LSU - EE - 4580

EEE 480HW # 6 SOLUTIONSProblem 5.21The root-locus asymptotes for pure-gain compensation are at 2:5. Since the required closed-loop polelocations are 1 j , we need to introduce a lag compensator. Observe that asymptotes at 1 are obtained forpoles at 0

LSU - EE - 4580

LSU - EE - 4580

LSU - EE - 4580

Solution to Homework 6 of EE4002 Spring 2004

LSU - EE - 4580

Problem 6.25Using the MATLAB command[GM PM]=margin(sys) we findGM = 1.57 and PM = 100.55From the Bode plot it can be observedthat the gain margin is so small that aslight increase in gain may result inunstable closed loop system. The phase onthe o

LSU - EE - 4580

EEE 480HW # 8 SOLUTIONSNOTE: Ballpark computations relating bandwidth (wBW ) and crossover frequency (wGC ).In classical compensator designs, it is often useful to convert closed-loop bandwidth specications to crossover specpications. The closed-loop

LSU - EE - 4580

Problem 6Problem 6.25Using theMATLABcommand[GM PM]=margin(sys)we findGM = 1.57 andPM = 100.55From the Bodeplot it can beobserved thatthe gainmargin is sosmall that aslight increasein gain mayresult inunstableclosed loopsystem. Thepha

LSU - EE - 4580

LSU - EE - 4580

LSU - EE - 4580

Course OutlineWeek28 JulyIntroduction204 Aug.Frequency Domain Modelling311 Aug.Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics601 Sept.Root Locus708 Sept.Root Locus 2815 Sept.Bode Plots9Dr. Stefan B.

LSU - EE - 4580

GAIN AND PHASE MARGIN-The gain margin is defined as the change in open loop gain required to make the systemunstable. Systems with greater gain margins can withstand greater changes in systemparameters before becoming unstable in closed loop.-The phas

LSU - EE - 4580

Course OutlineWeekDateContent128 JulyIntroductionAssignment notes2Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics601 Sept.Root Locus708 Sept.Root Locus 215 Sept.Bode Plots9Dr. Stefan B. WilliamsFreque

LSU - EE - 4580

Course ObjectivesMech 3800 : System ControlIntroduction To introduce the methods used for theanalysis and design of linear time invariantfeedback control systemsDr. Stefan B. WilliamsDr. Stefan B. WilliamsAdministrative DetailsAdministrative Deta

LSU - EE - 4580

Course OutlineWeekContent128 JulyIntroduction2Mech 3800 : System ControlRoot LocusDateAssignment notes04 Aug.Frequency Domain Modelling311 Aug.Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics01 Sept.Root

LSU - EE - 4580

Angles of departure and arrival Looking at the open loop poles and zerosbelow, the root locus starts at the poles andends at zeros but at what angles?j Taking a point () close to a complex pole.Assuming this point lies on the root locus, theangles

LSU - EE - 4580

Course OutlineWeek28 JulyIntroduction204 Aug.Frequency Domain Modelling311 Aug.Block Diagrams418 Aug.System Response525 Aug.Feedback System Characteristics601 Sept.Root Locus708 Sept.Root Locus 2815 Sept.Bode Plots9Dr. Stefan B.

LSU - EE - 4580

Root-Locus DesignThe root-locus can be used to determine the value of the loop gain K , which results in asatisfactory closed-loop behavior. This is called the proportional compensator orproportional controller and provides gradual response to deviatio

LSU - EE - 4580

S Plane ControlXia HongMay 14, 2004Books: Dorf and Bishop (1998) Modern ControlSystems Addison and Wesley Franklin, Powell and Emami-Naeini (1994)Feedback Control of Dynamic Systems Addison and Wesley. Kuo (1995) Automatic Control Systems Wiley.

LSU - EE - 4580

Routh-Hurwitz Criterion: Special CasesZero only in the rst columnWhen forming the Routh table, replace thezero with a small number and evaluate thest column for positive or negative values of .Problem: Determine the stability of the closedloop transf

LSU - EE - 4580

Stability Design via Routh-HurwitzGiven the system below nd the range of thegain K that will lead to stability, unstabilityand marginal stability.R(s)+E(s)KC(s)s(s+7)(s+11)Closing the loop:KT (s) = 3s + 18s2 + 77s + K(1)1s3177s218Ks1

LSU - EE - 4580

Root locus techniquesj5K=504321K=0K=255R(s)+E(s)KK=00C(s)s(s+10)K=50The root locus show the changes in the transient response as the gain ,K , is varied.K < 25 real poles, overdampedK = 25 multiple poles, critically dampedK > 25 Com

LSU - EE - 4580

Sketching the root locusWe can see that the root locus can be plotted by locating the points in the s-plane forwhich the angles add up to an odd multiple of180o. We can now construct rules for sketching a root locus, plotting any point of interestif r

