Solution: 10.8.1.9 The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of Figure 1. For ! small,  1  2 and are all Im(s) Re(s) θ 1θ 2 θ 3 α ω -1 -2-10 2 poles + Figure 1: Geometric interpretation of angle and magnitude approximately 0  . The twopoles 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 ;180  That is, the twoarbitrarily 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 ;180  to ;270  .At the same time the magnitude goes to zero. This can also be determined from Figure 1 As ! !1, ,  1 and  2 all approch90  ,Sothat lim !!1 6 GH(j!)=;270  : Atthe same time the magnitude of eachofthe vectors in Figure 1 becomes in nite in length. Since there is one large magnitude in the numerator and four equally large magnitudes in the denominator lim !!1 jGH(j!)j=0 The rst part of the plot of GH, namely the mapping of the positivehalf 1 -275 -250 -225 -200 -175 Phase in Degrees -160 -120 -80 -40 0 40 80 Magnitude in dB 0.01 0.1 1 10 100 1000 Frequency in Radians/sec. Phase in Degrees Magnitude in dB + Figure 2: Bode Magnitude and Phase of GH 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. To complete the contour, wemust evaluate GH along the semicircle of radius . Figure 5 depicts this evaluation. Note that the vectors drawn from s = ;10, s = ;1 and s = ;2tothesemicircle all have 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 eachofthe twopoles at the origin. Eachpoleatthe origin is thus represented byavector of arbitrarily small length and angle 2 Im(GH) Re(GH) + Figure 3: Plot of GH(I) .Thus, along the semicircle GH( 6 )  2K 10  1  6  6  = 1 6 ;2 As wetraverse the semicircle counterclockwise( ccw), evaluating GH, the plot of GH will rotate clockwise (cw) through twice the angle of 180  wetraverse along the semicircle, or 360  .The nal Nyquist plot is shown in Figure 6. We see from the nal Nyquist plot that there are always two clockwise encirclements of s = ;1forany gain K wechoose. Thus the Nqquist equation is Z = N +P = 2+0 = 2 This says that for any K>0 the closed loop system is unstable. The root locus is shown in Figure 7. Note that the twopoles at the origin migrate immediately into the righthalfofthes-plane and stay there. Thus the system is always unstable, in agreementwith the Nyquist analysis. 3 Im(GH) Re(GH) + Figure 4: Plot of GH(II) Re(s) Im(s) -10 -2 -1 ε θ + Figure 5: Plot of GH(III) 4 Im(GH) Re(GH) -1 Everthing in box is at s = - infinity + Figure 6: Final Nyquist Plot ; Re(s) Im(s) -10 -2 -1 + Figure 7: Root Locus 5