5
x
Imag Axis
o
0 -0.5 -1 -1.5 -2 -1
x o
x
-0.8
-0.6
-0.4
-0.2 Real Axis
0
0.2
0.4
1.76
1.758
1.756
1.754
p(s)
1.752 1.75 1.748 1.746 -0.8
-0.75
-0.7
-0.65
-0.6 s
-0.55
-0.5
-0.45
-0.4
FIGURE P7.27 2 105625 s3 +s (a) Root locus for 1 + K s +0.2 +1) = 0. (b) Plot of p(s) = s2 +0.105625 versus s. s(s
P7.28
The characteristic equation is 1 + L(s) = 1 + K (s2 + 12s + 20) =0. s3 + 10s2 + 25s
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308
CHAPTER 7
The Root Locus Method
The root locus is shown in Figure P7.28. The breakaway point is s = 5.0
6
4
2
Imag Axis
0
-2
-4
-6 -20
-15
-10 Real Axis
-5
0
FIGURE P7.28 (s2 +12s+20) Root locus for 1 + K s3 +10s2 +25s = 0.
and the entry point is s = 15.6. When K = 2, the roots are s1 = 1.07 s2,3 = 5.46 j 2.75 . When K = 2, the roots are s1 = 1.07 s2,3 = 4.36 j 1.68 . The predicted step response when K = 2 is Ts = 9 sec and P O 0%. P7.29 The characteristic equation is 1+K s2 + 8s + 25 =0. s2 (s + 4)
The root locus is shown in Figure P7.29. When = 0.707, the necessary
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
309
gain is K = 18.285. The corresponding roots are s1 = 14.285 and s2,3 = 4 j 4.
System: sys Gain: 18.2 Root Locus Pole: 4 + 4i Damping: 0.707 Overshoot (%): 4.34 Frequency (rad/sec): 5.66
5 4 3 2 Imaginary Axis 1 0 1 2 3 4 5 8
7
6
5
4 3 Real Axis
2
1
0
1
FIGURE P7.29 2 Root locus for 1 + K ss+8s+25 = 0. 2 (s+4)
P7.30
The transfer function is Z (s) = So, R r1 = + 2 R2 1 4
1 2
LCRs2 + Ls Rs2 + s =2 . 2 + CRs + 1 LCs s + Rs + 1
.
1 Thus, the nominal r1o = 2 . Simultaneously,
R r2 = 2
R2 1 4
1 2
.
Thus, the nominal r2o = 2. We see that there is a dierence by a factor
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310
CHAPTER 7
The Root Locus Method
of 4. Also,
r SRi
r1 = R
Ro
2 Ro Ro Ro = + 2 4
2 Ro 1 4
1 2
=
5 , 6
where Ro = 2.5. And
r SR2
r2 = R
2 Ro Ro Ro = 2 4 Ro
2 Ro 1 4
1 2
=
10 . 3
r r So, the magnitude of |SR2 | = 4|SR1 |.
P7.31
The characteristic equation is 1+K s+4 =0. s(s + 0.16)(s2 + 14.6s + 148.999)
The root locus is shown in Figure P7.31. When K = 1350, the roots are s1,2 = j 9.6 s3,4 = 7.4 j 1.9 .
20 15 10 5
x +
(+) K=326 -->
*
<-- K=1350 (*)
Imag Axis
+ *
0
*
o +
x x
-5
+
-10 -15 -20 -20
x
*
-15
-10
-5
0 Real Axis
5
10
15
20
FIGURE P7.31 s+4 Root locus for 1 + K s(s+0.16)(s2 +14.6s+148.999) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
311
When K = 326, the roots are s1,2 = 6.5 j 8.7 P7.32 The characteristic equation is 1+ K (s + 1)(s + 5) =0. s(s + 1.5)(s + 2) s3,4 = 0.9 j 3.2 .
4 3 2 1
*
*
Imag Axis
0 -1 -2 -3 -4 -10
o
x
x
o * *
x
*
*
-8
-6
-4 Real Axis
-2
0
2
4
FIGURE P7.32 (s+1)(s+5) Root locus for 1 + K s(s+1.5)(s+2) = 0.
K 1.57 3.48 2.35
TABLE P7.32
0.707 0.707 0.69
Ts (sec) 0.98 1.1 1.3
P.O. (%) 1.4 5.8 4.0
Step Response Results for K = 1.57, K = 3.48, and K = 2.35.
(a) The breakaway point is s = 1.73; the entry point is s = 8.62.
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312
CHAPTER 7
The Root Locus Method
(b) The damping ratio = 0.707 when K = 1.57 and again when K = 3.46. (c) The minimum damping ratio = 0.69 is achieved when K = 2.35. (d) The results are summarized in Table P7.32.The best choice of gain is K = 1.57. P7.33 (a) The root locus for the V-22 is shown in Figure P7.33a. The system is
2 1.5 1 0.5
Imag Axis
0 -0.5 -1 -1.5 -2 -3
x
o
o
xxx
-2.5
-2
-1.5
-1 Real Axis
-0.5
0
0.5
1
FIGURE P7.33 s2 +1.5s+0.5 (a) Root locus for 1 + K s(20s+1)(10s+1)(0.5s+1) = 0.
stable when 0 < K < 0.48 and K > 136.5. (b) The unit step input response (for K = 280) is shown in Figure P7.33b. The step response has a P.O. = 90% and Ts 50 sec. (c) The plot of y (t) for a unit step disturbance is shown in Figure P7.33b. The response to the disturbance is oscillatory, but the maximum value of oscillation is about 0.003; so it is negligible. (d) The eect of adding a prelter can be seen in Figure P7.33b. With the prelter we nd P O = 7% and Ts 40 sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
(i) Unit step input response 2 1.5 y(t) 1 0.5 0 y(t) w/o pre lter .... (dotted line) y(t) with pre lter ____ (solid line)
313
0 x 10
-3
10
20
40 50 Time (sec) (ii) Unit step disturbance response
30
60
70
80
4 3 y(t) 2 1 0 -1
0
10
20
30
40 Time (sec)
50
60
70
80
FIGURE P7.33 CONTINUED: (b) (i) Unit step input response with and without prelter; (ii) Unit step disturbance response.
P7.34
The characteristic equation is 1+ K (s + 2) =0. (s + 1)(s + 2.5)(s + 4)(s + 10)
The root locus is shown in Figure P7.34a. The roots, predicted and actual percent overshoot for K = 400, 500, and 600 are summarized in Table P7.34. The actual unit step input responses are shown in Figure P7.34b.
K 400 500 600
TABLE P7.34
roots -13.5,-1.00 5.71j,-1.98 -14.0,-0.75 6.24j,-1.98 -14.4,-0.53 6.71j,-1.98
0.173 0.120 0.079
predicted P.O. (%) 57.6 68.4 77.9
actual P.O. (%) 51.6 61.2 69.6
Summary for K = 400, 500, 600.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
314
CHAPTER 7
The Root Locus Method
Root Locus 20
15
10
Imaginary Axis
5
0
5
10
15
20 30
25
20
15
10 Real Axis
5
0
5
10
1.6
1.4
K=400 .... (dotted line) K=500 (dashed line)
1.2
K=600 ___ (solid line)
1
y(t)
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE P7.34 (a) Root locus for 1 + K (s+1)(s+2.s+2 +4)(s+10) = 0. (b) Unit step input responses for 5)(s K = 400, 500, 600.
P7.35
(a) The characteristic equation is 1+ K (s + 1)2 =0. s(s2 + 1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
315
The root locus is shown in Figure P7.35.
3
2
K=4.52
*
1
x
Imag Axis
0
o
*
x
-1
x
-2
*
-3 -5
-4
-3
-2 Real Axis
-1
0
1
FIGURE P7.35 (s+1)2 Root locus for 1 + K s(s2 +1) = 0.
(b) When K = 4.52, the roots are s1 = 0.58 s2,3 = 1.96 j 1.96 . The complex roots have = 0.707. (c) The entry point is s = 3.38 when K = 7.41. (d) The predicted P.O. = 4.5% ( = 0.707) and the actual P.O. = 17%. P7.36 The characteristic equation is 1+ K (s + 1)(s + 2)(s + 3) =0. s3 (s 1)
(a) The root locus is shown in Figure P7.36. (b) When K = 2.96, the roots are s1,2 = j 4.08 s3,4 = 0.98 j 0.33 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
316
CHAPTER 7
The Root Locus Method
8 6 4 2
Imag Axis
0 -2 -4 -6 -8 -10
o
o
o
x
x
-8
-6
-4 Real Axis
-2
0
2
FIGURE P7.36 (s+1)(s+2)(s+3) Root locus for 1 + K = 0. s3 (s1)
(c) When K = 20, the roots are s1 = 1.46 s2 = 1.07 s3,4 = 8.23 j 2.99 . When K = 100, the roots are s1 s2 s3 s4 = 92.65 = 3.51 = 1.82 = 1.01 .
(d) When K = 20, the damping ratio is = 0.94. Therefore, the predicted P.O. = 0.02%. The actual overshoot is P.O. = 23%. P7.37 Since we know that ess = 0 for a step input, we know that a = 0 or b = 0. Select a = 0. Also, n = 2/T = 20 rad/sec. The desired characteristic polynomial is (s + r1 )(s + j 20)(s j 20) = s3 + r1 s2 + 400s + 400r1 = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
317
The actual characteristic polynomial is 1+ 2K =0, s(s + b)(s + 40) or s3 + (40 + b)s2 + 40bs + 2K = 0 .
Comparing the coecients in the desired and actual characteristic polynomials, we determine that b = 10, r1 = 50, and K = 10000. P7.38 (a) The characteristic equation is 1+ K (s + 1) =0. s(s 3)
The system is stable for K > 3. When K = 3, the roots are s = j 3. (b) The root locus is shown in Figure P7.38a.
6
4
2
Imag Axis
0
o
x
x
-2
-4
-6 -6
-4
-2
0 Real Axis
2
4
6
FIGURE P7.38 (a) Root locus for 1 + K s(s+1 = 0. s3)
(c) When K = 10 , the roots are s 1 = 2 s 2 = 5 . Since both roots are real and stable, we expect that there will be
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318
CHAPTER 7
The Root Locus Method
zero overshoot. The actual response has a 40% overshoot, as seen in Figure P7.38b.
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.5
1
1.5 Time (secs)
2
2.5
3
FIGURE P7.38 CONTINUED: (b) Unit step response.
P7.39
The loop transfer function is Gc (s)G(s) = 22K . (s + 1)(s2 + 8s + 22)
When K = 0.529, the closed-loop poles are s1,2 = 3.34 1.83j and s3 = 2.32 and have the maximum damping = 0.877. The root locus is shown in Figure P7.39a. The step respose is shown in Figure P7.39b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
319
Root Locus 10 8 6 4 Imaginary Axis 2 0 2 4 6 8 10 14
12
10
8
6 Real Axis
4
2
0
2
Step Response 0.35
0.3
0.25
Amplitude
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5 2 Time (sec)
2.5
3
3.5
FIGURE P7.39 (a) Root locus for
22K (s+1)(s2 +8s+22)
= 0. (b) Unit step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
320
CHAPTER 7
The Root Locus Method
Advanced Problems
AP7.1 The characteristic equation is 1+K s+6 =0. s(s + 4)(s2 + 4s + 8)
The root locus is shown in Figure AP7.1. The gain at maximum is
10
5
x +
Imag Axis
0
o
x+ x
+ +
x
-5
-10 -10 -5 0 Real Axis 5 10
FIGURE AP7.1 s(s+4) Root locus for 1 + K s2 +2s+2 = 0.
K = 3.7 . The roots at K = 3.7 are s1 = 3.6424 s2,3 = 1.3395 +1.3553j s4 = 1.6786 .
Using Figure 5.13 in Dorf & Bishop, the predicted percent overshoot and settling time are P.O. = 5% and Ts = 3 sec ,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
321
since = 0.7 and a 6 = = 4.5 . n 1.9(0.7) The actual percent overshoot and settling time are P.O. = 1% AP7.2 The characteristic equation is 1+K (s + 1)(s + 3) =0. s(s 1)(s + 4)(s + 8) and Ts = 2.8 sec .
The root locus is shown in Figure AP7.2a. The selected gain is K = 48.
15
10
5
+
Imag Axis
0
+ x x o + ox x
-5
+
-10
-15 -15 -10 -5 0 Real Axis 5 10 15
FIGURE AP7.2 (s+1)(s+3) (a) Root locus for 1 + K s(s1)(s+4)(s+8) = 0.
Using Figure 5.13 in Dorf & Bishop, the predicted percent overshoot is P.O. = 30% . The actual percent overshoot (see Figure AP7.2b) is P.O. = 45% .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
322
CHAPTER 7
The Root Locus Method
1.6 1.4 1.2 1
Amplitude
0.8 0.6 0.4 0.2 0 0
0.5
1
1.5 Time (secs)
2
2.5
3
FIGURE AP7.2 CONTINUED: (b) Step response for K = 48.
AP7.3
The characteristic equation (with p as the parameter) is 1+p s3 s(s + 1) =0. + s2 + 10
The root locus is shown in Figure AP7.3.
5 4 3 2 1
x
Imag Axis
+
0 -1 -2 -3 -4 -5 -5 -4 -3
x
o +
o
x
-2
-1
0 Real Axis
1
2
3
4
5
FIGURE AP7.3 s(s+1) Root locus for 1 + p s3 +s2 +10 = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
323
When p = 21 the dominant roots have a damping ratio of = 0.707. AP7.4 The characteristic equation (with as the parameter) is 1+ s(s + 1) =0. s3 + s2 + 1
The root locus is shown in Figure AP7.4a.
3
2
1
x
Imag Axis
0
x
o
o
x
-1
-2
-3 -3
-2
-1
0 Real Axis
1
2
3
FIGURE AP7.4 s(s+1) (a) Root locus for 1 + p s3 +s2 +10 = 0.
The steady-state error is ess = lim sE (s) = lim
s0
1 =1 . s0 1 + G(s)
To meet the steady-state error specication, we require 0.9 < < 1.1 . The step responses for = 0.9, 1 and 1.1 are shown in Figure AP7.4b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
324
CHAPTER 7
The Root Locus Method
alpha=0.9 (solid); alpha=1.0 (dashed); alpha=1.1 (dotted) 1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0 0
5
10
15
20
25 Time (sec)
30
35
40
45
50
FIGURE AP7.4 CONTINUED: (b) Step responses for = 0.9, 1 and 1.1.
AP7.5
The root locus is shown in Figure AP7.5. When K = 20.45, = 0.707. The
5 4 3 2 1
Imag Axis
+
0 -1 -2 -3 -4 -5 -15 -10
+x
x +
x
-5 Real Axis
0
5
FIGURE AP7.5 1 Root locus for 1 + K s3 +10s2 +7s18 = 0.
r1 root sensitivity is SK r1 /(K/20.45) = 3.15 87.76o . When K = 88, =
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
325
the complex roots lie on the j -axisa 330% increase in the gain. AP7.6 A gain of K = 10 provides an acceptable response (see the root locus in Figure AP7.6).
Root Locus 2.5 2 1.5 1 Imaginary Axi s 0.5 0 -0.5 -1 -1.5 -2 -2.5 -2 -1.5 -1 Real Axi s -0.5 0
FIGURE AP7.6 s2 s+5 Root locus for 1 + K s3 +2+2+2s+1 = 0. s2
AP7.7
The root locus for the positive feedback system is shown in Figure AP7.7.
15
10
5
Imag Axis
0
x
x
-5
-10
-15 -15 -10 -5 0 Real Axis 5 10 15
FIGURE AP7.7 Root locus for 1 + K s2 +121 +32 = 0. s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
326 AP7.8
CHAPTER 7
The Root Locus Method
The closed-loop characteristic equation is 1+k s3 + 19s2 120s =0. + 34s + 120
The root locus is shown in Figure AP7.8a. When k = 0.448, all the roots of the characteristic equation are realthe step response is shown in Figure AP7.8b.
30
20
10
Imag Axis
x
0
x x
o
-10
-20
-30 -30
-20
-10
0 Real Axis
10
20
30
1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 Time (secs) 2 2.5 3
FIGURE AP7.8 (a) Root locus for 1 + k s3 +19s120s s+120 = 0. (b) Step response with k = 0.448. 2 +34
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
327
AP7.9
The root locus for each controller is shown in Figure AP7.9.
(a) (b) 5 Imaginary Axis 10 5 Real Axis (c) 0 5
5 Imaginary Axis
0
0
5 15
5 15
10
5 Real Axis (d)
0
5
15 Imaginary Axis Imaginary Axis 10 5 Real Axis 0 5 10 5 0 5 10 15 15
5
0
5 15
10
5 Real Axis
0
5
FIGURE AP7.9 Root locus for the various controllers.
AP7.10
The characteristic equation (with K as the parameter) is 1+K s2 + 7s + 20 =0. s(s2 + 7s + 10)
The root locus is shown in Figure AP7.10. The steady-state value of the step response for any K is 0.5. With K = 15 the closed-loop transfer function is T (s) = s3 10s + 150 . + 22s2 + 115s + 300
The step response has the following characteristics: P.O. = 4.8% and Ts = 2 seconds .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
328
CHAPTER 7
10 8 6 4 2 Imag Axis 0 -2 -4 -6 -8
The Root Locus Method
-10 -10
-8
-6
-4
-2
0 Real Axis
2
4
6
8
10
FIGURE AP7.10 2 Root locus for 1 + K s(s 2+7s+20 = 0. s +7s+10)
AP7.11
The root locus is shown in Figure AP7.11a. A suitable gain is K = 1525. The step response is shown in Figure AP7.11b.
20 15 10
+
5
Imag Axis
0 -5
+
x
x
x +o+ x x
+
-10 -15 -20 -30
-25
-20
-15 Real Axis
-10
-5
0
5
FIGURE AP7.11 (s+1)2 (a) Root locus for 1 + K s(s+8)(s+20)(s2 +3.2s+3.56) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
329
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.5
1
1.5
2
2.5 Time (secs)
3
3.5
4
4.5
5
FIGURE AP7.11 CONTINUED: (b) Step response with K = 1525.
AP7.12
The root locus is shown in Figure AP7.12a.
8
6
4
2 Imag Axis
0
-2
-4
-6
-8 -7
-6
-5
-4
-3 Real Axis
-2
-1
0
1
2
FIGURE AP7.12 (s+0.2) (a) Root locus for 1 + Kp s(s2 +7s+10) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
330
CHAPTER 7
The Root Locus Method
Step Response
From: U(1) 1.4
1.2
1
Amplitude
0.8 To: Y(1) 0.6 0.4 0.2 0 0
5
10
15
Time (sec.)
FIGURE AP7.12 CONTINUED: (b) Step response with Kp = 5.54.
The PI controller can be written as Gc (s) = Kp s + KI s
and setting KI = 0.2Kp , the characteristic equation can be weitten as 1 + Kp s(s2 (s + 0.2) =0 + 7s + 10)
A suitable gain is Kp = 5.55. The step response is shown in Figure AP7.12b. AP7.13 The characteristic equation is 1 + K1 K2 1 = 0. (s + 5)(s 1)
The root locus is shown in Figure AP7.12a. The fastest expected settling time is Ts = 4/n = 2 seconds since maximum |n | = 2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
331
Root Locus 4
3
2
Imaginary Axis
1
0
1
2
3
4 6
5
4
3
2 Real Axis
1
0
1
2
FIGURE AP7.13 1 Root locus for 1 + K1 K2 (s+5)(s1) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
332
CHAPTER 7
The Root Locus Method
Design Problems
CDP7.1
The closed-loop transfer function from the input to the output is (s) 26.035Ka =2 , R(s) s + (33.1415 + 26.035Ka K1 )s + 26.035Ka where we consider for the rst time the tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). The characteristic equation is 1 + K1 26.035Ka s =0. s2 + 33.1415s + 26.035Ka
The root locus is shown below. In accordance with the discussion in Chap30
20
10 Imag Axis
0
-10
-20
-30 -30
-20
-10
0 Real Axis
10
20
30
ter 5, we continue to use Ka = 22. This allows us to meet the overshoot specication (P.O. < 5%) without the tachometer feedback and to provides good steady-state tracking errors to a step input. To meet the design specications of both P.O. and Ts we want the closed-loop poles to the left of = 4/0.3 = 13.33 and > 0.69. A reasonable selection is K1 = 0.012. This places the closed-loop poles at s = 20 j 13. DP7.1 (a) The characteristic equation is 1+ (s2 18K (s + 0.015)(s + 0.45) =0. + 1.2s + 12)(s2 + 0.01s + 0.0025)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
333
Since we want a negative feedback system, we have Gc (s) = K . When n > 2 and = 0.15, the gain K = 0.12. The root locus is shown in Figure DP7.1a.
6
4
x
2
Imag Axis
0
o
x o x
-2
x
-4
-6 -4
-3.5
-3
-2.5
-2
-1.5 Real Axis
-1
-0.5
0
0.5
1
FIGURE DP7.1 18(s+0.015)(s+0.45) (a) Root locus for 1 + K (s2 +1.2s+12)(s2 +0.01s+0.0025) = 0.
(b) The unit step response is shown in Figure DP7.1b. The percent overshoot is P.O. = 100% .
(c) The characteristic equation with the anticipatory controller is 1+ 18K (s + 2)(s + 0.015)(s + 0.45) =0. (s2 + 1.2s + 12)(s2 + 0.01s + 0.002s)
The root locus is shown in Figure DP7.1c. If we select K = 9.2/18 , then the complex roots have a damping = 0.90. The roots are at s1 = 0.253 s2 = 0.019 s3,4 = 5.07 j 2.50 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
334
CHAPTER 7
The Root Locus Method
0.7
0.6
0.5
Amplitude
0.4
0.3
0.2
0.1
0 0
20
40
60
80
100 Time (secs)
120
140
160
180
200
FIGURE DP7.1 CONTINUED: (b) Unit step response for gain controller.
6
4
x
2
Imag Axis
0
o
x oo x
-2
x
-4
-6 -6 -4 -2 0 Real Axis 2 4 6
FIGURE DP7.1 18(s+2)(s+0.015)(s+0.45) CONTINUED: (c) Root locus for 1 + K (s2 +1.2s+12)(s2 +0.01s+0.0025) = 0.
(d) The unit step response for the system with the anticipatory controller is shown in Figure DP7.1d.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
335
1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 20 40 60 80 100 Time (secs) 120 140 160 180 200
FIGURE DP7.1 CONTINUED: (d) Unit step response for anticipatory controller.
DP7.2
The characteristic equation is 1+ 10K (s + 1) =0. s(s2 + 4.5s + 9)
(a) The root locus is shown in Figure DP7.2a. When K = 0.435, we have = 0.6 and the roots are s1 = 0.368 s2,3 = 2.1 j 2.75 . (b) The response to a step input is shown in Figure DP7.2b. The performance results are P.O. = 0% Tss = 10 sec ess = 0 . (c) We have = 0.41 when K = 1.51. The step response is shown in Figure DP7.2b. The performance results to the step input are P.O. = 0% Ts = 4 sec ess = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
336
CHAPTER 7
The Root Locus Method
5 4 3 2 1
x
Imag Axis
0 -1 -2 -3 -4 -5 -5 -4 -3 -2
x
o
x
-1
0 Real Axis
1
2
3
4
5
FIGURE DP7.2 10(s+1) (a) Root locus for 1 + K s(s2 +4.5s+9) = 0.
1 0.9 0.8 0.7 K=0.435 ____ (solid line) K=1.510 ---- (dashed line)
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 2 4 6 8 Time (sec) 10 12 14 16
FIGURE DP7.2 CONTINUED: (b) Unit step responses for K = 0.425, 1.51.
DP7.3
The characteristic equation is 1+ K (s2 + 6.5s + 12) =0. s(s + 1)(s + 2)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
337
(a) The root locus is shown in Figure DP7.3.
6
4
2
o
Imag Axis
0
o
x
x
x
-2
-4
-6 -6
-5
-4
-3 Real Axis
-2
-1
0
1
FIGURE DP7.3 s2 +6.5 Root locus for 1 + K s(s+1)(s+12 = 0. s+2)
(b) The percent overshoot is P.O. 1% when = 0.82 at K = 0.062. (c) Select K > 300. DP7.4 The characteristic equation is 1+K 10(0.01s + 1) =0. s(s2 + 10s + 10K1 )
When K = 41, the roots are s1 = 37.12 and s2,3 = 3.44 j 1.19 .
If we choose K1 = 2.5, then the root locus will start at s = 0, 5 and -5. This is shown in Figure DP7.4. The root locus then has a nice shape so that we can select K to place the complex poles where desired and the one real root will be farther in the left half-plane; thus the notion of dominant poles will be valid. So, if we desire a P.O. < 5%, we want > 0.69. This occurs when K 3. Thus, our design is K1 = 2.5 and K = 3 .
The unit step response is shown in Figure DP7.4. The settling time is less than 3.5 sec and the P O < 4%. The response to a unit step disturbance is also shown in Figure DP7.4. The steady-state error magnitude to the disturbance is 0.33.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
338
CHAPTER 7
4
The Root Locus Method
3
2 K=3 --> 1 Imag Axis
0
-1
-2
-3
-4 -20
1.4
-15
-10
-5 Real Axis
0
5
10
Input step response Disturbance step response
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE DP7.4 10(0.01s+1) (a) Root locus for 1 + K s(s2 +10s+25) = 0. (b) System response to step input and disturbance.
DP7.5
The characteristic equation is 1+K s+1 =0. s(s 1)(s2 + 10s + 41)
The root locus is shown in Figure DP7.5a.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
339
10 8 6 4 2
x +
Imag Axis
0 -2
+
+
o
x
x
+
-4 -6 -8 -10 -10 -8 -6
x
-4
-2
0 Real Axis
2
4
6
8
10
2.5
2
Amplitude
1.5
1
0.5
0 0
2
4
6
8
10
12
14
Time (secs)
FIGURE DP7.5 s+1 (a) Root locus for 1 + K s(s1)(s2 +10s+41) = 0. (b) Step response with K = 140.
The system is stable for 58.6 < K < 222.1 . The step response with K = 140 is shown in Figure DP7.5b. The percent overshoot and settling time are P.O. = 131% and Ts = 9.9 seconds; the aircraft response is highly oscillatory.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
340 DP7.6
CHAPTER 7
The Root Locus Method
The characteristic equation is 1+K s+2 =0. s(s + 10)(s 1)
The maximum damping is = 0.46 at K = 55. The root locus is shown in Figure DP7.6a; the step response is shown in Figure DP7.6b. The percent overshoot and settling time are P.O. = 61.3% and Ts = 2 seconds.
20 15 10 5
Imag Axis
+
0 -5 -10 -15 -20 -20
x
+
o +
xx
-15
-10
-5
0 Real Axis
5
10
15
20
1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0 0
0.5
1
1.5 Time (secs)
2
2.5
3
FIGURE DP7.6 s+2 (a) Root locus for 1 + K s(s+10)(s1) = 0. (b) Step response with K = 55.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
341
DP7.7
The closed-loop transfer function is T (s) = The dc gain is T (0) = G(0) 1 . 1 + KG(0) K Vo (s) G(s) = . V (s) 1 + KG(s)
The root locus is shown in Figure DP7.7. The maximum value of K for
x10 7 2 1.5 1 0.5
+
Imag Axis
0+ -0.5 -1 -1.5 -2 -2
x
x
+
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2 x10 7
FIGURE DP7.7 1017 Root locus for 1 + K (s3.142K1s+107 )2 = 0. +3142)(
stability is K = 0.062 . Therefore, the minimum dc gain is about 1/0.062=16. Selecting K = 0.05 yields R2 = 19R1 = 190 K . and R1 = 10 K
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
342 DP7.8
CHAPTER 7
The Root Locus Method
The closed-loop transfer function (with Gp (s) = 1 and K = 1) is T (s) = 2s3 + 6s2 + 14s + 10 . s4 + 6s3 + 13s2 + 26s + 6
So, if we select Gp (s) = 1/T (0) = 0.6, the step response (with K = 1) will have a zero steady-state tracking error. The root locus is shown in Figure DP7.8a. The step responses for K = 1, 1.5 and 2.85 are shown in Figure DP7.8b. For K = 1, we have P.O. = 0%, Tr = 7.8 and Ts = 13.9; for K = 1.5, we have P.O. = 0%, Tr = 5.4 and Ts = 9.6; and for K = 2.85,
8 6 4 2
x
Imag Axis
0 -2 -4 -6 -8 -8
x
o
x
x
-6
-4
-2
0 Real Axis
2
4
6
8
K=1 (solid); K=1.5 (dashed); K=2.85 (dotted) 1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE DP7.8 6(s+1) (a) Root locus for 1 + K s(s+4)(s2 +2s+5) = 0. (b) Step responses with K = 1, 1.5, 2.85.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
343
we have P.O. = 5.2%, Tr = 0.5 and Ts = 7.3. The best gain selection is K = 2.85. DP7.9 A suitable selection of the various parameters is = 0.5 and q = 3/5 .
With q = 3/5, the open-loop zeros are real and equal. Then, it follows that = 2q =3. 1q
The root locus is shown in Figure DP7.9. A reasonable choice of gain is K = 30 . The resulting step response is extremely fast with no overshoot. The closed-loop transfer function is approximately given by T (s) 1923 . s + 1923
6
4
x
2
Imag Axis
0
o
x
-2
x
-4
-6 -6 -4 -2 0 Real Axis 2 4 6
FIGURE DP7.9 4s2 Root locus for 1 + K 0.0625s3+4s.+1s2 +s = 0. +0 25
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
344 DP7.10
CHAPTER 7
The Root Locus Method
The characteristic equation (with K as the parameter) is 1+K 10(s2 + 10) =0. s3 + 20s
The root locus is shown in Figure DP7.10a. To maximize the closed-loop
5
4
3
2
1 Imag Axis
0
-1
-2
-3
-4
-5 -2
-1.5
-1
-0.5 Real Axis Step Response
From: U(1)
0
0.5
1
1.4
1.2
1
Amplitude
0.8 To: Y(1) 0.6 0.4 0.2 0 0
1
2
3
4
5
6
Time (sec.)
FIGURE DP7.10 10(s2 +10) (a) Root locus for 1 + K s3 +20s = 0. (b) Step response with K = 0.513.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
345
system damping we choose K = 0.513. The step response is shown in Figure DP7.10b. DP7.11 The characteristic equation is 1+K s + 1.5 =0. (s + 1)(s + 2)(s + 4)(s + 10)
The root locus is shown in Figure DP7.11a.
10 8 6 4 2
Imag Axis
0 -2 -4 -6 -8 -10 -15
x
x
xox
-10
-5 Real Axis
0
5
K=100 (solid); K=300 (dashed); K=600 (dotted) 1.6 1.4 1.2 1
Amplitude
0.8 0.6 0.4 0.2 0 0
1
2
3
4
5 Time (sec)
6
7
8
9
10
FIGURE DP7.11 s+1 5 (a) Root locus for 1 + K (s+1)(s+2)(s.+4)(s+10) = 0. (b) Step response with K = 100, 300, 600.
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346
CHAPTER 7
The Root Locus Method
The closed-loop system roots are: K = 100 : s1 = 11.38 K = 300 : s1 = 12.94 K = 600 : s1 = 14.44 DP7.12 s2,3 = 2.09 3.10j s2,3 = 1.29 5.10j s2,3 = 0.53 6.72j s4 = 1.45 s4 = 1.48 s4 = 1.49
The step responses are shown in Figure DP7.11b. The closed-loop transfer function is T (s) = A suitable choice of gains is Ka = 0.52 and K2 = 3 . s3 + s2 Ka . + Ka K2 s + Ka
The step response is shown in Figure DP7.12.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0
0
2
4
6
8
10 Time (secs)
12
14
16
18
20
FIGURE DP7.12 Step response with Ka = 0.52 and K2 = 3.
DP7.13
The characteristic equation is s2 + 10KD s + 10(KP 1) = 0 .
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Design Problems
347
In the Evans form we have 1 + KD 10(s + ) =0. s2 10
The root locus is shown in Figure DP7.13 for = 6. When 0 < <
10,
Root Locus 6
4
2 Imaginary Axis
0
2
4
6 25
20
15
10 Real Axis
5
0
5
FIGURE DP7.13 Root locus when = 6.
the closed-loop poles are both real numbers. The loop in Figure DP7.13 disappears. A viable controller is KP = 36 and KD = 6.
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348
CHAPTER 7
The Root Locus Method
Computer Problems
CP7.1 The root locus for parts (a)-(d) are shown in Figures CP7.1a - CP7.1d.
num=[10]; den=[1 14 43 30]; rlocus(sys) 20 15 10 Imaginary Axi s 5 0 -5 -10 -15 -20 -30 -20 -10 Real Axis
num=[1 20]; den=[1 5 20]; sys=tf(num,den); rlocus(sys) 20
0
10
15
10
5 Imag Axis
0
-5
-10
-15
-20
-45
-40
-35
-30
-25
-20 Real Axis
-15
-10
-5
0
5
FIGURE CP7.1 10 s+20 (a) Root locus for 1 + k s3 +14s2 +43s+30 = 0. (b) Root locus for 1 + k s2 +5s+20 = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
349
num=[1 1 1]; den=[1 5 10 0]; rlocus(sys)
2 1.5 1 Imaginary Axi s 0.5 0 -0.5 -1 -1.5 -2 -5 -4 -3 Real Axis
num=[1 4 6 8 6 4]; den=[1 2 2 1 1 10 1]; sys=tf(num,den); rlocus(sys) 2.5
-2
-1
0
2
1.5
1
0.5 Imag Axis
0
-0.5
-1
-1.5
-2
-2.5 -7
-6
-5
-4
-3 Real Axis
-2
-1
0
1
2
FIGURE CP7.1 2 +s+1 CONTINUED: (c) Root locus for 1 + k s(ss +5s+10) = 0. (d) Root locus for 1 + 2
+8s +6s+4 k s6s +4s +6s s3 +s2 +10s+1 = 0. +2s5 +2s4 +
5 4 3 2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
350 CP7.2
CHAPTER 7
The Root Locus Method
The maximum value of the gain for stability is k = 0.791. The m-le script and root locus is shown in Figure CP7.2.
Select a point in the graphics window num=[1 -2 2]; den=[1 3 2 0]; sys = tf(num,den); rlocus(sys) rloc nd(sys) selected_point = -0.0025 + 0.6550i ans = 0.8008
1
0.8
0.6
0.4
0.2 Imag Axis
0
0.2
0.4
0.6
0.8
1 3
2.5
2
1.5
1
0.5 Real Axis
0
0.5
1
1.5
2
FIGURE CP7.2 Using the rlocnd function.
The value of k = 0.8008 selected by the rlocnd function is not exact since you cannot select the j -axis crossing precisely. The actual value is determined using Routh-Hurwitz analysis. CP7.3 The partial fraction expansion of Y (s) is Y (s) = s+2 0.15 0.25 0.4 = + . + 6s + 6) s+5 s+1 s
s(s2
The m-le script and output is shown in Figure CP7.3.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
r= -0.1500 -0.2500 0.4000 p= -5 -1 0 k= []
351
num=[1 2]; den=[1 6 5 0]; [r,p,k]=residue(num,den)
FIGURE CP7.3 Using the residue function.
CP7.4
The characteristic equation is 1+p s2 s1 =0. + 4s + 6
The root locus is shown in Figure CP7.4. The closed-loop system is stable for 0<p<6.
num=[1 1]; den=[1 4 6]; sys=tf(num,den); rlocus(sys) 1.5
1
0.5
Imag Axis
0
-0.5
-1
-1.5 -4
-3
-2
-1 Real Axis
0
1
2
FIGURE CP7.4 s1 Root locus for 1 + p s2 +4s+6 = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
352 CP7.5
CHAPTER 7
The Root Locus Method
The characteristic equation is 1+k s+1 =0. s2
The root locus is shown in Figure CP7.5. For k = 2 we obtain s1,2 = 1 j , that is, we have = 0.707.
num=[1 1]; den=[1 0 0]; sys = tf(num,den); hold o , clf rlocus(sys); hold on plot([0 -2],[0 2*tan(acos(0.707))],'--') plot([0 -2],[0 -2*tan(acos(0.707))],'--') plot([-1 -1],[1 -1],'*')
2.5
2
1.5
1
0.5 Imag Axis
0
-0.5
-1
-1.5
-2
-2.5 -3
-2.5
-2
-1.5
-1 Real Axis
-0.5
0
0.5
1
FIGURE CP7.5 +1 Root locus for 1 + k ss2 = 0.
CP7.6
The m-le script to generate the root locus for each controller in parts (a)-(c) is shown in Figure CP7.6. The performance region is indicated on each root locus in Figures CP7.6b - CP7.6d. For part (a), the controller gain is found to be Gc (s) = 11.3920. The integral controller in part (b) is determined to be Gc (s) = 4.093 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
353
numg=[1]; deng=[1 5 6]; sysg = tf(numg,deng); t=[0:0.1:15]; % % Part (a) % Select a point in the graphics window sys1 = sysg; rlocus(sys1), grid selected_point = hold on plot([-0.4 -0.4],[-6 6],'--',... -2.5030 + 3.3380i [0 -6*tan(36.2*pi/180)],[0 6],'--',... [0 -6*tan(36.2*pi/180)],[0 -6],'--') ans = hold o [kp,poles] = rloc nd(sys1) 11.3920 % % Part (b) % numc=[1]; denc=[1 0]; sysc = tf(numc,denc); sys2 = series(sysc,sysg); Select a point in the graphics window gure rlocus(sys2), grid selected_point = hold on plot([-0.4 -0.4],[-6 6],'--',... -0.6690 + 0.8210i [0 -6*tan(36.2*pi/180)],[0 6],'--',... [0 -6*tan(36.2*pi/180)],[0 -6],'--') ans = hold o [ki,poles] = rloc nd(sys2) 4.0930 % % Part (c) % Plot performance region boundaries on graph. gure numc=[1 1]; denc=[1 0]; sysc = tf(numc,denc); Select a point in the graphics window sys3 = series(sysc,sysg); rlocus(sys3), grid selected_point = hold on plot([-0.4 -0.4],[-6 6],'--',... -2.0695+ 2.7387i [0 -6*tan(36.2*pi/180)],[0 6],'--',... [0 -6*tan(36.2*pi/180)],[0 -6],'--') ans = hold o [kpi,poles] = rloc nd(sys3) 9.2516 % % Part (d) % gure sys1_o = kp*sys1; sys1_cl = feedback(sys1_o,[1]); sys2_o = ki*sys2; sys2_cl = feedback(sys2_o,[1]); sys3_o = kpi*sys3; sys3_cl = feedback(sys3_o,[1]); [y1,t]=step(sys1_cl,t); [y2,t]=step(sys2_cl,t); [y3,t]=step(sys3_cl,t); plot(t,y1,t,y2,'--',t,y3,':'),grid xlabel('time [sec]'),ylabel('y(t)') title('Gc(s): proportional (solid), integral (dashed) & PI (dotted)')
FIGURE CP7.6 (a) Script to generate the root locus for each controller.
The proportional integral (PI) controller in part (c) is Gc (s) = 9.2516(s + 1) . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
354
CHAPTER 7
6
The Root Locus Method
4
2
Imag Axis
0
-2
-4
-6 -4
-3
-2
-1 Real Axis
0
1
2
FIGURE CP7.6 CONTINUED: (b) Root locus for proportional controller with selected K = 11.3920.
6
4
2
Imag Axis
0
-2
-4
-6 -4
-3
-2
-1 Real Axis
0
1
2
FIGURE CP7.6 CONTINUED: (c) Root locus for integral controller with selected K = 4.0930.
The proportional controller is stable for all K > 0 but has a signicant steady-state error. The integral controller has no steady-state error,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
6
355
4
2
Imag Axis
0
-2
-4
-6 -4
-3
-2
-1 Real Axis
0
1
2
FIGURE CP7.6 CONTINUED: (d) Root locus for PI controller with selected K = 9.2516.
but is stable only for K < 30. The PI controller has zero steady-state error and is stable for all K > 0. Additionally, the PI controller has a fast transient response. The step responses for each controller is shown in Figure CP7.6e.
Gc(s): proportional (solid), integral (dashed) & PI (dotted) 1.4
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
5 time [sec]
10
15
FIGURE CP7.6 CONTINUED: (e) Step responses for each controller.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
356 CP7.7
CHAPTER 7
The Root Locus Method
The loop transfer function can be written as Gc (s)G(s) = where K2 = K2 /J . The parameter of interest for the root locus is K2 . The root locus is shown in Figure CP7.7. The selected value of K2 = 7.1075 . Therefore, K2 = 7.1075 J and K1 = 35.5375 . J K1 + K2 s s+5 = K2 2 2 Js s
num=[1 5]; den=[1 0 0]; sys=tf(num,den); rlocus(sys); rloc nd(sys) 10 8 6
+
4 2
Imag Axis
0 -2 -4
o
x
+
-6 -8 -10 -10 -8 -6 -4 -2 0 Real Axis 2 4 6 8 10
FIGURE CP7.7 Root locus to determine K2 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
357
CP7.8
The value of K that results in a damping ratio of = 0.707 is K = 5.2. The root locus is shown in Figure CP7.8.
Root Locus 5 4 3 2 Imaginary Axis 1 0 1 2 3 4 5 10
s = -0.68 + 0.68j
s = -6.63 s = -0.68 - 0.68j
5 Real Axis
0
5
FIGURE CP7.8 Root locus for 1 + K s3 +8s21 s+1 = 0. +10
CP7.9
(a) The characteristic equation is s3 + (2 + k)s2 + 5s + 1 = 0 . (b) The Routh array is s3 s2 s1 so For stability we require 2 + k > 0 or and 5k + 9 > 0 or k > 9/5 . k > 2 1 2+k
5k +9 2+k
5 1
1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
358
CHAPTER 7
The Root Locus Method
Therefore, the stability region is dened by k > 1.8 . (c) Rearranging the characteristic equation yields 1+k s2 s3 + 2s2 + 5s + 1 = 0 .
The root locus is shown in Figure CP7.9. We see that the system is stable for all k > 0.
Root Locus 2 1.5 1 Imaginary Axi s 0.5 0 ?-0.5 ?-1 ?-1.5 ?-2 ?-2.5 ?-2 ?-1.5 ?-1 ?-0.5 0
Real Axi s
FIGURE CP7.9 2 Root locus for 1 + k s3 +2ss +5s+1 = 0. 2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
8
Frequency Response Methods
Exercises
E8.1 Given the loop transfer L(s) function = we determine that |L(j )| = 4 (1 + 2 ) and ( ) = 2 tan1 . 4 , (s + 1)2
The polar plot is shown in Figure E8.1.
3
2
1
Imag Axis
0
*
*
*
*
-1
-2
* *
-3 -0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
Real Axis
FIGURE E8.1 Polar plot for L(s) =
4 . (s+1)2
359
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
360
CHAPTER 8
Frequency Response Methods
The magnitude and phase angle for = 0, 0.5, 1, 2, 4, are summarized in Table E8.1.
0
0.5
1
2
4
0
|Gc G(j )| (deg)
4
3.2
2
0.8
0.23
0
-53
-90
-127
-152
-180
TABLE E8.1
Magnitude and phase for Gc (s)G(s) =
4 . (s+1)2
E8.2
The transfer function is G(s) = s2 2572 . + 386s + 15, 434
The frequency response plot is shown in Figure E8.2. The phase angle is computed from = tan1 tan1 . 45.3 341
The phase angles for = 10, 45, 100, 341 and 1000 are summarized in Table E8.2.
10
45
100
341
1000
(deg)
14.1
52.3
-82.0
-127.5
-158.6
TABLE E8.2
Phase angle for G(s) =
2572 s2 +386s+15,434 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
361
0
-20
* *
Gain dB
-40 -60 -80 -100 100 101 102 Frequency (rad/sec) 103 104
0
Phase deg
-50 -100 -150 -200 100 101 102 Frequency (rad/sec)
*
103
104
FIGURE E8.2 Frequency response for G(s) =
2572 s2 +386s+15434 .
E8.3
The loop transfer function is Gc (s)G(s) = The phase angle is computed via ( ) = 90o tan1 At = 28.3, we determine that = 90o 70.5o 35.3o + 15.8o = 180o . Computing the magnitude yields |Gc G(j )| =
300(100)(1 + ( 100 )2 ) 2 10(1 + ( 10 )2 ) 2 40(1 + ( 40 )2 ) 2
1 1 1
300(s + 100) . s(s + 10)(s + 40)
tan1 + tan1 . 10 40 100
= 0.75 ,
when = 28.3. We can also rewrite Gc (s)G(s) as Gc (s)G(s) =
s 75( 100 + 1) . s s s( 10 + 1)( 40 + 1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
362
CHAPTER 8
Frequency Response Methods
Then, the magnitude in dB is 20 log10 |Gc G| = 20 log 10 (75) + 10 log 10 (1 + ( 2 ) ) 10 log 10 (1 + ( )2 ) 100 10 10 log 10 (1 + ( )2 ) 20 log10 = 2.5 dB , 40
at = 28.3. E8.4 The transfer function is G(s) = Ks . (s + a)(s + 10)2
Note that = 0o at = 3, and that = +90o tan1 2 tan1 . a 10
Substituting = 3 and solving for a yields a=2. Similarly, from the magnitude relationship we determine that K = 400 . E8.5 The lower portion for < 2 is 20 log at = 8. Therefore, 20 log which occurs when K=8. We have a zero at = 2 and another zero at = 4. The zero at = 4 yields a = 0.25 . We also have a pole at = 8, and a second pole at = 24. The pole at = 24 yields b = 1/24 . K = 0 dB 8 K = 0 dB ,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
363
Therefore, G(s) = E8.6 8(1 + s/2)(1 + s/4) . s(1 + s/8)(1 + s/24)(1 + s/36)
The loop transfer function is L(s) = 10 . s(s/6 + 1)(s/100 + 1)
The Bode diagram is shown in Figure E8.6. When 20 log 10 |L| = 0 dB , we have = 6.7 rad/sec .
Bode Diagram 50 0 Magnitude (dB) Phase (deg) 50 100 150 200 90
135
180
225
270 1 10
10
0
10 10 Frequency (rad/sec)
1
2
10
3
10
4
FIGURE E8.6 Bode Diagram for L(s) =
10 s(s/6+1)(s/100+1) .
E8.7
The transfer function is T (s) = 4 . (s2 + s + 1)(s2 + 0.4s + 4)
(a) The frequency response magnitude is shown in Figure E8.7. The frequency response has two resonant peaks at r1 = 0.8 rad/sec and r2 = 1.9 rad/sec .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
364
CHAPTER 8
Frequency Response Methods
10 5
Gain dB
0 -5 -10 10-1 100 Frequency (rad/sec) 101
1.5
Amplitude
1 0.5 0 0
2
4
6
8
10 12 Time (secs)
14
16
18
20
FIGURE E8.7 (a) Bode Diagram for T (s) =
4 (s2 +s+1)(s2 +0.4s+4) .
(b) Unit step response.
(b) The percent overshoot is P.O. = 35% , and the settling time is Ts 16 sec . (c) The step response is shown in Figure E8.7. E8.8 (a) The break frequencies are 1 = 1 rad/sec, 2 = 5 rad/sec, and 3 = 20 rad/sec . (b) The slope of the asymptotic plot at low frequencies is 0 dB/decade. And at high frequencies the slope of the asymptotic plot is -20 dB/decade. (c) The Bode plot is shown in Figure E8.8.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
365
Bode Diagram 20 10 Magnitude (dB) Phase (deg) 0 10 20 30 180 135 90 45 0 45 90 2 10
1 0 1 2 3
10
10 10 Frequency (rad/sec)
10
10
FIGURE E8.8 Bode Diagram for Gc (s)G(s) =
100(s1) . s2 +25s+100
E8.9
The Bode diagram for G(s) is shown in Figure E8.9.
