Solution: 10.8.4.4 ?ω Im(s) Re(s) -2 -1 θ 1 θ 4 θ 2 θ 3 α β (a) 3 Im(s) Re(s) (b) ?3 2 ?20 ?10 γ Figure 1: Bode Magnitude and Phase Plots The angle contributions of eachofthe poles and zeros along GH(I  ) can be determined from part (a) the vector diagram of Figure 2. For ! small, ,  2 and  3 are each0  .Thetwopoles at the origin eachcontribute ;90  each at all values of !.Ifweevaluate GH at ! = j,where  is arbitrarily small, weobtain 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  1 ?ω Im(s) Re(s) -2 -1 θ 1 θ 4 θ 2 θ 3 α β (a) 3 Im(s) Re(s) (b) ?3 2 ?20 ?10 γ Figure 2: Vector Representation of Components of GH 2 Im(GH) Re(GH) Figure 3: Plot of GH(I) The alternativeistojust take the negatives of the angles on the Bode phase plot. As ! !;1, ,  1 ,and 2 all appproch ;90  ,Sothat lim !!1 6 GH(j!)=270  : At the same time the magnitude of eachofthe vectors in Figure 2 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 3 Wesee that the plot crosses the negativerealaxis twice. 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 wesimplytakethe 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. To complete the contour, wemust evaluate GH along the semicircle of radius . Figure 5 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 3 Im(GH) Re(GH) Figure 4: Plot of GH(I), GH(II), GH(III) XXXX q 2 Re(GH) Im(GH) Figure 5: Evaluation of GH(III) 4 Im(GH) Re(GH) I II III Figure 6: Completed Nyquist Plot semicircle GH( 6 )  20 6  1  6  6  = 1 6 ;2 As wetraverse the semicircle clockwise( cw), evaluating GH,theplot of GH will rotate counterclockwise (ccw) through twice the angle we traversed along the semicircle, or 360  .The nal Nyquist plot is shown in Figure 6. For a gain of 50, GH crosses the negativerealaxis twice. From the Bode plot we see that the rst crossing occurs for s = j2:5, and the magnitude is jGH(j2:5)j=10 ;15=20 =0:1778: The second crossing is at s = j10 with a magnitude of jGH)j10)j=10 ;34=20 =0:01995: Tomakethcrossing at ! =2:5occur at (;1;;;)inthe GH plane wemust increase the gain to 50 0:1778 =281: To makethe crossing at ! =10reach the point(;1;;0) wemust increase the gain to 50 0:01995 =2506: 5 For 0 <K;;281 the point(;1;;0) is in region I and the Nyquist equation is Z = ;2+P = ;2+4 = 2 This means there are twoclosed loop poles inside in the s-plane. Since encloses the left half of the s-plane, there are two unstable poles for gains below K<281. For 281 <K;;2506 the point(;1;;0) is in region II,where wehaveone clockwise and one counterclockwise encirclement, and the Nyquist equation is Z = N +4 = ;1+1+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 281K< 2505. For 2506 <K, the point(;1;;0) is in region III,wehavetwocounter- clockwise encirclementof(;1;;0). The Nyquist equation is Z = ;N +P = ;2+4 = 2 This means that for 2506 <Kthere are twoclosed loop poles inside the contour , that is tw0 closed loop poles in the left half plane. Note that the root locus crosses the imaginary axis at w  2:5, and !  10, as shown in Figure 7 6 -30 -25 -20 -15 -10 -5 0 -10 -5 0 5 10 Figure 7: Root Locus 7