Solution: 10.8.5.5 The Bode magnitude and phase plots are shown in Figure 1. The Nyquist -275 -250 -225 -200 -175 -150 Phase in Degrees -80 -60 -40 -20 0 20 40 60 Magnitude in Decibels 0.001 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH = K(s+1) s 2 (s+2)(s+10) For K =100 plot for the standard contour around the right half plane is shown in Figure 2 The Nyquist plot for the nonstandard contour of this problem is shown in Figure 3. The Nyquist plot for the nonstandard contour can be drawn by rst drawing the Nyquist plot for the standard contour and then reversing the direction of the arrows. This works because wegoaround segmentIII of clockwise on the standard contour and counterclockwise on the non standard contour. If this wer not the case wewould havetobecareful with segmentIII. For K =100the the nyquist plot crosses the negativerealaxisinthe 1 Re(GH) Im(GH) Figure 2: Nyquist plot for standard contour Re(GH) Im(GH) Figure 3: Nyquist plot for standard contour 2 GH-plane at GH = ;1. That is GH(j3) = 10 0=20 6 ;180  =1 6 ;180  This can be determined from the Bode magnitude and phase plots as shown in Figure 1 For 0 <K<100 The point GH = ;1isinregion I, and there is are no encirclements of the point ;1. For the chosen contour in the s-plane P =0. Thus Z = N +P = 0+0 = 0: SinceZ =0,sothat all no closed loop poles inside the contour in the s-plane, that is, in the righthalf plane for 0 <K<100. Thus the system is stable for 0 <K<100. For K>100, the point GH = ;1isinregion I. Here there are two counterclockwise encirclements 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>100. 3