Solution 4.6.4.1 The closed loop transfer function is K(s+1) s 2 1+ K(s+1) s 2 = K(s+1) s 2 + Ks+ K : The characteristic equation is s 2 +Ks+K =0: The MATLAB program K=2 p=[1 K K] roots(p) K=4 p=[1 K K] roots(p) K=20 p=[1 K K] roots(p) K=[2 40 20] gh = zpk([],[-1 -10],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl4641.eps generates the following output EDU>sm4641 K= 2 p= 1 2 2 1 ans = -1.0000+ 1.0000i -1.0000- 1.0000i K= 4 p= 1 4 4 ans = -2 -2 K= 20 p= 1 20 20 ans = -18.9443 -1.0557 K= 2 2 40 20 Zero/pole/gain: 1 ------------ (s+1) (s+10) R= -9.7720 -5.5000+ 4.4441i -6.0000 -1.2280 -5.5000- 4.4441i -5.0000 K= 2 40 20 EDU> EDU> The plot of the points is shown in Figure 1 For K =2,the poles of the closed loop systme are at s = ;1j.Thus, the closed loop transfer function is T c (s)= 2(s+1) (s+1;j)(s+1+j) : Then the step response is C(s)= 2(s+1) s(s+1;j)(s+1+j) = A s + M s+1;j + M  s+1+j : The residues A and M are determined bythefollowing MATLAB program, whichalso plots the step response, shown in Figure 2. K=2;; p0 = [1 0];; p1 = [1 1+j*1];; 3 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure 1: Plot of solutions p2 = [1 1-j*1];; B=2*[1 1] A=conv(p1,p2);; A=conv(A,p0) [R,P,K] = residue(B,A) M=R(1) absm =abs(M) abs2m = 2*abs(M) angm =angle(M) test = 1 + 2*abs(M)*cos(angle(M)) t=0 dt = 0.1 kount = 1 while t<3 c(kount) = 1+0.02728*exp(-1.1436*t) +2*absm*exp(-2.928*t)*cos(3*t + angm);; time(kount) = t;; t=t+dt;; kount = kount + 1;; 4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step Response for K =2 end plot(time,c) print -deps sr4641a.eps For K =4,therearetwo closed loop poles are at s = ;2. For K =2, the poles of the closed loop systme are at s = ;1j.Thus, the closed loop transfer function is T c (s)= 4(s+1) (s =2) 2 : Then the step response is C(s)= 4(s+1) s(s+2) 2 = A s + B s+2 + C (s+2) 2 : Then A =  4(s+1) (s+2) 2  s=0 = 1 5 B =  d ds  4(s+1) s  s=;2 =  d ds  ;4(s+1) s 2 + 4 s  s=;2 =  ;4s;4+4s (s+2) 2  s=;2 = ;1 C =  4(s+1) s  s=;2 = 2 The residues A, B, and C are determined bytheMATLAB program K=4;; p0 = [1 0] p1 = [1 2] p2 = [1 2] B=K*[1 1] A=conv(p1,p2) A=conv(A,p0) [R,P,K] = residue(B,A) A=R(3) B=R(1) C= R(2) t=0 dt = 0.1 kount = 1 while t<3 c(kount) = 1+B*exp(-2*t) +C*t*exp(-2*t);; time(kount) = t;; t=t+dt;; kount = kount + 1;; end plot(time,c) print -deps sr4641b.eps whichalso plots the step response, shown in Figure 3. For K =20, there 6 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response for K =4 are two closed loop poles are at s = ;1:0557 and s = ;18:9443 Thus, the closed loop transfer function is T c (s)= 20(s+1) (s+1:0557)(s+18:9443) : Then the step response is C(s)= 20(s+1) s(s+1:0557)(s+18:9443) = A s + B s+1:0557 + C s+18:9443 : The residues A, B, and C are determined by theMATLAB program K=20;; p0 = [1 0] p1 = [1 1.0557] p2 = [1 18.9443] B=K*[1 1] A=conv(p1,p2) A=conv(A,p0) 7 0 0.5 1 1.5 2 2.5 3 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 Figure 4: Step Response for K =4 [R,P,K] = residue(B,A) A=R(3) B=R(2) C= R(1) t=linspace(0,3,100);; A=ones(100,1) c=A+(B*exp(-1.0557*t))'+(C*exp(-18.9443*t))' plot(t,c) print -deps sr4641c.eps whichalso plots the step response, shown in Figure 4. 8