Solution 4.6.4.12 The closed loop transfer function is K s(s+1) 1+ K s(s+1)(s+20) = K s 3 +21s 2 +20s + K : The characteristic equation is s 3 +21s 2 +(20+K)s + K =0: The MATLAB program K=2 p=[1 21 20 K] roots(p) K=4.87 p=[1 21 20 K] roots(p) K=10 p=[1 21 20 K] roots(p) K=50 p=[1 21 20 K] roots(p) K=[2 4.87 10 50] gh = zpk([],[0 -1 -20],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl46412.eps produces the output EDU>sm46412 K= 2 p= 1 1 21 20 2 ans = -20.0053 -0.8813 -0.1134 K= 4.8700 p= 1.0000 21.0000 20.0000 4.8700 ans = -20.0128 -0.5108 -0.4764 K= 10 p= 1 21 20 10 ans = -20.0262 2 -0.4869+ 0.5121i -0.4869- 0.5121i K= 50 p= 1 21 20 50 ans = -20.1298 -0.4351+ 1.5148i -0.4351- 1.5148i K= 2.0000 4.8700 10.0000 50.0000 Zero/pole/gain: 1 -------------- s(s+1) (s+20) R= -20.0053 -20.0128 -20.0262 -20.1298 -0.8813 -0.5108 -0.4869- 0.5121i -0.4351- 1.5148i -0.1134 -0.4764 -0.4869+ 0.5121i -0.4351+ 1.5148i K= 3 -25 -20 -15 -10 -5 0 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 1: Plot of solutions 2.0000 4.8700 10.0000 50.0000 EDU> The plot of the points is shown in Figure 1 For K =2,the MATLAB dialogue EDU>g = zpk([],[0 -1 -20],2) Zero/pole/gain: 2 -------------- s(s+1) (s+20) EDU>h = 1 4 Time (sec.) A mp li tu d e Step Response 0 5 10 15 20 25 30 35 40 45 50 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 2: Step Response for K =2 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 2 ------------------------------- (s+20.01) (s+0.8813) (s+0.1134) EDU>step(tc) EDU>print -deps sr46412a.eps EDU> produces the step response shown in Figure 2. For K =4:87, the MATLAB dialogue EDU>g = zpk([],[0 -1 -20],4.87) 5 Zero/pole/gain: 4.87 -------------- s(s+1) (s+20) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 4.87 ------------------------------- (s+20.01) (s+0.5108) (s+0.4764) EDU>step(tc) EDU>print -deps sr46412b.eps EDU> produces the step response shown in Figure 3. For K = 10, the MATLAB dialogue EDU>g = zpk([],[0 -1 -20],10) Zero/pole/gain: 10 -------------- s(s+1) (s+20) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) 6 Time (sec.) A mp li tu d e Step Response 0 2 4 6 8 10 12 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 3: Step Response for K =4:87 7 Time (sec.) A mp li tu d e Step Response 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Step Response for K =10 Zero/pole/gain: 10 ---------------------------------- (s+20.03) (s^2 + 0.9738s + 0.4993) EDU>step(tc) EDU>print -deps sr46412c.eps EDU> produces the step response shown in Figure 4. For K = 50, the MATLAB dialogue EDU>g = zpk([],[0 -1 -20],50) Zero/pole/gain: 50 -------------- s(s+1) (s+20) 8 Time (sec.) A mp li tu d e Step Response 0 2 4 6 8 10 12 14 0 0.5 1 1.5 Figure 5: Step Response for K =50 EDU>h =1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 50 --------------------------------- (s+20.13) (s^2 + 0.8702s + 2.484) EDU>step(tc) EDU>print -deps sr46412d.eps EDU> produces the step response shown in Figure 5. 9