Unformatted text preview: ME375 Handouts Introduction to Feedback Control
• Control System Design
–
–
– Why Control?
OpenLoop vs ClosedLoop (Feedback)
OpenClosedWhy Use Feedback Control? • ClosedLoop Control System Structure
– Elements of a Feedback Control System
– ClosedLoop Transfer Functions (CLTF)
Closed • Performance Specifications
– Steady State Specifications
– Transient (Dynamic) Specifications
School of Mechanical Engineering
Purdue University ME375 Feedback Control  1 Control System Design
Control: verb, 1. To exercise authority or dominating influence over; direct;
verb,
regulate. 2. To hold in restraint.
Control is the process of causing a system to behave in a prescribed manner.
manner.
Specifically, control system design is the process of causing a system variable
(output) to conform to some desired input (reference).
output)
(reference).
Reference
Input Input R (s) U (s) System
System
(Plant)
(Plant)
GP(s)
GP(s) Output
Y(s) The objective of the control system is to control the output y by using the input u,
such that the output y follows a set of reference inputs r.
School of Mechanical Engineering
Purdue University ME375 Feedback Control  2 1 ME375 Handouts OpenLoop vs ClosedLoop
• OpenLoop Control
OpenThe control input u(t) (or U(s)) is synthesized based on the a priori knowledge
of the system (plant) and the reference input r(t) (or R(s)). The control system
does not measure the output, and there is no comparison of the output to make it
output
conform to the desired output (reference input).
Reference Input
(Command)
R(s) System Output C(s) GP(s) U(s)
Control Input Y(s) Plant or System Q: Ideally, if we want Y(s) to follow R(s) (i.e. want Y(s) = R(s)), how would you design the controller
R( (i.e.
Y(
C(s) ? School of Mechanical Engineering
Purdue University ME375 Feedback Control  3 OpenLoop Control Example
Static Cruise Control W The vehicle speed model can be
approximated by a static gain between
the throttle angle (input) and the vehicle
speed (output). From experiment, on
level road, at 55 mph, 1o of throttle angle
causes 10 mph change in speed. When
the road grade changes by 1%, 1o of
throttle angle will only change vehicle
speed by 5 mph. Design an openloop
opencruise controller for this vehicle.
R
U
Y
W : reference speed, mph
: throttle angle, degree
: actual speed, mph
: road grade, % R U − Y +
Cruise
Controller Speed
Model Q: What are potential problems with this cruise control ? School of Mechanical Engineering
Purdue University ME375 Feedback Control  4 2 ME375 Handouts OpenLoop vs ClosedLoop
• ClosedLoop (Feedback) Control
ClosedThe control input u(t) (or U(s)) is synthesized based on the a priori knowledge
of the system (plant), the reference input r(t) (or R(s)) and the measurement of
the actual output y(t) (or Y(s)). For example the temperature control of this
classroom:
Disturbance D(s) Heater Room
Temperature Room Actuator Y(s) Plant or System School of Mechanical Engineering
Purdue University ME375 Feedback Control  5 ClosedLoop Control Example
Static Cruise Control
Same vehicle system as the previous example. The vehicle speed is measured and fed back.
Design a closedloop cruise control that uses the measured vehicle speed and the reference
closedspeed.
W R − U Y +
Speed Model Q: How would road grade, plant uncertainty affect the closedloop performance ?
closedQ: How is the steady state performance ? Will you have any steady state error ?
School of Mechanical Engineering
Purdue University ME375 Feedback Control  6 3 ME375 Handouts ClosedLoop Control Example
Static Cruise Control (ClosedLoop Control)
(Closed(a) Find the actual vehicle speed when the reference speed is 50 mph and the road grade is 1%
and 10%, respectively. (b) If the actual vehicle speed model is 1o of throttle angle corresponds to 9 mph change in
speed, what is the actual vehicle speed with the same cruise controller.
controller. (c) When there is no grade and the vehicle speed model is accurate, what is the actual output
accurate,
speed when a reference speed of 50 mph is desired. School of Mechanical Engineering
Purdue University ME375 Feedback Control  7 Why Feedback ?
Using feedback, we can change the closedloop system’s dynamic behavior (the
closedsystem’
ClosedLoop Transfer Function (CLTF) will be different from the original
Closedoriginal
system’s (openloop) transfer function). By using feedback to change the CLTF,
system’ (openwe can achieve the following:
– Stabilize Unstable Systems
For example, unstable plants such as inverted pendulum and DC motor
motor
positioning systems can be stabilized using feedback.
