Solution 6.8.1.12 The root locus is shown in Figure 1. Eventually two limbs of the root locus Im(s) Re(s) -30 -1 Figure 1: Root locus will cross into the righthalf plance, but near the origin the pole at s = ;30 has very little in uence on the root locus. Near the origin the root locus looks very like that of the simpler transfer function G(s)= K s(s +1) : So a good place to start looking for the roots is s = ;0:5j2. The MATLAB program p1 =0 p2 = 1 p3 = 30 1 2.1 2.15 2.2 79.6 79.8 180 80.2 80.4 80.6 80.8 181 81.2 Figure 2: Angle versus ! n zeta = 0.2 omegan =linspace(2.1,2.2,400);; 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 68112a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constantdamping at about s = ;0:4232 + j2:0733. Indeed, at s = ;0:4232+ j2:0733 the net angle is 180:000  . Then the gain to place closed loop poles at s = ;0:4232+ j2:0733 is K = jsks +1js +30jj s=;0:4232+j2:0733 =135:064: The natural frequency of the closed loop poles is ! n = p 0:4232 2 +2:0733 2 =2:116: The MATLAB program p1 =0 p2 = 1 p3 = 30 2 0 1 2 3 4 5 6 7 8 9 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Figure 3: Comparison of step responses of T c and T N2 for  =0:4 zeta = 0.2 omegan =linspace(2.1,2.2,400);; 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 68112a.eps s=-zeta*omegan(65) + j*omegan(65)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3)) omegn=omegan(65) sc = conj(s) tn2 = zpk([],[s sc],omegan(65)^2) g=zpk([],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,10,200);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr68112.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3