Solution 6.8.1.1 Referring to Figure 1 Wesee that the gain to place the closed loop poles at X X 3 -3 Re(s) Im(s) -1-5 Figure 1: Gain to place poles with  =1= p 2 s = ;3j3 is K = p 3 2 +2 2  p 3 2 +2 2 = 13: Then the closed loop transfer function is T c (s)= 13 (s +3;j3)(s+3+j3) : 1 Time (sec.) A mp li tu d e Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Figure 2: Unit step response of compensated system The MATLAB dialogue EDU>g = zpk([],[-1 -5],13) Zero/pole/gain: 13 ----------- (s+1) (s+5) EDU>tc = feedback(g,1) Zero/pole/gain: 13 --------------- (s^2 + 6s + 18) EDU>step(tc) EDU>print -deps sr6811.eps EDU> produces the step response shown in Figure 2 As can be seen there is a steady state error of about 22%. Wecan nd this is twoways. Since we 2 have unityfeedbackwecan nd K p = lim s!0 13 (s + 1)(s+5) = 13 5 : Then e ss = 1 1+K p = 1 1+(13=5) = 5 18 : The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 13 (s +3;j3)(s+3+j3) = 1; 13 18 = 5 18 : 3