– Improve System Performance (Achieve Performance Specifications)
• Steady State Performance  For example, reduce steady state error ...
• Transient Performance  For example, reduce rise time, reduce settling
time, reduce overshoot …
– Reduce (attenuate) the effect of modeling uncertainty (error) and
and
external disturbances
School of Mechanical Engineering
Purdue University ME375 Feedback Control  8 4 ME375 Handouts Example
More Realistic Cruise Control Problem
The relationship between a vehicle’s speed y and the throttle angle u is described by
vehicle’
a first order system with a steady state gain KC and a time constant of 3 sec. The
gain KC is affected by various operating conditions like the temperature and
temperature
humidity. Due to these effects, the actual value of KC is between 5 and 15. The
objective of the cruise control is to design a control law (strategy) to determine the
(strategy)
throttle angle u such that the vehicle’s steady state speed will stay within 2% of the
vehicle’
desired reference speed set by the driver.
Cruise
Controller
R Speed Model
Y U School of Mechanical Engineering
Purdue University Use a simple “proportional” feedback
proportional”
control, i.e. the control input u(t) is
u(
proportional to the regulation error
e(t) = r(t) − y(t). The control design
parameter is the proportional constant
between the input and the error. This
constant KP is usually called the
feedback gain or the proportional
gain.
gain.
ME375 Feedback Control  9 Example
Calculate ClosedLoop Transfer Function (CLTF):
ClosedCLTF): Select an appropriate feedback gain KP to satisfy the performance specification : Q: Will this proportional control law work for attenuating external disturbances ?
School of Mechanical Engineering
Purdue University ME375 Feedback Control  10 5 ME375 Handouts Elements of Feedback Control
Elements of a Feedback Control System:
• Plant (Process) GP(s)  The plant is the system (process) whose output is to be
controlled, e.g., the room in the room temperature control example.
example.
• Actuator  An actuator is a device that can influence the input to the plant, e.g. the heater
plant,
(furnace) in the room temperature control example.
• Disturbance d(t)  Disturbances are uncontrollable signals to the plant that tend to
adversely affect the output of the system, e.g., opening the windows in the room
windows
temperature control example.
• Sensor (Measurement System) H(s)  The transfer function (frequency response
function) of the device (system) that measures the system output, e.g., a thermocouple.
output,
• Controller GC (s)  The controller is the device that generates the controlled input that is
input
to affect the system output, e.g., the thermostat in the room temperature control example.
temperature
Controller Reference
Input Disturbance D(s)
Output GC (s) R(s) Heater Actuator
Sensor GP(s) Y(s) Plant
(Process) H(s)
School of Mechanical Engineering
Purdue University ME375 Feedback Control  11 ClosedLoop Transfer Function
Disturbance D(s)
Control
Input
U(s) Output GP(s) Y(s) Plant Plant Equation (Transfer function model that we all know how to obtain ?!): Control Law (Algorithm) (we will try to learn how to design): School of Mechanical Engineering
Purdue University ME375 Feedback Control  12 6 ME375 Handouts ClosedLoop Transfer Function
Disturbance D(s)
Reference
Input + Error GC (s)
− E(s) R(s) Control
Input + + U(s) Output GP(s) Y(s) Plant H(s) Y (s) = ⋅ R( s ) +
GYR ( s ) ⋅ D( s )
GYD ( s ) School of Mechanical Engineering
Purdue University ME375 Feedback Control  13 ClosedLoop Transfer Function
The closedloop transfer functions relating the output y(t) (or Y(s)) to the reference
closedinput r(t) (or R(s)) and the disturbance d(t) (or D(s)) are:
Y (s) = ⋅ R( s ) + GYR ( s )
ClosedLoop Transfer Function
From R ( s ) to Y ( s ) ⋅ D( s) GYD ( s )
ClosedLoop Transfer Function
From R ( s ) to Y ( s ) The objective of control system design is to design a controller GC (s), such that
certain performance (design) specifications are met. For example:
example:
• we want the output y(t) to follow the reference input r(t), i.e.,
for certain frequency range. This is equivalent to specifying that , • we want the disturbance d(t) to have very little effect on the output y(t) within the
frequency range where disturbances are most likely to occur. This is equivalent
to specifying that
School of Mechanical Engineering
Purdue University ME375 Feedback Control  14 7 ME375 Handouts Performance Specifications
Given an input/output representation, GCL (s), for which the output of the system
should follow the input, what specifications should you make to guarantee that the
system will behave in a manner that will satisfy its functional requirements?
