Solution 6.8.1.4 Referring to Figure 1 First note that X X Re(s) Im(s) -10 -5 -7.5 ζ = 0.8 ω d Figure 1: Gain to place poles with  =0:8 ! n =   ;; where, in this case  =7:5 Then ! d = ! n q 1; 2 =  p 1; 2  : Thus in the present case ! d = 7:5 p 1;0:8 2 0:8 =5:625 Then the dominantpoles will be at s = ;7:5j5:625: 1 The gain to place the poles at the desired locations is K = p 7:5+5:625 2  p 7:5 2 +5:625 2 = 37:8906: Then the closed loop transfer function is T c (s)= 37:8906 (s +7:5;j5:625)(s+7:5+j5:625) : The MATLAB dialogue EDU>g = zpk([],[-5 -10],37.8096) Zero/pole/gain: 37.8096 ------------ (s+5) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 37.8096 ------------------- (s^2 + 15s + 87.81) EDU>step(tc) EDU>print -deps sr6814.eps EDU> produces the step response shown in Figure 2 As can be seen there is a steady state error of about 55%. Wecan nd this is twoways. Since we have unityfeedbackwecan nd K p = lim s!0 37:8096 (s +5)(s +10) = 37:8096 50 = 0:7578: Then e ss = 1 1+K p 2 Time (sec.) A mp li tu d e Step Response 0 0.16 0.32 0.48 0.64 0.8 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Figure 2: Unit step response of compensated system 3 = 1 1+0:7578 = 0:5689: The other wayto nd the steady state error is to evaluae e ss = = 1; lim s!0 T c (s) = 1; lim s!0 37:8096 (s +7:5;j5:625)(s+7:5+j5:625) = 1;0:4311 = 0:5689: 4