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# Norman Nise Control Systems Engineering 6th Solultions.PDF

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& Sons, Inc. 0 5 10 15 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 SOLUTIONS TO DESIGN PROBLEMS 69. Writing the equation of motion, ( fvs + 2)X(s) = F(s). Thus, the transfer function is X(s) F(s) = 1/ fv s + 2 fv . Hence, Ts = 4a = 4 2 fv = 2 fv , or fv = Ts2 . 70. The transfer function is, F(s) = 1/ M s2 + 1 M s + K M . Now, 4 44 81Re 2 sT M M = = = = . Thus, 1 2 M = . Substituting the value of M in the denominator of the transfer function yields, 2 2 2s s K+ + . Identify the roots 1,2 1 2 1s j K= \u2212 ± \u2212 . Using the imaginary part and substituting into the peak time equation yields 1 Im 2 1p T K \u3c0 \u3c0= = = \u2212 , from which 5.43K = . Solutions to Design Problems 4-71 Copyright © 2011 by John Wiley & Sons, Inc. 71. Writing the equation of motion, (Ms2 + fvs +1)X(s) = F(s) . Thus, the transfer function is X(s) F(s) = 1/ M s2 + fv M s + 1 M . Since Ts =10 = 4\u3b6\u3c9n , \u3b6\u3c9n = 0.4 . But, fv M = 2\u3b6\u3c9n = 0.8.Also, from Eq. (4.39) 17% overshoot implies \u3b6 = 0.491. Hence, \u3c9n = 0.815. Now, 1/M = \u3c9n2 = 0.664. Therefore, M 1.51. Since 2 0.8, 1.21v n v f f M \u3b6\u3c9= = = . 72. Writing the equation of motion: (Js2+s+K)\u3b8(s) = T(s). Therefore the transfer function is \u3b8(s) T(s) = 1 J s2+ 1 Js+ K J . \u3b6 = - ln ( %OS 100 ) \u3c02 + ln2 (%OS100 ) = 0.358. Ts = 4 \u3b6\u3c9n = 4 1 2J = 8J = 3. Therefore J = 3 8 . Also, Ts = 3 = 4 \u3b6\u3c9n = 4 (0.358)\u3c9n . Hence, \u3c9n = 3.724. Now, K J = \u3c9n2 = 13.868. Finally, K = 5.2. 73. Writing the equation of motion [s2+D(5)2s+ 1 4 (10) 2]\u3b8(s) = T(s) The transfer function is \u3b8 (s) T(s) = 1 s2+25Ds+25 Also, \u3b6 = - ln ( %OS 100 ) \u3c02 + ln2 (%OS100 ) = 0.358 and 2\u3b6\u3c9n = 2(0.358)(5) = 25D Therefore D = 0.14. 4-72 Chapter 4: Time Response Copyright © 2011 by John Wiley & Sons, Inc. 74. The equivalent circuit is: where Jeq = 1+( N1 N2 ) 2 ; Deq = ( N1 N2 ) 2; Keq = ( N1 N2 ) 2. Thus, \u3b81(s) T(s) = 1 Jeqs2+Deqs+Keq . Letting N1 N2 = n and substituting the above values into the transfer function, \u3b81(s) T(s) = 1 1+n2 s2 + n2 1+n2 s + n2 1+n2 . Therefore, \u3b6\u3c9n = n 2 2(1+n2) . Finally, Ts = 4 \u3b6\u3c9n = 8(1+n2) n2 = 16. Thus n = 1. 75. Let the rotation of the shaft with gear N2 be \u3b8L(s). Assuming that all rotating load has been reflected to the N2 shaft, JeqLs 2 + DeqLs + K( )\u3b8 L(s) + F(s)r = Teq (s) , where F(s) is the force from the translational system, r = 2 is the radius of the rotational member, JeqL is the equivalent inertia at the N2 shaft, and DeqL is the equivalent damping at the N2 shaft. Since JeqL = 1(2)2 + 1 = 5 and DeqL = 1(2)2 = 4, the equation of motion becomes, 5s2 + 4s + K( )\u3b8L(s) + 2F(s) = Teq (s). For the translational system (Ms2 + s)X(s) = F(s) . Substituting F(s) into the rotational equation of motion, 5s2 + 4s + K( )\u3b8L(s) + Ms2 + s( )2X(s) = Teq (s). But,\u3b8 L(s) = X(s)r = X(s) 2 and Teq (s) = 2T (s). Substituting these quantities in the equation above yields (5 + 4M)s 2 + 8s + K( )X(s) 4 = T (s). Thus, the transfer function is X(s) T(s) = 4 /(5 + 4M) s2 + 8 (5 + 4M) s + K (5 + 4M) . Now, 4 415 (5 4 )8Re 2(5 4 ) sT M M = = = = + + . Hence, M = 5/2. For 10% overshoot, \u3b6 = 0.5912 from Eq. (4.39). Hence, Solutions to Design Problems 4-73 Copyright © 2011 by John Wiley & Sons, Inc. 82 0.5333 (5 4 )n M \u3b6\u3c9 = =+ . Solving for \u3c9n yields \u3c9n = 0.4510. But, 0.4510. (5 4 ) 15n K K M \u3c9 = = =+ Thus, K = 3.051. 