Solution 10.8.5.6 The Bode magnitude and phase plots are shown in Figure 1, for K =1000. The Nyquist plot for the standard contour is shown in Figure 2 -275 -250 -225 -200 -175 -150 -125 -100 -75 -50 -25 0 25 Phase in Degrees -120 -100 -80 -60 -40 -20 0 20 Magnitude in Decibels 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH = 5(s+2) (s+1+j)(s+1;j)(s+20)(s+40) The Nyquist plot for the nonstandard contour used in this problem is shown in Figure 3 For this particular nonstandard contour all wehavetodoisreverse the dirctions of the arrows. This works because wetraverse segmentIIIin opposite directions on the twocontours. For K = 100 the point`a'is GH(j30) = 10 ;33=20 6 ;180  =0:02239 6 ;180  1 a I II Im(GH) Re(GH) Figure 2: Plot in in GH-plane To make GH(j30) = 1:0 6 ;180  : The gain must be increased to K = 1000 0:02239 = 44;;670 For 0 <K<44;;670the point GH = ;1isinregion I, and there are no encirclements of ;1. For the chosen contour in the s-plane, all the poles of GH are outside the contour. Thus the Nyquist equation is Z = N + P = 0+0=0 Thus for 0 <K<44;;670 The system is stable. For K>44;;670 the point GH = ;1isinregion II, and there are thus twocounterclockwise encirclements of the point GH = ;1. Thus Z = N +P = 2+0 = 2 2 a Im(GH) Re(GH) I II Figure 3: Nyqusit plot for nonstandard contour Thus, there are twoclosed loop poles in the righthalf plane. 3