Solution: 10.8.3.3 The angle contributions of eachofthe poles and zeros can be determined from the vector diagram of Figure . For ! =0, =0  while  1 and  2 are 1 2 -3 θ 1 θ 2 α Im(s) Re(s) Figure 1: Vector Components of GH each 180  .Thus 6 GH(j0) = ;360  =0  : For K =10, jGH(j0)j=15: As ! !1, ,  1 ,and 2 all appproch90  ,Sothat lim !!1 6 GH(j!)=;90  : At the same time the magnitude of eachofthevectors in Figure becomes in nite in length. Since one of the vectors is in the numerator of GH and two are in the denominator, clearly lim !!1 jGH(j!)j=0 This can also be seen from the Bode plots in Figure 2, where 6 GH ap- proaches ;90  and the magnitude plot is descending at -20 db/decade for large !. The rst part of the plot is shown in Figure . Wesee that the plot crosses the negativereal axis of the GH-plane at ;10 10=20 = ;3:16. We obtain this value bygoing to the phase scale and nding ;180  .Wethen movehorizontally until wereachthe phase plot. We then movevertically to the magnitude plot and nally horizontally to 1 -400 -350 -300 -250 -200 -150 -100 -50 Phase in Degrees -40 -20 0 20 40 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 Plots of GH ! 0.5 1 2 2.8 2.9 3 3.2 5 jGH(j!)j 13.2 10 5.7 4.0 3.86 3.72 3.46 2.12 6 GH(j!) ;310  ;270  ;218  ;192  ;189  ;187  ;182  ;154  Table 1: Table of Values of GH(j!)forSelected Values of ! 2 -3.46 Re(GH) Im(GH) Figure 3: Plot of GH(I) the magnitude scale whichinthis case yields 10 db. These numbers are approximate because of the resolution of the Bode plot. Table 1 gives acutal computes values for jGH(j!)j and 6 GH(j!). These values are obtained by simply evaluating GH at selected values of s,namely s = j! for 0 <!<1. ABode plot is simply this information in logarithmic form. Weseefromthe Bode plot and the table that GH has angle ;180  for ! slightly greater than 3.0. The Bode plot indicates that !  3:4rad./s. The table shows the value to be close to 3.2 rad./s. From the Bode plot weget jGH(j3:2)j  10 10=20 = 3:16: From the table weobtain 3.46. These values are both approximate. In the larger scheme of things wearehappywith either number, because we don't wanttobeany place close to instablity! As weevaluate GH along part II of , weremainessentially at the origin. To draw the portion of the Nyquist plot that corresponds to evaluating GH along part I  of the contour wesimply take 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 I, II, and I  is shown in Figure . This completes the contour. Because GH has no poles at the origin, we do not havetodetour around the origin of the s-plane on a semicircle of radius . From the Nyquist plot, weseethat for a gain of 10, GH crosses the negative real axis at -3.16. Clearly,ifwedecrease the gain the plot will 3 Im(GH) Re(GH) III a Figure 4: Completeed Nyquist Plot cross to the rightof;1 instead of to the left. For a gain of K =10=3:16 = 3:16 the plot passes through ;1. For gains below3.16,the numberofencir- clements is N =0,andtheNyquist equation is Z = N +P = 0+0 = 0 This means there are no closed loop poles inside in the s-plane. Since encloses the left half of the s-plane, there are two unstable poles for K<3:16. For K>3:16 The point ;1intheGH-plane is in region I. In this case N =2.The sign of N is positivebecause both encirclements of ;1areinthe counterclockwise (ccw) direction, and in our evaluation of GH along we traveled counterclockwise(cw). Thus, Z = N +0 = 2+0 = 2 This means that for K>3:16 there are twoclosed loop poles inside the contour , that is twoclosed looppoles in the left half of the s-plane. The root locus is shown in Figure 5. The root locus crosses the imaginary axis at s = j3:2foragain of K =3:16. The root locus o the real axis is a circle centered at the zero at s = ;3. 4 -3 1 2 Re(s) Im(s) Figure 5: Root Locus 5