Solution: 10.8.1.20 The Bode magnitude and phase plots are shown in Figure 1, for K =100. The Nyquist plot is shown in Figure 2 For K =100thepoint`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 Nsothat none of the closed loop poles are inside the contour in the s-plane, that is, in the right half of the s plane for 0 <K<1585. Thus the system is stable for 0 <K<1585. For K>1585, the point GH = ;1isinregion I. Here there are two clockwise encirclements of GH = ;1. Since wewentaround the contour in the s-plane in the clockwise direction N =2.Thus z = N ;P =2;0=2 which means that there are twoclosed loop poles inside the contour in the s-plane, that is, two closed loop poles in the righthalfplane. Thus, the system in unstable for K>1585. 1 -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) 2 8 a II I Im(GH) Re(GH) Figure 2: Plot in in GH-plane 3