Solution 10.8.1.15 The Bode magnitude and phase plots are shown in gure 1 The pieces of -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 and the completed plot are shown in gure 2. From the Bode Magnitude and phase plots weseethat 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) Im(GH) I II a (c) Figure 2: Nyquist Plot, By Stages and Final For K<3162 the point ;1isregion I and there are noencirclements. The Nyquist equation is Z = N +P = 0+0 = 0 There are no closed loop poles in the righthalfofthes-plane, that is inside . For K>3162 point`a'is to the rightof;1intheGH-plane. Wenow havetwo clockwise encirclements. The Nyquist equation is now Z = N +P 2 = 2+0 = 2;; and there are two closed loop poles in the right half of the s-plane. Since there are four closed loop poles altogether, twomust be in the left half plane. The root locus is shown in gure 3. 2 poles -20 -6 -4 Re(s) Im(s) ω = 3 for K = 398 Figure 3: Nyquist Plot, By Stages and Final 3