(Solution Manual) Control Systems Engineering Exercises for Control Systems

Nise, 7

Engineering

Solution Manual

7th Edition

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Solution Manual for Control Systems Engineering Exercises for Control Systems, The downloadable file consists of 649 pages, and the total size of the file is 5.05 MB. Purchase now only for highly discounted price.

Author(s): Nise

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O N E Introduction ANSWERS TO REVIEW QUESTIONS 1. Guided m issiles, autom atic gain control in radio receivers, satellite tracking antenna 2. Yes - power gai n, rem ote cont rol, param eter conversi on; No - Expense, com plexity 3. Motor, low pass filter, inertia supported between two bearings 4. Closed-l oop sy stems com pensat e for di sturbances by measuri ng the response, com paring it to the input response (t he desi red out put), and t hen correct ing the out put response. 5. Under t he condi tion that the feedb ack elem ent is other than unity 6. Actuating si gnal 7. Multiple subsystem s can time share th e controller. An y adjustments to the controller can be implemented with simply so ftware ch anges. 8. Stability, transient respons e, and steady-state error 9. Steady-state, transient 10. It follows a growing transient response until th e steady-state response is no longer visible. The system will either destroy itself, reach an equilibrium state b ecause of saturation in driving amplifiers, o r hit lim it sto ps. 11. Transi ent response 12. True 13. Transfer function, state- space, differential equations 14. Transfer function - the Laplace transform of the differential equation State-space - represent ation of an nt h order di fferen tial equat ion as n si multaneous fi rst-order different ial equat ions Differen tial eq uation - Mo deling a system with its d ifferen tial eq uation SOLUTIONS TO PROBLEMS 1. Five turns y ields 50 v. Therefore K = 50 volts 5 x 2 πrad= 1.59 2 Chapter 1: Introduction 2. ThermostatAmplifier and valvesHeaterTemperature differenceVoltage differenceFuel flowActual temperatureDesired temperature + - 3. Desired roll angleInput voltage + -Pilot controlsAileron position controlError voltageAileron position Aircraft dynamicsRoll rate IntegrateRoll angle GyroGyro voltage 4. Speed Error voltageDesired speedInput voltage + -transducer AmplifierMotor and drive systemActual speed Voltage proportional to actual speedDancer position sensorDancer dynamics 5. Desired power Power Error voltageInput voltage + -Transducer AmplifierMotor and drive system Voltage proportional to actual powerRod position ReactorActual power Sensor & transducer Solutions to Problems 3 6. Desired student population+ -AdministrationPopulation errorDesired student rate AdmissionsActual student rate +-Graduating and drop-out rate Net rate of influx IntegrateActual student population 7. Desired volume + -TransducerVolume control circuitVoltage proportional to desired volumeVolume error RadioVoltage representing actual volume Actual volume -+ Transducer - SpeedVoltage