Solution 6.8.1.9 Referring to Figure 1 First note that X X Re(s) Im(s) -3 -1 -2 ζ = 0.9 ω d Figure 1: Gain to place poles with  =0:8 ! n =   ;; where, in this case  =2 Then ! d = ! n q 1; 2 =  p 1; 2  : Thus in the present case ! d = 2 p 1;0:9 2 0:9 =0:9686 Then the dominantpoles will be at s = ;2j0:9686: 1 The gain to place the poles at the desired locations is K = p 2 2 +0:9686 2  p 2 2 +0:9686 2 = 1:9383: Then the closed loop transfer function is T c (s)= 1:9383 (s +2;j0:9686)(s+2+j0:9686) : The MATLAB dialogue EDU>g = zpk([],[-1 -3],1.9383) Zero/pole/gain: 1.9383 ----------- (s+1) (s+3) EDU>tc = feedback(g,1) Zero/pole/gain: 1.9383 ------------------ (s^2 + 4s+4.938) EDU>step(tc) EDU>print -deps sr6819.eps EDU> produces the step response shown in Figure 2 As can be seen there is a steady state error of about 60%. Wecan nd this is twoways. Since we have unityfeedbackwecan nd K p = lim s!0 1:9383 (s + 1)(s+3) = 1:9383 3 = 0:6461: Then e ss = 1 1+K p 2 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Figure 2: Unit step response of compensated system 3 = 1 1+0:6461 = 0:6075: The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 1:9383 (s +2;j0:9686)(s+2+j0:9686) = 1;0:3925 = 0:6075: 4