Solution: 10.8.5.4 The Bode magnitude and phase plots are shown in Figure 1, for K =100. One waytodrawthe Nyquist plot for the nonstandard contour is to draw -275 -250 -225 -200 -175 -150 Phase in Degrees -100 -80 -60 -40 -20 0 20 40 60 80 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 = 100(s+1) s 2 (s+2)(s+40) it rst for the standard contour and then reverse the arrows. This work for this particular nonstandard contour because wetraverse segment III in the opposite direction on the nonstandard contour. If not wewould havetobe careful with that segment, but in the presentcase wedonot. 1 8 a II I Figure 2: NYquist plot for standard contour The Nyquist plot for the standard contour is shown in Figure 2 The Nyquist plot for the nonstandard contour is shown in Figure 3 For K = 100 the point`a'is GH(j6:5) = 10 ;24=20 6 ;180  =0:0631 6 ;180  To make GH(j6:5) = 1:0 6 ;180  : The gain must be increased to K = 100 0:0631 =1585 For 0 <K<1585 the point GH = ;1isinregion II, and there is are no encirclements. For the chosen contour in the s-plane P =0because the contour encloses none of the poles of GH. Then Z = N +P = 0+0 = 0 Then, Z =0,sothat none of the closed loop poles are inside the contour in the s-plane, that is, in the righthalfplane for 0 <K<1585. Thus the system is stable for 0 <K<1585. 2 Re(GH) j Im(GH) 8 a II I Figure 3: NYquist plot for nonstandard contour For K>1585, the point GH = ;1isinregion I. Here there are two counterclockwise encirclements of GH = ;1. Since wewentaround the contour in the s-plane in the counterclockwise direction N =2.Thus Z = N +P = 2+0 = 2;; which means that there are twoclosedlooppoles inside the contour in the s-plane, that is, twoclosedlooppoles in the right half plane. Thus, the system in unstable for K>1585. 3