FREQUENCYRESPONSE DESIGN


 Susan Lane
 4 years ago
 Views:
Transcription
1 ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins of a closedloop system based on openloop response. Now, we look at frequencyresponse based design methods which primarily aim at improving stability margins. Also, given the relationship between PM and performance, we have some idea of transient response as well. Start thinking of Bodemagnitude and Bodephase plots as LEGO to make the frequency response we want. PD compensation Compensator D(s) = K ( + TD s). We have seen from rootlocus that this has a stabilizing effect. Magnitude and phase effect: K 9 K. T D T D. T D T D Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli T D T D
2 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 2 PD controller increases phase for frequencies over ω./td.so, locate /T D < crossover so that phase margin at crossover is better. Problem: Magnitude response continues to increase as frequency increases. This amplifies highfrequency sensor noise. Lead compensation [ ] Ts + Compensator D(s) = K αts +, α <. Approximate PD control for frequencies up to ω = αt. Magnitude and phase effect: K α φ max K T ω max αt. T The phase contributed at frequency ω is T ω max αt αt φ = tan (T ω) tan (αt ω). To find where the maximum phase contribution is, we take the derivative of φ with respect to ω and set it to zero dφ dω = dtan (T ω) dω = T T 2 ω 2 + dtan (αt ω) dω αt α 2 T 2 ω 2 + = Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
3 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 3 α 2 T 3 ω 2 + T = αt 3 ω 2 + αt T 3 ω 2 α(α ) = T (α ) ω max = T α. For this frequency, the amount of phase added is φ max = tan (T ω max ) tan (αt ω max ) ( ) ( ) α = tan α tan α }{{}}{{} θ θ 2 To get a simpler expression, consider the triangles to the right. + α φ max θ θ 2 We can solve for φmax using the law of cosines. This gives: α α + α 2 α α ( α) 2 = ( + α) + (α + α 2 ) 2 + α α + α 2 cos (φ max ) 2α + α 2 = + 2α + α α α + α 2 cos (φ max ) 4α cos (φ max ) = 2 + α α + α = 2 α 2 ( + α). Now, we remember that cos 2 (θ) + sin 2 (θ) =, so sin 2 (φ max ) = 4α ( + α) 2 = ( + α)2 4α ( + α) 2 So, we conclude that = ( α)2 ( + α) 2. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
4 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 4 ( ) α φ max = sin + α We can invert this to find α to add a certain phase φmax α = sin(φ max) + sin(φ max ). We can show that ωmax occurs midway between T and on a log αt scale ( ) ( ) ( ) log(ω max ) = log T = log + log αt T αt. = 2 [ ( ) ( )] log + log. T αt 9 8 How much can we improve phase with a lead network? φmax If we need more phase improvement, we can use a doublelead compensator: D(s) = K [ ] Ts + 2. αts + Need to compromise between good phase margin and good sensor noise rejection at high frequency. /α Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
5 ECE45/ECE55, FREQUENCYRESPONSE DESIGN : Design method # for lead controllers Design specification include: Desired steadystate error. Desired phase margin. Compute steadystate error of compensated system. Set equal to desired steadystate error by computing K. Evaluate the PM of the uncompensated system using the value of K from above. ω max initially set to crossover frequency. Add a small amount of PM (5 to 2 )totheneededphaselead. (Lead network moves crossover point, so need to design conservatively): Gives φ max. Determine α = sin(φ max ) + sin(φ max ). Determine T = ω max α. Iterate if necessary (choose a slightly different ωmax or φ max ). EXAMPLE: Plant G(s) = s(s + ) Specification : Want steadystate error for ramp input <.. Specification 2: Want overshoot Mp < 25%. Start by computing steadystate error [ ] e ss = lim s R(s) s + D(s)G(s) [ ] = lim = s s + D(s) D(). (s+) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
6 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 6 [ ] Ts + D(s) = K αts + So, D() = K so D() = K. =.... K =. In Bode plot below, red line is original plant. Green line is plant with added gain of K =, butnoleadcompensator. We see that the new crossover is at 3 rad s,soω max = 3rads. Overshoot specification Mp < 25% gives PM > 45. Evaluating (red line) Bode diagram of KG(s) at crossover, we see we have a phase margin of 8.Needabout27 more to meet spec. Note, addition of pole and zero from lead network will shift crossover frequency somewhat. To be safe, design for added phase of 37 instead of 27. α = sin(37 ) + sin(37 ) =.25. T =.25(3) =.667. So, D(s) = [ 2s/3 + 2s/2 + ]. In plots below, the blue line is the Bode plot of the compensated system KD(s)G(s). Also, uncompensated root locus is plotted topright; compensated root locus is plotted bottomright. We find that we just missed specifications: PM = 44.Iterateif desired to meet specs. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
