Problem 1 (Analysis of a Feedback System - Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function 1.

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1 1 EEE480 Final Exam, Spring 2016 A.A. Rodriguez Rules: Calculators permitted, One sheet, closed notes/books, open minds GWC 352, Problem 1 (Analysis of a Feedback System - Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function L = PK = 10(s [ ] 4 [ ] +1.93)(s s ) [ 10 7 ][ ] s s s 2 (s 1) 2 s s [ ][ with series compensator K = 10(s+1.93)(s s+2.591) s s+400] and reference command pre-filter [ ][ ] W = s s s (a) Sketch approximate Bode magnitude and phase plots. (b) Determine the approximate 10 (±20 db) frequencies associated with L. (c) Sketch an approximate root locus plot. Compute angles of asymptotes. (d) Sketch a Nyquist plot (only partial polar plot is required). (e) Provide a complete stability summary - indicating Nyquist N cw values for open k i intervals. (f) Compute all margins ( GM, GM, PM, DM), their associated frequencies and bounds for peak S, T. EXTRA CREDIT: g, h, i (g) What is the approximate closed loop transfer function T ry and associated closed loop poles? (h) Approximate the output y when r = 10 1(t). What is (approximate) overshoot? settling time? (i) Determine r such that y ss =10+0.1sin(0.1t +10 )whend i =10sin(0.01t +10 ) + 30 cos(400t +40 ). Problem 2 (Control System Design: Design for Bandwidth and Phase Margin) Suppose [ ] [ 2Ω 1 s 0.5Ωs π P = s ] s 0.5 s +0.5Ω s s + 2Ωs π where Ω s = kω g (k>0) represents a (digital) sampling frequency. The first term in brackets represents an anti-aliasing filter. The second term in brackets represents a Pade approximation to a zero order hold half sample time delay. Suppose that k = 20. (a) Design a feedback control system such that the closed loop system (1) is stable, (2) exhibits constant steady state error to ramp input disturbances d i, (3) ω g = 2 rad/sec, (4) PM =60, (5) impact of high frequency sensor noise on controls is addressed, (6) overshoot to step commands is addressed. (b) Determine the approximate gain margins associated with your design.

2 2 Here is an mfile for Problem # 1 (Spring 2016 Final Exam). EEE480 SPRING 2016 FINAL A.A. Rodriguez (All rights reserved) 4th ORDER CLS L = ( c3 s^3 + c2 s^2 + c1 s + co ) / ( s^2 (s - p1) (s - p2) ) Phi_CL = s^4 + (c3 - p1 - p2)s^3 + (c2 +p1 p2) s^2 + c1 s + co Phi_CL_desired = (s^2 + a1 s + ao) (s^2 + b1 s + b0) = s^4 +(a1 + b1)s^3 + (ao + a1b1 + bo)s^2 + (ao b1 + a1 bo)s + aobo c3 = a1 + b1 + p1 + p2 c2 = ao + a1 b1 + bo - p1 p2 c1 = ao b1 + a1 bo co = ao bo PLANT P = 1 / ( (s - p1) (s - p1) ) NOTE: DATA GIVEN IN PROBLEM STATEMENT RESULTS IN THE FOLLOWING NICE DOMINANT CLOSED LOOP POLES! -1 +/- j 1-3 +/- j 4 PLEASE SET FLAG PROPERLY BEFORE RUNNING MFILE FLAG IS SET TO NOMINAL VALUE OF 1 - to get 1 unity gain crossover FLAG = 1 to get 1 unity gain crossover for nominal data given FLAG = 3 to get 3 unity gain crossovers for nominal data given PLEASE SET HIGH FREQUENCY POLE APPROXIMATION FLAG PROPERLY BEFORE RUNNING MFILE POLE APPROXIMATION FLAG IS CURRENTLY SET TO ITS NOMINAL VALUE FLAGAPPROX = 1 FLAGAPPROX = 0 NOMINAL: USE THIS TO APPROXIMATE HIGH FREQUENCY POLES AND GET NICER CLOSED LOOP POLE NUMBERS YOU SHOULD COMPUTE DURING EXAM USE THIS TO NOT APPROXIMATE HIGH FREQUENCY POLES PLANT DATA - PLANT = 1 / (s - p1) (s - p2)