40 20
Gain dB
0 -20 -40 10-1 100 101 Frequency (rad/sec) 102 103
50
Phase deg
0 -50 10-1 100 101 Frequency (rad/sec]) 102 103
FIGURE E8.9 Bode Diagram for G(s) =
(s/5+1)(s/20+1) . (s+1)(s/80+1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
366 E8.10
CHAPTER 8
Frequency Response Methods
The frequency response has two peaks; the rst peak at f 1.8 and the second peak at f 3.1. One possible G(j ) is G(j ) = (j + 1) 1 + where = 1 , 2 (0.2) n2 = 2 (3.1 103 ) . 1
21 n1
j +
j n1
2
1+
22 n2
j +
j n2
2
,
1 = 0.15;
n1 = 2 (1.8 103 ) 2 = 0.15;
The damping ratios are estimated using Figure 8.10 in Dorf & Bishop. E8.11 The Bode plot is shown in Figure E8.11. The frequency when 20 log10 |GC G( )| = 0 is = 9.9 rad/sec.
Bode Diagram 20 0 Magnitude (dB) Phase (deg) 20 40 60 80 100 120 0 45 90 135 180 225 270 1 10
0 1 2 3
10
10 Frequency (rad/sec)
10
10
FIGURE E8.11 Bode Diagram for Gc (s)G(s) =
1000 . (s2 +10s+100)(s+2)
E8.12
(a) The transfer function is G(s) = C(sI A)1 B + D = (b) The Bode plot is shown in Figure E8.12. 5(s 1) . s2 + 3s + 2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
Bode Diagram 10 Magnitude (dB) 0 -10 -20
367
Phase (deg)
270 180 90 -2 10
10
-1
10 Frequency (rad/sec)
0
10
1
10
2
FIGURE E8.12 Bode Diagram for G(s) =
5(s1) . s2 +3s+2
E8.13
The closed-loop transfer function is T (s) = s3 + 11s2 100 . + 20s + 110
The Bode plot of T (s) is shown in Figure E8.13, where B = 4.9 rad/sec.
Bode Diagram 50 Magnitude (dB) 0 -50 -100
-3 dB
b=4.9
-45 -90 -135 -180 -225 -270 -1 10
Phase (deg)
10
0
10 Frequency (rad/sec)
1
10
2
10
3
FIGURE E8.13 Bode Diagram for T (s) =
100 s3 +11s2 +20s+110 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
368 E8.14
CHAPTER 8
Frequency Response Methods
The loop transfer function is L(s) = 20 . (s2 + 1.4s + 1)(s + 10)
The Bode plot of L(s) is shown in Figure E8.14. The frequency when 20 log10 |L( )| = 0 is = 1.32 rad/sec.
Bode Diagram 50
Magnitude (dB) Phase (deg)
0
50
100
150 0 45 90 135 180 225 270 2 10
1 0 1 2 3
10
10 10 Frequency (rad/sec)
10
10
FIGURE E8.14 Bode Diagram for L(s) =
20 . (s2 +1.4s+1)(s+10)
E8.15
The closed-loop transfer function is T (s) = 3s . s2 + 2s + K + 5
The bandwidth as a function of K is shown in Figure E8.15. The bandwidth as a function of K is: (a) K = 1 and b = 2.77 rad/sec. (b) K = 2 and b = 3.63 rad/sec. (c) K = 10 and b = 9.66 rad/sec. The bandwidth increases as K increases.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
369
18
16
14
12
b (rad/s)
10
8
6
4
2
0
0
2
4
6
8
10 K
12
14
16
18
20
FIGURE E8.15 Bandwith of T (s) =
3s s2 +2s+K +5 .
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370
CHAPTER 8
Frequency Response Methods
Problems
P8.1 (a) The transfer function is Gc (s)G(s) = and Gc G(j ) = 1 . (1 2 ) + j 2.5 1 , (1 + 0.5s)(1 + 2s)
The polar plot is shown in Figure P8.1a. A summary of the magnitude and phase angles for = 0, 0.5, 1, 2, 5 and can be found in Table P8.1a.
0.8 0.6 0.4 0.2
Imag Axis
0 -0.2 -0.4 -0.6 -0.8 -0.2
0
0.2
0.4 Real Axis
0.6
0.8
1
FIGURE P8.1 (a) Polar plot for Gc (s)G(s) =
1 . (1+0.5s)(1+2s)
|GH (j )| (deg )
TABLE P8.1
0 1 0
o
0.5 0.686 59
o
1 0.40 90
o
2 0.172 121
o
5 0.037 152.6
o
0 180o
(a) Magnitudes and phase angles for Gc (s)G(s) =
1 . (1+0.5s)(1+2s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
371
(b) The transfer function is Gc (s)G(s) = and Gc G(j ) = 10 (1 2 ) + 1.4j . (1 2 ) 2j 10(s2 + 1.4s + 1) (s 1)2
The polar plot is shown in Figure P8.1b. A summary of the magnitude and phase angles for = 0, 0.25, 0.5, 1, 2, 8, 16 and can be found in Table P8.1b.
Nyquist Diagram 10 8 6 4 Imaginary Axis 2 0 2 4 6 8 10 8
6
4
2
0 2 Real Axis
4
6
8
10
FIGURE P8.1 CONTINUED: (b) Polar plot for Gc (s)G(s) =
10(s2 +1.4s+1) . (s1)2
|GH (j )| (deg )
TABLE P8.1
0 10 0
o
0.25 9.4 48.5
o
0.5 8.2 96.2
o
1 7 180
o
2 8.2 96.2
o
8 9.8 24.3
o
16 9.96 12.2
o
10 0o
CONTINUED: (b) Magnitudes and phase angles for Gc (s)G(s) =
10(s2 +1.4s+1) . (s1)2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
372
CHAPTER 8
Frequency Response Methods
(c) The transfer function is Gc (s)G(s) = (s2 (s 10) . + 6s + 10)
We have complex poles at n = 10 and = 3 . The polar plot 10 is shown in Figure P8.1c. A summary of the magnitude and phase angles for = 0, 1, 2, 3, 4, 5, 6, can be found in Table P8.1c.
Nyquist Diagrams
From: U(1) 0.8
0.6
0.4
0.2
Imaginary Axis
To: Y(1)
0
-0.2
-0.4
-0.6
-0.8
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Real Axis
FIGURE P8.1 CONTINUED: (c) Polar plot for Gc (s)G(s) =
s10 . s2 +6s+10
|GH (j )| (deg )
TABLE P8.1
0 1 180
o
1 0.93 140.6
o
2 0.76 105.3
o
3 0.58 76.5
o
4 0.44 54.2
o
5 0.33 36.9
o
6 0.26 23.2
o
0 90o
CONTINUED: (c) Magnitudes and phase angles for Gc (s)G(s) =
s10 . s2 +6s+10
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
373
(d) The transfer function is Gc (s)G(s) = 30(s + 8) . s(s + 2)(s + 4)
The polar plot is shown in Figure P8.1d. A summary of the magnitude and phase angles for = 1, 0.1, 0.8, 1.6, 3.2, 12.8, can be found in Table P8.1d.
20 15 10 5
Imag Axis
0 -5 -10 -15 -20 -20
-15
-10 Real Axis
-5
0
5
FIGURE P8.1 CONTINUED: (d) Polar plot for Gc (s)G(s) =
30(s+8) . s(s+2)(s+4)
|GH (j )| (deg )
TABLE P8.1
0 90
0.1 299.6
o
0.8 34.3 117.4
o
1.6 13.9 139.1
o
3.2 4.4 163.8
o
12.8 0.20 185.8
o
0 180o
93.6
o
CONTINUED: (d) Magnitudes and phase angles for Gc (s)G(s) =
30(s+8) . s(s+2)(s+4)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
374 P8.2
CHAPTER 8
Frequency Response Methods
(a) The Bode plot is shown in Figure P8.2a. A summary of the magnitude and phase angles for = 0.25, 0.5, 1, 2, 4, 8, 16 can be found in Table P8.2a.
0 -10
Gain dB
-20 -30 -40 -50 10-2 10-1 Frequency (rad/sec) 0 100 101
Phase deg
-50 -100 -150 -200 10-2 10-1 Frequency (rad/sec) 100 101
FIGURE P8.2 (a) Bode plot for Gc (s)G(s) =
1 . (1+0.5s)(1+2s)
|GH (j )| dB (deg )
TABLE P8.2
0.25 -1.03 34
o
0.5 -3.27 59
o
1.0 -8.0 90
o
2.0 -15.3 121
o
4.0 -25.1 146
o
8.0 -36.4 162
o
16.0 -48.2 171o
(a) Magnitudes and phase angles for Gc (s)G(s) =
1 . (1+0.5s)(1+2s)
(b) The transfer function is Gc (s)G(s) = 1 + 0.5s . s2
The Bode plot is shown in Figure P8.2b. A summary of the magnitude
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
375
and phase angles for = 0.25, 0.5, 1, 2, 4, 8, 16 can be found in Table P8.2b.
40 20
Gain dB
0 -20 -40 10-1 100 Frequency (rad/sec) -100 101
Phase deg
-120 -140 -160 -180 10-1 100 Frequency (rad/sec) 101
FIGURE P8.2 CONTINUED: (b) Bode plot for Gc (s)G(s) =
1+0.5s . s2
|GH (j )| dB (deg )
TABLE P8.2
0.25 24.1 173
o
0.5 12.3 166
o
1.0 0.98 153
o
2.0 -9.02 135
o
4.0 -17.1 117
o
8.0 -23.9 104
o
16.0 -30.2 97o
CONTINUED: (b) Magnitudes and phase angles for Gc (s)G(s) =
1+0.5s . s2
(c) The transfer function is Gc (s)G(s) = (s2 (s 10) . + 6s + 10)
The Bode plot is shown in Figure P8.2c. A summary of the magnitude and phase angles for = 0.6, 1, 2, 3, 4, 5, 6, can be found in Table P8.2c.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
376
CHAPTER 8
Frequency Response Methods
Bode Diagrams
From: U(1)
0 -5 -10 -15 -20
Phase (deg); Magnitude (dB)
-25 -30 -35 -40 200 150 100 To: Y(1) 50 0 -50 -100 -1 10
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE P8.2 CONTINUED: (c) Bode plot for Gc (s)G(s) =
s10 . s2 +6s+10
|GH (j )| dB (deg )
TABLE P8.2
0.6 -0.23 156.1o
1 -0.64 140.6o
2 -2.38 105.6o
3 -4.74 76.5o
4 -7.22 54.2o
5 -9.54 36.9o
6 -11.61 23.2o
120 90o
CONTINUED: (c) Magnitudes and phase angles for Gc (s)G(s) =
s10 s2 +6s+10 .
(d) A summary of the magnitude and phase angles for = 0.2, 0.8, 3.2, 6.4, 12.8, 25.6, 51.2 can be found in Table P8.2d. The Bode plot is shown in Figure P8.2d.
|GH (j )| dB (deg )
TABLE P8.2
0.2 43.5 97.1o
0.8 30.7 117.4o
3.2 12.4 163.8o
6.4 -0.5 182o
12.8 -13.8 185.8o
25.6 -26.5 184o
30(s+8) . s(s+2)(s+4)
51.2 -38.8 182.2o
CONTINUED: (d) Magnitudes and phase angles for Gc (s)G(s) =
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
377
50
Gain dB
0
-50 -100 10-1
100 Frequency (rad/sec)
101
102
-50
Phase deg
-100
-150 -200 10-1
100 Frequency (rad/sec)
101
102
FIGURE P8.2 CONTINUED: (d) Bode plot for Gc (s)G(s) =
30(s+8) . s(s+2)(s+4)
P8.3
(a) The bridged-T network we found has zeros at s = jn and poles at s= n n 1/Q2 1 . Q
The frequency response is shown in Figure P8.3 for Q = 10. (b) For the twin-T network, we evaluate the magnitude at = 1.1n or 10% from the center frequency (see Example 8.4 in Dorf & Bishop). This yields |G| 2.1 0.1 3.9 1.1 = 0.05 .
Similarly, for the bridged-T network |G| = 2.1 0.1 2.1 0.14 = 0.707 .
The bridged-T network possesses a narrower band than the twin-T network.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
378
0 -10
CHAPTER 8
Frequency Response Methods
Gain dB
-20 -30 -40 10-1 100 w/wn 100 101
Phase deg
50 0 -50 -100 10-1 100 w/wn 101
FIGURE P8.3 Bode plot for G(s) =
2 s2 +n 2, s2 +(2n /Q)s+n
where = 1/Q = 0.1.
P8.4
The transfer function is G(s) = Gc G1 H (s) 1 s 30000(2s + 1) = . s(s + 10)(s + 20)(s2 + 15s + 150)
A summary of the magnitude and phase angles can be found in Table P8.4. The Bode plot is shown in Figure P8.4.
|G(j | dB (deg )
TABLE P8.4
1 6.95 40.89o
3 5.78 52.39o
5 5.08 77.28o
8 3.38 118.41o
10 1.59 145.99o
15 -5.01 203.52o
24 -17.56 258.57o
Magnitudes and phase angles for GH (s) =
30000(2s+1) . s(s+10)(s+20)(s2 +15s+150)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
Bode Diagram 50 Magnitude (dB) 0 -50 -100 -150 0 Phase (deg) -90 -180 -270 -360 -2 10
-1 0 1 2 3
379
10
10 10 Frequency (rad/sec)
10
10
FIGURE P8.4 Bode plot for GH (s) =
30000(2s+1) . s(s+10)(s+20)(s2 +15s+150)
P8.5
The Bode plot is shown in Figure P8.5.
Bode Diagram 50
Magnitude (dB) Phase (deg)
0
50
100
150 0
90
180
270
360 1 10
10
0
10 Frequency (rad/sec)
1
10
2
10
3
FIGURE P8.5 Bode plot for G(s) =
10 (s/5+1)(s+1)(s/15+1)(s/75+1) .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
380 P8.6
CHAPTER 8
Frequency Response Methods
(a) The transfer function is GH (s) = 3.98(1 + s/1) . s(1 + s/10)2
We have a zero at = 1 and two poles at = 10.0. The low frequency approximation is K/s and at = 1 we have 20 log Therefore, K = 3.98 at = 1 (an approximation). The phase plot is shown in Figure P8.6a. K = 12dB .
(a) -40 -60 -80 Phase deg -100 -120 -140 -160 -180 -2 10 10
-1
10
0
10
1
10
2
(b) 100
50 Phase deg
0
-50
-100 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
10
3
FIGURE P8.6 Phase plots for (a) G(s) =
3.98(s/1+1) . s(s/10+1)2
(b) G(s) =
s . (s/10+1)(s/50+1)
(b) The transfer function is GH (s) = s . (1 + s/10)(1 + s/50)
The poles are located by noting that the slope is 20 dB/dec. The
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
381
low frequency approximation is Ks, so 20 log K = 0dB . At = 1 we determine that K=1. The phase plot is shown in Figure P8.6b. P8.7 The loop transfer function is L(s) = Kv . s(s/ + 1)2
(a) Set Kv = 2 . The Bode plot is shown in Figure P8.7a.
40
Gain dB
20
0
-20 10-1
100 Frequency (rad/sec)
101
-80 -100
Phase deg
-120 -140 -160 -180 10-1 100 Frequency (rad/sec) 101
FIGURE P8.7 (a) Bode plot for L(s) =
Kv , s(s/ +1)
where Kv = 2 .
(b) The logarithmic magnitude versus the phase angle is shown in Figure P8.7b.
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382
CHAPTER 8
Frequency Response Methods
40
30
20
Gain dB
10
0
-10
-20 -170
-160
-150
-140
-130 Phase deg
-120
-110
-100
-90
FIGURE P8.7 CONTINUED: (b) Log-magnitude-phase curve for L(j ).
P8.8
The transfer function is T (s) = s2 K . + 7s + K
(a) When P.O. = 10%, we determine that = 0.59 by solving 2 0.10 = e/ 1 .
2 So, 2n = 7 implies that n = 5.93, hence K = n = 35.12. Also,
Mp = (2 1 2 )1 = 1.05 . (b) For second-order systems we have r = n 1 2 2 = 0.55n = 3.27 when = 0.59 and n = 5.93. (c) We estimate B to be B (1.19 + 1.85)n = 1.14n = 6.8 rad/s . P8.9 The log-magnitude phase curves are shown in Figure P8.9.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
383
(a) 0 40 (b)
-5 -10
30
20 -15
Gain dB
-25 -30
Gain dB
-150 -100 Phase deg -50 0
-20
10
0
-10 -35 -40 -45 -200 -20
-30 -180
-160
-140 Phase deg
-120
-100
FIGURE P8.9 Log-magnitude-phase curve for (a) Gc (s)G(s) = 1+0.5s . s2
1 (1+0.5s)(1+2s)
and (b) Gc (s)G(s) =
P8.10
The governing equations of motion are F (s) = Kf If (s) and If (s) = Vf (s) . Rf + L f s
Without loss of generality we can let Kf = 1.0. Also, we have F (s) = (M s2 + bs + K )Y (s) . Therefore, the transfer function is GH (s) = KKf 50K = . 2 + bs + K ) (Rf + Lf s)(M s (s + 0.5)(s2 + 2s + 4)
This is a type 0 system, therefore Kp = 25K . (a) If we allow a 1% error , we have ess = |R|/(1 + Kp ) = 0.01|R|. Thus Kp = 25K = 99. Select K=4. (b) The Bode plot is shown in Figure P8.10a. (c) The log-magnitude phase curve is shown in Figure P8.10b.
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384
CHAPTER 8
Frequency Response Methods
40
Gain dB
20
0 -20 10-2
10-1 Frequency (rad/sec)
100
101
0
Phase deg
-100
-200 -300 10-2
10-1 Frequency (rad/sec)
100
101
FIGURE P8.10 (a) Bode plot for GH (s) =
200 . (s2 +2s+4)(s+0.5)
40
30
20
Gain dB
10
0
-10
-20 -300
-250
-200
-150 Phase deg
-100
-50
0
FIGURE P8.10 CONTINUED: (b) Log-magnitude-phase curve for GH (s) =
200 . (s2 +2s+4)(s+0.5)
(d) The closed-loop transfer function Bode plot is shown in Figure P8.10c. We determine from the plot that Mp = 1.6, r = 4.4 and B = 6.8.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
385
5 0
Gain dB
-5 -10 -15 10-1 100 Frequency (rad/sec) 100 101
Phase deg
0
-100 -200 10-1
100 Frequency (rad/sec)
101
FIGURE P8.10 CONTINUED: (c) Bode plot for closed-loop T (s) = Y (s)/R(s).
P8.11
The Bode plot is shown in Figure P8.11.
200 100 0
Gain dB
-100 10-4
10-3
10-2
10-1
100
101
Frequency (rad/sec) 100
Phase deg
0
-100 -200 10-4
10-3
10-2
10-1
100
101
Frequency (rad/sec)
FIGURE P8.11 Bode plot for G(s) =
0.164(s+0.2)(s+0.32) . s2 (s+0.25)(s0.009)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
386 P8.12
CHAPTER 8
Frequency Response Methods
The three transfer functions are G1 (s) = 10 G2 (s) = 1 s(s/0.6 + 1) G3 (s) = 3s .
(a) When G3 (s) is out of the loop, the characteristic equation is 10 =0 s(s/0.6 + 1) or s2 + 0.6s + 6 = 0. Thus, = 0.6/(2 6) = 0.12. (b) With G3 (s), the characteristic equation is 1 + G1 G2 (s) = 1 + 1 + G1 G2 (s) + G2 G3 (s) = 1 + or s2 + 2.4s + 6 = 0 . Thus, = 2.4/(2 6) = 0.49. P8.13 By inspection of the frequency response, we determine L(s) = Gc (s)G(s)H (s) = K . s(s/100 + 1)(s/1000 + 1)2 1.85 6 + =0, s(s + 0.6) s(s + 0.6)
For small , we have 20 log K/ = 40 dB at = 10. So, K = 1000. P8.14 The data we have are R1 = R2 = 1000, c1 = 107 farad and c2 = 106 farad. The governing equations are V2 (s) C1 s = , V1 (s) R1 + C1 s 1 and Vo (s) KR2 = . V2 (s) R2 + C1 s 2 So Vo (s) KR2 C2 s 109 s = = . V1 (s) (R1 C1 s + 1)(R2 C2 s + 1) (s + 107 )(s + 1000) (a) The Bode plot is shown in Figure P8.14. (b) The mid-band gain is = 40 dB. (c) The -3 dB points are (rad/sec): low 7 and high 1.5 109 .
1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
387
40
Gain dB
20
0
-20 100
101
102
103
104
105
106
107
108
109
1010
Frequency (rad/sec) 100
Phase deg
0
-100
-200 100
101
102
103
104
105
106
107
108
109
1010
Frequency (rad/sec)
FIGURE P8.14 Bode plot for G(s) =
109 s . (s+107 )(s+103 )
P8.15
The data are plotted in Figure P8.15, denoted by an asterisk (*).
50
* *
0 -50 -100 10-1
*
*
*
*
*
*
* * *
100
101
102
-50 -100 * -150 -200 -250 -300 10-1 100 101
* * * * * * * * * *
102
FIGURE P8.15 Bode plot for G(s) =
809.7 ; s(s2 +6.35s+161.3)
tabular data is indicated by an asterick (*).
The low frequency slope is -20 dB/dec and the initial low frequency is
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388
CHAPTER 8
Frequency Response Methods
90o , so we have an integrator of the form K/s. The initial phase is 90o and the nal phase 270o , so we have a minimum phase G(s). Now, |G| is 0.97 at = 8 and = 10 indicating two complex poles. We postulate a transfer function of the form G(s) = K s
s2 2 n
+
2s n
+1
.
The phase angle = 180o at = n . Then, from Figure 8.10 in Dorf & Bishop, we determine that n = 12.7. At = 8, n = 0.63 and , due to o (subtract 90o due to the integrator). Again, the complex poles is 30 from Figure 8.10 in Dorf & Bishop, we estimate = 0.25. To determine K , note that when n 0.1, the eect of the complex poles on magnitude is negligible, so at = 1 we have K |G| = 5.02 . = j1 So K = 5.02. Therefore, G(s) = P8.16 5.02 s
s2 161.3
+
0.5s 12.7
+1
=
s(s2
809.7 . + 6.35s + 161.3)
(a) The unit step input response is shown in Figure P8.16. The closed-
Step Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Amplitude
0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE P8.16 Unit step input response for T (s) =
53.5 s2 +14.1s+53.5 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
389
loop transfer function is T (s) = Y (s) 53.5 =2 . R(s) s + 14.1s + 53.5
(b) The system bandwidth is B = 4.95 rad/sec. P8.17 The transfer function is Gc (s)G(s) = 4(0.5s + 1) . s(2s + 1)(s2 /64 + s/20 + 1)
(a) The Bode plot is shown in Figure P8.17.
50
Gain dB
0
-50 -100 10-1
100 Frequency (rad/sec)
101
102
-50 -100
Phase deg
-150 -200 -250 -300 10-1 100 Frequency (rad/sec) 101 102
FIGURE P8.17 Bode plot for Gs (s)G(s) =
4(0.5s+1) . s(2s+1)(s2 /64+s/20+1)
(b) When the magnitude is 0 dB, we have 1 = 1.6 rad/sec and when = 180o , we have 2 = 7.7 rad/sec . P8.18 The transfer function is Gc (s)G(s) = 12(s + 0.5) 0.2(2s + 1) = . (s + 3)(s + 10) (s/3 + 1)(s/10 + 1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
390
CHAPTER 8
Frequency Response Methods
The Bode plot is shown in Figure P8.18. Near 0 dB, the frequency is = 5.4 rad/sec.
0 -5
Gain dB
-10 -15 -20 10-1 100 Frequency (rad/sec) 50 0 101 102
Phase deg
-50 -100 -150 -200 10-1 100 Frequency (rad/sec) 101 102
FIGURE P8.18 Bode plot for Gc (s)G(s) =
12(s+0.5) . s2 +13s+30
P8.19
Examining the frequency response, we postulate a second-order transfer function
2 (s) n =2 . 2 I (s) s + 2n s + n
From the data we see that = 90o at = 2. Using Figure 8.10 in Dorf & Bishop, we determine that n = = 2. We also estimate = 0.4 from Figure 8.10. Thus, (s) 4 =2 . I (s) s + 1.6s + 4 P8.20 The transfer function is Gc (s)G(s) = s2 781(s + 10) . + 22s + 484
The Bode plot is shown in Figure P8.20. The maximum value of 20 log 10 |Gc G| = 31.85 dB occurs at = 21.08 rad/sec and the corresponding phase is = 20.5o .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
Bode Diagrams
From: U(1) 35
391
30
25
Phase (deg); Magnitude (dB)
20
15 50
0 To: Y(1) -50 100 10
-1
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE P8.20 Bode plot for Gc (s)G(s) =
781(s+10) s2 +22s+484 .
P8.21
The Bode plot is shown in Figure P8.21. The gain is 24 dB when = 180o
40 20
Gain dB
0 -20 -40 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-100 -200 -300 10-1
100 Frequency (rad/sec)
101
102
FIGURE P8.21 Bode plot for Gc (s)G(s) =
200s2 s3 +14s2 +44s+40 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
392 P8.22
CHAPTER 8
Frequency Response Methods
The transfer function is G(s) = 10000(s + 1)(s + 80) . s(s + 300)(s + 9000)
P8.23
The transfer function is G(s) = 100(s + 20)(s + 8000) . (s + 1)(s + 80)(s + 500)
The system is type 0 and the steady-state error to a unit step input is ess = since Kp = lim G(s) = 400 .
s0
1 = 0.0025 1 + Kp
P8.24
(a) From the Bode plot we see that 20 log 10 Mp = 12 or Mp = 3.981. For a second-order system we know that Mp = (2 1 2 )1 . Solving for (with Mp = 3.981) yields = 0.12. Also, from the Bode plot, r = 0.9rad/sec . So, n = r = 0.91 . 1 2 2
Therefore, the second-order approximate transfer function is T (s) =
2 n 0.83 =2 . 2 s2 + 2n s + n s + 0.22s + 0.83
(b) The predicted overshoot and settling time are P.O. = 68% and Ts = 37 sec. P8.25 The transfer function is G(s) = 100(s + 10) . s2 (s + 100)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
393
P8.26
The transfer function is T (s) = Vo (s) 1 + R1 /R2 = . V (s) 1 + RCs
Substituting R = 10k, C = 1F , R1 = 9k, and R2 = 1k yields T (s) = 10 . 1 + 0.01s
The frequency response is shown in Figure P8.26.
Bode Diagrams
20
15
10 Phase (deg); Magnitude (dB)
5
0 0 -20 -40 -60 -80 -100 1 10
10
2
10
3
Frequency (rad/sec)
FIGURE P8.26 Bode plot for T (s) =
1+R1 /R2 1+RCs
P8.27
The frequency response is shown in Figure P8.27.
K |L(j )|j=0 , dB b , rad/s c , rad/s
TABLE P8.27
0.75 3.52 8.3 3.5
1 12.04 14.0 8.7
10 26.02 33.4 22.9
System performance as K varies.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
394
CHAPTER 8
Frequency Response Methods
Bode Diagram 10 0 Magnitude (dB) 10 20 30 40 50 60 0
K increases
K decreases
Phase plot remains unchanged as K varies
Phase (deg) 45
90
135 1 10
10
0
10 Frequency (rad/sec)
1
10
2
10
3
FIGURE P8.27 Bode plot for K = 1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
395
Advanced Problems
AP8.1 The spring-mass-damper system is described by mx + bx + kx = p . Taking the Laplace transform (with zero initial conditions) yields X (s) 1 = . 2 + bs + k P (s) ms From Figure AP8.1(b) in Dorf & Bishop, we determine that 20 log Solving for k yields k = 19.96 N/m .
2 2 Also, n = k/m implies m = k/n , where n = corner frequency = 3.2 rad/sec. So,
X (j 0) 1 = 20 log = 26dB . P (j 0) k
m = 1.949 kg . Comparing Figure AP8.1(b) in Dorf & Bishop to the known standard Bode plot of a second-order system, we estimate 0.32. Therefore, b = 2mn = 2(1.949)(0.32)(3.2) = 3.992 N s/m . AP8.2 The closed-loop transfer function is T (s) = With K = 2, we have T (s) = The sensitivity is
T Sb =
Y (s) Kb = . R(s) s + 1 + 0.5Kb
2b . s+1+b
dT (s) b s+1 = . db T (s) s+1+b
With the nominal value of b = 4, we have
T Sb =
s+1 0.2(s + 1) = . s+5 0.2s + 1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
396
CHAPTER 8
Frequency Response Methods
The sensitivity plot is shown in Figure AP8.2.
0
-2
-4
20*log(mag) (dB)
-6
-8
-10
-12
-14 10-1
100 Frequency (rad/sec)
101
102
FIGURE AP8.2 T Bode plot for Sb (s) =
0.2(s+1) 0.2s+1 .
AP8.3
The equation of motion is mx + bx + Kx = br + Kr . Taking Laplace transforms yields X (s) bs + K = . 2 + bs + K R(s) ms Then, given the various system parameters m = 1 kg, b = 4 Ns/m, K = 18 N/m, we obtain the transfer function: X (s) 4s + 18 =2 . R(s) s + 4s + 18 Also, n = corner frequency = K/m = 18 = 4.243 rad/s and = damping ratio = b/m 4 = = 0.471 . 2n 2(4.243)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
397
The Bode plot is shown in Figure AP8.3.
10 0
Gain dB
-10 -20 -30 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-50 -100 -150 -200 10-1 100 Frequency (rad/sec) 101 102
FIGURE AP8.3 Bode plot for G(s) =
4s+18 . s2 +4s+18
AP8.4
The Bode plot is shown in Figure AP8.4.
0
Gain dB
-50 -100
-150 10-1
100 Frequency (rad/sec)
101
102
0
Phase deg
-100
-200
-300 10-1
100 Frequency (rad/sec)
101
102
FIGURE AP8.4 Bode plot for GH (s) =
0.0625 . (0.5s+1)(0.0625s2 +0.2s+1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
398 AP8.5
CHAPTER 8
Frequency Response Methods
The closed-loop transfer function with unity feedback is given by T (s) = Gc (s)G(s) 10(s + 1) =2 . 1 + Gc (s)G(s) s + 9s + 10
(a) Solving for Gc (s)G(s) yields Gc (s)G(s) = 10(s + 1) . s(s 1)
(b) A summary of the plot data (see Figure AP8.5) is presented in Table AP8.5. (c) The open-loop system is unstable; the closed-loop system is stable.
40
30
20
20 log|GcG(j )|, dB
10
0
10
20
30
40 100
120
140
160
180 200 Phase, degrees
220
240
260
280
FIGURE AP8.5 Log-magnitude-phase curve for Gc G(j ).
20 log |Gc G| phase (deg)
TABLE AP8.5
1 40 101.42
10 4.85 250.17
50 -13.33 267.53
110 -20.61 268.93
500 -33.94 269.77
Summary of magnitude and phase for = 1, 10, 50, 110, 500.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
399
AP8.6
The transfer function is given by T (s) = s2 1/m . + (b/m)s + (k/m)
Selecting k = 1 and b = 2 results in the Bode plot magnitude always less than 0 dB. Choosing b = 2/2 leads to a peak response with a sinusoidal input at = 0.66 rad/s. Figure AP8.6a shows the Bode plot and Figure AP8.6b shows the response to a sinusiodal input with frequency = 1 rad/s is less than 1 in the steady-state, as desired.
Bode Diagram 10
System: sys Peak gain (dB): 6.3 At frequency (rad/sec): 0.661
0
10 Magnitude (dB)
20
30
40
50 2 10
10
1
10 Frequency (rad/sec)
0
10
1
Impulse Response 1
0.5
Amplitude
0
0.5
1
1.5
0
100
200
300
400 Time (sec)
500
600
700
800
FIGURE AP8.6 Bode plot for b/m = 1 and k/m = 1.
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400
CHAPTER 8
Frequency Response Methods
Design Problems
CDP8.1
With the PI controller in the loop, the closed-loop transfer function from the input to the output is (s) 26.035K (s + 2) =2 , R(s) s + (33.1415 + 26.035K )s + 52.07K where we switch o the tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). The Bode plot is shown below for K = 40. From the step response we determine that P.O. = 0 and Ts = 0.19. With K = 40, the closed-loop poles are both real roots with values of s1 = 1072.6 and s2 = 1.9.
60
40 Gain dB
20
0 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
0
Phase deg
-30
-60
-90 10
-1
10
0
10 Frequency (rad/sec)
1
10
2
DP8.1
The loop transfer function is L(s) = Gc (s)G(s) = K (s + 2) . s2 (s + 12)
(a,b) Let K = 1. The Bode plot of the loop transfer function and the closed-loop transfer functions are shown in Figure DP8.1a and Figure DP8.1b, respectively. (c) Let K = 50. The Bode plot of the loop transfer function and the closed-loop transfer functions are shown in Figure DP8.1c and Figure DP8.1d, respectively.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
50
401
0 Gain dB
-50
-100 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
-120
Phase deg
-140
-160
-180 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE DP8.1 (a) Bode plot for the loop transfer function Gc (s)G(s) =
(s+2) . s2 (s+12)
50
0 Gain dB
-50
-100 -2 10
10
-1
10 Frequency (rad/sec)
0
10
1
10
2
0
Phase deg
-90
-180 10
-2
10
-1
10 Frequency (rad/sec)
0
10
1
10
2
FIGURE DP8.1 CONTINUED: (b) Bode plot for the closed-loop T (s) =
(s+2) . s3 +12s2 +s+2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
402
CHAPTER 8
Frequency Response Methods
100
50 Gain dB
0
-50 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
-120
Phase deg
-140
-160
-180 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE DP8.1 CONTINUED: (c) Bode plot for the loop transfer function Gc (s)G(s) =
20 0 Gain dB -20 -40 -60 -1 10
50(s+2) . s2 (s+12)
10
0
10 Frequency (rad/sec)
1
10
2
0
Phase deg
-90
-180 10
-1
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE DP8.1 CONTINUED: (d) Bode plot for the closed-loop T (s) =
50(s+2) . s3 +12s2 +50s+100
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
403
(d) The peak value of Mp 2 occurs for 14 K 350. The maximum bandwidth is achieved for the largest gain K . Thus, we select K = 350 and the corresponding bandwidth is B = 29 rad/sec. (e) The system is type 2the steady-state error is zero for a ramp input. DP8.2 The open-loop transfer function is Gc (s)G(s) = 20(s + 1) . s(s + 5)(s2 + 2s + 10)
(a) The phase angle is = 180o when = 4 rad/sec. The magnitude is 0 dB when = 0.4 rad/sec. (b) The closed-loop transfer function is T (s) = 20(s + 1) . + 20s2 + 70s + 20
s4
+
7s3
The closed-loop Bode plot is shown in Figure DP8.2.
0 -20
Gain dB
-40 -60 -80 -100 10-2 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-100
-200 -300 10-2
10-1
100 Frequency (rad/sec)
101
102
FIGURE DP8.2 Bode plot for closed-loop T (s) =
20(s+1) . s4 +7s3 +20s2 +70s+20
(c) When K = 20, we have Mp = none , r = n/a , and B = 3.96 rad/sec .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
404
CHAPTER 8
Frequency Response Methods
When K = 40, we have Mp = 9.4 dB , (d) Select K = 40. DP8.3 The closed-loop transfer function is T (s) = s3 + 7s2 K (s + 5) . + 12s + 10 + 5K r = 3.72 rad/sec , and B = 4.69 rad/sec .
When K = 4.2, we have 10 log 10 Mp = 3 dB. The system bandwidth is b = 3.7178 rad/sec. The steady-state tracking error to a unit step input is ess = lim sE (s) = lim 1 T (s) .
s0 s0
So, ess = 1 5K = 0.322 , 10 + 5K
when K = 4.2. Since the system is unstable when K > 14.8, the steadystate error does not exist after K = 14.8. The Bode plot is shown in Figure DP8.3.
20 0
Gain dB
-20 -40 -60 -80 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-50 -100 -150 -200 10-1 100 Frequency (rad/sec) 101 102
FIGURE DP8.3 Bode plot for closed-loop T (s) =
K (s+5) , s3 +7s2 +12s+10+5K
where K = 4.2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
405
DP8.4
We have a second-order loop transfer function Gc (s)G(s) = K . (0.2s + 1)(0.5s + 1)
With Mp = 1.5, we determine that Mp = (2 1 2 )1 Now the characteristic equation is s2 + 7s + 10(1 + K ) = 0 . So, solving 2n = 7 yields n = 9.81. Therefore, K=
2 n 10 = 8.6 . 10
or
= 0.3568 .
The closed-loop transfer function is T (s) = K 10(K + 1) . 2 + 7s + 10(K + 1) K +1s
So, the overall gain of the standard second-order system will be attenuated by the factor K/(K + 1). To compensate, we amplify the gain by a small factor. Thus we choose K = 10.8. The bandwidth is b = 15.6 rad/sec. DP8.5 From the Bode plot of G(s) we nd that there exists two poles, at approximately = 1 rad/sec and = 10 rad/sec. Then, by examining the Bode plot we estimate G(s) = 10 . (s + 1)(s + 10)
We use a scale factor of 10 because at low frequency the Bode plot has magnitude 0 dB (or a DC gain of 1). With G(s) as above, we can utilize the controller Gc (s) = yielding a crossover c = 12.9 rad/sec and a magnitude of at least 25 dB for < 0.1 rad/sec. Figure DP8.5 shows the compensator Bode plot of Gc (s)G(s). 500 s + 20
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
406
CHAPTER 8
Frequency Response Methods
Bode Diagram 50
Magnitude (dB)
0 -50 -100 -150 0 -45 -90 -135 -180 -225 10
-2
25 dB
c=12.9
Phase (deg)
10
-1
10 10 Frequency (rad/sec)
0
1
10
2
10
3
FIGURE DP8.5 Bode Diagram for G(s)Gc (s) =
5000 . (s+1)(s+10)(s+20)
DP8.6
Let K = 1 to meet the steady-state tracking error requirement and p = 2 , where = 0.69 to obtain a 5% overshoot. The system is given by x = Ax + Bu where A=
0
1
1 1.38
The characteristic polynomial is
,
B=
1 0
, and
C=
01
.
s2 + 1.38s + 1 = 0 . The associated damping ratio is = 0.69 and the natural frequency is n = 1 rad/s. Using the approximation b = (1.19 + 1.85)n we obtain b 1.028 rad/s. The Bode plot is shown in Figure DP8.6. The bandwidth is b = 1.023 rad/s.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
407
Bode Diagram 20 0 Magnitude (dB) Phase (deg) 20 40 60 80 0
45
90
135
180 2 10
10
1
10 Frequency (rad/sec)
0
10
1
10
2
FIGURE DP8.6 Bode diagram for K = 1 and p = 1.38.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
408
CHAPTER 8
Frequency Response Methods
Computer Problems
CP8.1 The m-le script and Bode plot are shown in Figure CP8.1. The script automatically computes Mp and r .
num=[25]; den=[1 1 25]; sys = tf(num,den); w=logspace(0,1,400); [mag,phase]=bode(sys,w); [y,l]=max(mag); mp=20*log10(y), wr=w(l) bode(sys,w); mp = 14.0228 wr = 4.9458
Bode Diagrams
From: U(1) 15
10
5
0
Phase (deg); Magnitude (dB)
-5
- 10 0
- 50
To: Y(1)
- 100
- 150
- 200 0 10
10
1
Frequency (rad/sec)
FIGURE CP8.1 Generating a Bode plot with the bode function.
CP8.2
The m-le script to generate the Bode plots is shown in Figure CP8.2a. The Bode plots are presented in Figures CP8.2b-CP8.2e. The transfer functions are (a) : G(s) = 1 ; (s + 1)(s + 10) (b) : G(s) = s + 10 ; (s + 2)(s + 40)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
409 1 ; + 2s + 50 (d) : G(s) = s7 . (s + 2)(s2 + 12s + 50)
(c) : G(s) =
s2
% Part (a) num=[1]; den=conv([1 1],[1 10]); sys1=tf(num,den); sys = tf(num,den); gure(1), bode(sys1), grid % Part (b) num=[1 10]; den=conv([1 2],[1 40]); sys2=tf(num,den); sys = tf(num,den); gure(2), bode(sys2), grid % Part (c) num=[1]; den=[1 2 50]; sys3=tf(num,den); sys = tf(num,den); gure(3), bode(sys3), grid % Part (d) num=[1 -7]; den=conv([1 2],[1 12 50]); sys4=tf(num,den); sys = tf(sys); gure(4), bode(sys4), grid
FIGURE CP8.2 (a) Script to generate the four Bode plots.
Bode Diagram 20 40 Magnitude (dB) Phase (deg) 60 80 100 120 0
45
90
135
180 2 10
10
1
10 10 Frequency (rad/sec)
0
1
10
2
10
3
FIGURE CP8.2 CONTINUED: (b) Bode plot for G(s) =
1 (s+1)(s+10) .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
410
CHAPTER 8
Frequency Response Methods
Bode Diagram 10 20 Magnitude (dB) Phase (deg) 30 40 50 60 0
45
90 1 10
10
0
10 Frequency (rad/sec)
1
10
2
10
3
FIGURE CP8.2 CONTINUED: (c) Bode plot for G(s) =
s+10 . (s+2)(s+40)
Bode Diagram
20 30 Magnitude (dB) Phase (deg) 40 50 60 70 80 0
45
90
135
180 0 10
10 Frequency (rad/sec)
1
10
2
FIGURE CP8.2 CONTINUED: (d) Bode plot for G(s) =
1 . s2 +2s+50
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
411
Bode Diagram 20 30 Magnitude (dB) Phase (deg) 40 50 60 70 80 180
90
0
90
180 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE CP8.2 CONTINUED: (e) Bode plot for G(s) =
s7 . (s+2)(s2 +12s+50)
CP8.3
The Bode plots are shown in Figure CP8.3(a-d) with the transfer functions listed in the caption. The crossover frequency for (a) is 24 rad/sec.
Bode Diagram 20
Magnitude (dB) Phase (deg)
0
20
40
60 0
45
90
135
180 0 10
10
1
10 Frequency (rad/sec)
2
10
3
FIGURE CP8.3 (a) Bode plot for G(s) =
1000 . (s+10)(s+30)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
412
CHAPTER 8
Frequency Response Methods
The crossover frequency for (b) is 5.97 rad/sec.
Bode Diagram 40 20 Magnitude (dB) Phase (deg) 0 20 40 60 80 0 45 90 135 180 225 270 2 10
1 0 1 2
10
10 Frequency (rad/sec)
10
10
FIGURE CP8.3 CONTINUED: (b) Bode plot for G(s) =
100 . (s+0.2)(s2 +s+20)
The crossover frequency for (c) is 70.7 rad/sec.
Bode Diagram 40 30 Magnitude (dB) Phase (deg) 20 10 0 10 20 30 0
45
90
135 2 10
10
1
10 10 Frequency (rad/sec)
0
1
10
2
10
3
FIGURE CP8.3 CONTINUED: (c) Bode plot for G(s) =
50(s+100) . (s+1)(s+50)
The crossover frequency for (d) is 3.1 rad/sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
413
Bode Diagram 20 10 Magnitude (dB) Phase (deg) 0 10 20 30 40 0
45
90 1 10
10
0
10 10 Frequency (rad/sec)
1
2
10
3
10
4
FIGURE CP8.3 CONTINUED: (d) Bode plot for G(s) =
100(s2 +14s+50) . (s+1)(s+2)(s+500)
CP8.4
The m-le script and Bode plot are shown in Figure CP8.4a and b. The bandwidth is b = 10 rad/sec.
Bode Diagrams
10 0 -10 -20 Phase (deg); Magnitude (dB) -30 -40 -50 0
-50
-100
-150
-200 0 10
10
1
10
2
Frequency (rad/sec)
FIGURE CP8.4 (a) Bode plot for T (s) =
50 . s2 +5s+50
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
414
CHAPTER 8
Frequency Response Methods
numg=[50]; deng=[1 5 0]; sys_o = tf(numg,deng); sys_cl = feedback(sys_o,[1]) bode(sys_cl)
FIGURE CP8.4 CONTINUED: (b) M-le script to obtain the closed-loop Bode plot.
CP8.5
The Bode plot of the closed-loop system is shown in Figure CP8.5. The closed-loop transfer function is T (s) = s2 100 . + 6s + 100
(a) From the Bode plot we determine that Mp 5 dB and r 9 rad/sec . (b) From Equations (8.36) and (8.37) in Dorf & Bishop, we nd that 0.28 and r /n 0.92
Bode Diagrams
From: U(1) 20
0
- 20
- 40
Phase (deg); Magnitude (dB)
- 60
- 80 0
- 50
To: Y(1)
- 100
- 150
- 200 -1 10
10
0
10
1
10
2
10
3
Frequency (rad/sec)
FIGURE CP8.5 Closed-loop system Bode plot.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
415
which implies that n = r /0.92 = 9.8 rad/sec . (c) From T (s) we nd that n = 10 rad/sec and = 0.3 .
The actual values and the estimated values compare very well. CP8.6 The open-loop and closed-loop Bode plots are shown in Figure CP8.6a and b. The open-loop and closed-loop transfers functions are Gc (s)G(s) = and T (s) = Gc (s)G(s) 50 = . 2 + s + 50) + 50 1 + Gc (s)G(s) (s + 1)(s 50 (s + 1)(s2 + s + 50)
Loop transfer function; bode(syso) 0 20 Magnitude (dB) Phase (deg) 40 60 80 100 0 45 90 135 180 225 270 2 10
1 0 1 2
10
10 Frequency (rad/sec)
10
10
FIGURE CP8.6 (a) Open-loop system Bode plot for
50 . (s+1)(s2 +s+50)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
416
CHAPTER 8
Frequency Response Methods
Closedloop system; bode(sys l)
c
40 20 Magnitude (dB) Phase (deg) 0 20 40 60 80 100 0 45 90 135 180 225 270 1 10
0 1 2
10
10 Frequency (rad/sec)
10
FIGURE CP8.6 CONTINUED: (b) Closed-loop system Bode plot
10 . (s+1)(s2 +s+50)+50
CP8.7
The m-le script and plot of b versus p are shown in Figure CP8.7a and b.
p=[0:0.001:1]; w=logspace(-1,1,1000); n=length(p); for i=1:n num=[1]; den=[1 2*p(i) 0]; sys = tf(num,den); sys_cl = feedback(sys,[1]); [mag,phase,w]=bode(sys_cl,w); a= nd(mag<0.707); wb(i)=w(a(1)); end plot(p,wb) xlabel('p'), ylabel('Bandwidth (rad/sec)')
FIGURE CP8.7 (a) M-le script to generate plot of b versus p.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
1.6 1.5 1.4 1.3 Bandwidth (rad/sec) 1.2 1.1 1 0.9 0.8 0.7 0.6
417
0
0.1
0.2
0.3
0.4
0.5 p
0.6
0.7
0.8
0.9
1
FIGURE CP8.7 CONTINUED: (b) Plot of b versus p.
CP8.8
The transfer function from Td (s) to (s) is (s)/Td (s) = s3 + 10s2 0.01(s + 10) . + (0.01K 10.791)s 107.91 + 0.05K 0.1 . 107.91 + 0.05K
Using the nal value theorem and Td (s) = 1/s, we determine that
s0
lim s (s) =
The design specications require that |ess | < 0.1o . So, solving for K yields K > 3300 . We can select K = 3300 as the initial value of K for the design. The m-le script is shown in Figure CP8.8a. For the design shown, the nal selection for the gain is K = 6000. The disturbance response is shown in Figure CP8.8b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
418
CHAPTER 8
Frequency Response Methods
Mb=100; Ms=10; L=1; g=9.81; a=5; b=10; % K=6000; % Final design value of K % numg=[-1/Mb/L]; deng=[1 0 -(Mb+Ms)*g/Mb/L]; sysg = tf(numg,deng); numc=-K*[1 a]; denc=[1 b]; sysc = tf(numc,denc); % % Part (a) % sys = feedback(sysg,sysc); w=logspace(0,1,400); bode(sys,w) [mag,phase]=bode(sys,w); [M,l]=max(mag); MpDb=20*log10(M)-20*log10(mag(1)) % Mpw in decibels wr=w(l) % Mpw and peak frequency % % Part (b) % % From Eqs. (8.35) and (8.37) Mpw=10^(MpDb/20);zeta=sqrt((1-sqrt(1-(1/Mpw^2)))/2); wn=wr/sqrt(1-2*zeta^2); ts=4/zeta/wn po=100*exp(-zeta*pi/sqrt(1-zeta^2)) % % Part (c) % t=[0:0.1:10]; [y,x]=step(sys,t); plot(t,y*180/pi) xlabel('time [sec]') ylabel('theta [deg]') grid
MpDb = 4.0003 wr = 4.7226
meets specs
ts = 2.23 po = 32.75
0 -0.005 -0.01 -0.015
theta [deg]
-0.02 -0.025 -0.03 -0.035 -0.04
0
1
2
3
4
5 time [sec]
6
7
8
9
10
FIGURE CP8.8 (a) Design script. (b) Disturbance response - meets all specs!