Input
R(s) Output GCL (s) r(t) Y(s) y(t) Time Time School of Mechanical Engineering
Purdue University ME375 Feedback Control  15 Unit Step Response
1.6 yMAX
1.4 Unit Step Response OS
1.2 ± X% 1
0.8
0.6
0.4
0.2
0 tP Time tS tr
School of Mechanical Engineering
Purdue University ME375 Feedback Control  16 8 ME375 Handouts Performance Specifications
• Steady State Performance → Steady State Gain of the Transfer Function
Specifies the tracking performance of the system at steady state. Often it is
state.
specified as the steady state response, y(∞) (or ySS(t)), to be within an X% bound of
the reference input r(t), i.e., the steady state error eSS(t) = r(t) − ySS(t) should be
within a certain percent. For example:
r ( t ) − y SS ( t )
≤ 2% = 0.02
r (t ) ⇔ y SS ( t )
≥ 98% = 0.98
r (t ) ⇔ To find the steady state value of the output, ySS(t):
– Sinusoidal references: use frequency response, i.e.
references: – General references: use FVT, provided that
references: is stable, ... School of Mechanical Engineering
Purdue University ME375 Feedback Control  17 Performance Specifications
• Transient Performance (Transient Response)
Response)
Transient performance of a system is usually specified using the unit step response
of the system. Some typical transient response specifications are:
are:
– Settling Time (tS): Specifies the time required for the response to reach and
stay within a specific percent of the final (steadystate) value. Some typical
(steadysettling time specifications are: 5%, 2% and 1%. For 2nd order systems, the
specification is usually: ⎧ 4
⎪ ζ ω for 2% bound
⎪ n
⎨
⎪ 5 for 1% bound
⎪ ζω n
⎩ ≤ Desired Settling Time (t S ) ⇒
– % Overshoot (%OS):(2nd order systems)
(%OS):(2nd
%OS = 100e −π ζω n ω n 1−ζ 2 = 100e −π ζ
1−ζ 2 ≤ X% Q: How can we link this performance specification to the closedloop transfer function?
closedfunction?
(Hint) What system characteristics affect the system performance ?
Hint)
School of Mechanical Engineering
Purdue University ME375 Feedback Control  18 9 ME375 Handouts Performance Specifications
Transient Performance Specifications and CLTF Characteristic Poles
Poles
Recall that the positions of the system characteristic poles directly affect the system
directly
output. For example, assume that the closedloop transfer function of a feedback
closedKω n2
control system is:
G (s) =
s 2 + 2ζω ns + ω n2 CL The characteristic poles are:
s1 , 2 = − ζ ω n ± jω 1 − ζ 2 = −ζω n n ± jω d = − ± j⋅ Settling Time (2%): → Puts constraint on the real part of the dominating closedloop poles.
(2%):
closedpoles.
4
ζω tS ( 2 % ) = = 4 n %OS: → Puts constraint on the imaginary part of the dominating closedloop poles.
%OS:
closedpoles. %OS = 100e − πζ
1−ζ 2 = 100e − π⋅ ζω n ω n 1−ζ 2 = 100e − π⋅ School of Mechanical Engineering
Purdue University ME375 Feedback Control  19 Performance Specification → CL Pole Positions
Transient Performance Specifications and CLTF Pole Positions
Transient performance specifications can be interpreted as constraints on the
constraints
positions of the poles of the closedloop transfer function. Let a pair of closedclosedclosedloop poles be represented as: p 1 , 2 = − σ ± j ω
Img.
Transient Performance Specifications:
jω
−σ + jω
– Settling Time (2 %) ≤ TS
tS ( 2 % ) = 4 σ ≤ TS – %OS ≤ X %
%OS = 100e −π⋅ σ
ω ⇒ σ ≥ 4
TS
Real σ = π⋅
100
100
≤ X% ⇒ e ω ≥
πσ
ω
X
e −σ − jω ⇒ − jω School of Mechanical Engineering
Purdue University ME375 Feedback Control  20 10 ME375 Handouts Example
A DC motor driven positioning system can be
modeled by a second order transfer function:
GP ( s ) = Find closedloop transfer function:
closedfunction: 3
s ( s + 6) A proportional feedback control is proposed
and the proportional feedback gain is chosen
to be 16/3. Find the closedloop transfer
closedfunction, as well as the 2% settling time and
the percent overshoot of the closed loop
system when given a step input.
