Solution 10.8.7.8 The characteristic equation is 1+GH(s)=0: If we let GH(s)= (s) (s) ;; then we can write 1+ (s) (s) =0;; or (s)+ (s) (s) =0: Clearly,another waytowrite the characteristic equation is (s)+ (s)=0: We next divide both sides of 1+GH(s)=0 by GH(s)toobtain 1 GH(s) +1=0: We then rewrite this last expression using GH(s)= (s) (s) ;; to obtain 1 GH(s) +1 = 1 (s) (s) +1 = () (s) +1 = (s)+ (s) (s) : Thus, we see that, as must be the case, 1+GH(s)=0;; 1 Im(s) Re(s) III I* Figure 1: Root locus and GH(s) ;1 +1=0 are simply di erentways of writing the characteristic equation. If we use GH(s) ;1 +1=0;; Then wesimply reverse the roles of the poles and zeros of GH(s), but ev- erything else remains the same. Now consider GH(s)= K(s+2) s 3 (s+6) Then GH(s) ;1 = s 3 (s+10) K Wenow draw the Nyquist plot for the contour shown in Figure 1 Note that the phase will be 180  as weapproach s =0,sincewe get an angle of 90  from each zero at the origin. The Bode phase and magnitude plots are shown in Figure 2. Note that the phase starts at 270  goes backto240  ,grows again to 270  as the magnitude of GH(j!)becomes in nitely large. Thus, the Nyquist plot for part I of the contour is shown in Figure 3 As weevaluate GH ;1 along the semicircle of ini nite radius, the mag- nitude of GH ;1 remains in nite, but the angle rotates through 540  in the clockwise direction. This is because wehavefour zeros contributing 180  2 10 -1 10 0 10 1 10 2 10 3 -60 -40 -20 0 20 40 60 10 -1 10 0 10 1 10 2 10 3 260 280 300 320 340 360 380 Figure 2: Bode magnitude and phase plots 3 Re(s) Im(s) Figure 3: B GH(I) eachintheclockwise direction and a pole contributing ;180  in the clock- wise direction. The evaluation of GH ;1 along I  is just the mirror image of GH(I). The completed Nquist plot is shown in Figure 4. Wenowcan complete the analysis. One thing to remember. The gain K is in the denominator so that as we increase K the Nyquist plot shrinks and as we decrease the gain the Nyquist plot expands. This does not, however, change the fact that this particular plot has in nite extentforany gain. Since GH(s) ;1 has no poles inside the righthalf plane, weseethat for all values of K there are two clockwise encirclements of the point(;1;;0) in the GH ;1 plane. The twoclockwise encirclements are positivebecause we wentaround the right half of the GH ;1 plane in the clockwise direction. Thus, Z = N +P = 2+0 = 2;; and there are twoclosed loop poles in the righthalfplane for all positive values of gain. Thus the system is always unstable. This can, of course, be seen from the root locus shown in Figure 5. 4 Re(s) Im(s) _ 8 Figure 4: Completed Nyquist plot 5 -10 -8 -6 -4 -2 0 2 4 -5 -4 -3 -2 -1 0 1 2 3 4 5 Real Axis Imag Axis Figure 5: Root locus 6