Solution 8.8.12 The overall forward loop transfer function is G c (s)G p (s)= 15K c (s +1) s(s +3:2)(s+ ) : The rst assigned task it nd the gain that critically damps the system for =12and =15.Figure 1 shows howwe make the calculation of gain for apointonthereal axis between s = ;3:2ands = ; . The gain is Re(s) Im(s) X X X -3.2 s + 3.2 s s + γ -1 s + 1 ?γ Figure 1: Calculating gain along negativerealaxis K =15K c =  jsjjs +3:2jjs+ j js +1j  : Wehaveelected to compute thecomposite gains 15K c . The MATLAB program z=3 p1 =0 p2 = 3.2 p3 = 12 x=linspace(-10,-4,600);; K1 = (abs(x + p1).*abs(x+p2).*abs(x+p3))./abs(x + z);; p3 = 15 K2 = (abs(x + p1).*abs(x+p2).*abs(x+p3))./abs(x + z);; plot(x,K1,'k-',x,K2,'k--') print -deps bo8812a.eps 1 -10 -9 -8 -7 -6 -5 -4 15 20 25 30 35 40 45 50 55 γ = 15 γ = 12 Figure 2: Break out points for =12and 15 K12 = K1(366) s12 = x(366) K15 = K2(225) s15 = x(225) plots the gains along the negativerealaxisfor both =12and =15.The plot is shown in Figure 2. For =12thebreak out is at s = ;7:3957 and K =22:3388. For =15thebreak out is at s = ;8:8581 and K =39:1739. Wenowhavetwovalues of 15K c for whichtodoaroot contour. We rst nd the characteristic polynomial. Wehave 1+G c G p = 1+ 15K c (s +1) s(s +3:2)(s+ ) = s(s +3:2)(s+ )+15K c (s +1) s(s +3:2)(s+ ) = s 3 +(3:2+ )s 2 +3:2 s+15K c (s +1) s(s +3:2)(s+ ) = s 3 +(3:2+ )s 2 +(3:2 +15K c )s +15K c s(s +3:2)(s+ ) : 2 The characteristic equation is then s 3 +(3:2+ )s 2 +(3:2 +15K c )s +15K c =0;; whichwe can rewrite as (s 3 +3:2s 2 +15K c s +15K c )+ s 2 +3:2 s=0: Dividing through by s 3 +3:2s 2 +15K c s +15K c ;; weobtain p(s)=1+ s(s +3:2) s 3 +3:2s 2 +15K c s +15K c : If wechoose one of the twovalues of 15K c wecan factor the denominator and drawtheroowlocus with playing the role of the variable gain. Now, for 15K c =22:3388, we obtain p(s) = 1+ s(s+3:2) s 3 +3:2s 2 +15K c s +15K c = 1+ s(s +3:2) (s +1:1162)(s+1:0419;j4:3505)(s+1:0419+ j4:3505) : The full root contour for 0 < <1 is shown in Figure 3. If werestrict the range of so that 12 < <15, then wegetthe root contour shown in Figure 4 Now, for 15K c =39:1739, we obtain p(s) = 1+ s(s +3:2) s 3 +3:2s 2 +15K c s +15K c = 1+ s(s +3:2) (s +1:0615)(s+1:0692;j5:98)(s+1:0692+ j5:98) : The full root contour for 0 < <1 is shown in Figure 5. If werestrict the range of so that 12 < <15, then wegetthe root contour shown in Figure 6 The complete root contour for 12 < <15 for both K =22:3382 and K =39:1739 is shown in Figure 7. 3 -40 -35 -30 -25 -20 -15 -10 -5 0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Figure 3: Full root contour for 0 < <1, K =22:3388. 4 -14 -12 -10 -8 -6 -4 -2 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure 4: Full root contour for 12 < <15, K =22:3388. 5 -50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0 -8 -6 -4 -2 0 2 4 6 8 Figure 5: Full root contour for 0 < <1, K =39:1739. 6 -50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0 -8 -6 -4 -2 0 2 4 6 8 Figure 6: Root contour for 12 < <15, K =39:1739. 7 -14 -12 -10 -8 -6 -4 -2 0 -4 -3 -2 -1 0 1 2 3 4 Figure 7: Full root contour for 12 < <15,K =22:3382 and K =39:1739. 8