Solution 9.10.5.2 T c (s)= K(s + ) (s +  ;j! d )(s +  + j! d )(s + ) :  = ! n =0:85=4 ! d = ! n q 1; 2 =50:6=3: = 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  = 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 ): (a) Then, for part(a)  = ( 2 + ! 2 d ) 2 +(1;)( 2 + ! 2 d ) 20(4 2 +3 2 ) 2420 + (1;0)(4 2 +3 2 ) =2:703 K = 2 +(1;)( 2 + ! 2 d ) = 2420 + 3 2 +4 2 =185: The MATLAB program omegan = 5 zeta = 0.8 sigma = omegan * zeta gamma = 20 eps = 0.0 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 sr91052.eps t=0:0.01:1;; u=t;; lsim(tc,u,t) print -deps rr91052.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 0.5 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.) 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: Ramp response of closed loop system 3 (b) The MATLAB program omegan = 5 zeta = 0.8 sigma = omegan * zeta gamma = linspace(10,30,100);; eps = 0.0 omegad = omegan * sqrt(1 - zeta^2) delta =(sigma^2 + omegad^2).* gamma;; delta = delta./(2*sigma.*gamma +(1 -eps)* (sigma^2 + omegad^2) );; K=2*sigma*gamma + (1 -eps)*( sigma^2 + omegad^2);; plot(gamma,delta);; grid on print -deps 91052gvda.eps plot(gamma,K) grid on print -deps 91052gvka.eps produces the plots shown in Figures 3 The plot of K versus is perfectly linear. The plot of delta versus is slightly nonlinear. (c) It is nowatrivial matter to modify the MATLAB code to omegan = 5 zeta = 0.8 sigma = omegan * zeta gamma = linspace(10,30,100);; eps = 0.1 omegad = omegan * sqrt(1 - zeta^2) delta =(sigma^2 + omegad^2).* gamma;; delta = delta./(2*sigma.*gamma +(1 -eps)* (sigma^2 + omegad^2) );; K=2*sigma*gamma + (1 -eps)*( sigma^2 + omegad^2);; plot(gamma,delta);; grid on print -deps 91052gvdb.eps plot(gamma,K) grid on print -deps 91052gvkb.eps 4 10 12 14 16 18 20 22 24 26 28 30 2.35 2.4 2.45 2.5 2.55 2.6 2.65 2.7 2.75 2.8 2.85 δ γ Figure 3:  versus gamma for K v =100 and produce the graphs shown in Figure 5 and 6. These curves have the same as their predecessors. 5 10 12 14 16 18 20 22 24 26 28 30 100 120 140 160 180 200 220 240 260 280 γ K Figure 4: K versus gamma for K v =100 6 10 12 14 16 18 20 22 24 26 28 30 2.4 2.45 2.5 2.55 2.6 2.65 2.7 2.75 2.8 2.85 2.9 δ γ Figure 5:  versus for K v =100 7 10 12 14 16 18 20 22 24 26 28 30 100 120 140 160 180 200 220 240 260 280 δ γ Figure 6: K versus for K v =100 8