Solution: 10.8.1.16 The Bode magnitude and phase plots are shown in Figure 1, for K =100. The Nyquist plot is shown in Figure 2. For K = 100 GH(j8:5) = 10 ;22=20 6 ;180  = 0:0794 6 ;180  Togetpoint`a'to be 1 6 ;180  The gain must be K = 100 0:0794 = 1258:9 1259 For K<1259 there are no closed loop poles inside the contour in the s-plane, which encloses the righthalfofthe s-plane. Thus N =0 = Z ;P = Z ;0 or Z =0;; and the system is stable for K<1259. For K>1259 the point ;1isinregion I. In this region there is two clockwise encirclements of (;1;;0) in the GH plane. Thus. N =2 = Z ;P = Z ;0 or Z =2: Thus, for K>1259 there are twoclosed loop poles in the right half of the s-plane. 1 -300 -250 -200 -150 -100 Phase in Degrees -140 -120 -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 = K(s+1) s 2 (s+4)(s+20) For K =100 2 a III Im(GH) Re(GH) Figure 2: Plot in in GH-plane 3