Solution 6.8.1.2 Referring to Figure 1 Wesee that the gain to place the closed loop poles at X X -3 Re(s) Im(s) -1-5 4 Figure 1: Gain to place poles with  =1= p 2 s = ;3j4 is K = p 4 2 +2 2  p 4 2 +2 2 = 20: Then the closed loop transfer function is T c (s)= 20 (s +3;j4)(s+3+j4) : The MATLAB dialogue EDU>g = zpk([],[-1 -5],20) Zero/pole/gain: 20 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 0.9 Figure 2: Unit step response of compensated system ----------- (s+1) (s+5) EDU>tc = feedback(g,1) Zero/pole/gain: 20 --------------- (s^2 + 6s+25) EDU>step(tc) EDU>print -deps sr6812.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 20 (s + 1)(s+5) = 4: Then e ss = 1 1+K p = 1 1+4 = 0:2: The other wayto nd the steady state error is to evaluae e ss = 1; lim s!0 T c (s) = 1; lim s!0 20 (s +3;j4)(s+3+j4) = 1; 20 25 = 0:2: 3