Im(s) Re(s) -50 -40 -3 -2 θ 2 θ 1 θ 3 α β ω Figure 1: Computation of Magnitude and Phase of GH solution: 10.8.2.7 The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of Figure 1. For ! small, , ,  1 and  2 are each 0  . The three poles at the origin eachcontribute 90  at all values of !. If weevaluate GH at ! = j, where  is arbitrarily small, weobtain lim !0 GH(j)=1 6 ;270  That is, the three arbitrarily short vectors in the denominator of GH send the magnitude of GH to in nityforany xed gain K. The next question is how GH(j)approaches the origin of the GH-plane. That is, does the plot of GH go into the third quadrant and then return to the second quadrant, or does it stayinthe second quadrant. Figure 2 shows the Bode magnitude and phase plots for K =1.The phase plot shows that 6 GH equals ;180  twice, rst at ! =2rad./sec. and again at ! =40 rad./sec. In between these frequencies ;180  < 6 GH < ;90  .Sothe plot of GH actually starts in the second quadrant, enters the third quadrant and then returns to the second quadrant. As ! !1, , ,  1 , and  2 all appproch90  ,Sothat lim !!1 6 GH(j!)=;270  : Atthe same time the magnitude of eachofthe vectors in Figure 1 becomes in nite in length. Since twoofthe vectors are in the numerator of GH and 1 ve are 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 ;270  and the magnitude plot is descending at -60 dB/decade for large !. The rst part of the plot of GH,namely the mapping of the positivehalf of the imaginary axis of the s-plane into the GH-plane is shown in Figure 3. Wesee that the plot crosses the negative real axis at s = ;10 ;70=20 = ;0:0003162: and again at s = ;10 ;102=20 = ;7:9410 ;6 Weobtain these values bygoing to the phase scale and nding ;180  . We then movehorizontally until wereach the phase plot. We then move vertically to the magnitude plot and nally horizontally to the magnitude scale whichinthis case yields ;70 db and ;102 dbat! =2rad./sec. and ! =40rad./sec. respectively. 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,andI  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. Thus, along the semicircle GH( 6 )  6 200  1  6  6  6  = 1 6 ;3 2 -300 -250 -200 -150 -100 Phase in Degrees -140 -120 -100 -80 -60 -40 -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 Plots of GH = K(s+2)(s+3) s 3 (s+40)(s+50) For K =1 3 Im(GH) Re(GH) GH(j2) = -0.000316 GH(j40) = -0.00000794 Figure 3: Plot of GH(I) Im(GH) Re(GH) Figure 4: Plot of GH(I), GH(II), GH(I  ) 4 -3 θ -2 -50 -40 Re(s) Im(s) Figure 5: Contour of Aribitrarily Small Radius Around Origin As we traverse the semicircle counterclockwise( ccw), through 180  eval- uating GH,theplot of GH will rotate clockwise (cw) with eachpoleatthe origin contributing 180  of rotation. Thus, the total clockwise rotation of GH will be 540  .The nal Nyquist plot is shown in Figure 6. For a gain of 1, GH crosses the negativerealaxisat-0.0003162. and ;7:9410 ;6 .Thus the point GH = ;1isinregion I in Figure 6, and we havehave Z = N + P = ;2+5 = 3 This says that as the gain is increased from zero twopoles go immediately into the righthalfofthes-plane, since only three are in the left half plane, If 3162 <K<125;;000, the point GH = ;1isinregion II We then havetwoencirclements, one counterclockwise and one clockwise. Since we traversed the contour in the s-plane in the counterclockwise direction, weas- sign a minus sign to the clockwise encirclement. Thus the Nyquist equation is Z = N +5 = ;1+1+5 = 5 5 Re(GH) Im(GH) I II III Figure 6: Completed Nyquist Plot 6 Thus for 3162 <K<125;;000, there are no closed loop poles in the right half of the s-plane and the system is stable, all ve are inside the contour that encloses the left half of the s plane. For K>125;;000, the point GH = ;1isinregion III.Inthis case we again havetwoclockwise encirclements, and the Nyquist equation is again Z = N + P = ;2+5 = 3 Again three closed loop poles are in the left half of the s plane and twoin the right half of the S plane. The root locus is shown in Figure 7. Note that the two branches of the root locus initially in the righthalfofthe s-plane cross backinto the left half plane at s = j2. This can be determined from the Bode plots, because 6 GH(j2) = ;180  The two branches pass backinto the righthalfplane at s = j40, again because 6 GH(j40) = ;180  : This follows from the fact that at every pointontheroot locus, 6 GH(s)=;180  : Wealsoknowthatevery pointontheroot locus jGH(s)j=1;; whichiswhywe adjust the magnitude of GH to one in order to nd the related gain. 7 -3 -2 -50 -40 ω = 2 ω = 40 Im(s) Re(s) K = 3162 K = 125,00 Figure 7: Rpot Locus 8