Solution 10.8.7.10 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, as must be the case, that 1+GH(s)=0;; 1 j Im(s) Re(s) III I* r = 8 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(s+ 1)(s+10)(s+40) Then GH(s) ;1 = s(s +1)(s+10)(s+40) K(s+2) 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 90  and goes to 270  as the magnitude of GH(j!)be- comes in nitely large. Thus, the Nyquist plot for part I of the contour is shown in Figure 3 2 10 -1 10 0 10 1 10 2 10 3 -40 -20 0 20 40 60 80 100 120 10 -1 10 0 10 1 10 2 10 3 100 150 200 250 Figure 2: Bode magnitude and phase plots Re(s) Im(s) a Figure 3: B GH(I) 3 Re(s) Im(s) a II I 8 Figure 4: Completed Nyquist plot 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 counterclockwise direction. This is because wehave three zeros contributing 180  eachintheclockwise direction and a pole contributing ;180  in the clockwise 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 ! =20,For s = j20 jGH(s) ;1 (j20)j = jsjjs+1jjs+10jjs+40j Kjs+2j s=j20 = 28 dB = 25:118;; for K = 1000. Dividing both sides by25,118 weobtain jsj 2 js+2jjs+3j 25;;118js+1j s=j1 =1: 4 Thus, if If we increase the gain byafactor of 25:11 the point `a' will pass through the point(;1;;0) in the GH ;1 plane. Thus, fore 0 <K<25;;118 the point(;1;;0) will be region I. For 25;;118K<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;; 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<25;;118. This can, of course, be seen from the root locus shown in Figure 5. 5 -50 -40 -30 -20 -10 0 10 -25 -20 -15 -10 -5 0 5 10 15 20 25 Real Axis Imag Axis Figure 5: Root locus 6