Solution 10.8.7.9 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 2 (s+1)(s+10) Then GH(s) ;1 = s 2 (s+2)(+3) K(s+1) 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 180  goes backto160  ,grows again to 180  and then nally goes 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 500  in the clockwise direction. This is because wehavethree zeros contributing 180  2 10 -1 10 0 10 1 10 2 10 3 -40 -20 0 20 40 60 80 10 -1 10 0 10 1 10 2 10 3 160 180 200 220 240 260 280 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 wedecrease the gain the Nyquist plot expands. The value of gain K that makes GH ;1 =1can be determined from the Bode plots. The phase starts at 180  ,goesto160  then to 180  again before going to 270  . The phase is 180  for ! =1,and the magnitude for K = 100 isj1 Thus the only place where the Nyquist plot touches the real axis is for s =0. Itiseasy enough to see that for s =0 jGH(s) ;1 (j1)j = jsj 2 js+2jjs+3j Kjs+1j s=j1 = ;25 dB = 0:0562: Dividing both sides by0.0562 weobtain jsj 2 js+2jjs+3j 5:62js+1j s=j1 =1: 4 Re(s) Im(s) _ 8 Figure 4: Completed Nyquist plot Thus if If we decrease the gain byafactor of 1 0:0562 =17:78 the point `a' will pass through the point(;1;;0) in the GH ;1 plane. Thus, fore 0 <K<5:62: the point(;1;;0) will be region I. For 5:62K<1, the point(;1;;0) will be region II. Since GH(s) ;1 has no poles inside the righthalf plane, in region I the Nyquist equation is Z = N + P = (;1+1)+0 = 0: N =0becausewehaveonecounterclockwise and one counterclockwise encirclement. Thus there are no closed loop poled in the righthalfplane. In region II the Nyquist equation is Z = N +P = 2+0 = 0;; 5 -12 -10 -8 -6 -4 -2 0 2 4 -10 -8 -6 -4 -2 0 2 4 6 8 10 Real Axis Imag Axis Figure 5: Root locus and there are twoclosedlooppoles in the righthalf plane. The circlements are positivebecause they are clockwise and wewentaround the righthalf plane in the s plane in the clockwise direction. Thus, the the system is stable for K<5:62. This can, of course, be seen from the root locus shown in Figure 5. 6