Draw block diagram:
diagram: School of Mechanical Engineering
Purdue University ME375 Feedback Control  21 Example
Find closedloop poles:
closedpoles: 2% settling time:
time: %OS:
%OS: School of Mechanical Engineering
Purdue University ME375 Feedback Control  22 11 ME375 Handouts Example
A DC motor driven positioning system can be
modeled by a second order transfer function:
GP ( s ) = Find closedloop transfer function:
closedfunction: 3
s ( s + 6) A proportional feedback control is proposed.
It is desired that:
– for a unit step response, the steady state
position should be within 2% of the
desired position,
– the 2% settling time should be less than 2
sec, and
– the percent overshoot should be less than
10%.
Find (1) the condition on the proportional gain
such that the steady state performance is
satisfied; (2) the allowable region in the
complex plane for the closedloop poles.
closed Write down the performance specifications:
specifications: School of Mechanical Engineering
Purdue University ME375 Feedback Control  23 Example
Steady state performance constraint:
constraint: Percent Overshoot (%OS)
%OS) Img. jω Transient performance constraint:
2% Settling Time
Real − jω
School of Mechanical Engineering
Purdue University ME375 Feedback Control  24 12 ME375 Handouts Feedback Control Design Process
A typical feedback controller design process involves the following steps:
following
(1) Model the physical system (plant) that we want to control and obtain its I/O transfer
obtain
function GP(s). (Sometimes, certain model simplification should be performed.)
performed.)
(2) Determine sensor dynamics (transfer function of the measurement system) H(s) and
actuator dynamics (if necessary).
(3) Draw the closedloop block diagram, which includes the plant, sensor, actuator and
closedand
controller GC (s) transfer functions.
(4) Obtain the closedloop transfer function GCL (s).
closed(5) Based on the performance specifications, find the conditions that the CLTF, GCL (s), has to
that
satisfy.
(6) Choose controller structure GC (s) and substitute it into the CLTF GCL (s).
(7) Select the controller parameters (e.g. the proportional feedback gain of a proportional
control law) so that the design constraints established in (5) are satisfied.
are
(8) Verify your design via computer simulation (MATLAB) and actual implementation.
implementation. School of Mechanical Engineering
Purdue University ME375 Feedback Control  25 In Class Exercise
You are the young engineer that is in charge of
designing the control system for the next
generation inkjet printer (refer the example
discussed in lecture notes 1020 to 1023).
1010During the latest design review, the following
plant parameters are obtained:
LA = 10 mH
RA = 10 Ω
KT = 0.06 Nm/A
JE = 6.5 × 106 Kg m2
BE = 1.4 × 105 Nm/(rad/sec)
The drive roller angular position is sensed by a
rotational potentiometer with a static sensitivity
of KS = 0.03 V/deg. The design (performance)
specifications for the paper positioning system
are: – The steady state position for a step
input should be within 5% of the
desired position.
– The 2% settling time should be less
than 200 msec, and
– the percent overshoot should be less
than 5%.
You are to design a controller that satisfies
the above specifications: School of Mechanical Engineering
Purdue University ME375 Feedback Control  26 13 ME375 Handouts In Class Exercise
(1) Model the physical system (plant) that we want to control and obtain its I/O transfer
obtain
function GP(s). (Sometimes, certain model simplification should be performed.)
performed.)
DC Motor
+ eLa − + eRa −
+
ei(t)
_ R iA A LA +
Eemf
_ τm θ, ω N2 θL, ωL
JL JA
BL N1
B From previous example, the DC motor driven paper positioning system can be modeled by
system
+
− 2 1
LA s + RA KT 1
JE s + BE ⎛1⎞
JE = JA + ⎜ ⎟ ⋅ JL
⎝N⎠
2 ⎛1⎞
BE = B + ⎜ ⎟ ⋅ BL
⎝N⎠ Kb School of Mechanical Engineering
Purdue University ME375 Feedback Control  27 In Class Exercise
The plant transfer function GP(s) can be derived to be:
be:
GP ( s ) = θ (s)
KT
=
Ei ( s ) s ( LA J E s 2 + ( BE LA + RA J E ) s + ( RA BE + K b KT )) As discussed in the previous example, we can further simplify the plant model by neglecting
the
the electrical subsystem dynamics (i.e., by letting LA = 0 ):
GP ( s ) =
= θ (s)
KT
=
=
Ei ( s ) s ( RA J E s + ( RA BE + K b KT ))
KM
s (τ M s + 1) Substituting in the numerical values, we have our plant transfer function:
GP ( s ) = KM
=
s (τ M s + 1)
School of Mechanical Engineering
Purdue University ME375 Feedback Control  28 14 ME375 Handouts In Class Exercise
(2) Determine sensor dynamics (transfer function of the measurement system) H(s) and
actuator dynamics (if necessary). (3) Draw the closedloop block diagram, which includes the plant, sensor, actuator and
closedand
controller GC (s) transfer functions.
