Solution: 10.8.5.3 The Bode magnitude and phase plots are shown in Figure 1, for K =100. -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) One waytodraw the Nyquist plot for this nonstandard contour, is to drawthecontour for the standard contour and then reverse the direction of the arrows. The Nyquist plot for the standard contour is shown in Figure 2 The Nyquist plot for the nonstandard contour of this problem is shown in Figure 3. 1 a II I Re (GH) Im(GH) Figure 2: Nyquist plot for standard contour 2 a II I Re (GH) Im(GH) Figure 3: Nyquist plot for nonstandard contour 3 For K = 100 the point GH = ;1isinregion II.Thatis 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 no encirclements of GH = ;1, Then Z = N +P = 0+0 = 0 Then Z =0implying that there are no closed loop poles are inside the contour in the s-plane, that is, in the =righthalfplane for 0 <K<5;;623. For K>5;;623, the point GH; = ;1isinregion I. Here there are two counterclockwise one encirclementof`a'since wewent around the contour in the s-plane in the counterclockwise direction N =2.Thus Z = N +P = 2+0 = 2 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. 4