Problem: 7.9.9.10 There are a couple of ways to work this problem. One is to nd the equivalenttransfer function for the inner feedbacklook. The other wayis to put the inner summer past the blockwith K s in it. Weelect the rst method. Then G p (s) = 60 (s;1)(s+1:2)(s+10) 1+ 60K r s (s;1)(s +1:2)(s+10) = 60 (s;1)(s+1:2)(s+10)+60K r s Now nding the overall closed loop transfer function wehave G p (s) = 60K s (s;1)(s+1:2)(s+10)+60K r s 1+ 60K s (s;1)(s +1:2)(s+10)+60K r s = 60K s (s;1)(s+1:2)(s+10)+60K r s +60K s Todotheroot locus analysis wewrite (s;1)(s +1:2)(s+10)+60K r s +60K s =1+ 60K r (s + K s =K r ) (s;1)(s+1:2)(s+10) : In this form wesee that, at least as far as drawing the root locus is concerned, the e ect of the inner rate feedbackloopplusthe cascade gain K s is to create azero. Nowallwehavetodoispick the ratio of K s =K r , thereby xing the zero. We next havetodecide where to place the zero, and what e ect the zero location has on the performance of the missile. Figure 1 shows the root locus for K s K r =0:5;; and 0 < 60K r  60. Wehaveclearly achieved our goal of a damping ratio >0:5. However, a related question of some interest is the steady state error to a step input. If wechoose 60K r =60,orK r =1,then K s =0:5, and the closed loop transfer function is T c (s) = 60K s (s +0:311)(s+4:944;j5:747)(s+4:944+ j5:747) = 30 (s +0:311)(s+4:944;j5:747)(s+4:944+ j5:747) 1 -10 -8 -6 -4 -2 0 2 6 4 2 0 2 4 6 O X XX Figure 1: Root Locus for K s =K r =0:5 2 -10 -8 -6 -4 -2 0 2 -3 -2 -1 0 1 2 3 X X XO Figure 2: Root Locus for K s =K r =0:5 Then e ss = 1;T c (0) = 1; 30 (0:311)(4:944 2 +5:747 2 ) = 1;1:68 = ;0:68 Figure 2 shows the root locus for K s K r =2;; and 0 < 60K r  40. If wechoose 60K r =40,then K r = 2 3 ;; and K s = 4 3 : 3 -10 -8 -6 -4 -2 0 2 -8 -6 -4 -2 0 2 4 6 8 X XXO Figure 3: Root Locus for K s =K r =2 The closed loop transfer function is then T c (s) = 60K s (s +4:482)(s+2:859;j2:645)(s+2:859+ j2:645) = 80 (s +4:482)(s+2:859;j2:645)(s+2:859+ j2:645) Then e ss = 1;T c (0) = 1; 80 (4:482)(2:859 2 +2:645 2 ) = 1;1:18 = ;0:18 This is an improvement from 68% steady state error to 18% steady state error. If weincrease the gain to 60K r =80,then wegetthe root locus shown in Figure 3 In this case K r = 4 3 ;; 4 and K s = 8 3 : The closed loop transfer function is then T c (s) = 60K s (s +2:38)(s+3:91;j6:8484)(s+3:91+ j6:8484) = 160 (s +2:38)(s+3:91;j6:8484)(s+3:91+ j6:8484) Then e ss = 1;T c (0) = 1; 160 (2:38)(3:91 2 +6:8484 2 ) = 1;1:08 = ;0:08 The damping ratio is now 0.5. To improve the accuracy wewould haveto lower the damping ratio. If wechange the location of the zero in the root locus and repeat the analysis, wewill nd that wearevery close to the best performance we can achieve. 5