3 3 p1 = 1; 1 Nominal plant pole p2 = 1; 2 Nominal plant pole plant = tf(1, [1 -(p1+p2) p1*p2 ]) zpk(plant) w = logspace(-2,3,2000); Vector of frequencies [plant_mag, plant_phase] = bode(plant, w); figure(10) semilogx(w,20*log10(plant_mag(1,:)) ) title( PLANT: P ) figure(11) semilogx(w,plant_phase(1,:) ) title( PLANT: P ) ylabel( Phase (degrees) ) DESIRED CLOSED LOOP CHARACTERISTIC EQUATION Phi_CL_desired = (s^2 + a1 s + ao) (s^2 + b1 s + b0) = s^4 +(a1 + b1)s^3 + (ao + a1b1 + bo)s^2 + (ao b1 + a1 bo)s + aobo SET NOMINAL CLOSED LOOP POLES a1 = 2 ao = 2 s^2 + 2s \pm j 1 if FLAG == 1 b1 = 6; 6 bo = 25; 225/16 s^2 + 6s + 25 = 0 yields faster CLPs = -2 \pm j 3/2 Yields 1 unity gain crossover for nominal data given! end if FLAG == 1 b1 = 12 bo = 225/4 s^2 + 12s + 225/16 = 0 yields faster CLPs = -6 \pm j 9/2 Yields 1 unity gain crossover for nominal data given! end if FLAG == 1 b1 = 1 bo = 0.5 s^2 + s = 0 yields CLPs = -0.5 \pm j 0.5 Yields 1 unity gain crossover for nominal data given! end

4 4 if FLAG==3 b1 = 2 bo = 25/16 s^2 + 2s + 25/16 = 0 yields slower CLPs = -1 \pm j 3/4 yields 3 unity gain crossovers for nominal data given! YUK!!!! end COMPENSATOR: K = ( c3 s^3 + c2 s^2 + c1 s + co ) / s^2 c3 = a1 + b1 + p1 + p2 c2 = ao + a1*b1 + bo - p1*p2 c1 = ao*b1 + a1*bo co = ao*bo hfp = High frequency poles for roll off in controller Actual value of pole used in problem statement if FLAGAPPROX ==1 hfp = hfp High frequency poles for roll off in controller USE THIS TO GET NICER CLOSED LOOP POLE NUMBERS; i.e. approximation that should be made during the exam SET FLAGAPPROX = 0 TO GET NUMBERS IN PROBLEM STATEMENT end comp_num = [ c3 c2 c1 co] comp_zeros = roots(comp_num) Compensator zeros comp_den = [ 1 0 0] comp = tf (comp_num,comp_den) roll_off = tf(hfp, [1 hfp] ) num_rolloff_terms = 4; Number of Roll Off Terms for end rolloff_counter = 1:1: num_rolloff_terms comp = series(comp, roll_off) add high frequency roll off to comp NOTE: This just includes 5 roll off terms. Does NOT include all pass term in Plant P given on Exam#2 Spring 2016 Problem 1 statement. PM computed in this mfile will therefore be bigger!...off by about 2*atan(7/100)*180/pi = degrees zpk(comp) [comp_mag, comp_phase] = bode(comp, w); figure(20)

5 5 semilogx(w,20*log10(comp_mag(1,:)) ) title( CONTROLLER: K ) figure(21) semilogx(w,comp_phase(1,:),w, -180, r-. ) title( CONTROLLER: K ) ylabel( Phase (degrees) ) OPEN LOOP: L = PK loop = series(plant,comp) zpk(loop) [loop_mag, loop_phase] = bode(loop, w); figure(30) semilogx(w,20*log10(loop_mag(1,:)) ) title( LOOP: PK ) figure(31) semilogx(w,loop_phase(1,:), w, -180, r-. ) title( LOOP: PK ) ylabel( Phase (degrees) ) figure(32) bode(loop, w) allmargin(loop) SENSITIVITY: S = 1 / (1 + PK) sen = feedback(1,loop) Form sensitivity eig(sen) closed loop poles [sen_mag, sen_phase] = bode(sen, w); figure(40) semilogx(w,20*log10(sen_mag(1,:)),w, 6, r-- ) title( SENSITIVITY: S = 1 / (1 + L) ) axis([ ])

