Solution 10.8.4.7 The Bode magnitude and phase plots are shown in gure 1 One waytodraw -275 -250 -225 -200 -175 -150 Phase in Degrees -40 -20 0 20 40 Magnitude in Decibels 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Figure 1: Bode Plots of GH = K(s+2) s 2 (s+5)(s+30) For K =100 the Nyquist plot for this nonstandard contour is to rs draw the plot for the standard contour, as shown in Figure 2, and then reverse the direction of the arrows as shown in Figure 3. From the Bode Magnitude and phase plots we see that for a gain of K =100 GH(j9) = 10 ;30=20 6 ;180  =0:03162 6 ;180  : Thus for K> 100 0:03162 =3162;; point `a' is to the left of the point ;1inthe GH-plane. Thus wehavetwo stability cases. 1 Re(GH) Im(GH) GH(I) Re(GH) Im(GH) GH(I,II,I*) (a) (b) Re(GH) j Im(GH) I II a (c) Figure 2: Nyquist Plot, standard contour 2 Re(GH) Im(GH) GH(I) Re(GH) Im(GH) GH(I,II,I*) (a) (b) Re(GH) j Im(GH) I II a (c) Figure 3: Nyquist Plot, standard contour 3 For K<3162 the point ;1isregion I and there are no encirclements. The Nyquist equation is Z = N +P = 0+4 = 4 There are four closed loop poles in the left half of the s-plane, that is inside . For K>3162 point`a'is t0 the rightof;1intheGH-plane. Wenow havetwo counterclockwise encirclements. The Nyquist equation is now Z = N + P = ;2+4 = 2;; and there are twoclosedlooppoles in the left half of the s-plane. Since there are four closed loop poles altogether, twomust be in the righthalf plane. The root locus is shown in gure 4. 2 poles -20 -6 -4 Re(s) Im(s) ω = 3 for K = 398 Figure 4: Root locus 4