Solution 6.8.2.10 T c (s)= K(s + ) (s +  ;j! d )(s +  + j! d )(s + ) :  = ! n =0:98=7:2 ! d = ! n q 1; 2 =80:4359 = 3:4871 = 20: We rst nd  to set the desired steady state error to a ramp, namely 0.01. Wehave e ss = 1  ;j! d + 1  + j! d + 1 ; 1  = 2  2 + ! 2 d + 1 ; 1  = 2 +  2 + ! 2  2 + ! 2 d ; 1  : For nonzero steady state error to a ramp wehave 2 +  2 + ! 2 d ( 2 + ! 2 d ) ; 1  = ;; or 2 +  2 + ! 2 d ( 2 + ! 2 d ) ; = 1  ;; or 2 +  2 + ! 2 d ;( 2 + ! 2 d ) ( 2 + ! 2 d ) = 1  : Then nally  = ( 2 + ! 2 d ) 2 +(1;)( 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;; 1 or K ( 2 + ! 2 d ) =1;; yielding K = ( 2 + ! 2 d )  substituting our value for  obtained earlier wehave K = ( 2 + ! 2 d ) ( 2 + ! 2 d ) 2 +(1;)( 2 +! 2 d ) =2 +(1;)( 2 + ! 2 d ): Then, for this problem  = 20(3 2 +4 2 ) 2320 + (1;0:01)(3 2 +4 2 ) =3:643 K = 2 +(1;)( 2 + ! 2 d ) = 27:220 + 7:2 2 +3:4871 2 = 351:36: The MATLAB program omegan = 8 zeta = 0.9 sigma = omegan * zeta gamma = 20 eps = 0.01 omegad = omegan * sqrt(1 - zeta^2) delta = gamma*(sigma^2 + omegad^2);; delta = delta/(2*sigma*gamma +(1 -eps)* (sigma^2 + omegad^2) ) K=2*sigma*gamma + (1 -eps)*( sigma^2 + omegad^2) tc = zpk([-delta],[-sigma+j*omegad -sigma-j*omegad -gamma],K) step(tc) print -deps sr68210.eps t=0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr68210.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 0.16 0.32 0.48 0.64 0.8 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