6 6 COMPLEMENTARY SENSITIVITY: T = L / (1 + PK) compsen = feedback(loop,1) Form complementary sensitivity eig(compsen) closed loop poles [compsen_mag, compsen_phase] = bode(compsen, w); figure(50) semilogx(w,20*log10(compsen_mag(1,:)),w, 6, r-- ) title( COMPLEMENTARY SENSITIVITY: T = L / (1 + L) ) axis([ ]) KS ksen = feedback(comp,plant) Form KS eig(ksen) closed loop poles [ksen_mag, ksen_phase] = bode(ksen, w); figure(60) semilogx(w,20*log10(ksen_mag(1,:)), w, 20, r--, w, 40, r-- ) title( KS ) axis([ ]) SP psen = feedback(plant, comp) Form SP eig(psen) closed loop poles [psen_mag, psen_phase] = bode(psen, w); figure(70) semilogx(w,20*log10(psen_mag(1,:)), w,-40, r-- ) title( SP ) axis([ ]) CLOSED LOOP STEP RESPONSE figure(100)

7 7 t = 0:.001:10; Vector of time points num_tpoints = size(t)*[0 1] ref_command = ones(1, num_tpoints); prefilter = tf(co/c3, [ 1 c2/c3 c1/c3 co/c3 ] ) zpk(prefilter) filtref_command = lsim(prefilter,ref_command, t); Filtered Reference Command y = lsim(compsen,ref_command, t); u = lsim(ksen, ref_command, t); y_filt = lsim(compsen,filtref_command, t); u_filt = lsim(ksen, filtref_command, t); plot(t,y,t,y_filt) title( Output Response to Step Reference (Filtered and Unfiltered) ) xlabel( Time (sec) ) ylabel( Output: y ) figure(101) plot(t,u,t,u_filt) title( Control Response to Step Reference (Filtered and Unfiltered) ) xlabel( Time (sec) ) ylabel( Control: u ) axis([ ]) ROOT LOCUS figure(200) k = 0:.001:20; rlocus(loop,k) axis([ ]) figure(210) rlocus(loop,k) axis([ ]) ************************************************************************** NOMINAL LOOP AT LOW FREQUENCIES compnom_num = [ c3 c2 c1 co] compnom_den = [ 1 0 0]

8 8 compnom = tf (compnom_num,compnom_den) [compnom_mag, compnom_phase] = bode(compnom, w); figure(20) semilogx(w,20*log10(comp_mag(1,:)), w,20*log10(compnom_mag(1,:)) ) title( CONTROLLER: K ) figure(21) semilogx(w,comp_phase(1,:),w, -180, r-., w,compnom_phase(1,:) ) title( CONTROLLER: K ) ylabel( Phase (degrees) ) OPEN LOOP: L = PK loopnom = series(plant, compnom); zpk(loopnom) allmargin(loopnom) damp(eig(1/(1+loopnom))) [loopnom_mag, loopnom_phase] = bode(loopnom, w); figure(30) semilogx(w,20*log10(loop_mag(1,:)), w,20*log10(loopnom_mag(1,:)), w, -20, r--, w, 0, r--, w, 20, r-- title( LOOP: PK ) figure(31) semilogx(w,loop_phase(1,:), w,loopnom_phase(1,:), w, -180, r-. ) title( LOOP: PK ) ylabel( Phase (degrees) ) SENSITIVITY: S = 1 / (1 + PK) sennom = feedback(1,loopnom) Form sensitivity eig(sennom) closed loop poles

9 9 [sennom_mag, sennom_phase] = bode(sennom, w); figure(40) semilogx(w,20*log10(sen_mag(1,:)),w, 6, r--, w,20*log10(sennom_mag(1,:)), w, -20, r-- ) title( SENSITIVITY: S = 1 / (1 + L) ) axis([ ]) COMPLEMENTARY SENSITIVITY: T = L / (1 + PK) compsennom = feedback(loopnom,1) Form complementary sensitivity eig(compsennom) closed loop poles [compsennom_mag, compsennom_phase] = bode(compsennom, w); figure(50) semilogx(w,20*log10(compsen_mag(1,:)),w, 6, r--, w,20*log10(compsennom_mag(1,:)), w, -20, r-- ) title( COMPLEMENTARY SENSITIVITY: T = L / (1 + L) ) axis([ ]) [Gm,Pm,Wcg,Wcp] = margin(loop) figure(55) semilogx(w,20*log10(sen_mag(1,:)),w,20*log10(compsen_mag(1,:)), w, 20*log10(Gm), r--, w,20*log10(s title( SENS AND COMP SENSITIVITY: S = 1 / (1 + L), T = L / (1 + L) ) axis([ ]) KS ksennom = feedback(compnom,plant) Form KS eig(ksennom) closed loop poles [ksennom_mag, ksennom_phase] = bode(ksennom, w); figure(60) semilogx(w,20*log10(ksen_mag(1,:)), w, 20, r--, w, 40, r--, w,20*log10(ksennom_mag(1,:)) ) title( KS ) axis([ ])

10 10 SP psennom = feedback(plant, compnom) Form SP eig(psennom) closed loop poles [psennom_mag, psennom_phase] = bode(psennom, w); figure(70) semilogx(w,20*log10(psen_mag(1,:)), w,20*log10(psennom_mag(1,:)), w,-40, r-- ) title( SP ) axis([ ]) CLOSED LOOP STEP RESPONSE ynom = lsim(compsennom,ref_command, t); unom = lsim(ksennom, ref_command, t); CANT SIMULATE WITHOUT ROLLOFF IN K ynom_filt = lsim(compsennom,filtref_command, t); unom_filt = lsim(ksennom, filtref_command, t); CANT SIMULATE WITHOUT ROLLOFFIN K figure(100) plot(t,y,t,y_filt, t,ynom,t,ynom_filt) title( Output Response to Step Reference (Filtered and Unfiltered) ) xlabel( Time (sec) ) ylabel( Output: y ) figure(101) plot(t,u,t,u_filt) title( Control Response to Step Reference (Filtered and Unfiltered) ) xlabel( Time (sec) ) ylabel( Control: u ) axis([ ]) DESIGN DATA zpk(loop) damp(comp_zeros) allmargin(loop) eig(sen) damp(eig(sen)) FLAG FLAGAPPROX c3 c2

11 11 c1 co zpk(loopnom) allmargin(loopnom) damp(eig(1/(1+loopnom))) ************************************************************************* ************************************************************************* ************************************************************************* EEE480 FINAL SPRING PROBLEMS 1-4 DID NOT APPROXIMATED HIGH FREQUENCY POLES LOOP Zero/pole/gain: (s+1.93) (s^ s ) s^2 (s+400)^4 (s-1)^2 = 10 (s+1.93) (s^ s ) (400)^ s^2 (s-1)^2 (s+400)^4 CONTROLLER ZEROS Eigenvalue Damping Freq. (rad/s) -1.93e e e e e+000i 5.81e e e e+000i 5.81e e+000 ALLMARGINS GainMargin: [ ] GMFrequency: [ e+002] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 CLOSED LOOP POLES 1.0e+002 *

12 i i i i i i CLOSED LOOP POLES Eigenvalue Damping Freq. (rad/s) -5.47e e e e e+002i 9.37e e e e+002i 9.37e e e e e e e+000i 6.14e e e e+000i 6.14e e e e+000i 7.05e e e e+000i 7.05e e+000 FLAG = 1 FLAGAPPROX = 0 CONTROLER COEFFICIENTS c3 = 10 c2 = 38 c1 = 62 co = 50 ************************************************************************* ************************************************************************* ************************************************************************* EEE480 FINAL SPRING 2016 NOMINAL LOOP Zero/pole/gain: 10 (s+1.93) (s^ s ) s^2 (s-1)^2 ALL MARGINS GainMargin: [ ] GMFrequency: [ ]

