Solution 10.8.5.1 The Bode magnitude and phase plots are shown in gure 1 One wayto -275 -250 -225 -200 -175 -150 Phase in Degrees -100 -80 -60 -40 -20 0 20 40 60 80 Magnitude in Decibels 0.001 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH(s)= 100(s+3)(s+7) s 2 (s+5)(s+8)(s+20) draw the Nyquist plot for this nonstandard contour is to rst drawforthe standard contour and then reverse the direction of the arrows. The pieces and the completed Nyquist plot for the standards contour are shown in gure 2. The Nyquist plot for the nonstandard contour used in this problem is shown in Figure 3 1 Re(GH) Im(GH) GH(I) Re(GH) Im(GH) GH(I,II,I*) (a) (b) Re(GH) Im(GH) I II a (c) Figure 2: Nyquist Plot, for standard contour 2 Re(GH) Im(GH) GH(I ) Re(GH) Im(GH) GH(I,II,I*) (a) (b) Re(GH) Im(GH) I II a (c) * Figure 3: Nyquist Plot, for nonstandard contour 3 From the Bode Magnitude and phase plots we see that for a gain of K =100 GH(j6) = 10 ;20=20 6 ;180  =10 ;1 6 ;180  : Thus for K> 100 10 ;1 =1000;; point `a' is to the left of the point ;1inthe GH-plane. Thus wehavetwo stability cases. For K<10;;000 the point ;1isregion I and there are no encirclements. The Nyquist equation is Z = N +P = 0+0 = 0 There are no closed loop poles in the righthalfofthe s-plane. For K>10;;000 There are twocounterclockwise encirclements of the point ;1intheGH-plane and the Nyquist equation becomes Z = N +P = 2+0 = 2;; and there are twoclosed loop poles in the left half of the s-plane, and two in the righthalfplane. The root locus is shown in gure 4. 4 2 poles -20 Re(s) Im(s) ω = 6 for K = 1000 -3-5-7-8 Figure 4: Root Locus 5