Solution 4.6.1.7 The characteristic equation is 1+ K (s +5) 3 =0;; or s 3 +15s 2 +75s +(125+K) (s +5) 3 =0;; or equivalently s 3 +15s 2 +75s +(125+K)=0: The MATLAB program K=10 p=[1 15 75 125+K] roots(p) K=100 p=[1 15 75 125+K] roots(p) K=400 p=[1 15 75 125+K] roots(p) K=580 p=[1 15 75 125+K] roots(p) K=1000 p=[1 15 75 125+K] roots(p) K=[10 100 400 580 1000] gh = zpk([],[-5 -5 -5],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl4617.eps generates the following output EDU>sm4617 K= 1 10 p= 1 15 75 135 ans = -7.1544 -3.9228+ 1.8658i -3.9228- 1.8658i K= 100 p= 1 15 75 225 ans = -9.6416 -2.6792+ 4.0197i -2.6792- 4.0197i K= 400 p= 2 1 15 75 525 ans = -12.3681 -1.3160+ 6.3809i -1.3160- 6.3809i K= 580 p= 1 15 75 705 ans = -13.3396 -0.8302+ 7.2223i -0.8302- 7.2223i K= 1000 p= 1 15 75 1125 ans = -15.0000 3 -0.0000+ 8.6603i -0.0000- 8.6603i K= 10 100 400 580 1000 Zero/pole/gain: 1 ------- (s+5)^3 R= Columns 1 through 4 -7.1544 -9.6416 -12.3681 -13.3396 -3.9228+ 1.8658i -2.6792+ 4.0197i -1.3160+ 6.3809i -0.8302+ 7.2223i -3.9228- 1.8658i -2.6792- 4.0197i -1.3160- 6.3809i -0.8302- 7.2223i Column 5 -15.0000 -0.0000+ 8.6603i -0.0000- 8.6603i K= 10 100 400 580 1000 EDU> 4 The plot of the points is shown in Figure ?? -15 -10 -5 0 -10 -8 -6 -4 -2 0 2 4 6 8 10 Figure 1: Plot of solutions 5