Solution 6.8.2.7 T c (s)= K(s + ) (s +  ;j! d )(s +  + j! d )(s + ) :  = ! n =0:65=3 ! d = ! n q 1; 2 =50:8=4: = 20: We rst nd  to set the desired steady state error to a ramp, namely zero. Wehave e ss = 1  ;j! d + 1  + j! d + 1 ; 1  = 2  2 = ! 2 d + 1 ; 1  : For zero steady state error to a ramp wehave 2 +  2 + ! 2 d ( 2 + ! 2 d ) ; 1  =0;; or  = ( 2 + ! 2 d ) 2 +  2 + ! 2 d For zero steady state error to a step input wemust have T c (0) = 1;; or K(s + ) (s +  ;j! d )(s +  + j! d )(s + ) s=0 =1;; or K ( 2 + ! 2 d ) =1;; yielding K = ( 2 + ! 2 d )  1 substituting our value for  obtained earlier wehave K = ( 2 + ! 2 d ) ( 2 + ! 2 d ) 2 +  2 + ! 2 d =2 +  2 + ! 2 d : Then, for this problem  = 20(3 2 +4 2 ) 2320+ 3 2 +4 2 =3:4483 K = 2 +  2 + ! 2 d = 2320+ 3 2 +4 2 =145 The MATLAB program omegan = 5 zeta = 0.6 sigma = omegan * zeta gamma = 20 omegad = omegan * sqrt(1 - zeta^2) delta = gamma*(sigma^2 + omegad^2);; delta = delta/(2*sigma*gamma+sigma^2 + omegad^2) K=2*sigma*gamma + sigma^2 + omegad^2 tc = zpk([-delta],[-sigma+j*omegad -sigma-j*omegad -gamma],K) step(tc) print -deps sr6827.eps t =0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr6827.eps generates the step and ramp responses shown in Figure 1 and 2 respectively. 2 Time (sec.) A mp li tu d e Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 1: Step response of closed loop system Time (sec.) A mp li tu d e Linear Simulation Results 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Ramp response of closed loop system 3