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
419
CP8.9
A viable lter is G(s) = 0.7 (s + 1000)(s + 1) . (s + 100)(s + 10)
The Bode plot is shown in Figure CP8.9
Bode Diagram 20 Magnitude (dB) Phase (deg) 15 10 5 0 -5 90 45 0 -45 -90 -2 10
0 2 4
10
10
10
Frequency (rad/sec)
FIGURE CP8.9 (s+1000)(s+1) Bode plot for G(s) = 0.7 (s+100)(s+10)
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
9
Stability in the Frequency Domain
Exercises
E9.1 The Bode plot for the transfer function Gc (s)G(s) is shown in Figure E9.1, where Gc (s)G(s) = 2.5(1 + s/5) . s(1 + 2s)(1 + s/7 + s2 /49)
The gain and phase margins are G.M. = 20.6 dB and P.M. = 28o .
Bode Diagram Gm = 20.6 dB (at 4.57 rad/sec) , Pm = 28 deg (at 1.08 rad/sec) 50
Magnitude (dB) Phase (deg)
0
50
100 90
135
180
225
270 2 10
10
1
10 Frequency (rad/sec)
0
10
1
10
2
FIGURE E9.1 Bode Diagram for Gc (s)G(s) =
2.5(1+s/5) . s(2s+1)(s2 /49+s/7+1)
420
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
421
E9.2
The loop transfer function is Gc (s)G(s) = 8.1(1 + s/6) . s(1 + s/3)(1 + s/12)
The Bode plot is shown in Figure E9.2. The phase margin is P.M. = 48.1o at c = 5 rad/sec.
Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 48.1 deg (at 5 rad/sec) 40 20 Magnitude (dB) Phase (deg) 0 20 40 60 80 100 90
135
180 1 10
10
0
10 Frequency (rad/sec)
1
10
2
10
3
FIGURE E9.2 Bode Diagram for Gc (s)G(s) =
8.1(1+s/6) . s(s/3+1)(s/12+1)
E9.3 E9.4
The phase margin P.M. 75o at 200 kHz. We estimate the 180o phase angle at 2 MHz, so the gain margin is G.M. 25 dB. The loop transfer function is Gc (s)G(s) = 100 . s(s + 10)
The Nichols diagram is shown in Figure E9.4. When the gain is raised by 4.6 dB, Mp = 3 and the resonant frequency is R = 11 rad/sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
422
CHAPTER 9
Stability in the Frequency Domain
40 30 0.5 20 10 3 6 1 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40 K=171 ----------- K=100
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE E9.4 Nichols Diagram for Gc (s)G(s) =
K) , s(s+10)
where K = 100 and K = 171.
E9.5 E9.6
(a) The G.M. 5 dB and the P.M. 10o . (b) Lower the gain by 10 dB to obtain P.M. 60o . The Bode plot of the closed-loop transfer function is shown in Figure E9.6. The value of Mp = 3 dB. The phase margin is P.M. = 40o when K = 50.
5 0 -5 -10 -15
Gain dB
-20 -25 -30 -35 -40 -45 10-1 100 Frequency (rad/sec) 101 102
FIGURE E9.6 Closed-loop Bode Diagram for T (s) =
50(s+100) s3 +50s2 +450s+5000 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
423
E9.7
The Nyquist plot is shown in Figure E9.7 for K = 2; the plot is a circle with diameter= K/2. For K > 2, we have P = 1 and N = 1 (ccw as
Nyquist Diagram 0.5 0.4 0.3 0.2 Imaginary Axis 0.1 0 0.1 0.2 0.3 0.4 0.5 1
0.8
0.6
0.4 0.2 Real Axis
0
0.2
0.4
FIGURE E9.7 Nyquist Diagram for Gc (s)G(s) =
K s2 ,
where K = 2.
shown). So Z = N + P = 1 + 1 = 0 and the system is stable for K > 2. E9.8 (a) When K = 4, the G.M. = 3.5 dB. This is illustrated in Figure E9.8.
Bode Diagram Gm = 3.52 dB (at 1.41 rad/sec) , Pm = 11.4 deg (at 1.14 rad/sec) 50
Magnitude (dB) Phase (deg)
0
50
100
150 90
135
180
225
270 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE E9.8 Bode Diagram for Gc (s)G(s) =
K , s(s+1)(s+2)
where K = 4.
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424
CHAPTER 9
Stability in the Frequency Domain
(b) The new gain should be K = 1 for a gain margin G.M. = 16 dB. E9.9 For K = 5, the phase margin P.M. = 5o as shown in Figure E9.9.
Bode Diagram Gm = 1.58 dB (at 1.41 rad/sec) , Pm = 5.02 deg (at 1.29 rad/sec) 100 Magnitude (dB) Phase (deg) 50 0 -50 -100 -150 -90 -135 -180 -225 -270 -2 10
-1 0 1 2
10
10 Frequency (rad/sec)
10
10
FIGURE E9.9 Bode Diagram for Gc (s)G(s) =
K , s(s+1)(s+2)
where K = 5.
E9.10
The Bode plot is shown in Figure E9.10a. The closed-loop frequency
100 50
Gain dB
0 -50 -100 10-2 10-1 GM=12.35 dB
100 Frequency (rad/sec)
101
102
0
Phase deg
-100 PM=23.14 deg -200 -300 10-2 10-1 100 Frequency (rad/sec) 101 102
FIGURE E9.10 (a) Bode Diagram for Gc (s)G(s) =
326s+1304 s4 +14.76s3 +151.3s2 +23.84s .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
425
10 0 -10 -20
Gain dB
-30 -40 -50 -60 -70 10-1
100 Frequency (rad/sec)
101
102
FIGURE E9.10 CONTINUED: (b) Closed-loop frequency response: B = 6 rad/sec.
response is shown in Figure E9.10b. The bandwidth is B = 6 rad/sec. E9.11 The Bode plot is shown in Figure E9.11. The system is stable.
Bode Diagram Gm = 3.91 dB (at 3.74 rad/sec) , Pm = 14.4 deg (at 2.76 rad/sec) 100
Magnitude (dB) Phase (deg)
50
0
50
100 90
135
180
225
270 2 10
10
1
10 Frequency (rad/sec)
0
10
1
10
2
FIGURE E9.11 Bode Diagram for Gc (s)G(s) =
10(1+0.4s) . s(1+2s)(1+0.24s+0.04s2 )
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
426 E9.12
CHAPTER 9
Stability in the Frequency Domain
We select the gain K = 10 to meet the 10% steady-state tracking error specication for a ramp input. The Bode plot and Nichols chart are shown in Figures E9.12a and E9.12b, respectively.
50 0
Gain dB
GM=14.82 dB -50 -100 -150 10-1 100 101 Frequency (rad/sec) 102 103
0
Phase deg
-100 -200 -300 10-1 100 101 Frequency (rad/sec) 40 30 0.5 20 10 3 6 8 1 -1 -3 -6 -12 -20 0.25 102 103
PM=31.79 deg
0
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE E9.12 (a) Bode Diagram for Gc (s)G(s) = 10 . s(0.02s+1)(0.2s+1)
10 . s(0.02s+1)(0.2s+1)
(b) Nichols chart for Gc (s)G(s) =
E9.13
(a) The Nichols diagram is shown in Figure E9.13a and Mp = 7.97 dB. (b) The closed-loop Bode plot is shown in Figure E9.13b. The bandwidth B = 18.65 rad/sec and the resonant frequency is r = 11.69 rad/sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
427
40 30 0.5 20 10 3 6 8 1 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg) 10 0
Gain dB
-10 -20 -30 -40 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-50 -100 -150 -200 10-1 100 Frequency (rad/sec) 101 102
FIGURE E9.13 (a) Nichols Diagram for Gc (s)G(s) = 150 . s2 +5s+150
150 . s(s+5)
(b) Closed-loop Bode Diagram for T (s) =
(c) From Mp = 8 dB, we estimate = 0.2, so the expected P.O. = 52%. E9.14 (a) The peak resonance Mp = 6 dB. (b) The resonant frequency is r = 2 = 3 rad/sec. (c) The bandwidth is B = 4 = 10 rad/sec. (d) The phase margin is P.M. = 30o .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
428 E9.15
CHAPTER 9
Stability in the Frequency Domain
The loop transfer function is Gc (s)G(s) = and the closed-loop transfer function is T (s) = 100 . s + 110 100 , s + 10
The magnitude plot for the closed-loop system is shown in Figure E9.15. With bandwidth dened as frequency at which the magnitude is reduced
Bode Diagram 0
5
Magnitude (dB)
10
15
20
25 1 10
10 Frequency (rad/sec)
2
10
3
FIGURE E9.15 Magnitude plot for the closed-loop T (s) =
100 s+110 .
by 0.707 from the dc value, we determine that B = 109.7 rad/sec. E9.16 The transfer function of the approximation is G(j ) = and the magnitude is |G(j )| = (1 j )2 =1, 1 + 2 1 j , 1 + j
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
429
which is equivalent to the actual time delay magnitude. The phase approximation is = tan1 + tan1 ( ) = 2 tan1 and the actual phase is = 2 . The phase plots are shown in Figure E9.16. The approximation is accurate for < 0.5 rad/sec.
Actual _______ & Approximation ------0
-20
-40
Phase deg
-60
-80
-100
-120 10-2
10-1 Frequency (rad/sec)
100
FIGURE E9.16 Phase plots for time delay actual vs approximation.
E9.17
(a,b) The phase angle for P.M. = 30 is = 90o + tan1 2 tan1 = 150o . 2 15 2
Solving for yields = 4.7. Then, at = 4.7, we have K = 10.82 when |Gc G(j )| = K ( 2 + 4) 2 ((2 2 )2 + (15 2 )2 ) 2
1 1
=1.
The Bode plot is shown in Figure E9.17.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
430
CHAPTER 9
Stability in the Frequency Domain
Bode Diagrams Gm=3.5545 dB (at 4.3301 rad/sec), Pm=40 deg. (at 3.5147 rad/sec)
50
0
Phase (deg); Magnitude (dB)
- 50
- 100 - 50
- 100
- 150
- 200
- 250 -1 10
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE E9.17 Bode Diagram for Gc (s)G(s) =
K (s+2) , s3 +2s2 +15s
where K = 10.82.
(c) The steady-state error for a ramp is ess = where R(s) = A/s2 . E9.18 (a) The gain crossover is at c = 486 Hz, and the phase margin P.O. = 36.2o . So, 0.36. Then, the expected percent overshoot to a step input is 2 P.O. = 100e/ 1 = 30% , where = 0.36 . (b) The estimated bandwidth is B 2 (600). (c) Approximate n r = 2 (480) . Then, Ts = 4 4 = 4 ms . n (0.36)2 (480) A A = 10K = 0.60A , Kv 15
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
431
E9.19
The Bode plot is shown in Figure E9.19 for K = 16.75. The phase and gain margins are P M = 50.0o and GM = 2.72 dB.
Gm=2.7233 dB (at 20.618 r ad/sec), Pm=50 deg(at 13.434ad/sec) . r
10
0
- 10
- 20
Pha (deg); Magitude (dB) se n
- 30
- 40 0
- 100
- 200
- 300
- 400
- 500 0 10
10
1
10
2
10
3
Frequency (r d/sec) a
FIGURE E9.19 0.1s Bode Diagram for Gc (s)G(s) = K es+10 , where K = 16.75.
E9.20
The system response for both drivers is shown in Figure E9.20.
T=1 sec (solid line) & T=1.5 sec (dashed line) 1 0 -1
Automobile velocity change
-2 -3 -4 -5 -6 -7 -8
0
1
2
3
4
5 Time (sec)
6
7
8
9
10
FIGURE E9.20 Change in automobile velocity due to braking for two drivers.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
432 E9.21
CHAPTER 9
Stability in the Frequency Domain
The Bode plot is shown in Figure E9.21.
50 0
Gain dB
-50 -100 -150 10-1 100
GM=12.04 dB
101 Frequency (rad/sec)
102
103
0
Phase deg
-100 -200 -300 10-1 100 101 Frequency (rad/sec) 102 103
PM=16.85 deg
FIGURE E9.21 Bode Diagram for Gc (s)G(s) =
1300 . s(s+2)(s+50)
E9.22 E9.23
When K = 8, the P.M. = 20o ; the system is stable. Increasing the gain to K = 11.1 results in a P.M. = 45o . The Nichols chart is shown in Figure E9.23.
40 30 0.5 20 10 3 6 8 1 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE E9.23 Nichols chart for Gc (s)G(s) =
438 . s(s+2)(s+50)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
433
The actual values are Mp = 1.6598 (4.4 dB) r = 2.4228 rad/sec E9.24 Using the Nyquist criterion, we have P =1 which implies Z =N +P =1 . Therefore, the system has one root in the right half-plane. E9.25 The Bode plot is shown in Figure E9.25.
50 PM=27.73 deg at wc=8.29 rad/sec
B = 4.5834 rad/sec .
and N = 0
Gain dB
0
-50 -100 10-1
100 Frequency (rad/sec)
101
102
0
Phase deg
-100 -200 -300 10-1 100 Frequency (rad/sec) 101 102
FIGURE E9.25 Bode plot for Gc (s)G(s) =
11.7 . s(0.05s+1)(0.1s+1)
E9.26
The Nichols chart for Gc (s)G(s) = 11.7 s(0.05s + 1)(0.1s + 1)
is shown in Figure E9.26, where we nd that Mp = 6.76 dB r = 8.96 rad/sec B = 13.73 rad/sec .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
434
CHAPTER 9
Stability in the Frequency Domain
40 30 0.5 20 10 3 6 8 1 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE E9.26 Nichols chart for Gc (s)G(s) =
11.7 . s(0.05s+1)(0.1s+1)
E9.27
The Bode plot for G(s) with K = 122.62 is shown in Figure E9.27.
K=122.63
50
Gm=10.938 dB (at 6 rad/sec), Pm=40 deg. (at 2.7978 rad/sec)
0
Phase (deg); Magnitude (dB)
- 50
- 100 - 50
- 100
- 150
- 200
- 250
- 300 -1 10
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE E9.27 Bode plot for Gc (s)G(s) =
K , s(s+6)2
with K = 122.62.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
435
The phase margin is P.M. = 40.0o and the gain margin is G.M. = 10.94 dB . E9.28 The phase margin is P.M. = 28o . The estimated damping is = P.M. = 0.28 . 100
The estimated percent overshoot is 2 P.O. = 100e/ 1 = 40% . The actual overshoot is P.O. = 44.43%. E9.29 The F (s)-plane contour is shown in Figure E9.29, where F (s) = 1 + Gc (s)G(s) = s+3 . s+2
F(s)-plane 0.6
*
0.4
0.2
*
*
Im
0
*
*
*
-0.2
*
-0.4
*
-0.6
1
1.1
1.2
1.3
1.4
1.5 Re
1.6
1.7
1.8
1.9
2
FIGURE E9.29 F (s)-plane contour, where F (s) = 1 + Gc (s)G(s) =
s+3 s+2 .
E9.30
The Bode plot is shown in Figure E9.30.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
436
CHAPTER 9
Stability in the Frequency Domain
Bode Diagram 50 Magnitude (dB) Phase (deg) 0
-50
-100 0 -45 -90 -135 -180 10
-2 0 2 4
10
10
10
Frequency (rad/sec)
FIGURE E9.30 Bode plot for G(s) = C [sI A]1 B + D =
1000 . s2 +100s+10
E9.31
The Bode plot is shown in Figure E9.31. The phase margin is P.M. = 50.6 deg.
Bode Diagram Gm = Inf , Pm = 50.6 deg (at 0.341 rad/sec) 80 Magnitude (dB) Phase (deg) 60 40 20 0 -20 -40 -90
-120
-150 -3 10
10
-2
10
-1
10
0
10
1
Frequency (rad/sec)
FIGURE E9.31 Bode plot for L(s) = G(s)H (s) =
2s+1 10s2 +s .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
437
E9.32
The Bode plot is shown in Figure E9.32. The phase margin is P.M. = 45 .
Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 45 deg (at 2.56 rad/sec) 20 0 Magnitude (dB) Phase (deg) 20 40 60 80 0
45
90
135
180 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE E9.32 Bode plot for G(s) = C [sI A]1 B + D =
3.62 . s2 +s+4
E9.33
The Bode plot is shown in Figure E9.33. The phase margin is P.M. = 17.7 and the gain margin is G.M. = 5.45 dB.
Bode Diagram Gm = 5.45 dB (at 5.68 rad/sec) , Pm = 17.7 deg (at 4.24 rad/sec) 50
Magnitude (dB) Phase (deg)
0
50
100
150 0 45 90 135 180 225 270 1 10
0 1 2 3
10
10 Frequency (rad/sec)
10
10
FIGURE E9.33 Bode plot for L(s) =
200 . (s2 +2.83s+4)(s+10)
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438
CHAPTER 9
Stability in the Frequency Domain
Problems
P9.1 (a) The loop transfer function is Gc (s)G(s) = 1 . (1 + 0.5s)(1 + 2s)
P = 0, N = 0; therefore Z = N + P = 0. The system is stable. (Note: See P8.1 for the polar plots.) (b) The loop transfer function is 1 + 0.5s . s2 P = 0, N = 0, therefore Z = N + P = 0. The system is stable. (c) The loop transfer function is s2 (d) The loop transfer function is 30(s + 8) . s(s + 2)(s + 4) P = 0, N = 2 therefore Z = P + N = 2. Therefore, the system has two roots in the right half-plane, and is unstable. P9.2 (a) The loop transfer function is Gc (s)G(s) = and Gc G(j ) = K K [ 2 j (4 2 )] == . j ( 2 + j + 4) [(4 2 )2 2 + 4 ] Im{Gc G(j )} = 0 = K (4 2 ) or =2. s(s2 K , + s + 4) s+4 . + 5s + 25
P = 0, N = 0, Z = N + P = 0. Therefore, the system is stable.
To determine the real axis crossing, we let
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
439
Then, Re{Gc G(j )}=2 = K 2 4 =
=2
K . 4
(b) The loop transfer function is
So, K/4 > 1 for stability. Thus K < 4 for a stable system. K (s + 2) . s2 (s + 4)
Gc (s)G(s) =
The polar plot never encircles the -1 point, so the system is stable for all gains K (See Figure 10 in Table 9.6 in Dorf & Bishop). P9.3 (a,b) The suitable contours are shown in Figure P9.3.
jw q =cos z q Gs r r approaches in nity s -s 1 jw Gs r r approaches in nity s
(a)
(b)
FIGURE P9.3 Suitable contours s for (a) and (b).
(c) Rewrite the characteristic equation as 1+ 96 =0. s(s2 + 11s + 56)
In this case, 1 = 1. Therefore, we have one pole inside the contour at s = 0, so P = 1. The polar plot yields N = 1, so Z = N + P = 0. Therefore, all three roots have real parts less than -1. In fact, the roots are s1 = 3, and s2,3 = 4 j 4. P9.4 (a) P = 0, N = 2, therefore Z = 2. The system has two roots in the right hand s-plane. (b) In this case, N = +1 1 = 0, so Z = 0. Therefore the system is stable.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
440 P9.5
CHAPTER 9
Stability in the Frequency Domain
(a) The loop transfer function is L(s) = Gc (s)G(s)H (s) = The steady-state error is ess = |R| . 1+K K . (s + 1)(3s + 1)(0.4s + 1)
We require ess = 0.1|R|, so K > 9. (b) Use K = 9. The Nyquist plot is shown in Figure P9.5. We determine that P = 0 and N = 0. Therefore, Z = 0 and the system is stable.
8 6 4 2
Imag Axis
0 -2 -4 -6 -8 -2
0
2
4 Real Axis
6
8
10
FIGURE P9.5 Nyquist Diagram for L(s) = Gc (s)G(s)H (s) =
9 . (s+1)(3s+1)(0.4s+1)
(c) The phase and gain margins are P.M. = 18o and G.M. = 5 dB. P9.6 The rotational velocity transfer function is (s) = G(s) = R(s) 1+ K
s 3.7(2 ) s 68(2 )+1
.
At low frequency, we have the magnitude near 35 dB, so 20 log K = 35 dB
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
441
and K = 56. Since the frequency response plot is for rotational velocity (s), and we are interested in position control, we add an integrator. The characteristic equation is 1 56(23)(427) 1 + G(s) = 1 + =0. s s(s + 23)(s + 427) The roots are s1 = 430 and s2,3 = 10 j 35 .
Thus, n = 36 and = 0.28. The time constant of the closed-loop system is = P9.7 The loop transfer function is L(s) = Gc (s)G(s)H (s) = 10K1 s(s + 7) . (s + 3)(s2 + 0.36) 1 = 99.6 msec . n
(a) The Bode plot is shown in Figure P9.7 for K1 = 2.
100 50 0 -50 10-1
Gain dB
100 Frequency (rad/sec)
101
102
100 50
Phase deg
0 -50 -100 -150 10-1 100 Frequency (rad/sec) 101 102
FIGURE P9.7 Bode Diagram for Gc (s)G(s)H (s) =
10K1 s(s+7) , (s+3)(s2 +0.36)
where K1 = 2.
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442
CHAPTER 9
Stability in the Frequency Domain
(b) The phase margin P.M. = 80o and the gain margin G.M. = , since never crosses = 180o . (c) The transfer function from Td (s) to (s) is (s) = G(s) Td (s) . 1 + Gc (s)G(s)H (s)
Then, for a step disturbance () = lims0 s (s) = G(0) = 10/0.36 = 27.8, since H (0) = 0. (d) The system is so highly damped, there is very little resonant peak. (e) The estimated = P.M./100 = 0.80. The actual = 0.97. P9.8 (a) The loop transfer function is Gc (s)G(s)H (s) =
s2 2 1
+
21 s 1 s2 2 2
+1
22 s 2
(0.02s + 1)
+
+1
,
where 1 = 20 = 62.8, 2 = 14 = 43.9, 1 = 0.05 and 2 = 0.05. The Bode plot is shown in Figure P9.8a. The phase margin is P.M. = 9o . Therefore, the system is unstable.
20
Gain dB
0
-20 -40 100
101 Frequency (rad/sec)
102
103
0
Phase deg
-50 -100 -150 -200 100 101 Frequency (rad/sec) 102 103
FIGURE P9.8 (a) Bode Diagram for Gc (s)G(s)H (s) = and 2 = 14 .
2 s2 /1 +(0.1/1 )s+1 , 2 (0.02s+1)(s2 /2 +(0.1/2 )s+1)
where 1 = 20
(b) In this case 2 = 0.25, with all other parameters the same as before.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
443
10 0
Gain dB
-10 -20 -30 -40 100 101 Frequency (rad/sec) 0 102 103
Phase deg
-50 -100 -150 -200 100 101 Frequency (rad/sec) 102 103
FIGURE P9.8 CONTINUED: (b) Bode Diagram for Gc (s)G(s)H (s) = where 1 = 20 and 2 = 14 .
2 s2 /1 +(0.1/1 )s+1 , 2 (0.02s+1)(s2 /2 +(0.5/2 )s+1)
The Bode plot is shown in Figure P9.8b. The phase margin is P.M. = 86o . Therefore, the system is now stable. P9.9 (a) The Bode plot is shown in Figure P9.9a The phase margin is P.M. = 83o and the gain margin is G.M. = . (b) With the compensator, the loop transfer function is Gc (s)G(s)H (s) = K1 where K2 /K1 = 0.5 . Let K1 = 1. The Bode plot is shown in Figure P9.9b. The phase margin is P.M. = 80o and the gain margin is G.M. = , essentially the same as in (a). But the system in (b) is a type one, so that ess = 0 to a step input or disturbance. We cannot achieve a G.M. = 10 dB by increasing or decreasing K1 . 0.30(s + 0.05)(s2 + 1600)(s + 0.5) , s(s2 + 0.05s + 16)(s + 70)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
444
CHAPTER 9
Stability in the Frequency Domain
40 20
Gain dB
0 -20 -40 -60 10-3 10-2 10-1 100 Frequency (rad/sec) 0 101 102 103
Phase deg
-100 -200 -300 10-3 10-2 10-1 100 Frequency (rad/sec) 101 102 103
FIGURE P9.9 (a) Bode Diagram for Gc (s)G(s)H (s) =
0.3(s+0.05)(s2 +1600) . (s+70)(s2 +0.05s+16)
40 20
Gain dB
0 -20 -40 -60 10-3 10-2 10-1 100 101 Frequency (rad/sec) 102 103
0
Phase deg
-100 -200 -300 10-3 10-2 10-1 100 Frequency (rad/sec) 101 102 103
FIGURE P9.9 CONTINUED: (b) Bode Diagram for Gs (s)G(s)H (s) = K1 = 1.
0.15K1 (s+0.05)(s2 +1600)(s+0.5) , (s+70)(s2 +0.05s+16)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
445
P9.10
The equations of motion are F (s) = 3I (s) So, F (s) = 30 Eo (s) . (2s + 1) and I (s) = Eo (s) Eo (s) = . R + Ls 0.1 + 0.2s
The actuator without the spring (see Table 2.7, Number 9 in Dorf & Bishop) is modeled via X (s) 1 Ka = = . 2 + Bs Y (s) Ms a s 2 + s With the spring, we have Ka X (s) = 2+s+K Y (s) a s s or GA (s) = 0.4s2 1 . + s + 1.5
Then, the loop transfer function is L(s) = 30K1 . (2s + 1)(0.4s2 + s + 1.5)
(a) The Bode plot for K1 = 0.2 in Fig. P9.10 shows the P.M. = 30o .
20 0
Gain dB
-20 -40 -60 10-2 10-1 Frequency (rad/sec) 0 100 101
Phase deg
-100 -200 -300 10-2 10-1 Frequency (rad/sec) 100 101
FIGURE P9.10 Bode Diagram for L(s) =
30K1 , (2s+1)(0.4s2 +s+1.5)
where K1 = 0.2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
446
CHAPTER 9
Stability in the Frequency Domain
(b) For K1 = 0.2, we determine that Mp = 7.8 dB, r = 1.9 rad/sec, and B = 2.8 rad/sec. (c) The estimated percent overshoot is P.O. = 51% and the estimated settling time is Ts = 10 sec. This is based on = 0.21 and n r = 1.9 rad/sec. P9.11 The loop transfer function is Gc (s)G(s) = (a) Let K1 = K2 = 1. Then Gc (s)G(s) = 5(s + 1) 1.5s e . s(5s + 1) 5(K1 s + K2 )e1.5s . s(5s + 1)
The Bode plot is shown in Figure P9.11a. The phase margin is P.M. = 48o . The system is unstable. (b) Let K1 = 0.1 and K2 = 0.04. Then, the loop transfer function is Gc (s)G(s) = 5(0.1s + 0.04)e1.5s . s(5s + 1)
The Bode plot shown in Figure P9.11b shows P.M. = 45o . Thus, the system is stable.
60 40
Gain dB
20 0 -20 10-2 10-1 Frequency (rad/sec) 0 -200 100 101
Phase deg
-400 -600 -800
-1000 10-2
10-1 Frequency (rad/sec)
100
101
FIGURE P9.11 (a) Bode Diagram for Gc (s)G(s) =
5(s+1)esT s(5s+1)
, where T = 1.5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
447
40 20
Gain dB
0 -20 -40 10-2 10-1 Frequency (rad/sec) 0 -200 100 101
Phase deg
-400 -600 -800
-1000 10-2
10-1 Frequency (rad/sec)
100
101
FIGURE P9.11 CONTINUED: (b) Bode Diagram for Gc (s)G(s) =
5(0.1s+0.04)esT s(5s+1)
, where T = 1.5.
(c) When K2 = 0.1394, the phase margin is P.M. = 0o and G.M. = 0 dB. So, for stability we require K2 0.1394 when K1 = 0. P9.12 (a) The Bode plot is shown in Figure P9.12.
Bode Diagram Gm = 6.47 dB (at 2.89 rad/sec) , Pm = 29.6 deg (at 2 rad/sec) 20 Magnitude (dB) Phase (deg) 0 -20 -40 -60 -80 -100 0 -45 -90 -135 -180 -225 -270 -2 10
10
-1
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE P9.12 Bode Diagram for Gc (s)G(s) =
3.8 . (0.6s+1)3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
448
CHAPTER 9
Stability in the Frequency Domain
The loop transfer function (without the time delay) is Gc (s)G(s) = 3.8 . (0.6s + 1)3
The phase margin is P.M. = 29.6o . (b) With the delay, the loop transfer function is Gc (s)G(s)H (s) = 3.8e0.15s . (0.6s + 1)3
The phase margin is now P.M. = 12.4o . So the 0.15 sec time delay has reduced the phase margin by 50%. P9.13 The loop transfer function is Gc (s)G(s) = Ka (Ks + 1) 1.2s e . s
(a) Let Ka = K = 1. Without the time delay, the system has innite phase and gain margin. However, with the time delay, the system has a negative gain margin, hence it is unstable. (b) A plot of phase margin versus Ka is shown in Figure P9.13.
100 80 60
Phase margin deg
40 20 0 -20 -40 -60 0
0.1
0.2
0.3
0.4
0.5 Ka
0.6
0.7
0.8
0.9
1
FIGURE P9.13 Phase margin as a function of Ka for Gc (s)G(s) =
Ka (s+1)e1.2s . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
449
Let K = 1, and nd Ka for a stable system. Then, Gc (s)G(s) = Ka (s + 1)e1.2s . s
If Ka = 0.8, then the phase margin is P.M. = 50o . P9.14 The loop transfer function is Gc (s)G(s) = Ke0.2s . s(0.1s + 1)
(a) The Nichols diagram is shown in Figure P9.14 for K = 2.5.
40 30 0.5 20 10 1 2 3 6 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE P9.14 Nichols diagram for Gc (s)G(s) =
Ke0.2s , s(0.1s+1)
for K = 2.5.
It can be seen that Mp = 2.0 dB . The phase and gain margins are P.M. = 48.5o and G.M. = 7.77 dB. (b) We determine that = 0.43 (based on Mp = 2 dB) and = 0.48 (based on the phase margin P.M. = 48.5o ). (c) The bandwidth is B = 5.4 rad/sec .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
450 P9.15
CHAPTER 9
Stability in the Frequency Domain
(a) The ship transfer function is G(s) = 0.164(s + 0.2)(s 0.32) . s2 (s + 0.25)(s 0.009)
The closed-loop system is unstable; the roots are s1 = 0.5467 s2,3 = 0.2503 0.1893j s4 = 0.1949 Therefore the ship will not track the straight track. (b) The system cannot be stabilized by lowering the gain; this is veried in the root locus in Figure P9.15, where it is seen that the locus has a branch in the right half-plane for all K > 0. (c) Yes, the system can be stabilized. (d) When the switch is closed, we have a derivative feedback, which adds 90o phase lead. This is not enough to stabilize the system. Additional lead networks are necessary.
0.6
0.4
0.2
Imag Axis
0
x
o
xx
o
-0.2
-0.4
-0.6 -0.6
-0.4
-0.2
0 Real Axis
0.2
0.4
0.6
FIGURE P9.15 0.164(s+0.2)(s0.32) Root locus for 1 + GH (s) = 1 + K s2 (s+0.25)(s0.009) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
451
P9.16
The loop transfer function is Gc (s)G(s) = K . (s/10 + 1)(s2 + s + 2)
When K = 3.2, the phase margin is P.M. 30o . The Bode plot is shown in Figure P9.16.
Gm=10.88 dB, (w= 3.464) Pm=29.91 deg. (w=2.083) 50
0 Gain dB
-50
-100 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
0 -90 Phase deg -180 -270 -360 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE P9.16 Bode plot for Gc (s)G(s) =
K , (s/10+1)(s2 +s+2)
where K = 3.2.
P9.17
(a) We require ess 0.05A, and we have ess = or Kp > 19. But Kp = lim G1 (s)G2 (s)G3 (s)G4 (s) = lim
s0
A < 0.05A 1 + Kp
20K1 s0 (0.5s + 1)
0.1 1 + 4s
2
= 0.2K1 .
So, Kp = 0.2K1 > 19, or K1 > 95.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
452
CHAPTER 9
Stability in the Frequency Domain
(b) Given 1 s+1 G1 (s) = K1 (1 + ) = K1 s s ,
we require 1.05 < MPt < 1.30, or 0.70 > > 0.36, or 70o > P.M. > 36o . Then, G1 (s)G2 (s)G3 (s)G4 (s) = 0.2K1 (s + 1) . s(0.5s + 1)(4s + 1)2
When K1 = 0.8, the P.M. = 40o . The Bode plot is shown in Figure P9.17a.
50
Gain dB
0
-50
-100 10-2
10-1 Frequency (rad/sec)
100
101
0
Phase deg
-100 -200 -300 10-2 10-1 Frequency (rad/sec) 100 101
FIGURE P9.17 (a) Bode plot for G1 (s)G2 (s)G3 (s)G4 (s) = P.M. = 40o .
0.2K1 (s+1) , s(0.5s+1)(4s+1)2
where K1 = 0.8 and
(c) For part (a), we had G1 (s)G2 (s)G3 (s)G4 (s) = The characteristic equation is s3 + 2.5s2 + 1.06s + 2.50 = (s + 2.48)(s2 + 0.02s + 1.013) . The dominant complex roots are lightly damped since = 0.01 and n = 0.01. 2.375 . (s + 2)(s + 0.25)2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
453
Thus, Ts = For part (b), we had G1 (s)G2 (s)G3 (s)G4 (s) = The characteristic equation is 8s4 + 20s3 + 8.5s2 + 1.16s + 0.16 = 0 . The roots are s1 = 2, s2 = 0.4 and s3,4 = 0.05 j 0.15. Thus = 0.16 and n = 0.05. So, Ts = 4 4 = = 75 sec . n 0.05 (0.2)(0.8)(s + 1) . s(0.5s + 1)(4s + 1)2 4 = 400 sec . n
(d) Let U (s) be a unit step disturbance and R(s) = 0. Then Y (s) G3 (s)G4 (s) = = U (s) 1 + G1 (s)G2 (s)G3 (s)G4 (s) 1+
2 0.1 1+4s 20K1 (s+1) s(0.5s+1)(4s+1)2
.
The disturbance response is shown in Figure P9.17b.
x10 -3 6 5 4 3
Amplitude
2 1 0 -1 -2
0
10
20
30
40
50 Time (secs)
60
70
80
90
100
FIGURE P9.17 CONTINUED: (b) System response to a unit disturbance U (s).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
454 P9.18
CHAPTER 9
Stability in the Frequency Domain
The transfer function is Gc (s)G(s)H (s) = 5.3(s2 + 0.8s + 0.32)eT s . s3
The Bode plot is shown in Figure P9.18.
80 60
Gain dB
40 20 0 -20 10-1 100 Frequency (rad/sec) 300 T=0 solid ___ & T=0.1 dashed ---- & T=0.2674 dotted .... 101
Phase deg
250 200 150 100 10-1 100 Frequency (rad/sec) 101
FIGURE P9.18 K (s2 +0.8s+0.32)esT , where T = 0 (solid line), Bode diagram for Gc (s)G(s)H (s) = s3 T = 0.1 (dashed line), and T = 0.2674 (dotted line).
The following results are veried in the gure. (a) The phase margin is P.M. = 81o at = 5.3 when T = 0. (b) For T = 0.1, the added phase is = T (in radians). The phase margin is P.M. = 51o at = 5.3 when T = 0.1. (c) The system is borderline stable when T = 0.2674 sec. The phase margin is P.M. = 0o at = 5.3. P9.19 The transfer function is Gc (s)G(s) = 0.25 . s(1 + 4s)(3 + s)
(a) The Nichols diagram is shown in Figure P9.19. The actual gain margin is G.M. = 31.8 dB. (b) The phase margin is P.M. = 70.9o and Mp = 0 dB. The bandwidth is 0.11 rad/sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
455
Nichols Chart 40 20 0 20 40 60 80 100 120 140 160 360 0.25 dB 0.5 dB 1 dB 3 dB 6 dB 0 dB 1 dB 3 dB 6 dB 12 dB 20 dB 40 dB 60 dB 80 dB 100 dB 120 dB 140 dB 160 dB 0
OpenLoop Gain (dB)
315
270
225 180 135 OpenLoop Phase (deg)
90
45
FIGURE P9.19 Nichols diagram for Gc (s)G(s) =
0.25 . s(4s+1)(s+3)
P9.20
(a) Let K = 100. The Bode plot is shown in Figure P9.20a. The loop transfer function is Gc (s)G(s) = K (s2 + 1.5s + 0.5) . s(20s + 1)(10s + 1)(0.5s + 1)
100 50 0 -50 10-3
Gain dB
10-2
10-1 Frequency (rad/sec)
100
101
0
Phase deg
-100 -200 -300 10-3 10-2 10-1 Frequency (rad/sec) 100 101
FIGURE P9.20 (a) Bode diagram for Gc (s)G(s) =
K (s2 +1.5s+0.5) , s(20s+1)(10s+1)(0.5s+1)
where K = 100.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
456
CHAPTER 9
Stability in the Frequency Domain
(b) The phase margin is P.M. = 3.5o and the gain margin is G.M. = 2.7 dB. (c) You must decrease K below 100 to achieve a P.M. = 40o . For K = 0.1, the phase margin P.M. = 37.9o . (d) The step response is shown in Figure P9.20b for K = 0.1.
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
50
100
150
200 Time (secs)
250
300
350
400
FIGURE P9.20 CONTINUED: (b) Unit step response K = 0.1.
P9.21
The loop transfer function is Gc (s)G(s) = K . s(s + 1)(s + 4)
(a) The Bode plot is shown in Figure P9.21 for K = 4. (b) The gain margin is G.M. = 14 dB . (c) When K = 5, the gain margin is G.M. = 12 dB . (d) We require Kv > 3, but Kv = K . So, we need K > 12. This gain can 4 be utilized since K < 20 is required for stability.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
457
50 0
Gain dB
-50 -100 -150 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-100 -200 -300 10-1 100 Frequency (rad/sec) 101 102
FIGURE P9.21 Bode diagram for Gc (s)G(s) =
K , s(s+1)(s+4)
where K = 4.
P9.22
(a) The resonant frequency r = 5.2 rad/sec is point 6 on the Nichols chart. (b) The bandwidth is between points 8 and 9. We estimate the bandwidth to be B = 7.5 rad/sec. (c) The phase margin P.M. = 30o . (d) The gain margin G.M. = 8 dB. (e) Since we have P.M. = 30o , then we estimate = 0.3. We can also approximate n r = 5.2 . Thus, Ts = 4 4 = = 2.5sec . n 1.56
P9.23
The phase margin is P.M. = 60 deg when K = 97. The gain margin is G.M. = 17.8 dB . The Bode plot is shown in Figure P9.23.
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458
CHAPTER 9
Stability in the Frequency Domain
Bode Diagram Gm = 17.8 dB (at 7.07 rad/sec) , Pm = 60 deg (at 1.8 rad/sec) 40 20 Magnitude (dB) Phase (deg) 0 20 40 60 80 100 90
135
180
225
270 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE P9.23 Bode diagram for Gc (s)G(s) =
K , s(s+5)(s+10)
where K = 97.
P9.24
When K = 14.1, then P.M. = 45 deg, G.M. = dB and B = 29.3 rad/sec.
Gm=356.59 dB (at 0 rad/sec), Pm=60 deg. (at 17.321 rad/sec)
100
50
Phase (deg); Magnitude (dB)
0
- 50 - 80
- 100
- 120
- 140
- 160
- 180 -1 10
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE P9.24 Bode diagram for G(s) =
K (s+20) , s2
where K = 14.1.
P9.25
The phase margin is P.M. = 60 deg when K = 2.61 and T = 0.2 second. The Bode plot is shown in Figure P9.25.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
K=2.61; PM=60.09 at wc=2.61 rad/sec 40 20 Gain dB 0 -20 -40 -1 10
459
10
0
10 Frequency (rad/sec)
1
10
2
0
Phase deg
-500
-1000
-1500 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE P9.25 Bode diagram for Gc (s)G(s) =
Ke0.2s , s
where K = 2.61.
P9.26
The loop transfer function is Gc (s)G(s) = K . s(0.25s + 1)(0.1s + 1)
The Bode plot is shown in Figure P9.26a for K = 10. The Nichols chart is shown in Figure P9.26b. The phase and gain margins are P.M. = 9o and G.M. = 3 dB .
The system bandwidth is B = 8 rad/sec. From the P.M. = 9o , we estimate = 0.09. Therefore, the predicted overshoot is P.O. = 100e
/
1 2
= 75% , where = 0.09 .
The resonant peak occurs at r = 5.5 rad/sec. If we estimate n r = 5.5 rad/sec, then the settling time is Ts = 4 = 8 sec . n
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
460
CHAPTER 9
Stability in the Frequency Domain
50
Gain dB
0
-50 -100 10-1
100 Frequency (rad/sec)
101
102
0
Phase deg
-100 -200 -300 10-1 100 Frequency (rad/sec) 40 30 0.5 20 10 1 2 3 16 -1 -3 -6 -12 -20 0.25 101 102
0
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE P9.26 (a) Bode diagram for Gc (s)G(s) = for Gc (s)G(s) =
K , s(0.25s+1)(0.1s+1)
K , s(0.25s+1)(0.1s+1)
where K = 10. (b) Nichols chart
where K = 10.
P9.27
The loop transfer function is L(s) = Gc (s)G(s)H (s) = 4K . (s2 + 2s + 4)(s + 1)
The plot of the phase margin versus the gain K is shown in Figure P9.27. As the gain increases towards Kmax = 3.5, the phase margin decreases
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
461
towards zero.
180
160
140
120 Phase margin (deg)
100
80
60
40
20
0
1
1.5
2 K
2.5
3
3.5
FIGURE P9.27 Phase margin versus the gain K .
P9.28
The loop transfer function is Gc (s)G(s) = KP . s(s + 1)
When KP = 1.414, we have P.M. 45 . Using the approximation that P.M./100 we estimate that = 0.45. Then using the design formula 2 P.O. = 100e/ 1 = 20.5% . The actual overshoot is 23.4%. The step input response is shown in Figure ??. The actual damping ratio is = 0.42. This shows that the approximation P.M./100 is quite applicable and useful in predicting the percent overshoot.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
462
CHAPTER 9
Stability in the Frequency Domain
Step Response 1.4
1.2
1
System: syscl Peak amplitude: 1.23 Overshoot (%): 23.3 At time (sec): 2.97
Amplitude
0.8
0.6
0.4
0.2
0
0
5 Time (sec)
10
15
FIGURE P9.28 Step response showing a 23.3% overshoot.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
463
Advanced Problems
AP9.1 The loop transfer function is L(s) = Gc (s)G(s)H (s) = (a) The Bode plot for
2 n = 15267
236607.5(s + 10)(s + 5) . 2 s(s + 2)(s2 + 100s + n )(s + 1)
is shown in Figure AP9.1a.
150 100
Gain dB
50 0 -50 -100 10-3 10-2 10-1 100 101 Frequency (rad/sec) 102 103
0
Phase deg
-100 -200 -300 10-3 10-2 10-1 100 Frequency (rad/sec) 101 102 103
FIGURE AP9.1 (a) Bode Diagram for L(s) =
236607.5(s+10)(s+5) , 2 s(s+2)(s2 +100s+n )(s+1)
2 where n = 15267.
The phase and gain margins are P.M. = 48.6o and G.M. = 15.5 dB .
2 (b) The Bode plot for n = 9500 is shown in Figure AP9.1b. The gain and phase margins are
P.M. = 48.5o
and G.M. = 10.9 dB .
Reducing the natural frequency by 38% has the eect of reducing the gain margin by 30%.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
464
CHAPTER 9
Stability in the Frequency Domain
150 100
Gain dB
50 0 -50 -100 10-3 10-2 10-1 100 Frequency (rad/sec) 0 101 102 103
Phase deg
-100 -200 -300 10-3 10-2 10-1 100 Frequency (rad/sec) 101 102 103
FIGURE AP9.1 CONTINUED: (b) Bode Diagram for L(s) =
236607.5(s+10)(s+5) , 2 s(s+2)(s2 +100s+n )(s+1)
2 where n = 9500.
AP9.2
(a) The Bode plot with T = 0.05 sec is shown in Figure AP9.2a. The phase margin is P.M. = 47.7o and the gain margin is G.M. = 11.2 dB.
40 20
Gain dB
0 -20 -40 100 101 Frequency (rad/s) -100 102
Phase deg
-200
-300 -400 100
101 Frequency (rad/s)
102
FIGURE AP9.2 (s+5) (a) Bode Diagram for Gc (s)G(s)H (s) = 8 s(s+2) esT , where T = 0.05s.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
465
(b) The Bode plot with T = 0.1 sec is shown in Figure AP9.2b. The
40 20
Gain dB
0 -20 -40 100 101 Frequency (rad/s) 102
0
Phase deg
-200 -400 -600 -800 100 101 Frequency (rad/s) 102
FIGURE AP9.2 (s+5) CONTINUED: (b) Bode Diagram for Gc (s)G(s)H (s) = 8 s(s+2) esT , where T = 0.1s.
phase margin is P.M. = 22.1o and the gain margin is G.M. = 4.18 dB. A 100% increase in time delay T leads to a 50% decrease in phase and gain margins. (c) The damping ratio P.M./100 and 2 P.O. 100e/ 1 .
So, for T = 0.05 sec, 0.47 and P.O. 18.7%. Also, for T = 0.1 sec, 0.22 and P.O. 49.2%. AP9.3 The loop transfer function is L(s) = Gc (s)G(s)H (s) = 66K (1 + 0.1s) . (1 + 0.01s)(1 + 0.01s)(1 + 1.5s)(1 + 0.2s)
(a) When K = 1, the gain and phase margins are G.M. = 18.4 dB and P.M. = 55o . (b) When K = 1.5, the gain and phase margins are G.M. = 14.9 dB and P.M. = 47.8o .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
466
CHAPTER 9
Stability in the Frequency Domain
(c,d) The bandwidth and settling time with K = 1 are B = 233.6 rad/sec and Ts = 0.4 second. When K = 1.5, we determine that B = 294.20 rad/sec and Ts = 0.33 second. AP9.4 The loop transfer function is L(s) = Gc (s)G(s) = K (s + 50) . s3 + 30s2 + 200s
The gain K = 22.5 satises the specications. The actual gain and phase margins are G.M. = 22.5 dB and P.M. = 55.6o . The system bandwidth is B = 8.56 rad/sec. The step response is shown in Figure AP9.4.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (secs)
FIGURE AP9.4 Closed-loop system step response.
AP9.5
The loop transfer function is L(s) = Gc (s)G(s) = K s4 s + 0.4 . + 9s3 + 18s2
The Bode plot for K = 1 is shown in Figure AP9.5. From the phase response, we determine that the maximum P.M. 41o . From the magnitude response (for K = 1), we nd that the gain needs to be raised to
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
467
K = 14 to achieve maximum phase margin at = 0.826 rad/sec. The gain and phase margin with K = 14 are G.M. = 19.3 dB and P.M. = 40.9o . Also, the overshoot is P.O. = 38.3%.