13 13 PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 NOMINAL CLOSED LOOP POLES Eigenvalue Damping Freq. (rad/s) -3.00e e+000i 6.00e e e e+000i 6.00e e e e+000i 7.07e e e e+000i 7.07e e+000 ************************************************************************* ************************************************************************* ************************************************************************* EEE480 FINAL SPRING 2016 APPROXIMATED HIGH FREQUENCY POLES - using BIG HIGH FREQ POLE NUMBERS LOOP Zero/pole/gain: (s+1.93) (s^ s ) s^2 (s+1.04e004)^4 (s-1)^2 10 (s+1.93) (s^ s ) (10400)^4 = s^2 (s-1)^2 (s+10400)^4 CONTROLLER ZEROS Eigenvalue Damping Freq. (rad/s) -1.93e e e e e+000i 5.81e e e e+000i 5.81e e+000 ALL MARGINS GainMargin: [ e+002] GMFrequency: [ e+003] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1

14 14 CLOSED LOOP POLES 1.0e+004 * i i i i i i CLOSED LOOP POLES (NICE "EXAM" NUMBERS!!!!) Eigenvalue Damping Freq. (rad/s) -1.22e e e e e+003i 9.85e e e e+003i 9.85e e e e e e e+000i 6.00e e e e+000i 6.00e e e e+000i 7.07e e e e+000i 7.07e e+000 FLAG = 1 FLAGAPPROX = 1 CONTROLER COEFFICIENTS c3 = 10 c2 = 38 c1 = 62 co = 50 ************************************************************************* ************************************************************************* ************************************************************************* EEE480 EXAM#2 SPRING PROBLEMS 1-4 APPROXIMATED HIGH FREQUENCY POLES - using BIG HIGH FREQ POLE NUMBERS LOOP

15 Zero/pole/gain: (s+1.304) (s^ s ) s^3 (s+1.01e004)^5 (s-1) 7 (s+1.304) (s^ s ) (1.01e004)^5 = s^3 (s-1) (s+1.01e004)^5 COMPENSATOR ZEROS Eigenvalue Damping Freq. (rad/s) -1.30e e e e e+000i 4.09e e e e+000i 4.09e e+000 ALL MARGINS GainMargin: [ Inf] GMFrequency: [ e+003 Inf] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 CLOSED LOOP POLES 1.0e+004 * i i i i i i i i CLOSED LOOP POLES Eigenvalue Damping Freq. (rad/s) -1.20e e+003i 9.94e e e e+003i 9.94e e e e+003i 9.72e e e e+003i 9.72e e e e e e e+000i 8.01e e

16 e e+000i 8.01e e e e+000i 6.00e e e e+000i 6.00e e+000 FLAG = 1 FLAGAPPROX = 1 CONTROLLER COEFFICIENTS c3 = 7 c2 = c1 = co = ************************************************************************* ************************************************************************* ************************************************************************* EEE480 EXAM#2 SPRING PROBLEMS 1-4 DID NOT APPROXIMATE HIGH FREQUENCY POLES LOOP Zero/pole/gain: (s+1.304) (s^ s ) s^3 (s+100)^5 (s-1) 7 (s+1.304) (s^ s ) (100)^5 = s^3 (s-1) (s+100)^5 COMPENSATOR ZEROS Eigenvalue Damping Freq. (rad/s) -1.30e e e e e+000i 4.09e e e e+000i 4.09e e+000 ALL MARGINS GainMargin: [ Inf] GMFrequency: [ Inf] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1

