Solution: 10.8.1.14 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 importantstepismapping 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 Figure 2: Vector Components of GH ε 8 Re(s) Im(s) Re(GH) Im(GH) a 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. 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 radius is 360  . The completed Nyquist plot is shown in Figure 4. For K =100thepoint 2 Re(GH) Im(GH) III a Figure 4: Completed Nyquist Plot `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 =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<2;;512. Thus the system is stable for 0 <K<2;;512. For K>2;;512, ;1isinregion II.Herethere are twoclockwise en- circlements of ;1. Since wewentaround the contour in the s-plane in the 3 Re(s) Im(s) -20 -10 -4 ω = 9 K = 2,512 Figure 5: Root Locus clockwise direction N =2.Thus N =2 = Z ;P = Z 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 righthalf plane 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