Solution 10.9.1.11 For GH(s)= K(s+1) s(s+ 4)(s+10)(s+40) the Bode magnitude and phase plots are shown in Figure 1, for K =100. The most importantstepismapping 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. 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, the Bode 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 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  1 -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 2 ω Re(s) Im(s) -40 -4-10 Figure 2: Vector Components of GH 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 none of the poles of GH.Then N =0 = Z ;P = Z Then, Z =0,sothat none of the four closed loop poles are inside the contour inthes-plane, that is, in the righthalf 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 twoclockwise encirclements of ;1. Since wewentaround the contour in the s-plane in the clockwise direction N =2.Thus N =2 = Z ;P = Z 3 j ε j 8 Re(s) Im(s) Re(GH) Im(GH) a Figure 3: Mapping of GH(I). Re(GH) Im(GH) I II a Figure 4: Completed Nyquist Plot 4 Re(s) Im(s) -40 -10 -4 -1 j 25 K = 31,623 Figure 5: Root Locus Thus Z =2 which means that there are twoclosed loop poles inside the contour in the s-plane, that is, two closed loop poles in the right half plane. Since there are four closed loop poles altogether, twomust be in the left half 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. 5