Bode Diagram Gm = 42.3 dB (at 3.79 rad/sec) , Pm = 16.7 deg (at 0.154 rad/sec) 50
Magnitude (dB)
0
50
100
150 135
System: sys Frequency (rad/sec): 0.865 Phase (deg): 139
Phase (deg)
180
225
270 2 10
10
1
10 Frequency (rad/sec)
0
10
1
10
2
FIGURE AP9.5 s Bode plot for L(s) = K s4 +9+0.4 s2 with K = 1. s3 +18
AP9.6 AP9.7
With D > 2m, the gain can be increased up to K = 100, while still retaining stability. The loop transfer function is L(s) = Gc (s)G(s) = We select K=2 2 for P.M. = 45o . The system bandwidth is B = 5.88 rad/sec . The disturbance response is shown in Figure AP9.7. The maximum output due to a disturbance is y (t) = 0.11. K (s + 4) . s2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
468
CHAPTER 9
Stability in the Frequency Domain
0.12
0.1
0.08
Amplitude
0.06
0.04
0.02
0 0
0.5
1
1.5
2 Time (secs)
2.5
3
3.5
4
FIGURE AP9.7 Closed-loop system disturbance response.
AP9.8
A reasonable choice for the gain is K = 1450. The phase margin is P.M. = 43.2o and the percent overshoot is P.O. = 17.4%. The Nichols chart is shown in Figure AP9.8.
40 30 0.5 20 10 3 6 8 1 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE AP9.8 Nichols chart.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
469
AP9.9
The loop transfer function is L(s) = Gc (s)G(s) = Kp (s + 0.2) 2 (s2 + 7s + 10) s .
At the maximum phase margin, Kp = 4.9 for P.M. = 48.6o . The Bode diagram is shown in Figure AP9.9.
Bode Diagrams Gm=21.788 dB (at 2.9326 rad/sec), Pm=48.457 deg. (at 0.50782 rad/sec)
100
50
0
-50
Phase (deg); Magnitude (dB)
-100
-150 -100
-150
-200
-250
-300 -3 10
10
-2
10
-1
10
0
10
1
10
2
Frequency (rad/sec)
FIGURE AP9.9 Phase and gain margin.
AP9.10
The closed-loop transfer function is T (s) = s2 K . + 3s + 1
We require K = 1 a zero steady-state tracking error to a unit step. The step response is shown in Figure AP9.10. Computing T (j ) = 0.707 it follows that (j )2 1 = 0.707 + 3j + 1 or 4 + 7 2 1 = 0 .
Solving for yields = 0.37 rad/s. This is the bandwidth of the system.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
470
CHAPTER 9
Stability in the Frequency Domain
Step Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Amplitude
0
2
4
6
8 Time (sec)
10
12
14
16
FIGURE AP9.10 Unit step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
471
Design Problems
CDP9.1
The plant model with parameters given in Table CDP2.1 in Dorf and Bishop is given by: 26.035 (s) = , Va (s) s(s + 33.142) where we neglect the motor inductance Lm and where we switch o the tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). The closedloop system characteristic equation is 1+ 26.035Ka =0. s(s + 33.142)
The phase margin is P.M. = 70.4 when Ka = 16. The step response with K = 16 is shown below.
1.2
1
0.8 Amplitude
0.6
0.4
0.2
0
0
0.05
0.1
0.15 Time (secs)
0.2
0.25
0.3
DP9.1
(a) The gain and phase margins are G.M. = 7 dB and P.M. = 60o . (b) The resonant peak and frequency are Mp = 2 dB and r = 5 rad/sec. (c) We have B = 20 rad/sec. From Mp = 2 dB we estimate = 0.45 (Figure 8.11 in Dorf & Bishop). Also, r /n = 0.8, so n = 6.25. Thus, Ts = 1.4. (d) We need P.O. = 30o or = 0.3 or P.M. 30o . So, we need to raise the gain by 10 dB or K = 3.2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
472 DP9.2
CHAPTER 9
Stability in the Frequency Domain
The loop transfer function is L(s) = Gc (s)G(s) = s2 (s2 K (s + 0.5) . + 7.5s + 9)
When K = 6.25, we have the maximum phase margin. The phase margin maximum is P.M. = 23o . The plot of P.M. versus K is shown in Figure DP9.2a.
24 22 20 18 Phase Margin deg 16 14 12 10 8 6 4
0
1
2
3
4 K
5
6
7
8
9
FIGURE DP9.2 (a) Phase margin versus K for L(s) =
K (s+0.5) . s2 (s2 +7.5s+9)
The predicted damping is = 0.23. It then follows that the predicted percent overshoot is 2 P.O. = 100e/ 1 = 48% . The actual overshoot is 65%. The step input response is shown in Figure DP9.2b. The resonant peak occurs at r = 0.75 rad/sec. Approximating n r = 0.75 rad/sec, we can estimate the settling time as Ts = 4 = 23 sec . n
The actual settling time is 20 sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
473
1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0 0
5
10
15 Time (secs)
20
25
30
FIGURE DP9.2 CONTINUE: (b) Closed-loop unit step response.
DP9.3
We want to select the gain K as large as possible to reduce the steady-state error, but we want a minimum phase margin of P.M. = 45o to achieve good dynamic response. A suitable gain is K = 4.2, see Figure DP9.3.
K=4.2; PM=45.34 at wc=0.102 rad/sec 20
0 Gain dB
-20
-40 -2 10
10
-1
10 Frequency (rad/sec)
0
10
1
0 -100 Phase deg -200 -300 10
-2
10
-1
10 Frequency (rad/sec)
0
10
1
FIGURE DP9.3 Bode plot for G(s) =
Ke10s 40s+1 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
474 DP9.4
CHAPTER 9
Stability in the Frequency Domain
We are given the loop transfer function L(s) = Gc (s)G(s) = which can be written as Gc (s)G(s) = Kv . s(s + 1)(0.5s + 1) K s(s + 1)(s + 2)
The performance results are summarized in Table DP9.4.
Kv
P.M. (deg)
G.M. (dB)
B (rad/sec)
n (rad/sec)
P.O. (%)
Ts (sec)
0.325 0.450
64 56
19.3 16.5
0.54 0.74
0.54 0.63
0.71 0.56
4 12
10.5 11.4
TABLE DP9.4
Summary for Kv = 0.325 and Kv = 0.45.
When Kv = 0.45, we have ess 1 = = 2.22 , A 0.45 or twice the magnitude of the ramp. This system would be acceptable for step inputs, but unacceptable for ramp inputs. DP9.5 (a) With a time delay of T = 0.8 second, we determine that the proportional controller Gc (s) = K = 7 provides a suitable response with P.O. = 8.3 % ess = 12.5 % Ts = 4.38 sec .
(b) A suitable proportional, integral controller is Gc (s) = K1 + K2 /s = 6 + 0.6/s .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
475
The response to a unit step is P.O. = 5.14 % ess = 0 % Ts = 6.37 sec .
The Nichols chart is shown in Figure DP9.5.
40 30 0.5 20 10 3 6 8 1 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE DP9.5 Nichols chart for Gc (s)G(s) =
(K1 s+K2 )e0.8s , s(10s+1)
where K1 = 6 and K2 = 0.6.
DP9.6
With K = 170, at the two extreme values of b, we have b = 80 b = 300 P.M. = 91.62o P.M. = 75.23o G.M. = 13.66 dB G.M. = 25.67 dB .
Since reducing the value of K only increases the P.M. and G.M., a value of K = 170 is suitable to meet P.M. = 40o and G.M. = 8 dB for the range of b. DP9.7 A suitable gain is K = 0.22 . This results in P.M. = 60.17o and G.M. = 13.39 dB. The step reponse is shown in Figure DP9.7.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
476
CHAPTER 9
Stability in the Frequency Domain
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8
10 Time (secs)
12
14
16
18
20
FIGURE DP9.7 Lunar vehicle step response.
DP9.8
A gain of K = 315000 will satisfy the P.O. specication, while giving the fastest response. The step response is shown in Figure DP9.8.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.1
0.2
0.3
0.4
0.5 Time (secs)
0.6
0.7
0.8
0.9
1
FIGURE DP9.8 Steel rolling mill step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
477
DP9.9
The closed-loop transfer function is Ts (2) = where G1 (s) = and G2 (s) = 0.01 . (10s + 1)(50s + 1) 1 (10s + 1)(50s + 1) G1 (s) Gc (s)G2 (s) To (s) + T2d (s) . 1 + Gc (s)G2 (s) 1 + Gc (s)G2 (s)
The steady-state error (with Gc (s) = 500) to a unit step 2A (and after the system has settled out subsequent to a step of magnitude A) is ess = 2(0.167) = 0.33 . The step response is shown in Figure DP9.9.
Gc=500 (solid); Gc=1/s (dashed); Gc=600+6/s (dotted) 2.5
2
1.5
T2/A
1 0.5 0 0
200
400
600
800 Time (sec)
1000
1200
1400
1600
FIGURE DP9.9 Two tank temperature control step response.
A suitable integral controller is Gc (s) = 1 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
478
CHAPTER 9
Stability in the Frequency Domain
In this case, the steady-state tracking error is zero , since the system is a type 1. The system response is shown in Figure DP9.9. With the integral controller, the settling time is about Ts = 438 seconds and the P.O. = 7%. A suitable PI controller is Gc (s) = 600 + 6 . s
With the PI controller, the settling time is about Ts = 150 seconds and the P.O. = 10%. DP9.10 The system is given by x = Ax + Br y = Cx where A=
0
1
2 K1 3 K2
The associated transfer function is T (s) = s2
,
B=
0 1
, and
C=
10
.
1 . + (K2 3)s + K1 2
The characteristic polynomial is s2 + (K2 3)s + K1 2 = 0 . If we select K1 = 3, then we have a zero-steady error to a unit step response R(s) = 1/s, since
s0
lim s [1 T (s)] R(s) = lim
s2 + (K2 3)s =0. s0 s2 + (K2 3)s + K1 2 3 4.3 .
Let K=
The step response is shown in Figure DP9.10a. The bandwidth is b = 1.08 rad/s, as seen in Figure DP9.10b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
479
Step Response 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
1
2
3
4 5 Time (sec)
6
7
8
9
Bode Diagram 5
0
System: sys Frequency (rad/sec): 1.08 Magnitude (dB): 3
5
10 Magnitude (dB)
15
20
25
30
35
40 1 10
10 Frequency (rad/sec)
0
10
1
FIGURE DP9.10 Step response with K = [3 4.3] and closed-loop Bode plot.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
480
CHAPTER 9
Stability in the Frequency Domain
Computer Problems
CP9.1 The m-le script to generate the Bode plot (from which the gain and phase margin can be determined) is shown in Figure CP9.1. The transfer function is G(s) = The gain margin is G.M. = and the phase margin is P.M. = 24o . s2 100 . + 4s + 10
num=100; den=[1 4 10]; sys = tf(num,den); margin(sys);
Gm=NaN dB, (w= NaN) Pm=23.91 deg. (w=10.08) 50
Gain dB
0
-50 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
0 -90 Phase deg -180 -270 -360 -1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE CP9.1 Gain and phase margin with the margin function.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
481
CP9.2
The Nyquist plots are shown in Figures CP9.2a-c.
num=[2]; den=[1 2]; sys=tf(num,den); nyquist(sys) 0.5 0.4 0.3 0.2 Imaginary Axis 0.1 0 0.1 0.2 0.3 0.4 0.5 1
0.8
0.6
0.4
0.2
0 Real Axis
0.2
0.4
0.6
0.8
1
FIGURE CP9.2 (a) Nyquist plot for G(s) =
2 s+2 .
num=[25]; den=[1 8 16]; sys=tf(num,den); nyquist(sys) 1.5
1
0.5 Imaginary Axis
0
0.5
1
1.5 1
0.5
0
0.5 Real Axis
1
1.5
2
FIGURE CP9.2 CONTINUED: (b) Nyquist plot for G(s) =
25 . s2 +8s+16
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
482
CHAPTER 9
Stability in the Frequency Domain
num=[5]; den=[1 3 3 1]; sys=tf(num,den); nyquist(sys) 4
3
2
Imaginary Axis
1
0
1
2
3
4 2
1
0
1 Real Axis
2
3
4
5
FIGURE CP9.2 CONTINUED: (c) Nyquist plot for G(s) =
5 . s3 +3s2 +3s+1
CP9.3
The m-le script to generate the Nichols chart for part (a) is shown in Figure CP9.3a. The Nichols charts for (b) and (c) are similiarly generated; all plots are in Figure CP9.3a-c.
Nichols Chart 40 0 dB 30 0.25 dB 0.5 dB OpenLoop Gain (dB) 20 1 dB 1 dB
num = [1]; den = [1 0.2]; sys = tf(num,den); nichols(sys) ngrid
10
3 dB 6 dB 3 dB
0
6 dB
10
12 dB
20 dB 20 360 315 270 225 180 135 OpenLoop Phase (deg) 90 45 0
FIGURE CP9.3 (a) M-le script and Nichols chart for G(s) =
1 s+0.1 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
483
The gain and phase margin for each transfer function are as follows: (b) G.M. = and P.M. = (a) G.M. = and P.M. = 102o (c) G.M. = 20 dB and P.M. =
Nichols Chart 40 0 dB 30 20 10 0 10 20 30 40 50 60 360 60 dB 0 40 dB 1 dB 3 dB 6 dB 0.25 dB 0.5 dB 1 dB 3 dB 6 dB 12 dB 20 dB
OpenLoop Gain (dB)
315
270
225 180 135 OpenLoop Phase (deg)
90
45
FIGURE CP9.3 CONTINUED: (b) Nichols chart for G(s) =
1 s2 +2s+1 .
Nichols Chart 40 0 dB 20 0.25 dB 0.5 dB 1 dB 3 dB 6 dB 0 OpenLoop Gain (dB)
1 dB 3 dB 6 dB 12 dB
20
20 dB
40
40 dB
60
60 dB
80
80 dB
100 360
315
270
225 180 135 OpenLoop Phase (deg)
90
45
100 dB 0
FIGURE CP9.3 CONTINUED: (c) Nichols chart for G(s) =
24 s3 +9s2 +26s+24 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
484 CP9.4
CHAPTER 9
Stability in the Frequency Domain
To obtain a phase margin P.M. = 45 we select K=9 when T = 0.1 second. The variation in the phase margin for 0 T 0.2 is shown in Figure CP9.4.
T=[0:0.01:0.2]; K=9; num=K;den=[1 1]; sys = tf(num,den); for i=1:length(T) [mag,phase,w]=bode(sys); ph=phase(:,:,i)-w*T(i)*180/pi; [Gm,Pm,Wcg,Wcp]=margin(mag,ph,w); PMo(i)=Pm; end plot(T,PMo), grid xlabel('Time delay (sec)') ylabel('Phase margin (deg)')
100
80
60 Phase margin (deg)
40
20
0
-20
0
0.02
0.04
0.06
0.08 0.1 0.12 Time delay (sec)
0.14
0.16
0.18
0.2
FIGURE CP9.4 Variation in the phase margin for 0 T 0.2 with K = 9.
CP9.5
The loop transfer function is L(s) = Gc (s)G(s) = K (s + 50) . s(s + 20)(s + 10)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
485
The plot of system bandwidth versus the gain K is shown in Figure CP9.7.
K=[0.1:1:50]; w=logspace(-2,3,2000); den=[1 30 200 0]; for i=1:length(K) num=K(i)*[1 50]; sys = tf(num,den); sys_cl = feedback(sys,[1]); [mag,phase,w]=bode(sys_cl,w); L= nd(mag<0.707); wb(i)=w(L(1)); end plot(K,wb), grid xlabel('Gain K') ylabel('Bandwidth (rad/sec)')
15
10 Bandwidth (rad/sec) 5 0 0
5
10
15
20
25 Gain K
30
35
40
45
50
FIGURE CP9.5 Variation in the system bandwidth for 0 K 50.
CP9.6
The m-le script and Bode plot are shown in Figure CP9.6. The gain and phase margin and c are determined to be G.M. = 2.23, P.M. = 26o and c = 12.6 rad/sec. So, the maximum value of bo is found to be bomax = 2.13bo = 1.11 . In this problem, there is also a minimum value of bo . Using the Routh-
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
486
CHAPTER 9
Stability in the Frequency Domain
gm = 2.2238 pm = 26.3187 wg = 26.1155 wc = 12.6487
50
numg = -0.5*[1 0 -2500]; deng = [1 47 850 -3000]; sysg = tf(numg,deng); numc = 10*[1 3]; denc = [1 0]; sysc = tf(numc,denc); sys_o = series(sysc,sysg); bode(sys_o) [mag,phase,w] = bode(sys_o); [gm,pm,wg,wc] = margin(mag,phase,w)
Gain dB
0
-50 10-1
100
101 Frequency (rad/sec)
102
103
250
Phase deg
200 150 100 50 10-1 100 101 Frequency (rad/sec) 102 103
FIGURE CP9.6 Using the margin function to compute stability margins.
Hurwitz method, we determine that (for stability) the range of bo is 0.14 < bo < 1.11 . CP9.7 The m-le script is shown in Figure CP9.7a. Since we do not have a value for J , we write the loop transfer function as Gc (s)G(s) = K1 + K2 s s2
where K1 = K1 /J and K2 = K2 /J . We work with K1 and K2 , then we can always compute K1 and K2 whenever J is specied. A PD controller which meets the specs is given by Gc (s) = 0.04 + 0.3s . The step response is shown in Figure CP9.7b. The Bode plot is shown in
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
487
% % Part (a) % numc = [0.3 0.04]; denc = [1]; sysc = tf(numc,denc); numg = [1]; deng = [1 0 0]; sysg = tf(numg,deng); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); step(sys_cl), pause % % Part (b) % w = logspace(-1,1,400); [mag,phase] = bode(sys_o,w); [gm,pm,w1,w2] = margin(mag,phase,w); margin(mag,phase,w), pause % % Part (c) % T = [1:0.1:5]; for i = 1:length(T) [numd,dend] = pade(T(i),2); sysd = tf(numd,dend); sys_o1 = series(sysd,sys_o); sys_cl1 = feedback(sys_o1,sysd); p(:,i) = pole(sys_cl1); end plot(real(p),imag(p),'*');grid xlabel('Real Axis'); ylabel('Imag Axis')
FIGURE CP9.7 Script to assist in all three parts of the problem.
Figure CP9.7c. The phase margin is P.M. = 67.7o at = 0.32 rad/sec. The loop transfer function is Gc (s)G(s)H (s) = K1 + K2 s 2T s e 2 s
where T is the one-way time delay. If the phase lag introduced by the delay is greater than 67.7o at = 0.32 rad/sec, then the system will become unstable. So, since the phase lag due to the time delay T is ( ) = T we have 67.7o /180 = 0.32(2T ) where T = 2T . Solving for T yields T = 1.82 seconds. This is the maximum allowable one-way time delay. Executing the third part of the m-le script in Figure CP9.7a generates the plot illustrating the movement of the closed-loop system roots as the time delay is varied. The plot is shown in Figure CP9.7d. Examining the root locations, we nd that when T = 1.9, the closed-loop roots
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
488
CHAPTER 9
Stability in the Frequency Domain
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
5
10
15
20 Time (secs)
25
30
35
40
FIGURE CP9.7 CONTINUED: (b) Step response without time delays meets specs.
20
Gain dB
0 -20
-40 10-1
100 Frequency (rad/sec)
101
0
Phase deg
-100 -200 -300 10-1 100 Frequency (rad/sec) 101
FIGURE CP9.7 CONTINUED: (c) System Bode plot shows P.M. = 67.7o .
are s1 = 4.56, s2,3 = 0.94 2.02j , s4 = 0.19, and s5,6 = 0.32j . Therefore, the system is marginally stable when T = 1.9, and is unstable as the time delay increases.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
489
4
*
3 2 1
* * * * *
*
** ** ** ** ** ** ** ** ** * ** ** * * ***** ****** * * * *** ** * * ****** **** * * ** ** * ** ** ** * ** ** ** ** **
Imag Axis
0 -1 -2 -3
*
*
*
*
*
*** * * * * * * * * * * * * * * *******************
* * *
*
*
*
*
*
-4 -8
-7
-6
-5
-4
-3
-2
-1
0
1
Real Axis
FIGURE CP9.7 CONTINUED: (d) Closed-loop root locations as the time delay varies.
CP9.8
The Nyquist plot and associated Matlab code are shown in Figure CP9.8.
Nyquist Diagram 150 100 50 0 -50 -100 -150 -50
a=[0 1;-1 -10]; b=[0;22]; c=[10 0]; d=[0]; sys=ss(a,b,c,d); nyquist(sys)
Imaginary Axis
0
50
100 Real Axis
150
200
250
FIGURE CP9.8 Using the Nyquist function to obtain a Nyquist plot.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
490 CP9.9
CHAPTER 9
Stability in the Frequency Domain
The Nichols chart is shown in Figure CP9.9. The phase and gain margins are 37.1 degrees and dB, respectively.
a=[0 1;-1 -10]; b=[0;22]; c=[10 0]; d=[0]; sys=ss(a,b,c,d); nichols(sys) ngrid
Nichols Chart 60 40 Open-Loop Gain (dB) 20 0 -20 -40 -60 -360 0.25 dB 0.5 dB 1 dB 3 dB 6 dB
0 dB ?-1 dB ?-3 dB ?-6 dB ?-12 dB ?-20 dB ?-40 dB ?-60 dB
-315
-270
-225
-180
-135
-90
-45
0
O pen-Loop Phase ( de g)
FIGURE CP9.9 The Nichols chart for the system in CP9.8.
CP9.10
(a) The Nyquist plot is shown in Figure CP9.10. The phase margin is P.M. = 18o . (b) When the time delay is T = 0.05 seconds, the phase margin is P.M. = 9o . (c) When the time delay is T = 0.1 seconds, the system is marginally stable. So, for T > 0.1 seconds, the system is unstable.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
491
Nyquist Diagram 200 150 100 Imaginary Axi s 50 0 -50 -100 -150 -200 -10 -8 -6 -4 Real Axi s -2 0 -1 point
FIGURE CP9.10 Nyquist plot for G(s)H (s) =
10 . s(s+1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
10
The Design of Feedback Control Systems
Exercises
E10.1 From the design specications, we determine that our desired = 0.69 and n = 5.79. The characteristic equation is 1 + Gc (s)G(s) = 1 + or s2 + (2 + K )s + Ka = 0 . Our desired characteristic polynomial is
2 s2 + 2n s + n = s2 + 8s + 33.6 = 0 .
K (s + a) =0, s(s + 2)
Thus, K + 2 = 8, or K=6 and Ka = 33.6, so a = 5.6. The actual percent overshoot and settling time will be dierent from the predicted values due to the presence of the closed-loop system zero at s = a. In fact, the actual percent overshoot and settling time are P.O. = 12.6% and Ts = 0.87s, respectively. E10.2 The characteristic equation is 1 + Gc (s)G(s) = 1 + or 1 + K1 492 400s =0. s3 + 40s2 + 400 400 1 K1 + s(s + 40) s =1+ 400(K1 s + 1) =0, s2 (s + 40)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
493
We desire = 0.45 for an overshoot of 20%. The root locus is shown in Figure E10.2. We select a point slightly inside the performance region (dened by = 0.45 ) to account for the zero. Thus, K1 = 0.5 and the closed-loop poles are s1 = 35 s2,3 = 2.7 j 2 . The actual P.O. = 20.7% .
50 40 30 20
Imag Axis
10 0 -10 -20 -30 -40 -50 -50 -40 -30 -20 Real Axis -10 0
x * * * x o x
FIGURE E10.2 400 Root locus for 1 + K1 s3 +40s2s +400 = 0.
E10.3
The step response is shown in Figure E10.3 for = 1 and K = 0.5. It can be seen that the P.O. = 4% , so this is a valid solution.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
494
CHAPTER 10
The Design of Feedback Control Systems
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0
-0.2
0
1
2
3
4 Time (secs)
5
6
7
8
FIGURE E10.3 Step response for K = 0.5 and = 1.
E10.4
The Bode plot is shown in Figure E10.4. The phase and gain margins are marked on the plot, where it can be seen that P.M. = 75.4o and G.M. = 28.6 dB.
Bode Diagram Gm = 28.6 dB (at 11.8 rad/sec) , Pm = 75.4 deg (at 0.247 rad/sec) 150 Magnitude (dB) Phase (deg) 100 50 0 -50 -100 -150 -45 -90 -135 -180 -225 -270 -4 10 10
-2
10
0
10
2
Frequency (rad/sec)
FIGURE E10.4 Bode plot for Gc (s)G(s) =
100(s+0.15)(s+0.7) . s(s+5)(s+10)(s+0.015)(s+7)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
495
E10.5
We require that Kv 2.7, = 0.5 and n = 3 for the dominant roots. We want to place a zero to left of the pole at -2, so the complex roots will dominate. Set the zero at s = 2.2. Then for the desired roots nd the location of pole p in compensator Gc (s) = K1 (s + 2.2) (s + p)
to satisfy 180o phase at the desired roots. This yields p = 16.4. Using root locus methods, we nd that KK1 = 165.7, so with K1 = 7.53, we determine that K = 22, and Gc (s) = Then Kv = 2.78 . E10.6 The closed-loop transfer function is T (s) = 326(s + 4) Gc (s)G(s) =4 . 3 + 151.3s2 + 349.8s + 1304 1 + Gc (s)G(s) s + 14.76s 7.46(s + 2.2) . (s + 16.4)
The roots are s1,2 = 0.87 j 3.2 s3,4 = 6.5 j 8.7 . Assuming s1,2 dominates, then we expect overshoot P.O. = 43% and Ts = 4.6 sec .
The discrepencies with the actual P.O. and Ts are due to the poles s3,4 and the zero at s = 4. E10.7 The open-loop transfer function is G(s) = Ke0.5s . s(s + 7)
A plot of P.M. as a function of K is shown in Figure E10.7. It can be seen that P.M. = 45o when K = 20.88.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
496
CHAPTER 10
The Design of Feedback Control Systems
phase margin versus K (PM=45, K=20.88) 90
80
70
60 Phase Margin deg
50
40
30
20
10
0
0
5
10
15 K
20
25
30
FIGURE E10.7 Plot of phase margin versus K .
E10.8
The open-loop transfer function is G(s) = and the compensator is Gc (s) = K1 (s + z ) , s 2257 806071.4 = , s(0.0028s + 1) s(s + 357.14)
where z = K2 /K1 . The characteristic equation is s3 + 357.14s2 + K1 s + K2 = 0 . Using Routh-Hurwitz methods, the system is stable for 0 < K2 < 357.14 K1 or K2 /K1 < 357.14. Select the zero z at s = 10, then using root locus methods we determine that K1 = 0.08 and K2 = 0.8. The roots of the characteristic equation are s1 = 10.6 and s2,3 = 175 j 175 ,
and = 0.707, as desired. The step response is shown in Figure E10.8.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
497
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.05
0.1
0.15
0.2
0.25 Time (secs)
0.3
0.35
0.4
0.45
0.5
FIGURE E10.8 Step response with K1 = 0.08 and K2 = 0.8.
E10.9
The loop transfer function is L(s) = Gc (s)G(s) = and Kv = lim sGc (s)G(s) = K2 .
s0
K1 (s + K2 /K1 ) , s(s + 1)
Select K2 = 5. The characteristic equation is s2 + (K1 + 1) + K2 = 0 , and we want
2 s2 + 2n s + n = 0 . Equating coecients yields n = K2 = 5. Also, since we want P.O. = 5%, we require = 0.69. Thus,
implies K1 = 2.08 . The step response with K1 = 2.08 and K2 = 5 yields a P.O. > 5%. This
2n = K1 + 1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
498
CHAPTER 10
The Design of Feedback Control Systems
is due to the zero at s = 1.08 . So, we raise the gain K1 = 3 and then the P.O. = 5%. The step response is shown in Figure E10.9.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.5
1
1.5 Time (secs)
2
2.5
3
FIGURE E10.9 Step response with K1 = 3 and K2 = 5.
E10.10
The loop transfer function is Gc (s)G(s) = (K1 s + K2 ) . s(s + 1)(s + 2)
Let K2 = 4. Then, the plot of the phase margin as a function of K1 is shown in Figure E10.10, where it can be seen that P.M. = 53.4o is the maximum achievable phase margin. This occurs when K1 = 5.5. For K1 = 5.5 and K2 = 4 we have P.O. = 13.3% and Tp = 1.5 sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
499
55
50
Phase Margin (deg)
45
40
35
30
25 1
2
3
4
5 K1
6
7
8
9
10
FIGURE E10.10 Phase margin versus K1 with K2 = 4.
E10.11
The Nichols diagram and the closed-loop Bode plot are shown in Figures E10.11a and E10.11b, respectively.
40 30 0.5 20 10 1 2.3 0.25
0
-1 -3 -6 -12 -20
Gain dB
0 -10 -20 -30 -40
-350
-300
-250
-200
-150
-100
-50
-40 0
Phase (deg)
FIGURE E10.11 (a) Nichols diagram for Gc (s)G(s) =
1350(1+0.25s) . s(s+2)(s+30)(1+0.025s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
500
CHAPTER 10
The Design of Feedback Control Systems
Bode Diagram 20 0 Magnitude (dB) Phase (deg) 20 40 60 80 100 0 45 90 135 180 225 270 0 10
1 2 3
10
10 Frequency (rad/sec)
10
FIGURE E10.11 CONTINUED: (b) Closed-loop Bode plot.
E10.12
The loop transfer function is L(s) = Gc (s)G(s) = When KK1 = 5.12, the roots are s1,2 = 0.58 j 0.58 s3 = 3.84 . The complex poles have = 0.707 and the predicted settling time is Ts = 4/0.58 = 6.89 sec . The actual settling time is Ts = 6.22 s. KK1 s +
1 2
s2 (s + 5)
.
E10.13
For the cascade compensator, we have T1 (s) = Gc (s)G(s) 8.1(s + 1) = , 1 + Gc (s)G(s) (s + r1 )(s + r1 )(s + r2 )
where r1 = 1 + j 2 and r2 = 1.67. For the feedback compensator, we
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
501
have T2 (s) = where G(s) = and Gc (s) = s+1 . s + 3.6 8.1 s2 G(s) 8.1(s + 3.6) = , 1 + Gc (s)G(s) (s + r1 )(s + r1 )(s + r2 )
The response of the two systems dier due to dierent value of the zero of T1 and T2 , however, both systems have the same characteristic equation. E10.14 The Bode plot (with the lag network) is shown in Figure E10.14; the phase margin is P.M. = 46o .
Bode Diagram Gm = 21.9 dB (at 1.84 rad/sec) , Pm = 46.4 deg (at 0.344 rad/sec) 100 50 Magnitude (dB) Phase (deg) 0 50 100 150 90
135
180
225
270 4 10
10
3
10
2
10 Frequency (rad/sec)
1
10
0
10
1
10
2
FIGURE E10.14 Bode plot for Gc (s)G(s) =
5(7.5s+1) s(s+1)(0.25s+1)(110s+1)
= 0.
E10.15
At the desired crossover frequency c = 10 rad/sec, we have 20 log |Gc (j 10)G(j 10)| = 8.1 dB and Gc (j 10)G(j 10) = 169o .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
502
CHAPTER 10
The Design of Feedback Control Systems
Therefore, the phase margin is P.M. = 11o . So, = 30o 11o = 19o E10.16 and M = 8.1 dB .
Since > 0 and M > 0, a lead compensator is required. At the desired crossover frequency c = 2 rad/sec, we have 20 log |Gc (j 2)G(j 2)| = 17 dB and Gc (j 2)G(j 2) = 134o .
Therefore, the phase margin is P.M. = 46o . So, = 30o 46o = 16o M = 17 dB . Since < 0 and M < 0, a lag compensator is required. E10.17 Using a prelter Gp (s) = the closed-loop transfer function is T (s) = s2 KI . + (KP + 1)s + KI KI KP s + KI
The required coecients for a deadbeat system are = 1.82 and Ts = 4.82. Therefore,
2 KI = n KP = n 1 .
Since we desired a settling time less than 2 seconds, we determine that n = Ts /2 = 4.82/2 = 2.41 . Then, the gains are KP = 3.39 KI = 5.81 . The step response (with the prelter) is shown in Figure E10.17. The percent overshoot is P.O. = 0.098% and the settling time is Ts = 1.99 seconds.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
503
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.5
1
1.5
2
2.5 Time (secs)
3
3.5
4
4.5
5
FIGURE E10.17 Step response for the deadbeat system.
E10.18
Consider the PI controller Gc (s) = Kp + and the prelter Gp (s) = 10 . Then, the closed-loop system is T (s) = s2 300s + 3000 . + 280s + 3000 Kp s + KI 30s + 300 KI = = s s s
The percent overshoot is P.O. = 9.2% and the settling time Ts = 0.16 seconds. The steady-state tracking error to a unit step is zero, as desired. E10.19 Consider the PID controller Gc (s) = 29 s2 + 10s + 100 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
504
CHAPTER 10
The Design of Feedback Control Systems
The closed-loop transfer function is T (s) = 29(s2 + 10s + 100) . s3 + 24s2 + 290s + 2900
The settling time to a unit step is Ts = 0.94 seconds. E10.20 Consider the PD controller Gc (s) = KD s + Kp = 2s + 1 . The loop transfer function is L(s) = Gc (s)G(s) = 2s + 1 . s(s 1)
The Bode plot is shown in Figure E10.20. The phase margin is P.M. = 45.8 . This is a situation where decreasing the gain leads to instability. The Bode plot shows a negative gain margin indicating that the system gain can be decreased up to -6 dB before the closed-loop becomes unstable.
Bode Diagram Gm = 6.02 dB (at 0.707 rad/sec) , Pm = 45.8 deg (at 1.82 rad/sec) 50 40 Magnitude (dB) Phase (deg) 30 20 10 0 10 20 90
135
180
225
270 2 10
10
1
10 Frequency (rad/sec)
0
10
1
FIGURE E10.20 Bode plot for the loop transfer function L(s) =
2s+1 s(s1) .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
505
E10.21
The transfer function from Td (s) to Y (s) is T (s) = s2 1 . + 4.4s + K
The tracking error is E (s) = R(s) Y (s). When R(s) = 0, then E (s) = Y (s). The nal value of the output to a unit step disturbance is ess = 1/K . If we want the tracking error to be less than 0.1, then we require K > 10. When K = 10, we have the disturbance response shown in Figure E10.21.
Step Response 0.12
0.1
0.08 Amplitude
0.06
0.04
0.02
0
0
0.5
1
1.5 Time (sec)
2
2.5
3
FIGURE E10.21 Disturbance response for K = 10.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
506
CHAPTER 10
The Design of Feedback Control Systems
Problems
P10.1 (a) The loop transfer function is L(s) = Gc (s)G(s)H (s) = (1 + s)K1 K2 . (1 + s)(Js2 )
We desire = 0.6, Ts 2.5 or n 1.6. The uncompensated closedloop system is T (s) = K , s2 + K
2 where K = K1 K2 /J and K = n . We can select K = 20, and then n > 1.6. First, plot the Bode diagram for
G(s)H (s) =
20 s2
where K1 K2 /J = 20. The phase margin of the uncompensated system is 0o . We need to add phase at c . After several iterations, we choose to add 40o phase at c , so sin 40o = Therefore, = 4.6. Then, 10 log = 10 log 4.6 = 6.63dB . We determine the frequency where magnitude is -6.63 dB to be m = 6.6 rad/sec. Then, p = n = 14.1 and z = p/ = 3.07 . The compensated loop transfer function (see Figure P10.1a) is Gc (s)G(s)H (s) = 20 s2
s 3.07 s 14.1
1 = 0.64 . +1
+1 . +1
(b) Since we desire n 1.6, we place the compensator zero at z = 1.6. Then, we place the compensator pole far in the left half-plane; in this case, we selected p = 20. Thus, the compensator is Gc (s) = s + 1.6 . s + 20
The root locus is shown in Figure P10.1b. To satisfy the = 0.6 requirement, we nd K = 250, and the compensated loop transfer
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
507
100
Gain dB
50
0 -50 10-1
100 Frequency (rad/sec)
101
102
-140
Phase deg
-150 -160 -170 -180 10-1 100 Frequency (rad/sec) 101 102
FIGURE P10.1 (a) Compensated Bode plot for Gc (s)G(s)H (s) =
20(s/3.07+1) . s2 (s/14.1+1)
function is Gc (s)G(s)H (s) = 20 250(s + 1.6) =2 2 (s + 20) s s
s 1.6 + 1 s 20 + 1
.
20 15
*
10 5
Imag Axis
0 -5 -10
x
*o
x
*
-15 -20 -25
-20
-15
-10 Real Axis
-5
0
5
FIGURE P10.1 CONTINUED: (b) Root locus for Gc (s)G(s)H (s) = 1 + K s2s+1.6 . (s+20)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
508 P10.2
CHAPTER 10
The Design of Feedback Control Systems
The transfer function of the system is G(s) = s3 1.0e + 14 , + 2000s2 + 1e + 11s
where we use the system parameters given in P7.11 with the following modications: 1 = 1 = 0 and K1 = 1. Also we have scaled the transfer function so that the time units are seconds. The parameters in P7.11 are given for time in milliseconds. A suitable compensator is Gc (s) = s + 500 . s+1
The closed-loop system response is shown in Figure P10.2. The percent overshoot is P.O. 20% and the time to settle is Ts < 0.01 second.
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.005
0.01
0.015
0.02
0.025 0.03 Time (secs)
0.035
0.04
0.045
0.05
FIGURE P10.2 Step response.
P10.3
The loop transfer function is Gc (s)G(s) = 16(s + 1) K (s + z ) . s(s2 + 2s + 16) (s + p)
We desire dominant roots with Ts < 5 sec and P.O. < 5%, so use = 0.69 and n = 0.8. One solution is to select z = 1.1 (i.e. to the left of the
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
509
existing zero at s = 1) and determine the pole p and gain K for dominant roots with = 0.69. After iteration, we can select p = 100, so that the root locus has the form shown in Figure P10.3. Then, we select K = 320,
200
150
100
50 Imag Axis
0
-50
-100
-150
-200 -200
-150
-100
-50
0 Real Axis
50
100
150
200
FIGURE P10.3 16(s+1)(s+1.1) Root locus for 1 + K s(s2 +2s+16)(s+100) = 0.
so that = 0.69. The nal compensator is Gc (s) = 320(s + 1.1) . s + 100
The design specications are satised with this compensator. P10.4 The uncompensated loop transfer function is G(s) = 1
1 s2 ( 40 s +
1)
=
s2 (s
40 . + 40)
We desire 10% < P.O. < 20%, so 0.58 < < 0.65, and Ts < 2 implies n < 2. We will utilize a PD compensator Ka (s + a). We select a = 2, to obtain the root locus shown in Figure P10.4. Then with Ka = 23.5, we have the desired root location, and Gc (s) = 23.5(s + 2) . The design specications are satised with the PD compensator.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
510
CHAPTER 10
The Design of Feedback Control Systems
30
+ *
20
10
Imag Axis
0
x
+ o *
x
-10
-20
+ *
-30 -50
-40
-30
-20 Real Axis
-10
0
FIGURE P10.4 40(s+2) Root locus for 1 + Ka s2 (s+40) = 0.
P10.5
We desire P.O. < 10% and Ts < 1.5 sec. The compensator is a PI-type, given by Gc (s) = K2 + K2 s + K3 K2 (s + a) K3 = = s s s
where a = K3 /K2 . So, ess = 0 for a step input and G(s) = 3.75Ka 25Ka = . (s + 0.15)(0.15s + 1) (s + 0.15)(s + 6.67)
The loop transfer function is Gc (s)G(s) = 25Ka K2 (s + a) . s(s + 0.15)(s + 6.67)
Using root locus methods, we select a = 0.2 (after several iterations) and determine Ka K2 to yield = 0.65. This results in Ka K2 = 1. The root locus is shown in Figure P10.5. The design specications are met. The actual percent overshoot and settling time are P.O. = 7.4% and Ts = 1.3 s. The controller is Gc (s) = 1 + 0.2 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
511
20 15 10 5
+ *
Imag Axis
0 -5 -10 -15 -20 -20
x
x + o x *
+ *
-15
-10
-5
0 Real Axis
5
10
15
20
FIGURE P10.5 25(s+0.2) Root locus for 1 + Ka K2 s(s+0.15)(s+6.67) = 0.
P10.6
As in P10.5, using root locus we nd that placing z = 15 and p = 30 yields a root locus shape (see Figure P10.6) where the loop transfer function is
60
40
20
Imag Axis
+ *
0
x
+ *
o
x + *
x
-20
-40
-60 -60
-40
-20
0 Real Axis
20
40
60
FIGURE P10.6 25(s+15) Root locus for 1 + Ka (s+0.15)(s+6.67)(s+30) = 0.
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512
CHAPTER 10
The Design of Feedback Control Systems
Gc (s)G(s) =
25Ka (s + z ) . (s + p)(s + 0.15)(s + 6.67)
and where z, p and Ka are the parameters to be determined. Properly choosing the parameter values allows us to increase n of the dominant roots (compared to the PI compensator of P10.5). Then, with Ka = 3.7, the dominant roots have = 0.65. The design specications are met with the compensator. P10.7 The loop transfer function is Gc (s)G(s) = K2 (bs + 1) e0.4s s(0.2s + 1)
where a = K1 /K2 . We desire P.O. < 20%, or = 0.4 or P.M. 40o . Let K2 = 1 and nd b. A plot of P.M. versus b is shown in Figure P10.7.
90 80 70
Phase Margin (deg)
60 50 40 30 20 10 0
0.2
0.4
0.6 b
0.8
1
1.2
1.4
FIGURE P10.7 Phase margin versus b with K2 = 1.
So, many values of b will suce. We select b = 0.1. Then, Gc (s)G(s) = (0.1s + 1)e0.4s . s(0.2s + 1)
The actual P.M. = 62o . Thus, we estimate = 0.62 and P.O. = 8.3%.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
513
The actual P.O. = 4.2% and Ts = 3.25 sec. P10.8 The plant transfer function is G(s) = The steady-state error is ess = A < 0.1A . 1 + Kp e50s . (40s + 1)2
Therefore, Kp > 9. Insert an amplier with the compensator with a dc gain = 9, as follows Gc (s)G(s) = 9e50s (s + 2) . (40s + 1)2 (s + p)
The system is unstable without compensation, and it is very dicult to compensate such a time delay system with a lead compensator. Consider a lag network Gc (s) = s+z s+p
where z > p. Let z = 10p. Then, a plot of the P.M. versus p is shown in Figure P10.8a. Suitable system performance can be obtained with
(a)
150
Phase Margin (deg)
100 50 0 -50 0 0.5 1 1.5 p (b) 2 2.5 3 x10 -3
1.5
Amplitude
1 0.5 0 0
100
200
300
400 Time (secs)
500
600
700
800
FIGURE P10.8 (a) Phase margin versus p. (b) Step response with p = 0.0001 and z = 0.001.
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514
CHAPTER 10
The Design of Feedback Control Systems
P.M. > 45o , so choose p = 0.0001. The Bode plot of the compensated and uncompensated systems is shown in Figure P10.8c, where we have selected z = 0.001 and p = 0.0001. The compensated system has P.M. = 62o and Ts = 9 minutes .
The step response is shown in Figure P10.8b.
20 10
Gain dB
0 -10 -20 -30 10-4 10-3 Frequency (rad/sec) 0 -100 10-2 10-1
Phase deg
-200 -300 -400 -500 10-4 10-3 Frequency (rad/sec) 10-2 10-1
FIGURE P10.8 CONTINUED: (c) Bode plot for the compensated system (solid lines) and the uncompensated system (dashed line).
P10.9
The transfer function is G(s) = 5000 . s(s + 10)2
To meet the steady-state accuracy, we need Kv > 40. The uncompensated Kv = 50, so the steady-state accuracy can be met. (a) Using the Bode method, we need P.M. = 70% (to meet P.O. < 5% specication). Let Gc (s) = bs + 1 . as + 1
The plot of P.M. versus b is shown in Figure P10.9a, where we set a = 50b. Choosing b = 20 should satisfy the P.O. specication. The Bode plot is shown in Figure P10.9c. Thus,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
(a)
515
80
Phase Margin (deg)
70 60 50 40 30 0 5 10 15 b (b) 20 25 30
1.5
Amplitude
1 0.5 0 0
5
10
15
20
25 Time (secs)
30
35
40
45
50
FIGURE P10.9 (a) Phase margin versus b; (b) Step response for lag compensator designed with Bode where a = 1000 and b = 20.
150 100
Gain dB
50 0 -50 -100 10-4 10-3 10-2 10-1 100 Frequency (rad/sec) 101 102
0
Phase deg
-100 -200 -300 10-4 10-3 10-2 10-1 Frequency (rad/sec) 100 101 102
FIGURE P10.9 CONTINUED: (c) Bode plot for the compensated system with Gc (s) =
20s+1 1000s+1 .
Gc (s)G(s) =
5000(20s + 1) . s(s + 10)2 (1000s + 1)
The step response is shown in Figure P10.9b.
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516
CHAPTER 10
The Design of Feedback Control Systems
(b) We require that = 0.7 to meet the P.O. specications. Let Gc (s) = K (bs + 1) . (as + 1)
Using root locus methods, we x a and b, and then determine K for = 0.7. Let a = 50b and select b = 10 (other values will work). The root locus is shown in Figure P10.9d. We nd K = 2.5 when = 0.7.
20 15 10 5
Imag Axis
*
0 -5 -10 -15 -20 -20
*
x *
o x *
-15
-10
-5
0 Real Axis
5
10
15
20
FIGURE P10.9 5000(10s+1) CONTINUED: (d) Root locus for 1 + K s(s+10)2 (500s+1) .
Now, Kv = 125, so the steady-state accuracy requirement is satised for the step response as shown in Figure P10.9e.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
517
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8
10 Time (secs)
12
14
16
18
20
FIGURE P10.9 CONTINUED: (e) Step response for lag compensator designed with root locus methods, where K = 2.5.
P10.10
We desire a small response for a disturbance at 6 rad/sec. The Bode plot of Gc (s)G(s) is shown in Figure P10.10a where we consider a compensator
0 -50 -100 -150 10-1
Gain dB
100
101 Frequency (rad/sec)
102
103
0
Phase deg
-100 -200 -300 10-1 100 101 Frequency (rad/sec) 102 103
FIGURE P10.10 (a) Bode plot for the compensated system with Gc (s) =
10(s2 +4s+10) . s2 +36
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
518
CHAPTER 10
The Design of Feedback Control Systems
of the form Gc (s) = K (s2 + as + b) . s2 + 36
Notice that the magnitude is large at = 6, as desired. We select a = 4, b = 10 and K = 10 .
The response to a sinusoidal disturbance at 6 rad/sec is shown in Figure P10.10b. Notice that the eect of the disturbance is virtually eliminated in steady-state.
0.02 0.015 0.01 0.005
Amplitude
0 -0.005 -0.01 -0.015 -0.02
0
10
20
30
40
50 Time (secs)
60
70
80
90
100
FIGURE P10.10 CONTINUED: (b) Disturbance response for a sinusoidal disturbance at 6 rad/sec.
P10.11
The step response with Gc (s) = 1 is shown in Figure P10.11. A suitable lag compensator is Gc (s) = s + 0.05 . s + 0.005
The step response of the compensated system is also shown in Figure P10.11. The settling time of the compensated system is Ts = 28 seconds .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
519
Compensated system (solid) & Uncompensated system (dashed) 30
25
20
Amplitude
15
10
Input (dotted line)
5
0 0
5
10
15 Time (sec)
20
25
30
FIGURE P10.11 Step response of uncompensated and compensated systems.
