TEST BANK FOR Control Systems Engineering 7th Edition By Norman S. Nice (Solutions Manual to ISBN 9781118170519)

Exam (elaborations) TEST BANK FOR Control Systems Engineering 7th Edition By Norman S. Nice (Solutions Manual to ISBN 0519) O N E Introduction ANSWERS TO REVIEW QUESTIONS 1. Guided missiles, automatic gain control in radio receivers, satellite tracking antenna 2. Yes - power gain, remote control, parameter conversion; No - Expense, complexity 3. Motor, low pass filter, inertia supported between two bearings 4. Closed-loop systems compensate for disturbances by measuring the response, comparing it to the input response (the desired output), and then correcting the output response. 5. Under the condition that the feedback element is other than unity 6. Actuating signal 7. Multiple subsystems can time share the controller. Any adjustments to the controller can be implemented with simply software changes. 8. Stability, transient response, and steady-state error 9. Steady-state, transient 10. It follows a growing transient response until the steady-state response is no longer visible. The system will either destroy itself, reach an equilibrium state because of saturation in driving amplifiers, or hit limit stops. 11. Transient response 12. True 13. Transfer function, state-space, differential equations 14. Transfer function - the Laplace transform of the differential equation State-space - representation of an nth order differential equation as n simultaneous first-order differential equations Differential equation - Modeling a system with its differential equation SOLUTIONS TO PROBLEMS 1. Five turns yields 50 v. Therefore K = 50 volts 5 x 2π rad = 1.59 2 Chapter 1: Introduction 2. Thermostat Amplifier and valves Heater Temperature difference Voltage difference Fuel flow Actual temperature Desired temperature + - 3. Desired roll angle Input voltage + - Pilot controls Aileron position control Error voltage Aileron position Aircraft dynamics Roll rate Integrate Roll angle Gyro Gyro voltage 4. Speed Error voltage Desired speed Input voltage + - transducer Amplifier Motor and drive system Actual speed Voltage proportional to actual speed Dancer position sensor Dancer dynamics 5. Desired power Power Error voltage Input voltage + - Transducer Amplifier Motor and drive system Voltage proportional to actual power Rod position Reactor Actual power Sensor & transducer Solutions to Problems 3 6. Desired student population + - Administration Population error Desired student rate Admissions Actual student rate + - Graduating and drop-out rate Net rate of influx Integrate Actual student population 7. Desired volume + - Transducer Volume control circuit Voltage proportional to desired volume Volume error Radio Voltage representing actual volume Actual volume - + Transducer - Speed Voltage proportional to speed Effective volume 4 Chapter 1: Introduction 8. a. R +V -V Differential