Solution 10.8.4.3 nyq8: The Bode magnitude and phase plots are shown in Figure 1, for K =100. -300 -250 -200 -150 -100 -50 Phase in Degrees -160 -140 -120 -100 -80 -60 -40 -20 0 20 40 Magnitude in Decibels 0.01 0.1 1 10 100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH = K(s+1) s 2 (s+2)(s+40) For K =100 The most important step is mapping GH along part I  of the contour . Note that 1. As ! ! 0, GH(j!)!1 6 ;90. This can be seen from the Bode plots, or from Figure 2. Since lim !!0 GH(j!) = lim !!0 K(j!+1) (j!)(j!+4)(j!+10)(j!+40) = 1 6 90  In other words the limit is dominated by the pole at the origin. 1 ?ω Re(s) Im(s) -40 -4-10 Figure 2: Vector Components of GH 2. As As ! !;1, GH(j!) ! 0 6 270. Again this can be determined from the Bode plots or from Figure 2, since lim !!1 GH(j!) = lim !!0 K(1 6 ;90  ) (1 6 ;90  )(1 6 ;90  )(1 6 ;90  )(1 6 ;90  ) = 0 6 270  3. In between ! =0and ! = ;1,theBode plots showthe magnitude decreasing monotonically as 6 G(j!)swings rst backtowards 50  then back through ;180  and nally to 270  . Thus the mapping is that shown in Figure 3. 4. Once GH(I  ) has been plotted the rest of the plot can be completed quickly. (a) Along Part II of , jGH(j!)j=0and 6 GH(j!)swings through 180  in the clockwise direction. (b) GH(I)isjust the mirror image through the real axis of GH(I). (c) jGH(III)j= 1 and 6 GH(III)sweeps out 180  in the counter- clockwise direction. The completed Nyquist plot is shown in Figure 4. For K =100thepoint `a' is GH(j25) = 10 ;50=20 6 ;180  =0:00316 6 ;180  2 ε Re(s) Im(s) Re(GH) Im(GH) 8 - - a Figure 3: Mapping of GH(I  ). Re(GH) j Im(GH) I II a Figure 4: Completed Nyquist Plot 3 To make GH(j25) = 1:0 6 ;180  : The gain must be increased to K = 100 0:00316 = 31;;623 For 0 <K<31;;623, the point ;1inthe GH-plane is in region I, and there is are no encirclements of ;1. For the chosen contour in the s-plane P =0 because the contour encloses all of the poles of GH. Then Z = N +P = 0+4 = 4 Then, Z =4,sothat all of the four closed loop poles are inside the contour inthes-plane, that is, in the left half plane for 0 <K<31;;623. Thus the system is stable for 0 <K<31;;623. For K>31;;623, point ;1intheGH-plane is in region II,and there are twocounterclockwise encirclements of ;1. Since wewent around the contour in the s-plane in the clockwise direction N = ;2. Thus Z = N + P = ;2+4 = 2 Thus Z =2 which means that there are twoclosedlooppoles inside the contour in the s-plane, that is, twoclosed loop poles in the left half plane. Since there are four closed loop poles altogether, twomust be in the righthalf plane for K>31;;623. Thus, the system in unstable for K>31;;623. The root locus is shown in Figure 5. Note that the root locus agrees with the Nyquist analysis, as it must. 4 Re(s) Im(s) -40 -10 -4 -1 ω = 25 K = 31,623 Figure 5: Root Locus 5