Solution 6.8.1.3 Referring toe Figure 1 Weseethat the gain to place the closed loop poles X X Re(s) Im(s) -5 5 -10 -7.5 Figure 1: Gain to place poles with  =1= p 2 at s = ;7:5j5 is K = p 2:5 2 +5 2  p 2:5 2 +5 2 = 31:25: Then the closed loop transfer function is T c (s)= 31:25 (s +7:5;j5)(s+7:5+j5) : 1 The MATLAB dialogue EDU>g = zpk([],[-5 -10],31.25) Zero/pole/gain: 31.25 ------------ (s+5) (s+10) EDU>tc = feedback(g,1) Zero/pole/gain: 31.25 ------------------- (s^2 + 15s + 81.25) EDU>step(tc) EDU>print -deps sr6813.eps EDU> produces the step response shown in Figure 2 As can be seen there is a steady state error of about 60%. Wecan nd this is twoways. Since we have unityfeedbackwecan nd K p = lim s!0 31:25 (s +5)(s +10) = 4: Then e ss = 1 1+K p = 1 1+0:625 = 0:6154: The other wayto nd the steady state error is to evaluae e ss = 1; lim s!0 T c (s) = 1; lim s!0 31:25 (s +7:5;j5)(s+7:5+j5) = 1; 31:25 81:25 = 0:6154: 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 Figure 2: Unit step response of compensated system 3