ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =


 Prudence Horton
 3 years ago
 Views:
Transcription
1 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. We are asked to determine the value of K that maximizes the phase margin, then, with that gain, to determine the overshoot resulting from a step input. Solution: The closed loop transfer function is T(s) = K(s +.2) s 2 (s 2 + 7s + 1) + Ks +.2K. (2) I don t see a straightforward calculation we can do to express the phase margin in terms of the gain K. Accordingly, we will have to write a Matlab script that searches for this maximum. First, we ll have to determine what values of K we should search over. The Routh array, shown in Table 1, leads us to conclude that we should vary K from to 6.2. (Obviously we don t want K to be exactly zero. We ll start with, say, K =.1.) Table 1: Routh array for AP9.5 s 4 : 1 1.2K s 3 : 7 K s 2 1 : 7 (K 7) 1 7 ( 1.4K) s 1 7 : K 7 (1.4K + K(K 7)/7) s 1 : 7 ( 1.4K) Next, using the Matlab script below, we obtain optimal gain Km = 4.9, with gain margin Gm = at Wcg = 2.932, phase margin Pm = , at Wcp =.578, and step response overshoot of 29. The Bode plot and step response for the closed loop system with K = 3.7 are shown in Figures 1 and 2. AP9_5.m solves part of problem AP9_5 from Dorf and Bishop 1th ed. 17 April 5 sk for K = [.1:.1:6.1]; oltf = tf(k*[1.2],[1 7 1 ]); [Gm,Pm,Wcg,Wcp] = margin(oltf); pmar(round(k*1)) = Pm; end;
2 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 1: Closed Loop Bode Diagram for AP9.5 with K = 3.7 Closed Loop Bode Plot for AP9.5 with K = 4.9 Magnitude (db) Phase (deg) Frequency (rad/sec) [pmm,loc]=max(pmar); Km = loc/1 oltf = tf(km*[1.2],[1 7 1 ]); cltf = feedback(oltf,1); figure(5) bode(cltf) grid title([ Closed Loop Bode Plot for AP9.5 with K =,num2str(km)]) print deps ap9_5a [Gm,Pm,Wcg,Wcp] = margin(oltf) open loop not closed loop xfer fn! [y,t] = step(cltf); figure(6) plot(t,y) grid
3 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 2: Step Response for AP9.5 with K = 3.7 Step Response for AP9.5 with K = y(t) Time (s) xlabel( Time (s) ) ylabel( y(t) ) title([ Step Response for AP9.5 with K =,num2str(km)]) print deps ap9_5aa overshoot = max(y) AP9.7 We are given a unity negative feedback system with compensator transfer function K(s + 4) and plant transfer function 1/s 2 cascaded in the forward path. A disturbance input occurs between the compensator and the plant. We are asked to find a K that results in phase margin of 45 degrees, then find the bandwidth and peak amplitude M p (not M pω ) of the response to a unit step disturbance. Solution: The closed loop transfer function is T(s) = Y (s) K(s + 4) = R(s) s 2 + Ks + 4K (3)
4 ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 4 so the natural frequency of the closedloop system is ω n = 2 K and the damping coefficient is ζ = K/4. We ll use an iterative search similar to that used in AP9.5, except that rather than maximizing the phase margin, we want to minimize the difference between the phase margin and the design specified phase margin of 45 degrees. This time it is not so easy to determine the range of K to use. The poles of the system are K/2 ± K 2 16K/2, so, assuming that K is positive, the only way for a pole to have a positive real part is for the square root to be real and greater than K. But, when real, the square root of K 2 16K is always less than K. In other words, the gain does not affect the stability. That doesn t help us decide what range of K to check. Let s start with the range < K < 1 and, if necessary, we ll increase the upper bound. Using the Matlab script below, we obtain optimal gain Km = 2.8, with infinite gain margin and phase margin Pm = , at Wcp = Using the interactive feature of the Matlab Bode plot we determine that the system bandwidth is