Solution: 7.9.6.7 For the system of Figure 1 wehave R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback G p = 1 (s +1) 2 and G c (s)= K c (s +1)(s+ z 1 ) s(s +20) : Part a. G c (s)G p (s) = K c (s + z 1 ) s(s +1)(s +20) Satisfaction of the angle condition is shown in Fig. 2 The angle equation that must be solved at s = ;8+j6is 6 s + z 1 ; 6 s; 6 s +1; 6 s +20=;180  or 6 (s + z 1 ) = 6 (s)+ 6 (s +1)+ 6 (s +20);180  = 143:1  +139:4  +26:6  ;180  = 129:1  Im(s) -20 -z 1 -1 -8 6 θ 3 θ 2 θ 1 α s + 20 s + 1 s s + z 1 Figure 2: Satisfaction of Angle Condition 1 Then z 1 =8; 6 tan(51  ) =3:125: Then, asshown in the gure, K c = jsjjs +1jjs +20j js +3:125j s=;8+j4 = p 100 p 85 p 180 p 59:77 = 160 Thus C c (s)= 160(s+ 1)(s+3:125) s(s +20) : The unit step response is c(t)=[1+1:333e ;5t +3:0732e ;8t cos(6t; 2:433)]1(t): The following MATLAB dialogue validates these results and plots the unit step response shown in Figure 3 EDU>gcgp = zpk([-3.125],[0 -1 -20],160) Zero/pole/gain: 160 (s+3.125) -------------- s(s+1) (s+20) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 160 (s+3.125) ----------------------- (s+5) (s^2 + 16s + 100) EDU>step(tc,4) EDU>print -deps sr7967.eps EDU> 2 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Unit step response of compensated system 3