Solution 10.8.7.2 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, 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+1)(s;2) Then GH(s) ;1 = (s+ 1)(s;2) 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 righthalfplane, GH ;1 will haveazeroin the righthalfplane and as aresult the phase goes from 180  to 200  and then swings backto90  as ! goes from zero to in nity. Since GH ;1 has twozeros and one pole its 2 10 -1 10 0 10 1 10 2 10 3 -80 -60 -40 -20 0 20 10 -1 10 0 10 1 10 2 10 3 100 120 140 160 180 200 Figure 2: Bode magnitude and phase plots 3 Re(s) Im(s) a b Figure 3: B GH(I) 4 Re(s) Im(s) III II Figure 4: Completed Nyquist plot 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 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;2jjs+1j 100js+4j s=0 = ;44 dB = 0:006: 5 If We divide both sides by 0.006 weobtain jGH ;1 (j0)j= js;2jjs+1j 0:6js+4j s=o =1: For this same gain of 100 the point'b'is at jGH ;1 (j1:5)j= ;40 dB = 0:01: If we decrease the gain byafactor of 100 to 1, the pointbwill pass through the point(;1;;0) in the GH plane. If wedecrease the gain byafactor of 44 dB, or 158.48, the point 'a' will pass through the point(;1;;0). Remember we are working with the inverse of the loop transfer function and K is in the denominator, so that as K increases the points `a' and `b' movetothe right. Thus, for 0 <K<0:6thepoint(;1;;0)will be region I. For 0:6 <K<1, the point(;1;;0) will be region II. Finally,forK>1thethepoint(;1;;0) will be region III. Recalling the GH(s) ;1 has no pole in the righthalf -planem 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 = 2+0 = 2 Thus there are twoclosedlooppoles in the right half plane. Finally,forK>1inregion III there are no encirclements and the Nyquist equation is z = N + P = 0+0 = 0 and there are no closed loop poles in the right half plane. 6 Re(s) Im(s) XX Figure 5: Root locus 7 Thus, the the system is stable for K>1. 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 = ;1 reaches the origin. The gain to achievethis is K =  js;2jjs+1j js+4j  s=j1:5 =1:055;; whichiscloseto the value wecomputed from the Bode plot and Nyquist analysis. 8