7 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 7 Magnitude Phase ω (rad/sec) ω (rad/sec) Imag Axis Imag Axis Real Axis Real Axis Amplitude Step Response Amplitude Ramp Response Time (sec.) Summary of design example Time (sec.). Determine dcgain so that steadystate errors meet spec. 2. Select α and T to achieve an acceptable PM at crossover. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
8 ECE45/ECE55, FREQUENCYRESPONSE DESIGN : Design method #2 for lead controllers Design specifications Desired closedloop bandwidth requirements (rather than e ss ). Desired PM requirements. Choose openloop crossover frequency ωc to be half the desired closedloop bandwidth. Evaluate KG = G( jω c ) and φ G = G( jω c ). Compute phase lead required φmax = PM φ G 8. Compute α = sin(φ max ) + sin(φ max ). Compute T = αωc. Compensation is [ ] Ts + D(s) = K αts + α D( jω c ) =K + α α + = K α + α α + = K. α Openloop gain at ωc is designed to be : D( jω c ) G( jω c ) = K α G( jω c ) = Our design is now complete. K = α K G. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
9 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 9 EXAMPLE: Desired gain crossover frequency of radians per second. Desired PM of 6. G(s) = s(s + ). KG = j( j + ) =., φ G = φmax = = sin(φ max ) α = + sin(φ max ) =.37. T = α =.35. K D = α K G = Matlab code to automate this procedure: % [K,T,alpha]=bod_lead(np,dp,wc,PM) % Computes the lead compensation of the plant np/dp to have % phase margin PM (in degrees) at crossover frequency wc (in radians/sec). % e.g., [K,T,alpha]=bod_lead([],[ ],5,6); function [K,T,alpha]=bod_lead(np,dp,wc,PM) % first compute the plant response at wc. [magc,phc]=bode(np,dp,wc); % now, compute the needed phase lead at wc and convert to radians % for use with "sine" phir=(8+pmphc)*pi/8; if abs(phir)pi/2, fprintf('a simple phase lead/lag cannot change the phase by more\n'); fprintf('than +/ 9 degrees.\n'); error('aborting.'); end; % Compute alpha, T, and K for compensator alpha=(sin(phir))/(+sin(phir)); T=/(wc*sqrt(alpha)); K=sqrt(alpha)/magc; Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
10 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 % compute the new openloop system by convolving the plant polynomials % with the compensator polynomials. nol=conv(np,k*[t ]); dol=conv(dp,[alpha*t ]); % check the solution by plotting the Bode plot for the new openloop % polynomials. Include the frequency w= % to get the full resonance response to show the gain margin. Also, plot % the uncompensated Bode response. w=logspace(2,)*wc; w(34)=wc; clf; [mag,ph]=bode(np,dp,w); [mag2,ph2]=bode(nol,dol,w); subplot(2); semilogx(w/wc,2*log(mag),''); hold on; semilogx(w/wc,2*log(mag2)); grid; plot(w/wc,*w+,'g'); ylabel('magnitude (db)'); title('phaselead design (uncompensated=dashed; compensated=solid)'); subplot(22); semilogx(w/wc,ph,''); hold on; semilogx(w/wc,ph2); grid; plot(w/wc,*w8+pm,'g'); ylabel('phase'); xlabel('frequency/wc (i.e., ""=wc)'); Matlab results: Magnitude (db) Phase Phase lead design (uncompensated=dashed; compensated=solid) Frequency/wc (i.e., ""=wc) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
11 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 9.4: PI and lag compensation PI compensation In many problems it is important to keep bandwidth low, and also reduce steadystate error. PI compensation used here. K D(s) = K [ + T I s ]..ωt I ωt I ωt I K 9 ωt I ωt I Infinite gain at zero frequency Reduces steadystate error to step, ramp, etc. But also has integrator antiwindup problems. Adds phase below breakpoint. We want to keep breakpoint frequency very low to keep from destabilizing system. Lag compensation T I ω c. Approximates PI, but without integrator overflow. [ ] Ts + D(s) = α α>. αts + Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
12 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 2 Primary objective of lag is to add 2 log α db gain to low frequencies without changing PM. α αt T αt T 9 Steadystate response improves with little effect on transient response. Typical process Assumption is that we need to modify (increase) the dcgain of the loop transfer function. If we apply only a gain, then ω c typically increases and the phase margin decreases. NOT GOOD. Instead, use lag compensation to lower highfrequency gain. Assume plant has gain K (adjustable), or that we insert a gain K into the system. Adjust the openloop gain K to meet phase margin requirements (plus about 5 slop factor) at crossover without additional compensation. Draw Bode diagram of system using the gain K from above. Evaluate lowfrequency gain. Determine α to meet lowfrequency gain requirements. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
