Solution 10.8.4.5 The Bode magnitude and phase plots are shown in gure 1 As can be seen -275 -250 -225 -200 -175 Phase in Degrees -60 -40 -20 0 20 40 60 Magnitude in Decibels 0.1 110100 1000 Frequency in Radians/Sec. nyq10abode.dat Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH = K(s+4) s 2 (s+6)(s+20) For K =100 from the phase plot the Nyquist plot in the GH just barely gets into the third quadrantbefore crossing into the third quadrant. Wewanttodraw the Nyquist plot in reverse order. That is wewanttodrwGH(jI  ) rst, starting at ;j.Todothatwehavetoreverse the sign of all the angles on the bode phase plot. The pieces of the Nyquist plot and the completed plot are shown in gure 2. From the Bode Magnitude and phase plots wesee 1 Re(GH) Im(GH) GH(I) Re(GH) Im(GH) GH(I,II,I*) (a) (b) Re(GH) j Im(GH) I II a (c) Figure 2: Nyquist Plot, By Stages and Final that for a gain of K =100 GH(j3) = 10 ;12=20 6 ;180  =0:2511 6 ;180  : Thus for K> 100 0:2511 =398;; point `a' is to the left of the point ; in the GH-plane. Thus wehavetwo stability cases. For K<398 the point ;1isregion I and there are no encirclements. The Nyquist equation is Z = N +P = 0+4 2 = 4 There are no closed loop poles in the righthalfofthe s-plane. For K>398 There are twocounterclockwise encirclements of the point ;1intheGH-plane and the Nyquist equation becomes Z = N + P = ;2+4 = 2;; and there are twoclosedlooppoles in the left half of the s plande and two in the righthalfofthe s-plane. The root locus is shown in gure 3. 2 poles -20 -6 -4 Re(s) Im(s) j 3 for K = 398 Figure 3: Root Locus 3