Solution: 10.8.2.2 The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of Figure 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 plot starts on the negativerealaxis of the GH-plane at ;2 for K =1.From the Bode plot we see that the plot moves into the second quadrantbefore recrossing the imaginary axis in the GH-plane and nally approaching the origin at an angle of ;90  . 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 2 Re(GH) Im(GH) -2 for K = 1 Figure 3: Plot of GH(I) Re(GH) Im(GH) -2 for K = 1 I II III a b Figure 4: Completed Nyquist Plot 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, point `a' of the plot will cross the imaginary axis rightat;1. Thus for 0 <K<0:5point `a' will be to the rightof;1, placing ;1inregion I. As the gain increases beyond K =0:5, the point ;1willbeinregion II. Further increasing the gain will eventually put point `b' at ;1. The gain to do this can be obtained from the bode plot. That is, for K =1GH(j1:3) = ;1 6 ;180  .Thus, ;1 enters reqion III when k =1. 3 We then havethree cases: 1. For 0 <K<0:5, the point ;1isinregion I and there are no encir- clements. The Nyquist equation is Z = N +P = 0+1 = 1 This means there is one closed loop pole inside in the s-plane. Since encloses the left half of the s-plane, there is one stable pole and one unstable pole for 0 <K<0:50. 2. For 0:5 <K<1thepoint ;1isinregion II and there is one clockwise encirclementof;1. Since wewent around in the counterclockwise direction, wehaveone negativeencirclement. Thus the Nyquist equa- tion is Z = N +P = ;1+1 = 0 Thus for 0:5 <K<1 there are no closed loop poles in the left half plane. 3. For K>1, the point ;1isinregion III and wehaveonecounter- clockwise encirclement leading to one positiveencirclement. Thus the Nyquist equation is Z = N +P = 1+1 = 2 Thus for K>1there are no unstable poles. The root locus is shown in Figure 5. The break-out pointistothe left of the origin in the s-plane. 4 Re(s) Im(s) -4 2 -1 ω = 1.3 for K = 1 Figure 5: Root Locus 5