proportional to speedEffective volume 4 Chapter 1: Introduction 8. a. R+V -VDifferential amplifierDesired level + -Power amplifierActuatorValve FloatFluid inpu t DrainTankR+V -V b. Desired level AmplifiersActuator and valveFlow rate in IntegrateActual level Flow rate outPotentiometer+ -+ Drain Float Potentiometer-voltage in voltage outDisplacement Solutions to Problems 5 9. Desired force Transducer Amplifier ValveActuator and loadTire Load cellActual force + -Current Displacement Displacement 10. Commanded blood pressure Vaporizer PatientActual blood pressure+ -Isoflurane concentration 11. + -Controller & motorGrinderForce Feed rate IntegratorDesired depthDepth 12. + -Coil circuitSolenoid coil & actuatorCoil current Force Armature & spool dynamicsDesired position DepthTransducerCoil voltage LVDT 13. a. Ldi dt+ Ri = u(t) 6 Chapter 1: Introduction b. Assum e a steady-state solution i ss = B. Su bstituting this into the differen tial eq uation yield s RB = 1, from which B = 1 R. The characteristic equation is LM + R = 0, from which M = -R L. Thus, the total solution is i(t) = A e-(R/L)t +1 R. Solving for the arb itrary co nstants, i(0 ) = A + 1 R = 0. Thus, A = - 1 R. The final solution is i(t) = 1 R -- 1 Re-(R/L)t = 1 R(1−e−(R/L)t). c. 14. a. Writing the loop equat ion, Ri+Ldi dt+1 Cidt+vC(0) ∫=v(t) b. Differen tiatin g and substituting values, d2i dt2+2di dt+30i=0 Writing the charact eristic equat ion and fact oring, M2+2M+30=M+1+29iM+ 1-29i. The general form of the solution and i ts derivative is i=e-tcos 29tA+Bsin 29te-t =-A+29Be-tcos 29t-2 9A+Be-tsin 29tdi dt Using i(0)=0;di dt(0)=vL(0) L=1 L=2 i0 A==0 di dt(0)=−A+29B=2 Thus, and A=0 B=2 29. The solution is Solutions to Problems 7 i=2 2929e-tsin 29t c. i t 15. a. Assum e a part icular sol ution of Substitute into the differen tial eq uation and obtain Equating like coefficients, From which, C = 35 53 and D = 10 53 . The characteristic polynom ial is Thus, the total so lution is Solving for t he arbi trary const ants, x(0) = A +35 53 = 0. Therefore, A = - 35 53 . Th e final solution is b. Assum e a part icular sol ution of xp = Asi n3t + Bcos3t 8 Chapter 1: Introduction Substitute into the differen tial eq uation and obtain (18A −B)cos(3t) −(A+18B)sin(3t) =5sin(3t ) Therefore, 18A – B = 0 and –(A + 18B ) = 5. Sol ving for A and B we obt ain xp = (-1/ 65)si n3t + (-18/ 65)cos3t The characteristic polynom ial is M2+6M+ 8=M+ 4M+ 2 Thus, the total so lution is x=Ce-4t+De-2t+-18 65cos3t-1 65sin3t Solving for the arb itrary co nstants, x(0)=C+D−18 65=0. Also , the derivative of the solution is =-3 65cos3t+54 65sin3t-4Ce-4t-2De-2t dx dt Solving for the arb itrary co nstants, x. (0) −3 65−4C−2D=0, or C = −3 10and D = 15 26. The final solution is x=-18 65cos3t-1 65sin3t-3 10e-4t+15 26e-2t c. Assum e a part icular sol ution of xp = A Substitute into the differen tial eq uation and obtain 25A = 10, or A = 2 /5. The characteristic polynom ial is M2+8M+25=M+4+3iM+4-3i Thus, the total so lution is x=2 5+e-4tBsin3t+Ccos3t Solving for t he arbi trary const ants, x(0) = C + 2/5 = 0. Therefore, C = -2/ 5. Also, the deri vative of t he solution is =3 B-4Ccos3t-4B+3Csin3te-4t dx dt Solutions to Problems 9 Solving for the arb itrary co nstants, x. (0) = 3B – 4C = 0. Therefore, B = -8/ 15. The fi nal solution is x(t)=2 5−e−4t8 15sin(3 t)+2 5cos(3t)⎛ ⎝ ⎞ ⎠ 16. a. Assum e a part icular sol ution of Substitute into the differen tial eq uation and obtain Equating like coefficients, From which, C = - 1 5 and D = - 1 10 . The characteristic polynom ial is Thus, the total so lution is Solving for t he arbi trary const ants, x(0) = A - 1 5 = 2. Therefore, A = 11 5. Also , the derivative of the solution is dx dt Solving for the arb itrary co nstants, x. (0) = - A + B - 0.2 = -3. Therefore, B = −3 5. The final solution is x(t)=−1 5cos(2t)−1 10sin(2 t)+e−t11 5cos(t)−3 5sin(t)⎛ ⎝ ⎞ ⎠ b. Assum e a part icular sol ution of xp = Ce-2t + Dt + E Substitute into the differen tial eq uation and obtain 10 Chapter 1: Introduction Equat ing like coeffi cients, C = 5, D = 1, and 2D + E = 0. From which, C = 5, D = 1, and E = - 2. The characteristic polynom ial is Thus, the total so lution is Solving for t he arbi trary const ants, x(0) = A + 5 - 2 = 2 Therefore, A = -1. Al so, t he deri vative of t he solution is dx dt=(−A+B)e−t−Bte−t−10e−2t+1 Solving for the arb itrary co nstants, x. (0) = B - 8 = 1. Therefore, B = 9. The fi nal solution is c. Assum e a part icular sol ution of xp = Ct2 + Dt + E Substitute into the differen tial eq uation and obtain Equating like coefficients, C = 1 4 , D = 0, and 2C + 4E = 0. From which, C = 1 4 , D = 0, and E = - 1 8 . The characteristic polynom ial is Thus, the total so lution is Solving for t he arbi trary const ants, x(0) = A - 1 8 = 1 Therefore, A = 9 8 . Also, the derivative of the solution is dx dt Solving for the arb itrary co nstants, x. (0) = 2B = 2. Therefore, B = 1. The fi nal solution is Solutions to Problems 11 17. + -Input transducerDesired forceInput voltage Controller ActuatorPantograph dynamicsSpringFupSpring displacement Fout Sensor T W O Modeling in the Frequency Domain SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transfer Functions Finding each transfer function: Pot: Vi(s) θi(s) = 10 π ; Pre-Am p: Vp(s) Vi(s) = K; Power Am p: Ea(s) Vp(s) = 150 s+150 Motor: J m = 0.05 + 5 (50 250 )2 = 0.25 D m =0.01 + 3 (50 250 )2 = 0.13 Kt Ra = 1 5 KtKb Ra = 1 5 Therefore: θm(s) Ea(s) = Kt RaJm s(s+1 Jm(Dm+KtKb Ra)) = 0.8 s(s+1.32) And: θo(s) Ea(s) = 1 5 θm(s) Ea(s) = 0.16 s(s+1.32) Transfer Function of a Nonlinear Electrical Netw ork W riting the differential equation, d(i0+δi) dt+2(i0+δi)2−5=v(t). Lineari zing i2 about i0, (i0+δi)2-i02=2i⎮ i=i 0δi=2i0δi..Thus,(i0+δi)2=i02+2i0δi.Solutions to Problems 13 Substituting into the diffe rential equation y ields, dδi dt + 2i 02 + 4i 0δi - 5 = v(t). B ut, the resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since the voltage across the induct or is zero at dc. Hence, 2i 02 = 5, or i 0 = 1.58. Substituting into the linearized differential equation, dδi dt + 6.32 δi = v(t). C onverting to a transfer function, δi(s) V(s) = 1 s+6.32 . Using