Solution 4.6.1.10 The characteristic equation is 1+ K(s +1) s 2 (s +4)(s +10) =0;; or s 4 +14s 3 +40s 2 + Ks+ K s 2 (s +4)(s +10) =0;; or equivalently s 4 +14s 3 +40s 2 + Ks+ K =0: The MATLAB program K=1 p=[1 14 40 K K] roots(p) K=5 p=[1 14 40 K K] roots(p) K=10 p=[1 14 40 K K] roots(p) K=100 p=[1 14 40 K K] roots(p) K=200 p=[1 14 40 K K] roots(p) K=[1 5 10 100 200] gh = zpk([-1],[0 0 -4 -10],1) [R,K] = rlocus(gh,K) plot(R,'kd') print -deps rl46110.eps generates the following output EDU>sm46110 K= 1 1 p= 1 14 40 1 1 ans = -10.0149 -3.9687 -0.0082+ 0.1584i -0.0082- 0.1584i K= 5 p= 1 14 40 5 5 ans = -10.0736 -3.8437 -0.0414+ 0.3570i -0.0414- 0.3570i K= 10 p= 2 1 14 40 10 10 ans = -10.1446 -3.6869 -0.0842+ 0.5102i -0.0842- 0.5102i K= 100 p= 1 14 40 100 100 ans = -11.1435 -0.6963+ 2.3759i -0.6963- 2.3759i -1.4639 K= 200 p= 1 14 40 200 200 3 ans = -11.9349 -0.4469+ 3.7559i -0.4469- 3.7559i -1.1713 K= 1 5 10 100 200 Zero/pole/gain: (s+1) ---------------- s^2 (s+4) (s+10) R= Columns 1 through 4 -10.0149 -10.0736 -10.1446 -11.1435 -3.9687 -3.8437 -3.6869 -0.6963+ 2.3759i -0.0082+ 0.1584i -0.0414+ 0.3570i -0.0842+ 0.5102i -1.4639 -0.0082- 0.1584i -0.0414- 0.3570i -0.0842- 0.5102i -0.6963- 2.3759i Column 5 -11.9349 -0.4469+ 3.7559i -1.1713 -0.4469- 3.7559i K= 1 5 4 10 100 200 EDU> The plot of the points is shown in Figure 1 -12 -10 -8 -6 -4 -2 0 -4 -3 -2 -1 0 1 2 3 4 Figure 1: Plot of solutions 5