amplifier Desired level - + Power amplifier Actuator Valve Float Fluid input Drain Tank R +V -V b. Desired level Amplifiers Actuator and valve Flow rate in Integrate Actual level Flow rate out Potentiometer + - + Drain Potentiometer Float - voltage in voltage out Displacement Solutions to Problems 5 9. Desired force Transducer Amplifier Valve Actuator and load Tire Load cell Actual + force - Current Displacement Displacement 10. Commanded blood pressure Vaporizer Patient Actual blood + pressure - Isoflurane concentration 11. + - Controller & motor Grinder Force Feed rate Integrator Desired depth Depth 12. + - Coil circuit Solenoid coil & actuator Coil current Force Armature & spool dynamics Desired position Depth Transducer Coil voltage LVDT 13. a. L di dt + Ri = u(t) 6 Chapter 1: Introduction b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields 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) = Ae-(R/L)t + 1 R . Solving for the arbitrary constants, i(0) = A + 1 R = 0. Thus, A = - 1 R . The final solution is i(t) = 1 R -- 1 R e-(R/L)t = 1 R (1 − e−( R/ L)t ). c. 14. a. Writing the loop equation, Ri + L di dt + 1 C ∫ idt + vC (0) = v(t) b. Differentiating and substituting values, d2i dt 2 + 2 di dt + 30i = 0 Writing the characteristic equation and factoring, M2 + 2 M+ 30 = M+ 1 + 29 i M +1 - 29 i . The general form of the solution and its derivative is i = e-t cos 29 t A + B sin 29 t e- t di = - A + 29 B e-t cos 29 t - 29 A + B e- t sin 29 t dt Using i(0) = 0; di dt (0) = vL(0) L = 1 L = 2 i 0 = A =0 di dt (0) = −A + 29B=2 Thus, A = 0 and B = 2 29 . The solution is Solutions to Problems 7 i = 2 29 29 e- t sin 29 t c. i t 15. a. Assume a particular solution of Substitute into the differential equation and obtain Equating like coefficients, From which, C = 35 53 and D = 10 53 . The characteristic polynomial is Thus, the total solution is Solving for the arbitrary constants, x(0) = A + 35 53 = 0. Therefore, A = - 35 53 . The final solution is b. Assume a particular solution of xp = Asin3t + Bcos3t 8 Chapter 1: Introduction Substitute into the differential equation and obtain (18A − B)cos(3t) − (A + 18B)sin(3t) = 5sin(3t) Therefore, 18A – B = 0 and –(A + 18B) = 5. Solving for A and B we obtain xp = (-1/65)sin3t + (-18/65)cos3t The characteristic polynomial is M2 + 6 M+ 8 = M+ 4 M+ 2 Thus, the total solution is x =C e- 4 t + De- 2 t+ - 18 65 cos 3 t - 1 65 sin 3 t Solving for the arbitrary constants, x(0) = C + D − 18 65 = 0 . Also, the derivative of the solution is = - 3 65 cos 3 t + 54 65 dx sin 3 t - 4 C e- 4 t - 2 D e- 2 t dt Solving for the arbitrary constants, x . (0) − 3 65 − 4C − 2D = 0 , or C = − 3 10 and D = 15 26 . The final solution is x =- 18 65 cos 3 t - 1 65 sin 3 t - 3 10 e- 4 t + 15 26 e- 2 t c. Assume a particular solution of xp = A Substitute into the differential equation and obtain 25A = 10, or A = 2/5. The characteristic polynomial is M2 + 8M+ 25 = M+ 