Solution 6.8.3.6 Im(s) Re(s) -20 -0.2-1 Figure 1: Root locus The root locus is shown in Figure 1. The MATLAB program z=1 p1 =0 p2 = 0.2 p3 = 20 zeta = 1/sqrt(2) omegan =linspace(1.2,1.4,200);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + z)-angle(s + p1)-angle(s+p2)-angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6836a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the 1 1.2 1.22 1.24 1.26 1.28 1.3 1.32 1.34 1.36 1.38 1.4 -186 -185 -184 -183 -182 -181 -180 -179 -178 -177 Figure 2: Angle versus real part of s rayofconstantdamping ratio  =1= p 2ats = ;0:9338+ j0:9338 Indeed, at s = ;0:9338+ j0:9338, where the net angle is 180:01  . Then the gain to place closed loop poles at s = ;0:9338+ j0:9338 is K = jsjjs+2js +20jj s=;0:9338+j0:9338 =31:981: The natural frequency of the closed loop poles is ! n = p 0:9338 2 +0:9338 2 =1:3206: The MATLAB program z=1 p1 =0 p2 = 0.2 p3 = 20 zeta = 1/sqrt(2) omegan =linspace(1.2,1.4,200);; s=-zeta*omegan + j* omegan *sqrt(1-zeta^2);; ang = (angle(s + z)-angle(s + p1)-angle(s+p2)-angle(s+p3))*180/pi;; plot(omegan,ang) print -deps 6836a.eps 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 T N2 T C Figure 3: Comparison of step responses s=-zeta*omegan(121) + j*omegan(121)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3))/abs(s+z) omegana = omegan(121) sc = conj(s) tn2 = zpk([],[s sc],omegana^2) g=zpk([-z],[-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 sr6836.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3