Solution 4.6.4.3 The closed loop transfer function is K(s+1) s 2 (s+10) 1+ K(s+1) s 2 (s+10) = K(s+1) s 3 +10s 2 +Ks+K : The characteristic equation is s 3 +10s 2 +Ks+K =0: The MATLAB program K=20 p=[1 10 K K] roots(p) K=32 p=[1 10 K K] roots(p) K=50 p=[1 10 K K] roots(p) K=[20 32 50] gh = zpk([-1],[0 0 -10],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl4643.eps produces the output EDU>sm4643 K= 20 p= 1 10 20 20 1 ans = -7.75310919791344 -1.12344540104328 + 1.14781561520580i -1.12344540104328 - 1.14781561520580i K= 32 p= 1 10 32 32 ans = -4.00000000000000 + 0.00000012019357i -4.00000000000000 - 0.00000012019357i -2.00000000000000 K= 50 p= 1 10 50 50 ans = -4.35509743609111 + 4.44956990188039i -4.35509743609111 - 4.44956990188039i -1.28980512781779 2 K= 20 32 50 Zero/pole/gain: (s+1) ---------- s^2 (s+10) R= Columns 1 through 2 -7.75310919791344 -4.00000000000000 + 0.00000012019357i -1.12344540104328 + 1.14781561520580i -2.00000000000000 -1.12344540104328 - 1.14781561520580i -4.00000000000000 - 0.00000012019357i Column 3 -4.35509743609111 + 4.44956990188039i -1.28980512781779 -4.35509743609111 - 4.44956990188039i K= 20 32 50 EDU> The plot of the points is shown in Figure 1 For K = 20, the poles of the closed loop systme are at s = ;1j.Thus, 3 -8 -7 -6 -5 -4 -3 -2 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 Figure 1: Plot of solutions the closed loop transfer function is T c (s)= 20(s+1) s+7:7531)(s+1:1234454;j1:12782)(s+1:1234454;j1:12782) : Then the step response is C(s) = 20(s+1) s(s+7:7531)(s+1:1234454;j1:12782)(s+1:1234454;j1:12782) = D 1 s + D 2 s =7:7531 + M s +1:1234454;j1:12782 + M  s +1:1234454+j1:12782 : The residues D 1 , D 2 and M are determined bythefollowing MATLAB program, whichalso plots the step response, shown in Figure 2. K=20;; p0 = [1 0] p1 = [1 7.753109] p2 = [1 1.1234454+j*1.1278156152 ] p3 = conj(p2) B=K*[1 1] 4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step Response for K =2 A=conv(p0,p1) A=conv(A,p2) A=conv(A,p3) [R,P,K1] = residue(B,A) M=R(3) absm =abs(M) abs2m = 2*abs(M) angm =angle(M) t=0 dt = 0.025 kount = 1 while t < 5 c(kount) = R(4)+ R(1)*exp(-7.7531*t) + abs2m*exp(-1.123445*t) *cos(1.12781*t + angm);; time(kount) = t;; t=t+dt;; kount = kount + 1;; end plot(time,c) print -deps sr4643a.eps The step response can also be generated bytheMATLAB dialogue: 5 EDU>g = zpk([-1],[0 0 -10],20) Zero/pole/gain: 20 (s+1) ---------- s^2 (s+10) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 20 (s+1) ------------------------------- (s+7.753) (s^2 + 2.247s + 2.58) EDU>step(tc) EDU>print -deps sr4643aa.eps EDU> The response, shown in Figure 3 is the same as we obtained by partial fraction expansion. For K =32, there are twoclosed loop poles are at s = ;2. Thus, the closed loop transfer function is T c (s)= 32(s+1) (s+2)(s+4) 2 : Then the step response is C(s) = 32(s+1) s(s+2)(s+4) 2 = D 1 s + D 2 s +4 + D 3 (s+4) 2 + D 4 s+2 : Then D 1 =  32(s+1) (s+ 2)(s+4) 2  s=0 6 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step Response for K =20generated byMATLAB 7 = 1 D 2 =  d ds  32(s+1) s(s+2)  s=;4 =  d ds  ;32(s+1)(2s+2) [s(s+ 2)] 2 + 32 s(s+2)  s=;4 = " ;64(s 2 +2s+1)+32s 2 +64s [s(s+ 2)] 2 # s=;4 = " ;32s 2 ; 64s;64 [s(s+2)] 2 # s=;4 =  ;32(16)+ 64(4);64 [;4(;2)] 2  s=;4 = ;5 D 3 =  32(s+1) s(s+2)  s=;4 = ;12 D 4 =  32(s+1) s(s +4) 2  s=;2 = 4 The residues D 1 , D 2 , D 3 ,and D 4 can also be determined by the MATLAB program K=32;; p0 = 0 p1 = 2 p2 = 4 p3 = 4 v0 = [1 p0] v1 = [1 p1] v2 = [1 p2] v3 = [1 p3] B=K*[1 1] A=conv(v0,v1) A=conv(A,v2) 8 0 1 2 3 4 5 6 7 8 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step Response for K =32 A=conv(A,v3) [R,P,K1] = residue(B,A) t=0 dt = 0.01 kount = 1 while t<8 c(kount) = R(4) + R(3)*exp(-p1*t) +R(1)*exp(-p2*t) + R(2)*t*exp(-p2*t);; time(kount) = t;; t=t+dt;; kount = kount + 1;; end plot(time,c) print -deps sr4643b.eps whichalso plots the step response, shown in Figure 4. For K = 50, there are twoclosed loop poles are at s = ;1:0557 and s = ;18:9443 Thus, the closed loop transfer function is T c (s)= 32(s+1) (s+1:2898)(s+4:3551;j 4:44957)(s+4:3551+j 4:44957) : 9 Then the step response is C(s) = 32(s+1) s(s+1:2898)(s+4:3551;j4:44957)(s+4:3551+j4:44957) = D 1 s + D 2 s+1:2898 + M s +4:3551;j4:44957 + M  s+4:3551+ j4:44957 : The residues D 1 , D 2 , and M are determined by the MATLAB program K=50;; p0 = 0 p1 = 1.289805 p2 = 4.3551- j*4.44957 p3 = conj(p2) v0 = [1 p0] v1 = [1 p1] v2 = [1 p2] v3 = [1 p3] B=K*[1 1] A=conv(v0,v1) A=conv(A,v2) A=conv(A,v3) [R,P,K1] = residue(B,A) M=R(2) absm =abs(M) abs2m = 2*abs(M) angm =angle(M) sigma = real(p2) omega = abs(imag(p2)) t=0 dt = 0.01 kount = 1 while t<6 c(kount) = R(4) + R(3)*exp(-p1*t) + abs2m*exp(-sigma*t)*cos(omega*t+angm);; time(kount) = t;; t=t+dt;; kount = kount + 1;; end plot(time,c) print -deps sr4643c.eps whichalso plots the step response, shown in Figure 5. 10 0 1 2 3 4 5 6 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 5: Step Response for K =20 11