13 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 3 Choose one corner frequency ω = (the zero) to be about one T decade below crossover frequency. (This way, the phase added by the lag compensator will minimally affect PM. Thephaseaddedat crossover will be about 5,henceourpreviousslopfactor). The other corner frequency is at ω = αt. Iterate design to meet spec. EXAMPLE: K G(s) = (.5 s + ) (s + ) ( 2 s + ). Bode plot for plant is red line, below. Uncompensated root locus is topright plot. We want to design compensator for PM 25, K p = 9.. Set K = 4.5 for PM = 3.Thisgivescrossoveratω c.2 rads/sec. (Green line on Bode plot.) 2. Lowfrequency gain now about 3 db or (3/2) = Should be raised by a factor of 2 to get K p = 9. So,α = Choose corner frequency at ω =.2 rads/sec. =.2, ort = 5. T 5. We then know other corner frequency ω = αt =..4 Step Response From: U() Altogether, we now have the compensator [ ] 5s + D(s) = 2. s + Amplitude To: Y() Time (sec.) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
14 ECE45/ECE55, FREQUENCYRESPONSE DESIGN Magnitude Phase ω (rad/sec) ω (rad/sec) Imag Axis Imag Axis Real Axis Real Axis Please understand that the recipes in these notes are not meant to be exhaustive! And, we often require both lead and lag to meet specs. Many other approaches to control design using frequencyresponse methods exist. In fact, once you are comfortable with Bode plots, you can add poles and zeros like LEGOs to build whatever frequency response you might like (within the limitations of Bode s gainphase theorem). Our next section gives some guidance on this. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
15 ECE45/ECE55, FREQUENCYRESPONSE DESIGN : Design based on sensitivity Goal: Develop conditions on Bode plot of loop transfer function D(s)G(s) that will ensure good performance with respect to sensitivity, steadystate errors and sensor noise. Steadystate performance = lower bound on system lowfreq. gain. Sensor noise = upper bound on highfrequency gain. Consider also the risk of an unmodeled system resonance: Magnitude may go over. Can cause instability. Must ensure that highfrequency gain is low so magnitude does not go over. Magnitude Frequency Together, these help us design a desired loop frequency response: Magnitude must be high at low frequencies, and Low at high frequencies. Magnitude of D(s)G(s) Steadystate Error Boundary Sensor noise and plant sensitivity boundary Sensitivity functions The presence of noises also enters our design considerations. w(t) r(t) D(s) G(s) y(t) v(t) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
16 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 6 Y (s) = W (s) + G(s)D(s)[R(s) V (s) Y (s)] [ + G(s)D(s)] Y (s) = W (s) + G(s)D(s)[R(s) V (s)] or, Y (s) = G(s)D(s) W (s) + [R(s) V (s)]. + G(s)D(s) + G(s)D(s) Tracking error = R(s) Y (s) G(s)D(s) E(s) = R(s) W (s) [R(s) V (s)] + G(s)D(s) + G(s)D(s) = + G(s)D(s) [R(s) W (s)] + G(s)D(s)V (s) + G(s)D(s) Define the sensitivity function S(s) to be S(s) = + G(s)D(s) which is the transfer function from r(t) to e(t) and from w(t) to e(t). The complementary sensitivity function T (s) = S(s) S(s) = G(s)D(s) + G(s)D(s) = T (s) which is the transfer function from r(t) to y(t). If V =, then and Y (s) = S(s)W (s) + T (s)r(s) E(s) = S(s)[R(s) W (s)]. The sensitivity function here is related to the one we saw several weeks ago SG T = T G G T = = + D(s)G(s) D(s)G(s) [ + D(s)G(s)] 2 + D(s)G(s) = S(s). G(s)[ + D(s)G(s)] G(s) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
17 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 7 So S(s) is the sensitivity of the transfer function to plant perturbations. Recall that T (s) + S(s) =, regardlessofd(s) and G(s). We would like T (s) =. ThenS(s) = ; Disturbanceiscancelled, design is insensitive to plant perturbation, steadystate error. BUT, forphysicalplants,g(s) for high frequencies (which forces S(s) ). Furthermore, the transfer function between V (s) and E(s) is T (s). To reduce high frequency noise effects, T (s) as frequency increases, and S(s). Typical sensitivity and complementary sensitivity (closedloop transfer) functions are: S( jω) Performance Spec. T ( jω) Stability Spec. Another view of sensitivity: So, + D( jω)g( jω) is the distance between the Nyquist + D( jω)g( jω) curve to the point, and S( jω) =. D( jω)g( jω) + D( jω)g( jω) I(s) R(s) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
