Solution: 10.8.1.21 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 For K =100the the nyquist plot crosses the negativerealaxisinthe 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. 1 Re(GH) Im(GH) Figure 2: Nyquist plot for standard contour Thus Z = N ;P = 0;0 = 0: Since Z =0,sothat 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 clockwise encirclements. Thus Z = N ;P = 2;0;; 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. 2