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

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