18 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 8 Alargevalueof S( jω) indicates a nearly unstable Nyquist plot. The maximum value of S is a more accurate measure of stability than PM or GM. So,wewantmax S( jω) small. How small? Note: E( jω) = S( jω)r( jω) E( jω) = S( jω)r( jω) S( jω) R( jω) put a frequencybased error bound Let W (ω) = R( jω)/e b.then, E( jω) S( jω) R( jω) e b S( jω) W (ω). EXAMPLE: Aunityfeedbacksystemistohaveanerrorlessthan.5 for all unityamplitude sinusoids below 5 rads/sec. Draw W ( jω) for this design. Spectrum of R( jω) is unity for ω 5. Since eb =.5, W (ω) =.5 = 2 5 for this range Magnitude 2 5 Frequency (rads/sec.) We can translate this requirement into a loopgain requirement. When errors are small, loop gain is high, so S( jω) D( jω)g( jω) and D( jω)g( jω) W (ω) or, D( jω)g( jω) W (ω) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
19 ECE45/ECE55, FREQUENCYRESPONSE DESIGN : Robustness Typically there is some uncertainty in the plant transfer function. We want our design to be robustly stable, and to robustly give good performance (often called H design). Uncertainty often expressed as multiplicative G( jω) = G n ( jω)[ + W 2 ( jω) ( jω)] W2 (ω) is a function of frequency expressing uncertainty, or size of possible error in transfer function as a function of frequency. W2 (ω) is almost always small at low frequencies. W2 (ω) increases at high frequencies as unmodeled structural flexibility is common. Typical W2 Magnitude Frequency (rads/sec.) ( jω) expresses uncertainty in phase. The only restriction is Design ( jω). Assume design for nominal plant Gn (s) is stable. Thus, + D( jω)g n ( jω) = ω. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
20 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 2 For robust stability, + D( jω)g n ( jω) + D( jω)g n ( jω) }{{} = by assumption + D( jω)g( jω) = ω + D( jω)g n ( jω)[ + W 2 (ω) ( jω)] = + D( jω)g n( jω) + D( jω)g n ( jω) W 2(ω) ( jω) = recall, T ( jω) = D( jω)g n( jω) + D( jω)g n ( jω), [ + T ( jω)w 2 (ω) ( jω)] = so, T ( jω)w 2 (ω) ( jω) < or, T ( jω) W 2 (ω) <. For high freqencies D( jω)gn ( jω) is typically small, so T ( jω) D( jω)g n ( jω). Thus D( jω)g n ( jω) W 2 (ω) < D( jω)g n ( jω) < W 2 (ω). EXAMPLE: The uncertainty in a plant model is described by a function W 2 (ω) which is zero until ω = 3 rads/sec, and increases linearly from there to a value of at ω =, rads/sec. It remains constant at for higher frequencies. Plot constraint on D( jω)g n ( jω). Where W2 (ω) =, thereisnoconstraintonthemagnitudeoftheloop gain. Above ω = 3, /W 2 (ω) is a hyperbola from to. at,. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
21 ECE45/ECE55, FREQUENCYRESPONSE DESIGN 9 2 Magnitude Frequency (rads/sec.) Combining W (ω) and W 2 (ω) requirements, 8 Log Magnitude Frequency (rads/sec.) Limitation: Crossover needs to be with slope. So,cannotmake constraints too strict, or design will be unstable. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli
ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationCDS 101/110a: Lecture 81 Frequency Domain Design
CDS 11/11a: Lecture 81 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve
More informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
More informationDIGITAL CONTROLLER DESIGN
ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steadystate accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some
More informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
More informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationChapter 9: Controller design
Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer function from disturbances to the output 9.2.2. Feedback
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationEECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8 am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control
More information= rad/sec. We can find the last parameter, T, from ωcg new
EE572 Solution to HW#22. Keep working on your project!! 1. Consider the following system: W(s) + T s =1 msec G lead (z) G zoh (z) 8 ( s+ 4)  a) Design a lead compensator, G lead (z), which meets the following
More informationFREQUENCYRESPONSE ANALYSIS
ECE450/550: Feedback Control Systems. 8 FREQUENCYRESPONSE ANALYSIS 8.: Motivation to study frequencyresponse methods Advantages and disadvantages to rootlocus design approach: ADVANTAGES: Good indicator
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
More informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
More informationThe loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)
Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)
More informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
More informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #11 Wednesday, January 28, 2004 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Relative Stability: Stability
More informationBoise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain
Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type servomechanism:
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationBASIC PROPERTIES OF FEEDBACK
ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationLoop shaping exercise
Loop shaping exercise Excerpt 1 from Controlli Automatici  Esercizi di Sintesi, L. Lanari, G. Oriolo, EUROMA  La Goliardica, 1997. It s a generic book with some typical problems in control, not a collection
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationModule 5: Design of Sampled Data Control Systems Lecture Note 8
Module 5: Design of Sampled Data Control Systems Lecture Note 8 Laglead Compensator When a single lead or lag compensator cannot guarantee the specified design criteria, a laglead compensator is used.
More informationActive Control? Contact : Website : Teaching
Active Control? Contact : bmokrani@ulb.ac.be Website : http://scmero.ulb.ac.be Teaching Active Control? Disturbances System Measurement Control Controler. Regulator.,,, Aims of an Active Control Disturbances
More informationEngraving Machine Example
Engraving Machine Example MCE44  Fall 8 Dr. Richter November 24, 28 Basic Design The Xaxis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional
More informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Boundedinput boundedoutput (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationTopic # Feedback Control Systems
Topic #19 16.31 Feedback Control Systems Stengel Chapter 6 Question: how well do the large gain and phase margins discussed for LQR map over to DOFB using LQR and LQE (called LQG)? Fall 2010 16.30/31 19
More information6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson
Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closedloop behavior what we want it to be. To review:  G c (s) G(s) H(s) you are here! plant For
More informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
More informationAn Internal Stability Example
An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several polezero cancellations between the controller and
More informationECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =
ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
More informationLast week: analysis of pinionrack w velocity feedback
Last week: analysis of pinionrack w velocity feedback Calculation of the steady state error Transfer function: V (s) V ref (s) = 0.362K s +2+0.362K Step input: V ref (s) = s Output: V (s) = s 0.362K s
More informationLecture 11. Frequency Response in Discrete Time Control Systems
EE42  Discrete Time Systems Spring 28 Lecturer: Asst. Prof. M. Mert Ankarali Lecture.. Frequency Response in Discrete Time Control Systems Let s assume u[k], y[k], and G(z) represents the input, output,
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More informationLINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad
LINEAR CONTROL SYSTEMS Ali Karimpour Associate Professor Ferdowsi University of Mashhad Controller design in the frequency domain Topics to be covered include: Lag controller design 2 Dr. Ali Karimpour
More informationControl Systems. Control Systems Design LeadLag Compensator.
Design LeadLag Compensator hibum@seoulteh.a.kr Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design. Frequeny Domain Design Frequeny response
More informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pickandplace robot to move the link of a robot between two positions. Usually
More informationECE382/ME482 Spring 2005 Homework 8 Solution December 11,
ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 1 Solution to HW8 P1.21 We are given a system with open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1) and unity negative feedback. We are
More informationChapter 6  Solved Problems
Chapter 6  Solved Problems Solved Problem 6.. Contributed by  James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the ZN tuning strategy for the nominal
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationLecture 1: Feedback Control Loop
Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
More informationAutomatic Control (TSRT15): Lecture 7
Automatic Control (TSRT15): Lecture 7 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13282226 Office: Bhouse extrance 2527 Outline 2 Feedforward
More information