4 + 3 i M +4 - 3 i Thus, the total solution is x = 2 5 + e- 4 t B sin 3 t + C cos 3 t Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of the solution is dx = 3 B -4 C cos 3 t - 4B+ 3 C sin 3 t e- 4 t dt Solutions to Problems 9 Solving for the arbitrary constants, x . (0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution is x(t) = 2 5 − e−4t 8 15 sin(3t) + 2 5 cos(3t ) ⎛ ⎝ ⎞ ⎠ 16. a. Assume a particular solution of Substitute into the differential equation and obtain Equating like coefficients, From which, C = - 15 and D = - 1 10 . The characteristic polynomial is Thus, the total solution is Solving for the arbitrary constants, x(0) = A - 15 = 2. Therefore, A = 11 5 . Also, the derivative of the solution is dx dt Solving for the arbitrary constants, x . (0) = - A + B - 0.2 = -3. Therefore, B = − 3 5 . The final solution is x(t) = − 1 5 cos(2t) − 1 10 sin(2t) + e−t 11 5 cos(t) − 3 5 sin(t) ⎛ ⎝ ⎞ ⎠ b. Assume a particular solution of xp = Ce-2t + Dt + E Substitute into the differential equation and obtain 10 Chapter 1: Introduction Equating like coefficients, C = 5, D = 1, and 2D + E = 0. From which, C = 5, D = 1, and E = - 2. The characteristic polynomial is Thus, the total solution is Solving for the arbitrary constants, x(0) = A + 5 - 2 = 2 Therefore, A = -1. Also, the derivative of the solution is dx dt = (−A+ B)e− t − Bte−t −10e−2t +1 Solving for the arbitrary constants, x . (0) = B - 8 = 1. Therefore, B = 9. The final solution is c. Assume a particular solution of xp = Ct2 + Dt + E Substitute into the differential equation and obtain Equating like coefficients, C = 14 , D = 0, and 2C + 4E = 0. From which, C = 14 , D = 0, and E = - 18 . The characteristic polynomial is Thus, the total solution is Solving for the arbitrary constants, x(0) = A - 18 = 1 Therefore, A = 98 . Also, the derivative of the solution is dx dt Solving for the arbitrary constants, x . (0) = 2B = 2. Therefore, B = 1. The final solution is Solutions to Problems 11 17. + - Input transducer Desired force Input voltage Controller Actuator Pantograph dynamics Spring Fup Spring 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-Amp: Vp(s) Vi(s) = K; Power Amp: Ea(s) Vp(s) = 150 s+150 Motor: Jm = 0.05 + 5( 50 250 )2 = 0.25 Dm =0.01 + 3( 50 250 )2 = 0.13 Kt Ra = 15 KtKb Ra = 15 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) = 15 θm(s) Ea(s) = 0.16 s(s+1.32) Transfer Function of a Nonlinear Electrical Network Writing the differential equation, d(i0 + δi) dt + 2(i0 +δi)2 − 5 = v(t) . Linearizing i2 about i0, (i0+δi) 2 - i0 2 = 2i ⎮ i=i 0 δi = 2i0 .