Solution 6.8.1.5 Referring to Figure 1 For ! d =10, The dominantpoles willbeat X X Re(s) Im(s) -40 -20-30 Figure 1: Gain to place poles with  =1:0 s = ;30j0: The damping ratio is thus  = cos h tan ;1 (0=30) i = 1 The gain to place the poles at the desired locations is K = js +20jjs +40j =100: Then the closed loop transfer function is T c (s)= 100 (s +30)(s+30) : The MATLAB dialogue EDU>g = zpk([],[-20 -40],100) Zero/pole/gain: 100 ------------- (s+20) (s+40) 1 Time (sec.) A mp li tu d e Step Response 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0 0.02 0.04 0.06 0.08 0.1 0.12 Figure 2: Unit step response of compensated system EDU>tc = feedback(g,1) Zero/pole/gain: 100 -------- (s+30)^2 EDU>step(tc) EDU>print -deps sr6815.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!0 100 (s +20)(s+40) = 100 800 = 0:125: 2 Then e ss = 1 1+K p = 1 1+0:125 = 0:8889: The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 210 (s +30)(s+30) = 1; 0:1111 = 0:8889: 3