Solution 6.8.1.10 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 = ;0:5 j0:5. 1 0.55 0.6 0.65 0.7 0.75 0.8 172 174 176 178 180 182 184 186 188 190 Figure 2: Angle versus jsj = ! n The MATLAB program p1 =0 p2 = 1 p3 = 30 zeta = 1/sqrt(2) omegan =linspace(0.6,0.8,200);; 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 68110a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the line of constantdamping at about s = ;0:4924 + j0:4924. Indeed, at s = ;0:4924+ j0:4924 the net angle is 180:09  . Then the gain to place closed loop poles at s = ;0:4924+ j0:4924 is K =[jsks +1js +30j] s=;0:4924+j0:4924 =14:536: The natural frequency of the closed loop poles is ! n = p 0:4924 2 +0:4925 2 =0:6965: 2 The MATLAB program p1 =0 p2 = 1 p3 = 30 zeta = 1/sqrt(2) omegan =linspace(0.6,0.8,200);; 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 68110a.eps s=-zeta*omegan(97) + j*omegan(97)*sqrt(1-zeta^2) K=abs(s + p1)*abs(s+p2)*abs(s+p3) omegana = omegan(97) sc = conj(s) tn2 = zpk([],[s sc],omegana^2) g=zpk([],[-p1 -p2 -p3],K) tc = feedback(g,1) T=linspace(0,8,200);; [Yn2,T] = step(tn2,T);; plot(T,Yn2,'k-') print -deps sr68110.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 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Comparison of step responses 4