the linearized i about i 0, and the fact that v r(t) is 5 volts at equilibrium , the linearized v r(t) is v r(t) = 2i2 = 2(i0+δi)2 = 2(i 02+2i0δi) = 5+6.32 δi. For excursions aw ay from equilibrium , vr(t) - 5 = 6.32 δi = δvr(t). Therefore, m ultiply ing the transfer function by 6.32, y ields, δVr(s) V(s) = 6.32 s+6.32 as the transfer function about v(t) = 0. ANSWERS TO REVIEW QUESTIONS 1. Transfer function 2. Linear tim e-invariant 3. Laplace 4. G(s) = C (s)/R(s), where c(t) is the output and r(t) is the input. 5. Initial conditions are zero 6. Equations of m otion 7. Free body diagram 8. There are direct analogies between the electrical va riables and com ponents and the m echanical variables and com ponents. 9. Mechanical advantage for rotating sy stems 10. Arm ature inertia, arm ature dam ping, load inertia, load dam ping 11. Multiply the transfer function by the gear ra tio relating arm ature position to load position. 12. (1) R ecognize the nonlinear com ponent, (2) Write the nonlinear differential e quation, (3) Select the equilibrium solution, (4) Linearize the nonlinear di fferential equation, (5) Take the Laplace transform of the linearized differential equati on, (6) Find the transfer function. SOLUTIONS TO PROBLEMS 1. a. F(s)=e−stdt 0∞ ∫=−1 se−st 0∞ =1 s b. F(s)=te−stdt 0∞ ∫=e−st s2(−st−1)0∞=−(st+1) s2est 0∞ 14 Chapter 2: Modeling in the Frequency Domain Using L' Hopital' s Rule F(s)t→∞=−s s3est t→∞=0. Therefore, F(s)=1 s2. c. F(s)=sinωt e−stdt 0∞ ∫=e−st s2+ω2(−ssinωt−ωcosωt) 0∞ =ω s2+ω2 d. F(s)=cosωt e−stdt 0∞ ∫=e−st s2+ω2(−scosωt+ωsinωt) 0∞ =s s2+ω2 2. a. Using the frequency shift theorem and the Laplace transform of sin ωt, F(s) = ω (s+a)2+ω2 . b. Using the frequency shift theorem and the Laplace transform of cos ωt, F(s) = (s+a) (s+a)2+ω2 . c. Using the integration theorem , and su ccessively integrating u(t) three tim es, ⌡⌠dt = t; ⌡⌠tdt = t2 2 ; ⌡⌠t2 2dt = t3 6 , the Laplace transform of t3u(t), F(s) = 6 s4 . 3. a. The Laplace transform of the differential e quation, assum ing zero initial conditions, is, (s+7)X(s) = 5s s2+22 . Solving for X(s) and expa nding by partial fractions, Or, Taking the inverse Laplace transform , x(t) = - 35 53 e-7t + (35 53 cos 2t + 10 53 sin 2t). b. The Laplace transform of the differential equa tion, assum ing zero initial conditions, is, (s2+6s+8)X(s) = 15 s2+9. Solving for X(s) X(s) =15 (s2+9)(s2+6s+8) and expanding by partial fractions, X(s) =−3 656s+1 99 s2+9−3 101 s+4+15 261 s+2 Solutions to Problems 15 Taking the inverse Laplace transform , x(t)=−18 65cos(3t) −1 65sin(3t) −3 10e−4t+15 26e−2t c. The Laplace transform of the differential e quation is, assum ing zero initial conditions, (s2+8s+25)x(s) = 10 s. Solving for X(s) Xs =10 ss2+8s+25 and expanding by partial fractions, Xs =2 51 s-2 51s+4+4 99 s+42+9 Taking the inverse Laplace transform , x(t)=2 5−e−4t8 15sin(3 t)+2 5cos(3 t)⎛ ⎝ ⎞ ⎠ 4. a. Taking the Laplace transform with initial conditions, s2X(s)-2s+3+2sX(s)-4+2X(s) = 2 s2+22 . Solving for X(s), X(s) = 2s3+s2+8s+6 (s2+4)(s2+2s+2). Expanding by partial fractions