76. The transfer function for the capacitor voltage is VC(s) V(s) = 1 Cs R+Ls+ 1 Cs = 106 s2+Rs+106 . For 20% overshoot, \u3b6 = - ln ( %OS 100 ) \u3c02 + ln2 (%OS100 ) = 0.456. Therefore, 2\u3b6\u3c9n = R = 2(0.456)(103) = 912\u3a9. 77. Solving for the capacitor voltage using voltage division, VC (s) = Vi(s) 1/(CS) R + LS + 1 CS . Thus, the transfer function is VC (s) Vi(s) = 1/(LC) s2 + R L s + 1 LC . Since Ts = 4Re =10 \u22123, Re = R 2L = 4000. Thus R = 8 K\u3a9 . Also, since 20% overshoot implies a damping ratio of 0.46 and 2\u3b6\u3c9n = 8000, \u3c9n = 8695.65 = 1LC . Hence, C = 0.013 \u3bcF. 78. Using voltage division the transfer function is, VC (s) Vi(s) = 1 Cs R + Ls + 1 Cs = 1 LC s2 + R L s + 1 LC Also, 3 4 4 87 10 Re 2 s LT x R R L \u2212= = = = . Thus, 1143R L = . Using Eq. (4.39) with 15% overshoot, \u3b6 = 0.5169. But, 2\u3b6\u3c9n = R/L. Thus, 5 1 11105.63 (10 )n LC L \u3c9 \u2212= = = . Therefore, L = 81.8 mH and R = 98.5 \u3a9. 4-74 Chapter 4: Time Response Copyright © 2011 by John Wiley & Sons, Inc. 79. For the circuit shown below R1 = L = i1(t) i2(t) o write the loop equations as R 1 L s+ I 1 s R 1 I 2 s\u2212 V i s= R 1 I 1 s\u2212 R 1 R 2 1 C s + + I 2 s+ 0= Solving for I2(s) I 2 s R 1 L s+ V i s R 1\u2212 0 R 1 L s+ R 1\u2212 R 1\u2212 R 1 R 2 1C s+ + = ( ) But, V o s 1 C s I 2 s= . Thus, V o s V i s R 1 R 2 R 1+ C L s 2 C R 2 R 1 L+ s R 1+ + = Substituting component values, Vo (s) Vi(s) = 1000000 1 (R2 + 1000000)C s2 + (1000000CR2 +1) (R2 +1000000)C s +1000000 1 (R2 +1000000)C For 8% overshoot, \u3b6 = 0.6266. For Ts = 0.001, \u3b6\u3c9n = 40.001 = 4000. Hence, \u3c9n = 6383.66. Thus, 2 2 11000000 6383.66 ( 1000000)R C =+ or, 2 10.0245 1000000 C R = + (1) Also, Solutions to Design Problems 4-75 Copyright © 2011 by John Wiley & Sons, Inc. 1000000 C R 2 1+ R 2 1000000+ C 8000= (2) Solving (1) and (2) simultaneously, 2 8023R = \u3a9, and C = 2.4305 x 10-2 \u3bcF. 80. sI \u2212 A = s 0 0 s \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 \u2212 (3.45 \u221214000Kc) \u22120.255x10\u22129 0.499x1011 \u22123.68 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 = s \u2212 (3.45 \u221214000Kc) 0.255x10 \u22129 \u22120.499x1011 s + 3.68 \u23a1 \u23a3 \u23a2 \u23a4 \u23a6 \u23a5 sI \u2212 A = s2 + (0.23 + 0.14x105 Kc )s + (51520Kc + 0.0285) (2\u3b6\u3c9n )2 = [2* 0.9]2 *(51520Kc + 0.0285) = (0.23 + 0.14x105 Kc)2 or Kc 2 \u2212 8.187x10\u22124 Kc \u2212 2.0122x10\u221210 = 0 Solving for Kc, Kc = 8.189x10\u22124 81. a. The transfer function from Chapter 2 is, Yh(s) \u2212 Ycat (s) Fup(s) = 0.7883(s + 53.85) (s2 + 15.47s + 9283)(s2 +8.119s + 376.3) The dominant poles come from s 2 + 8.119s + 376.3. Using this polynomial, 2\u3b6\u3c9n = 8.119, and \u3c9n2 = 376.3. Thus, \u3c9n = 19.4 and \u3b6 = 0.209. Using Eq. (4.38), %OS = 51.05%. Also,Ts = 4\u3b6\u3c9n = 0.985 s, and Tp = \u3c0 \u3c9n 1\u2212 \u3b6 2 = 0.166 s . To find rise time, use Figure 4.16. Thus,\u3c9nTr = 1.2136 or Tr = 0.0626 s. b. The other poles have a real part of 15.47/2 = 7.735. Dominant poles have a real part of 8.119/2 = 4.06. Thus, 7.735/4.06 = 1.91. This is not at least 5 times. c. Program: syms s numg=0.7883*(s+53.85); deng=(s^2+15.47*s+9283)*(s^2+8.119*s+376.3); 'G(s) transfer function' G=vpa(numg/deng,3); pretty(G) numg=sym2poly(numg); deng=sym2poly(deng); G=tf(numg,deng) step(G) 4-76 Chapter 4: Time Response Copyright © 2011 by John Wiley & Sons, Inc. Computer response: ans = G(s) transfer function .788 s + 42.4 ------------------------------------------ 2 2 (s + 15.5 s + 9280.) (s + 8.12 s + 376.) Transfer function: