Solution: 10.8.1.19 The Bode magnitude and phase plots are shown in Figure 1, for K =100. The Nyquist plot is shown in Figure 2 For K =100the point GH = ;1is in region II. That is GH(j15) = 10 ;35=20 6 ;180  =0:0178 To make GH(j15) = 1:0 6 ;180  : The gain must be increased to K = 100 0:0178 = 5;;623 For 0 <K<5;;623 the point GH = ;1isinregion II, and there are no encirclements of GH = ;1, For the chosen contour P =0because the contour does not enclose the poles of GH.Then N =0 = Z ;P = Z ;0 Then Z =0implying that none of the closed loop poles are inside the contour in the s-plane, that is, in the righthalf plane for 0 <K<5;;623. For K>5;;623, 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 N =2 = Z ;P = Z ;0 Thus Z =2 implying that there are twoclosed loop poles inside the contour in the s- plane, that is, twoclosedlooppoles in the righthalfplane. Thus, the system in unstable for K>5;;623. 1 -300 -250 -200 -150 -100 -50 Phase in Degrees -80 -60 -40 -20 0 20 Magnitude in Decibels 0.01 0.1 1 10 100 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH = 100(s+2) s(s+4)(s+8)(s+20) 2 a II I Re (GH) Im(GH) Figure 2: Plot in in GH-plane 3