Solution: 10.8.4.2 The nyquist plot is shown in Figure 1. I II Re(GH) Im(GH) Figure 1: Nyquist Plot The gain that makes 6 GH(j!)=;180  is shown in Figure 3. For K  1000 10 ;26=20 = 19;;952 The there is one counterclockwise encirclementsincewetraversed the contour in the clockwise direction, weattachanegative sign to this encir- clement. The Nyquist equation is then z = N + P = 0+4 = 4 and since the contour in the s-plane encircles the left half of the s-plane, there are four closed loop poles in the left half of the s-plane. For K>19;;952 there are twocounterclockwise encirclements and the system is unstable. That is, z = N +P = ;2+4 = 2 1 and since the contour in the s-plane encircles the left half of the s-plane, there are two closed loop poles in the left half of the s-plane, and twointhe right half plane making the system unstable. 2 -300 -250 -200 -150 -100 -50 0 Phase in Degrees -80 -60 -40 -20 0 20 40 Magnitude in Decibels 0.001 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 2: Bode Magnitude and Phase Plots of GH(s)= K(s+2) s(s+1)(s+10)(s+40) for K = 1000 3 -300 -250 -200 -150 -100 -50 0 Phase in Degrees -80 -60 -40 -20 0 20 40 Magnitude in Decibels 0.001 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 3: Bode Magnitude and Phase Plots of GH(s)= K(s+2) s(s+1)(s+10)(s+40) for K = 1000 4