Solution 10.8.7.4 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 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+3) (s;1)(s+1) Then GH(s) ;1 = (s;1)(s;1) K(s+3) Wenow draw the Nyquist plot for the contour shown in Figure 1 The Bode phase and magnitude plots are shown in Figure 2. SInce GH has on pole in the righthalfplane, GH ;1 will have one zero in the righthalf plane and as a result the phase goes from 180  to 90  as ! goes from zero to in nity. Since GH ;1 has twozeros and one pole its magnitude will increase without bound as ! !1. 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 180  in the clockwise direction. This is because wehavetwozeros contributing 180  2 10 -1 10 0 10 1 10 2 10 3 -60 -40 -20 0 20 40 10 -1 10 0 10 1 10 2 10 3 80 100 120 140 160 180 200 Figure 2: Bode magnitude and phase plots 3 Re(s) Im(s) Figure 3: B GH(I) 4 Re(s) Im(s) _ 8 a III Figure 4: Completed Nyquist plot 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  and goes to 90  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 (0)j = js;1jjs+1j js+3j s=0 = 0:3333: The Bode magnitude plot was drawn for K =100,and we see from the 5 Bode magnitude plot jGH ;1 (j0)j;50 dB = 0:003162 That is js;1jjs+1j 100js+3j s=0 =0:003162: Dividing both sides by0.03162 weobtain js;1jjs+1j 0:3162js+3j s=0 =1 Thus if If we decrease the gain byafactor of 1 0:003162 =316:23 the point`a' will pass through the point(;1;;0) in the GH plane. The value for K wehave obtained from the Bode plots, namely 0.3162, is pretty close to the precise value of one third. Remember we are working with the inverse of the loop transfer function and K is in the denominator. Thus, fore 0 <K<0:3333: the point(;1;;0) will be region I. For 0:333K<1,thepoint(;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+0 = 1 Thus there is one closed loop pole in the righthalfplane. In region II the Nyquist equation is Z = N +P = 0+0 = 0;; and there are no closed loop poles in the right half plane. Thus, the the system is stable for K>0:03333. This can, of course, be seen from the root locus shown in Figure 5. 6 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 -3 -2 -1 0 1 2 3 Real Axis Imag Axis Figure 5: Root locus 7