Solution 4.6.4.7 The closed loop transfer function is T(c(s) = K (s+2)(s+4) 1+ K (s+2)(s+4) = K s 2 +6s +(8+K) : The characteristic equation is s 2 +6s+(8+K)=0: The MATLAB program K=0.5 p=[1 6 8+K] roots(p) K=1 p=[1 6 8+K] roots(p) K=2 p=[1 6 8+K] roots(p) K=10 p=[1 6 8+K] roots(p) K=[0.5 1 2 10] gh = zpk([],[-2 -4],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl4647.eps generates the following output EDU>sm4647 K= 0.50000000000000 1 p= 1.00000000000000 6.00000000000000 8.50000000000000 ans = -3.70710678118655 -2.29289321881345 K= 1 p= 1 6 9 ans = -3 -3 K= 2 p= 1 6 10 ans = 2 -3.00000000000000 + 1.00000000000000i -3.00000000000000 - 1.00000000000000i K= 10 p= 1 6 18 ans = -3.00000000000000 + 3.00000000000000i -3.00000000000000 - 3.00000000000000i K= 0.50000000000000 1.00000000000000 2.00000000000000 10.00000000000000 Zero/pole/gain: 1 ----------- (s+2) (s+4) R= Columns 1 through 2 -3.70710678118655 -3.00000000000000 -2.29289321881345 -3.00000000000000 Columns 3 through 4 3 -3.8 -3.6 -3.4 -3.2 -3 -2.8 -2.6 -2.4 -2.2 -3 -2 -1 0 1 2 3 Figure 1: Plot of solutions -3.00000000000000 + 1.00000000000000i -3.00000000000000 + 3.00000000000000i -3.00000000000000 - 1.00000000000000i -3.00000000000000 - 3.00000000000000i K= 0.50000000000000 1.00000000000000 2.00000000000000 10.00000000000000 EDU> The plot of the points is shown in Figure 1 For K =0:5, the closed loop transfer function is T c (s)= 0:5 (s+2:29289)(s+3:7071) : 4 Then the step response is C(s)= 0:5 s(s+2:29289)(s+3:7071) = D 1 s + D 2 s2:29289 + D 3 s +3:7071 : The residues A and M are determined bythefollowing MATLAB program, whichalso plots the step response, shown in Figure 2. K=0.5 p0 = 0 p1 = 2.29289 p2 = 3.7071 v0 = [1 p0] v1 = [1 p1] v2 = [1 p2] B=K A=conv(v0,v1) A=conv(A,v2) [R,P,K1] = residue(B,A) t=0 dt = 0.01 kount = 1 while t<6 c(kount) = R(3) + R(2)*exp(-p1*t) +R(1)*exp(-p2*t);; time(kount) = t;; t=t+dt;; kount = kount + 1;; end plot(time,c) print -deps sr4647a.eps The step response can also be generated by the MATLAB dialogue EDU>g = zpk([],[-2 -4],0.5) Zero/pole/gain: 0.5 ----------- (s+2) (s+4) EDU>tc = feedback(g,1) 5 0 1 2 3 4 5 6 0 0.01 0.02 0.03 0.04 0.05 0.06 Figure 2: Step Response for K =0:5 Zero/pole/gain: 0.5 ------------------- (s+3.707) (s+2.293) EDU>step(tc) EDU>print -deps sr4647aa.eps EDU> For K =1,therearetwo closed loop poles are at s = ;3. For K =2, the poles of the closed loop systme are at s = ;1j.Thus, the closed loop transfer function is T c (s)= 1 (s+3) 2 : Then the step response is C(s) = 1 s(s +3) 2 = D 1 s + D 2 s+3 + M  (s+3) 2 : 6 Then D 1 =  1 (s+3) 2  s=0 = 1 9 D 2 =  d ds  1 s  s=;3 =  ;1 s 2  s=;3 = ;1 9 D 3 =  1 s  s=;3 = ;1 3 The residues D 1 , D 2 and D 3 can also be determined by the MATLAB pro- gram K=1 p0 = 0 p1 = 3 p2 = 3 v0 = [1 p0] v1 = [1 p1] v2 = [1 p2] B=K A=conv(v0,v1) A=conv(A,v2) [R,P,K1] = residue(B,A) t=0 dt = 0.01 kount = 1 while t<6 c(kount) = R(3) + R(1)*exp(-p1*t) +R(2)*t*exp(-p1*t);; time(kount) = t;; t=t+dt;; kount = kount + 1;; 7 0 1 2 3 4 5 6 0 0.02 0.04 0.06 0.08 0.1 0.12 Figure 3: Step Response for K =1 end plot(time,c) print -deps sr4647b.eps whichalso plots the step response, shown in Figure 3. The step response can also be found using the MATLAB dialogue: EDU>clear all EDU>g = zpk([],[-2 -4],1) Zero/pole/gain: 1 ----------- (s+2) (s+4) EDU>tc = feedback(g,1) Zero/pole/gain: 1 ------- 8 (s+3)^2 EDU>step(tc) EDU>print -deps sr4647bb.eps EDU> For K =2,the closed loop transfer function is T c (s)= 2 (s+3;j)(s+3+j) : Then the step response is C(s)= 2 s(s +3;j)(s+3+j) = D 1 s + M s+3;j + C s +3+j : The residues D 1 and M are determined by the MATLAB program K=2 p0 = 0 p1 = 3 + j p2 = conj(p1) v0 = [1 p0] v1 = [1 p1] v2 = [1 p2] B=K A=conv(v0,v1) A=conv(A,v2) [R,P,K1] = residue(B,A) M=R(1) absm =abs(M) abs2m = 2*abs(M) angm =angle(M) sigma = real(p1) omega = imag(p1) t=0 dt = 0.01 kount = 1 while t<6 c(kount) = R(3) + +2*absm*exp(-sigma*t)*cos(omega*t + angm);; time(kount) = t;; t=t+dt;; 9 0 1 2 3 4 5 6 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 Figure 4: Step Response for K =2 kount = kount + 1;; end plot(time,c) whichalso plots the step response, shown in Figure 4. For K = 10, the closed loop transfer function is T c (s)= 10 (s+3;j3)(s+3+j3) : Then the step response is C(s)= 10 s(s+3;j3)(s+3+j3) = D 1 s + M s+3;3j + C s +3+3j : The residues D 1 and M are determined by the MATLAB program K=10 p0 = 0 p1 = 3 + j*3 p2 = conj(p1) 10 v0 = [1 p0] v1 = [1 p1] v2 = [1 p2] B=K A=conv(v0,v1) A=conv(A,v2) [R,P,K1] = residue(B,A) M=R(1) absm =abs(M) abs2m = 2*abs(M) angm =angle(M) sigma = real(p1) omega = imag(p1) t=0 dt = 0.01 kount = 1 while t<6 c(kount) = R(3) + +2*absm*exp(-sigma*t)*cos(omega*t + angm);; time(kount) = t;; t=t+dt;; kount = kount + 1;; end plot(time,c) print -deps sr4647d.eps whichalso plots the step response, shown in Figure 5. 11 0 1 2 3 4 5 6 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 Figure 5: Step Response for K =2 12