Solution 6.8.2.4 T c (s)= K(s + ) (s +  ;j! d )(s +  + j! d ) :  = ! n = p 2=25=3:5355 ! d = ! n q 1; 2 =5 p 2=2=3:5355 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  = 2  2 + ! 2 d ; 1  : For nonzero steady state error to a ramp wehave 2  2 + ! 2 d ; 1  = ;; where  is the steady state error. Then 1  = 2  2 + ! 2 d ; = 2 ;( 2 + ! 2 d )  2 + ! 2 d ;; 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=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  = 3:5355 2 +3:5355 2 23=5355;0:1(3:5355 2 +3:5355 2 ) =4:2947 K = 2;( 2 + ! 2 d )=23:5355;0:05(3:5355 2 +3:5355 2 )=5:8211 The MATLAB program omegan = 5 zeta = sqrt(2)/2 eps = 0.05 sigma = omegan * zeta omegad = omegan * sqrt(1 - zeta^2) delta = (sigma^2 + omegad^2 )/ (2*sigma - eps*(sigma^2 + omegad^2)) K=2*sigma - eps*(sigma^2 + omegad^2) tc = zpk([-delta],[-sigma+j*omegad -sigma-j*omegad],K) step(tc) print -deps sr6824.eps t=0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr6824.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 0.4 0.8 1.2 1.6 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 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 2: Step response of closed loop system 3