X(s) =−1 5⎛ ⎝ ⎞ ⎠ s+1 44 s2+4+1 5⎛ ⎝ ⎞ ⎠ 11(s +1)−3 11 (s+1)2+1 Therefore, x(t) = -0.2 cos2t - 0.1 sin2t +e-t (2.2 cost - 0.6 sint). b. Taking the Laplace transform with initial conditions, s2X(s)-2s-1+2sX(s)-4+X(s) = 5 s+2 + 1 s2 . Solving for X(s), Therefore, x(t) = 5e-2t - e-t + 9te-t - 2 + t. c. Taking the Laplace transform with initial conditions, s2X(s)-s-2+4X(s) = 2 s3 . Solving for X(s), 16 Chapter 2: Modeling in the Frequency Domain Therefore, x(t) = 9 8 cos2t + sin2t - 1 8 + 1 4 t2. 5. Program: syms t f=5*t^2*cos(3*t+45); pretty(f) F=laplace(f); F=simple(F); pretty(F) 'b' f=5*t*exp(-2*t)*sin(4*t+60); pretty(f) F=laplace(f); F=simple(F); pretty(F) Computer response: ans = a 2 5 t cos(3 t + 45) 3 2 s cos(45) - 27 s cos(45) - 9 s sin(45) + 27 sin(45) 10 ----------------------------------------------------- 2 3 (s + 9) ans = b 5 t exp(-2 t) sin(4 t + 60) sin(60) ((s + 2) sin(60) + 4 cos(60)) (s + 2) -5 ------------- + 10 ------------------------------------- 2 2 2 (s + 2) + 16 ((s + 2) + 16) 6. Program: syms s 'a' G=(s^2+3*s+7)*(s+2)/[(s+3)*(s+4)*(s^2+2*s+100)]; pretty(G) g=ilaplace(G); pretty(g) 'b' G=(s^3+4*s^2+6*s+5)/[(s+8)*(s^2+8*s+3)*(s^2+5*s+7)]; pretty(G) g=ilaplace(G); pretty(g) Computer response: ans = a 2 Solutions to Problems 17 (s + 3 s + 7) (s + 2) -------------------------------- 2 (s + 3) (s + 4) (s + 2 s + 100) 11 4681 1/2 1/2 - 7/103 exp(-3 t) + -- exp(-4 t) - ----- exp(-t) 11 sin(3 11 t) 54 61182 4807 1/2 + ---- exp(-t) cos(3 11 t) 5562 ans = b 3 2 s + 4 s + 6 s + 5 ------------------------------------- 2 2 (s + 8) (s + 8 s + 3) (s + 5 s + 7) 299 1367 1/2 - --- exp(-8 t) + ---- exp(-4 t) cosh(13 t) 93 417 4895 1/2 1/2 - ---- exp(-4 t) 13 sinh(13 t) 5421 232 1/2 1/2 - ----- exp(- 5/2 t) 3 sin(1/2 3 t) 12927 272 1/2 - ---- exp(- 5/2 t) cos(1/2 3 t) 4309 7. The Laplace transform of the differential equa tion, assum ing zero initial conditions, is, (s3+3s2+5s+1)Y(s) = (s3+4s2+6s+8)X(s). Solving for the transfer function, Y(s) X(s)= s3+4s2+6s+8 s3+3s2+5s+1. 8. a. Cross m ultiplying, (s2+2s+7)X(s) = F(s). Taking the inverse Laplace transform , d2x dt2+ 2dx dt+ 7x = f(t). b. Cross m ultiply ing after expanding the denom inator, (s2+15s+56)X(s) = 10F(s). Taking the inverse Laplace transform , d2x dt2+ 15dx dt+ 56x =10f(t). c. Cross m ultiply ing, (s3+8s2+9s+15)X(s) = (s+2)F(s). Taking the inverse Laplace transform , d3x dt3+ 8d2x dt2+ 9dx dt+ 15x = df(t) dt+2f(t). 9. = s5+2s4+4s3+s2+3 s6+7s5+3s4+2s3+s2+3. The transfer function is C(s) R(s) 18 Chapter 2: Modeling in the Frequency Domain Cross m ultiply ing, (s6+7s5+3s4+2s3+s2+3)C (s) = (s5+2s4+4s3+s2+3)R (s). Taking the inverse Laplace transform assum ing zero initial conditions, d6c dt6+ 7d5c dt5+ 3d4c dt4+ 2d3c dt3+ d2c dt2+ 3c = d5r dt5+ 2d4r dt4+ 4d3r dt3+ d2r dt2+ 3r. 10. = s4+2s3+5s2+s+1 s5+3s4+2s3+4s2+5s+2. The transfer function is C(s) R(s) Cross m ultiply ing, (s5+3s4+2s3+4s2+5s+2)C (s) = (s4+2s3+5s2+s+1)R (s). Taking the inverse Laplace transform assum ing zero initial conditions, d5c dc5+ 3d4c dt4+ 2d3c dt3+ 4d2c dt2+ 5dc dt+ 2c = d4r dt4+ 2d3r dt3+ 5d2r dt2+ dr dt+ r. Substituting r(t) = t3, d5c dc5+ 3d4c dt4+ 2d3c dt3+ 4d2c dt2+ 5dc dt+ 2c = 18δ(t) + (36 + 90 t + 9t2 + 3t3) u(t). 