Solution 6.8.1.6 Referring to Figure 1 For ! d =10, The dominantpoles willbeat X X Re(s) Im(s) -20 10 -40 -30 s + 40 s + 20 Figure 1: Gain to place poles with  =1:0 s = ;30 j10: The damping ratio is thus  = cos h tan ;1 (10=30) i = The gain to place the poles at the desired locations is K = js +20jjs +40j =200: Then the closed loop transfer function is T c (s)= 200 (s +30;j10)(s+30+j10) : The MATLAB dialogue EDU>g = zpk([],[-20 -40],200) 1 Time (sec.) A mp li tu d e Step Response 0 0.05 0.1 0.15 0.2 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Figure 2: Unit step response of compensated system Zero/pole/gain: 200 ------------- (s+20) (s+40) EDU>tc = feedback(g,1) Zero/pole/gain: 200 ------------------ (s^2 + 60s + 1000) EDU>step(tc) EDU>print -deps sr6816.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 2 feedbackwe can nd K p = lim s!0 200 (s +20)(s+40) = 200 800 = 0:25: Then e ss = 1 1+K p = 1 1+0:25 = 0:8: The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 200 (s +30; j10)(s+30+j10) = 1;0:2 = 0:8: 3