LSU - EE - 4580

Rening the root locus sketchWe have covered the rules for a rapid sketchof the root locus. For full use of the sketch weneed to calculate the points of interest of thesketch, point on the real axis where the rootlocus enters or leaves the complex pla

LSU - EE - 4580

Angles of departure and arrival Looking at the open loop poles and zerosbelow, the root locus starts at the poles andends at zeros but at what angles?j Taking a point () close to a complex pole.Assuming this point lies on the root locus, theangles

LSU - EE - 4580

Designing via root locus We have seen how a root locus graphicallydisplays stability information, in addition rootlocus can also display the transient response ofa system.jclosed loop poleopen loop poleA and B represent the actual and desired tran

LSU - EE - 4580

Improving transient response Again, we will study two techniques, placinga pure dierentiator in the forward path, andplacing a zero and more distant pole.Ideal derivative compensation (PD)The transient response of a system can be selected by choosing

LSU - EE - 4580

Types of electrical noises(i) Johnson NoiseJohnson noise occurs in any device that dissipates power, such as resistor,hence it is very common. Principle of equipartition of energy states: For asystem in thermal equilibrium with its surroundings : The

LSU - EE - 4580

SignalsSignal a transmitted effect conveying a messageA essential characteristic of a signal is that of change, since it must becapable of carrying information.The change must be partly unpredictable.It is also necessary to discuss the sinusoidal or

LSU - EE - 4580

EE4580 Test Fall 2004Suppose that the plant model is given byG(s) =10(s + 1)(s + 10).s(s2 + s + 25)1. Sketch straightline approximation for magnitude Bode plot.2. Find errors at corner frequencies, and make corrections with dashed line.3. Repeat 1

LSU - EE - 4580

Solution to Midterm IProblem 2: For departure angle, we have 2 = dep ,1 = 135o ,3 = 90o ,1 = 225o ,and 2 = 90o + tan1 (3/7) = 113.2o . Hencedep = 1 + 2 1 3 180o = 113.2o 180o = 66.8o .For arrival angle, we have 1 = tan1 (5), 2 = 45o ,3 = tan1 (7/3

LSU - EE - 4580

LSU - EE - 4580

LSU - EE - 4580

LSU - EE - 4580

LSU - EE - 4580

Jason McAllisterNovember 29, 2004UNCOMPENSATED SYSTEMStep Response1.41.2System: Closed Loop: r to yI/O: r to ySettling Time (sec): 9.961System: Closed Loop: r to yI/O: r to yPeak amplitude: 1Overshoot (%): 0.314At time (sec): 10.9Amplitude

LSU - EE - 4580

The take home part of the second midterm is:Deasign dynamic compensator for (s + 5)G(s) = - s(s^2 + s + 10)such that the closed loop system has overshot no more than 15%, settling time no more than5s, and ess \leq 0.1 for the unit ramp inpiut.The

LSU - EE - 4580

Analysis of the Continuous Linear SystemThrough the design in project 1, the plant with the servo system was calculated as12.88. Figure 1 shows the frequency response of this system.s + 6.5s 3 + 20.7 s 24Figure 1Due to the required response time of

LSU - EE - 4580

A1Appendix ADesign of Continuous-Time SystemTransfer Function and Kp/Kv Derivation:KpAmAmKpAmKpTm= Tm * s + AmKv + 1 ==2AmKpAmKv + 1 AmKpdTm * s + ( AmKv + 1) * s + AmKp1+s 2 + s()+Tm * s + AmKv + 1TmTmn =2AmKpTmAND 2 n = 4.8 ~ 8

LSU - EE - 4580

Page: 1Project 1EE 4002Alicia CorbettMichael DantinScott HammattPatrick QuebedeauxIntroductionThe first logical step in designing a feedback controller for the ball and beamsystem is to analyze the DC motor, the part of the system that causes the

LSU - EE - 4580

Page: 1Project 1EE 4002Alicia CorbettMichael DantinScott HammattPatrick QuebedeauxIntroductionThe first logical step in designing a feedback controller for the ball and beamsystem is to analyze the DC motor, the part of the system that causes the

LSU - EE - 4580

J ohnnyLaraEE4002DesignProject2:DigitalControlofthe BallBeamSystemM ay3,2004IntroductionThe purpose of this project is to gain experience and knowledge of theperformance of an uncompensated system and how it reflects our design if the system isun

LSU - EE - 4580

Page: 1Karrie HugghinsPatrick QuebedeauxMatthew BullEE 4580Report 1 of Design Project 1September 12, 2003Executive SummaryThe first logical step in designing a feedback controller for the ball and beam systemis to analyze the DC motor, the part o

LSU - EE - 4580

Page: 1Project 1EE 4002Alicia CorbettMichael DantinScott HammattPatrick QuebedeauxIntroductionThe first logical step in designing a feedback controller for the ball and beamsystem is to analyze the DC motor, the part of the system that causes the

