Solution: 10.8.1.1 For GH(s)= K(s+4) (s+2)(s;1) The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of gure 1. For ! =0, =  2 =0  while  1 =180  . θ 1 θ 2α Im(s) Re(s) -4 -2 1 Figure 1: Vector Components of GH Thus 6 GH(j0) = ;180  : For K =1, jGH(j0)j= 4 21 =2 As ! !1, ,  1 ,and 2 all appproch90  ,Sothat lim !!1 6 GH(j!)=;90  : At the same time the magnitude of eachofthe vectors in Figure 1 be- comes in nite in length. Since one of the vectors is in the numerator of GH and twoare in the denominator, clearly lim !!1 jGH(j!)j=0 This can also be seen from the Bode plots in Figure 2, where 6 GH ap- proaches ;90  and the magnitude plot is descending at -20 db/decade for large !. The rst part of the nyquist plot is shown in Figure 3. Weseethat the 1 -200 -175 -150 -125 -100 -75 Phase in Degrees -60 -40 -20 0 20 Magnitude in dB 0.01 0.1 1 10 100 1000 Frequency in Radians/sec. Phase in Degrees Magnitude in dB Figure 2: Bode Magnitude and Phase Plots of GH Re(GH) Im(GH) -2 for K = 1 Figure 3: Plot of GH(I) 2 Re(GH) Im(GH) I II -2 for K = 1 Figure 4: Completed Nyquist Plot plot starts on the negativerealaxisofthe GH-plane at ;2forK =1. As weevaluate GH along part II of , weremainessentially at the origin. To draw the portion of the Nyquist plot that corresponds to evaluating GH along part I  of the contour wesimply take the mirror image through the real axis of the rst part of the Nyquist contour shown above. Thus, the contour generated byevaluating GH along portions I, II, and I  is shown in Figure 4. This completes the contour. Because GH has no poles at the origin, we do not havetodetour around the origin of the s-plane on a semicircle of radius . From the Nyquist plot, weseethat for a gain of 1, GH crosses the negativereal axis at ;2. Clearly,ifwedecrease the gain to 0.5, the plot will cross the imaginary axis rightat;1. Thus, for K>0:5 the plot will cross the imaginary axis of the GH-plane to the left of ;1. For K<0:5, the point ;1isinregion I, and the numberofencirclements is N =0.The Nyquist equation is Z = N +P = 0+1 = 1 This means there is one closed loop poles inside in the s-plane. Since encloses the right half of the s-plane, there is one unstable pole for K<0:5. For K>0:5 The point ;1inthe GH-plane is in region II. In this case N = ;1. The sign of N is negativebecause the encirclement of ;1isinthe counterclockwise (ccw) direction, and in our evaluation of 3 1 Re(s) Im(s) -4 -2 Figure 5: Root Locus GH along we traveled clockwise(cw). Thus, Z = N + P = ;1+1 = 0 This means that forK>0:5there are no closed loop poles inside the contour , that is twoclosed loop poles in the left half of the s-plane. The root locus is shown in Figure 5. The break-out pointistothe left of the origin in the s-plane. 4