Solution: 7.9.3.4 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehavelet G p = 5 s(s +1)(s+10) ;; and G c = K c (s +a) s +b : Part a. Toachieve t p =0:785 s wesolve t p =  ! d for ! d . That is ! d =  t p = 3:14159 0:785 =4rad/s. Since ! d = ! n p 1; 2 ,and =0:8, wehave ! n = ! d p 1; 2 = 4:0 p 0:36 =6:667 rad/s. Then the real part of the complex roots is ;! n = ;0:86:667 = ;5:33: Part b. Figure 2 shows the vector evaluation of G c G p at s = ;5:33 + j4. Togetthe root locus to pass through this pointwemust have 6 G c G p (s) j s=;5:33+j4 = ;180  : Eachofthe vectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;5:33 + j4. That is the vector V 1 is the polar repre- sentation of the factor s in the denominator of G c G p ,thevector V 2 the polar representation of the factor s +1,thevector V 3 the polar representation of the factor s+10,thevector V 4 the polar representation of the factor s+b, and the vector V 5 the polar representation of the vector s + a.Thus 1 β α Re(s) Im(s) -1 -b -5.33 4.0 -a -10 V 1 V 5 V 2 V 3 θ 1 θ 2 θ 3 V 4 Figure 2: Satisfaction of Angle Condition at s = ;5:33+ j4:0 G c G p j s=;5:33+j4 = 5K c jV 5 j 6 jV 1 j 6  1 jV 2 j 6  2 jV 3 j 6  3 jV 4 j 6 = 5K c jV 5 j jV 1 jjV 2 jjV 3 jjV 4 j 6 ( ; ; 1 ; 2 ; 3 ) The evaluation of G c G p at s = ;5:33 + j4 has nowbeenbroken down into a composite magnitude and a composite angle. For the root locus to pass through s = ;5:33+ j4wemust have ; ; 1 ; 2 ; 3 = ;180  : The gain that places a closed loop poles at s = ;5:33j4isobtained bysolving 5K c jV 5 j jV 1 jjV 2 jjV 3 jjV 4 j =1;; (1) or K c = jV 1 jjV 2 jjV 3 jjV 4 j 5jV 5 j : (2) It should be clear that it is the angle condition that drives this whole business. The angle condition will be used to nd b.Once b is determined then K can easily be calculated. The angle contribution of the plantis 6 G p = ;[ 1 +  2 +  3 ] = ;f[180  ;tan ;1 (4=5:33)]+ [180  ;tan ;1 (4=4:33)+ tan ;1 (4=4:67)g = ;f143:11  + 137:27  +40:58  g = ;321  Thus ; =  1 +  2 + 3 ;180  =321  ;180  =141:06  : For >0wemust have >141  .Thisplaces the zero of the lead compensator between the pole at s = ;1and the pole at the origin. Wecanbemoreprecise by noting that tan(180  ;39  )=tan(39  )= 4 5:33;a : 2 Re(s) Im(s) -1 -5.33 j 4.0 -a -10 α ?186 V 5 Figure 3: Determinining Location of Zero of Lead This can be rewritten as a =5:33; 4 tan(39  ) =0:38 So ;a has to be to the rightofs = ;0:38. Part c. If the pole of the lead compensator is placed at s = ;186, then = tan ;1 (4=180:67) = 1:27  : Then =141:06  + = 142:33  : Wenowuse simple trigonometry to nd a.Refering to Figure 3 a = 5:33; 4 tan(180  ;14:33  ) = 5:33; 4 tan(37:67  ) = 0:15 Wecannow ndthe gain using equation(2). K c = jV 1 jjV 2 jjV 3 jjV 4 j jV 5 j = 6:675:896:14180:7 56:528 =1334:5 Thus the complete compensator is G c = 1334:5(s+0:15) s +186 : Part d. 3 Re(s) Im(s) -1 -b -10 ?0.15 V 5 V 3 V 2 V 4 V 1 Figure 4: Calculation of K Near Zero s -0.12 ;0:122 ;0:121 ;0:1207 K 1297 1409 11351 1334.3 Table 1: Search for K =1334:5 Near Zero at s = ;0:15 To nd the other twopoles of the closed loop system, wesimply look along the appropriate parts of the real axis for the same gain that we used to place the dominantcomplex poles. The