approximately 5.8 rad/s. The maximum value of the closed loop step disturbance response is The Bode plot and step response for the closed loop system with K = 2.8 are shown in Figures 3 and 4. AP9_7.m solves part of problem AP9_7 from Dorf and Bishop 1th ed. 17 April 5 sk iterate for closed loop gain margin for K = [.1:.1:1]; oltf = tf(k*[1 4],[1 ]); [Gm,Pm,Wcg,Wcp] = margin(oltf); pmar(round(k*1)) = Pm; end; [foo,loc]=min(abs(pmar45*ones(size(pmar)))); Km = loc/1 oltf = tf(km*[1 4],[1 ]); cltf = feedback(oltf,1); figure(7) bode(cltf) grid title([ Closed Loop Bode Plot for AP9.7 with K =,num2str(km)]) print deps ap9_7a [Gm,Pm,Wcg,Wcp] = margin(oltf) dcltf = tf(1,[1 Km 4*Km]) [y,t] = step(cltf); figure(8)
5 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 3: Closed Loop Bode Diagram for AP9.7 with K = 2.8 Closed Loop Bode Plot for AP9.7 with K = Magnitude (db) Phase (deg) Frequency (rad/sec) plot(t,y) grid xlabel( Time (s) ) ylabel( y(t) ) title([ Closed Loop Response to unit step disturbance for AP9.7,... with K =,num2str(km)]) print deps ap9_7aa max(y) AP9.9 We are given a system with unity negative feedback and forward path transfer function G c (s)g(s) = K P s + K I s 2 (s 2 + 7s + 1) (4) where K I /K P =.2 and we are asked to find the value of K P that maximizes the phase margin.
6 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 4: Closed Loop Step Disturbance Response for AP9.7 with K = 2.8 Closed Loop Response to unit step disturbance for AP9.7 with K = y(t) Time (s) Solution: With K I /K P =.2, the closed loop transfer function is T(s) = K P (s +.2) s 4 + 7s 3 + 1s 2 + K P (s +.2). (5) Conveniently, this is the same transfer function that we had in problem AP9.5. The answer will thus be the same, i.e. K p = 4.9 will yield the maximum phase margin, etc. (Ok, so I wasn t paying enough attention when I assigned this one. I trust there will be no complaints. :) DP9.3 We are given a system having unity negative feedback and transfer function G(s) = Ke 1s 4s + 1 in the forward path. We are asked to choose a value for K to keep the output within a narrow range while maintaining good dynamic response. Solution: First, let s generate a Bode diagram for the open loop system. The delay will add φ(ω) = ωt where T = 1 to the phase shift while not affecting the magnitude. We (6)
7 ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 7 use the commands shown in the Matlab script below to generate the Bode diagram shown in Figure 5. This doesn t really seem very useful. Next, we ll use a Pade approximation to Figure 5: Open Loop Bode Diagram for DP9.3 with K = 1 Open Loop Bode Plot for DP9.3 with K=1 5 Magnitude (db) Phase (deg) Frequency (rad/s) express the time delay as a transfer function so that we can analyze the time and frequency response of the closed loop system. As shown in the Matlab script, the pade command yields the following second order transfer function approximation to time the delay: G d (s) = s2.6s +.12 s 2 +.6s (7) To get some idea what range of values of the gain K are allowable for a stable system, we can do a root locus of the system including the approximate time delay. The root locus plot is shown in Figure 6. We find that the complex valued loci cross the imaginary axis when K 7. Were this a real application, we d ask the physicians or biomedical engineers on our design team to be a bit more specific about what they mean by maintains narrow deviation for blood pressure while achieving a good dynamic response. That not being an option, I m going to use the working hypothesis that overshoot of more than 5 percent in a step response would be bad. I used a loop to generate step responses for different gains
