solution: 10.8.3.6 The shape of the Bode plots can be determined qualitatively byrepre- senting GH in vector form as shown in Figure 1. θ 1 θ 2 α ω Im(s) Re(s) Figure 1: Vector Representation of GH As ! ! 0, =  2 ! 0  : The three poles at the origin eachcontribute 90  at all values of !.Ifwe evaluate GH at ! = j, where  is arbitrarily small, we obtain lim !0 6 GH(j)=;270  As ! !1,the magnitude of GH shrinks to zero and the phase goes to ;270  . This can be seen from Figure 1, since the magnitude of eachvector increases to 1 and the angle of eachvector approaches 90  . The only remaining question is whether the Nyquist plot will cross the negativereal axis. By looking at the Bode plots weseethat 6 GH never reaches ;180  . If wedonot wanttogenerate detailed Bode plots, wecaninstead evaluate the phase of GH along the j!-axis using Figure 1. Table 1 below gives some values of jGH(j!)j and 6 GH(j!)which help us sketchinthe rst portion of the Nyquist plot. As ! !1, and  2 appproch90  ,Sothat ! 0.5 1 2 3 4 5 6 10 jGH(j!)j 8.63 1.83 0.27 0.01 - - - - 6 GH(j!) ;260  ;252  ;243  ;240  ;240  ;241:6  ;244  ;250  Table 1: Calculation of Angles Along Imaginary Axis lim !!1 6 GH(j!)=;270  : 1 Im(GH) Re(GH) Figure 2: Plot of Gh(I) 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. Weseethat the plot never crosses the negative real axis. As weevaluate GH along part II of , we remain essentially 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 3. To complete the contour, wemust evaluate GH along the semicircle of radius , that is, Part III of the contour. 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. Thus, along the semicircle GH( 6  1 )  10 6  1  6  1  6  1  6  1 = 1 6 ;3 1 As wetraverse the semicircle clockwise( cw), evaluating GH,theplot of GH will rotate counterclockwise (ccw) through thrice the angle we traversed 2 Im(GH) Re(GH) Figure 3: Im(s) Re(s) θ θ 1 2 α X Figure 4: Evaluation of GH Along Segment III 3 Im(GH) Re(GH) Figure 5: Nyquist Plot for Alternative Contour along the semicircle, or 540  .The nal Nyquist plot is shown in Figure 5. Note that the overall traverse of is in the s-plane is counterclockwise, but locally on part III of , werotate clockwise Note that for all K>0 there is one encirclementof;1inthe GH-plane, but that the encirclementisin the counterclockwise (ccw) direction. Since in evaluating GH along in the s-plane we also travel ccw around , the sign of the encirclementispositive. However, three of the poles of GH are putside , namely the poles at s =0, and hence P =1.The Nyquist equation is thus Z = N +P = 1+1 = 2 Thus, there are two closed loop poles inside , that is, in the left half plane for all K>0. Since there are a total of four closed loop poles, the other twomust be in the righthalfplane, for all K>0. This is same result we obtained with the other contour. The root locus is shown in Figure 6. 4 -2-6 Im(s) Re(s) Figure 6: Root Locus 5