Solution 9.10.5.6 ΣΣ G G P c + + + C D R For the system shown above, we assume the disturbance is a constant posi- tiveo set D, and G p (s)= K s : Wehavealinear system, wecan use superposition. The closed loop transfer function between D and C can be derived as follows. With R =0, C d =(D;C d G c )G p ;; which can be rewritten as C d + C d G c G p = DG p ;; or C d (1+ G c G p )=DG p ;; or nally T d = C d D = G p 1+G c G p The transfer function between R and C with D =0is T c = C d D = G c G p 1+G c G p If G p (s)= K s ;; then the system, from R to C is type one and it doesn't matter what G c is as long as it doesn't cancel the pole of the plant and doesn't make the system unstable. Thus anypositive nite G c will do. 1 However, for the system between D and C,wehave T d (s) = K=s 1+G c (K=s) = K s +KG c We will get no steady state error due to T c (s). That means wemust drive the output due to T d to the step disturbance of magnitude D to some value .That is, lim s!0 sC d (s)=: Then, lim s!0 sC d (s) = lim s!0 s D s T d (s) = lim s!0 DT d (0) = DK KG c = D G c For zero steady state error wemust have D G c =0;; or G c = 1: However, for any nite error, wehave D G c = ;; or G c = D  : For 2% error wethenget G c = D 0:02 =50D: 2