Solution 6.8.3.1 The root locus is shown in Figure 1. Eventually two limbs of the root locus Im(s) Re(s) -30 -2 -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 +2) : So a good place to start looking for the roots is s = ;1  j3. Indeed, the contribution of the pole at s = ;30 at s = ;1+j3is8:9826  . 1 -0.8 -0.78 -0.76 -0.74 -0.72 -0.7 -0.68 178.5 179 179.5 180 180.5 181 181.5 182 182.5 Figure 2: Angle versus real part of s The MATLAB program p1 =0 p2 = 2 p3 = 20 x =linspace(-0.7,-0.8,19) s=x+j*3 ang = (angle(s + p1)+angle(s+p2)+angle(s+p3))*180/pi plot(x,ang) print -deps 6831a.eps creates the plot shown in Figure 2. Wesee that the root locus crosses the horizontal line through s = j3atabouts = ;0:7422. Indeed, at s = ;0:7422+ j3 the net angle is 180:0041  . Then the gain to place closed loop poles at s = ;0:7422+ j3is K = jsks +2js +20jj s=;0:7422+j3 = 195:9385: The natural frequency of the closed loop poles is ! n = p 0:7422 2 +3 2 =3:0904: The damping ratio is  = cos[tan ;1 (3=0:83)] = 0:2402: 2 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 T N2 T c Figure 3: Comparison of step responses The MATLAB program s=-0.7422 + j*3 K=abs(s + p1)*abs(s+p2)*abs(s+p3) zeta = cos(atan(3/0.7422)) omegana = abs(s) tn2 = zpk([],[s conj(s)],omegana^2) g=zpk([],[0 -2 -20],K) tc = feedback(g,1) T=linspace(0,4,100);; [Yn2,T] = step(tn2,T);; [Ytc,T] = step(tc,T);; plot(T,Yn2,'k-',T,Ytc,'k--') print -deps sr6831.eps Prints and plots the step responses of T c (t) and T N2 (t) shown in Figure 3 3