Solution 10.8.9.5 The MATLAB program % % Note when matlab takes the transpose %ofarow matrix with complet elements %itCHANGES THE SIGN of the complex part % x=linspace(-1,-4,100);; s=x+j*x;; y=linspace(0.001,1);; s1 = -1 - j*y;; z=linspace(4,0,100);; s2 = -4 -j*z;; s = s';; s1 = s1';; s2 = s2';; s=[s1;; s;;s2] K=10 z=2 p1 = 0 p2 = 1 p3 = 10 g=(K*(s + z))./((s + p1).*(s + p2).*(s+p3)) realg = real(g) imagg = imag(g) figure(1) plot(realg,imagg) grid on axis([-2 0 -0.15 1]) print -deps 10895polara.eps figure(2) plot(realg,imagg) grid on axis([-0.8 -0.15 -0.05 0.05]) print -deps 10895polarb.eps K=linspace(0 ,50,1000);; gcgp = zpk([-z],[-p1 -p2 -p3],10) [R,K] = rlocus(gcgp,K);; figure(3) plot(R,'k.') grid on axis([-9,1,-5,5]) print -deps 10895rl.eps 1 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Figure 1: Large scale polar plot K4 = 10 s4 = -1 + j*0.001 g4 = (K4*(s4 + z))/((s4 + p1)*(s4 + p2)*(s4+p3)) Draws the polar plots shown in Figures 1 and 2, for the contour shown in Figure 3. The complete Nyquist plot is shown in Figure 4 For K =10 pointa = ;0:485 pointb = ;0:275 pointc = ;0:24 For K< 10 0:485 =20:6185;; there is one clockwise encirclement. The pole at s = ;1isinside the contour : Thus Z = N + P 2 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05 Figure 2: Small scale polar plot Re(s) Im(s) Figure 3: Contour 3 Im(G G ) cp Re( )G G cp a b c 8- Figure 4: Complete Nyquist plot 4 = ;1+1 = 0;; and none of the closed loop poles are inside the contour. For 20:6185<K< 10 0:275 =36:36 There is one counterclockwise encirclementand Z = N +P = 1+1 = 2;; and there are two closed loop poles inside the contour. This means that the closed loop poles that started at s = ;1ands =0are now inside . The closed loop pole that originated at s = ;10 is still outside . For 36:36 <K<10=0:24 = 41:67 there are two counterclockwise encirclements and so Z = N +P = 2+1 = 3;; and the all three closed loop poles are now inside the contour again. For 41:67 <K weagain havenoencirclements and Z = N +P = 0+1 = 1;; and the pole that originated at s = ;10 is still inside the but the dominant poles that started at s =0and s = ;1nolonger are. Thus the dominantpoles are inside the triangular area for 20:6185 <K<40:67: All this can be seen from the root locus in Figure 5. 5 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 -5 -4 -3 -2 -1 0 1 2 3 4 5 Figure 5: Root locus 6