Solution 4.6.4.9 The closed loop transfer function is T c (s) = K (s +0:1)(s+4)(s+ 10) 1+ K (s +0:1)(s+4)(s +10) = K s 3 +14:1s 2 +41:4s+(K +4) : The characteristic equation is s 3 +14:1s 2 +41:4s+(K +4)=0: The MATLAB program K=10 p=[1 14.1 41.1 4+K] roots(p) K=30.68 p=[1 14.1 41.1 4+K] roots(p) K=68 p=[1 14.1 41.1 4+K] roots(p) K=60 p=[1 14.1 41.1 4+K] roots(p) K=100 p=[1 14.1 41.1 4+K] roots(p) gh = zpk([],[-0.1 -4 -10],1) K1 = [10 30.68 60 100] [R,K] = rlocus(gh,K1) plot(R,'kd') print -deps rl4649.eps produces the output K= 10 1 p= 1.00000000000000 14.10000000000000 41.10000000000000 14.00000000000000 ans = -10.20815647672320 -3.50000000000000 -0.39184352327679 K= 30.68000000000000 p= 1.00000000000000 14.10000000000000 41.10000000000000 34.68000000000000 ans = -10.50039600358139 -1.79980199820931 + 0.25188360163417i -1.79980199820931 - 0.25188360163417i K= 68 p= 1.00000000000000 14.10000000000000 41.10000000000000 72.00000000000000 2 ans = -10.94617029430882 -1.57691485284559 + 2.02261773677814i -1.57691485284559 - 2.02261773677814i K= 60 p= 1.00000000000000 14.10000000000000 41.10000000000000 64.00000000000000 ans = -10.85749836877930 -1.62125081561034 + 1.80723258634298i -1.62125081561034 - 1.80723258634298i K= 100 p= 1.0e+02 * 0.01000000000000 0.14100000000000 0.41100000000000 1.04000000000000 ans = -11.27239055897766 3 -1.41380472051116 + 2.68835232899338i -1.41380472051116 - 2.68835232899338i Zero/pole/gain: 1 -------------------- (s+0.1) (s+4) (s+10) K1 = 1.0e+02 * 0.10000000000000 0.30680000000000 0.60000000000000 1.00000000000000 R= Columns 1 through 2 -10.16131396237127 -10.45858281898964 -3.55065176580416 -1.82070859050518 - 0.03093642467982i -0.38803427182457 -1.82070859050518 + 0.03093642467982i Columns 3 through 4 -10.82056537115050 -11.23989884266038 -1.63971731442475 - 1.79610421381808i -1.43005057866981 - 2.68471742538630i -1.63971731442475 + 1.79610421381808i -1.43005057866981 + 2.68471742538630i K= 1.0e+02 * 0.10000000000000 0.30680000000000 0.60000000000000 1.00000000000000 4 -12 -10 -8 -6 -4 -2 0 -3 -2 -1 0 1 2 3 Figure 1: Plot of solutions EDU> The plot of the points is shown in Figure 1 For K = 10, the MATLAB dialogue EDU>g = zpk([],[-0.1 -4 -10],10) Zero/pole/gain: 10 -------------------- (s+0.1) (s+4) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 10 ----------------------------- (s+10.16) (s+3.551) (s+0.388) 5 Time (sec.) A mp li tu d e Step Response 0 5 10 15 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Figure 2: Step Response for K =30:68 EDU>step(tc) EDU>print -deps sr4649a.eps EDU> produces the step response shown in Figure 2. For K =30:68, the MATLAB dialogue EDU>g = zpk([],[-0.1 -4 -10],30.68) Zero/pole/gain: 30.68 -------------------- (s+0.1) (s+4) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 30.68 6 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 3: Step Response for K =4:87 -------------------------------- (s+10.46) (s^2 + 3.641s + 3.316) EDU>step(tc) EDU>print -deps sr4649b.eps EDU> produces the step response shown in Figure 3. For K = 60, the MATLAB dialogue EDU>g = zpk([],[-0.1 -4 -10],60) Zero/pole/gain: 60 -------------------- (s+0.1) (s+4) (s+10) EDU>tc = feedback(g,1) 7 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 1 Figure 4: Step Response for K =60 Zero/pole/gain: 60 -------------------------------- (s+10.82) (s^2 + 3.279s + 5.915) EDU>step(tc) EDU>print -deps sr4649c.eps EDU> produces the step response shown in Figure 4. For K = 100, the MATLAB dialogue EDU>g = zpk([],[-0.1 -4 -10],100) Zero/pole/gain: 100 -------------------- (s+0.1) (s+4) (s+10) 8 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 5: Step Response for K =100 EDU>tc = feedback(g,1) Zero/pole/gain: 100 ------------------------------- (s+11.24) (s^2 + 2.86s + 9.253) EDU>step(tc) EDU>print -deps sr4649d.eps EDU> produces the step response shown in Figure 5. 9