Solution 6.8.4.3 The root locus is shown in Figure 1. Eventually two limbs of the root locus Im(s) Re(s) -10 -4 Figure 1: Root locus will cross into the right half plane. The MATLAB program p1 =0 p2 = 4 p3 = 10 zeta = 1/sqrt(2) omegan =linspace(2.2,2.35,200);; 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 6834a.eps creates the plot shown in Figure 2. We see that the root locus crosses the ray 1 2.2 2.25 2.3 2.35 177.5 178 178.5 179 179.5 180 180.5 181 181.5 182 Figure 2: Angle versus real part of s of constantdamping ratio  =1=sqrt(2) at s = ;1:6153+ j1:6153 Indeed, at s = ;1:6153+ j1:6153, where the net angle is 180:000  . Then the gain to place closed loop poles at s = ;1:6153+ j1:6153 is K = jsjjs+4js +20jj s=;1:6153+j1:6153 =56:1834: The natural frequency of the closed loop poles is ! n = p 1:6153 2 +1:6153 2 =2:2075: The MATLAB program p1 =0 p2 = 4 p3 = 10 zeta = 1/sqrt(2) omegan =linspace(2.2,2.35,200);; 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 6843a.eps s=-zeta*omegan(113) + j*omegan(113)*sqrt(1-zeta^2) 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 K=abs(s + p1)*abs(s+p2)*abs(s+p3) omegana = omegan(11) 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 sr6843.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 Even with the pole ve times closer to the origin, the actual response to a step is still very close to that for the step response of T N2 (s). 3