Solution: 10.8.4.6 The Bode magnitude and phase plots are shown in Figure 1, for K =100. -275 -250 -225 -200 -175 -150 Phase in Degrees -160 -140 -120 -100 -80 -60 -40 -20 0 20 40 60 Magnitude in Decibels 0.1 110100 1000 Frequency in Radians/Sec. Phase in Degrees Magnitude in Decibels Figure 1: Bode Plots of GH = K(s+4) s 2 (s+10)(s+20) 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 ; 180. This can be seen from the Bode plots, or from Figure 2. Since lim !!0 GH(j!) = lim !!0 K(j!+4 (j!) 2 (j!+10)(j!+ 40) = 1 6 180  In other words the limit is dominated by the twopolesat 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  1 ?ω Re(s) Im(s) -20 -4-10 x 2 Figure 2: Vector Components of GH Re(s) Im(s) Im(GH) Re(GH) Figure 3: Plot iof GH(I) 3. In between ! =0and! = infty, the Bode plots show the magnitude decreasing monotonically as 6 G(j!)swings rst backtowards ;170  then back through ;180  and nally to ;270  .Thus the mapping is that shown in Figure 3. If wewant the plot of GH(I  )wehaveto reverse the sign of all the angles on the Bode plot. Thus as wegofrom ;j to ;j1 The angle rst swings towards 170  then back through ;180  and nally to ;270  . 4. OnceGH(I)has been plotted the rest of the plot can be completed quickly. (a) Along Part II jGH(j!)j=0and 6 GH(j!)swingsthrough 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 360  in the clockwise direction. That is, eachpole at the origin causes the plot of GH(III)torotate clockwise through 180  . Since there are two poles at the origin the total angle swept out along an arc of in nite 2 Re(GH) Im(GH) I II a Figure 4: Completed Nyquist Plot radius is 360  . The completed Nyquist plot is shown in Figure 4. For K =100thepoint `a' is GH(j18) = 10 ;28=20 6 ;180  =0:0398 6 ;180  To make GH(j18) = 1:0 6 ;180  : The gain must be increased to K = 100 0:0398 = 2;;512 For 0 <K<2;;512 the point ;1intheGH-plane is in region I, and there is are no encirclements of ;1. For the chosen contour in the s-plane P =4 because the contour encircles 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 inthe s-plane, that is, in the left half plane for 0 <K<2;;512. Thus the system is stable for 0 <K<2;;512. 3 Re(s) j Im(s) -20 -10 -4 j 9 K = 2,512 Figure 5: Root Locus For K>2;;512, ;1isinregion II.Herethere are twocounterclockwise encirclements of ;1. Since wewentaround the contour in the s-plane in the clockwise direction N = ;2, and Z = N + P = ;2+4 = 2: Thus, there are twoclosed loop poles inside the contour in the s-plane, that is, two closed loop poles in the left half plane. Since there are four closed loop poles altogether, twomust be in the righthalfplane for K>2;;512. Thus, the system in unstable for K>2;;512. The root locus is shown in Figure 5. Note that the root locus agrees with the Nyquist analysis, as it must. 4