Solution 6.8.1.11 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 = ;0:5j1. The MATLAB program p1 =0 p2 = 1 p3 = 30 1 1.1 1.12 1.14 1.16 1.18 1.2 1.22 1.24 1.26 1.28 1.3 176 177 178 179 180 181 182 183 184 Figure 2: Angle versus ! n zeta = 0.4 omegan =linspace(1.1,1.3,400);; 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 68111a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constantdamping at about s = ;0:4757 + j1:0899. Indeed, at s = ;0:4757+ j1:0899 the net angle is 180:000  . Then the gain to place closed loop poles at s = ;0:4924+ j0:4924 is K = jsks +1js +30jj s=;0:4757+j1:0899 =42:4954: The natural frequency of the closed loop poles is ! n = p 0:4757 2 +1:0899 2 =1:1892: The MATLAB program p1 =0 p2 = 1 2 p3 = 30 zeta = 0.4 omegan =linspace(1.1,1.3,400);; 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 68111a.eps s=-zeta*omegan(179) + j*omegan(179)*sqrt(1-zeta^2) K=(abs(s + p1)*abs(s+p2)*abs(s+p3)) omegn=omegan(179) sc = conj(s) tn2 = zpk([],[s sc],omegan(179)^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 sr68111.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3 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 Figure 3: Comparison of step responses of T c and T N2 for  =0:4 4