Solution: 7.9.3.5 For the system of Figure 1 Let Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback G p = 10 s(s+1) : and G c = K c (s+3) s+18 : K=7 The following MATLAB programwill generate the step response for K c =7. g=zpk([] ,[0 -1],10) gc = zpk([-3],[-18],7) gcgp = series(g,gc) h=1 tc = feedback(gcgp,h) step(tc,3) The output of the MATLAB program is Zero/pole/gain: 10 ------- s (s+1) Zero/pole/gain: 7 (s+3) ------- (s+18) Zero/pole/gain: 70 (s+3) -------------- s (s+18) (s+1) 1 h= 1 Zero/pole/gain: 70 (s+3) -------------------------------- (s+13.69) (s^2 + 5.306s + 15.34) EDU> Thus T c (s)= 70(s+3) (s+13:69)(s+2:65;j2:88)(s+2:65+ j2:88) Then c(t)=[1+Ae ;13:69t +2jMje ;2:65t cos(2:88t+ )]1(t): Wenow nd the residues of the partial fraction expansion. A =  (s+13:69) 70(s+3) s(s+13:69)(s+2:65;j2:88)(s+2:65+ j2:88)  s=;13:69 = 0:42 M =  (s +2:65;j2:88) 70(s+3) s(s+13:69)(s+2:65;j2:88)(s+2:65+ j2:88)  s=;2:65+j2:88 = ;0:70995;j0:34416 = 0:789 6 ;2:69 Then c(t)=[1+0:42e ;13:69t +1:578e ;2:65t cos(2:88t;2:69)]1(t): K=7.8 The procedures are the same. T c (s)= 78(s+3) (s+13)(s+3;j3)(s+3+j3) 2 Then c(t)=[1+Ae ;13t +2jMje ;3t cos(3t+ )]1(t): A =  (s+13) 78(s+3) s(s+13)(s+3;j3)(s+3+j3)  s=;13 = 0:55 M =  (s+3; j3) 78(s+3) s(s+13)(s+3;j3)(s+3+j3)  s=;3+j3 = ;0:77523; j0:4174 = 0:8805 6 ;2:648 Then c(t)=[1+0:55e ;13t +1:76e ;3t cos(3t; 2:648)]1(t): K=8.6 T c (s)= 86(s+3) (s+12:22)(s+3:392; j3:1)(s+3:392+j3:1) Then c(t)=[1+Ae ;12:22t +2jMje ;3:392t cos(3:1t+)]1(t): A =  (s+12:22) 86(s+3) s(s+12:22)(s+3:392; j3:1)(s+3:392+ j3:1)  s=;12:22 = 0:742 M =  (s +3:392;j3:1) 86(s+3) s(s+12:22)(s+3:392; j3:1)(s+3:392+ j3:1)  s=;3:392+j3:1 = ;0:8709; j0:5085 = 1:008 6 ;2:613 Then c(t)=[1+0:742e ;12:22t +2:017e ;3:392t cos(3:1t; 2:613)]1(t): Fig. 2 shows the three responses. As can be seen they are very close. AMATLAB program to nd all the residues and plot the responses follows. 3 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 123 Time in Seconds K = 8.6 K = 7.8 K = 7 Figure 2: Time Response for Nominal Gain 10% g=zpk([] ,[0 -1],10) gc = zpk([-3],[-18],7) gcgp = series(g,gc) h=1 tc = feedback(gcgp,h) step(tc,3) a=3 Kc = 7 Kp = 10 K=Kp*Kc B=K*[1 a] B=[00B] c=[10] d=[11] e=[118] A=conv(c,d) A=conv(A,e) roots(A) A=A+B roots(A) A=conv(A,c) roots(A) roots(B) [R,P,K] = residue(B,A) 4 realm = real(R(2)) imagm = imag(R(2)) m=realm + j*imagm absm = abs(m) absm2 =2*absm angm = angle(m) t0=1 + R(1) + absm2*cos(angm) Kc = 7.8 Kp = 10 K=Kp*Kc B=K*[1 a] B=[00B] c=[10] d=[11] e=[118] A=conv(c,d) A=conv(A,e) roots(A) A=A+B roots(A) A=conv(A,c) roots(A) roots(B) [R,P,K] = residue(B,A) realm = real(R(2)) imagm = imag(R(2)) m=realm + j*imagm absm = abs(m) absm2 =2*absm angm = angle(m) t0=1 + R(1) + absm2*cos(angm) Kc = 8.6 Kp = 10 K=Kp*Kc B=K*[1 a] B=[00B] c=[10] d=[11] e=[118] A=conv(c,d) A=conv(A,e) roots(A) A=A+B roots(A) A=conv(A,c) 5 roots(A) roots(B) [R,P,K] = residue(B,A) realm = real(R(2)) imagm = imag(R(2)) m=realm + j*imagm absm = abs(m) absm2 =2*absm angm = angle(m) t0=1 + R(1) + absm2*cos(angm) 6