δi. Thus, (i0+δi) 2 = i0 2 + 2i0 δi. Solutions to Problems 13 Substituting into the differential equation yields, dδi dt + 2i02 + 4i0δi - 5 = v(t). But, the resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since the voltage across the inductor is zero at dc. Hence, 2i02 = 5, or i0 = 1.58. Substituting into the linearized differential equation, dδi dt + 6.32δi = v(t). Converting to a transfer function, δi(s) V(s) = 1 s+6.32 . Using the linearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i2 = 2(i0+δi)2 = 2(i02+2i0δi) = 5+6.32δi. For excursions away from equilibrium, vr(t) - 5 = 6.32δi = δvr(t). Therefore, multiplying the transfer function by 6.32, yields, 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 time-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 motion 7. Free body diagram 8. There are direct analogies between the electrical variables and components and the mechanical variables and components. 9. Mechanical advantage for rotating systems 10. Armature inertia, armature damping, load inertia, load damping 11. Multiply the transfer function by the gear ratio relating armature position to load position. 12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select the equilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform of the linearized differential equation, (6) Find the transfer function. SOLUTIONS TO PROBLEMS 1. a. F(s) = e− stdt 0 ∞ ∫ = −1 s e−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 s 3est t →∞ = 0. Therefore, F(s) = 1 s2 . c. F(s) = sinωt e− stdt 0 ∞ ∫ = e− st s2 + ω 2 (−s sinωt − ω cosωt) 0 ∞ = ω s2 +ω 2 d. F(s) = cosωt e− stdt 0 ∞ ∫ = e− st s2 + ω 2 (−s cosω t + ω sinωt) 0 ∞ = s s2 +ω 2 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 successively integrating u(t) three times, ⌡⌠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 equation, assuming zero initial conditions, is, (s+7)X(s) = 5s s2+22 . Solving for X(s) and expanding 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 equation, assuming 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 65 6s + 1 9 9 s2 + 9 − 3 10 1 s + 4 + 15 26 1 s + 2 Solutions to Problems 15 Taking the inverse Laplace transform, x(t) = − 18 65 cos(3t) − 1 65 sin(3t) − 3 10 e−4t + 15 26 e−2t c. The Laplace transform of the differential equation is, assuming zero initial conditions, (s2+8s+25)x(s) = 10 s . Solving for X(s) X s = 10 s s2+ 8 s+ 25 and expanding by partial fractions, X s = 25 1s - 25 1 s+4 + 4 9 9 s + 42 + 9 Taking the inverse Laplace transform, x(t) = 2 5 − e−4t 8 15 sin(3t) + 2 5 cos(3t ) ⎛ ⎝ ⎞ ⎠ 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 4 4 s2 + 4 + 1 5 ⎛ ⎝ ⎞ ⎠ 11(s + 1) − 3 1 1 (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) = 98 cos2t + sin2t - 18 + 14 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) /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 equation, assuming 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 multiplying, (s2+2s+7)X(s) = F(s). Taking the inverse Laplace transform, d2 x dt2 + 2 dx dt + 7x = f(t). b. Cross multiplying after expanding the denominator, (s2+15s+56)X(s) = 10F(s). Taking the inverse Laplace transform, d2 x dt2 + 15 dx dt + 56x =10f(t). c. Cross multiplying, (s3+8s2+9s+15)X(s) = (s+2)F(s). Taking the inverse Laplace transform, d3 x dt3 + 8 d2 x dt2 + 9 dx dt + 15x = df (t) dt +2f(t). 