Solution 7.9.6.6 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 = ;5+j4is 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  = 141:3  +135  +14:9  ;180  = 111:3  Then z 1 =5; 4 tan(68:7  ) =5;1:55 = 3:44: Then, asshown in the gure, K c = jsjjs +1jjs +20j js +3:44j s=;5+j4 = p 41 p 32 p 241 p 18:425 = 131 1 Im(s) -20 -z 1 -1 -5 4 θ 3 θ 2 θ 1 α s + 20 s + 1 s s + z 1 Figure 2: Satisfaction of Angle Condition Thus G c (s)= 131(s+1)(s+3:44) s(s +20) : The unit step response is c(t)=[1+1:731e ;11t +3:0445e ;5t cos(4t +2:6836)]1(t): The following MATLAB dialogue generates the unit step response shown in Figure 3. EDU>gcgp = zpk([-3.44],[0 -1 -20],131) Zero/pole/gain: 131 (s+3.44) -------------- s(s+1) (s+20) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 131 (s+3.44) -------------------------------- (s+10.99) (s^2 + 10.01s + 40.99) EDU>step(tc,4) 2 EDU>print -deps sr7966.eps EDU> 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: Step response of compensated system 3