Solution: 7.9.2.8 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 (s +2)(s +6) The PI compensator can be put in the form: G c = k pi + k i s = k pi s + k i s = k pi (s + k i =k pi ) s : The constant k pi serves as the gain of the compensator and the ratio k i =k pi determines the zero of the compensator. In this problem we rstelect to make k i k pi =2;; and the zero of the PI compensator cancels the pole of the plantats = ;2 As as consequence G c (s)G p (s)= K pi s(s +6) : The root locus is shown in Figure 2 Toachieve t p < 1sec;; wemust have ! d >  1 =3:2 rad/sec: The gain to place the closed loop poles at s = ;3j! d is K pi = jsjjs +6jj s=;3+j! d = ! 2 n : Since we are allowed 20% overshoot, our next goal is to determine ! d . The following MATLAB dialogue shows that wecanlet  =0:45. 1 Im(s) Re(s) -3 ω s + 6 s -6 d Figure 2: Root Locus of Gain Compensated System EDU>x =-0.45*pi/sqrt(1-0.45^2) x= -1.583058779296969e+00 EDU>exp(x) ans = 2.053460284220887e-01 EDU>wn = 3/0.45 wn = 6.666666666666666e+00 EDU>wd = wn*sqrt(1-0.45^2) wd = 2 5.953523699830583e+00 EDU>k = 3^2 + wd^2 k= 4.444444444444444e+01 EDU>kv = k/6 kv = 7.407407407407406e+00 EDU> Weseethat wecan just meet the speci cation on steady state accuracy. Based on these calculations wechoose Then the compensator is G c (s)= 44:44(s+2) s : We can try to improveonthis result bymoving the zero of the PI com- pensator around, but as the following analysis shows it is hard to get much improvementonthe pole/zero cancellation approach. The following Matlab program nds a, and K v over a range of acceptable values of ! d . Because wedonot haveperfect cancellation, wehave increased  to 0.5. p1 = 2 p2 = 6 kplant=1 omegan = 6 zeta = 0.5 dw = 0.05 i=0 while omegan <10 s=-omegan*zeta + j*omegan*sqrt(1-zeta^2);; x=omegan*zeta;; i=i+1;; alpha(i) = angle(s) + angle(s + p1) +angle(s + p2) - pi;; a(i) = x + ( x / tan(alpha(i) ) );; 3 5 5.5 6 6.5 7 7.5 8 8.5 9 3 4 5 6 7 8 9 Figure 3: K v as a function of ! d xp(i) = omegan*sqrt(1-zeta^2);; k(i) = ( abs(s)*abs(s + p1)*abs(s + p2) )/ (kplant*abs(s + a(i)) );; kv(i) = (kplant*k(i)*a(i))/(p1*p2);; ess(i) = 1 / ( kv(i));; omegan = omegan + dw;; end plot(xp,kv) print -deps pi8c.eps Figure 3 shows K v as a function of ! d .Thefollowing Matlab dialogue shows determines the step response for a system with K v =8:24. EDU>xp(29) ans = 6.40858798800484 4 EDU>kv(29) ans = 8.24248135292352 EDU>wn = xp(29)/sqrt(1-zeta^2) wn = 7.40000000000000 EDU>a(29) ans = 1.96569850494106 EDU>k(29) ans = 50.31787732780915 EDU>g = zpk([-1.966],[0 -2 -6],50.32) Zero/pole/gain: 50.32 (s+1.966) --------------- s (s+2) (s+6) EDU>h=1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 5 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step Response of Compensated System 50.32 (s+1.966) ------------------------------ (s+1.96) (s^2 + 6.04s + 50.48) EDU>step(tc,3) EDU>print -deps pi8step.eps EDU> Then the compensator is G c (s)= 50:32(s+1:966) s ;; The step response is shown in Figure 4. 6