P10.12
The root locus is shown in Figure P10.12 where a suitable lead-lag com-
300
200
100
Imag Axis
+
0
x
+ +
oo x +
-100
-200
-300 -300
-200
-100
0 Real Axis
100
200
300
FIGURE P10.12 160(s+17)(s+10) Root locus for 1 + K s2 (s+170)(s+1) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
520
CHAPTER 10
The Design of Feedback Control Systems
pensator is Gc (s) = K s + 10 s + 17 . s + 1 s + 170
The selected gain is K = 57, so that the damping of the complex roots is about = 0.7. For this particular design, the closed-loop system zeros will aect the system response and the percent overshoot specication may not be satised. Some design iteration may be necessary or a prelter can be utilized. A suitable prelter is Gp (s) = 17 . s + 17
The acceleration constant is Ka = 9120. P10.13 We choose K = 10. This yields a velocity constant Kv = 20K = 200, as desired. A suitable two-stage lead compensaator is Gc (s) = (0.05s + 1)(0.05s + 1) . (0.0008s + 1)(0.0008s + 1)
The Bode plot is shown in Figure P10.13. The phase margin is P.M. = 75.06o .
100 50 Phase margin=75.06 deg
Gain dB
0 -50 -100 10-1 100 101 102 103 104
Frequency (rad/sec) 0
Phase deg
-100 -200 -300 10-1 100 101 102 103 104
Frequency (rad/sec)
FIGURE P10.13 200(0.05s+1)2 Bode plot for s(0.1s+1)(0.05s+1)(0.0008s+1)2 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
521
P10.14
(a) When Gc (s) = K = 0.288 , the phase margin is P.M. = 49.3o and the bandwidth is B = 0.95 rad/sec. (b) A suitable lag compensator is Gc (s) = 25s + 1 . 113.6s + 1
The compensated system phase margin is P.M. = 52.21o and Kv = 2, as desired. P10.15 A suitable lead compensator is Gc (s) = 1.155s + 1 . 0.032s + 1
The compensated system phase margin is P.M. = 50o and Kv = 2, as desired. The settling time is Ts = 3.82 seconds. P10.16 One possible solution is Gc (s) = K (s + 12)(s + 15) , (s + 120)(s + 150)
where K = 900. The disturbance response is shown in Figure P10.16.
Step Response 0.1
0.08
Amplitude
0.06
0.04
0.02
0
0
0.1
0.2
0.3 Time ( sec )
0.4
0.5
0.6
FIGURE P10.16 Compensated system disturbance response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
522 P10.17
CHAPTER 10
The Design of Feedback Control Systems
The PI controller is given by Gc (s) = K (s + b) , s
where K and b are to be determined. To meet the design specications, we need = 0.6 and n = 6.67 rad/sec .
The closed-loop transfer function is T (s) = K (s + b) . s2 + Ks + bK
2 Solving for the gains yields K = 2n = 8 and b = n /K = 5.55. A suitable prelter is
Gp (s) =
5.55 . s + 5.55
The step response, with and without the prelter, is shown in Figure P10.17.
Without pre lter (solid) & with pre lter (dashed) 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.5
1
1.5 Time (sec)
2
2.5
3
FIGURE P10.17 Compensated system response with and without a prelter.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
523
P10.18
The plant transfer function is G(s) = K . s(s + 10)(s + 50)
We desire n > 10 to meet Ts < 0.4 sec and = 0.65 to meet P.O. < 7.5%. Try a pole at s = 120. The root locus is shown in Figure P10.18. The gain K = 6000 for = 0.65. Thus, Gc (s)G(s) = 6000(s/15 + 1) s(s + 10)(s + 50)(s/120 + 1) and Kv = 6000 = 12 . 500
200 150 100 50
Imag Axis
0 -50 -100 -150 -200 -200
*x
x
*
* ox x *
-150
-100
-50
0 Real Axis
50
100
150
200
FIGURE P10.18 s/15+1 Root locus for 1 + K s(s+10)(s+50)(s/120+1) .
P10.19
(a) The loop transfer function is L(s) = K1 e2T s 0.25s + 1
where T = 1.28. The phase angle is = 2.56 tan 0.25 . So, = 1.12 rad/sec when = 180o . However, the break frequency
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
524
CHAPTER 10
The Design of Feedback Control Systems
is 4 rad/sec. Therefore, you cannot achieve P.M. = 30o and have the system be stable for K1 < 1. The steady-state error is ess = since K1 = Kp . (b) Set K1 = 20, then Kp = 20 and this yields a 5% steady-state error. Without compensation, the system is now unstable. Let Gc (s) = s/b + 1 s/a + 1 A A = 1 + Kp 1 + K1
where b = 5 and a = 0.01. Then, the system is stable with P.M. = 63o . The system response is shown in Figure P10.19.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0
-0.2
0
2
4
6
8
10 Time (secs)
12
14
16
18
20
FIGURE P10.19 Unit step response with Gc (s) =
20(s/5+1) s/0.01+1 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
525
P10.20
(a) The open-loop transfer function is G(s) = KesT , (s + 1)(s + 3)
where T = 0.5 sec. We desire P.O. < 30%, thus > 0.36. We will design for = 0.4, which implies P.M. = 40o . Then = tan1 tan1 0.5 (57.3o ) . 3
At c = 1.75, the phase margin is P.M. = 40o , and solving |G(j )| = K [(3 2 )2 + (4 )2 ] 2
1
=1
at = 1.75 yields K = 7. Then ess = 0.3. (b) We want ess < 0.12, so use ess = 0.10 as the goal. Then Gc (s)G(s) = and ess = where Kp =
2K 3b .
Ke0.5s (s + 2) , (s + 1)(s + 3)(s + b)
1 1 + Kp
If b = 0.1 then Kp = 6.7K and ess = 1 . 1 + 6.7K
So, we need 6.7K = 9, or K = 1.35. We need a lag compensator (i.e. b < 2) to meet ess < 12% and have stability. P10.21 We desire Kv = 20, P.M. = 45o and B > 4 rad/sec. Thus, we set K = 20, and G(s) = s
s 2
20 +1
s 6
+1
.
Then, the Bode plot yields P.M. = 21o uncompensated at c = 5.2 rad/sec. The phase lead compensator must add 66 o plus phase lead to account for the shift of the crossover to a higher frequency with the phase lead compensator. Consider Gc (s) = 1 + s 1 + s
2
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
526
CHAPTER 10
The Design of Feedback Control Systems
One solution is to use = 10 = 1/67 . Then Gc (s) = 100(s + 6.7)2 . (s + 67)2
The compensator has two zeros at = 6.7, two poles at = 67 yielding P.M. = 47o , c = 7.3 and B = 12 rad/sec. P10.22 We desire Kv = 20, P.M. = 45o and B 2. The lag compensator is Gc (j ) = 1 + j 1 + j
where > 1. From the Bode plot, = 135o at 1.3. So, at = 1.3, = we need to lower the magnitude by 22 dB to cause = 1.3 to be c , the new crossover frequency. Thus, solving 22 = 20 log
yields = 14. We select the zero one decade below c or Therefore, 1
= 0.13.
1 0.13 = = 0.0093 . 14 Then, the lag compensator is given by Gc (s) = 1 + 0.s s + 0.13 13 . = s 1 + 0.0093 14(s + 0.0093)
The new crossover is c = 1.3, and B = 2.14 rad/sec.
P10.23
We desire P.M. = 45o , Kv = 20 and 2 B 10. The lead-lag compensator is Gc (s) =
s 1 + s 1 + 10a b s s. 1 + 10b 1 + a
Since B 1.5c , we design for a new crossover frequency c so that = 1.4 < c < 7 . Try for c = 4. The phase = 190o at = 4, so we need to add phase lead of 55o plus phase to account for lag part of network at c . Use = 10 and bracket = 4 with the lead network. Put the zero at = 0.8 = b
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
527
and the pole at = 8. For the lag compensator, put the zero at a lower frequency than c /10. So try a zero at = 0.2 = 10a and a pole at = 0.02 = a. Then, the lead-lag compensator is Gc (s) = 1 + 0s8 1 + 0s2 . . . s 1 + 8 1 + 0.s 02
The compensated Bode plot yields
c = 3.5 rad/sec,
P.M. = 50o
and B = 6.2 rad/sec .
P10.24
The steady-state error is ess = 1 1 = = 0.05 . 1 + Kp 1 + K/16
So, we need K/16.geq 19 or K 304. One possible solution is Gc (s) = 4s + 1 . 12s + 1
The compensated Bode plot is shown in Figure P10.24. The phase margin is P.M. = 45.8o .
Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 45.8 deg (at 9.24 rad/sec) 30 20 Magnitude (dB) Phase (deg) 10 0 10 20 30 40 0
45
90
135
180 3 10
10
2
10 10 Frequency (rad/sec)
1
0
10
1
10
2
FIGURE P10.24 Bode plot for GGc (s) =
304(4s+1) . (s+4)2 (12s+1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
528 P10.25
CHAPTER 10
The Design of Feedback Control Systems
The arm-rotating dynamics are represented by G(s) = 100 s
s2 6400 s 50
+
+1
.
We desire Kv = 20, and P.O. < 10%. One possible solution is the lead-lag compensator Gc (s) = (s + 50)(s + 0.48) . 5(s + 400)(s + 0.06)
With this compensator, we have P.O. = 9.5% P10.26 and Kv = 20 .
Neglect the pole of the airgap feedback loop at s = 200. The characteristic equation is 1+K where K= K K1 + K2 K2 b c= . K1 + K2 (s + 20)(s + c) =0, s3
Choose c = 10 to attain the root locus structure shown in Figure P10.26. The gain K = 38.87 insures the damping ratio of = 0.5. Then, solving for K1 and b yields K1 = and b= 0.1K . 38.87K2 K K2 38.87
For given values of K and K2 (unspecied in the problem), we can compute K1 and b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
529
40 30
+
20 10
Imag Axis
0 -10 -20
o
o+
x
+
-30 -40 -40
-30
-20
-10
0 Real Axis
10
20
30
40
FIGURE P10.26 (s+20)(s+10) = 0. Root locus for 1 + K s3
P10.27
The loop transfer function is Gc (s)G(s) = 0.15K (10as + 1) , s(s + 1)(5s + 1)(as + 1)
where K and a are to be selected to meet the design specications. Suitable values are K = 6.25 and a = 0.15 .
Then, the phase margin is P.M. = 30.79o and the bandwidth is B = 0.746 rad/sec. The lead compensator is Gc (s) = 6.25 P10.28 1.5s + 1 . 0.15s + 1
(a) Let Gc (s) = K = 11. Then the phase margin is P.M. = 50o and the performance summary is shown in Table P10.28. (b) Let Gc (s) = K (s + 12) , (s + 20)
where K = 32. Then, the phase margin is P.M. = 50o and the performance summary is given in Table P10.28.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
530
CHAPTER 10
The Design of Feedback Control Systems
compensator
P.M.
P.O.
Tp
Ts
Mp
B
Gc (s) = K = 11
32(s+12) s+20
50o 50o
18%
0.34 sec
0.78 sec
1.5 dB
13.9 rad/sec
Gc (s) =
18%
0.20 sec
0.47 sec
1.5 dB
26.3 rad/sec
TABLE P10.28
Performance Summary.
P10.29
The loop transfer function is Gc (s)G(s) = K (as + 1) , s(s + 10)(s + 14)(10as + 1)
where K and a are to be selected to meet the design specications, and we have set = 10. The root locus is shown in Figure P10.29a. To satisfy
30
20
10
Imag Axis
*
0
*
x
x *
*ox
-10
-20
-30 -30
-20
-10
0 Real Axis
10
20
30
FIGURE P10.29 1400(s+1) (a) Root locus for 1 + K s(s+10)(s+14)(10s+1) = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
531
the steady-state tracking error we must select K > 1400 . Suitable values for the lag compensator are K = 4060 and a = 1 .
Then, the percent overshoot is P.O. = 31% and the settling time is Ts = 2.34 sec. The lag compensator is Gc (s) = s+1 . 10s + 1
The step response is shown in Figure P10.29b.
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.5
1
1.5 Time (secs)
2
2.5
3
FIGURE P10.29 CONTINUED: (b) Step response.
P10.30
The plant transfer function is G(s) = The lead network Gc (s) = 16(s + 0.7) (s + 9) 10e0.05s . s2 (s + 10)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
532
CHAPTER 10
The Design of Feedback Control Systems
provides Mp = 3.4 dB and r = 1.39 rad/sec. The step response is shown in Figure P10.30. The overshoot is P.O. = 37% and Ts = 3.5 sec.
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
1
2
3 Time (secs)
4
5
6
FIGURE P10.30 Unit step response with Gc (s) =
16(s+0.7) . s+9
P10.31
The vehicle is represented by G(s) = K K . s(0.04s + 1)(0.001s + 1) s(0.04s + 1)
For a ramp input, we want ess 1 = 0.01 = . A Kv So, let G(s) = 100 . s(0.04s + 1)
The uncompensated P.M. = 28o at c = 47 rad/sec. We need to add 17o . Case (1) Phase lead compensation: Gc (s) = 1 + 0.021s . 1 + 0.01s
The phase margin is P.M. = 45o .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
533
Case (2) Phase lead compensation: Gc (s) = 1 + 0.04s . 1 + 0.005s
The phase margin is P.M. = 65o . For Case 1, we have P.O. = 25% , For Case 2, we have P.O. = 4% , P10.32 Ts = 0.04 sec and Tp = 0.03 sec . Ts = 0.13 sec and Tp = 0.05 sec .
As in P10.31, the plant is given by G(s) = 100 . s(0.04s + 1)
The uncompensated P.M. = 28o . We need P.M. = 50o . The phase lag compensator Gc (s) = 1 + 0.5s 1 + 2.5s
results in P.M. = 50o . The P.O. = 21%, Ts = 0.72 sec and Tp = 0.17 sec. P10.33 (a) To obtain Kv = 100, we have Gc (s)G(s) = 43.33(s + 500) . s(s + 0.0325)(s2 + 2.57s + 6667)
With K = 43.33, we have P.M. = 1.2o , Mp = 26 dB , r = 1.8 rad/sec and B = 3.7 rad/sec .
The Bode plot is shown in Figure P10.33. (b) Let Gc (s) = 0.35s + 1 , 0.001s + 1
and K = 43.33 (as before). Then, P.M. = 36o , Mp = 5.4 dB , r = 1.7 rad/sec and B = 3.0 rad/sec .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
534
CHAPTER 10
The Design of Feedback Control Systems
100 50
Gain dB
0 -50 -100 -150 10-2 10-1 100 101 102 103
Frequency (rad/sec) -100 -150
Phase deg
-200 -250 -300 -350 10-2 10-1 100 101 102 103
Frequency (rad/sec)
FIGURE P10.33 Bode plot with Gc (s) = K = 43.33.
P10.34
The step response is shown in Figure P10.34, where Gc (s) = 10(s + 0.71)(s + 0.02) . (s + 0.0017)(s + 10)
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.5
1
1.5
2
2.5 Time (secs)
3
3.5
4
4.5
5
FIGURE P10.34 Step response with the lead-lag compensator Gc (s) =
10(s+0.71)(s+0.02) . (s+0.0017)(s+10)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
535
Then, Kv = 80 and P.O. = 17%, Ts = 1.8 sec, and = 0.54. P10.35 The process model is G(s) = s2 (s 1 , + 10)
and we consider the lead compensator Gc (s) = K 1 + s , 1 + s
where = 100, = 0.4 and K = 0.5. Then, P.M. = 46.4o . The step response is shown in Figure P10.35. The system performance is P.O. = 22.7% Ts = 5.2 sec Tp = 1.72 sec .
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5 Time (secs)
6
7
8
9
10
FIGURE P10.35 40 Step response with the lead compensator Gc (s) = 0.5 0.4s+1 . s+1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
536 P10.36
CHAPTER 10
The Design of Feedback Control Systems
The phase margin is shown in Figure P10.36. As the time delay increases, the phase margin decreases. The system is unstable when T > 2.1843 s.
140
120
100
Phase margin (deg)
80
60
40
Stability boundary
20
0
20
0
0.5
1 Time delay (s)
1.5
2
T=2.1843 s
2.5
FIGURE P10.36 Step response with Gc (s)G(s) =
2s+0.54 T s e , s(s+1.76)
where 0 T 2.5.
P10.37
One possible solution is the integral controller Gc (s) = 2/s. The step response is shown in Figure P10.37. The steady-state tracking error to a
1.6 1.4 1.2 1
Amplitude
0.8 0.6 0.4 0.2 0 -0.2
0
1
2
3
4
5 Time (secs)
6
7
8
9
10
FIGURE P10.37 Step response with the integral controller Gc (s) = 2/s.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
537
step input is zero since the system is type-1. The phase margin is P.M. = 32.8 and the bandwidth is B = 4.3 rad/s . P10.38 A suitable PI controller is Gc (s) = K s+a s + 0.1 = 1720 . s s
The roots are s1 = 29.15, s2,3 = 5.37 5.37j , and s4 = 0.10. The step response is shown in Figure P10.38. The settling time is Ts = 1.09 s and the time to peak is Tp = 0.618 s.
Step Response 1.4
System: sysa Peak amplitude: 1.06 Overshoot (%): 6.39 At time (sec): 0.618
1.2
1
System: sysa Settling Time (sec): 1.09
Amplitude
0.8
0.6
0.4
0.2
0
0
1
2
3
4 5 Time (sec)
6
7
8
9
FIGURE P10.38 Step response with the PI controller.
P10.39
One possible solution is Gc (s) = 51.78(s + 0.1) . s + 0.025
The overshoot to a unit step is P.O. = 0% and the steady-state error to a step input is ess = 5%. The system bandwidth is B = 6.5 rad/sec. P10.40 The lead compensator is Gc (s) = 2.88(s + 2.04) . s + 5.88
The Bode plot is shown in Figure P10.40. The phase margin is P.M. = 30.4o at c = 9.95 rad/sec and the bandwidth is B = 17.43 rad/sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
538
CHAPTER 10
The Design of Feedback Control Systems
60 40
Gain dB
20 0 -20 -40 10-1 100 Frequency (rad/sec) 0 101 102
Phase deg
-100 -200 -300 10-1 100 Frequency (rad/sec) 101 102
FIGURE P10.40 Bode plot for Gc (s)G(s) =
115.29(s+2.04) . s(s+2)(s+5.88)
P10.41
The lag compensator is Gc (s) = 1 + 1.48s . 1 + 11.08s
100
Gain dB
50 0
-50 10-3
10-2
10-1 Frequency (rad/sec)
100
101
0
Phase deg
-100 -200 -300 10-3 10-2 10-1 Frequency (rad/sec) 100 101
FIGURE P10.41 Bode plot for Gc (s)G(s) =
40(1+1.48s) . s(s+2)(1+11.08s)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
539
The Bode plot is shown in Figure P10.41. The steady-state error specication is satised since Kv = 20. The phase margin is P.M. = 28.85o at c = 2 rad/sec and the bandwidth is B = 3.57 rad/sec. P10.42 The lag compensator is Gc (s) = 2.5(1 + 1.64s) . 1 + 30.5s
The Bode plot is shown in Figure P10.42. The steady-state error specication is satised since Kv = 50 . The phase margin is P.M. = 28.93o at c = 1.98 rad/sec and the bandwidth is B = 3.59 rad/sec.
100
Gain dB
50
0
-50 10-3
10-2
10-1 Frequency (rad/sec)
100
101
0
Phase deg
-100 -200 -300 10-3 10-2 10-1 Frequency (rad/sec) 100 101
FIGURE P10.42 Bode plot for Gc (s)G(s) =
100(1+1.64s) . s(s+2)(1+30.5s)
P10.43
We use Table 10.2 in Dorf & Bishop to determine the required coecients = 1.9 Also, n Tr = 4.32 implies n = 4.32 , and = 2.2 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
540
CHAPTER 10
The Design of Feedback Control Systems
since we require Tr = 1 second. The characteristic equation is s3 + 8.21s2 + 41.06s + 80.62 = s3 + (1 + p)s2 + (K + p)s + Kz = 0 . Equating coecients and solving yields p = 7.21 P10.44 K = 33.85 z = 2.38 .
From Example 10.4 in Dorf & Bishop, we have the closed-loop transfer function T (s) = A suitable prelter is Gp (s) = 4 . s+4 (s2 96.5(s + 4) . + 8s + 80)(s + 4.83)
The step response (with and without the prelter) is shown in Figure P10.44.
With pre lter (solid) & without pre lter (dashed) 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE P10.44 Step response with and without the prelter.
P10.45
Let K = 100. The Bode plot is shown in Figure P10.45a and the response to a simusoidal noise input with = 100 rad/s is shown in Figure P10.45b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
541
Bode Diagram 60
40
20
Magnitude (dB)
0
20
System: sysg Frequency (rad/sec): 100 Magnitude (dB): 40.1
40
60
80 1 10
10
0
10 Frequency (rad/sec)
1
10
2
10
3
0.07 0.06 0.05 0.04 Amplitude 0.03 0.02 0.01 0 0.01
0.02
0
1
2
3
4 Time (sec)
5
6
7
8
FIGURE P10.45 (a) Bode magnitude plot. (b) Response to a noise input.
P10.46
For K > 0.635, the system is unstable. The percent overshoot is shown in Figure P10.46 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
542
CHAPTER 10
The Design of Feedback Control Systems
90 80 70 60 Percent Overshoot 50 40 30 20 10 0 10
0
0.1
0.2
0.3 K
0.4
0.5
0.6
0.7
FIGURE P10.46 Percent overshoot.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
543
Advanced Problems
AP10.1 (a) With Gc (s) = K , the closed-loop transfer function is T (s) = s3 + 5s2 K . + 4s + K
When K = 2.05, the characteristic equation is s3 + 5s2 + 4s + 2.05 = 0 with poles at s = 4.1563 and s = 0.4219 j 0.5615. Therefore = 0.6, and the predicted overshoot is P.O. = 100e0.6/
10.62
= 9.5% < 13% .
The actual overshoot is P.O. = 9.3% and Ts = 8.7 seconds. (b) When Gc (s) = 82.3(s + 1.114) s + 11.46
the closed-loop transfer function is T (s) = 82.3(s + 1.114) + + 61.3s2 + 128.14s + 91.6822 82.3(s + 1.114) = . (s + 1.196)(s + 12.26)(s + 1.5 j 2) s4 16.46s3
Therefore = 0.6 and the predicted overshoot is P.O. = 9.5% < 13%. The actual overshoot is P.O. = 12% and Ts = 2.5 seconds. AP10.2 The lag network is given by Gc = K (s + a1 ) . s + a2
The closed-loop transfer function is T (s) = K s4 + (5 + a2 )s3 s + a1 . + (4 + 5a2 )s2 + (4a2 + K )s + Ka1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
544
CHAPTER 10
The Design of Feedback Control Systems
Computing the steady-state tracking error yields ess = lim s4 + (5 + a2 )s3 + (4 + 5a2 )s2 + 4a2 s s0 s5 + (5 + a2 )s4 + (4 + 5a2 )s3 + (4a2 + K )s2 + Ka1 s 4a2 = < 0.125 . a1 K
If we select K = 2.05 (as in AP10.1), then a1 > 15.61a2 . So, take a2 = a1 /16. The lag compensator can now be written as Gc (s) = 2.05 s + a1 . s + a1 /16
Select a1 = 0.018. Then, the closed-loop transfer function is T (s) = s4 + 5.0011s3 2.05(s + 0.018) . + 4.0056s2 + 2.0545s + 0.0369
The performance results are P.O. = 13% and Ts = 29.6 seconds for a step input, and ess = 0.12 for a ramp input. AP10.3 The plant transfer function is G(s) = and the PI controller is given by Gc (s) = Kp s + KI . s 1 s(s + 1)(s + 4)
The closed-loop transfer function is T (s) = s4 + 5s3 Kp s + KI . + 4s2 + Kp s + KI
For a unit ramp, the steady-state tracking error is ess = lim s4 + 5s3 + 4s2 =0. s0 s5 + 5s4 + 4s3 + Kp s2 + KI s
Any KI > 0 and Kp > 0 (such that the system is stable) is suitable and will track a ramp with zero steady-state error. Since we want P.O. < 13%, the damping of the dominant roots should be 0.6. One suitable
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
545
solution is to place the zero at s = 0.01 and select the PI controller Gc (s) = 2.05(s + 0.01) . s
Therefore, Kp = 2.05 and KI = 0.0205. The closed-loop transfer function is T (s) = s4 + 5s3 2.05(s + 0.01) . + 4s2 + 2.05s + 0.0205
The performance results are P.O. = 11.5% and Ts = 9.8 seconds for a step input, and ess = 0 for a unit ramp. AP10.4 The closed-loop transfer function is T (s) = 10K1 . s2 + 10(1 + K1 K2 )s + 10K1
From the performance specications, we determine that the natural frequency and damping of the dominant poles should be n = 5.79 and = 0.69. So,
2 s2 + 10(1 + K1 K2 )s + 10K1 = s2 + 2n s + n = s2 + 7.99s + 33.52 .
Solving for the gains yields K1 = 3.35 and K2 = 0.06. The closed-loop transfer function is T (s) = 33.52 . s2 + 7.99s + 33.52
The performance results are P.O. = 5% and Ts = 1 second. AP10.5 (a) From the overshoot specication P.O. = 16%, we require the dominant poles to have a damping 0.5. Neglecting the pole at s = 100, the plant transfer function is G(s) = 1 . s(s + 1)
Let Gp = 1. A suitable compensator is Gc = K s+2 . s + 10
Using root locus methods, we determine that K = 86.2 yields = 0.5. The closed-loop poles are s = 4.37 j 7.57 and s = 2.254. (b) The closed-loop transfer function is T (s) = s4 + 111s3 8620(s + 2) . + 1110s2 + 9620s + 17240
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
546
CHAPTER 10
The Design of Feedback Control Systems
The step response is shown in Figure AP10.5. The overshoot and settling time are P.O. = 30% and Ts = 1.2 seconds. (c) A suitable prelter is Gp (s) = The closed-loop transfer function is T (s) = s5 + 114s4 25860(s + 2) . + 1443s3 + 12950s2 + 46100s + 51720 3 . s+3
The step response is shown in Figure AP10.5. The overshoot and settling time are P.O. = 2% and Ts = 1.15 seconds.
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.2
0.4
0.6
0.8
1 Time (sec)
1.2
1.4
1.6
1.8
2
FIGURE AP10.5 Step response with prelter (dashed line) and without prelter (solid line).
AP10.6
From Example 10.12 in Dorf & Bishop, we have the relationship n Ts = 4.04 . Thereore, minimizing Ts implies maximizing n . Using Table 10.2 in Dorf & Bishop, we equate the desired and actual characteristic polynomials
2 3 q (s) = s3 + 1.9n s2 + 2.2n s + n = s3 + (1 + p)s2 + (K + p)s + Kz .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
547
Comparing coecients yields (1 + p) = 1.9n , K + p = 2.2 1+p 1.9
2
,
3 Kz = n .
So, from the rst relationship we see that maximizing n implies maximizing p. Solving for p while maintaining K < 52 K= we determine that 9.3643 < p < 9.005 . The largest p = 9. Therefore, K = 51.94 and z = 2.81. The step response is shown in Figure AP10.6. The settling time is Ts = 0.77 second. 2.2 2 (p + 2p + 1) p < 52 3.61
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.2
0.4
0.6
0.8
1 Time (secs)
1.2
1.4
1.6
1.8
2
FIGURE AP10.6 Step response with minimum settling time.
AP10.7
Let Gp = 1. The closed-loop transfer function is T (s) = K (s + 3) . s4 + 38s3 + 296s2 + (K + 448)s + 3K
When K = 311, the characteristic equation s4 + 38s3 + 296s2 + 759s + 933 = 0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
548
CHAPTER 10
The Design of Feedback Control Systems
has poles at s = 1.619 j 1.617 ( = 1/ 2), s = 6.25, and s = 28.51. (a) When Gp (s) = 1 and K = 311, the closed-loop transfer function is T (s) = 311(s + 3) . s4 + 38s3 + 296s2 + 759s + 933
The step input performance is P.O. = 6.5%, Ts = 2.5 seconds, and Tr = 1.6 seconds. With the prelter Gp (s) = 3 s+3
and K = 311, the closed-loop transfer function is T (s) = s4 + 38s3 933 . + 296s2 + 759s + 933
In this case, the step response is P.O. = 3.9%, Ts = 2.8 seconds, and Tr = 1.3 seconds. (b) Now, consider the prelter Gp (s) = 1.8 s + 1.8
and K = 311. The closed-loop transfer function is (s) = s5 + 39.8s4 559.8(s + 3) . + 364.4s3 + 1291.8s2 + 2299.2s + 1679.4
The step input response is P.O. = 0.7%, Ts = 2.14 seconds and Tr = 1.3 seconds. AP10.8 The plant transfer function is G(s) = 250 . s(s + 2)(s + 40)(s + 45)
The performance specications are P.O. < 20%, Tr < 0.5 second, Ts < 1.2 seconds and Kv 10. A suitable lead compensator is Gc = 1483.7 The closed-loop transfer function is T (s) = 250(1483.7)(s + 35) s(s + 2)(s + 40)(s + 45)(s + 33.75) + 250(1483.7)(s + 3.5) s + 3.5 . s + 33.75
The actual step input performance (see Figure AP10.8) is P.O. = 18%, Ts = 0.88 second, Tr = 0.18 second, and Kv = 10.7.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
549
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.1
0.2
0.3
0.4
0.5 Time (secs)
0.6
0.7
0.8
0.9
1
FIGURE AP10.8 Step response with lead compensator.
AP10.9
The frequency response is shown in Figure AP10.9.
Bode Diagrams Gm=12.4 dB (Wcg=20.9); Pm=42.0 deg. (Wcp=9.0) 150 100 50 0 Phase (deg); Magnitude (dB) -50 -100 -150 -50 -100 -150 -200 -250 -300 -3 10
10
-2
10
-1
10
0
10
1
10
2
10
3
Frequency (rad/sec)
FIGURE AP10.9 Bode plot with Gc (s) =
(s+2.5)(s+0.9871) (s+36.54)(s+0.0675)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
550
CHAPTER 10
The Design of Feedback Control Systems
One lead-lag compensator that satises the specications is Gc (s) = (s + 2.5)(s + 0.9871) . (s + 36.54)(s + 0.0675)
The gain and phase margins are Gm = 12.35 dB and P m = 41.8 , respectively. The velocity error constant is Kv = 100. Therefore, all specications are satised. AP10.10 The transfer function is T (s) = Vo (s) Vi (s) 1 + R2 C2 s . = 1 + R1 C1 s 1 + 0.01s . 1 + 0.001s
Substituting C1 = 0.1 F ,C2 = 1 mF , R1 = 10 k, and R2 = 10 yields T (s) =
The frequency response is shown in Figure AP10.10.
Bode Diagrams
20
15
10 Phase (deg); Magnitude (dB)
5
0 60 50 40 30 20 10 0 1 10
2 3 4
10
10 Frequency (rad/sec)
10
FIGURE AP10.10 Bode plot for T (s) =
1+0.01s 1+0.001s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
551
Design Problems
CDP10.1
The plant model with parameters given in Table CDP2.1 in Dorf and Bishop is given by: (s) 26.035 = , Va (s) s(s + 33.142) where we neglect the motor inductance Lm and where we switch o the tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). With a PD controller the closed-loop system characteristic equation is s2 + (33.142 + 26.035KD )s + 26.035Kp = 0 . Using Table 10.2 in Dorf and Bishop we determine that for a second-order system with a deadbeat response we have = 1.82 and n Ts = 4.82. Since we desire Ts < 0.25 seconds, we choose n = 19.28. Equating the actual characteristic equation with the desired characteristic equation we obtain
2 s2 + n s + n = s2 + (33.142 + 26.035KD )s + 26.035Kp .
Solving for Kp and KD yields the PD controller: Gc (s) = 14.28 + 0.075s . The step response is shown below. The settling time is Ts = 0.24 second.
1 0.9 0.8 0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0
0
0.05
0.1
0.15 Time (secs)
0.2
0.25
0.3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
552 DP10.1
CHAPTER 10
The Design of Feedback Control Systems
The plant is given as G(s) = One possible lead compensator is Gclead (s) = 10(s + 2) . s + 20 4 . s (s + 0.5)
Similarly, a suitable lag compensator is Gclag (s) = s + 0.1 . s + 0.01
The loop transfer function with the lead-lag compensator is Gc (s)G(s) = 40(s + 2)(s + 0.1) . s (s + 0.5) (s + 20)(s + 0.01)
The step response for the compensated and uncompensated systems is shown in Figure DP10.1. The velocity constant is Kv = 80, so the steadystate error specication is satised.
With lead-lag (solid) & uncompensated (dashed) 1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0 0
1
2
3
4
5 Time (sec)
6
7
8
9
10
FIGURE DP10.1 Step response for compensated and uncompensated systems.
DP10.2
(a) When Gc (s) = K , we require K > 20 to meet the steady-state tracking specication of less than 5%.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
553
(b) The system is unstable for K > 20. (c) A single stage lead compensator is Gc1 (s) = 1 + 0.49s . 1 + 0.0035s
With this compensator, the bandwidth is B = 68.9 rad/sec and the phase margin is P.M. = 28.57o . (d) A two stage lead compensator is Gc2 (s) = (1 + 0.0185s)(1 + 0.49s) . (1 + 0.00263s)(1 + 0.0035s)
With the two stage compensator, the bandwidth is B = 83.6 rad/sec and the phase margin is P.M. = 56.79o . The step response for the two compensators is shown in Figure DP10.2.
Single stage (solid) & two stage (dashed) 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE DP10.2 Step response for one- and two-stage lead compensators.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
554 DP10.3
CHAPTER 10
The Design of Feedback Control Systems
The mast ight system is modeled as G(s) = 4 . s(s + 1)(s + 2)
Consider the proportional controller Gc (s) = K = 0.26 . The system step response is shown in Figure DP10.3. The percent overshoot is P.O. = 15.6%, the rise time is Tr = 3.75 seconds, and the phase margin is P.M. = 52.26o .
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE DP10.3 Step response for the mast ight system.
DP10.4
The loop transfer function is Gc (s)G(s) = 2K1 (s + a) s(s + 1)(s + 2)
where a = K2 /K1 . One possible choice for the PI controller parameters is a = 1.1 and K1 = 0.8182. The step response is shown in Figure DP10.4. The performance results are P.O. = 4.59% Ts = 4.93 sec Kv = 0.9 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
555
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
1
2
3 Time (sec)
4
5
6
7
FIGURE DP10.4 Step response for the robot and vision system.
DP10.5
One possible compensator is Gc (s) = 5682 s + 12.6 . s + 87.3
The step response is shown in Figure DP10.5.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE DP10.5 Step response for the high speed train system.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
556
CHAPTER 10
The Design of Feedback Control Systems
The performance results are P.O. = 4.44% DP10.6 Ts = 0.36 sec Kv = 14.1 .
One solution is the lead-lag controller Gc (s) = 8(s + 0.01)(s + 5.5) . (s + 0.0001)(s + 6.5)
The step response is shown in Figure DP10.6. The performance results are P.O. = 1.23% Ts = 1.67 sec Kv = 135.4 .
The disturbance response is also shown in Figure DP10.6.
Step response 1.2 0.16 0.14 1 0.12 0.8 0.1 Disturbance response
Amplitude
0.6
Amplitude
5 Time (sec)
0.08 0.06
0.4 0.04 0.2 0.02 0 0 0 0
5 Time (sec)
10
FIGURE DP10.6 Input and disturbance response for the antenna control system.
DP10.7
The stringent design specications are Kv > 200; Ts < 12 ms and percent overshoot P.O. < 10%. A suitable compensator is Gc (s) = K where K = 1.9476e + 13. Then, P.O. = 9.5% Ts = 10 ms Kv = 560 . s + 403 , s + 2336
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
557
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8
10 Time (ms)
12
14
16
18
20
FIGURE DP10.7 Step response for the tape transport system.
DP10.8
A solution to the problem is the PI controller Gc (s) = 4.21s + 1.2 . s
The step response is shown in Figure DP10.8.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
1
2
3 Time (sec)
4
5
6
FIGURE DP10.8 Step response for the engine control system.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
558
CHAPTER 10
The Design of Feedback Control Systems
The performance results are P.O. = 8.8% and Ts = 2.14 .
The system is a type-1, so the steady-state error for a step input is zero, as desired. DP10.9 The jet aircraft roll angle motion is represented by the transfer function G(s) = 5 . (s + 12)(s2 + 2s + 4)
A good controls solution is obtained with a PID controller Gc (s) = 20s2 + 40s + 80 . s
The system is type-1, so the steady-state tracking error is zero for a step input. The performance results are P.O. = 9.48% and Ts = 0.59 .
1.2
1
0.8 Amplitude
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1 Time (sec)
1.2
1.4
1.6
1.8
2
FIGURE DP10.9 Step response for the jet aircraft roll control system.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
559
DP10.10
One good solution is obtained with the following PI controller Gc (s) = 27.35(s + 2) . s
The system is type-1, so the steady-state tracking error is zero for a step input. The step response is shown in Figure DP10.10.
Step Response
From: U(1) 1.4
1.2
1
Amplitude
0.8 To: Y(1) 0.6 0.4 0.2 0 0
0.2
0.4
0.6
0.8
1
1.2
Time (sec.)
FIGURE DP10.10 Step response for the windmill radiometer.
DP10.11
Consider the PID controller Gc (s) = Kp + KD s + and the lead-lag controller Gc (s) = K s+a s+b s+c s+d = 6.04 (s + 10)(s + 2) . (s + 1)(s + 5) KI 1.554s2 + 1.08s + 1 = s s
Both are stabilizing in the presence of a T = 0.1 second time delay. For the PID controller the phase margin is P.M. = 40o . For the lead-lag controller the phase margin is P.M. = 45o . We nd (for these particular designs)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
560
CHAPTER 10
The Design of Feedback Control Systems
that the lead-lag controller is more able to remain stable in the process of increasing time delay. For a time-delay of T = 0.2 seconds, the lead-lag compensator has a phase margin of P.M. = 22o , while the PID controller is unstable. DP10.12 One solution is Gc (s) = 50(s + 0.01) . s+2
The Bode magnitude is shown in Figure DP10.12. You want high gain at
Bode Diagram 80
60
System: sys Frequency (rad/sec): 0.101 Magnitude (dB): 26.9
40
20 Magnitude (dB)
0
20
System: sys Frequency (rad/sec): 10 Magnitude (dB): 26.9
40
60
80
100 4 10
10
3
10
2
10 Frequency (rad/sec)
1
10
0
10
1
10
2
FIGURE DP10.12 Step response for the windmill radiometer.
low frequency to improve disturbance rejection and decrease sensitivity to plant changes and low gain at high frequency to attenuate measurement noise.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
561
Computer Problems
CP10.1 The m-le script and step response is shown in Figure CP10.1. The phase margin and percent overshoot are P.M. = 24o P.O. 50% , respectively.
numc=[5.5]; denc=[1 0]; sysc = tf(numc,denc); numg=[1]; deng=[1 1]; sysg = tf(numg,deng); syss = series(sysc,sysg); [gm,pm]=margin(syss); pm % sys_cl = feedback(syss,1); y=step(sys_cl); step(sys_cl); grid ymax=max(y))
pm = 24.0476 ymax = 1.5024
Step Response 1.6
1.4
1.2
1 Amplitude
0.8
0.6
0.4
0.2
0
0
2
4
6 Time (sec)
8
10
12
FIGURE CP10.1 Phase margin and step response for the closed-loop system.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
562 CP10.2
CHAPTER 10
The Design of Feedback Control Systems
Using a proportional controller the closed-loop characteristic equation is 1+K s2 17.5 . + 5s + 17.5
A simple m-le script which computes the P.M. as a function of the gain K yields the proportional controller gain K = 3. Checking the phase margin of the system reveals that P.M. = 45 , as desired.
n=17.5; d=[1 5 17.5]; sys = tf(n,d); K=3; margin(K*sys), grid
Bode Diagram Gm = Inf dB (at Inf rad/sec) , Pm = 44.8 deg (at 7.4 rad/sec) 10 0 Magnitude (dB) Phase (deg) 10 20 30 40 50 0
45
90
135
180 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE CP10.2 Bode plot with a proportional controller K = 3 in the loop.
CP10.3
The uncompensated system is type-1. To realize a zero steady-state error to a ramp input we need to increase the system type by one. One controller that does this is the PI controller: Gc (s) = KP s + KD . s
The step response is shown in Figure CP10.3 where it can be seen in the tracking error plot that the settling time is Ts < 5 seconds. The actual settling time is Ts = 3.6 seconds .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
563
KP=20; KD=10; nc=[KP KD]; dc=[1 0]; sysc = tf(nc,dc); n=1; d=[1 2 0]; sys = tf(n,d); sys_o = series(sysc,sys); sys_cl = feedback(sys_o,[1]); t=[0:0.001:10]; sys1 = tf([1],[1 0]); sys_cl1 = series(sys_cl,sys1); subplot(121) y=step(sys_cl1,t); plot(t,y,t,t,'--'), grid xlabel(' Time (sec)'), ylabel('Ramp response') e=y-t'; L= nd(abs(e)>0.02); Ts=t(L(length(L))) subplot(122) plot(t,e,[0 10],[0.02 0.02],':',[0 10],[-0.02 -0.02],':') xlabel(' Time (sec)'), ylabel(' Track ing error') grid
10 9 0.05 8 7 Ramp response 6 5 4 3 2 -0.2 1 0 -0.25 0 0.1
Tracking error 0 5 Time (sec) 10
-0.05
-0.1
-0.15
0
5 Time (sec)
10
FIGURE CP10.3 Ramp response with a PI controller Gc (s) =
20s+10 s
in the loop.
CP10.4
From the percent overshoot spec we determine that P.O. < 10% implies > 0.6. So, we target a phase margin P.M. = 100 = 60o . The m-le script which generates the uncompensated Bode plot is shown in Figure CP10.4a.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
564
CHAPTER 10
The Design of Feedback Control Systems
numg = 100*conv([1 1],[1 0.01]); deng = conv([1 10],conv([1 2 2],[1 0.02 0.0101])); sysg = tf(numg,deng) w=logspace(-1,2,200); [mag,phase,w]=bode(sysg,w); [Gm,Pm,Wcg,Wcp]=margin(mag,phase,w); % Phi=60-Pm Pm Phi=(60-Pm)*pi/180; alpha=(1+sin(Phi))/(1-sin(Phi)) M=-10*log10(alpha)*ones(length(w),1); [mag,phase,w]=bode(sysg,w); for i = 1:length(w), magdB(i) = 20*log10(mag(1,1,i)); end semilogx(w,magdB,w,M), grid xlabel('Frequenc y (rad/sec)'), ylabel('mag [dB]') title('Uncompensated Bode Plot') hold on semilogx([.56072 5.6072 56.072 560.72],[20 0 -20 -40],'--')
Uncompensated Bode Plot 60
Phi = 56.2111 Pm = 3.7889 alpha = 10.8408
40
20
mag [dB]
0
-20
-40
-60
-80 10-1
100 Frequency (rad/sec)
101
102
FIGURE CP10.4 (a) Uncompensated Bode plot.
We assume that K = 1 and raise the gain at a later step to meet settling time requirement. The uncompensated phase margin is P.M. = 3.7o , so that the lead compensator needs to add = 56.2o . The script also calculates = 10.84. Following the design procedure outlined in Dorf & Bishop, we locate the compensator zero at = 2 rad/sec (see dashed line in Figure CP10.4a). Then, p = z implies p = 21.68. After several iter-
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
565
ations, we converge on K = 4 as a good value. The lead compensator is Gc (s) = 4 s+2 . s + 22
The step response is shown in Figure CP10.4b. The compensated Bode is shown in Figure CP10.4c.
K=4; numg = 100*conv([1 1],[1 0.01]); deng = conv([1 10],conv([1 2 2],[1 0.02 0.0101])); sysg = tf(numg,deng) numc=K*[1 2]; denc=[1 22]; sysc = tf(numc,denc); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); t=[0:0.01:5]; f=10*pi/180; [y,t,x]=step(f*sys_cl,t); plot(t,y*180/pi), grid xlabel(' Time (sec)') ylabel('Attitude rate (deg/sec)'), pause w=logspace(-1,2,200); [mag,phase,w]=bode(sys_o,w); [Gm,Pm,Wcg,Wcp]=margin(mag,phase,w); bode(sys_o) title(['Gain Margin = ',num2str(Gm),' Phase Margin = ',num2str(Pm)])
12
10
Attitude rate (deg/sec)
8
6
4
2
0 0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE CP10.4 CONTINUED: (b) Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
566
CHAPTER 10
The Design of Feedback Control Systems
Gain Margin = 14.96 Phase Margin = 60.49
50 0 -50 -100 10-3
Gain dB
10-2
10-1 100 Frequency (rad/sec)
101
102
100
Phase deg
0 -100 -200 -300 10-3 10-2 10-1 100 101 102
Frequency (rad/sec)
FIGURE CP10.4 CONTINUED: (c) Bode plot with lead compensator.
CP10.5
The closed-loop transfer function is (s)/d (s) = s2 K1 + K2 s + K2 s + K1
where K1 = K1 /J and K2 = K2 /J . A percent overshoot P.O. 20% requires > 0.45. Select as the initial damping = 0.7 (initial selection) .
For a second-order system with = 0.7, we nd that /n 0.9 when | (s)/d (s)| = 0.7. So, we select n = B /0.9 as a starting choice. Therefore, since B = 10, we have n = 11 . The m-le script is shown in Figure CP10.5a. After several iterations, we nd a set of good values for = 0.8 and n = 4.5 (nal selection) .
The step response and closed-loop Bode plot are shown in Figures CP10.5b and CP10.5c.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
567
% Par t (a) wn=4.5; zeta=0.8; K2=2*zeta*wn; K1=wn^2; % Par t (b) num=[K2 K1]; den=[1 0 0]; sys = tf(num,den); sys_cl = feedback(sys,[1]); f=10*pi/180; % set-up for 10 deg step input t=[0:.05:3]; [y,t,x]=step(f*sys_cl,t); plot(t,y*180/pi), xlabel(' time [sec]'), ylabel(' theta [deg]'), grid, pause % Par t (c) w=logspace(-1,2,400); [mag,phase,w]=bode(sys_cl,w); for i = 1:length(w), magdB(i) = 20*log10(mag(1,1,i)); end semilogx(w,magdB,[w(1) w(length(w))],[-3 -3]), grid xlabel('Frequenc y (rad/sec)') ylabel('Gain dB')
FIGURE CP10.5 (a) Script to generate the step response and the closed-loop Bode plot.
12
10
8
theta [deg]
6
4
2
0 0
0.5
1
1.5 time [sec]
2
2.5
3
FIGURE CP10.5 CONTINUED: (b) Step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
568
CHAPTER 10
The Design of Feedback Control Systems
5
0
-5
Gain dB
-10
-15
-20
-25 10-1
100 Frequency (rad/sec)
101
102
FIGURE CP10.5 CONTINUED: (c) Closed-loop Bode plot.
CP10.6
The settling time and phase margin specications require that the dominant closed-loop poles have natural frequency and damping of 0.45 and n 1.78. The uncompensated roots locus is shown in Figure CP10.6a.
