Solution: 7.9.1.2 For the system of Figure 1 wehave R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback G p = 10 (s + 3)(s +30) and G c = K(s + b) s +0:1 Figure 2 shows the root locus for the plant G p . The break out pointisfarto the left. Simple gain compensation yields a closed loop system with poles at s = ;16:5j16:5: The gain required to place the closed loop poles at this location is K = 13:5 2 +13:5 2 10 =45:45: Then K p =lim s!0 1045:45 (s +3)(s +30) =5:05;; and e ss = 1 1+5:05 =0:165: Thus wedonot meet the speci cation on steady state error. If we add the compensator, then wehavethesituation shown in Figure 3. For the root locus to be on the desired line of constantdamping ratio, wemust have ; 1 ; 2 ; 3 = ;180  : The following program searches along the line of constantdamping ratio  =1= p 2, computing a, K, K p ,ande ss . 1 Im(s) Re(s) -30 -3 -16.5 Figure 2: Root Locus for Gain Compensation of G p Re(s) Im(s) -3 3 XX XO 1 2 3 Θ Θ Θ α -30 -0.1-a Figure 3: Satisfaction of Angle Condition Along Line of ConstantDamping Ratio 2 x=2.0 dx = 0.1 i=0 while x < 16 s=-x + j*x;; i=i+1;; alpha(i) = angle(s + 0.1) + angle(s + 3) +angle(s + 30) - pi;; a(i) =x+(x/tan(alpha(i) ) );; xp(i) = x;; k(i) = ( abs(s + 0.1)*abs(s+3)*abs(s + 30) )/ (10*abs(s + a(i)) );; kp(i) = (10*k(i)*a(i))/(0.1*30*3);; ess(i) = 1/(1+kp(i));; x=x+dx;; end figure(1) subplot(2,1,1), plot(xp,ess) subplot(2,1,2), plot(xp,k,'r:',xp,kp,'k') print -deps lag2a.eps figure(2) plotyy(xp,ess,'k',xp,a,'k') Figure 4 is a graph of e ss and a versus x the magnitude of the real and imaginaryparts of the complexpoles. The step response is shown in Figure 5. As can be seen, the steady state error is minimized for 10 <x<12: Weseethat wecanget one percent steady state error with a very modest gain. For instance, if weplacethe dominantpolesat s = ;4j4 Then a =5:693 and K =13:95 Thus, a very good design is G c (s)= 13:95(s +5:693) s +0:1 : The step response is shown in Figure 5. 3 2 4 6 8 10 12 14 16 0 0.01 0.02 0.03 0.04 0 2 4 6 8 e ss e ss a a x Figure 4: e ss and a as a function of jRe(x)j 4 Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 5: Step Response of Compensated System 5