8 ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 8 Figure 6: Root Locus for DP9.3 with 2nd Order Pade Approximation to Time Delay.25 Root Locus for DP System: oltf Gain: 7.1 Pole: i Damping:.12 Overshoot (): 1 Frequency (rad/sec): Imaginary Axis Real Axis less than 5. (5 = 7 minus a large margin for error.) I found that the rise and settling time decreases with increasing K (as we would expect). I found that a value of K = 1.8 resulted in 4.23 overshoot, meeting my interpretation of the design requirements. The final value of the step response was.6429 so I would need to add a gain block after the feedback path to boost the output. To be really thorough, I d have to check to see whether this gain were frequency dependent and, if necessary, use a filter rather than a simple amplifier. Here s my Matlab script. DP9_3.m solves part of problem DP9_3 from Dorf and Bishop 1th ed. 17 April 5 sk figure(3) K = 1; oltf = tf([k],[4 1]) [mag,phase,w] = bode(oltf);
9 ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 9 nphase = squeeze(phase)  w.*1; subplot(2,1,1) magdb = 2*log1(squeeze(mag)); semilogx(w,magdb) grid ylabel( Magnitude (db) ) title( Open Loop Bode Plot for DP9.3 with K=1 ) subplot(2,1,2) semilogx(w,nphase, .,w,squeeze(phase),  ) grid ylabel( Phase (deg) ) xlabel( Frequency (rad/s) ) print deps dp9_3a [dnum,dden]=pade(1,2) oltf = series(tf(dnum,dden),tf([k],[4 1])) figure(4) rlocus(oltf) title( Root Locus for DP9.3 ) print deps dp9_3rl t=[:.1:1] ; np=max(size(t)) for K = [.1:.1:5] oltf = series(tf(dnum,dden),tf([k],[4 1])); cltf = feedback(oltf,1); [y,t]= step(cltf,t); po(round(k*1)) = (max(y)y(np))/y(np); end K=1.8 oltf = series(tf(dnum,dden),tf([k],[4 1])); cltf = feedback(oltf,1); [y,t]= step(cltf,t); MP9.2 We are asked to use the Matlab nyquist command to analyze the following three systems: G a (s) = G b (s) = G c (s) = 1 s + 1 (8) 2 s 2 + 8s + 14 (9) 1 (s + 1) 3 (1) Solution:
10 ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 The Nyquist plots shown in Figures 7 through 9 are generated using the Matlab script below. Note that in the first two Nyquist diagrams the point ( 1,) is not encircled, indicating that if we close the loop with unity negative feedback the systems will still be stable. On the other hand, the third Nyquist diagram has two clockwise encirclements. Having no zeros, we have P = Z N = ( 2) = 2 poles of the closed loop system in the right half plane. This is shown in the following Matlab transcript fragment. >> clf = feedback(tf([1],[ ]),1) Transfer function: s^3 + 3 s^2 + 3 s + 11 >> roots([ ]) ans = i i MP9_2.m solves part of problem MP9_2 from Dorf and Bishop 1th ed. 17 April 5 sk figure(2) nyquist(tf([1],[1 1])); title( Nyquist diagram for MP9.2a G(s) = 1/(s+1) ) print deps mp9_2a figure(3) nyquist(tf([2],[1 8 14])); title( Nyquist diagram for MP9.2b G(s) = 2/(s^2+8s+14) ) print deps mp9_2b figure(4) nyquist(tf([1],[ ])); title( Nyquist diagram for MP9.2c G(s) = 1/(s+1)^3 ) print deps mp9_2c MP9.6 We are given a unity negative feedback system with forward path transfer function G c (s)g(s) = 1(s + 3)( b )(s 2 25) s(s 3)(s 2 + 5s + 1) where b =.5. We are asked to (a) use the Matlab margin command to find the phase and gain margins and crossover frequencies, (b) determine from the results of part (a), the (11)
11 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 7: Nyquist plot for MP9.2a Nyquist diagram for MP9.2a G(s) = 1/(s+1) Imaginary Axis Real Axis maximum value of b for which stability is retained, and (c) verify our answer to part (b) using the RouthHurwitz criterion. Solution: (a) The gain margin is 2.23 at rad/s and the phase margin is deg at the crossover frequency rad/s. The Matlab commands used to generate this information are shown below. b =.5 Gc = tf(1*[1 3],[1 ]) G = tf(b*[125],conv([13],[1 5 1])) oltf = series(gc,g) [Gm,Pm,Wcg,Wcp] = margin(oltf) (b) The gain margin being 2.23, we can increase the gain from b =.5 to b.
12 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 8: Nyquist plot for MP9.2b 1 Nyquist diagram for MP9.2b G(s) = 2/(s 2 +8s+14) Imaginary Axis Real Axis (c) The closed loop transfer function is (after some algebra) 1b (s + 3)(s + 5)(s 5) T(s) = s 4 + (47 1b )s 3 + (1 15 3b )s 2 + ( b )s + 75b (12) so the Routh array is as shown in Table 2. I used Maple to do the calculations for the Routh array which are quite messy. The Maple commands I used were as follows. (I think that there is supposed to be a way to do this in Matlab as well, but Maple is better suited for this type of computation so I use it instead.) > # maple script for computing RouthHurwitz array for MP9.6 > with(linalg): > r1 := array([1,853*b,75*b]); > r2 := array([471*b,3+25*b,]); > r3 := array([simplify(1/r2[1]*(r1[1]*r2[2]r1[2]*r2[1])), 1/r2[1]*(r1[1]*r2[3]r1[3]*r2[1]),]); > r4 := array([simplify(1/r3[1]*(r2[1]*r3[2]r2[2]*r3[1])), 1/r3[1]*(r2[1]*r3[3]r2[3]*r3[1]),]);
13 ECE382/ME482 Spring 25 Homework 7 Solution April 17, Figure 9: Nyquist plot for MP9.2c 8 Nyquist diagram for MP9.2c G(s) = 1/(s+1) Imaginary Axis Real Axis > r5 := array([simplify(1/r4[1]*(r3[1]*r4[2]r3[2]*r4[1])), 1/r4[1]*(r3[1]*r4[3]r3[3]*r4[1]),]); Now, analyzing the Routh array, we see from the second row that we need b < 4.7. Next, using the Matlab command >> roots([ ]) ans = and noting that if b < 4.7, the denominator of the first element of the third row will be negative and we have a negative in front of the fraction, so we want the factor in the parentheses to be positive, which it is if b < or b > The second option would conflict with b < 4.7 so we can summarize our constraints so far as b < From the fourth row, using the Matlab command
14 ECE382/ME482 Spring 25 Homework 7 Solution April 17, >> roots([ ]) ans = and then using the Maple commands > subs(b=.13,r4[1]); > subs(b=.15,r4[1]); > subs(b=1.2,r4[1]); we determine that we need.1436 < b < Thus we have finally determined that for stability we need.1444 < b < The upper bound on b agrees with that determined from the gain margin and that determined using the RouthHurwitz criterion. Note, however, that the gain margin approach did not tell us that there is also a minimum acceptable value for b to retain stability. Table 2: Routh array for MP9.6c s 4 : b 75b s 3 : 47 1b b s 2 : 1( b +3b 2 ) 47+1b 75b s 1 : 5(1663b b +2577) 3b b s : 75b
ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
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 informationHomework Assignment 3
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full
More informationECE382/ME482 Spring 2005 Homework 1 Solution February 10,
ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 1 Solution to HW1 P2.33 For the system shown in Figure P2.33 on p. 119 of the text, find T(s) = Y 2 (s)/r 1 (s). Determine a relationship that
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 informationMAE 143B  Homework 9
MAE 143B  Homework 9 7.1 a) We have stable firstorder poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straightline
More informationFrequency (rad/s)
. The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer
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 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 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 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 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 information] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command prefilter [ 0.
EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965372 Problem (Analysis of a Feedback System) Consider the feedback system
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 informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
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 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 informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
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 informationSTABILITY OF CLOSEDLOOP CONTOL SYSTEMS
CHBE320 LECTURE X STABILITY OF CLOSEDLOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 101 Road Map of the Lecture X Stability of closedloop control
More informationProblem 1 (Analysis of a Feedback System  Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function 1.
1 EEE480 Final Exam, Spring 2016 A.A. Rodriguez Rules: Calculators permitted, One 8.5 11 sheet, closed notes/books, open minds GWC 352, 9653712 Problem 1 (Analysis of a Feedback System  Bode, Root Locus,
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 informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More informationELECTRONICS & 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 informationControls Problems for Qualifying Exam  Spring 2014
Controls Problems for Qualifying Exam  Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
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 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 informationMAE 143B  Homework 7
MAE 143B  Homework 7 6.7 Multiplying the first ODE by m u and subtracting the product of the second ODE with m s, we get m s m u (ẍ s ẍ i ) + m u b s (ẋ s ẋ u ) + m u k s (x s x u ) + m s b s (ẋ s ẋ u
More information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
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 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 informationSECTION 5: ROOT LOCUS ANALYSIS
SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More informationFREQUENCYRESPONSE DESIGN
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
More informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
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 information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
More informationMAE 143B  Homework 9
MAE 43B  Homework 9 7.2 2 2 3.8.6.4.2.2 9 8 2 2 3 a) G(s) = (s+)(s+).4.6.8.2.2.4.6.8. Polar plot; red for negative ; no encirclements of, a.s. under unit feedback... 2 2 3. 4 9 2 2 3 h) G(s) = s+ s(s+)..2.4.6.8.2.4
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationK(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s
321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify
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 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 informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
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 informationECE382/ME482 Spring 2004 Homework 4 Solution November 14,
ECE382/ME482 Spring 2004 Homework 4 Solution November 14, 2005 1 Solution to HW4 AP4.3 Intead of a contant or tep reference input, we are given, in thi problem, a more complicated reference path, r(t)
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 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 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 informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More informationx(t) = x(t h), x(t) 2 R ), where is the time delay, the transfer function for such a e s Figure 1: Simple Time Delay Block Diagram e i! =1 \e i!t =!
1 TimeDelay Systems 1.1 Introduction Recitation Notes: Time Delays and Nyquist Plots Review In control systems a challenging area is operating in the presence of delays. Delays can be attributed to acquiring
More informationCHAPTER # 9 ROOT LOCUS ANALYSES
F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closedloop system is closely related to the location of the closedloop poles. If the system
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : RouthHurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
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 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 informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
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 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 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 informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad  500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III
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 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 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 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 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 informationChapter 7 : Root Locus Technique
Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci
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 informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
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 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 informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
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 informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
More informationTopic # Feedback Control
Topic #4 16.31 Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests. Fall 2007 16.31 4 2 Frequency Stability
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 informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
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 information2.010 Fall 2000 Solution of Homework Assignment 8
2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)
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 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 informationSome special cases
Lecture Notes on Control Systems/D. Ghose/2012 87 11.3.1 Some special cases Routh table is easy to form in most cases, but there could be some cases when we need to do some extra work. Case 1: The first
More informationEC6405  CONTROL SYSTEM ENGINEERING Questions and Answers Unit  I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
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 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 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 informationCh 14: Feedback Control systems
Ch 4: Feedback Control systems Part IV A is concerned with sinle loop control The followin topics are covered in chapter 4: The concept of feedback control Block diaram development Classical feedback controllers
More informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More informationChemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University
Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this
More informationCompensation 8. f4 that separate these regions of stability and instability. The characteristic S 0 L U T I 0 N S
S 0 L U T I 0 N S Compensation 8 Note: All references to Figures and Equations whose numbers are not preceded by an "S"refer to the textbook. As suggested in Lecture 8, to perform a Nyquist analysis, we
More informationPM diagram of the Transfer Function and its use in the Design of Controllers
PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude
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 informationEEE 184 Project: Option 1
EEE 184 Project: Option 1 Date: November 16th 2012 Due: December 3rd 2012 Work Alone, show your work, and comment your results. Comments, clarity, and organization are important. Same wrong result or same
More information