= rad/sec. We can find the last parameter, T, from ωcg new

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1 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 specs: ess ramp = 1/5 gain margin > 5 db phase margin > 3 o Solution: If we let G s K Ts + 1 lead ()= c, we see that we need a type 1 system to meet the steady-state error αt + 1 s specifications. Since we only have a type one system, we must increase the system type. This is accomplished by letting K c =K/s. We can find the appropriate value of K from K = 5 = lim sg G G( s) = K(8) / 4 = 2K. 2 Thus, K=5/2=2.5 and K c =2.5/s. Next, we must make a Bode plot of K c G zoh G(s)=. The Bode ss ( + 4)( 1+ s/ 2) plot is shown below: 6 EE572 - HW#21: Bode Plot of KcGzohG(jw) -5 v desired s lead zoh ωcg=13.84 r/s g.m. = db p.m. = ωcp= r/s w (rad/sec) Next, we can determine how much more phase angle we need to meet specifications from: φ m = p.m. desired - p.m. current + fudge factor = 35 o o +1 o = o. We can use this value to solve for α : 1 α sin( φm ) = 1 φ or α = sin( m ) = Recall that the lead compensator adds a gain of 1/ α = 1.92= α 1+ sin( φ ) m db. Thus, the gain crossover frequency will shift to the right where K c G zoh G =-5.66 db. From the Bode plot, we see 1 that this occurs at ω cgnew = rad/sec. We can find the last parameter, T, from ωcg new = = or T=.992. α T Hence, G s K Ts 1.. s lead () = c T ++ = We can check our answer by making a Bode plot of G lead G zoh G(s): α 1 s. 269s + 1 s

2 1 Gm=19.65 db, (w= 77.46) Pm=41.14 deg. (w=19.35) Gain db Frequency (rad/sec) Phase deg Frequency (rad/sec) As we can see, the compensated system has a phase margin of o at a new gain cross-over frequency of rad/sec and a gain margin of db at a phase cross-over frequency of rad/sec (well within our specs!!!) z z Finally, using the Bilinear transformation, we find G lead (z) to be Glead () z = z. 9989z 2. a) Repeat problem 1 using lag compensation techniques Solution: If we let G s K Ts + 1 lag ()= c, again we see that we need a type 1 system to meet the steady-state error βt + 1 s specifications. Since we only have a type one system, we must increase the system type. This is accomplished by letting K c =K/s. We can find the appropriate value of K from K = 5 = lim sg G G( s) = K(8) / 4 = 2K. 2 Thus, K=5/2=2.5 and K c =2.5/s. Next, we must make a Bode plot of K c G zoh G(s)=. The Bode ss ( + 4)( 1+ s/ 2) plot is the same as shown in problem 1. Next, we need to find a new gain cross-over frequency where the system has sufficient phase angle to meet our phase margin specifications. That is, we need to find ω cgnew where K G G( jω ) = p. m fudge factor = = 14. From the Bode plot, we can see that c zoh cg desired new this occurs at ω cgnew = 3.85 rad/sec. At this frequency, the magnitude of K c G zoh G =19.42 db = Since our lag compensator will introduce an attenuation of 1/β at high frequencies, we set β = KG c zohg( jωcg new ) = Finally, to insure that we get the desired attenuation of 1/β, we must set the zero of our lag compensator (located at s=-1/t) a decade or so below ω cgnew. In other words, T = 1 / ω cg new = Hence, G s K Ts 1.. s lag () = c T ++ = We can check our answer by making a Bode plot of G lag G zoh G(s): β 1 s s + 1 s v desired s lag zoh

3 1 Gm=3.84 db, (w= 27.2) Pm=39.82 deg. (w=3.862) Gain db Frequency (rad/sec) Phase deg Frequency (rad/sec) As we can see, the compensated system has a phase margin of o at a new gain cross-over frequency of rad/sec and a gain margin of 3.84 db at a phase cross-over frequency of 27.2 rad/sec (well within our specs!!!) z z Finally, using the Bilinear transformation, we find G lag (z) to be Glag () z = z. 3456z b) Find the step response and the closed-loop Bode plot for both of your designs. What is the bandwidth of each closed-loop compensated system? Solution: The following is a closed-loop Bode plot of both the lead and lag compensated systems: 2 EE572 - HW#21: Closed-loop Lead and Lag Bode Plot -2 Lag Lead w (rad/sec) The bandwidth of the lag compensated system is about 6 rad/sec while the lead compensated system has a bandwidth of about 3 rad/sec. We can also find the closed-loop step response for both systems:

4 EE572 - HW#21: Lead and Lag Closed-loop Step Response Lead Lag time (sec) Note that the settling time for the lead step response is about.4 seconds while the lag settling time is nearly 3 seconds! c) Suppose our input is corrupted by periodic additive noise such that the actual input is w(t)+2.5cos2t. For both closed-loop systems, find the magnitude of the noise at the output (i.e., the magnitude of y 2.5cos2t ). (Hint:use your closed-loop Bode plot from part b)). Which system does a better job at attenuating the effects of the noise? Solution: From the closed-loop lead Bode plot, at a frequency of 2 rad/sec the magnitude of the lead system is about 3.2 db = 1.44 while the phase is about -75 o. Thus, the steady-state output of the lead system due to the noise will be about 3.6cos(2t -75 o ). From the closed-loop lag Bode plot, at a frequency of 2 rad/sec the magnitude of the lag system is about -25 db =.562 while the phase is about -175 o. Thus, the steadystate output of the lead system due to the noise will be about.146cos(2t -175 o ). Obviously, the lag system does a far better job of attenuating the additive input noise! APPENDIX - Matlab Code for HW#21» num=2;» den=conv([1 4 ],[1/2 1]) den = » w=logspace(-1,3,2);» [mg,ph]=bode(num,den,w);» semilogx(w,2*log1(mg))» grid;title('ee572 - HW#21: Bode Plot of KcGzohG(jw)')» margin(num,den)» semilogx(w,ph)» grid;xlabel('w (rad/sec)')» semilogx(w,ph,[.1 1],[-18-18],'--')» grid;xlabel('w (rad/sec)')» phim= phim =

5 32.85» phim*pi/ » alpha = (1-ans)/(1+ans) alpha =.2712» 1/sqrt(alpha) 1.923» 2*log1(ans) » margin(1/sqrt(alpha)*num,den)» T=1/(sqrt(alpha)*19.35) T =.992» alpha*t.269» numc=2.5*[t 1] numc = » denc=[alpha*t 1 ] denc = » [numd,dend]=bilinear(numc,denc,1/1) numd =

6 dend = » numd/numd(1) » denlead=conv([alpha*t 1],den) denlead = » numlead=conv([t 1],num) numlead = » margin(numlead,denlead)» [g,p,w,w2]=margin(mg,ph-45,w) g =.169 p = w = w2 = » beta=1/g beta =

7 » 2*log1(beta) » T=1/w T = » beta*t » numl=conv([t 1],num) numl = » denl=conv([beta*t 1],den) denl = » margin(numl,denl) Warning: Divide by zero» numc=2.5*[t 1] numc = » denc=[beta*t 1 ] denc = » [numd,dend]=bilinear(numc,denc,1/1) numd = dend =

8 » numd/numd(1) » numclag=2.5*[t 1] numclag = » denclag=[beta*t 1 ] denclag = » T=1/(sqrt(alpha)*19.35) T =.992» numclead=2.5*[t 1] numclead = » [numldcl,denldcl]=cloop(numlead,denlead) numldcl = denldcl = » [numlgcl,denlgcl]=cloop(numl,denl) numlgcl = denlgcl =

9 » [mgld,phld]=bode(numldcl,denldcl,w);» [mglg,phlg]=bode(numlgcl,denlgcl,w);» semilogx(w,2*log1(mgld),w,2*log1(mglg))» w=logspace(-1,3,2);» [mgld,phld]=bode(numldcl,denldcl,w);» [mglg,phlg]=bode(numlgcl,denlgcl,w);» semilogx(w,2*log1(mgld),w,2*log1(mglg))» grid;title(ee572 - HW#21: Closed-loop Lead and Lag Bode Plot')??? d;title(ee572 - HW# Improper function reference. A "," or ")" is expected.» grid;title('ee572 - HW#21: Closed-loop Lead and Lag Bode Plot')» semilogx(w,phld,w,phlg)» grid;xlabel('w (rad/sec)')» ylg=step(numldcl,denldcl);» yld=step(numldcl,denldcl);» ylg=step(numlgcl,denlgcl);» plot(yld)» t=[:.1:2];» yld=step(numldcl,denldcl,t);» ylg=step(numlgcl,denlgcl,t);» plot(t,ylg,t,yld)» t=[:.1:3];» yld=step(numldcl,denldcl,t);» ylg=step(numlgcl,denlgcl,t);» plot(t,ylg,t,yld,[ 3],[.98.98],'--',[ 3],[ ],'--')» grid;title('ee572 - HW#21: Lead and Lag Closed-loop Step Response')» xlabel('time (sec)')» semilogx(w,2*log1(mgld),w,2*log1(mglg))» grid» max(mgld) » 2*log1(ans) » semilogx(w,phld,w,phlg)» grid» semilogx(w,2*log1(mgld),w,2*log1(mglg))» grid» 1^(-25/2).562

10 » 1.44* » 1^(-25/2).562» ans*

EE 4343/ Control System Design Project LECTURE 10

EE 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 Phase-lead and Phase-lag compensators using