Solution 4.6.4.10 The closed loop transfer function is K (s+1)(s+4)(s+10) 1+ K = K s 3 +15s 2 +54s +(K + 40) : The characteristic equation is s 3 +15s 2 +54s +(K +40)=0: The MATLAB program p1 =1 p2 =4 p3 =10 v1 = [1 p1] v2 = [1 p2] v3 =[1 p3] p=conv(v1,v2) p=conv(p,v3) K=10 p=[1 15 54 K+40] roots(p) K=17.04 p=[1 15 54 K+40] roots(p) K=50 p=[1 15 54 K+40] roots(p) K=100 p=[1 15 54 K+40] roots(p) K=[10 17.04 50 100] gh = zpk([],[-1 -4 -10],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl46410.eps produces the output 1 EDU>sm46410a p1 = 1 p2 = 4 p3 = 10 v1 = 1 1 v2 = 1 4 v3 = 1 10 p= 1 5 4 p= 1 15 54 40 2 K= 10 p= 1 15 54 50 ans = -10.1764 -3.3622 -1.4613 K= 17.0400 p= 1.0000 15.0000 54.0000 57.0400 ans = -10.2915 -2.3623 -2.3462 K= 50 3 p= 1 15 54 90 ans = -10.7582 -2.1209+ 1.9666i -2.1209- 1.9666i K= 100 p= 1 15 54 140 ans = -11.3229 -1.8386+ 2.9973i -1.8386- 2.9973i K= 10.0000 17.0400 50.0000 100.0000 Zero/pole/gain: 1 ------------------ (s+1) (s+4) (s+10) 4 -12 -10 -8 -6 -4 -2 0 -3 -2 -1 0 1 2 3 Figure 1: Plot of solutions R= -10.1764 -10.2915 -10.7582 -11.3229 -3.3622 -2.3623 -2.1209- 1.9666i -1.8386- 2.9973i -1.4613 -2.3462 -2.1209+ 1.9666i -1.8386+ 2.9973i K= 10.0000 17.0400 50.0000 100.0000 EDU> The plot of the points is shown in Figure 1 For K = 10, the MATLAB dialogue 5 EDU>g = zpk([],[-1 -4 -10],10) Zero/pole/gain: 10 ------------------ (s+1) (s+4) (s+10) EDU>h =1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 10 ----------------------------- (s+10.18) (s+3.362) (s+1.461) EDU>step(tc) EDU>print sr46410a.eps EDU> produces the step response shown in Figure 2. For K =17:04, the MATLAB dialogue EDU>g = zpk([],[-1 -4 -10],17.04) Zero/pole/gain: 17.04 ------------------ (s+1) (s+4) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 17.04 ----------------------------- (s+10.29) (s+2.362) (s+2.346) 6 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.05 0.1 0.15 0.2 0.25 Figure 2: Step Response for K =10 7 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Figure 3: Step Response for K =4:87 EDU>step(tc) EDU>print -deps sr46410b.eps produces the step response shown in Figure 3. For K = 50, the MATLAB dialogue EDU>g = zpk([],[-1 -4 -10],50) Zero/pole/gain: 50 ------------------ (s+1) (s+4) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 50 -------------------------------- (s+10.76) (s^2 + 4.242s + 8.366) 8 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Figure 4: Step Response for K =10 EDU>step(tc) EDU>print -deps sr46410c.eps EDU> produces the step response shown in Figure 4. For K = 100, the MATLAB dialogue EDU>g = zpk([],[-1 -4 -10],100) Zero/pole/gain: 100 ------------------ (s+1) (s+4) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 100 9 Time (sec.) A mp li tu d e Step Response 0 0.6 1.2 1.8 2.4 3 3.6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Figure 5: Step Response for K =100 -------------------------------- (s+11.32) (s^2 + 3.677s + 12.36) EDU>step(tc) EDU>print -deps sr46410d.eps EDU> produces the step response shown in Figure 5. 10