calculation for the closed loop pole between s = ;0:17 and s =0isshown in Figure 4. As always the vectors representing each term in G c G p are drawn to the pointwhere G c G p is to be evaluated. Then K = jV 1 jjV 2 jjV 3 jjV 4 j 5jV 5 j Note that the plantgain of 5 has to be divided out. Table 1 shows the search for the gain of 1335. Note that wedonot nd a gain of 1334.3 exactly. The gain changes very rapidly in this region because K goes from zero when s =0to1 as s approaches -0.15. So wesettle for a closed loop pole location approximately at s = ;0:1207. The search for the closed loop pole to the left of s = ;186 proceeds in the same way. Figure 5 shows the vectors used in the calculation. The calculation is the same, but the location of s has changed. In this case weuse the following MATLAB dialogue to nd the other closed loop pole EDU>g = zpk([-0.15],[0 -1 -10 -186],6672.4) Zero/pole/gain: 6672.4 (s+0.15) 4 -1 -10 ?0.15-186 V 1 V 5 V 3 V 2 Im(s) Re(s) V 4 Figure 5: Calculation of K to Left of s = ;186 ---------------------- s (s+1) (s+10) (s+186) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 6672.4 (s+0.15) ------------------------------------------ (s+186.2) (s+0.1208) (s^2 + 10.67s + 44.5) EDU> In this case the closed loop pole is very close to s = ;186. There is a reason for this. For K small the closed loop pole is near the pole of G c G p at s = ;186. As K !1the closed loop pole must travel all the waytos = ;1.Thus the closed loop pole has plentyofroomtomove in, so a gain of 1334.3 doesn't move it very far. It is just the opposite of the situation near the zero at s = ;0:15. Solution Part e. We can nowwrite down the closed loop transfer function byinspection. T c (s) = 6672:4(s+0:15) (s +5:33;j4)(s +5:33+ j4)(s +0:1207)(s+186:2) 5 The output C(s) for a step is C(s) = 1 s T c (s) = 6672:4(s+0:15) s(s +5:33;j4)(s +5:33+ j4)(s +0:1207)(s+186:2) = 1 s + B s +0:1207 + D s +186:2 + M s+5:33;j4 + M  s +5:33+ j4 At this pointweknowthegeneral form of c(t). That is, c(t)=  1+Be ;0:1207t + De ;186:2t +2jMje ;5:33t cos(4t+ 0  1(t);; where  = 6 M.Soweknowquite a bit. What remains to be done is to nd the constants B, D and M. B = (s +0:1207)C(s) j s=;0:1207 = (s +0:1207)  6672:4(s+0:15) s(s +5:33;j4)(s +5:33+j4)(s +0:1207)(s+186:2)  j s=;0:1207 = ;0:2007!;0:2 D = (s +186:2)T c (s) j s=;186:2 = (s +186:2)  6672:4(s+0:15) s(s +5:33;j4)(s +5:33+ j4)(s +0:1207)(s+186:2)  j s=;186:2 = ;0:0012 M = (s +5:33;j4)T c (s) j s=;5:33+j4 = (s +5:33;j4)  6672:4(s+0:15) s(s +5:33;j4)(s +5:33+j4)(s +0:1207)(s+186:2)  j s=;5:33+j4 = 0:6883 6 2:1893 Thus, nally wehave c(t)=  1;0:2e ;0:1207t ;0:0012e ;186:2t +1:378e ;5:33t cos(4t+2:1893)  1(t) The time response is shown in Figure 6. The time response of the pole at s = ;186:2isnot shown because it is almost instantly zero. Note that the pole at s = ;0:1207 causes the system to creep towards its nal steady state value. This is because the choice of complex poles forced the zero location of the lead compensator in close to the origin where it attracts the pole at the origin. 6 -1.00 -0.50 0.00 0.50 1.00 012345678 Time in Seconds Total Response 1.379exp(-5.33t)cos(4t +2.18) -0.1937exp(-0.138t) Figure 6: Step Response of Compensated System 7