solution: 10.8.3.8 The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of Figure 1. For ! small  2 is approximately 0  . Im(s) Re(s) ω θ 1 θ 2 3 poles Figure 1: Computation of Magnitude and Phase of GH 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) = K 10 ( 6 90  )( 6 90  )( 6 90  ) = 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. By examining the Bode mangitude and phase plots of Figure 2 Weseethat as ! increases the angle of GH moves steadily from ;270  to ;360  .At the same time the magnitude goes to zero. The value of gain chosen is K =1, but anyvalue will work just as well. The asymptotic behavior of the plot of GH can also be determined from Figure 1. As ! !1,  2 appproches 90  ,Sothat lim !!1 6 GH(j!)=;360  : Atthe same time the magnitude of eachofthe vectors in Figure 2 becomes in nite in length. Since all four of the vectors are in the denominator, clearly lim !!1 jGH(j!)j=0 1 -375 -350 -325 -300 -275 -250 Phase in Degrees -220 -180 -140 -100 -60 -20 20 60 Magnitude in dB 0.1 110100 1000 Frequency in Radians/sec. Phase in Degrees Magnitude in dB Figure 2: Bode Magnitude and Phase of GH 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 never crosses the negativerealaxis. 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 1, 2, and 1  is shown in Figure 4. 2 Re(GH) Im(GH) Figure 3: Plot of GH(I) Re(GH) Im(GH) Figure 4: Plot of GH(I),GH(II),GH(I  ) 3 Re(s) ε θ ?10 3 poles Im(s) Figure 5: Plot of GH(I),GH(II),GH(I  ) To complete the contour, wemust evaluate GH along the semicircle of radius . Figure 5 depicts this evaluation. Note that the vector drawn from s = ;10 to the semicircle has nite length, and an associated angle of zero. The angles are zero, because the radius of the semicircle is arbitrarily small. The vector drawn from the origin to the semicircle represents eachofthepoles at the origin. Eachpoleatthe origin is thus represented byavector of arbitrarily small length and angle .Thus, along the semicircle GH( 6 )  K 10  1  6  6  6  = 1 6 ;3 As we traverse the semicircle clockwise( cw), evaluating GH, the plot of GH will rotate counterclockwise (ccw) through thrice the angle of 180  wetraverse along the semicircle, or 540  .The nal Nyquist plot is shown in Figure 6. We see from the nal Nyquist plot that there is always one counterclockwise encirclementofGH = ;1forany gain K wechoose. Thus the Nyquist equation is Z = N +P = 1+1 = 2 This says that for any K>0 the closed loop system is unstable becasue only two closed loop poles are inside the left half of the s plane. 4 Im(GH) Re(GH) Figure 6: Complete Nyquist Plot 5 Im(s) Re(s) 3 poles Figure 7: Root Locus The root locus is shown in Figure 7. Note that the two branches of the root locus go immediately into the righthalfofthes-plane and stay there. Thus the system is always unstable, in agreementwith the Nyquist analysis. 6