Solution: 7.9.6.3 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 let G p = 1 (s +2)(s +5) and G c (s)= K c (s + 2)(s+5) s(s +10) Figure 2 shows the root locus under perfect cancellation. The gain re- quired to place the dominantpolesass = ;5j4is K c = js +10jjsjj s=;5+j4 = p 41 p 41 = 41: The closed loop system is exactly T NS (s)sotheexact response can be ob- tained from the chapter on speci cations that is, c(t)= 1; e ;! n t p 1;  2 sin(! d t + );; ! 1(t);; with  = cos h tan ;1 (4=5) i =0:781 ! n = p 4 2 +5 2 =6:4  = tan ;1 "p 1; 2  # =tan ;1 "p 1; 0:781 2 0:781 # =0:615 rad. 1 Im(s) Re(s) 4 -10 s + 10 s Figure 2: Root Locus for Perfect Cancellation (b) If wedonot achieveperfect cancellation then the closed loop transfer function is T c (s) = 41(s +5:1) s(s +5)(s+10) 1+ 41(s +5:1) s(s +5)(s+10) = 41(s+5:1) s(s +5)(s+10)+41s +209:1 = 41(s +5:1) s 3 +15s 2 +91s +209:1 = 41(s+5:1) (s +5:255)(s+4:872;j4:006)(s+4:872+ j4:006) (c) The partial fraction expansion for a step input is C(s)= 1 s + 0:0748 s +5:255 + 0:8089 6 2:2967 s +4:872;j4:006 + 0:8089 6 ;2:2967 s +4:872;j4:006 : 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 01234 Time in Seconds Time Respnse Part (b) Time Response Part (a) Figure 3: Root Locus of Compensated System Thus c(t)=[1+0:0748e ;5:255t +1:62e ;4:872t cos(4t +2:2974)]1(t): A comparison of the time responses is shown in Fig. 3 3