Solution 10.8.7.1 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 = (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+4) (s+2)(s;1) Then GH(s) ;1 = (s+ 2)(s;1) K(s+4) 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 a pole in the right half plane, GH ;1 will haveazero in the right half plane and as a 2 -200 -175 -150 -125 -100 -75 Phase in Degrees -60 -40 -20 0 20 Magnitude in dB 0.01 0.1 1 10 100 1000 Frequency in Radians/sec. Phase in Degrees Magnitude in dB Figure 2: Bode magnitude and phase plots 3 Re(s) Im(s) Figure 3: B GH(I) 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  eachintheclockwise direction and a pole contriubting ;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 byevaluating GH ;1 (0), because 4 Re(s) Im(s) Figure 4: Completed Nyquist plot 5 it is for ! =0that the angle of GH ;1 = ;180  .Thus, the gain that causes the Nyquist plot to pass through (;1;;0) in the GH plane is jGH ;1 (0)j = j0+2jj0;1j Kj0+4j = 1 2K =1: Thus K =0:5: The for K<0:5 the Nyquist plot encircles the point(;1;;0) in the GH plane and the Nyquist equation is Z = N +P = 1+0 = 1 Thus there is one closed loop pole in the righthalfplane. For K>0:5 there are no encirclements and Z = N +P = 0+0 = 0 The the system is stable for K>0. 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 pole that originates from the loop transfer function pole at s = ;1reaches the origin. The gain to achieve this is K = js;1jjs+2j js+4j s=0 =0:5 6 Re(s) Im(s) XX Figure 5: Root locus 7