17 17 CLOSED LOOP POLES 1.0e+002 * i i i i i i i i CLOSED LOOP POLES Eigenvalue Damping Freq. (rad/s) -1.45e e+001i 9.79e e e e+001i 9.79e e e e+001i 8.42e e e e+001i 8.42e e e e e e e+000i 9.48e e e e+000i 9.48e e e e+000i 5.86e e e e+000i 5.86e e+000 FLAG = 1 FLAGAPPROX = 0 CONTROLLER COEFFICIENTS c3 = 7 c2 = c1 = co = ************************************************************************* ************************************************************************* ************************************************************************* EEE480 EXAM#2 FALL 2015 Zero/pole/gain: (s+1) (s^ s )

18 s^2 (s+90)^5 (s^2-2s + 2) 7 (s+1) (s^ s ) 90^5 = s^2 (s^2-2s + 2) (s+90)^5 Eigenvalue Damping Freq. (rad/s) -1.00e e e e e-001i 6.30e e e e-001i 6.30e e-001 ans = GainMargin: [ Inf] GMFrequency: [ Inf] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 ans = 1.0e+002 * i i i i i i i i Eigenvalue Damping Freq. (rad/s) -1.31e e+001i 9.79e e e e+001i 9.79e e e e+001i 8.36e e e e+001i 8.36e e e e e e e+000i 6.01e e e e+000i 6.01e e e e-001i 7.04e e e e-001i 7.04e e-001 FLAG =

19 19 1 FLAGAPPROX = 0 c3 = 7 c2 = c1 = 15 co = ************************************************************************* ************************************************************************* ************************************************************************* EEE480 FINAL SPRING 2015 Zero/pole/gain: (s ) (s^ s ) s^2 (s+70)^5 (s-2) (s-1) 10 (s ) (s^ s ) 70^5 = s^2 (s+70)^5 (s-2) (s-1) (s+70)^5 Eigenvalue Damping Freq. (rad/s) -5.18e e-001i 5.59e e e e-001i 5.59e e e e e-001 ans =

20 20 GainMargin: [ Inf] GMFrequency: [ Inf] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 ans = 1.0e+002 * i i i i i i i i Eigenvalue Damping Freq. (rad/s) -1.06e e+001i 9.76e e e e+001i 9.76e e e e+001i 8.03e e e e+001i 8.03e e e e+000i 4.21e e e e+000i 4.21e e e e e e e-001i 7.01e e e e-001i 7.01e e-001 FLAG = 1 FLAGAPPROX = 0 Zero/pole/gain: (s ) (s^ s ) s^2 (s+70)^4 (s-2) (s-1)

21 10 (s ) (s^ s ) 70^4 = s^2 (s-2) (s-1) (s+70)^4 Eigenvalue Damping Freq. (rad/s) -5.18e e-001i 5.59e e e e-001i 5.59e e e e e-001 ans = GainMargin: [ ] GMFrequency: [ ] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 ans = 1.0e+002 * i i i i i i Eigenvalue Damping Freq. (rad/s) -1.08e e e e e+001i 8.81e e e e+001i 8.81e e e e+000i 6.58e e e e+000i 6.58e e e e e e e-001i 7.02e e e e-001i 7.02e e-001 FLAG = 1 FLAGAPPROX = 21

22 22 0 Zero/pole/gain: (s ) (s^ s ) s^2 (s+1.007e004)^4 (s-2) (s-1) = 10 (s ) (s^ s ) (1.007e004)^ s^2 (s-2) (s-1) (s+1.007e004)^4 Eigenvalue Damping Freq. (rad/s) -5.18e e-001i 5.59e e e e-001i 5.59e e e e e-001 ans = GainMargin: [ ] GMFrequency: [ e+003] PhaseMargin: PMFrequency: DelayMargin: DMFrequency: Stable: 1 ans = 1.0e+004 * i i i i i i Eigenvalue Damping Freq. (rad/s) -1.18e e e e e+003i 9.85e e e e+003i 9.85e e e e e e e+000i 8.01e e e e+000i 8.01e e e e-001i 7.07e e-001

23 -5.00e e-001i 7.07e e-001 FLAG = 1 FLAGAPPROX = 1 23

] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command pre-filter [ 0.

] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command pre-filter [ 0. EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965-372 Problem (Analysis of a Feedback System) Consider the feedback system