10
+
numg=[1 10]; deng=[1 2 20]; sysg = tf(numg,deng); axis([-15,1,-10,10]); rlocus(sysg); hold on % zeta=0.45; wn=1.7778; x=[-10:0.1:-zeta*wn]; y=-(sqr t(1-zeta^2)/zeta)*x; xc=[-10:0.1:-zeta*wn]; c=sqr t(wn^2-xc.^2); plot(x,y,':',x,-y,':',xc,c,':',xc,- c,':') rlo c nd(sysg), hold o
8 6 4
K=10
x
Imag Axis
2 0 -2 -4 -6 -8
+ x o
-10
-14
-12
-10
-8 Real Axis
-6
-4
-2
0
FIGURE CP10.6 (a) Uncompensated root locus.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
569
From the nal value theorem, we determine that
s0
lim = sE (s) 0.1A
implies
A = 0.1A . 1 + GGc (s)
Therefore, the compensated Kpcomp 9. With the compensator Gc (s) = K we nd that Kpcomp = Kz Kpuncomp . p s+z s+p
But Kpuncomp = 0.5 and (from the uncompensated root locus) a gain of K = 10 results in roots of the characteristic equation in the desired region. Solving for z 1 Kpcomp = 2. p K Kpuncomp Select z = 0.5 to minimize changing the root locus. Then, p = 0.25, and the compensator is Gc (s) = 10 s + 0.5 . s + 0.25
The compensated root locus is shown in Figure CP10.6b and the step response is shown in Figure CP10.6c. The phase margin of the compensated
numg=[1 10]; deng=[1 2 20]; sysg = tf(numg,deng); numc=[1 0.5]; denc=[1 0.25]; sysc = tf(numc,denc); sys_o = series(sysc,sysg); axis([-15,1,-10,10]); rlocus(sys_o); hold on % zeta=0.45; wn=1.7778; x=[-10:0.1:-zeta*wn]; y=-(sqr t(1-zeta^2)/zeta)*x; xc=[-10:0.1:-zeta*wn]; c=sqr t(wn^2-xc.^2); plot(x,y,':',x,-y,':',xc,c,':',xc,- c,':') rlo c nd(sys_o) hold o
10
+
8 6 4
x
Imag Axis
2 0 -2 -4 -6 -8
+ x o ox +
-10
-14
-12
-10
-8 Real Axis
-6
-4
-2
0
FIGURE CP10.6 CONTINUED: (b) Compensated root locus.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
570
CHAPTER 10
The Design of Feedback Control Systems
system is P.M. = 62.3o and the settling time Ts < 5 seconds.
numg=[1 10]; deng=[1 2 20]; sysg = tf(numg,deng); numgc=10*[1 0.5]; dengc=[1 0.25]; sysc = tf(numgc,dengc); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); t=[0:0.1:5]; step(sys_cl,t) [mag,phase,w]=bode(sys_o); [gm,pm,w1,w2]=margin(mag,phase,w); pm
>> pm = 62.3201
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.5
1
1.5
2
2.5 Time (secs)
3
3.5
4
4.5
5
FIGURE CP10.6 CONTINUED: (c) Step response and phase margin verication.
CP10.7
Both design specications can be satised with an integral controller Gc (s) = K1 + K2 10 = . s s
The simulation results and m-le script are shown in Figures CP10.7a and b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
Unit Step Response
571
1.5
Phi dot
1
0.5 0 0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
0
Unit Ramp Response
Tracking error
-0.05
-0.1 -0.15
0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE CP10.7 (a) Simulation results.
K1=0; K2=10; numc=[K1 K2]; denc=[1 0]; sysc = tf(numc,denc); numg=[23]; deng=[1 23]; sysg = tf(numg,deng); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); t=[0:0.01:1]; ys=step(sys_cl,t); subplot(211) plot(t,ys), xlabel(' Time (sec)'), ylabel('Phi dot') title('Unit Step Response'), grid u=t; yr=lsim(sys_cl,u,t); subplot(212) plot(t,yr-u','--') xlabel(' Time (sec)'), ylabel(' Track ing error') title('Unit Ramp Response'), grid
FIGURE CP10.7 CONTINUED: (b) M-le design script.
CP10.8
From Example 10.3, we have that the loop transfer function is Gc (s)G(s) = 8.1(s + z ) , s2 (s + 3.6)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
572
CHAPTER 10
The Design of Feedback Control Systems
where z = 1. We want to determine a value of z so the the percent overshoot is reduced from 46% to less than 32%. A valid design is Gc (s)G(s) = 8.1(s + 0.45) . s2 (s + 3.6)
The m-le script and step response are shown in Figure CP10.8. The percent overshoot is P.O.=27.7 %.
Step Response
From: U(1) 1.4
1.2
K1 = 8.1; numg = [K1]; deng = [1 0 0]; sysg = tf(numg,deng); numc = [1 0.45]; denc = [1 3.6]; sysc = tf(numc,denc); sys_o = series(sysc,sysg); sys_cl = feedback(sys_o,[1]); step(sys_cl) y=step(sys_cl); po=100*(max(y)-1)
1
Amplitude
0.8 To: Y(1) 0.6 0.4 0.2 0 0
1.6
3.2
4.8
6.4
8
Time (sec.)
FIGURE CP10.8 Response of system with new lead compensator design.
CP10.9
From AP10.10, we have the transfer function is T (s) = Vo (s) Vi (s) 1 + R2 C2 s = . 1 + R1 C1 s 1 + 0.01s . 1 + 0.001s
Substituting C1 = 0.1 F ,C2 = 1 mF , R1 = 10 k, and R2 = 10 yields T (s) =
The frequency response is shown in Figure CP10.9.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
Bode Diagrams
573
20
15
10 Phase (deg); Magnitude (dB)
c1=0.0000001; c2=0.001; r1=10000; r2=10; n=[c2*r2 1]; d=[c1*r1 1]; sys=tf(n,d) bode(sys)
5
0 60 50 40 30 20 10 0 1 10
2 3 4
10
10 Frequency (rad/sec)
10
FIGURE CP10.9 Op-amp circuit frequency response.
CP10.10
The plot of K versus phase margin is shown in Figure CP10.10. The value of K that maximizes the phase margin is K = 4.15.
60
K=[0.1:0.01:10]; T=0.2; [np,dp]=pade(T,6); sysp=tf(np,dp); for i=1:length(K) ng=K(i)*[1 0.2]; dg=[1 6 0 0]; sysg=tf(ng,dg); [gm,pm]=margin(sysg*sysp); PM(i)=pm; end plot(K,PM), grid [P,n]=max(PM); K(n) xlabel('K'), ylabel('P.M.')
55
50
45
40
P.M.
35 30 25 20 15 0
1
2
3
4
5 K
6
7
8
9
10
FIGURE CP10.10 Plot of K versus phase margin.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
11
The Design of State Variable Feedback Systems
Exercises
E11.1 The system is given by x = Ax + Bu u = Kx where A=
01 10
B=
10 01
and
K=
1
0
0 k
.
Then, with u = Kx, we have x= The characteristic equation is det[sI A] = det
1
1
1 k
x .
(s + 1) 1
1 (s + k)
= s2 + (1 + k )s + k 1 = 0 .
We desire critical damping (i.e. = 1). So, rewrite the characteristic equation as (s2 + s 1) + ks + k = 0 574
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
575
or 1+ k(s + 1) k(s + 1) =1+ =0. s2 + s 1 (s + 1.62)(s 0.62)
The root locus shows that two equal roots are not possible, however it is stable for k > 1. E11.2 Let u = k1 x1 k2 x2 + r . Then, x=
0
1
3 k1 k2
and
x+
0 1
r ,
det(sI A) = s2 + k2 s + k1 3 = 0 . We want = 1 and Ts = 2. So, Ts = 4 =2 n implies n = 2 .
2 Solving for the gain yields k1 3 = n = 4 or k1 = 7 and k2 = 2n = 2n = 4.
E11.3
The controllability matrix is Pc = B AB =
0
1
1 3
,
and det Pc = 0, therefore the system is controllable. The observability matrix is Po =
C CA
and det Po = 0; therefore the system is unobservable. E11.4 The controllability matrix is Pc = B AB =
=
0
2
0 6
,
0
0
1 1
,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
576
CHAPTER 11
The Design of State Variable Feedback Systems
and the det Pc = 0; therefore the system is uncontrollable. The observability matrix is Po =
C CA
and det Po = 0; therefore the system is also unobservable. E11.5 The controllability matrix is Pc = B AB =
=
10 4 0
,
1 1 1 1
and det Pc = 0; therefore the system is uncontrollable. The observability matrix is Po =
,
C CA
and det Po = 0; therefore the system is observable. E11.6 The controllability matrix is Pc = B AB =
=
10 01
,
0
1
1 2
and det Pc = 0; therefore the system is controllable. The observability matrix is Po =
,
C CA
and det Po = 0; therefore the system is observable.
=
10 01
,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
577
E11.7
The block diagram is shown in Fig. E11.7.
2 U(s) 10
+
--
1 s 7 2
1 s
2
+
-
Y(s)
FIGURE E11.7 The block diagram for E11.7.
E11.8
The block diagram is shown in Fig. E11.8.
10 8
U(s)
4
+
- --
1 s 1 3
+
-
1 s
1 s
2
+
++ Y(s)
9
FIGURE E11.8 The block diagram for E11.8.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
578 E11.9
CHAPTER 11
The Design of State Variable Feedback Systems
The controllability matrix is Pc = B AB =
k1
k1 k2
k2 k1 + k2
,
2 2 and det Pc = k1 + k2 . So, the condition for complete controllability is 2 = k2 . k1 2
E11.10
A matrix dierential equation representation is
0 0
1 0
x=
10 6 3
0 0 1 x + 0 u
1
y = [3 4 E11.11 The system is given by
2]x + [0]u .
x = Ax + Bu y = Cx + Du where
A=
01 00
2 0 7
0 0 1 , B = 0 , C =
120
, and D = [1] .
1
The controllability matrix is
Pc =
B AB A B
2
and det Pc = 1 = 0; therefore the system is controllable. The observability matrix is
C Po = CA =
0 = 0
0
1 7
1 1 7 , 49
12 01
CA2
4 0 13
0 2,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
579
and det Po = 29 = 0; therefore the system is observable. E11.12 The transfer function is G(s) = s2 2 . + 3s + 2
The response of the system to a unit step is y (t) = 1 2et + e2t . The step response is shown in Figure E11.11
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Step Response
0
1
2
3
4 Time (s)
5
6
7
8
FIGURE E11.11 Unit step response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
580
CHAPTER 11
The Design of State Variable Feedback Systems
Problems
P11.1 Consider the system x=x+u u = kx . So, x = x kx = (1 k)x and x(t) = e(1k)t x(0) . The system is stable if k > 1. Computing the value of J (assuming k > 1) yields J=
0
e2(1k)t x2 (0)dt =
1 . k1
Thus, J is minimum when k . This is not physically realizable. Select k = 35. Then, the value of the performance index J is J= 1 . 34
The system is not stable without feedback. P11.2 (a) The performance index is given J=
0
(x2 + u2 )dt .
The system is x=x+u u = kx . So, J=
0 0 0
(x2 + k2 x2 )dt =
(1 + k2 )x2 dt = (1 + k2 )
x2 dt .
Carrying out the integration (assuming k > 1) yields J = (1 + k2 ) 1 . k1
We want to determine k that minimizes J . Taking the partial of J
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
581
with respect to k and setting the result to zero yields J k2 2k 1 = =0, k (k 1)2 or k2 2k 1 = 0 . Solving for k yields k =1+ 1+ 1 ,
(b) For = 2, we determine that k = 2.2 and Jmin = 8.9. P11.3 The system is given by x=
where we reject the solution k = 1
1 1 + , since we require k > 1.
10 1 2
u = k(x1 + x2 ) = k[1 1]x .
x +
1 1
u
Then, with feedback applied, the system is x= Solving HT P + PH = I yields 2p11 (1 k) 2p12 (k + 1) = 1 p12 (3 2k) p11 k p22 (k + 1) = 0 2kp12 + 2p22 (2 k) = 1 . Solving for p11 , p12 and p22 yields p11 = p12 (2k2 6k + 7) 4(4k2 8k + 3) 2k2 2k 1 = 4(4k2 8k + 3)
(1 k)
k
(1 + k) (2 k)
x .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
582
CHAPTER 11
The Design of State Variable Feedback Systems
p22 =
(2k2 6k + 3) . 4(4k2 8k + 3)
The performance index is computed to be J = xT (0)Px(0) = p11 + 2p12 + p22 = 1 , 2k 1
when x(0) = [1 1]T . So as k , J 0. The system is unstable without feedback. P11.4 The performance index is J = xT (0)Px(0) = p11 + 2p12 + p22 . From Example 11.6 in Dorf and Bishop, we determine that J= 2k + 1 2 k+1 4k + 1 + + =1+ . 2k 2k 2k2 2k2
So, when k , the performance index J 0. The plot of J versus k is shown in Figure P11.4.
80 70 60 50 40 30 20 10 0 0
J
1
2
3
4
5 k
6
7
8
9
10
FIGURE P11.4 The performance index J versus k.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
583
P11.5
The system is given by x=
01 00
u = k x . The performance index is J=
0
x +
00 11
u
(xT x + uT u)dt =
0
(1 + k2 )(xT x)dt .
First, we solve HT P + PH = (1 + k2 )I , yielding, p12 = p22 p11 (1 + k2 ) 2k 3 + k2 + k + 1 k = 2k2 3 + k 2 + 2k + 1 2k = . 2k
The performance index is then given by J = p11 + 2p12 + p22 = 2k4 + 4k3 + 3k2 + 4k + 1 . 2k2
Taking the partial of J with respect to k, setting the result to zero and solving for k yields J 2k4 + 2k3 2k 1 = =0 k k3 or 2k4 + 2k3 2k 1 = 0 . Solving for k yields k = 0.90. The plot of J versus k is shown in Figure P11.5. The value of the performance index is J = 6.95 when k = 0.90.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
584
CHAPTER 11
The Design of State Variable Feedback Systems
8.8 8.6 8.4 8.2 8 7.8 7.6 7.4 7.2 7 6.8 0.5 0.6 0.7 0.8 0.9 k 1 1.1 1.2 1.3 1.4
FIGURE P11.5 The performance index J versus k.
P11.6
(a) For P11.3, we have J= 1 . 2k 1
So, as k , then J 0. But k = is not a practical solution, so select k = 10. Then, J = 1/19, and x=
J
9 10 11 8
x = Ax .
The closed-loop system roots are determined by solving det[sI A] = s2 + 17s 38 = 0 , which yields s = 19 and s2 = 2. The system is unstable. The original system was unstable, and it remains unstable with the feedback. In general, x=
(1 k)
k
(1 + k) (2 k)
and det[sI A] = s2 + s(2k 3) + (2 4k) = 0. A Routh-Hurwitz analysis reveals that the system is unstable for all k.
x = Ax
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
585
(b) For P11.4, we have x= and det[sI A] = s2 + ks + k = 0 . The performance index was found to be J =1+ 4k + 1 . 2k2
0
1
k k
x = Ax ,
As k , we have J 0. But k = is not a practical choice for k. Select k = 10. Then, det[sI A] = s2 + 10s + 10 = (s + 1.13)(s + 8.87) . The closed-loop system is stable. (c) In P11.5, we found that k = 0.90 for Jmin . We are given x= and det[sI A] = s2 + ks + k = s2 + 0.9s + 0.9 = (s + 0.45 + j 0.835)(s + 0.45 j 0.835) . P11.7 The closed-loop system is x= and
2 det[sI H] = s2 + k2 s + k1 = s2 + 2n s + n = 0 .
0
1
k k
x
0
1
k1 k2
x = Hx ,
We desire n = 2, so set k1 = 4. With xT (0) = [1, 0], we have J = p11 , and solving HT P + PH = I
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
586
CHAPTER 11
The Design of State Variable Feedback Systems
yields
0
4
1 k2
p11 p12 p12 p22
+
p11 p12 p12 p22 1 0 0 1
0
1
4 k2
and p11 = Select
=
,
20 k2 + 20 k2 + =2 . 8 8k2 8k2
k2 = for Jmin , where Jmin =
5 2.
20
Then 20s + 4 = 0 ,
det[sI H] = s2 + P11.8
and n = 2 and = 1.12. The system is overdamped. From Example 11.11 in Dorf and Bishop, we have P=
2 k2 +2 2k2 1 2
1 2 1 k2
So,
.
2 k2 + 2 2k2
J = xT (0)Px(0) =
when xT (0) = [1 0]. Taking the partial of J with respect to k2 and setting the result to zero yields J k2 + 2 =1 2 2 =0 . k2 2k2 Solving for the optimum value of k2 yields k2 = 2 . P11.9 Let x1 = and x2 = . We have that = d . dt
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
587
The state equations are x1 = x2 x2 = Ku . Select a feedback such that u = x1 K1 x2 + r when r (t) is the reference input. Then, x= and det[sI A] = s2 + K1 Ks + K .
4 so that the overshoot is 4%. Since Ts = 1 = n , we require n = 4 or n = 4 2. Then, s2 + 8s + 32 = s2 + K1 Ks + K , or 8 K = 32 and K1 = 32 = 1 . 4
0
1
K KK1
x+
0 K
r ,
We desire =
1 , 2
P11.10
The system with feedback is given by x = Ax =
8 16 1 0
x ,
where x1 (0) = 1, and x2 (0) = 0. The characteristic equation is det[sI A] = det s + 8 16 1 s
= s(s + 8) + 16 = s2 + 8s + 16 = 0 .
The roots are s1,2 = 4. The solution is x(t) =
11 12 21 22
x(0) =
11 21
since x1 (0) = 1 and x2 (0) = 0. We compute the elements of the state transition matrix as follows: 21 (t) = L1 s2 1 + 8s + 16 ,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
588
CHAPTER 11
The Design of State Variable Feedback Systems
therefore x2 (t) = te4t . Similarly, 11 (t) = L1 Therefore, x1 (t) = (1 4t)e4t . P11.11 Let u = k1 x1 k2 x2 + r where r (t) is the command input. A state variable representation of the plant is x= y=
s s2 + 8s + 16
.
5 2 2 01 0
x+
x+
0.5 0
0
u.
u
The closed-loop transfer function is T (s) = s2 . + (k1 /2 + 5)s + 4 + k2
To meet the performance specications we need n = 4.8 and = 0.826. Therefore, the desired characteristic polynomial is q (s) = s2 + 2(0.826)4.8s + 23 = s2 + 8s + 23 . Equating coecients and solving for k1 and k2 yields k2 = 19 and k1 = 6. Select = 23 to obtain zero steady-state error to a step input. P11.12 A state variable representation of the dc motor is
= x
3 2 0.75 0 0 3 0 0 0 0 2 0 0 00 00 10
020
0 0 x + 0 u 0 0 0
1
0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
589 y = [0 0 0 0 2.75]x .
The controllability matrix is
and the det Pc = 0, so the system is controllable. The observability matrix is
0 Po = 0 0
0 Pc = 0 0
1 3 0 0 0
3
4.5 18 9
3 9 0 0
6 18 6 0
0
13.5 18 18
12
0
0 0 0 11 0
0 0 5.5 0 0
0 5.5 0 0 0
2.75 0 0 0 0
33
and the det Po = 0, so the system is observable. P11.13
,
To meet the Kv = 35 specication, we need K = 2450. A state variable representation is x=
0
1
0 70
x +
0 2450
u
y = [1 0]x . Let u = k1 x1 k2 x2 . Then, the closed-loop characteristic equation is q (s) = s2 + (2450k2 + 70)s + 2450k1 = 0 . The desired characteristic polynomial is s2 + 72.73s + 2644.63 = 0 where we select = 0.707 and n = 51.42 to meet the performance
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
590
CHAPTER 11
The Design of State Variable Feedback Systems
specications. Equating coecients and solving for the gains yields k1 = 1.08 and k2 = 0.0011. P11.14 Let u = k1 x1 k2 x2 k3 r where r (t) is the command input. Then, the closed-loop system in state variable form is x=
10 k1 k2 1 0
x+
1 0
r
y = [0 1]x . To meet the performance specications, we want the closed-loop characteristic polynomial to be q (s) = s2 + 8s + 45.96 = 0 where = 0.59 and n = 6.78. The actual characteristic polynomial is det(sI A) = s2 + (10 + k1 )s + k2 = 0 . Equating coecients and solving for the gains yields k2 = 45.96 and k1 = 2. Select k3 = k2 = 45.96 to obtain a zero steady-state error to a step input. This results in a settling time of Ts = 0.87 s and a percent overshoot of P.O. = 10%. P11.15 The transfer function is G(s) = C(sI A)1 B = 1 . s+1
The system is not controllable and not observable. P11.16 Let u = Kx . Then, Ackermanns formula is K = [0, 0, ..., 1]P 1 q (A) c where q (s) is the desired characteristic polynomial, which in this case is q (s) = s2 + 4s + 8 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
591
A state-space representation of the limb motion dynamics is x= The controllability matrix is Pc = [B AB] = and P 1 = c Also, we have q (A) = A2 + 4A + 8I = Using Ackermanns formula, we have K = [2 8] . P11.17 The system is either uncontrollable or unobservable if a = 3 or a = 4. Both of these values correspond to system real poles. So, if a takes on either value, a pole-zero cancellation occurs in the transfer function. A matrix dierential equation representation is x=
2 0 10
x+
1 0
u .
1 2 0 1
12 01
.
40 28
.
P11.18
0
1
1 2 y = [1
x+
0 1
u
0]x .
Let u(t) = k1 x1 k2 x2 . Then, the closed-loop characteristic equation is q (s) = s2 + (2 + k2 )s + 1 + k1 = 0 . We desire the characteristic equation s2 + 2 2s + 2 = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
592
CHAPTER 11
The Design of State Variable Feedback Systems
Equating coecients and solving for the gains yields k1 = 1 and k2 = 2 2 2 = 0.828. P11.19 A state space representation is x=
0
1
3 2 y = [3
x+
0 1
r
1]x .
The controllability matrix is Pc =
0
1
1 2
,
and det Pc = 0, so the system is controllable. The observability matrix is Po = P11.20 31 31
,
and the det Po = 0, so the system is not observable. The characteristic equation associated with A is s2 (s2 + 0.2s + 0.0015) = 0 . There are two roots at the origin, so the system is unstable. The system can be stabilized with = k1 x1 k3 x3 = 20x1 10x3 . P11.21 (a) Let x1 = i1 , x2 = i2 and u = v . Then, the state equation is x= Also, y = vo , but y = [R3 R3 ]x .
(R1 +R3 ) L1 R3 L2 R3 L1 (R3 +R2 ) L2
x +
1 L1
0
u .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
593
(b) The observability matrix is Po = and det Po = So, when R2 R1 = , L1 L2 det Po = 0 and the system is not observable. (c) Let a= and b= Then det[sI A] = det
C CA
=
R3
2 1 RLR3 R3 1 1 L1
+
1 L2
R2 R3 L2
2 + R3
R3
1 L1
+
1 L2
R2 R1 2 R3 . L2 L1
R1 + R3 , L1
R3 + R2 . L2
(s + a)
3 R2 L
3 R1 L
(s + b)
= (s + a)(s + b) +
2 R3 = (s + r )2 L1 L2 2 R3 . L1 L2
= s2 + (a + b)s + ab + The system has two equal roots when (a + b)2 4 ab + or R1 + R3 R3 + R2 + L1 L2
2
2 R3 L1 L2
4
2 (R1 + R3 )(R3 + R2 ) + R3 =0. L1 L2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
594 P11.22
CHAPTER 11
The Design of State Variable Feedback Systems
(a) Without state feedback the state dierential equation is given by x=
0.4 1 1 01 0 x.
y=
x+
1 0
u
The step response is shown in Figure P11.22a.
2 1.5 (a) Without state feedback
x2
1 0.5 0 0 2 4 6 8 10 Time (sec) 1.5 (b) With state feedback 12 14 16 18 20
1
x2
0.5 0 0
0.2
0.4
0.6
0.8
1 Time (sec)
1.2
1.4
1.6
1.8
2
FIGURE P11.22 Step response (a) without state feedback, and (b) with state feedback.
(b) Consider state feedback u = K (ax2 + bx1 ) + cr where r is the reference input and K, a, b and c are to be determined. Then, the state dierential equation is x=
0.4 Kb 1 Ka 1 01 x, 0
y=
x+
c 0
r
and det(sI A) = s2 + (0.4 + Kb)s + (1 + Ka) = 0. Our specications are P.O. = 5% and Ts = 1.35 sec. So, = 0.69 and n = 1435 = 4.3. .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
595
Solving for K, a and b yields
2 Ka = n 1
and Kb = 2n 0.4 . Select K = 1. Then, a = 17.49 and b = 5.53. Select c = 1 + Ka to achieve a zero steady-state tracking error. (c) The step response is shown in Figure P11.22b for the system with state feedback. P11.23 Using the internal model design method for step inputs, we have
0 1 0 0 e e = 0 0 1 + 0 z z 000 1
where we choose
w ,
w = K1 e K2 z . To place the poles at s = 10 and s = 2 j we use Ackermanns formula to compute K1 = 50 K2 = [45
14] .
The compensator has the form shown in Figure 11.14 in Dorf and Bishop. P11.24 Using the internal model design method for ramp inputs, we have
0 1 0 0 e e 0 0 1 0 e = e 0 0 0 1 z z 0000 0 0
where we choose
+ w 0
1
w = K1 e K2 e K3 z . To place the poles at s = 20 and s = 2 2j we can use Ackermanns formula. We also need an additional pole (must be a stable pole); select
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
596
CHAPTER 11
The Design of State Variable Feedback Systems
s = 20 as the fourth pole. Then, K1 = 3200 K2 = 1920 K3 = [568 44] . The compensator has the form shown in Figure 11.16 in Dorf and Bishop. P11.25 The observability matrix is Po =
C CA
and det Po = 48 = 0; therefore the system is completely observable. The desired poles of the observer are s1,2 = 1. This implies that the desired characteristic polynomial is pd (s) = s2 + 2s + 1 . The actual characteristic polynomial is det |I (A LC)| = det 1 + L1 5 + L2 4 4L1 10 4L2
=
1
4
21 36
,
= 2 + (L1 4L2 11) + 10L1 + 8L2 + 30 = 0 . Solving for L1 and L2 yields L=
L1 L2
Checking we nd that det(I (A LC)) = s2 + 2s + 1. The response of the estimation error is shown in Figure P11.25, where e(0) = [ 1 1 ]T .
=
0.25 3.3125
.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
597
Response to Initial Conditions 2.5 2 To: Out(1) Amplitude To: Out(2) 1.5 1 0.5 0 1 0.5 0 -0.5 ?-1 -1.5 0 1 2 3 Time ( sec ) 4 5 6
FIGURE P11.25 Estimation error response to an initial condition.
P11.26
The observability matrix is
C Po = CA = 2 0
4 2 8
CA
2
4
14
0 4 .
The det Po = 184 = 0, hence the system is observable. The gain matrix
5.45 L = 0.48
1.24
results in the observer poles at s1,2 = 1 2j and s3 = 10, as desired. P11.27 The observability matrix is Po =
C CA
=
1
0
10
.
The det Po = 0, hence the system is not completely observable. So, we cannot nd an observer gain matrix that places the observer poles at the desired locations.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
598 P11.28
CHAPTER 11
The Design of State Variable Feedback Systems
Selecting K = 16 yields a zero steady-state error to a unit step input. The step response is shown in Figure P11.28.
Step Response 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Amplitude
0
0.5
1 Time (sec)
1.5
2
2.5
FIGURE P11.28 Estimation error response to an initial condition.
P11.29
The system transfer function is Y (s) = 2 U (s) . s+3
The associated state variable model is x = 3x + 2u y=x.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
599
Advanced Problems
AP11.1 The closed loop system in state-space form is given by
x1 x = 2 x3
0 0
1 1
0 2
2KK1 2KK2 4 2KK3
x1 0 x + 0 u 2
x3
2K
y=
100
x1 x . 2
x3
The closed-loop transfer function is T (s) = s3 + (2KK3 + 5)s2
4K . + (4KK2 + 2KK3 + 4)s + 4KK1
Setting the steady-state error to zero, we determine that ess = 1 T (0) = 1 Solving for K1 yields K1 = 0.5 . Choosing K2 = 0.5 and K3 = 1.5 1 . K1
results in a percent overshoot of P.O. = 2.82%. AP11.2 A state variable representation is given by x = Ax + Bu where
3 A= 4
0
1.75 1.25 0 0, 1 0
2 B= 0 .
0
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600
CHAPTER 11
The Design of State Variable Feedback Systems
Let u = Kx . Then, with K= 5 6.125 9.375 ,
the closed-loop system poles are s = 4, 4, and 5. AP11.3 Given A=
01 20
we compute the determinant of the controllability matrix as det Pc = det[B AB] = 2b2 b2 . 1 2 The system is controllable if and only if the determinant is non-zero. So, for the system to be controllable, we require that b2 = 2b2 . 2 1 AP11.4 Consider the state variable feedback law u = Kx . Using Ackermanns formula, we determine that K = [14.2045 17.0455 94.0045 31.0455] results in the closed-loop system characteristic roots at s = 2 j , s = 5 and s = 5. AP11.5 The closed-loop transfer function for the system is T (s) = s3 + (9 + 2K3 )s2 2Kp . + (26 + 2K2 + 10K3 )s + (26 + 6K2 + 12K3 )
,
and
B=
b1 b2
,
Setting the steady-state error for a step input to zero yields ess = 1 2Kp =0. 26 + 6K2 + 12K3
Solving for Kp in terms of K2 and K3 yields Kp = 13 + 3K2 + 12K3 . Now, choosing K2 = 5
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
601 K3 = 2
results in the closed-loop characteristic roots at s 1 = 4 s 2 = 4 s 3 = 5 . Also, the prelter gain is Kp = 52 . AP11.6 (a) A state variable representation is given by A= C=
0
1
1 2 10 .
,
B=
0 1
,
Since the determinant of the controllability matrix det[B AB] = 0, the system is controllable. (b) The state variable representation is x = Ax + Bu , or
x1 x2
The determinant of the controllability matrix
=
0
1
1 2
x1 x2
+
1 1
u .
det Pc = det[B AB] = 0 . Therefore, the system is uncontrollable. AP11.7 The closed-loop transfer function is T (s) = s3 + (10 + 60K3 )s2 120 . + (16 + 120(K3 + K2 ))s + 120
The state feedback gains K2 = 0.283 and K3 = 0.15
place the poles at the desired locations. The plot of the roll output for a unit step disturbance is shown in Figure AP11.7.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
602
CHAPTER 11
The Design of State Variable Feedback Systems
0.35
0.3
0.25
Amplitude
0.2
0.15
0.1
0.05
0 0
0.5
1
1.5
2 Time (secs)
2.5
3
3.5
4
FIGURE AP11.7 Roll angle response to a step disturbance.
AP11.8
The state equations are (using the parameters of P3.36 in Dorf and Bishop) 8 1 h = x1 = [80 50h] = x1 + x2 50 5 = x2 = = x3 Km Km Kb Km Ka 353 25000 = x3 = ia = + vi = x3 + vi . J JRa JRa 30 3 In state variable form we have (without feedback)
8 1 5 x= 0 0
0 1
0
0 353 30
(a) In this case we have vi = kh + ar = kx1 + ar , where k and a are the parameters to be determined and r is the reference input. With the feedback of h(t) we have
x +
0 0
25000 3
vi .
x=
1 0
8 5
0 1
0
25000 k 0 353 3 30
x+
0 0 a 25000 3
r .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
603
Since we only have one parameter to adjust, namely k, we will probably not be able to simultaneously meet both design specications, In fact with k = 0.00056 we obtain the percent overshoot P.O. = 9.89%. The settling time criterion cannot simultaneously be metthe best that can be obtained is Ts 7.5 seconds. In this case, we choose a = 0.00056 to make the steady-state value of h(t) = 1. (b) In this case we have vi = k1 h k2 + ar = k1 x1 k2 x2 + ar , where k1 , k2 , and a are the parameters to be determined and r is the reference input. Since we have two parameter to adjust, namely k1 and k2 we will probably be able to simultaneously meet both design specications. In fact with k1 = 0.00056 and k2 = 0.001
we obtain the percent overshoot P.O. = 4.35%. The settling time criterion is easily met Ts 5 seconds. In this case, we choose a = 0.0012 to make the steady-state value of h(t) = 1. AP11.9 (a) The state vector dierential equation is
01 2 0 x= 00
1 0 1 0
0 0 1 0 0
where x1 = z , x2 = z , x3 = y and x4 = y . (b) The characteristic equation is
0 0 u , x + 0 1
1
s4 + 3s2 + 1 = (s + j 0.618)(s j 0.618)(s + j 1.618)(s j 1.618) = 0 . So, the system is oscillatory. (c) Let u = kx4 . Then characteristic equation is s4 + ks3 + 3s2 + 2ks + 1 = 0 which is stable if k > 0. (d) Rewrite the characteristic equation as 1+ ks(s2 + 2) =0. s4 + 3s2 + 1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
604
CHAPTER 11
The Design of State Variable Feedback Systems
The root locus is shown in Figure AP11.9. A reasonable solution for k is k = 1.35.
3
2
x o
1
x
Imag Axis
0
o
x
-1
o x
-2
-3 -3
-2
-1
0 Real Axis
1
2
3
FIGURE AP11.9 s(s2 +2) Root locus for 1 + k s4 +3s2 +1 = 0.
AP11.10
The state dierential equation is y = ky + u where k and depend on the system parameters, such as mass and length. The transfer function is y =2 u s k which is unstable at the top of the arc. Since we can only use y for feedback, we have y s =2 . u s k Let Gc (s) = K1 s + K2 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
605
Then GGc (s) = (K1 s + K2 ) (s2 k)
and the closed-loop characteristic equation is K1 s + K2 + s2 k = 0 or s2 + K1 s + K2 k = 0 . Select K2 k > 0 and K1 > 0 for stability. AP11.11 The state-space representation of the plant is x = Ax + Bu y = Cx where A=
0
1
2 3
With the intermediate variables dened as
,
B=
0 1
, and
C=
10
.
z = x and w = u we have 1 0 0 e = 0 0 1 z 0 2 3
0 e + 0 w z 1
where
e=yr . To meet the design specications, we require the closed-loop poles to lie to the left of the line in the complex plane dened by s = 0.8. We choose K2 = [10 3] and Gc (s) = 8 . s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
606
CHAPTER 11
The Design of State Variable Feedback Systems
This places the closed-loop poles at s = 2, 2 and 2. The closed-loop transfer function with the internal model controller is T (s) = 8 . s3 + 6s2 + 12s + 8
The step response is shown on Figure AP11.11.
1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 2 4 6 8 10 12 14
Time (secs)
FIGURE AP11.11 Internal model controller step response.
AP11.12
The state-space representation of the plant is x = Ax + Bu y = Cx where A=
0
1
2 3
,
B=
0 1
, and
C=
10
.
With the intermediate variables dened as z = x and w = u
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
607
we have
where e = y r .
6
e e = z
0 1 0 0
0 e 1 0 e + 00 0 1 z 0 0 2 3
0
0 0 0 1
w
5
4
Amplitude
3
2
1
0 0
1
2
3 Time (secs)
4
5
6
FIGURE AP11.12 Internal model controller ramp response.
To meet the design specications, we require the closed-loop poles to lie to the left of the line in the complex plane dened by s = 0.67. We choose
e e K3 ] e = [16 32 22 5] e .
w = [K1 K2
z
z
Then,
Gc (s) =
16 + 32s K1 + K2 s = . 2 s s2
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
608
CHAPTER 11
The Design of State Variable Feedback Systems
The closed-loop transfer function with the internal model controller is T (s) = s4 8s3 32s + 16 . + 24s2 + 32s + 16
+
This places the closed-loop poles at s = 2, 2, 2 and 2. The ramp response is shown in Figure AP11.12. AP11.13 The controllability matrix is
Pc = and the observability matrix is
4 2 1 10
Po =
6 26
4 52
.
Computing the determinants yields det Pc = 38 = 0 and det P0 = 416 = 0 ,
hence the system is controllable and observable. The controller gain matrix K= 3.55 7.21
places the closed-loop poles at the desired locations. Similarly, the observer gain matrix
L= AP11.14 The controllability matrix is
0 Pc = 0
1.38 0.67
places the observer poles at the desired locations.
0 4
4
4 24
24
124
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
609
and the observability matrix is 2 2 9 Po = 8 8 21 Computing the determinants yields det Pc = 64 = 0 and
.
84
97
118
det P0 = 9358 = 0 ,
hence the system is controllable and observable. The controller gain matrix K= and the observer gain matrix
46.54 L = 0.46
0 0.25 0.5
31.60
yields the desired closed-loop system poles and observer poles, respectively. AP11.15 The state-variable representation of the system is x= The observability matrix is P0 =
0
1
7 2
y = [ 1 4 ]x + [0]u .
x+
0 1
u
1
4
28 7
and det P0 = 105 = 0, hence the system is observable. The observer gain matrix L=
,
7.18 6.29
places the observer poles at s1,2 = 10 10, as desired.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
610
CHAPTER 11
The Design of State Variable Feedback Systems
Design Problems
CDP11.1
A state variable representation is x= y=
0
1
0 33.14 10 x
x+
0 0.827
va
where x1 = x and x2 = x. Note that we are neglecting the motor induc tance and assuming that the position x(t) is the output. Assume that we have available for feedback the angle and angle rate (see CDP4.1), so that va = k1 k2 x1 x2 + au r r
where u(t) is the reference input (that is, the desired position x(t)), the gains k1 and k2 and the scaling parameter a are to be determined. Recall that x = r = 0.03175 . With the feedback in the loop we have x= y=
0
1
26.03k1 33.14 26.03k2 10 x
x+
0 0.827a
u
Choosing k1 = 50, k2 = 1 and a = 1574.1 results in P.O. = 1.1% and Ts = 0.11 second .
The closed-loop poles are s1,2 = 29.59 20.65j . DP11.1 The governing dierential equation is y 2055y = 18i . In state variable form, the system is described by x=
0
1
2055 0
x +
0 18
i .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
611
Consider the state feedback i = k1 x1 + k2 x2 + r where r (t) is the reference input and k1 , k2 and are to be determined. Then, the closed-loop system is x=
0
1
2055 18k1 18k2
x+
0 18
r .
The characteristic equation is s2 + 18k2 s 2055 + 18k1 = 0 . For stability, let 18k1 2055 > 0. Select k1 = 200. Then, n = 39.3 rad/sec, and k2 = 2n . 18
Let = 0.59 to meet 10% overshoot specication. Thus, k2 = 2(0.59)(39.3) = 2.58 . 15
The closed-loop transfer function is T (s) = Choose = 85.83 so that T (s) = The feedback law is i = 200x1 + 2.58x2 85.83r . DP11.2 The automobile engine control system (see DP10.8 in Dorf and Bishop) is modeled as G(s) = 2esT . (0.21s + 1)(4s + 1) s2 1545 . + 46.4s + 1545 s2 18 . + 6.4s + 1545
In this case, we will assume the delay is negligible. Therefore, T = 0. A
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612
CHAPTER 11
The Design of State Variable Feedback Systems
state variable representation of the system is x= Let r (t) = k1 x1 k2 x2 + k3 u where u(t) is the command input. Using ITAE methods, our desired characteristic polynomial is
2 q (s) = s2 + 1.4n s + n = 0 .
0
1
y = [1 0]x .
1.19 5.01
x +
0 1.19
r
Select n = 11.315 to obtain a settling time Ts < 0.5 seconds. The characteristic polynomial of the closed-loop system is s2 + (5.01 + 1.19k2 )s + (1.19 + 1.19k1 ) = 0 . Equating coecients and solving for the gains yields k1 = 106.59 and k2 = 9.235 .
Select k3 = 107.59 to yield a zero steady-state error to a step input.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time (secs)
FIGURE DP11.2 The step response of the engine control system.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
613
DP11.3
The compensator is x = [A BK LC] x + Ly + Mr u = Kx where A BK LC = N = 363.64 , K=
28.7
1
365.19 20
,
M=
0 200
,
344.55 15.82
, and L =
28.7 165.19
.
We selected the desired eigenvalues of A BK at p = 10 10j and the desired eigenvalues of A LC at q = 20 10j . For initial conditions we let x(0) = [1 1] and x(0) = [0 0].
1.5
Actual x1
1 x1
0.5
Estimated x1
0
0
0.1
0.2
0.3
0.4
0.5 Time (s)
0.6
0.7
0.8
0.9
1
6 4 x2 2 0
Estimated x2
2
Actual x2
0 0.1 0.2 0.3 0.4 0.5 Time (s) 0.6 0.7 0.8 0.9 1
FIGURE DP11.3 The step response showing the actual and estimated states.
DP11.4
The design specications are (a) Percent overshoot < 20% (b) Ts < 1.5s, and
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
614
CHAPTER 11
The Design of State Variable Feedback Systems
(c) steady-state error less than 20% of the input magnitude. The state dierential equation is x = Ax + Bu y = Cx where 1 0 0 A = 0 1 1
0 = 0
1 0.415 1.43
0
g 2 2
9.8
0.0198
0.0111 ,
0 0 B = n = 6.27
and C =
100
.
g
9.8
The transfer function is
ns + n2 1 g (s) =3 (s) s + (1 + 2 )s2 + (1 2 1 2 )s + 1 g 6.27s + 0.0154 =3 . s + 0.435s2 0.0077 + 0.109 Let u = K1 x1 K2 x2 K3 x3 . Then the closed-loop system matrix is 0 A BK = nK1
1 1 nK2
0
g gK1 2 gK2
1 nK3 ,
2 gK3
where K = [K1 K2 K3 ]. From the design specications, we have the desired roots at s3 +a2 s2 +a1 s+ao = s3 +36s2 +225s+1350 = (s+30)(s+3+j 6)(s+3j 6) = 0 . The actual characteristic equation is s3 + (gK3 + K2 n + 1 + 2 )s2 + (1 2 1 gK2 + K1 n 2 nK3 + gK3 1 + K2 n2 + 1 2 )s + 1 g 1 gK1 + gK3 n + 2 nK1 = 0 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
615
Comparing coecients yields
0 n 1 g + 2 n
a2 1 2 K1 n2 1 g 2 n + g1 K2 = a1 + 1 2 1 2 0 gn
n
g
K3
a0 1 g
where
a2 = 36 a1 = 225 a0 = 1350 . The solution for K is K = [53.11 28.64 21.96] . DP11.5 The controllability and observability matrices are Pc =
0.05
0.04 0
0.001 0.001 1 0.8 0.02
and
Computing the determinants yields det Pc = 1.002e 05 = 0
P0 =
, respectively.
and Po = 0.02 = 0 ,
hence the system is controllable and observable. The feedback gain matrix K = [ 3820 179620 ] yields the desired closed-loop poles. The observer gain matrix L=
120 180000
yields the desired observer poles. The integrated system is shown in Figure DP11.5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
616
CHAPTER 11
The Design of State Variable Feedback Systems
A=
System Model
-0.8 0.02 -0.02 0
B=
0.05 0.001
u
. x=Ax+Bu
x
C
y
Control Law
Observer
-K
^ x
. ^=Ax+Bu+L~ x^ y
~ y-Cx y= ^ C
+
C= 1 0
K= 3820 -179620 L=
FIGURE DP11.5 Integrated controller and observer.
120 180000
DP11.6
(a) The characteristic equation associated with the system matrix is q (s) = s2 + (12 + K2 )s + (36 + K1 ) = 0 , where we have assumed state feedback of the form u = K1 x1 K2 x2 . The deadbeat control characteristic equation is
2 s2 + n s + n = 0 ,
where = 1.82 and we use n = 9.64 to meet the settling time specication. Then, equating coecients and solving for the gains yields K1 = 56.93 and K2 = 5.54 .
(b) Since the closed-loop poles are located at s1,2 = 8.77 4, we can select the observer poles to be about ten times farther in the left-half plane, or s1,2 = 88, 88 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
617
Then the observer gains are
L=
164 5740
.
(c) The block diagram is shown in Figure DP11.6.
A=
System Model
01 -36 -12
B=
0 1
u
. x=Ax+Bu
x
C
y
Control Law
Observer
-K
^ x
. ^=Ax+Bu+L~ x^ y
~ y-Cx y= ^ C
+
C= 1 0
K= 56.93 5.54 L=
164 5740
FIGURE DP11.6 Block diagram for integrated controller and observer.
DP11.7
The compensator is x = [A LC] x + Ly + Bu u = Kx where 60 A LC = 1095
1 0
3750 5 10
0 1,
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
618
CHAPTER 11
The Design of State Variable Feedback Systems
N = 4000 ,
K=
3998 595 30
60 , and L = 1095 .
3748
We selected the desired eigenvalues of A BK at p1,2 = 10 10j , p3 = 20 and the desired eigenvalues of A LC at q1,2 = 20 10j , q3 = 30. For initial conditions we let x(0) = [1 1 1] and x(0) = [0 0 0]. The transfer function from r to y is T (s) = 4000s3 + 2.8e05s2 + 6.8e06s + 6e07 . s6 + 110s5 + 5100s4 + 1.29e05s3 + 1.9e06s2 + 1.58e07s + 6e07
The bandwidth is 11.7 rad/s.
1.5 1 0.5 0 20 10 0 10 200 100 0 100 200 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
FIGURE DP11.7 The step response showing the actual and estimated states.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
619
Computer Problems
CP11.1 The controllability and observablity matrices have nonzero determinants, as shown in Figure CP11.1. Therefore, the system is observable and controllable.
A=[-4 2 0;4 0 -6;-10 1 11]; b=[-3;0;1]; c=[ 1 0 1]; d=[0]; sys = ss(A,b,c,d); Co=ctrb(sys); dt_Co=det(Co) Ob=obsv(sys); dt_Ob=det(Ob)
>>
dt_Co = -11340 dt_Ob = 860
FIGURE CP11.1 Determining controllability and observability.
CP11.2
The system is controllable since the determinant of the controllability matrix is nonzero , as shown in Figure CP11.2.
a=[0 1;-2 -6.5]; b=[0;2]; c=[1 0]; d=[0]; sys_ss = ss(a,b,c,d); Pc=ctrb(sys_ss); det(Pc) sys_tf=tf(sys_ss) Transfer function: 2 --------------s^2 + 6.5 s + 2
ans = -4
FIGURE CP11.2 M-le script to determine controllability and to compute equivalent transfer function model.
CP11.3
The gain matrix (computed as shown in Figure CP11.3) is K= 0.5 0.5 .
a=[0 1;-1 -2]; b=[1;1]; c=[1 -1]; d=[0]; p=[-1;-2]; K=acker(a,b,p) K=
0.5000 0.5000
FIGURE CP11.3 M-le script to place the closed-loop system poles using state feedback.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
620 CP11.4
CHAPTER 11
The Design of State Variable Feedback Systems
The constant velocity guided missile is not controllable since the controllablity matrix, Co , has a zero determinant, as shown in Figure CP11.4. Using the tf function (see Figure CP11.4), we determine that the transfer function is G(s) = s5 5s . + 0.5s4 + 0.1s3
Cancelling common terms in the transfer function yields G(s) = 5 . + 0.5s3 + 0.1s2
s4
Then, using the ss function, we determine a state-space representation of G(s). As shown in Figure CP11.4, the state-space representation is x = Ax + Bu y = Cx
A=[0 1 0 0 0;-0.1 -0.5 0 0 0;0.5 0 0 0 0;0 0 10 0 0;0.5 1 0 0 0]; b=[0;1;0;0;0]; c=[0 0 0 1 0]; d=[0]; sys_ss = ss(A,b,c,d); Transfer function: % Part (a) dt_Co = 5s Co=ctrb(sys_ss); dt_Co=det(Co) 0 ----------------------% Part (b) s^5 + 0.5 s^4 + 0.1 s^3 sys_tf = tf(sys_ss) sys_new = minreal(sys_tf ); sys_new_ss=ss(sys_new) a= % Part (c) x1 x2 x3 x4 Co_new=ctrb(sys_new_ss); dt_Co_new=det(Co_new) x1 -0.50000 -0.10000 0 0 % Part (d) x2 1.00000 0 0 0 evalues=eig(sys_new_ss) x3 0 1.00000 0 0 dt_Co_new = x4 0 0 2.00000 0 32 b= u1 evalues = x1 2.00000 0 x2 0 0 x3 0 -0.2500 + 0.1936i x4 0 -0.2500 - 0.1936i c= x1 x2 x3 x4 y1 0 0 0 1.25000 d= u1 y1 0 Continuous-time system.
FIGURE CP11.4 Analysis of the constant velocity guided missile state-space model.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
621
where
0.5 0.1 0 0 1 0 0 0 2 0 0
A=
0 0
1 0
0 0 20
B=
and C =
0 0 0 1.25
.