LSU - EE - 4580

J ohnnyLaraEE4002DesignProject2:DigitalControlofthe BallBeamSystemM ay3,2004IntroductionThe purpose of this project is to gain experience and knowledge of theperformance of an uncompensated system and how it reflects our design if the system isun

LSU - EE - 4580

#Jason McAllister#J#a#s#o#n##M#c#A#l#l#i#s#t#e#r#0###

LSU - EE - 4580

Page: 1EE-4580 CONTROL DESIGNPROJECT 2- BALL AND BEAM SYSTEMJASON MCALLISTER438-49-472611-17-04Page: 2Project SummaryOur goal is to design two compensators: one for the actuator which is a D.C. Motor(inner-loop), and the other is for the ball and

LSU - EE - 4580

LSU - EE - 4580

LSU - EE - 4580

Design Project 1 for EE4580As in the distributed notes, the linearized model for inverted pendulum consists of twotransfer functions. We aim to design Gyu (s) and Gu (s) such that the dominant dynamics forthe closed-loop system have damping ratio 0.83,

LSU - EE - 4580

EE-4580 CONTROL DESIGNPROJECT 1- INVERTED PENDULUMJASON MCALLISTER438-49-472610-18-04Introduction:In this project we have analyzed an inverted pendulum. This is a system which has bothan angle and distance considerations. The control of such a syst

LSU - EE - 4580

LSU - EE - 4580

Design Project 1 of EE4580: Root Locus DesignThe plant model is a linearized pitch axis missile that has a transfer function = 2 + 8 7+ 25 = 2The design speci cations are given as follows::GsTransient response:P:O:Steady-state response:s 12,ess

LSU - EE - 4580

LSU - EE - 4002

,. meclrJo.V1iCQ\<';~E>ievYJbQ) lineaof' motioY)fo.~~f~ ~Hr~l~)Mmotion.!9UJc~'0.fdrtdV"'.tM'Mot). o.Md~clt.tfr 0. Dot.(=?u= ~"-~o.s'M+DFI.tdc.TIon It>)"rfI~'16 b ;nertiai V\plAt pv-Je,('Sef'fdc~d" LAt M~(-I-t)~1\V) " tdr \J.b) n

LSU - EE - 4002

load current Ia; Ia=Ia';t=Ia(:,1);Ia=Ia(:,2);plot(t,Ia,'k'),xlabel('time [s]'),ylabel('current Ia [A]'),gridload velocity; v=v';t=v(:,1);v=v(:,2);plot(t,v,'k'),xlabel('time [s]'),ylabel('velocity v [m/s]'),gridload torque T; T=T';t=T(:,1);T=T(:,2);

LSU - EE - 4002

EE 4002: Project #1Elevator DC MotorByCarranza J. Guidry III10/18/041.)Calculate the power of the motor and select (using Web Site of producer orvendor on internet, e.g. e-motorsonline.com) the appropriate separately excited DCmotor for driving th

LSU - EE - 4002

load speed speed;speed=speed';t=speed(:,1);speed=speed(:,2);plot(t,speed,'k'),xlabel('time [s]'),ylabel('car speed [km/h]'),grid

LSU - EE - 4002

I.CalculationsMotor Parameters:Ra = .066 Kt = 1.33 Nm/ALa = 6.5 mHKe = 1.33 V/(rad/s)Jm = 25 kg*mCar Parameters:M = 997 kgFad = .046Cw * A * uU = 50 km/hCw = 0.5Diameter of wheel = .6m1.) PWM converter parameter:a. Fs,I = 1kHzb. Kpwm = 12

LSU - EE - 4002

I.CalculationsMotor Parameters:Ra = .066 Kt = 1.33 Nm/ALa = 6.5 mHKe = 1.33 V/(rad/s)Jm = 25 kg*mCar Parameters:M = 997 kgFad = .046Cw * A * uU = 50 km/hCw = 0.5Diameter of wheel = .6m1.) PWM converter parameter:a. Fs,I = 1kHzb. Kpwm = 12

LSU - EE - 4002

r...-~.".,EE 400204.04.2003/ 0 Name: .JA1E~cfl;1/. tJYi?Y~".Q(f.MID SEMESTER TESTNO.2Answer 3 questions and solve one out of two numerical problemsQuestion 1 (2 pofuts):('j)a) Draw the simplified control circuit diagr~thtwo blocks: contr

LSU - EE - 4002

,-.,~",-sJ.~76.9EE 400204.12.2002Name: .MID SEMESTER TESTNO.2Answer 3 of 4 questions and solve the numerical problemQuestion 1 (2 points): For a switch-mode two-quadrant converter:a) Draw the circuit diagram of two-quadrant converter (with tr