Solution: 7.9.1.5 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 (s + 1)(s +10) and G c = K(s + a) s +0:1 Figure 2 shows the root locus for the plant G p .Thebreak out pointisfar to the left. Simple gain compensation yields a closed loop system with poles at s = ;5:5j4:125 The gain required to place the closed loop poles at this location is K = 4:5 2 +4:125 2 1 =37:266: Then K p =lim s!0 37:266 (s + 1)(s+10) =3:73;; and e ss = 1 1+3:73 =0:2116 If weaddthecompensator, then wehave the situation shown in Figure 3. For the root locus to be on the desired line of constant damping ratio, we must have ; 1 ; 2 ; 3 = ;180  : The following program searches along the line of constant damping ratio  =1= p 2, computing a, K, K p , and e ss . 1 Im(s) Re(s) -10 -5.5 -1 ζ = 0.8 Figure 2: Root Locus For Gain Compensation Re(s) Im(s) -1 XX XO 12 3 Θ ΘΘ α -10 -0.1 -a ζ = 0.8 Figure 3: Angle Contributions Along  =0:8 2 p1 = 1 p2 = 10 plag=0.1 kplant=1 omegan = 1 dw = 0.05 i=0 while omegan <6 s=-omegan*0.8 + j*omegan*0.6;; x=omegan*0.8;; i=i+1;; alpha(i) = angle(s + plag) + angle(s + p1) +angle(s + p2) - pi;; a(i) = x + ( x / tan(alpha(i) ) );; xp(i) = x;; k(i) = ( abs(s + plag)*abs(s + p1)*abs(s + p2) )/ (kplant*abs(s + a(i)) );; kp(i) = (kplant*k(i)*a(i))/(plag*p1*p2);; ess(i) = 1 / ( 1 + kp(i));; omegan = omegan + dw;; end subplot(2,1,1), plot(xp,ess) subplot(2,1,2), plot(xp,k,'r:',xp,kp,'k') print -deps lag2a.eps Figure 4 is a graph of e ss , K p ,andK versus x the magnitude of the real part of the complex poles. Figure 5 shows the zero location (absolute value of zero) as a function of the real part (absolute value) of the dominantpoles. For the three zero locations speci ed the real part is between 3.7 and 4.6. Exact correlations are: a = 0:5 K =34 x =4:65 a = 1:0 K =35:6 x =4:28 a = 1:5 K =34:5 x =3:75 Then for a =0:5, G c = 34(s +0:5) s +0:1 ;; while for a =1:0 G c = 34:6(s+1) s +0:1 ;; and for a =1:5 G c = 34:5(s+1:5) s +0:1 : 3 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 10 20 30 40 50 60 e ss K p K Figure 4: e ss and K p and K as a function of x 4 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 a x Figure 5: a as a function of x 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Time in Seconds a = 0.5 a = 1.0 a = 1.5 Step Response Figure 6: Step Responses For Three Values of a The step responses for the three choices of a are shown in Figure 6. The best choice is clearly a =1:0. 6