0
The reduced system is controllable but not stable, since there are two poles at the origin. Systems that are not controllable have too many states. After eliminating unnecessary states, a controllable system of minimal complexity (i.e. states) is obtained. In this case, the number of states is reduced from ve to four. CP11.5 The eigenvalues of A are e1 = 2.0727 e2 = 0.2354 e3,4 = 0.2761 0.2593j The system is unstable since there are two eigenvalues in the right halfplane, see Figure CP11.5. The characteristic polynomial is
A = [-0.0389 0.0271 0.0188 -0.4555; 0.0482 -1.0100 0.0019 -4.0208; >> 0.1024 0.3681 -0.7070 1.4200; 0 0 1 0]; evalues = b1 = [0.4422;3.5446;-6.0214;0]; 0.2761 + 0.2593i b2 = [0.1291;-7.5922;4.4900;0]; 0.2761 - 0.2593i % Part (a) -0.2354 evalues = eig(A) -2.0727 %part (b) p= p = poly(A) 1.0000 1.7559 -0.6431 0.0618 0.0700 r = roots(p) % Part (c) dt1 = Co1 = ctrb(A,b1); dt1 = det(Co1) -1.8451e+03 Co2 = ctrb(A,b2); dt2 = det(Co2)
dt2 = -90.6354
r= -2.0727 0.2761 + 0.2593i 0.2761 - 0.2593i -0.2354
FIGURE CP11.5 Analysis of the VTOL aircraft model.
p(s) = s4 + 1.7559s3 0.6431s2 + 0.0618s + 0.0700 . The roots of the characteristic equation are the same as the eigenvalues. Also, the system is controllable from either u1 or u2 . If the aircraft should lose the control of the vertical motion through u1 , then the control u2 can
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622
CHAPTER 11
The Design of State Variable Feedback Systems
be used to control both vertical and horizontal motion, and vice versa. CP11.6 The m-le script to analyze the translunar halo orbit problem is shown in Figure CP11.6. The translunar equilibrium point is not a stable point
A=[0 0 0 1 0 0;0 0 0 0 1 0; 0 0 0 0 0 1;7.3809 0 0 0 0 -2.1904 0 -2 0 0; 0 0 -3.1904 0 0 0]; c=[0 1 0 0 0 0];d=[0]; b1=[0;0;0;1;0;0]; b2=[0;0;0;0;1;0]; b3=[0;0;0;0;0;1]; sys_ss_1 = ss(A,b1,c,d); sys_ss_2 = ss(A,b2,c,d); dt1 = sys_ss_3 = ss(A,b3,c,d); 0 % Part (a) evalues=eig(A) dt2 = % Part (b) 0 Cb1=ctrb(sys_ss_1); dt1=det(Cb1) dt3 = % Part (c) 0 Cb2=ctrb(sys_ss_2); dt2=det(Cb2) % Part (d) Cb3=ctrb(sys_ss_3); dt3=det(Cb3) % Part (e) sys_tf = tf(sys_ss_2); sys_tf=minreal(sys_tf ) % Part (f ) sys_ss=ss(sys_tf ); Co=ctrb(sys_ss); dt_Co=det(Co) if dt_Co ~= 0 disp('System is completelly Controllable') else disp('System in uncontrollable') end % Part (g) P = [-1+i; -1-i;-10;-10]; [A,B]=ssdata(sys_ss); K = acker(A,B,P) dt_Co = 64
2 0;
evalues = 2.1587 -2.1587 0 + 1.8626i 0 - 1.8626i 0 + 1.7862i 0 - 1.7862i Transfer function: s^2 - 7.381 ---------------------------s^4 - 1.19 s^2 - 16.17
a= x1 x2 x3 x4 b= x1 x2 x3 x4 c= y1 d= y1 u1 0 x1 0 x2 0.50000 x3 0 x4 -0.92261 u1 1.00000 0 0 0 x1 x2 x3 x4 0 0.59525 0 2.02089 2.00000 0 0 0 0 2.00000 0 0 0 0 2.00000 0
System is completelly Controllable
FIGURE CP11.6 Analysis of the translunar satellite halo orbit.
as evidenced by the eigenvalues of A in the right half-plane; the system is not completely controllable from any ui individually. The transfer function
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
623
from u2 to is T (s) = s6 s4 4.191s2 23.55 . + 2s4 19.97s2 51.58
A careful analysis reveals that T (s) can be reduced by eliminating common factors. The common factors are s2 + 3.1834. The reduced transfer function is T (s) = Using state feedback u2 = Kx the gain matrix K which places the desired poles (using Ackermanns formula) is K= CP11.7 22 71.56 60 27.02 . s2 7.3815 . s4 1.1837s2 16.2030
The m-le script to determine the initial state is shown in Figure CP11.7a. Given three data points at t = 0, 2, 4, we construct the three equations
A=[0 1 0;0 0 1;-2 -4 -6]; b=[0;0;0]; c=[1 0 0]; d=[0]; sys=ss(A,b,c,d); % % Part (b) v1=c*expm(0*A); v2=c*expm(2*A); v3=c*expm(4*A); V=[v1;v2;v3]; Vi=inv( V ); n=[1;-0.0256;-0.2522]; x0=Vi*n % % Part (c) t=[0:0.1:4]; u=0.0*t; [y,x]=lsim(sys,u,t,x0'); plot(t,y,[0 2 4],[1;-0.0256;-0.2522],'*'), grid xlabel('Time (sec)'), ylabel('y(t)') title('Data points denoted by *')
FIGURE CP11.7 (a) Script to determine the initial state from three observations.
y (0) = 1 = Ce0A x0 y (2) = 0.0256 = Ce2A x0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
624
CHAPTER 11
The Design of State Variable Feedback Systems
y (4) = 0.2522 = Ce4A x0 or, in matrix form
Ce0A Ce4A
The problem is solvable if the matrix
Ce2A
1 x0 = 0.0256 0.2522
0A Ce Ce2A
.
Ce4A
is invertible. In this case, the inverse does exist and the solution is
x0 =
1 1 1.9998
The simulation is shown in Figure CP11.7b.
.
Data points denoted by * 1*
0.8
0.6
0.4
y(t)
0.2 0
*
-0.2
*
-0.4
0
0.5
1
1.5
2 Time (sec)
2.5
3
3.5
4
FIGURE CP11.7 CONTINUED: (b) System simulation using computed initial state.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
625
CP11.8
Suppose we are given A= and the feedback u = Kx = [K1 K2 ]x . Solving HT P + PH = I for P yields p11 = p12 p22
2 2 K2 + K1 + 3K1 + 2 2(K1 + 1)K2 1 = 2(K1 + 1) K1 + 2 = 2(K1 + 1)K2
0
1
1 0
B=
0 1
Then, with xo T = [1, 0] we nd that J = xo T Pxo = p11 . Computing the partial of J with respect to K2 yields J 1 1 K1 + 2 = 2 K2 2 K1 + 1 K2 Setting J =0 K2 and solving for K2 , we nd that K2 = (K1 + 2)(K1 + 1) . .
For a given value of K1 , the value of K2 that minimizes J can be computed via the above equation. With K2 given as above, we can compute J to be J= K1 + 2 . K1 + 1
A plot of J versus K1 (with K2 equal to the minimizing value) is shown in Figure CP11.8. As K1 increases, the performance index J decreases. However, we see that the rate of decrease slows considerably after K1 > 20. Also, K2 increases as K1 increases. We want to keep both gains as
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
626
CHAPTER 11
The Design of State Variable Feedback Systems
small as possible, while still having a small J . A reasonable selection is K1 = 20 and K2 = 21.5 .
1.5 1.4 1.3
Performance index J versus K1
J
1.2 1.1 1 0 5 10 15 20 25 K1 K2 versus K1 30 35 40 45 50 60 40
K2
20 0 0
5
10
15
20
25 K1
30
35
40
45
50
FIGURE CP11.8 Performance index as a function of K1 and K2 .
CP11.9
In this problem, A = 1 and B = 1. Computing Q yields Q = (1 + (k)2 ) = 1 + k2 . Dene H = A Bk = 1 k . Solving H T P + P H = Q yields p= 1 + k2 . 2(k + 1)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
627
0.5 0.49 0.48 0.47
J/x0^2
0.46 0.45 0.44 0.43 0.42 0.41
0
0.1
0.2
0.3
0.4
0.5 k
0.6
0.7
0.8
0.9
1
2.5
2
1.5
k min
1 0.5 0 0
1
2
3
4
5 lambda
6
7
8
9
10
FIGURE CP11.9 Plot of J/x2 versus k and the minimizing k versus . 0
The performance index is J = x2 p which implies 0 J/x2 = 0 1 + k2 . 2(k + 1)
The plot of J/x2 versus k is shown in Figure CP11.9. The minimum value 0 is achieved when k = 0.41. To arrive at this result analytically, take the
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628
CHAPTER 11
The Design of State Variable Feedback Systems
partial of J/x2 with respect to k, set the result to zero and solve for k: 0 J/x2 0 = 0 when k2 + 2K 1/ = 0 . k Solving for k yields k = 1 1 + 1/. So, when = 1, k = 0.41. The plot of kmin versus is shown in Figure CP11.9. CP11.10 The m-le is shown in Figure CP11.10.
A=[0 1;-19.04 -11.42]; B=[12.8;24.6]; C=[1 0]; D=[0]; % Controller Gains p=[-1;-1 ]; K=acker(A,B,p) % Observer Gains q=[-10+4*j;-10-4*j]; L = acker(A',C',q); L=L' >> K= -0.3742 -0.1882
L= 8.5800 -1.0236
FIGURE CP11.10 Using the acker function to compute the controller gains and the observer gains.
CP11.11
The m-le is shown in Figure CP11.11(a). The compensator can be repA=[0 1 0;0 0 1;-4.3 -1.7 -6.7]; B=[0;0;0.35]; C=[0 1 0]; D=[0]; % Controller Gains p=[-1.4+1.4*j;-1.4-1.4*j;-2]; K=acker(A,B,p) % Observer Gains q=[-18+5*j;-18-5*j;-20]; L = acker(A',C',q); L=L' % Simulation of closed-loop system with the observer Ac=[A -B*K;L*C A-B*K-L*C]; Bc=[zeros(6,1)]; Cc=eye(6); Dc=zeros(6,1); sys=ss(Ac,Bc,Cc,Dc); x0=[1;0;0;0.5;0.1;0.1]; t=[0:0.001:3.5]; [y,t]=initial(sys,x0,t); subplot(311) plot(t,y(:,1),t,y(:,4),'--'), grid subplot(312) plot(t,y(:,2),t,y(:,5),'--'), grid subplot(313) plot(t,y(:,3),t,y(:,6),'--'), grid >> K= 10.1143 22.3429 -5.4286
L= 1.0e+003 * -1.6223 0.0493 0.7370
FIGURE CP11.11 (a) M-le.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
629
resented as x = (A BK LC) + Ly x Since y = Cx, we can write x = (A BK LC) + LCx . x Similarly, with x = Ax + Bu we obtain x = Ax BKx . In matrix form, we have
and
u = Kx .
and u = Kx
x x
=
A
BK
x x
LC A BK LC
T
,
T
with initial conditions x(0)T x(0)T = 1 0 0 0.5 0.1 0.1 .
The response of the system is shown in Figure CP11.11(b).
5
x1 0 True state 5 0 (solid line) 1 x2 0
1 2 0 1 1
Estimated state (dashed line)
2
3
4
2
3
4
x3
0 2 0 1 2 Time (sec) 3 4
FIGURE CP11.11 CONTINUED: (b) Response of system to an initial condition.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
630 CP11.12
CHAPTER 11
The Design of State Variable Feedback Systems
The Simulink block diagram is shown in Figure CP11.12.
FIGURE CP11.12 Simulink block diagram.
CP11.13
The m-le to design the compensator is shown in Figure CP11.13(a). The Simulink simulation is shown in Figure CP11.13(b). The output shown on the x-y graph depicts the state x of the system. The initial conditions selected for the simulation are
1 0 0 0.5 0.1
x(0) =
0
and x(0) =
. 0.1
0.1
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
631
A=[0 1 0 0;0 0 1 0;0 0 0 1;-2 -5 -1 -13]; B=[0;0;0;1]; C=[1 0 0 0]; D=[0]; % Controller Gains p=[-1.4+1.4*j;-1.4-1.4*j;-2+j;-2-j]; K=acker(A,B,p) % Observer Gains q=[-18+5*j;-18-5*j;-20;-20]; L = acker(A',C',q); L=L' % Simulation of closed-loop system with the observer Ac=[A -B*K;L*C A-B*K-L*C]; Bc=[zeros(8,1)]; Cc=eye(8); Dc=zeros(8,1); sys=ss(Ac,Bc,Cc,Dc); x0=[1;0;0;0;0.5;0.1;0.1;0.1]; t=[0:0.001:10]; [y,t]=initial(sys,x0,t); subplot(311) 100 plot(t,y(:,1),t,y(:,4),'--'), grid subplot(312) 0 plot(t,y(:,2),t,y(:,5),'--'), grid subplot(313) ?100 0 2 plot(t,y(:,3),t,y(:,6),'--'), grid 2
0 ?2 10 0 ?10 0 2 4 6 8 10 0 2 4 6 8 10
>> K= 17.6000 24.6800 19.1200 -6.2000
L= 63 1369 10495 1479
4
6
8
10
FIGURE CP11.13 (a) M-le to design the compensator, including the observer.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
632
CHAPTER 11
The Design of State Variable Feedback Systems
. ^=[A-BK-LC]x+Ly ^ x ^ u=-Kx
FIGURE CP11.13 CONTINUED (b) The Simulink simulation.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
12
Robust Control Systems
Exercises
E12.1 The plant transfer function is G(s) = Try a PI controller, given by Gc = K1 + The ITAE characteristic equation is
2 s2 + 1.4n s + n ,
3 . s+3 K2 . s
where n = 30. Then K1 = 13 and K2 = 300 .
Without a prelter, the closed-loop system is Y (s) 39s + 900 =2 , R(s) s + 42s + 900 and with a prelter, the closed-loop system is Y (s) 900 =2 , R(s) s + 42s + 900 where Gp (s) = 23.07 . s + 23.07
The step response, with and without the prelter, is shown in Figure E12.1.
633
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
634
CHAPTER 12
Robust Control Systems
1.4 Without prefilter With prefilter 1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
0.05
0.1
0.15
0.2
0.25 Time (sec)
0.3
0.35
0.4
0.45
0.5
FIGURE E12.1 Step response: (a) w/o prelter (solid line), and (b) w/prelter (dashed line).
E12.2
The disturbance response is shown in Figure E12.2.
0.05
0.04
0.03
y(t)
0.02
0.01
0
0.01
0
0.05
0.1
0.15
0.2
0.25 Time (sec)
0.3
0.35
0.4
0.45
0.5
FIGURE E12.2 Disturbance response for system in E12.1.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
635
E12.3
The closed-loop transfer function is T (s) = and the sensitivity function is S (s) = s2 + bs , s2 + bs + 18 s2 18 , + bs + 18
where b = 5, nominally. The sensitivity of T to changes in p is determined to be
T Sb =
T b bs =2 . b T s + bs + 18
The plot of T (s) and S (s) is shown in Figure E12.3, where b = 5.
10
0
10
20 Gain dB
30
20log|S|
20log|T|
40
50
60 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE E12.3 Plot of T (s) and the sensitivity function S (s).
E12.4
The plant transfer function is G(s) = 1 , (s + 2)(s + 8)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
636
CHAPTER 12
Robust Control Systems
and the PID controller is given by Gc (s) = K3 (s + z )(s + z ) . s
Let z = 1 + j and let K3 = 180. Then, the closed-loop system is T (s) = 180(s2 + 2s + 2) . s3 + 190s2 + 376s + 360
A root locus analysis (with K3 as the parameter) will verify that z = 1 + j is a good choice for K3 = 180. The actual response rises to 0.95 and then asymptotically approaches 1, while the approximation reaches 1 in less than 0.1 sec. The plot is shown in Figure E12.4.
1.2
1
approximation actual
0.8
y(t)
0.6
0.4
0.2
0 0
0.05
0.1
0.15
0.2
0.25 Time (sec)
0.3
0.35
0.4
0.45
0.5
FIGURE E12.4 Step response for closed-loop actual and approximate transfer functions.
E12.5
The loop transfer function is Gc (s)G(s) = 1000K2 s +
KP KD
200
s(s + 10)(s + 20)
=
K s+
KP KD
s(s + 10)(s + 20)
,
where K = 2 105 KD . Select KP /KD = 10. Then Gc (s)G(s) = K , s(s + 20)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
637
and the closed-loop transfer function is T (s) = s2 K . + 20s + K
Let = 0.69, which implies P.O. < 5%. Also, 2n = 20, so n = 14.49. Thus,
2 K = n = 210 .
The settling time is Ts = 4/n = 0.4 s. The controller is Gc (s) = 0.00105(s + 10). E12.6 The loop transfer function with the PID controller is Gc (s)Gs(s) = 1 KD s2 + KP s + KI . s (s + 5)2
2 3 The ITAE step response requires s3 +1.75n s2 +2.15n s + n = s3 +(10+ KD )s2 + (25 + KP )s + KI . For n = 3 we estimate the normalized settling time to be n Ts 8 seconds. Thus, n 6, and KD = 0.5, KP = 52.4, and KI = 216. The step response is shown in Figure E12.6.
(a) Step response 1.4 12
x 10
3
(b) Disturbance response
1.2
10
1
8
0.8 y(t) y(t) 0.6
6
4
0.4
2
0.2
0
0
0
0.5
1 TIme (s)
1.5
2
2
0
0.5
1 TIme (s)
1.5
2
FIGURE E12.6 (a) Step response: w/o prelter (solid line) and w/prelter (dashed line); and (b) disturbance response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
638
CHAPTER 12
Robust Control Systems
The transfer function from the disturbance to the output is Y (s) G(s) s = =3 . 2 + 77.4s + 216 Td (s) 1 + Gc (s)G(s) s + 10.5s The disturbance response is shown in Figure E12.6. The system is eective in reducing the eects of the disturbance, and the maximum output is reduced by 1/100 for a step disturbance. E12.7 The plant transfer function is G(s) = and the PID controller is Gc (s) = K1 s + K2 + K3 s2 . s 1 , (s + 4)2
Using the ITAE criteria and selecting n = 10 yields K3 = 9.5 K2 = 1000 and K1 = 199 .
The step response is shown in Figure E12.7. The disturbance response is also shown in Figure E12.7. The maximum y (t) = 0.0041, so the system is eective in rejecting the step disturbance.
(a) step response 1.4 4.5 4 1.2 3.5 1 3 2.5 x10 -3 (b) disturbance
0.8
y(t)
y(t)
0.6 0.4 0.2 0 0 0.5 1 Time (sec) 1.5 2
2 1.5 1 0.5 0 -0.5 0 0.5 1 Time (sec) 1.5 2
FIGURE E12.7 (a) Step response: w/o prelter (solid line) and w/prelter (dashed line); and (b) disturbance response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
639
E12.8
The maximum n = 60. Then K1 = 3600 The maximum control input is max |u(t)| 80 . The plot of the step response and the control input u(t) is shown in Figure E12.8.
(a) step response 1.2 90 80 1 70 60 50 0.6 (b) control input u(t)
and
K2 = 80 .
0.8
u(t)
0.05 0.1 Time (sec) 0.15 0.2
y(t)
40 30
0.4
20 10 0
0.2
0 0
-10 0
0.05
0.1 Time (sec)
0.15
0.2
FIGURE E12.8 Step response w/o prelter; and (b) control input u(t).
E12.9
One possible PD controller is Gc (s) = 12.5 + 8.25s . When K=1, the system roots are s1,2 = 1.5 j 2 s 3 = 2 . The step response is shown in Figure E12.9 for K = 0.5, 1, and 1.5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
640
CHAPTER 12
Robust Control Systems
K=1 (solid); K=0.5 (dashed); and K=1.5 (dotted) 1.4
1.2
1
0.8
y(t)
0.6 0.4 0.2 0 0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE E12.9 Step response for K = 0.5, 1, and 1.5.
E12.10
One possible PI controller is Gc (s) = 2.05(s + 0.945) . s
When K = 1, the system roots are s1,2 = 0.365 j 0.514, s3 = 4.15, and s4 = 0.118. The step response is shown in Figure E12.10.
K=1 (solid); K=0.5 (dashed); and K=1.5 (dotted) 1.4
1.2
1
0.8
y(t)
0.6 0.4 0.2 0 0
5
10
15 Time (sec)
20
25
30
FIGURE E12.10 Step response for K = 0.5, 1, and 1.5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
641
E12.11
The plot is shown in Figure E12.11.
100 80 60 P.O. (%) 40 20 0 -20
0
1
2
3 k
4
5
6
FIGURE E12.11 Percent overshoot as a function of k in the interval 0.1 k 6.
E12.12
The controllability matrix is Pc =
c1
c2
c2 ac1 bc2
and
det Pc = c2 + [bc1 ]c2 + ac2 . 2 1 For controllability we require det Pc = 0, hence c2 + [bc1 ]c2 + ac2 = 0 2 1 implies c2 b = c1 2 (b/2)2 a
where (b/2)2 a 0. For real-valued c1 and c2 , if (b/2)2 a < 0, all real values of c1 and c2 are valid. Valid values of the constants are c1 = 0, c2 = 10, a = 10, and b = 3. The step response is shown in Figure E12.12.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
642
CHAPTER 12
Robust Control Systems
Step Response 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2 Time (sec)
2.5
3
3.5
4
FIGURE E12.12 Step response with c1 = 0, c2 = 10, a = 10, and b = 3.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
643
Problems
P12.1 The closed-loop transfer function is T (s) = and the sensitivity function is S (s) = s2 . s2 + 4s + 8 4(s + 2) s2 + 4s + 8
The plot of 20 log |T | and 20 log |S | is shown in Figure P12.1. The bandwidth is B = 6.31 rad/sec . Then
T |SK |B = 0.98 T |SK | B = 0.78
2
T |SK | B 4
= 0.30 .
10
0
-10 20log|T| 20log|S| -20 Gain dB
-30
-40
-50
-60 -1 10
10 Frequency (rad/sec)
0
10
1
FIGURE P12.1 Plot of T (s) and the sensitivity function S (s).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
644 P12.2
CHAPTER 12
Robust Control Systems
(a) The loop transfer function is given by Gc (s)G(s) = When K = 100 , the peak magnitude is Mp = 1.84 . (b) The plot of 20 log |T | and 20 log |S | is shown in Figure P12.2a.
20 20log|S| 0
K . s(0.02s + 1)(0.002s + 1)
-20 20log|T|
Gain dB
-40
-60
-80
-100
-120 101
102 Frequency (rad/sec)
103
104
FIGURE P12.2 (a) Plot of T (s) and the sensitivity function S (s).
(c) The bandwidth is B = 117 rad/sec , and
T |SK |B = 1.47 T |SK | B = 0.39
4 2
T |SK | B = 1.62 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
645
(c) The disturbance response is shown in Figure P12.2b.
x10 -7 8 7 6 5
Amplitude
4 3 2 1 0 0
0.05
0.1
0.15 Time (secs)
0.2
0.25
0.3
FIGURE P12.2 CONTINUED: (b) Disturbance response for K = 100.
P12.3
(a) The loop transfer function is Gc (s)G(s) = The characteristic equation is 1 + Gc (s)G(s) = 1 + or s3 + (14 + K )s2 + (44 + K )s + 40 20K = 0 . So, the system is stable for 8.25 < K < 2 . (b) The steady-state error is ess = 2 . 2K K (s 4)(s + 5) =0 (s + 2)2 (s + 12) K (s 4)(s + 5) . (s + 10)(s + 2)2
The steady-state error cannot be made zero for any stable K . Choose
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646
CHAPTER 12
Robust Control Systems
K close to -8.25 for minimal tracking error, although the transient response will be lightly oscillatory. (c,d) The plots of y (t) for K = 7 (nominal) K = 8.05 (+10%) K = 5.95 (10%) are shown in Figure P12.3.
K= 7 (solid) & K= 8.05 (dashed) & K= 5.95 (dotted) 2. 5 2 1. 5 1 y(t) 0. 5 0 -0. 5 -1 0 2 4 6 Time (sec) 8 10
FIGURE P12.3 Step input response for K = 7, K = 8.05 and K = 5.95.
P12.4
The loop transfer function and closed-loop transfer functions are G(s) = K (s + 5)(s + 1) and T (s) = s2 K . + 6s + 5 + K
(a,b) A summary of the system performance is shown in Table P12.4. (c) The best compromise is K = 10. (d) For K = 10, the transfer function from Td (s) to Y (s) is Y (s) s+5 =2 . Td (s) s + 6s + 15
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
647
K
n
P.O.
Ts
ess
r |SK |
1
2.45
1.22
0%
3.33
0.83
0.29
10
3.87
0.77
2.1%
1.38
0.33
2.03
20
5.00
0.60
9.5%
1.18
0.20
2.50
TABLE P12.4
Performance summary.
The plot of the disturbance response is shown in Figure P12.4. So, for Td (s) = 1/s, the response is reduced by a factor of 3.
Step Response 0.35
0.3
0.25
Amplitude
0.2
0.15
0.1
0.05
0
0
0.2
0.4
0.6
0.8
1 Time (sec)
1.2
1.4
1.6
1.8
2
FIGURE P12.4 Disturbance response for K = 10.
P12.5
(a) The plant is given by G= s
s 25
1 . +1
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648
CHAPTER 12
Robust Control Systems
We desire P.O. < 10% and Ts < 100 ms. Using a PD controller Gc (s) = 100 + 2.2s , we determine that P.O. = 7%, Ts < 100 ms and ess = input. The plot of y (t) is shown in Figure P12.5. (b) The sensitivity is
r |SK1 | = 27.95 A 100
for a ramp
when K1 = 1. (c) The plot of y (t) when K1 = 2 (the compensator Gc (s) is unchanged) is shown in Figure P12.5. (d) The disturbance response is shown in Figure P12.5.
(a) step response 1.2 0.012
(b) disturbance
1
0.01
0.8
0.008
y(t)
0.6
y(t)
0.05 0.1 Time (sec) 0.15 0.2
0.006
0.4
0.004
0.2
0.002
0 0
0 0
0.05
0.1 Time (sec)
0.15
0.2
FIGURE P12.5 (a) Step response: K1 = 1 (solid line) and K1 = 2 (dashed line); and (b) disturbance response.
P12.6
(a) The plant is given by G(s) = 1 s(s + p)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
649
where p = 2, nominally. One solution is Gc (s) = Then, T (s) = 18.7(s + 2.9) . (s + 3.41)(s + 2 + 2 3j )(s + 2 2 3j ) 18.7(s + 2.9) . (s + 5.4)
(b,d) The step responses are shown in Figure P12.6 for p = 2 and p = 1. (c,d) The disturbance responses are shown in Figure P12.6 for p = 2 and p = 1.
(a) step response 1.6 1.4 1.2 1 -0.06 0 (b) disturbance
-0.02
-0.04
y(t)
0.8 0.6 0.4 0.2 0 0 Time (sec)
y(t)
-0.08 -0.1 -0.12 5 -0.14
0 Time (sec)
5
FIGURE P12.6 (a) Step response: p = 2 (solid line) and p = 1 (dashed line); and (b) disturbance response: p = 2 (solid line) and p = 1 (dashed line).
P12.7
(a) The plant is given by G(s) = and the PID controller is Gc (s) = K (s + z )2 . s s(s2 1 , + 4s + 5)
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650
CHAPTER 12
Robust Control Systems
When z = 1.25 and K=4, all roots are s = 1 j 1.22 . Then, the closed-loop transfer function is T (s) = s4 + 4s3 4(s + 1.25)2 . + 9s2 + 10s + 6.25
(b,c) The step responses with and without a prelter are shown in Figure P12.7. (d) The disturbance response is shown in Figure P12.7.
(a) step response 1.6 1.4 0 1.2 -0.02 1 0.02 (b) disturbance
y(t)
0.8 0.6
y(t)
5 Time (sec) 10
-0.04
-0.06 0.4 -0.08 0.2 0 0 -0.1
0
5 Time (sec)
10
FIGURE P12.7 (a) Step response: w/o prelter (solid line) and w/prelter (dashed line); and (b) disturbance response.
P12.8
(a) The loop transfer function is Gc (s)G(s) = 10Ka (5s + 500 + 0.0475s2 ) . s3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
651
When Ka = 374.5 , the phase margin is P.M. = 40o . (b) The root locus is shown in Figure P12.8a.
150 100 50
* o
Imag Axis
0 -50
*
x
o
-100
*
-150 -150 -100 -50 0 Real Axis 50 100 150
FIGURE P12.8 10(0.0475s2 +5s+500) = 0. (a) Root locus for 1 + Ka s3
When Ka = 374.5 , the roots are s1 = 139.8 s2,3 = 19.1 j 114.2 . (c) The transfer function from Td (s) to Y (s) is Y (s) s =3 . 2 + 19150s + 1915000 Td (s) s + 182s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
652
CHAPTER 12
Robust Control Systems
The maximum is max |y (t)| = 0.0000389 . (d) The step responses, with and without a prelter, are shown in Figure P12.8b.
1.6 1.4 1.2 1
y(t)
0.8 0.6 0.4 0.2 0 0
0.05
0.1
0.15 Time (sec)
0.2
0.25
0.3
FIGURE P12.8 CONTINUED: (b) Step response: w/o prelter (solid line) and w/prelter (dashed line).
P12.9
The polynomial under investigation is s3 + 3s2 + 3s + 4 = 0 . From the uncertainty bounds on the coecients, we dene 0 = 4 0 = 5 1 = 1 1 = 4 2 = 2 2 = 4 Then, we must examine the four polynomials: s3 + 2s2 + 4s + 5 = 0 s3 + 4s2 + s + 4 = 0 s3 + 4s2 + 4s + 4 = 0 s3 + 2s2 + s + 5 = 0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
653
The fourth polynomial is not stabletherefore, the system is not stable for the uncertain parameters. P12.10 One possible PID controller is Gc (s) = 0.058s2 + 2.17s + 16.95 . s
A rst-order Pade approximation was used in the design to account for the delay system. The step input response is shown in Figure P12.10. A prelter should also be used with the PID controller. A suitable prelter is Gp (s) = K2 . K3 s2 + K1 s + K2
1.2
1
0.8
y(t)
0.6
0.4
0.2
0 0
0.05
0.1
0.15
0.2
0.25 Time (sec)
0.3
0.35
0.4
0.45
0.5
FIGURE P12.10 Step response with the PID controller and prelter.
P12.11
The PID controller is given by Gc (s) = KD s2 + KP s + KI . s
Using the ITAE method, we desire the characteristic polynomial to be
2 3 q (s) = s3 + 1.75n s2 + 2.15n s + n = 0 ,
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654
CHAPTER 12
Robust Control Systems
where we select n = 4 to obtain a peak time of Tp = 1 second. Here we use the approximation for ITAE third-order systems that n Tp 4 from Figure 5.30(c) in Dorf and Bishop. The actual characteristic equation is s3 + 25KD s2 + 25KP s + 25KI = 0 . Equating coecients and solving for the gains yields KP = 1.376 , KD = 0.28 , and KI = 2.56 .
The step response is shown in Figure P12.11, with the prelter Gp (s) = KI . + KP s + KI
KD
s2
Step Response 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5 Time (sec)
2
2.5
3
FIGURE P12.11 Step response with the PID controller and prelter.
P12.12
We will design for the case where K = 1 and p = 1. The design plant is G(s) = 1 . s(s + 1)(s + 4)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
655
The nominal plant is given by G(s) = and the PID controller is Gc (s) = KD s2 + KP s + KI . s 2.5 , s(s + 2)(s + 4)
Using the ITAE method, we desire the characteristic polynomial to be
2 3 4 q (s) = s4 + 2.1n s3 + 3.4n s2 + 2.7n s + n = 0 ,
where we select n = 2.38 to obtain a peak time around Tp = 3 seconds. The actual characteristic equation (with the worst-case plant) is s4 + 5s3 + (4 + KD )s2 + KP s + KI = 0 . Equating coecients and solving for the gains yields KP = 36.40, KI = 32.08, and KD = 15.26. The step response is shown in Figure P12.12, with the prelter Gp (s) = KI . KD s2 + KP s + KI
1.4 Worstcase plant Nominal plant 1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE P12.12 Step response with the prelter: nominal plant (dashed line) & worst-case plant (solid line).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
656 P12.13
CHAPTER 12
Robust Control Systems
The transfer function is G(s) = C(sI A)1 B = The sensitivity is
G SK =
20
s
3
5 s+K
0 1
=
s2
6 . + Ks + 5
G K Ks =2 . K G s + Ks + 5
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
657
Advanced Problems
AP12.1 Let Gp (s) = 1. A viable PID controller is Gc (s) = KP + KI 50s2 + 200s + 2.5 + KD s = . s s
The loop transfer function is Gc (s)G(s) = 40(50s2 + 200s + 2.5) . s(80s2 + 1)
where K = 40, so that Kv = 100. The step response is shown in Figure AP12.1.
Step Response 1.4
System: syscl Peak amplitude: 1.1 Overshoot (%): 9.91 At time (sec): 0.191
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6 Time (sec)
0.8
1
1.2
FIGURE AP12.1 Step response with PID controller.
AP12.2
For all three controllers, choose K = 1 as the design value. Also, use as the nominal points a = 2 and b = 5 for each design. ITAE methods were employed in all designs, although this did not work well for the PI controller. (a) PI controller: Let Gp (s) = 1 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
658
CHAPTER 12
Robust Control Systems
Not all specications could be met simultaneously with a PI controller. The best over-all results are achieved when using a = 3 and b = 4.5 as the design values. An acceptable PI controller is Gc (s) = 1.2 + 3.96 . s
Controller
P.O.
Ts
Tp
|u(t)|max 4.43
PI
0%
2.29s
n.a.
PD
4.6%
1.72s
1.26s
12.25
PID
1.97%
0.65s
0.47s
37.25
TABLE AP12.2
PI, PD, and PID controller performance summary.
The nal design is based on root locus methods since the ITAE methods did not produce an eective controller. The closed-loop transfer function is T (s) = (b) PD controller: Let Gp (s) = 12.25 . 7.25 + 2.9s s3 1.2s + 3.96 . + 3s2 + 5.7s + 3.96
The closed-loop transfer function is T (s) = s2 7.25 + 2.9s , + 4.9s + 12.25
where the PD controller (based on ITAE methods) is Gc (s) = 7.25 + 2.9s .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
659
(c) PID controller: Let Gp (s) = 15.5s2 1000 . + 210s + 1000
The closed-loop transfer function is T (s) = 15.5s2 + 210s + 1000 . s3 + 17.5s2 + 215s + 1000
And the PID controller (based on ITAE methods) is Gc (s) = 15.5s2 + 210s + 1000 . s
The performance of each controller is summarized in Table AP12.2. AP12.3 (a) The PID controller is Gc (s) = KD s2 +
KP KD s
+
KI KD
s
.
Since we want P.O. < 4% and Ts < 1s, we choose the dominant closed-loop poles to have n = 6 and = 0.8. Therefore, we place the zeros at s2 + KP KI s+ = s2 + 10s + 36 . KD KD
Solving for the constants yields, KP = 10 , KI KI = 36 . KD
Then, using root locus methods, we choose KD = 91 to place the roots near the zeros. The PID controller gains are computed to be KP = 910, KI = 3276 and KD = 91. (b) The loop transfer function is Gc (s)G(s) = KD s2 + KP s + KI . s2 (s2 + 5s + 4)
The closed-loop system characteristic equation is s3 + 5s2 + 4s + KD s2 + KP s + KI = 0 . Solving for the PID gains yields KP = 73.4, KI = 216 and KD = 5.5.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
660
CHAPTER 12
Robust Control Systems
Therefore, the controller is Gc (s) = Using the prelter Gp (s) = s2 39.3 , + 13.35s + 39.3 5.5(s2 + 13.35s + 39.3) . s
we obtain the closed-loop transfer function T (s) = s3 + 10.5s2 216 . + 77.4s + 216
The percent overshoot is P.O. 3.5% and the settling time is Ts 1.67 sec. AP12.4 The PID controller is Gc (s) = KD s2 +
KP KI KD s + KD
s
.
The bounds 1 a 2 and 4 b 12 imply that 2 n 3.46 and 0.5 n 1. One solution is to place the PID controller zeros at
1. 4 1. 2 1 Amplitude 0. 8 0. 6 0. 4 0. 2 0 0 0. 5 1 1. 5 Time (sec) 2 2. 5 3
FIGURE AP12.4 Family of step response with PID controller with nominal case (a, b) = (1.5, 9) denoted by the solid line.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
661
s = 1 j 8 (i.e. n = 1 and n = 3). So, s2 + KI KP 2 s+ = s2 + 2n s + n = s2 + 4s + 9 . KD KD
The nominal case for design is chosen to be a = 1.5 and b = 9. Using root locus, we select KD = 2.1 to place the closed-loop characteristic roots near the zeros. Then, the PID controller gains are computed to be KP = 8.4, KI = 18.9, and KD = 2.1. The plot of the response to a step input is shown in Figure AP12.4. The o-nominal cases shown in the simulations are (a, b) = (1.2, 4), (1.4, 6), (1.6, 10), and (1.8, 12). AP12.5 (a) With = 1, we nd that K = 2.5 for = 0.707. The roots are s1,2 = 0.5 j 0.5 and s3,4 = 3 j when K = 2.5. The magnitude of the sensitivity for roots s1,2 is 0.15 and for s3,4 the sensitivity magnitude is 0.38. (b) With = 1.25, we nd that K = 2 for = 0.707. The roots are s1,2 = 0.52 j 0.52 and s3,4 = 2.97 j 0.58. The magnitude of the sensitivity for roots s1,2 is 0.21 and for s3,4 the sensitivity magnitude is 0.54. (c) With = 1.5, we nd that K = 1.48 for = 0.707. The roots are s1,2 = 0.52 j 0.52, s3 = 3.4 and s4 = 2.5. The magnitude of the sensitivity is 0.31, 0.63, and 0.56 for the roots s1,2 , s3 and s4 , respectively. (d) With = 1.75, we nd that K = 1.1 for = 0.707. The roots are s1,2 = 0.51 j 0.51, s3 = 3.68 and s4 = 2.28. The magnitude of the sensitivity is 0.42, 0.33, and 0.27 for the roots s1,2 , s3 and s4 , respectively. (e) With = 2, we nd that K = 0.85 for = 0.707. The roots are s1,2 = 0.5 j 0.5, s3 = 3.81 and s4 = 2.18. The magnitude of the sensitivity is 0.55, 0.23, and 0.21 for the roots s1,2 , s3 and s4 , respectively. AP12.6 To obtain a phase margin of P.M. = 49.77o , select K = 1.5, b = 36 and choose Gp (s) = 1. The PID controller is Gc (s) = 1.5(s2 + 20s + 36) . s
When K1 = 0.75, the phase margin is reduced to P.M. = 45.45o ; and when K1 = 1.25, the phase margin is increased to P.M. = 52.75o . AP12.7 With the settling time Ts = 1 and percent overshoot P.O. < 10% specications, we target for dominant closed-loop poles with n = 10. Here we estimate n Ts 10 associated with the ITAE performance. The closed-
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
662
CHAPTER 12
Robust Control Systems
loop transfer function is T (s) = Gp (s) 1.5(KD s2 + KP s + KI ) , (1 + 1.5KD )s2 + 1.5KP s + 1.5KI
where we have neglected . Using the ITAE method, the desired characteristic polynomial is s2 +
2 2n s + n = s2 +
1.5Kp 1.5KI s+ . 1 + 1.5KD 1 + 1.5KD
Let KD = 0.25. Then solving for the remaining PID gains yields KP = 12.96 and KI = 91.67. The pre-lter is Gp (s) = 0.375s2 137.5 . + 19.45s + 137.5
Then the closed-loop transfer function (with = 0.001) is T (s) = 0.001s3 137.5 . + 1.375s2 + 19.45s + 137.5
The transfer function from the disturbance to the output is Y (s)/Td (s) = 0.001s3 1.5s . + 1.375s2 + 19.45s + 137.5
The step input response and disturbance response are shown in Figure AP12.7.
(a) 1.4 0.05 (b)
1.2
0.04
1 0.03 Amplitude 0.8 Amplitude
0.02
0.6 0.01 0.4
0.2
0
0
0
0.2
0.4 0.6 Time (sec)
0.8
1
0.01
0
0.2
0.4 0.6 Time (sec)
0.8
1
FIGURE AP12.7 (a) Input response; (b) Disturbance response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
663
AP12.8
The PI controller is given by Gc (s) = We will also use the prelter Gp (s) = KI . KP s + KI KP s + KI . s
2 Using the ITAE method, we determine that KP = 2n and KI = n . Let n = 2.2. Then KP = 3.11 and KI = 4.8. The step response and control u(t) are shown in Figure AP12.8.
(a) 1.2 1.2 (b)
1
1
0.8 0.8
Amplitude
Amplitude
5 Time (secs)
0.6
0.6
0.4
0.4 0.2 0.2
0
0 0
-0.2
0 Time (secs)
5
FIGURE AP12.8 (a) Input response; (b) Control history u(t).
AP12.9
Consider the PID controller Gc (s) = KD s2 + KP s + KI . s
Using the acceleration constant specication, we nd Ka = lim s2 Gc (s)G(s) = 3KI /5 = 2 .
s0
Therefore, KI = 3.33 satsies the acceleration constant specication. The
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
664
CHAPTER 12
Robust Control Systems
bandwidth requirement is satised with the PID gains KP = 5.5 The pre-lter is Gp (s) = 3.71s2 3.33 . + 5.5s + 3.33 KI = 3.33 KD = 3.71 .
The plot of the response to a step input is shown in Figure AP12.9.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
1
2
3
4
5 Time (secs)
6
7
8
9
10
FIGURE AP12.9 Step input response.
AP12.10
(a) A suitable PD controller is given by Gc (s) = 0.6 + 0.4s . The percent overshoot is P.O. = 18.8% and the peak time is Tp = 2.4 sec. (b) A suitable PI controller is given by Gc (s) = 0.15 + 0.01 . s
The percent overshoot is P.O. = 23.7% and the peak time is Tp = 7.8 sec.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
665
(c) A suitable PID controller is given by Gc (s) = 0.6 + 0.01 + 0.4s . s
The percent overshoot is P.O. = 19.9% and the peak time is Tp = 2.5 sec. (d) The PD or PID controllers are the best choices. AP12.11 A robust PID controller designed with ITAE methods will be a suitable controller. From the settling time specication we select n = 10, where we have used = 0.8. The worst case is a = 1 and K = 2 . The desired closed-loop transfer function is T (s) =
3 n 2 3 s3 + 1.75n s2 + 2.15n s + n
and the actual characteristic equation is q (s) = s3 + (2a + KKD )s2 + (a2 + KKP )s + KKI . Equating like terms, we nd that KP = 107 AP12.12 We use as the design plant G(s) = Select p1 = 2 and z1 = 3 s+2 . s(s + 3) KI = 500 KD = 7.75 .
to cancel a design plant pole and zero. Then, choose p2 = 0 to have zero steady-state error to a unit step. The remaining variables K and z2 are selected based on ITAE methods, where n = 100. A suitable compensator is Gc (s) = 141.42(s + 3)(s + 70.71) . s(s + 2)
A plot of the step responses for various values of p, q and r is shown in Figure AP12.12.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
666
CHAPTER 12
Robust Control Systems
1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
0.01
0.02
0.03
0.04
0.05 Time(sec)
0.06
0.07
0.08
0.09
0.1
FIGURE AP12.12 Family of step responses with the design plant (p, q, r ) = (3, 0, 2) denoted by the solid line.
AP12.13
A suitable compensator is Gc (s) =
1.4
1000(s + 1.8)(s + 3.5)(s + 5.5) . s(s + 600)
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
1
2
3
4
5 Time (sec)
6
7
8
9
10
FIGURE AP12.13 Step responses with nominal plant (solid line) and o-nominal plant with all poles reduced by 50% (dashed line).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
667
Design Problems
CDP12.1
The plant model with parameters given in Table CDP2.1 in Dorf and Bishop is given by: (s) 26.035 = , Va (s) s(s + 33.142) where we neglect the motor inductance Lm and where we switch o the tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). With a PID controller ,the closed-loop system characteristic equation is s3 + (33.142 + 26.035KD )s2 + 26.035KP s + 26.035KI = 0 . A suitable PID controller is Gc (s) = 50 + s + 0.1 . s
This PID controller places the closed-loop system poles to the left of the n line necessary to meet the settling time requirement. The step response is shown below. The settling time is Ts = 0.12 second. In the steady-state the error due to a step disturbance is zero.
1.2
1
0.8 Amplitude
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25 Time (secs)
0.3
0.35
0.4
0.45
0.5
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
668 DP12.1
CHAPTER 12
Robust Control Systems
The closed-loop transfer function is Y (s) Km Gc (s) =2 . R(s) s + (2 + Km K1 )s + Gc (s)Km (a) When Gc = K , we have T (s) = s2 15K , + (2 + 15K1 )s + 15K
where Km = 15. Using ITAE criteria and n = 10, we determine that K1 = 0.81 and K = 6.67. For the disturbance, we have 1 Y (s) =2 . TL (s) s + 14.14s + 100 The input and disturbance responses are shown in Figure DP12.1, without prelters.
(a) Step response 1.2 0 (b) Disturbance response
1
-0.002
0.8
-0.004
y(t)
0.6
y(t)
0.5 Time (sec) 1
-0.006
0.4
-0.008
0.2
-0.01
0 0
-0.012
0
0.5 Time (sec)
1
FIGURE DP12.1 (a) Step response: Gc (s) = K (solid line) and Gc (s) = KP + KD s (dashed line); and (b) disturbance response (same for both compensators).
(b) When Gc = KP + KD s, we have Y (s) 15(KP + KD s) =2 . R(s) s + (2 + 15K1 + 15KD )s + 15KP For n = 10 and with the ITAE criteria, we determine that (with
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
669
KD = 0.1) Y (s) 15(6.67 + 0.1s) =2 . R(s) s + 14.14s + 100 DP12.2 The nominal plant is given by G(s) = The closed-loop transfer function is T (s) = Let KP = 300 , KI = 500 , and KD = 100 . K (KD s2 + KP s + KI ) . s3 + (4 + KKD )s2 + KKP s + KKI 1 . s(s + 4)
A family of responses is shown in Figure DP12.2 a for various values of K . The percent overshoot for 0.1 K 2 is shown in Figure DP12.2b.
1.4
1.2
1
Step response
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5 Time (s)
3
3.5
4
4.5
5
FIGURE DP12.2 (a) Family of step responses for various values of K .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
670
CHAPTER 12
Robust Control Systems
12
10
8 Percent overshoot
6
4
2
0
0
0.2
0.4
0.6
0.8
1 K
1.2
1.4
1.6
1.8
2
FIGURE DP12.2 CONTINUED: (b) Percent overshoot for various values of K .
DP12.3
(a) The dexterous hand model is given by G(s) = Km , s(s + 5)(s + 10)
where Km = 1, nominally. The PID controller is Gc (s) = KD (s2 + 6s + 18) . s
The root locus is shown in Figure DP12.3a. If we select KD = 90 , the roots are s1,2 = 5.47 j 6.6 s3,4 = 2.03 j 4.23 . Thus, all roots have n > 4/3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
671
to meet the design specication Ts < 3 sec . (b) The step responses for Km = 1 and Km = 1/2 are shown in Figure DP12.3b. When K = 1/2 , an o-nominal value, the settling time specication is no longer satised.
20 15 10
*
5
Imag Axis
o
* x
0 -5
x
x o * *
-10 -15 -20 -20
-15
-10
-5
0 Real Axis
5
10
15
20
FIGURE DP12.3 s2 +6s (a) Root locus for 1 + KD s2 (s+5)(+18 = 0. s+10)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
672
CHAPTER 12
Robust Control Systems
1.6 1.4 1.2 1
y(t)
0.8 0.6 0.4 0.2 0 0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE DP12.3 CONTINUED: (b) Step response (without prelters): PID with K3 = 90 and Km = 1 (solid line) and PID with K3 = 90 and Km = 0.5 (dashed line).
