Solution: 10.8.4.8 The Bode magnitude and phase plots are shown in Figure 1, for K =100. -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 One waytodraw the Nyquist plot for the contour under consideration is to rst draw GH(I), as shown in Figure 2, and then reverse the direction of the arrows as shown in Figure 3. We can then proceed as before. Along segment II westayattheorgin. When we come down segment I wesimply mirror image GH(I  ). Since we 1 Re(GH) Im(GH) Figure 2: GH(I) Re(GH) Im(GH) Figure 3: GH(I  ) 2 Re(GH) I II Im(GH) Figure 4: Completed Nyquist plot go around segment III in the clockwise direction, wewill swing through an in nite arc in the counterclockwise direction, going 180  for eachpoleatthe origin. The completed Nyquist plot is shown in Figure 4. 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 the point ;1isinregion I. In this region there are no encir- clements of the point(;1;;0) in the GH plane. Then Z = N +P = 0+4 = 4: 3 Thus for K<1259 there are four closed loop poles inside the contour in the s-plane, whichencloses the left half of the s-plane. Thus the system is stable for K<1259. For K>1259 the point ;1isinregion II. In this region there is two counterclockwise encirclements of (;1;;0) . Thus. Z = N + P = ;2+4 = 2: Thus, for K>1259 there are two closed loop poles in the left half of the s-plane and twointherighthalfoftheS plane. 4