11. Taking the Laplace transform of the differential equation, s2X(s)-s+1+2sX(s)-2+3x(s)=R (s). Collecting term s, (s2+2s+3)X(s) = R(s)+s+1. Solving for X(s), X(s) = R(s) s2+2s+3+ s+1 s2+2s+3. The block diagram is then, 12. Program: 'Factored' Gzpk=zpk([-15 -26 -72],[0 -55 roots([1 5 30])' roots([1 27 52])'],5) 'Polynomial' Gp=tf(Gzpk) Computer response: ans = Factored Zero/pole/gain: 5 (s+15) (s+26) (s+72) -------------------------------------------- s (s+55) (s+24.91) (s+2.087) (s^2 + 5s + 30) ans = Solutions to Problems 19 Polynomial Transfer function: 5 s^3 + 565 s^2 + 16710 s + 140400 -------------------------------------------------------------------- s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s 13. Program: 'Polynomial' Gtf=tf([1 25 20 15 42],[1 13 9 37 35 50]) 'Factored' Gzpk=zpk(Gtf) Computer response: ans = Polynomial Transfer function: s^4 + 25 s^3 + 20 s^2 + 15 s + 42 ----------------------------------------- s^5 + 13 s^4 + 9 s^3 + 37 s^2 + 35 s + 50 ans = Factored Zero/pole/gain: (s+24.2) (s+1.35) (s^2 - 0.5462s + 1.286) ------------------------------------------------------ (s+12.5) (s^2 + 1.463s + 1.493) (s^2 - 0.964s + 2.679) 14. Program: numg=[-10 -60]; deng=[0 -40 -30 (roots([1 7 100]))' (roots([1 6 90]))']; [numg,deng]=zp2tf(numg',deng',1e4); Gtf=tf(numg,deng) G=zpk(Gtf) [r,p,k]=residue(numg,deng) Computer response: Transfer function: 10000 s^2 + 700000 s + 6e006 ----------------------------------------------------------------------------- s^7 + 83 s^6 + 2342 s^5 + 33070 s^4 + 3.735e005 s^3 + 2.106e006 s^2 + 1.08e007 s Zero/pole/gain: 10000 (s+60) (s+10) ------------------------------------------------ s (s+40) (s+30) (s^2 + 6s + 90) (s^2 + 7s + 100) r = -0.0073 0.0313 2.0431 - 2.0385i 2.0431 + 2.0385i -2.3329 + 2.0690i -2.3329 - 2.0690i 0.5556 p = 20 Chapter 2: Modeling in the Frequency Domain -40.0000 -30.0000 -3.5000 + 9.3675i -3.5000 - 9.3675i -3.0000 + 9.0000i -3.0000 - 9.0000i 0 k = [] 15. Program: syms s '(a)' Ga=45*[(s^2+37*s+74)*(s^3+28*s^2+32*s+16)]... /[(s+39)*(s+47)*(s^2+2*s+100)*(s^3+27*s^2+18*s+15)]; 'Ga symbolic' pretty(Ga) [numga,denga]=numden(Ga); numga=sym2poly(numga); denga=sym2poly(denga); 'Ga polynimial' Ga=tf(numga,denga) 'Ga factored' Ga=zpk(Ga) '(b)' Ga=56*[(s+14)*(s^3+49*s^2+62*s+53)]... /[(s^2+88*s+33)*(s^2+56*s+77)*(s^3+81*s^2+76*s+65)]; 'Ga symbolic' pretty(Ga) [numga,denga]=numden(Ga); numga=sym2poly(numga); denga=sym2poly(denga); 'Ga polynimial' Ga=tf(numga,denga) 'Ga factored' Ga=zpk(Ga) Computer response: ans = (a) ans = Ga symbolic 2 3 2 (s + 37 s + 74) (s + 28 s + 32 s + 16) 45 ----------------------------------------------------------- 2 3 2 (s + 39) (s + 47) (s + 2 s + 100) (s + 27 s + 18 s + 15) ans = Ga polynimial Transfer function: 45 s^5 + 2925 s^4 + 51390 s^3 + 147240 s^2 + 133200 s + 53280 -------------------------------------------------------------------------------- Solutions to Problems 21 s^7 + 115 s^6 + 4499 s^5 + 70700 s^4 + 553692 s^3 + 5.201e006 s^2 + 3.483e006 s + 2.75e006 ans = Ga factored Zero/pole/gain: 45 (s+34.88) (s+26.83) (s+2.122) (s^2 + 1.17s + 0.5964) ----------------------------------------------------------------- (s+47) (s+39) (s+26.34) (s^2 + 0.6618s + 0.5695) (s^2 + 2s + 100) ans = (b) ans = Ga symbolic 3 2 (s + 14) (s + 49 s + 62 s + 53) 56 ---------------------------------------------------------- 2 2 3 2 (s + 88 s + 33) (s + 56 s + 77) (s + 81 s + 76 s + 65) ans = Ga polynimial Transfer function: 56 s^4 + 3528 s^3 + 41888 s^2 + 51576 s + 41552 -------------------------------------------------------------------------------- s^7 + 225 s^6 + 16778 s^5 + 427711 s^4 + 1.093e006 s^3 + 1.189e006 s^2 + 753676 s + 165165 ans = Ga factored Zero/pole/gain: 56 (s+47.72) (s+14) (s^2 + 1.276s + 1.111) --------------------------------------------------------------------------- (s+87.62) (s+80.06) (s+54.59) (s+1.411) (s+0.3766) (s^2 + 0.9391s + 0.8119) 16. a. Writing the node equations, Vo−Vi s+Vo s+Vo=0. Solve for Vo Vi=1 s+2. b. Thevenizing, 22 Chapter 2: Modeling in the Frequency Domain Using voltage division, Vo(s)=Vi(s) 21 s 1 2+s+1 s. Thus, Vo(s) Vi(s)=1 2s2+s+2 17. a. Writing m esh equations (s+1)I 1(s) – I 2(s) = V i(s) -I1(s) + (s+2)I 2(s) = 0 But, I1(s) = (s+2)I 2(s). Substituting this in the first equation y ields, (s+1)(s+2)I 2(s) – I 2(s) = V i(s) or I2(s)/V i(s) = 1/(s2 + 3s + 1) But, VL(s) = sI 2(s). Therefore, V L(s)/V i(s) = s/(s2 + 3s + 1). b. Solutions to Problems 23 i1(t)i2(t) (2+2 s)I1(s)−(1+1 s)I2(s)=V(s) −(1+1 s)I1(s)+(2+1 s+2s)I2(s)=0 Solving for I 2(s): I2(s)=2(s+1) sV(s) −s+1 s0 2(s+1) s−s+1 s −s+1 s2s2+2s+1 s=V(s)s 4s2+3s+1 Therefore, VL(s) V(s) = 2s I2(s) V(s) = 2s2 4s2+3s+1 18. a. Writing m esh equations, (2s + 1)I 1(s) – I 2(s) = V i(s) -I1(s) + (3s + 1 + 2/s)I 2(s) = 0 Solving for I 2(s), 24 Chapter 2: Modeling in the Frequency Domain I2(s)=2s+1Vi(s) −10 2s+1 −1 −13s2+s+2 s Solving for I 2(s)/V i(s), I2(s) Vi(s)=s 6s3+5s2+4s+2 But V o(s) = I 2(s)3s. Therefore , G(s) = 3s2/(6s3 + 5s2 +4s + 2). b. Transform ing the network y ields, Writing the loop equations, (s+s s2+1)I1(s)−s s2+1I2(s)−sI3(s)=Vi(s) −s s2+1I1(s)+(s s2+1+1+1 s)I2(s)−I3(s)=0 −sI1(s)−I2(s)+(2s+1)I3(s)=0 Solving for I 2(s), I2(s)=s(s2+2s+2) s4+2s3+3s2+3s+2Vi(s) But, Vo(s) = I2(s) s = (s2+2s+2) s4+2s3+3s2+3s+2Vi(s). Therefore, Vo(s) Vi(s)=s2+2s+2 s4+2s3+3s2+3s+2 19. a. Writing the nodal equations y ields, Solutions to Problems 25 VR(s)−Vi(s) 2s+VR(s) 1+VR(s)−VC(s) 3s=0 −1 3sVR(s)+1 2s+1 3s⎛ ⎝ ⎞ ⎠ VC(s)=0 Rewriting and sim plifying, 6s+5 6sVR(s)−1 3sVC(s)=1 2sVi(s) −1 3sVR(s)+3s2+2 6s⎛ ⎝ ⎜ ⎞ ⎠ VC(s)=0 Solving for VR(s) and VC(s), VR(s)=1 2sVi(s) −1 3s 03s2+2 6s 6s+5 6s−1 3s −1 3s3s2+2 6s; VC(s)=6s+5 6s1 2sVi(s) −1 3s0 6s+5 6s−1 3s −1 3s3s2+2 6s Solving for Vo(s)/Vi(s) Vo(s) Vi(s)=VR(s)−VC(s) Vi(s)=3s2 6s3+5s2+4s+2 b. Writing the nodal equations y ields, (V1(s)−Vi(s)) s+(s2+1) sV1(s)+(V1(s)−Vo(s))=0 (Vo(s)−V1(s))+sVo(s)+(Vo(s)−Vi(s)) s=0 Rewriting and sim plifying, (s+2 s+1)V1(s)−Vo(s)=1 sVi(s) V1(s)+(s+1 s+1)Vo(s)=1 sVi(s)