Solution: 10.8.2.5 The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of Figure 1. θ 1 θ 2θ 3 α ω Im(s) Re(s) -3 -2 -1 Figure 1: Vector Representation of Components of GH For ! small, ,  2 and  3 are each0  .Thetwopoles at the origin each contribute 90  each at all values of !.Ifweevaluate GH at ! = j, where  is arbitrarily small, we obtain lim !0 6 GH(j)=;180  The only remaining question is how 6 GH(j) approaches ;180  . That is, does it approach ;180  from the second quadrantorthe third quadrant? The table below gives some values ofjGH(j!)jand 6 GH(j!)forK =20. These values help us sketchinthe rst portion of the Nyquist plot. ! 0.2 1 2 jGH(j!)j 84.4 4.0 1.09 6 GH(j!) ;178:2  ;180  ;195  As ! !1, ,  1 ,and 2 all appproch90  ,Sothat lim !!1 6 GH(j!)=;270  : 1 Im(GH) Re(GH) Figure 2: Plot of GH(I) Im(GH) Re(GH) Figure 3: Plot of GH(I;;IIandI  ) 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 four are in the denominator, clearly lim !!1 jGH(j!)j=0 The rst part of the plot is shown in Figure 2 Wesee that the plot crosses the negativerealaxis at about -4.0. As weevaluate GH along part 2 of , weremainessentially at the origin. To draw the portion of the Nyquist plot that corresponds to evaluating GH along part 1  of the contour wesimplytake 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 3. To complete the contour, wemust evaluate GH along the semicircle of radius . Figure 4 depicts this evaluation. Note that all the vectors except that drawn from the origin have nite length, and associated angles of zero. The angles are zero, because the radius of the semicircle is arbitrarily small. This doesn't appear to be the case because the semicircle has a large radius in the picture, while in reality this radius goes to zero. Thus, along the semicircle GH( 6 )  20 6  1  6  6  2 α -1-2-3 θ 1 θ 2 θ 3 Im(s) Re(s) Figure 4: Evaluation of GH on Segment III = 1 6 ;2 As we traverse the semicircle counterclockwise( ccw), evaluating GH, the plot of GH will rotate clockwise (cw) through twice the angle we traversed along the semicircle, or 360  .The nal Nyquist plot is shown in Figure 5. For a gain of 20, GH crosses the negative real axis at -4.0. Clearly,ifwe reduce the gain the plot will cross to the rightof;1 instead of to the left. For a gain of K =20=4=5;; the plot passes through ;1in the GH plane. This information can also be obtained from the Bode magnitude and phase plots shown in Figure 6. For gains below5,thepoint ;1intheGH-plane is in region I.asshown in Figure 5,the number of encirclements is N =0,and the Nyquist equation is Z = N +4 = 0+4 = 4 This means there are four closed loop poles inside in the s-plane. Since encloses the left half of the s-plane, there are no unstable poles for gains below K =2:5. For gains above K>5, ;1isinregion II.Inthis case N =2.Thesign of N is negativebecause both encirclements of ;1arein the clockwise (cw) 3 Im(GH) Re(GH) I II a Figure 5: Completed Nyquist Plot direction, and in our evaluation of GH along wetraveled counterclock- wise(cw). Thus, Z = N + P = ;2+4 = 2 This means that for K>5 there are twoclosedlooppolesinside the contour , that is twoclosed loop poles in the left half of the s plane, and two closed loop poles in the righthalf plane Note that the root locus crosses the j!-axis at w  1:0, as shown in Figure 7 4 -270 -260 -250 -240 -230 -220 -210 -200 -190 -180 -170 Phase in Degrees -160 -140 -120 -100 -80 -60 -40 -20 0 20 40 60 80 100 Magnitude in dB 0.01 0.1 1 10 100 1000 Frequency in Radians/sec. Phase in Degrees Magnitude in dB Figure 6: Bode Magnitude and Phase Plots 5 -1-2-3 Im(s) Re(s) Figure 7: Root Locus 6