Solution 9.10.5.1 T c (s)= K(s + ) (s +  ;j! d )(s +  + j! d ) :  = ! n =0:85=4;; ! d = ! n q 1; 2 =50:6=3: 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 ): (a) Then, for part(a) = 4 2 +3 2 24;0(4 2 +3 2 ) =3:125 K = 2 ;( 2 + ! 2 d ) = 24;0(4 2 +3 2 ) = 8: The MATLAB program omegan = 5 zeta = 0.8 eps = 0.0 sigma = omegan * zeta omegad = omegan * sqrt(1 - zeta^2) gamma = (sigma^2 + omegad^2 )/ (2*sigma - eps*(sigma^2 + omegad^2)) K=2*sigma - eps*(sigma^2 + omegad^2) tc = zpk([-gamma],[-sigma+j*omegad -sigma-j*omegad],K) step(tc) print -deps sr91051a.eps t=0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr91051a.eps generates the step and ramp responses shown in Figure 1 and 2 respectively. (b) Then, for part(b) 2 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 1: Step response of closed loop system Time (sec.) Amplitude 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 = 4 2 +3 2 24;0:01(4 2 +3 2 ) =3:2258 K = 2 ;( 2 + ! 2 d ) = 24;0:01(4 2 +3 2 ) = 7:75: The MATLAB program omegan = 5 zeta = 0.8 eps = 0.01 sigma = omegan * zeta omegad = omegan * sqrt(1 - zeta^2) gamma = (sigma^2 + omegad^2 )/ (2*sigma - eps*(sigma^2 + omegad^2)) K=2*sigma - eps*(sigma^2 + omegad^2) tc = zpk([-gamma],[-sigma+j*omegad -sigma-j*omegad],K) step(tc) print -deps sr91051b.eps t=0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr91051b.eps generates the step and ramp responses shown in Figure 3 and 4 respectively. (c) Then, for part(c) = 4 2 +3 2 24;0:1(4 2 +3 2 ) =4:5455 K = 2;( 2 + ! 2 d ) = 24;0:01(4 2 +3 2 ) = 5:5: The MATLAB program 4 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 3: Step response of closed loop system Time (sec.) Amplitude 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 4: Step response of closed loop system 5 Time (sec.) A mp li tu d e Step Response 0 0.5 1 1.5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 5: Step response of closed loop system omegan = 5 zeta = 0.8 eps = 0.1 sigma = omegan * zeta omegad = omegan * sqrt(1 - zeta^2) gamma = (sigma^2 + omegad^2 )/ (2*sigma - eps*(sigma^2 + omegad^2)) K=2*sigma - eps*(sigma^2 + omegad^2) tc = zpk([-gamma],[-sigma+j*omegad -sigma-j*omegad],K) step(tc) print -deps sr91051c.eps t=0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr91051c.eps generates the step and ramp responses shown in Figure 5 and 6 respectively. 6 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 Figure 6: Step response of closed loop system 7