Input
Ei (s) School of Mechanical Engineering
Purdue University GP(s) Output θ (s) ME375 Feedback Control  29 In Class Exercise
(4) Obtain the closedloop transfer function GCL (s).
closed School of Mechanical Engineering
Purdue University ME375 Feedback Control  30 15 ME375 Handouts In Class Exercise
(5) Based on the performance specifications, find the conditions that GCL (s) has to satisfy.
that
Steady State specification:
specification: Imag. jω Transient Specifications:
Settling Time Constraint:
Constraint: Real Overshoot Constraint:
Constraint:
− jω School of Mechanical Engineering
Purdue University ME375 Feedback Control  31 In Class Exercise
(6) Choose controller structure GC (s) and substitute it into the CLTF GCL (s).
Let’s try a simple proportional control, where the control input to the plant is proportional to
Let’
the current position error:
error: ei (t ) = K P ⋅ eθV (t ) = K P ⋅ (Vθ d (t ) − Vθ (t )) In sdomain (Laplace domain), this control law can be written as:
sdomain)
as: Substitute the controller transfer function into GCL (s): School of Mechanical Engineering
Purdue University ME375 Feedback Control  32 16 ME375 Handouts In Class Exercise
(7) Select the controller parameters (e.g., the proportional feedback gain of a proportional
feedback
control law) so that the design constraints established in (5) are satisfied.
are
Steady State Constraint:
Constraint: Transient Constraints:
Constraints:
To satisfy transient performance specifications, we need to choose KP such that the closedclosedchoose
loop poles are within the allowable region on the complex plane. To do this, we first need to
find an expression for the closedloop poles:
closedpoles: School of Mechanical Engineering
Purdue University ME375 Feedback Control  33 In Class Exercise Img. Axis For every KP , there will be two closedloop poles (closedloop characteristic roots). It’s
closedclosedroots) It’
obvious that the two closedloop poles change with the selection of different KP . For
closedexample:
example:
→ p1,2 =
KP =
30
→ p1,2 =
KP =
20
→ p1,2 =
KP =
KP =
→ p1,2 =
10
→ p1,2 =
KP =
0
→ p1,2 =
KP =
By inspecting the rootlocus, we can find
root locus,
10
that if
20 then the closedloop poles will be in the
closedallowable region and the performance
specifications will be satisfied. 30
60 50 School of Mechanical Engineering
Purdue University 40 30
20
Real Axis 10 0 ME375 Feedback Control  34 17 ME375 Handouts In Class Exercise
(8) Verify your design via computer simulation (MATLAB) and actual implementation.
implementation.
>>
>>
>>
>>
>> num = 16*Ks*Kp;
den = [tauM 1 16*Ks*Kp];
T = (0:0.0002:0.25)’;
(0:0.0002:0.25)’
y = step(num,den,T);
plot(T,y); Unit Step Response 1 KP = 100 KP = 40 0.8 KP = 29.93
0.6 KP = 15
0.4
0.2
0
0 0.05 0.1 0.15 0.2 0.25 Time (sec)
School of Mechanical Engineering
Purdue University ME375 Feedback Control  35 In Class Exercise
(9) Check the Bode Plots of the open loop and closed loop systems:
10
0
KP = 100 Phase (deg); Magnitude (dB) 20 KP = 40
40 KP = 29.93 60 KP = 15 80 Open Loop 0
KP = 100
45 KP = 40 90 KP = 29.93
KP = 15 135 Open Loop 180
10 1 10 0 1 10
Frequency (rad/sec) School of Mechanical Engineering
Purdue University 10 2 10 3 ME375 Feedback Control  36 18 ...
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This note was uploaded on 08/28/2010 for the course ME 375 taught by Professor Meckle during the Spring '10 term at Purdue.
 Spring '10
 Meckle

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