Solution 6.8.3.5 The root locus is shown in Figure 1. The real part of the complex poles is Im(s) Re(s) -2 s + 2 s Figure 1: Root locus -1 the imaginary part is ! d = ! n q 1;  2 = p 1;  2  = 1:333: ! n = 1 :6 =1:6667 The gain that places the poles at s = ;1 j1:3333 is K = jsjjs+2jj s=;1+j1:3333 =2:7778: The closed loop transfer function is exactly T N2 (s)sothe plots will be identical. The MATLAB dialogue EDU>g = zpk([],[0 -2 ], 2.7778) Zero/pole/gain: 2.7778 1 ------- s(s+2) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 2.7778 ------------------ (s^2 + 2s+2.778) EDU>step(tc) EDU>print -deps sr6835f.eps EDU> plots and saves the step response shown in Figure 2 shown in Figure 2 The partial fraction expansion for the step response is C(s)= 1 s + 0:625 6 2:4981 s +1+j1:3333 + 0:625 6 2:4981 s +1+j1:3333 Then the time response is c(t)=[1+1:25e ;t cos(1:3333t+2:4981)]1(t): 2 Time (sec.) A mp li tu d e Step Response 0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step response 3