9. = s5 + 2s 4 + 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 multiplying, (s6+7s5+3s4+2s3+s2+3)C(s) = (s5+2s4+4s3+s2+3)R(s). Taking the inverse Laplace transform assuming zero initial conditions, d6c dt6 + 7 d5c dt5 + 3 d 4c dt4 + 2 d3c dt3 + d2c dt2 + 3c = d5r dt5 + 2 d 4r dt4 + 4 d3r dt3 + d2r dt2 + 3r. 10. = s4 + 2s3 + 5s2 + s +1 s5 + 3s 4 + 2s3 + 4s2 + 5s + 2 The transfer function is . C(s) R(s) Cross multiplying, (s5+3s4+2s3+4s2+5s+2)C(s) = (s4+2s3+5s2+s+1)R(s). Taking the inverse Laplace transform assuming zero initial conditions, d5c dc5 + 3 d 4c dt4 + 2 d3c dt3 + 4 d2c dt2 + 5 dc dt + 2c = d 4r dt4 + 2 d3r dt3 + 5 d2r dt2 + dr dt + r. Substituting r(t) = t3, d5c dc5 + 3 d 4c dt4 + 2 d3c dt3 + 4 d2c dt2 + 5 dc dt + 2c = 18δ(t) + (36 + 90t + 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 terms, (s2+2s+3)X(s) = R(s)+s+1. Solving for X(s), X(s) = R(s) s 2 + 2s +3 + s +1 s 2 + 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 + -------------------------------------------------------------------- s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s 13. Program: 'Polynomial' Gtf=tf([],[ 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 + 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 + s^2 + s + 53280 -------------------------------------------------------------------------------- Solutions to Problems 21 s^7 + 115 s^6 + 4499 s^5 + 70700 s^4 + 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 + s^4 + 1.093e006 s^3 + 1.189e006 s^2 + s + 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) 2 1 s 1 2 + s + 1 s . Thus, Vo (s) Vi (s) = 1 2s2 + s + 2 17. a. Writing mesh equations (s+1)I1(s) – I2(s) = Vi(s) -I1(s) + (s+2)I2(s) = 0 But, I1(s) = (s+2)I2(s). Substituting this in the first equation yields, (s+1)(s+2)I2(s) – I2(s) = Vi(s) or I2(s)/Vi(s) = 1/(s2 + 3s + 1) But, VL(s) = sI2(s). Therefore, VL(s)/Vi(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 I2(s): I2 (s) = 2(s + 1) s V(s) − s +1 s 0 2(s + 1) s − s +1 s − s +1 s 2s2 + 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 mesh equations, (2s + 1)I1(s) – I2(s) = Vi(s) -I1(s) + (3s + 1 + 2/s)I2(s) = 0 Solving for I2(s), 24 Chapter 2: Modeling in the Frequency Domain I2 (s) = 2s +1 Vi (s) −1 0 2s + 1 −1 −1 3s2 + s + 2 s Solving for I2(s)/Vi(s), I2 (s) Vi (s) = s 6s3 + 5s2 + 4s + 2 But Vo(s) = I2(s)3s. Therefore , G(s) = 3s2/(6s3 + 5s2 +4s + 2). b. Transforming the network yields, Writing the loop equations, (s + s s2 + 1 )I1(s) − s s2 +1 I2 (s) − sI3(s) = Vi (s) − s s2 +1 I1(s) + ( s s2 + 1 +1 + 1 s )I2 (s) − I3 (s) = 0 − sI1(s) − I2 (s) +(2s + 1)I3 (s) = 0 Solving for I2(s), I2 (s) = s(s2 + 2s + 2) s4 + 2s3 + 3s2 + 3s + 2 Vi (s) But, Vo(s) = I2(s) s = (s2 + 2s + 2) s4 + 2s3 + 3s2 + 3s + 2 Vi (s). Therefore, Vo (s) Vi (s) = s2 + 2s + 2 s4 + 2s3 + 3s2 + 3s + 2 19. a. Writing the nodal equations yields, Solutions to Problems 