Solution: 7.9.3.7 Σ G C G P + R C Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 G p = 10 s(s+1) : and G c = K(s+1) s +6 : The MATLAB program: % % K=16 % k=16 b=k v1 = [1,0] v2 = [1,0] v3 = [1,6] %v4 = [1,18] denom = conv(v2,v3) %denom = conv(denom,v4) top = [0,0,b] bot = top + denom bot = conv(bot,v1) roots(bot) a=bot [w,s,v] = residue(b,a) % % K=18 % k=18 b=k v1 = [1,0] v2 = [1,0] v3 = [1,6] %v4 = [1,18] denom = conv(v2,v3) 1 %denom = conv(denom,v4) top = [0,0,b] bot = top + denom bot = conv(bot,v1) roots(bot) a=bot [w,s,v] = residue(b,a) % % K=20 % k=20 b=k v1 = [1,0] v2 = [1,0] v3 = [1,6] %v4 = [1,18] denom = conv(v2,v3) %denom = conv(denom,v4) top = [0,0,b] bot = top + denom bot = conv(bot,v1) roots(bot) a=bot [w,s,v] = residue(b,a) generates the output: k= 16 b= 16 v1 = 1 0 v2 = 1 0 2 v3 = 1 6 denom = 1 6 0 top = 0 0 16 bot = 1 6 16 bot = 1 6 16 0 ans = 0 -3.00000000000000 + 2.64575131106459i -3.00000000000000 - 2.64575131106459i a= 1 6 16 0 w= -0.50000000000000 + 0.56694670951384i -0.50000000000000 - 0.56694670951384i 1.00000000000000 3 s= -3.00000000000000 + 2.64575131106459i -3.00000000000000 - 2.64575131106459i 0 v= [] k= 18 b= 18 v1 = 1 0 v2 = 1 0 v3 = 1 6 denom = 1 6 0 top = 0 0 18 4 bot = 1 6 18 bot = 1 6 18 0 ans = 0 -3.00000000000000 + 3.00000000000000i -3.00000000000000 - 3.00000000000000i a= 1 6 18 0 w= -0.50000000000000 + 0.50000000000000i -0.50000000000000 - 0.50000000000000i 1.00000000000000 s= -3.00000000000000 + 3.00000000000000i -3.00000000000000 - 3.00000000000000i 0 v= [] k= 5 20 b= 20 v1 = 1 0 v2 = 1 0 v3 = 1 6 denom = 1 6 0 top = 0 0 20 bot = 1 6 20 bot = 1 6 20 0 ans = 6 0 -3.00000000000000 + 3.31662479035540i -3.00000000000000 - 3.31662479035540i a= 1 6 20 0 w= -0.50000000000000 + 0.45226701686665i -0.50000000000000 - 0.45226701686665i 1.00000000000000 s= -3.00000000000000 + 3.31662479035540i -3.00000000000000 - 3.31662479035540i 0 v= [] EDU> All that is left is to presentthedata. K=16 T c (s)= 16 (s+3;j2:646)(s+3+j2:646) Then c(t)=[1+2jMje ;2:65t cos(2:88t+ )]1(t): M =  (s+3; j2:646) 16 (s+3; j2:646)(s+3+j2:646)  s=;3+j2:646 = ;0:5+j0:567 = 0:756 6 2:293 7 Then c(t)=[1+1:512e ;3t cos(2:646t+2:293)]1(t): K=18 T c (s)= 18 (s+3;j3)(s+3+j3) Then c(t)=[1+2jMje ;3t cos(3t+)]1(t): M =  (s+3; j3) 18 s(s+3;j3)(s+3+j3)  s=;3+j3 = ;0:5+j0:5 = 0:707 6 ;2:356 Then c(t)=[1+1:414e ;3t cos(3t+2:356)]1(t): K=20 T c (s)= 20 (s+3;j3:317)(s+3+j3:317) Then c(t)=[1+2jMje ;3t cos(3:317t+ )]1(t): M =  (s+3; j3:317) 20 s(s+3; j3:317)(s+3+j3:317)  s=;3+j3:317 = ;0:5+j0:4523 = 0:674 6 2:406 Then c(t)=[1+1:348e ;3t cos(3:317t+2:406)]1(t): Fig. 2 shows the three responses. As can be seen they are reasonably close. 8 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.00 1.00 2.00 3.00 Time in Seconds K = 2.0 K = 1.8 K = 1.6 Figure 2: Time Response for Nominal Gain 10% 9