Solution 6.8.1.8 Referring to Figure 1 For ! d =2,The dominantpoleswill be at X X Re(s) Im(s) -1 2 -3 -2 s + 3 s + 1 Figure 1: Gain to place poles with ;2+j2 s = ;2 j2: The damping ratio is thus  = cos h tan ;1 (10=30) i = 0:7071 The gain to place the poles at the desired locations is K = js +1jjs +3j =5: Then the closed loop transfer function is T c (s)= 5 (s +2; j2)(s+2+j2) : The MATLAB dialogue EDU>g = zpk([],[-1 -3],5) 1 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Figure 2: Unit step response of compensated system Zero/pole/gain: 5 ----------- (s+1) (s+3) EDU>tc = feedback(g,1) Zero/pole/gain: 5 -------------- (s^2 + 4s+8) EDU>step(tc) EDU>print -deps sr6818.eps EDU> produces the step response shown in Figure 2 As can be seen there is a steady state error of about 35%. Wecan nd this is twoways. Since we 2 have unityfeedbackwecan nd K p = lim s!0 5 (s + 1)(s+3) = 5 3 = 1:6667: Then e ss = 1 1+K p = 1 1+1:6667 = 0:3750: The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 5 (s +2;j2)(s+2+2) = 1;0:625 = 0:375: 3