25 VR(s) −Vi (s) 2s + VR(s) 1 + VR (s) − VC (s) 3s = 0 − 1 3s VR(s) + 1 2 s + 1 3s ⎛ ⎝ ⎞ ⎠ VC (s) = 0 Rewriting and simplifying, 6s + 5 6s VR(s) − 1 3s VC (s) = 1 2s Vi (s) − 1 3s VR(s) + 3s2 + 2 6s ⎛ ⎝ ⎜ ⎞ ⎠ VC (s) = 0 Solving for VR(s) and VC(s), VR(s) = 1 2s Vi (s) − 1 3s 0 3s2 + 2 6s 6s + 5 6s − 1 3s − 1 3s 3s2 + 2 6s ; VC (s) = 6s + 5 6s 1 2s Vi (s) − 1 3s 0 6s + 5 6s − 1 3s − 1 3s 3s 2 + 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 yields, (V1(s) − Vi (s)) s + (s2 + 1) s V1(s) + (V1(s) − Vo (s)) = 0 (Vo (s) − V1(s)) + sVo(s) + (Vo (s) − Vi (s)) s = 0 Rewriting and simplifying, (s + 2 s + 1)V1(s) − Vo (s) = 1 s Vi (s) V1(s) + (s + 1 s + 1)Vo (s) = 1 s Vi (s) 26 Chapter 2: Modeling in the Frequency Domain Solving for Vo(s) Vo(s) = (s2 + 2s + 2) s4 + 2s3 + 3s2 + 3s + 2 Vi(s). Hence, Vo (s) Vi (s) = (s2 + 2s + 2) s4 + 2s3 + 3s2 + 3s + 2 20. a. Mesh: (2+2s)I1(s) - (1+2s)I2(s) - I3(s) = V(s) - (1+2s)I1(s) + (7+5s)I2(s) - (2+3s)I3(s) = 0 -I1(s) - (2+3s)I2(s) + (3+3s+ 5s )I3(s) = 0 Nodal: V1(s) - V(s) + V1(s) (1+2s) + (V1(s) - Vo(s)) 2+3s = 0 (Vo(s) - V1(s)) 2+3s + Vo(s) 4 + (Vo(s) - V(s)) 5s = 0 or 6s2 +12s + 5 6s2 + 7s + 2 V1(s) - 1 3s + 2 Vo(s) = V(s) − 1 3s + 2 V1(s) + 1 20 12s2 + 23s + 30 3s + 2 Vo(s) = s 5 V(s) b. Program: syms s V %Construct symbolic object for frequency Solutions to Problems 27 %variable 's' and V. 'Mesh Equations' A2=[(2+2*s) V -1 -(1+2*s) 0 -(2+3*s) -1 0 (3+3*s+(5/s))] %Form Ak = A2. A=[(2+2*s) -(1+2*s) -1 -(1+2*s) (7+5*s) -(2+3*s) -1 -(2+3*s) (3+3*s+(5/s))] %Form A. I2=det(A2)/det(A); %Use Cramer's Rule to solve for I2. G1=I2/V; %Form transfer function, G1(s) = I2(s)/V(s). G=4*G1; %Form transfer function, G(s) = V4(s)/V(s). 'G(s) via Mesh Equations' %Display label. pretty(G) %Pretty print G(s) 'Nodal Equations' A2=[(6*s^2+12*s+5)/(6*s^2+7*s+2) V -1/(3*s+2) s*(V/5)] %Form Ak = A2. A=[(6*s^2+12*s+5)/(6*s^2+7*s+2) -1/(3*s+2) -1/(3*s+2) (1/20)*(12*s^2+23*s+30)/(3*s+2)] %Form A. I2=simple(det(A2))/simple(det(A)); %Use Cramer's Rule to solve for I2. G1=I2/V; %Form transfer function, G1(s) = I2(s)/V(s). 'G(s) via Nodal Equations' %Display label. pretty(G) %Pretty print G(s) Computer response: ans = Mesh Equations A2 = [ 2+2*s, V, -1] [ -1-2*s, 0, -2-3*s] [ -1, 0, 3+3*s+5/s] A = [ 2+2*s, -1-2*s, -1] [ -1-2*s, 7+5*s, -2-3*s] [ -1, -2-3*s, 3+3*s+5/s] ans = G(s) via Mesh Equations 2 3 15 s + 12 s + 5 + 6 s 4 -------------------------- 2 3 120 s + 78 s + 65 + 24 s ans = Nodal Equations A2 = [ (6*s^2+12*s+5)/(2+7*s+6*s^2), V] [ -1/(2+3*s), 1/5*s*V] A = 28 Chapter 2: Modeling in the Frequency Domain [ (6*s^2+12*s+5)/(2+7*s+6*s^2), -1/(2+3*s)] [ -1/(2+3*s), (3/5*s^2+23/20*s+3/2)/(2+3*s)] ans = G(s) via Nodal Equations 2 3 15 s + 12 s + 5 + 6 s 4 -------------------------- 3 2 24 s + 78 s + 120 s + 65 21. a. Z1(s) = 5x105 + 106 s Z2 (s) = 105 + 106 s Therefore, - Z2 s Z1 s = -15 s + 10 s + 2 b. Z1 s 1 1×10−6 s = + = s+ 10 s Z2 s 1 1×10−6 s 1 + = + = s+ 