Solution: 7.9.6.5 For the system of Figure 1 wehave Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback G p (s)= 1 (s +1)(s+2)(s+20) and G c (s)= K c (s + 2)(s + z 1 ) s(s + 40) : Part a. Wehave G c G p = K c (s + z 1 ) s(s + 1)(s +20)(s+40) Wehaveto nd the crossing pointoftheline of constantdamping furthest from the origin. Todosowe use the MATLAB dialogue EDU>gcgp = zpk([-1.5],[0 -1 -20 -40],1) Zero/pole/gain: (s+1.5) --------------------- s(s+1) (s+20) (s+40) EDU>rlocus(gcgp) EDU>print -deps rl7965a.eps EDU> to generate the root locus shown in Figure 2. Then wehaveto ndthepoint where the equation ; 1 ; 2 ;  3 ; 4 = ;180  is satis ed alomg the raycorresponding to a damping ratio of  =1= p 2, as shown in Fig. 3. Wedothisbytrial and error, as summarized in the Table 1 1 -50 -40 -30 -20 -10 0 10 -25 -20 -15 -10 -5 0 5 10 15 20 25 Real Axis Imag Axis x x x Figure 2: Root locus with zero as s = ;1:5 Im(s) -40 -20 -1.5 -1 θ 2 θ 1θ 3 θ 4 α Re(s) Figure 3: Satisfaction of Angle Condition alomg Line of ConstantDamping 2 s -6.5 + j6.5 -7 + j7 -7.5 + j7.5 -7.33 + j7.33 6 G c G p ;174:4  ;177:7  ;181:2  ;180  Table 1: Thus the maximum damped frequency is 7.33 rad/s. The gain required to place the closed loop poles at s = ;7:33 j7:33 is K c = jsjjs +20jjs+40j js +1:5j s=;7:33+j7:33 =5253:6: The following MATLAB dialogue nds and save the step response shown in Figure 4. EDU>gcgp = zpk([-1.5],[0 -1 -20 -40],5253.6) Zero/pole/gain: 5253.6 (s+1.5) --------------------- s(s+1) (s+20) (s+40) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 5253.6 (s+1.5) ----------------------------------------- (s+44.7) (s+1.641) (s^2 + 14.66s + 107.4) EDU>step(tc,4) EDU>print -deps sr7965a.eps EDU> The unit step response is c(t)=[1+0:1218e ;1:641t ;0:0813e ;44:7t +1:8328e 07:33t cos(7:33t;2:17455)]1(t): part(b) The same as part (a) except wemove the free zero close to the origin at s = ;0:1. The following MATLAB dialogue creates the root loucs shown in Figure 5 Then the root locus crosses the line of constant damping for 3 Time (sec.) Step Response 0.5 1 1.5 2 2.5 3 3.5 4 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step response with zero at s = ;1:5 -50 -40 -30 -20 -10 0 10 -25 -20 -15 -10 -5 0 5 10 15 20 25 Real Axis Imag Axis x x Figure 5: Root locus with zero as s = ;1:5 4 s = ;8:1+j8:1foragain of K c =5134:8. The following MATLAB dialogue generates the step response shown in Figure 6. EDU>gcgp = zpk([-0.1],[0 -1 -20 -40],5134.8) Zero/pole/gain: 5134.8 (s+0.1) --------------------- s(s+1) (s+20) (s+40) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 5134.8 (s+0.1) ------------------------------------------ (s+44.74) (s+0.08763) (s^2 + 16.18s + 131) EDU>step(tc,40) EDU>print -deps sr7965b.eps EDU> The time response in this case is c(t)=[1;0:1253e ;0:0876t ;0:08146e ;44:74t +1:475e 8:1t cos(8:1t;2:13852)]1(t): The responses are remarably di erent, emphasizing the point that puttting the zero of the PI compensator close to the origin is not generally a good choice. 5 Time (sec.) Step Response 5 10 15 20 25 30 35 40 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 6: Root locus with zero as s = ;0:1 6