Solution 10.8.7.7 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 3 (s +40)(s+50) Then GH(s) ;1 = s 3 (s+40)(s+50) K(s+2)(s+3) 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 wehave vezeroscontributing 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 3 100 150 200 250 Figure 2: Bode magnitude and phase plots 3 Re(GH) Im(GH) Figure 3: B GH(I) eachintheclockwise direction and a twopole scontributing ;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 we decrease the gain the Nyquist plot expands. This does not, however, change the fact that this particular plot has in nite extentforany gain. For K =100 the Nyquist plot crosses the negativerealaxistwice, rst at ! =25rad/s with jGH(j25)j =30dB =31:62;; and again at ! =48rad/s with jGH(j25)j=65dB =1;;778:28: Then for ! =25rad/s, weseethat jGH(j25)j = jsj 3 js+40jjs+50j 100js+2jjs+3j s=j25 = 31:62: 4 Re(s) Im(s) 8 I II III Figure 4: Completed Nyquist plot If we divide both sides by31.62 weobtain jsj 3 js+40jjs+50j 3161:2js+2jjs+3j s=j25 =1: Similarly at ! =48rad/s if we divide by1778.28 we obtain jsj 3 js+40jjs+50j 177;;828js+2jjs+3j s=j48 =1: Note that GH(s) ;1 has no poles inside the righthalfplane, so P =0Then, for 0 <K<3161:2, the point(;1;;0) in the GH plane is in Region I and Z = N +P = 2+0 = 2;; and there are two closed loop poles in the righthalfplane. 5 For 3161:2<K<177828, the point(;1;;0)in the GH plane is in Region II and Z = N + P = (;1+1)+0 = 0;; and there are tno closed loop poles in the righthalfplane. Finally, for 177;;828<K, the point(;1;;0) in the GH plane is in Region III and Z = N +P = 2+0 = 2;; and there are tno closed loop poles in the righthalfplane. This can be seen from the root locus shown in Figure 5. 6 -60 -50 -40 -30 -20 -10 0 10 -20 -15 -10 -5 0 5 10 15 20 Real Axis Imag Axis Figure 5: Root locus 7