20 s+ 10 Therefore, G s Z2 Z1 = − = s+20 s s+10 2 − 22. a. Z1 s 1 1×10−6 s = + Z2 s 1 1×10−6 s = + Therefore, G s Z1 +Z2 Z1 = = 32 s 20 3 + s+ 5 . Solutions to Problems 29 b. Z1(s) = 2x105 + 5x1011 s 5x105 + 106 s Z2 (s) = 5x105 + 1011 s 105 + 106 s Therefore, Z1(s) + Z2 (s) Z1(s) = 7 2 (s + 3.18)(s + 11.68) (s + 7)(s + 10) 23. Writing the equations of motion, where x2(t) is the displacement of the right member of springr, (s2+s+1)X1(s) -X2(s) = 0 -X1(s) +X2(s) = F(s) Adding the equations, (s2+s)X1(s) = F(s) From which, X1(s) F(s) = 1 s(s + 1) . 24. Writing the equations of motion, (s 2 + s + 1)X1 (s) − (s + 1)X2 (s) = F(s) −(s + 1)X1(s) + (s2 + s + 1)X2(s) = 0 Solving for X2(s), X2 (s) = (s2 + s + 1) F(s) −(s + 1) 0 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ (s2 + s + 1) −(s + 1) −(s + 1) (s2 + s + 1) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = (s + 1)F(s) s2(s2 + 2s + 2) From which, X2 (s) F(s) = (s + 1) s2 (s2 + 2s + 2) . 25. Let X1(s) be the displacement of the left member of the spring and X3(s) be the displacement of the mass. Writing the equations of motion 30 Chapter 2: Modeling in the Frequency Domain 2x1(s) − 2x2 (s) = F(s) −2X1(s) + (5s + 2)X2(s) − 5sX3(s) = 0 −5sX2 (s) + (10s2 + 7s)X3(s) = 0 Solving for X2(s), X2(s) = ⎪ ⎪⎪ 5s2+10 -10 -10 5 1 s+10 ⎪ ⎪⎪ ⎪ ⎪⎪ 5s2+10 F(s) -10 0 ⎪ ⎪⎪ = s(s2+50s+2) 10F(s) Thus, X2 (s) F(s) = 1 10 (10s + 7) s(5s + 1) 26. s + 3 s+ 2 2 X1 s − s+ 1 X2 s = 0 − s+ 1 X1 s s2 + + 2 s+ 1 X2 s = F s Solving for X1(s); X1 0 − s+ 1 F s2 + 2 s+ 1 s2 + 3 s+ 2 − s+ 1 − s+ 1 s2 + 2 s+ 1 = = F s s3 4 s2 + +4 s+1 . Thus, X1 F s 1 s3 4 s2 + +4 s+1 = 27. Writing the equations of motion, (s2 + s + 1)X1 (s) − sX2 (s) = 0 −sX1(s) + (s2 + 2s + 1)X2 (s) − X3 (s) = F(s) −X2(s) + (s2 + s + 1)X3(s) = 0 Solving for X3(s), X3 (s) = (s2 + s + 1) − s 0 −s (s2 + 2s + 1) F(s) 0 −1 0 (s2 + s + 1) −s 0 −s (s2 + 2s + 1) −1 0 −1 (s2 + s + 1) = F(s) s(s3 + 3s2 + 3s + 3) Solutions to Problems 31 From which, X3 (s) F(s) = 1 s(s3 + 3s2 + 3s + 3) . 28. a. (s2 + 2s +1)X1(s) − 2sX2 (s) − X3 (s) = F(s) −2sX1(s) + (s2 + 4s)X2(s) − sX3(s) = 0 −X1(s) − sX2 (s) + (s +1)X3(s) = 0 Solving for X2(s), X2(s) = (s2 + 2s + 1) F(s) −1 −2s 0 −s −1 0 s+ 1 Δ = −F(s) −2s −s −1 s+1 Δ or, X2(s) F(s) = 2s + 3 s(s3 + 6s2 + 9s + 3) b. (4s2 + s+ 4)X1(s) − (s+ 1)X2(s) − 3X3(s) = 0 −(s+ 1)X1(s) + (2s2 + 5s+ 1)X2(s) − 4sX3(s) = F(s) −3X1(s) − 4sX2 (s) + (4s+ 3)X3(s) = 0 Solving for X3(s), X3(s) = (4s2 + s+ 4) −(s+ 1) 0 −(s+ 1) (2s2 + 5s+ 1) F(s) −3 −4s 0 Δ = −F(s) (4s2 + s+ 4) −(s+ 1) −3 −4s Δ or X3(s) F(s) = 16s3 + 4s2 +19s +3 32s5 + 48s4 +114 s3 +18s2 29. Writing the equations of motion, (s2 + 2s + 2)X1(s) − X2(s) − sX3 (s) = 0 −X1(s) + (s2 + s + 1)X2(s) − sX3 (s) = F(s) −sX1(s) − sX2 (s) + (s2 + 2s + 1) = 0 32 Chapter 2: Modeling in the Frequency Domain 30. a. Writing the equations of motion, (s2 + 9s + 8)θ 1 (s) − (2s + 8)θ 2(s) = 0 −(2s + 8)θ 1(s) + (s2 + 2s + 11)θ 2 (s) = T(s) b. Defining