Solution 6.8.3.7 The root locus is shown in Figure 1. Eventually two limbs of the root locus Im(s) Re(s) -30 -4 Figure 1: Root locus will cross into the right half plane. The MATLAB program p1 =0 p2 = 4 p3 = 30 zeta = 1/sqrt(2) omegan =linspace(2.6,2.7,400);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6837a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the 1 2.6 2.65 2.7 179 79.5 180 80.5 181 81.5 Figure 2: Angle versus ! n rayofconstantdamping ratio  =1= p 2ats = ;1:6874+ j1:8674 Indeed, at s = ;1:6874+ j1:8674, where the net angle is 180:00  . Then the gain to place closed loop poles at s = ;1:6874+ j1:8674 is K = jsjjs +4js +30jj s=;1:6874+j1:8674 =211:0605: The natural frequency of the closed loop poles is ! n = p 1:8674 2 +1:8674 2 =2:6409: The MATLAB program p1 =0 p2 = 4 p3 = 30 zeta = 1/sqrt(2) omegan =linspace(2.6,2.7,400);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6837a.eps s=-zeta*omegan(164) + j*omegan(164)*sqrt(1-zeta^2) K=abs(s + p1)*abs(s+p2)*abs(s+p3) 2 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Comparison of step responses omegana = omegan(164) sc = conj(s) tn2 = zpk([],[s sc],omegana^2) g=zpk([],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,5,200);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr6837.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3