DP12.4
The nominal plant is G(s) = and the PID controller is Gc (s) = KI (1 s + 1)(2 s + 1) . s s(s2 17640 , + 59.4s + 1764)
(a) Using ITAE methods, we determine that n = 28.29, KI = 36.28, 1 + 2 = 0.0954 and 1 2 = 0.00149. So, Gc (s) = 36.28(0.00149s2 + 0.0954s + 1) . s
(b) The step response for the nominal plant and the PID controller is shown in Figure DP12.4a, with and without a prelter. (c) The disturbance response is shown in Figure DP12.4b. (d) The o-nominal plant is G(s) = s(s2 16000 . + 40s + 1600)
The step response for the o-nominal plant is shown in Figure DP12.4a.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
673
(a) nominal plant 1.6 1.4 1.2 1 1.6 1.4 1.2 1
(b) o -nominal plant
y(t)
0.8 0.6 0.4 0.2 0 0
y(t)
0.5 Time (sec) 1
0.8 0.6 0.4 0.2 0 0
0.5 Time (sec)
1
FIGURE DP12.4 (a) Step response for (i) nominal plant: w/o prelter (solid line) and w/prelter (dashed line); and (ii) for o-nominal plant: w/o prelter (solid line) and w/prelter (dashed line).
disturbance response 0.3
0.25
0.2
0.15
y(t)
0.1 0.05 0 -0.05
0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE DP12.4 CONTINUED: (b) Disturbance response for the nominal plant.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
674 DP12.5
CHAPTER 12
Robust Control Systems
One possible solution is Gc (s) = 0.08 (0.01s + 1)(0.99s + 1) . s
The phase margin with this controller is P.M. = 45.5o . The step response is shown in Figure DP12.5 for the nominal plant (with and without a prelter); the step response for the o-nominal plant is also shown in Figure DP12.5. The prelter is Gp (s) = 13.97s2 1411 . + 1411s + 1411
(a) nominal plant 1. 4 1. 2 1 0. 8 y(t) 0. 6 0. 4 0. 2 0 0 10 Time (sec) 20 y(t) 0. 6 0. 4 0. 2 0 1. 4 1. 2 1 0. 8
(b) o nominal plant
0
10 Time (sec)
20
FIGURE DP12.5 (a) Step response for nominal plant: w/o prelter (solid line) and w/prelter (dashed line); and (b) for o-nominal plant: w/o prelter (solid line) and w/prelter (dashed line).
DP12.6
Using ITAE methods, three controllers are designed for the nominal plant: (i) PID controller: Gc (s) = 0.225s2 + 0.535s + 34.3 s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
675
(ii) PI controller: Gc (s) = (iii) PD controller: Gc (s) = 0.9s + 22.5 The step responses for each controller is shown in Figure DP12.6. The responses for the PID and PI controller are the same since the gains were selected to obtain the same ITAE characteristic equation. An appropriate prelter is used in all cases.
(a) nominal plant 1.2 1.2 (b) o -nominal plant
0.9s + 22.5 s
1
1
0.8
0.8
y(t)
0.6
y(t)
0.5 1 Time (sec) 1.5 2
0.6
0.4
0.4
0.2
0.2
0 0
0 0
0.5
1 Time (sec)
1.5
2
FIGURE DP12.6 (a) Step response for nominal plant: PID (solid line); PI (dashed line); and PD (dotted line); (b) for o-nominal plant: PID (solid line); PI (dashed line); and PD (dotted line).
DP12.7
The loop transfer function is G(s) = Ka Km K = (0.5s + 1)(f s + 1)s(s + 1) s(s + 2)(s + 1)
since f is negligible. A suitable PID controller is Gc (s) = 300(s2 + 2.236s + 2.5) KKD (s2 + as + b) = . s s
The step response is shown in Figure DP12.7. The percent overshoot is
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
676
CHAPTER 12
Robust Control Systems
P.O. = 4.6% and the settling time is Ts = 3.74 seconds.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
0.5
1
1.5
2
2.5 Time (secs)
3
3.5
4
4.5
5
FIGURE DP12.7 Step response for the elevator position control.
DP12.8
The system transfer function is Y (s) = We are given G(s) = esT where T = 1 second . G(s)Gc (s)Gp (s) R(s) . 1 + G(s)Gc (s)
Using a second-order Pade approximation yields G(s) s2 6s + 12 . s2 + 6s + 12
Three controllers that meet the specications are Gc1 (s) = 0.5 (Integral controller) s 0.04s + 0.4 Gc2 (s) = (PI controller) s 0.01s2 + 0.04s + 0.4 (PID controller) . Gc3 (s) = s
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
677
In all cases, the steady-state error is zero. Integral P.O.(%) Ts (sec) Tp (sec) |V (t)|max (volts) 4.05 6.03 4.75 1.04 PI 0 6.12 PID 0 6.02
N/A N/A 1 1
The prelter Gp (s) = 1 is used in all designs. To compute the voltage, the transfer function is V (s) = Gp (s)Gc (s) R(s) . 1 + Gc (s)G(s)
DP12.9
The space robot transfer function is G(s) = (a) Consider Gc (s) = K . Then T (s) = Gc (s)G(s) K =2 . 1 + Gc (s)G(s) s + 10s + K 1 . s(s + 10)
We determine that K = 50.73 for = 0.702. Thus, we expect P.O. < 4.5%. So, Gc (s) = 50.73 . (b) Consider the PD controller Gc (s) = KP + KD s . Then T (s) = KP + KD s . s2 + (10 + KD )s + KP
Using the ITAE method, we compute KP = 100 Thus, Gc (s) = 4s + 100 , and KD = 4 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
678
CHAPTER 12
Robust Control Systems
and the prelter is Gp (s) = (c) Consider the PI controller Gc (s) = KP + Then, T (s) = s3 KP s + KI . + 10s2 + KP s + KI KI KP s + KI = . s s 100 . 4s + 100
Using the ITAE method, we have n = 5.7 Thus, Gc (s) = 70.2 + 186.59/s , and the prelter is Gp (s) = (d) Consider the PID controller Gc (s) = Then, T (s) = KD s2 + KP s + KI . s3 + 10s2 + KD s2 + KP s + KI KD s2 + KP s + KI . s 186.59 . 70.2s + 186.59 KP = 70.2 and KI = 186.59 .
Using the ITAE method with n = 10, we have KD = 7.5 Thus, Gc (s) = and the prelter is Gp (s) = 7.5s2 1000 . + 215s + 1000 7.5s2 + 215s + 1000 , s KP = 215 and KI = 1000 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
679
A summary of the performance is given in Table DP12.7.
Gc (s)
P.O.
tp
ts
yss
max|y (t)| 0.026
K
4.5%
0.62 s
0.84 s
0
PD
5.2%
0.39 s
0.56s
0
0.010
PI
1.98%
0.81 s
1.32s
0
0.013
PID
1.98%
0.46 s
0.75 s
0
0.004
TABLE DP12.7
A summary of performance to a disturbance input.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
680
CHAPTER 12
Robust Control Systems
Computer Problems
CP12.1 The closed-loop transfer function is T (s) = s2 8K , + 2s + 8K
T and the sensitivity function, SK , is
S (s) =
s2 + s . s2 + 2s + 8K
The plot of T (s) and S (s) is shown in Figure CP12.1, where K = 10.
nt=[80]; dt=[1 2 80]; syst = tf(nt,dt); ns=[1 2 0];ds=[1 2 80]; syss = tf(ns,ds); w=logspace(-1,2,400); [magt,phaset]=bode(syst,w);magtdB(1,:) = 20*log10(magt(1,1,:)); [mags,phases]=bode(syss,w); magsdB(1,:) = 20*log10(mags(1,1,:)); semilogx(w,magtdB,w,magsdB,'--') legend('20log|T|','20log|S|') xlabel('Frequency (rad/sec)') ylabel('Gain dB') grid
20 20log|T| 20log|S| 10
0
10 Gain dB
20
30
40
50
60 1 10
10
0
10 Frequency (rad/sec)
1
10
2
FIGURE CP12.1 Plot of T (s) and the sensitivity function S (s).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
681
CP12.2
A reasonable value of the gain K = 5. The family of step responses is shown in Figure CP12.2.
p=[0.1:1:20]; K=5; t=[0:0.01:1]; for i=1:length(p) n=[K*p(i)]; d=[1 p(i)]; sys = tf(n,d); sys_cl = feedback(sys,[1]); y(:,i)=step(sys_cl,t); end plot(t,y) xlabel('Time (sec)'), ylabel('Step response')
0.9
p=20
0.8
0.7
p=1.1
0.6 Step response
0.5
0.4
0.3
0.2
p=0.1
0.1
0
0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE CP12.2 Family of step responses for 0.1 < p < 20.
CP12.3
The closed-loop characteristic equation is 1 + KD where a = KP /KD b = KI /KD . s2 + as + b =0 Js3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
682
CHAPTER 12
Robust Control Systems
We select a = 1 and b=2
to move the root locus into the left hand-plane (see Figure CP12.3a). Then, we choose KD = 30 from the root locus using the rlocnd function. The closed-loop Bode plot in Figure CP12.3b veries that the bandwidth B < 5 rad/sec. Also, the phase margin is P.M. = 53.4o , which meets the design specication. The plot of phase margin versus J is shown in Figure CP12.3c. We see that as J increases, the phase margin decreases.
J=10; a=1; b=2; num=[1 a b]; den=J*[1 0 0 0]; sys = tf(num,den); rlocus(sys), r loc nd(sys)
3
2
o +
1
Imag Axis
0
x
-1
o +
-2
-3 -3
-2
-1
0 Real Axis
1
2
3
FIGURE CP12.3 2 + (a) Root locus for 1 + KD s 10s+2 = 0. s3
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
683
a=1; b=2;J=10;k2=30; k1=a*k2; k0=b*k2; numc=[k2 k1 k0]; denc=[J 0 0 0]; sysc = tf(numc,denc); sys_cl = feedback(sysc,[1]); bode(sys_cl); [gm,pm,wg,wc]=margin(sysc); pm
20 10
pm = 53.4061
Gain dB
0 -10 -20 10-1 100 Frequency (rad/sec) 101
FIGURE CP12.3 CONTINUED: (b) Closed-loop Bode plot with B < 5 rad/sec.
Ji=[1:1:30]; for i=1:length(Ji) numc=[k2 k1 k0]; denc=[Ji(i) 0 0 0]; sysc = tf(numc,denc); [mag,phase,w]=bode(sysc); [gm,pm]=margin(mag,phase,w); Pm(i)=pm; end plot(Ji,Pm), grid xlabel('J'), ylabel('Phase Margin (deg)')
100
80
60
Phase Margin (deg)
o
40
20
0
-20
-40 0
5
10
15 J
20
25
30
FIGURE CP12.3 CONTINUED: (c) Phase margin versus J .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
684 CP12.4
CHAPTER 12
Robust Control Systems
The closed-loop characteristic equation is 1+K s2 1 =0 + bs + a
where a = 8 and the nominal value of b = 4. The root locus is shown in Figure CP12.4a.
clf, hold o a=8; b=4; num=[1]; den=[1 b a]; sys = tf(num,den); rlocus(sys), hold on zeta=0.59; wn=1.35; x=[-10:0.1:-zeta*wn]; y=-(sqrt(1-zeta^2)/zeta)*x; xc=[-10:0.1:-zeta*wn];c=sqrt(wn^2-xc.^2); plot(x,y,':',x,-y,':',xc,c,':',xc,-c,':') rloc nd(sys)
Select a point in the graphics window selected_point = -2.0165 + 2.5426i ans = 2.4659
K
4 3
+
2 1
x
Imag Axis
0 -1 -2 -3 -4 -4
x +
-3
-2
-1
0 Real Axis
1
2
3
4
FIGURE CP12.4 1 (a) Root locus for 1 + K s2 +4s+8 .
The performance region is specied by = 0.59 and n = 1.35 ,
which derives from the design specications Ts < 5 sec and P.O. < 10% .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
685
Using an m-le, the value of K = 2.5 is selected with the rlocnd function. The step responses for b = 0, 1, 4 and b = 40 are shown in Figure CP12.4b. When b = 0, the system is marginally stable; b = 1 results in a stable system with unsatisfactory performance. The nominal case b = 4 is stable and all performance specs are satised. When b = 40, the system is heavily damped: the percent overshoot specication is satised, but the settling time is too long.
0.5 0.45 0.4 0.35
b=0
b=1
Amplitude
0.3 0.25 0.2 0.15 0.1 0.05 0 0 b=40 1 2 3 4 5 Time (secs) 6 7 8 9 10 b=4
FIGURE CP12.4 CONTINUED: (b) Step responses for b = 0, 1, 4 and 40.
CP12.5
(a) An acceptable lead compensator (designed with root locus methods) is Gc (s) = K s+a s + 0.3 =5 . s+b s+2
The compensated root locus is shown in Figure CP12.5a, where K=5 is selected to place the closed-loop poles in the performance region. (b) The step responses for = 0, 0.005, 0.1 and 1 are shown in Figure CP12.5b.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
686
CHAPTER 12
Robust Control Systems
4 3
+
2
+
x o
1
Imag Axis
0 -1
x
+ o
x
+
-2
+
o x
-3 -4 -4
-3
-2
-1
0 Real Axis
1
2
3
4
FIGURE CP12.5 (a) Compensated root locus.
(c) You would like the actual structural damping to be greater than the design value, if it must be dierent at all.
zeta=0,0.005 (solid); zeta=0.1 (dashed); zeta=1 (dotted) 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0 0
2
4
6
8
10 Time (sec)
12
14
16
18
20
FIGURE CP12.5 CONTINUED: (b) Step responses for = 0, 0.005, 0.1 and 1.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
687
CP12.6
The m-le script which computes the phase margin as a function of the time delay (using the pade function) is shown in Figure CP12.6. The maximum time delay (for stability) is td = 4.3 seconds.
K=5; numg=K*[1]; deng=[1 10 2]; sysg = tf(numg,deng); time delay vector td=[0:0.1:5]; for i=1:length(td) [ndelay,ddelay]=pade(td(i),2); sysd = tf(ndelay,ddelay); sys = series(sysg,sysd); [mag,phase,w]=bode(sys); [gm,pm,w1,w2]=margin(mag,phase,w); pmv(i)=pm; end plot(td,pmv), grid xlabel('time delay [sec]') ylabel('phase margin [deg]')
120
100
80
phase margin [deg]
60
40
20
0
-20 0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
time delay [sec]
FIGURE CP12.6 Phase margin versus time delay.
CP12.7
The m-le script is shown in Figure CP12.7a. The steady-state error (shown in Figure CP12.7b) is zero when a = 0.5 and increases rapidly as a increases past a = 0.5. The maximum initial undershoot is shown in Figure CP12.7c. As a increases, the initial undershoot increases linearly. The gain margin is shown in Figure CP12.7d. It
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
688
CHAPTER 12
Robust Control Systems
can be seen that as a increases, the gain margin decreases very rapidly.
a=[0.01:0.01:0.99]; t=[0:0.1:30]; for i=1:length(a) num=a(i)*[1 -0.5]; den=[1 2 1]; sys_o = tf(num,den); [mag,phase,w]=bode(sys_o); [gm,pm,w1,w2]=margin(mag,phase,w); gain margin gmv(i)=gm; sys_cl = feedback(sys_o,[1]); [y,x]=step(-sys_cl,t); negative unit step input yf(i)=1-y(length(t)); steady-state tracking error ym(i)=-min(y)*100; max initial undershoot end gure(1), plot(a,gmv), grid, xlabel('a'), ylabel('gm') gure(2), plot(a,yf ), grid, xlabel('a'), ylabel('steady-state error') gure(3), plot(a,ym), grid, xlabel('a'), ylabel('maximum initial undershoot [%]')
FIGURE CP12.7 Script to generate all the plots.
1 0.9 0.8 0.7 steadystate error 0.6 0.5 0.4 0.3 0.2 0.1 0
0
0.1
0.2
0.3
0.4
0.5 a
0.6
0.7
0.8
0.9
1
FIGURE CP12.7 CONTINUED: (b) Steady-state tracking error.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
689
25
20 maximum initial undershoot [%]
15
10
5
0
0
0.1
0.2
0.3
0.4
0.5 a
0.6
0.7
0.8
0.9
1
FIGURE CP12.7 CONTINUED: (c) Maximum initial undershoot.
250
200
150 gm 100 50 0 0
0.1
0.2
0.3
0.4
0.5 a
0.6
0.7
0.8
0.9
1
FIGURE CP12.7 CONTINUED: (d) Gain margin.
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690 CP12.8
CHAPTER 12
Robust Control Systems
The plant (balloon and canister dynamics plus motor) is given by G(s) = and the PID controller is Gc (s) = KD (s2 + as + b) . s 1 , (s + 2)(s + 4)(s + 10)
Let a = 6. Then using the root locus methods, we determine that with KD = 12.5 we have the roots s1 = 8.4 s2 = 4.7 s3,4 = 1.43 j 1.05 . Thus, = 0.8. The plot of y (t) is shown in Figure CP12.8. The percent overshoot is less that 3%, as desired. and b = 10
1.4
1.2
With prefilter
1
0.8 y(t)
Without prefilter
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5 Time (sec)
3
3.5
4
4.5
5
FIGURE CP12.8 Simulation of the GRID device.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
CHAPTER
13
Digital Control Systems
Exercises
E13.1 (a) Elevation contours on a map are discrete signals. (b) Temperature in a room is a continuous signal. (c) A digital clock display is a discrete signal. (d) The score of a basketball game is a discrete signal. (e) The output of a loudspeaker is a continuous signal. (a) Using long-division we determine that Y (z ) = z 1 + 3z 2 + 7z 3 + 15z 4 + Therefore, with Y (z ) = we have y (0) = 0 y (T ) = 1 y (2T ) = 3 y (3T ) = 7 y (4T ) = 15 .
k =0
E13.2
y (kT )z k
(b) The exact solution is y (kT ) = ek ln 2 1 . E13.3 For the system response y (kT ) = kT where k 0, we have Y (z ) = E13.4 Tz . (z 1)2
The partial fraction expansion of Y (s) is Y (s) = 5 0.25 0.0625 0.3125 = + . s(s + 2)(s + 10) s s + 10 s+2 691
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
692
CHAPTER 13
Digital Control Systems
Then, using Table 13.1 in Dorf and Bishop, we determine that Y (z ) = 0.25 z z z + 0.0625 0.3125 10T z1 z e z e2T z z z = 0.25 + 0.0625 0.3125 , z1 z 0.135 z 0.670
where T = 0.1. E13.5 The Space Shuttle and robot arm control block diagram is shown in Figure E13.5. The human operator uses information from the computer generated data display and visual sensory data from the TV monitor and by looking out the window. He/she commands the robot arm via a joystick command to the computer.
data display
digital
measurement analog
A/D
joint angle & rate sensors
digital human operator joystick command
ref.
+
Computer
digital
analog
D/A
-
Robot arm & motors/gears
tip position
measurement
TV monitor & window view
FIGURE E13.5 The Space Shuttle/robot arm control block diagram.
E13.6
From Section 10.8 in Dorf and Bishop, we nd that the design resulted in the compensator Gc (s) = Using the relationships A = eaT , we compute B = ebT , and C 1A a =K , 1B b 6.66s + 1 s + 0.15 = 0.1 . 66.6s + 1 s + 0.015
A = e0.15(0.001) = 0.99985 , B = e0.015(0.001) = 0.999985 , and C = 0.1 .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
693
Therefore, D (z ) = C E13.7 zA z 0.99985 = 0.1 . zB z 0.999985
Using long-division, we determine that Y (z ) = 1 + 3.5z 1 + 5.75z 2 + 6.875z 3 + Therefore, with Y (z ) = we have y (0) = 1 y (T ) = 3.5 y (2T ) = 5.75 y (3T ) = 6.875 .
k =0
y (kT )z k
E13.8
The closed-loop system with T (z ) = z2 z + 0.2z 0.4
is stable since the poles of the transfer function (z = 0.74 and z = 0.54) both lie within the unit circle in the z-plane. E13.9 (a) Using long-division we determine that Y (z ) = z 1 + z 2 + z 3 + z 4 + Therefore, with Y (z ) = we have y (0) = 0 y (T ) = 1 y (2T ) = 1 y (3T ) = 1 (b) The exact solution is y (kT ) = 1 (k) where (k) = 1 when k = 0 and (k) = 0 when k = 0. E13.10 We compute T / = 1.25. (a) Using Figure 13.19 in Dorf and Bishop, we determine that K = 0.8 which implies K = 100. (b) Using Figure 13.21 in Dorf and Bishop, we determine that ess = 0.75. y (4T ) = 1 .
k =0
y (kT )z k
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
694
CHAPTER 13
Digital Control Systems
(c) Using Figure 13.20 in Dorf and Bishop, we determine that K = 0.7 which implies K = 88. E13.11 (a) The transfer function (including the zero-order hold) is Go (s)Gp (s) = 100(1 esT ) . s(s2 + 100)
Expanding into partial fractions yields G(z ) = (1 z 1 )Z 1 s 2 s s + 100 z z (z cos 10T ) = (1 z 1 ) 2 z 1 z 2 cos 10T z + 1 0.1224(z + 1) . 1.7552z + 1
.
When T = 0.05 we ha,ve G(z ) = z2
(b) The system is marginally stable since the system poles, z = 0.8776 0.4794j , are on the unit circle. (c) The impulse response and sinusoidal input response are shown in Figure E13.11.
0.5
Amplitude
0
-0.5
0
2
4
6
8 No. of Samples
10
12
14
16
40 20 0 -20 -40
Amplitude
0
10
20
30
40
50
60
70
80
90
100
No. of Samples
FIGURE E13.11 Impulse and sinusoidal (natural frequency) input response.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Exercises
695
E13.12
The partial fraction expansion of X (s) is X (s) = s2 s+1 2 1 = . + 5s + 6 s+3 s+2
Then, with T = 1, we have X (z ) = E13.13 2z z 2z z = . z e3 z e2 z 0.0498 z 1353
The root locus is shown in Figure E13.13. For stability: 0.25 < K < 3.75.
Root Locus 1 0.8 0.6 0.4 Imaginary Axis 0.2 0 0.2 0.4 0.6 0.8 1 3 K=3.75 K=0.25
2
1 Real Axis
0
1
2
FIGURE E13.13 Root locus with unit circle (dashed curve).
E13.14
Given Gp (s), we determine that (with K = 5) G(z ) = 5(1 e1 )z . z (z e1 )
The closed-loop characteristic equation is z 2 + 1.792z + 0.368 = 0 and the system is unstable, since there is a pole at z = 1.55. The
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
696
CHAPTER 13
Digital Control Systems
system is stable for 0 < K < 4.32 . E13.15 The transfer function G(z ) is G(z ) = z2 0.1289z + 0.02624 . 0.3862z + 0.006738
The sampling time is T = 1 s. E13.16 The transfer function G(z ) is G(z ) =
z2
The sampling time is T = 0.5 s.
0.2759z + 0.1982 . 1.368z + 0.3679
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
697
Problems
P13.1 The plot of the input to the sampler and the output r (t) is shown in Figure P13.1.
1 0.8 0.6 0.4 0.2
r(t), r*(t)
0 -0.2 -0.4 -0.6 -0.8 -1 0 0.2 0.4 0.6 0.8 1 Time (sec) 1.2 1.4 1.6 1.8 2
FIGURE P13.1 Plot of r (t) = sin(t) and r (t).
P13.2
The plot of the input and the output is shown in Figure P13.2.
1 0.8 0.6 0.4 0.2
r(t)
0 -0.2 -0.4 -0.6 -0.8 -1 0 0.2 0.4 0.6 0.8 1 Time (sec) 1.2 1.4 1.6 1.8 2
FIGURE P13.2 Plot of r (t) = sin(t) and output of sample and hold.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
698 P13.3
CHAPTER 13
Digital Control Systems
The transfer function Y (z )/R (z ) = G(z ) = The ramp input is represented by R(z ) = Tz . (z 1)2 z . z eT
The output Y (z ) = G(z )R(z ) is obtained by long division: Y (z ) = T z 1 + T (2 + eT )z 2 T (1 + 2eT ) (2 + eT )2 z 3 + T eT + (1 + 2eT )(2 + eT ) (2 + eT ) (1 + 2eT ) (2 + eT )2 P13.4 The transfer function Y (s)/R (s) = 1 esT . s(s + 2) z 4 +
The partial fraction expansion (with T = 1) yields G(z ) = (1 z 1 )Z = P13.5 0.4323 . z 0.1353 z . z1 0.5 0.5 s s+2 = (1 z 1 ) 0.5z 0.5z z 1 z 0.1353
The step input is R(z ) = Also, T (z ) = So, Y (z ) = T (z )R(z ) = 0.6321 z 0.6321z =2 . z + 0.2643 z 1 z 0.7357z 0.2643 G(z ) 0.6321 = . 1 + G(z ) z + 0.2643
Using long-division we determine that Y (z ) = 0.6321z 1 + 0.4650z 2 + 0.5092z 3 + 0.4975z 4 + 0.5006z 5 +
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
699
Therefore, with Y (z ) =
k =0
y (kT )z k
we have y (0) = 0, y (T ) = 0.6321, y (2T ) = 0.4650, y (3T ) = 0.5092, y (4T ) = 0.4975, and y (5T ) = 0.5006. P13.6 Using the nal value theorem (see Table 13.1 in Dorf and Bishop), we determine that (for a step input) Yss = lim (z 1)Y (z ) = lim (z 1)
z 1 z 1
0.6321 z 0.6321 = = 0.5 . z + 0.2643 z 1 1.2643 0.6321 z =0. z + 0.2643 z 1
And using the initial value theorem, we compute Yo = lim Y (z ) = lim
z z
P13.7
Using Figures 13.19 and 13.21 in Dorf and Bishop, we determine that the performance specications are satised when K = 0.5 and T = 2. Computing K and T (with = 0.5) yields K = 1 and T = 1. We can select K = 1 and r = 0.2. The step responses for the compensated and uncompensated systems are shown in Figure P13.8.
P13.8
1.2
1 Uncompensated 0.8 Amplitude
0.6
Compensated
0.4
0.2
0 0
10
20
30 40 No. of Samples
50
60
70
FIGURE P13.8 Plot of compensated and uncompensated systems.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
700 P13.9
CHAPTER 13
Digital Control Systems
Consider the compensator Gc (s) = K s+a . s+b
Then, using Bode methods we can select a = 1, b = 4, and K = 1. The compensated system phase margin is P.M. = 50o and the gain margin is G.M. = 15dB. The crossover frequency is c = 2.15 rad/sec. Utilizing the Gc (s)-to-D (z ) method and selecting T = 0.01 second, we determine D (z ) = C We use the relationships A = eaT , zA z 0.99 = . zB z 0.96 and C 1A a =K , 1B b
B = ebT ,
to compute A = e0.01 = 0.99, B = e0.04 = 0.96, and C = 1. P13.10 (a) The transfer function G(z )D (z ) is G(z )D (z ) = K z2 0.0037z + 0.0026 . 1.368z + 0.3679
(b) The closed-loop system characteristic equation is 1+K z2 0.0037z + 0.0026 =0. 1.368z + 0.3679
(c) Using root locus methods, the maximum value of K is found to be Kmax = 239. (d) Using Figure 13.19 in Dorf and Bishop for T / = 1 and a maximum overshoot of 0.3, we nd that K = 75. (e) The closed-loop transfer function (with K = 75) is T (z ) = z2 0.2759z + 0.1982 . 1.092z + 0.5661
The step response is shown in Figure P13.10. (f) The closed-loop poles with K = 119.5 are z = 0.4641 0.6843j . The overshoot is 0.55. (g) The step response is shown in Figure P13.10 (for K = 119.5).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
701
1.5 K=75
Amplitude
1
0.5
0
0
2
4
6
8 No. of Samples
10
12
14
16
2 1.5 1 0.5 0
K=119.5
Amplitude
0
2
4
6
8
10
12
14
16
18
20
No. of Samples
FIGURE P13.10 Step response for K = 75 and K = 119.5.
P13.11
(a) Consider the compensator Gc (s) = K s+a . s+b
Then, using Bode methods we can select a = 0.7, b = 0.1, and K = 150. The compensated system overshoot and steady-state tracking error (for a ramp input) are P.O. = 30% and ess < 0.01. (b) Utilizing the Gc (s)-to-D (z ) method (with T = 0.1 second), we determine D (z ) = C We use the relationships A = eaT , to compute A = e0.007 = 0.9324 , zA z 0.9324 = 155.3 . zB z 0.99 and C 1A a =K , 1B b and C = 155.3 .
B = ebT ,
B = e0.01 = 0.99 ,
(c) The step response for the continuous system with Gc (s) in part(a) and for the discrete system with D (z ) in part (b) is shown in Figure P13.11a. (d) Utilizing the Gc (s)-to-D (z ) method (with T = 0.01 second), we de-
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
702
CHAPTER 13
Digital Control Systems
T=0.1 sec 1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0
0
2
4
6
8
10
12
14
16
18
20
No. of Samples
FIGURE P13.11 (a) Step response for continuous and discrete systems (T=0.1s) in Parts (a) and (b).
termine D (z ) = C We use the relationships z 0.993 zA = 150 . zB z 0.999
to compute
A = eaT B = ebT 1A a C =K 1B b A = e0.07 = 0.993 B = e0.001 = 0.999 C = 150 .
The step response for the continuous system with Gc (s) in and for the discrete system with D (z ) in part (d) is shown ure P13.11b. (e) The ramp response for the continuous system with Gc (s) in and for the discrete system with D (z ) in part (b) is shown ure P13.11c.
part(a) in Figpart(a) in Fig-
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
T=0.01 sec 1.4
703
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
20
40
60
80
100
120
140
160
180
200
No. of Samples
FIGURE P13.11 CONTINUED: (b) Step response for continuous and discrete systems (T=0.01s) in Parts (a) and (d).
T=0.1 sec 2 1.8 1.6 1.4
Amplitude
1.2 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14 16 18 20 Ramp input (dashed line)
No. of Samples
FIGURE P13.11 CONTINUED: (c) Ramp response for continuous and discrete systems (T=0.1s) in Parts (a) and (b).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
704 P13.12
CHAPTER 13
Digital Control Systems
The root locus is shown in Figure P13.12. For stability: 0 < K < 2.5.
Root Locus 2
1.5
Unit circle (dashed line)
1
Imaginary Axis
0.5
0
System: sysz Gain: 2.5 Pole: 1 Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 1
0.5
1
1.5
2 2
1.5
1
0.5
0 Real Axis
0.5
1
1.5
2
FIGURE P13.12 2 Root locus for 1 + K zz(+0.1) = 0. z
P13.13
The root locus is shown in Figure P13.13. When K = 0.027, the characteristic equation has two equal roots: z1,2 = 0.7247 and z3 = 0.2593.
2 1.5 1 0.5 Unit circle (dashed line)
Imag Axis
0 -0.5 -1 -1.5 -2 -2
o
o
x x
x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE P13.13 z 2 +1 z 0. Root locus for 1 + K z 3 1.7358.1206.87110364.1353 = 0. z 2 +0 z 0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
705
P13.14
The root locus is shown in Figure P13.14. When K = 9.5655 105 , the two real roots break away from the real axis at z = 0.99. For stability: K < 9.7 105 .
2 1.5 1 0.5 Unit circle (dashed line)
Imag Axis
x
0 -0.5 -1 -1.5 -2 -2
o
o
x x
x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE P13.14 z 3 +10.3614 2 758z 8353 Root locus for 1 + K z 4 3.7123z 3 +5.z +9.2 3.+0.z +0.7408 = 0. 1644z 195
P13.15
Given Gp (s) = 20 s5
and the sample and hold (T=0.1s) as shown in Figure 13.18 in Dorf and Bishop, we determine that G(z ) = 2.595 . z 1.649
Then, with R(z ) = z/(z 1), we have Y (z ) = 2.595z . (z 1)(z + 0.9462)
Therefore, Y (z ) = 2.59z 1 + 0.14z 2 + 2.46z 3 + 0.26z 4 + .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
706 P13.16
CHAPTER 13
Digital Control Systems
Given Gp (s) and the sample and hold (T=1s) as shown in Figure 13.18 in Dorf and Bishop, we determine that G(z ) = z2 0.22775z + 0.088984 . 1.0498z + 0.049787
Then, with R(z ) = z/(z 1), we have Y (z ) = z2 0.22775z + 0.088984 z . 0.82203z + 0.13877 z 1
The plot of y (kT ) is shown in Figure P13.16.
1 0.9 0.8 0.7 0.6 y(kT) 0.5 0.4 0.3 0.2 0.1 0
1
2
3
4 kT
5
6
7
8
FIGURE P13.16 Plot of y (kT ) for a step input.
P13.17
The root locus is shown in Figure P13.17 for 1+K z2 0.42612z + 0.36082 =0. 1.6065z + 0.60653
The limiting value of the gain for stability is K = 1.09.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Problems
707
Root Locus 2
1.5
Unit circle (dashed line)
1
Imaginary Axis
0.5
0
0.5
System: sys Gain: 1.09 Pole: 0.568 0.818i Damping: 0.571 Overshoot (%): 888 Frequency (rad/sec): 0.996
1
1.5
2 6
5
4
3
2 Real Axis
1
0
1
2
FIGURE P13.17 0.42612z Root locus for 1 + K z 2 1.6065+0.36082 = 0. z +0.60653
P13.18
The plot of the step responses for 0 T 1.2 is shown in Figure P13.18. The overshoot and settling time summary is given in Table P13.18.
T
0
0.2
0.4
0.6
0.8
1.0
1.2
P.O.
16.3%
20.6%
25.6%
31.3%
36.9%
40.0%
51.0%
Ts
8.1
8.4
8.8
11.4
14.4
16.0
19.2
TABLE P13.18
Performance summary.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
708
CHAPTER 13
1.6
Digital Control Systems
1.4
1.2
1 Amplitude
0.8
0.6
0.4
0.2
0 0
10
20
30
40 50 60 No. of Samples
70
80
90
100
FIGURE P13.18 Step responses for 0 T 1.2.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
709
Advanced Problems
AP13.1 Given the sample and hold with Gp (s), we determine that G(z ) = 10.5K (z 0.9048) . (z 1)2
The root locus is shown in Figure AP13.1. For stability: 0 < K < 0.2.
2 1.5 1 0.5 Unit circle (dashed line)
Imag Axis
0 -0.5 -1 -1.5 -2 -2
ox
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE AP13.1 10.5(z 0.9048) = 0 with unit circle (dashed line). Root locus for 1 + K (z 1)2
AP13.2
The root locus is shown in Figure AP13.2a. The loop transfer function is
2 1.5 1 0.5 Unit circle (dashed line)
Imag Axis
0 -0.5 -1 -1.5 -2 -2
o
x
x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE AP13.2 0.0379 (a) Root locus for 1 + K (z 1)(z 0z368) = 0. .
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
710
CHAPTER 13
Digital Control Systems
G(z )D (z ) = K
0.0379z . (z 1)(z 0.368)
For stability: Kmax = 72. We select K = 8.2. The step response is shown in Figure AP13.2b.
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
1
2
3
4
5 No. of Samples
6
7
8
9
10
FIGURE AP13.2 CONTINUED: (b) Step response with K = 8.2.
AP13.3
The root locus is shown in Figure AP13.3a.
Root Locus 2
The maximum gain for
1.5
Unit circle (dashed line)
1
Imaginary Axis
0.5
0
0.5
System: sysz Gain: 52.8 Pole: 0.709 0.702i Damping: 0.00255 Overshoot (%): 99.2 Frequency (rad/sec): 19.5
1
1.5
2 3
2.5
2
1.5
1
0.5 Real Axis
0
0.5
1
1.5
2
FIGURE AP13.3 . (a) Root locus for 1 + K 0z005769z +0.005185 = 0. 2 1.726z +0.7261
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Advanced Problems
711
stability is Kmax = 52.8. We select K = 4.5. The step response is shown in Figure AP13.3b.
Step Response 1.4
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.5 Time (sec)
1
1.5
FIGURE AP13.3 CONTINUED: (b) Step response with K = 4.5.
AP13.4
The loop transfer function is G(z ) = 10(1 eT ) , z eT
and the closed-loop transfer function is T (z ) = For stability, we require |11eT 10| < 1 . Solving for T yields 0 < T < 0.2 . Selecting T = 0.1s provides a stable system with rapid response; the settling time is Ts = 0.2s. The step response is shown in Figure AP13.4. 10(1 eT ) . z (11eT 10)
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
712
CHAPTER 13
Digital Control Systems
1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 2 No. of Samples 2.5 3 3.5 4
FIGURE AP13.4 Step response with T = 0.1s.
AP13.5
The maximum gain for stability is Kmax = 63.15.
Root Locus 2
1.5
Unit circle (dashed line)
1
Imaginary Axis
0.5
0
0.5
System: sysz Gain: 63.2 Pole: 0.725 0.686i Damping: 0.00308 Overshoot (%): 99 Frequency (rad/sec): 7.58
1
1.5
2 3
2.5
2
1.5
1
0.5 Real Axis
0
0.5
1
1.5
2
FIGURE AP13.5 . Root locus for 1 + K 0z004535z +0.004104 = 0. 2 1.741z +0.7408
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
713
Design Problems
CDP13.1
The plant model with parameters given in Table CDP2.1 in Dorf and Bishop is given by: Gp (s) = 26.035 , s(s + 33.142)
where we neglect the motor inductance Lm and where we switch o the tachometer feedback (see Figure CDP4.1 in Dorf and Bishop). Letting G(z ) = Z G ( )G ( ) we obtain G(z ) = A suitable controller is D (z ) = 20(z 0.5) . z + 0.25 1.2875e 05(z + 0.989) . (z 1)(z 0.9674)
The step response is shown below. The settling time is under 250 samples. With each sample being 1 ms this means that Ts < 250 ms, as desired. Also, the percent overshoot is P.O. < 5%.
1.2
1
0.8 Amplitude
0.6
0.4
0.2
0 0
50
100
150 No. of Samples
200
250
300
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
714 DP13.1
CHAPTER 13
Digital Control Systems
(a) Given the sample and hold with Gp (s), we determine that KG(z ) = K 0.1228 . z 0.8465
The root locus is shown in Figure DP13.1a. For stablity: 0 K < 15.
1.5
Unit circle (dashed line)
1
0.5
Imag Axis
0
x
-0.5
-1
-1.5 -1.5 -1 -0.5 0 Real Axis 0.5 1 1.5
FIGURE DP13.1 0 1228 (a) Root locus for 1 + K z .0.8465 = 0 with unit circle (dashed line).
(b) A suitable compensator is Gc (s) = 15(s + 0.5) . s+5
Utilizing the Gc (s)-to-D (z ) method (with T = 0.5 second), we determine D (z ) = C We use the relationships A = eaT , to compute A = e0.5(0.5) = 0.7788 , B = e0.5(5) = 0.0821 , and C = 6.22 . B = ebT , and C 1A a =K , 1B b zA z 0.7788 = 6.22 . zB z 0.0821
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
715
(c) The step response is shown in Figure DP13.1b.
0.8 0.7 0.6 0.5
Amplitude
0.4 0.3 0.2 0.1 0
0
2
4
6
8
10
12
14
16
18
20
No. of Samples
FIGURE DP13.1 CONTINUED: (b) Closed-loop system step response.
DP13.2
With the sample and hold (T=10ms), we have G(z ) = A suitable compensator is D (z ) = K z 0.75 , z + 0.5 0.00044579z + 0.00044453 . z 2 1.9136z + 0.99154
where K is determined so that of the system is 1/ 2. The root locus is shown in Figure DP13.2. We choose K = 1400.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
716
CHAPTER 13
Digital Control Systems
Root Locus 1.5 Curve of constant zeta=0.707 (dashed line) 1
0.5 Imaginary Axis
0
0.5
1
1.5 5
4
3
2 1 Real Axis
0
1
2
FIGURE DP13.2 75 Root locus for 1 + K zz0..5 0.00044579z +0.00044453 = 0. +0 z 2 1.9136z +0.99154
DP13.3
The root locus is shown in Figure DP13.3a.
2 Curve of constant zeta=0.707 (dashed line) 1.5 1 0.5
Imag Axis
0 -0.5 -1 -1.5 -2 -2
o
x
x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE DP13.3 z +1 (a) Root locus for 1 + K (z 1)(z 0.5) = 0.
The gain for = 0.707 is K = 0.0627. The step response is shown in Figure DP13.3b. The settling time is Ts = 14T = 1.4s and P.O. = 5%.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
717
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8
10
12
14
16
18
20
No. of Samples
FIGURE DP13.3 CONTINUED: (b) Step response with K = 0.0627.
DP13.4
With the sample and hold (T=1s), we have G(z ) = 0.484(z + 0.9672) . (z 1)(z 0.9048)
2 Curve of constant zeta=0.5 (dashed line) 1.5 1 0.5
Imag Axis
0 -0.5 -1 -1.5 -2 -2
o
x
ox x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE DP13.4 88 0.484(z +0.9672) (a) Root locus for 1 + K zz0..5 (z 1)(z 0.9048) = 0. +0
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
718
CHAPTER 13
Digital Control Systems
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
2
4
6
8 No. of Samples
10
12
14
16
FIGURE DP13.4 CONTINUED: (b) Step response for K = 12.5.
A suitable compensator is D (z ) = K z 0.88 , z + 0.5
where K is determined so that of the system is 0.5. The root locus is shown in Figure DP13.4a. We choose K = 12.5. The step response is shown in Figure DP13.4b. Also, Kv = 1, so the steady-state error specication is satised. DP13.5 Select T = 1 second. With the sample and hold, we have G(z ) = z2 0.2838z + 0.1485 . 1.135z + 0.1353
The root locus is shown in Figure DP13.5. To meet the percent overshoot specication, we choose K so that of the system is 0.7. This results in K = 1. The step response has an overshoot of P.O. = 4.6%. Also, from Figure 13.21 in Dorf and Bishop, we determine that the steady-state error to a ramp input is ess = 2 (since T / = 2, and K = 0.3).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Design Problems
719
2 Curve of constant zeta=0.7 (dashed line) 1.5 1 0.5
Imag Axis
0 -0.5 -1 -1.5 -2 -2
o
x
x
-1.5
-1
-0.5
0 Real Axis
0.5
1
1.5
2
FIGURE DP13.5 0.2838z Root locus for 1 + K z 2 1.135+0.1485 = 0. z +0.1353
DP13.6
With the sample and hold at T = 1 , we have G(z ) = 0.2492z + 0.2483 . z 2 1.99z + 0.99
1.5
1
0.5 Imaginary Axis
0
0.5
1
1.5 5
4
3
2 Real Axis
1
0
1
2
FIGURE DP13.6 .9 2492z +0.2483 Root locus for 1 + K z 0.6 0.2 1.99z +0.99 = 0. z +0 z
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
720
CHAPTER 13
Digital Control Systems
Consider the digital controller Dz ) = K z 0.9 . z + 0.6
The root locus is shown in Figure DP13.6. To meet the percent overshoot specication, we choose K so that of the system is greater than 0.52. We select K = 2.3. The step response has an overshoot of P.O. = 12.7% and the settling time is Ts = 19s.
Step Response 1.4
System: syscl Peak amplitude: 1.13 Overshoot (%): 12.7 At time (sec): 4 System: syscl Settling Time (sec): 19
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
5
10
15 20 Time (sec)
25
30
35
FIGURE DP13.6 CONTINUED: (b) Step response for K = 2.3.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
721
Computer Problems
CP13.1 The m-le script and unit step response are shown in Figure CP13.1.
num=[0.2145 0.1609]; den=[1 -0.75 0.125]; sysd = tf(num,den,1); step(sysd,0:1:50)
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0 0
5
10
15
20
25
30
35
40
45
50
No. of Samples
FIGURE CP13.1 Step response.
CP13.2
The m-le script utilizing the c2d function is shown in Figure CP13.2.
% Part (a) num = [1]; den = [1 0]; T = 1; sys = tf(num,den); sys_d = c2d(sys,T,'zoh') % % Part (b) num = [1 0]; den = [1 0 2]; T = 1; sys = tf(num,den); sys_d=c2d(sys,T,'zoh')
Transfer function: 1 ----z-1 Transfer function: 0.6985 z - 0.6985 -----------------z^2 - 0.3119 z + 1
FIGURE CP13.2 Script utilizing the c2d function for (a) and (b).
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
722
CHAPTER 13
Digital Control Systems
% Part (c) num = [1 4]; den = [1 3]; T = 1; sys = tf(num,den); sys_d = c2d(sys,T,'zoh') % % Part (d) num = [1]; den = [1 8 0]; T = 1; sys = tf(num,den); sys_d = c2d(sys,T,'zoh')
Transfer function: z + 0.267 ----------z - 0.04979 Transfer function: 0.1094 z + 0.01558 ------------------z^2 - z + 0.0003355
FIGURE CP13.2 CONTINUED: Script utilizing the c2d function for (c) and (d).
CP13.3
The continuous system transfer function (with T = 0.1 sec) is T (s) = s2 13.37s + 563.1 . + 6.931s + 567.2
The step response using the dstep function is shown in Figure CP13.3a. The contrinuous system step response is shown in Figure CP13.3b.
1.8 1.6 1.4 1.2
Amplitude
1 0.8 0.6 0.4 0.2 0
0
2
4
6
8
10
12
14
No. of Samples
FIGURE CP13.3 (a) Unit step response using the dstep function.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
723
1.8
*
1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0* 0
* * * *
*
* *
* * *
*
*
*
0.2
0.4
0.6
0.8
1
1.2
1.4
FIGURE CP13.3 CONTINUED: (b) Continuous system step response (* denote sampled-data step response).
CP13.4
The root locus in shown in Figure CP13.4. For stablility: 0 < K < 0.88.
1.5
1
0.5 Imag Axis
0
-0.5
-1
-1.5 -1.5 -1 -0.5 0 Real Axis 0.5 1 1.5
FIGURE CP13.4 z Root locus for 1 + K z 2 z +0.1 = 0.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
724 CP13.5
CHAPTER 13
Digital Control Systems
The root locus in shown in Figure CP13.5. For stablility: 0 < K < .
1 0.8 0.6 0.4 Imag Axis 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 Real Axis 0.4 0.6 0.8 1
FIGURE CP13.5 (z 0.2)(z +1) Root locus for 1 + K (z 0.08)(z 1) = 0
CP13.6
The root locus is shown in Figure CP13.6.
Root Locus 1.5 1 0.5 0 0.5 1 1.5 1.5
Imaginary Axis
1
0.5
0 Real Axis
0.5
1
1.5
FIGURE CP13.6 Root locus.
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Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
Computer Problems
725
We determine the range of K for stability is 0.4 < K < 1.06.
% Part (a) num=[1 4 4.25 ]; den=[1 -0.1 -1.5]; sys = tf(num,den); rlocus(sys), hold on xc=[-1:0.1:1];c=sqrt(1-xc.^2); plot(xc,c,':',xc,-c,':') hold off % % Part (b) rlocfind(sys) rlocfind(sys)
Select a point in the graphics window selected_point = -0.8278 + 0.5202i ans = 0.7444
Kmax
Select a point in the graphics window selected_point = -0.9745 - 0.0072i ans = 0.3481
Kmin
FIGURE CP13.6 CONTINUED: Using the rlocus and rlocnd functions.
CP13.7
Using root locus methods, we determine that an acceptable compensator is Gc (s) = 11.7 s+6 . s + 20
With a zero-order hold and T = 0.02 sec, we nd that
1.2
1
* * *
*
******************************** **
0.8
* * *
Amplitude
0.6
* *
*
0.4
* *
0.2
* *
0* * 0
0.1
0.2
0.3
0.4
0.5 Time (sec)
0.6
0.7
0.8
0.9
1
FIGURE CP13.7 System step response (* denotes sampled-data response).
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Solutions Manual to Accompany Modern Control Systems, Eleventh Edition, by Richard C Dorf and Robert H. Bishop. ISBN-13: 9780132270298. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
726
CHAPTER 13
Digital Control Systems
D (z ) =
The closed-loop step response is shown in Figure CP13.7.
11.7z 10.54 . z 0.6703
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