θ 1(s) = rotation of J1 θ 2 (s) = rotation between K1 and D1 θ 3(s) = rotation of J3 θ 4 (s) = rotation of right - hand side of K2 the equations of motion are (J1s2 + K1) θ 1(s) − K1θ 2 (s) = T(s) −K1θ 1(s) + (D1s + K1 )θ 2 (s) − D1sθ 3(s) = 0 −D1s θ 2 (s) + (J2s2 + D1s + K2 )θ 3(s) − K2θ 4(s) = 0 −K2θ 3(s) + (D2s + (K2 + K3))θ 4 (s) = 0 31. Writing the equations of motion, (s 2 + 2s + 1)θ 1 (s) − (s + 1)θ 2 (s) = T(s) −(s + 1)θ 1(s) + (2s + 1)θ 2(s) = 0 Solving for θ 2 (s) θ 2 (s) = (s2 + 2s + 1) T(s) −(s + 1) 0 (s2 + 2s + 1) −(s + 1) −(s + 1) (2s + 1) = T(s) 2s(s + 1) Hence, θ 2 (s) T(s) = 1 2s(s + 1) 32. Reflecting impedances to θ3, Solutions to Problems 33 (Jeqs2+Deqs)θ3(s) = T(s) ( N4N2 N3N1 ) Thus, θ3 (s) T(s) = N4N2 N3N1 Jeq s2 + Deqs where Jeq = J4+J5+(J2+J3) N4 N3 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ 2 + J1 N4N2 N3N1 ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ 2 , and Deq = (D4 + D5 ) + (D2 + D3)( N4 N3 )2 + D1( N4N2 N3N1 )2 33. Reflecting all impedances to θ2(s), {[J2+J1(N2 N1 )2 +J3 (N3 N4 )2]s2 + [f2+f1(N2 N1 )2 +f3(N3 N4 )2]s + [K(N3 N4 )2]}θ2(s) = T(s) N2 N1 Substituting values, {[1+2(3)2+16(14 )2]s2 + [2+1(3)2+32(14 )2]s + 64(14 )2}θ2(s) = T(s)(3) Thus, θ2(s) T(s) = 3 20s2+13s+4 34. Reflecting impedances to θ2, 200 + 3 50 5 ⎛ ⎜⎝ ⎞ ⎟⎠ 2 + 200 5 25 x 50 5 ⎛ ⎜⎝ ⎞ ⎟⎠ ⎡ 2 ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ s2 + 1000 5 25 x 50 5 ⎛ ⎜⎝ ⎞ ⎟⎠ ⎡ 2 ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ s + 250 + 3 50 5 ⎛ ⎜⎝ ⎞ ⎟⎠ ⎡ 2 ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 50 5 ⎛ ⎜⎝ ⎞ ⎟⎠ T(s) Thus, θ 2 (s) T(s) = 10 1300s2 + 4000s + 550 35. Reflecting impedances and applied torque to respective sides of the spring yields the following equivalent circuit: 34 Chapter 2: Modeling in the Frequency Domain Writing the equations of motion, θ2(s) - θ3(s) = 4T(s) -θ2(s) + (s+1)θ3(s) = 0 Solving for θ3(s), θ 3 (s ) = 1 4T(s) -1 0 1 -1 -1 (s+ 1 ) = 4T(s) s Hence, θ3(s) T(s) = 4s . But, θ4(s) = 15 θ3(s). Thus, θ4(s) T(s) = 4/5 s . 36. Reflecting impedances and applied torque to respective sides of the viscous damper yields the following equivalent circuit: Writing the equations of motion, (s2 + s)θ 2 (s) − sθ 3(s) = 10T(s) − sθ 2 (s) + (s + 1)θ 3 (s) −θ 4 (s) = 0 − θ 3 (s) + (s + 1)θ 4 (s) = 0 Solving for θ 4 (s) , Solutions to Problems 35 θ 4 (s) = s(s + 1) −s 10T(s) −s (s + 1) 0 0 −1 0 s(s + 1) −s 0 −s (s + 1) −1 0 −1 (s + 1) = s10T(s) s(s + 1) −s 0 −s (s + 1) −1 0 −1 (s + 1) Thus, θ 4 (s) T(s) = 10 s(s + 1)2 But, θ L(s) = 5θ4 (s) . Hence, θ 4 (s) T(s) = 50 s(s + 1)2 37. Reflect all impedances on the right to the viscous damper and reflect all impedances and torques on the left to the spring and obtain the following equivalent circuit: Writing the equations of motion, (J1eqs2+K)θ2(s) -Kθ3(s) = Teq(s) -Kθ2(s)+(Ds+K)θ3(s) -Dsθ4(s) = 0 -Dsθ3(s) +[J2eqs2 +(D+Deq)s]θ4(s) = 0 where: J1eq = J2+(Ja+J1)(N2 N1 )2 ; J2eq = J3