Solution 10.8.7.3 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+3) (s;1)(s;2) Then GH(s) ;1 = (s;)(s;2) 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 twopoles in the right half plane, GH ;1 will havetwozeros in the righthalfplane and as a result the phase goes from 380  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 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 3 100 150 200 250 300 350 Figure 2: Bode magnitude and phase plots 3 Re(s) Im(s) Figure 3: B GH(I) 4 Re(s) Im(s) _ 8 a I II Figure 4: Completed Nyquist plot counterclockwise direction. This is because wehavetwozeros 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  and reaches 180  for the second time when ! =1:5rad/s. The Bode magnitude plot was drawn for K =100,and we see from the Bode magnitude plot that the point'a'isat jGH ;1 (j0)j = js 1 jjs 2 j 100js+3j s=0 5 = ;30 dB = 0:03162 Then if we divide both sides by0.03162 weobtain js 1 jjs 2 j 3:162js+3j s=0 =1: Thus, the pointawillpass through the point(;1;;0) in the GH plane for K =3:162. Remember weareworking with the inverse of the loop transfer function and K is in the denominator. Thus, for 0 <K<3:162 the point(;1;;0) will be region I. For 3:162K< 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 = 2+0 = 2 Thus there are twoclosedlooppoles in the right half plane. 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>3:12. This can, of course, be seen from the root locus shown in Figure 5. It is easy calculate the gain for which the system becomes stable. This occurs when the closed loop poles reach the imaginary axis at about ! =3:2. that originates from the loop transfer function pole at s = ;1reaches the origin. The gain to achieve this is K = js;2jjs;1j js+3j s=j3:2 =2:88;; whichiscloseto the value wecomputed from the Bode plot and Nyquist analysis. 6 -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 7