Solution 6.8.1.7 Referring to Figure 1 For ! d =10, The dominantpoles willbeat X X Re(s) Im(s) -1 10 -3 -2 s + 3 s + 1 Figure 1: Gain to place poles with ;2+j10 s = ;2j10: The damping ratio is thus  = cos h tan ;1 (10=30) i = 0:1961 The gain to place the poles at the desired locations is K = js +1jjs +3j =101: Then the closed loop transfer function is T c (s)= 100 (s +2;j10)(s+2+j10) : 1 The MATLAB dialogue EDU>g = zpk([],[-1 -3],101) Zero/pole/gain: 101 ----------- (s+1) (s+3) EDU>tc = feedback(g,1) Zero/pole/gain: 101 ---------------- (s^2 + 4s + 104) EDU>step(tc) EDU>print -deps sr6817.eps EDU> produces the step response shown in Figure 2 As can be seen there is a steady state error of 80%. Wecan nd this is twoways. Since wehave unity feedbackwe can nd K p = lim s!0 101 (s + 1)(s+3) = 101 3 = 33:6667: Then e ss = 1 1+K p = 1 1+33:6667 = 0:0288: The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 101 (s +2; j10)(s+2+j10) = 1; 0:9712 = 0:0288: 2 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 Figure 2: Unit step response of compensated system 3