Solution manual of Control Systems Engineering by Norman Nise [ 6th , 7th , 8th

1. https://gioumeh.com/product/control-systems-engineering-solution/https://gioumeh.com/product/control-systems-engineering-solution/ cl i ck h ere t o d ow nl oad 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. Natural response 12. Determine the transient response performance of the system. 13. Determine system parameters to meet the transient response specifications for the system. 14. True 15. Transfer function, state-space, differential equations 16. 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. 50 15 turns yield 50V, therefore ? = 15×2?= 0.53 @solutionmanual1

2. 1-2 Chapter 1: Introduction https://gioumeh.com/product/control-systems-engineering-solution/ cl i ck h ere t o d ow nl oad 0.53 ?(?) ??(?) 2. Voltage difference Desired temperature Temperature difference Actual temperature Fuel flow + Amplifier and valves Heater Thermostat - @solutionmanual1

3. https://gioumeh.com/product/control-systems-engineering-solution/https://gioumeh.com/product/control-systems-engineering-solution/ Solutions to Problems 1-3 cl i ck h ere t o d ow nl oad 3 Voltage proportional to desired volume Voltage representing actual volume Volume error Actual volume Desired volume + Volume Radio Transducer control circuit - Effective volume + Voltage proportional to speed - Transducer - Speed 4 Desired force Actual force Current Displacement Displacement + Actuator and load Transducer Amplifier Valve Tire - Load cell @solutionmanual1

4. 1-4 Chapter 1: Introduction https://gioumeh.com/product/control-systems-engineering-solution/ cl i ck h ere t o d ow nl oad 5 Desired depth + Controller & motor Force Feed rate Depth Grinder Integrator - 6 Nervous system electrical impulses a. Internal eye muscles Brain Desired Light Intensity Retina’s Light Intensity + - Retina + Optical Nerves b. Nervous system electrical impulses Desired Light Intensity Internal eye muscles Brain Retina’s Light Intensity + - External Light Retina + Optical Nerves If the narrow light beam is modulated sinusoidally the pupil’s diameter will also vary sinusoidally (with a delay see part c) in problem) c.If the pupil responded with no time delay the pupil would contract only to the point where a small amount of light goes in. Then the pupil would stop contracting and would remain with a fixed diameter. @solutionmanual1

5. https://gioumeh.com/product/control-systems-engineering-solution/https://gioumeh.com/product/control-systems-engineering-solution/ Solutions to Problems 1-5 cl i ck h ere t o d ow nl oad 7. HT’s Amplifier Desired + Actual - + Motors Gyroscopic sensors 8. Desired Body Temperature Light Stimulus Measured Body Temperature + Circadian Cycle Model Controller Astronaut - 9. Vision System Student Position Instructor Position + Tactile System Student - @solutionmanual1

6. 1-6 Chapter 1: Introduction https://gioumeh.com/product/control-systems-engineering-solution/ cl i ck h ere t o d ow nl oad 10. F M Disturbance equivalence + + Y f Fu Plant Controller + -1 + -1 -1 @solutionmanual1

7. https://gioumeh.com/product/control-systems-engineering-solution/https://gioumeh.com/product/control-systems-engineering-solution/ Solutions to Problems 1-7 cl i ck h ere t o d ow nl oad 11 a. Photo-Voltaic Solar Battery Voltage, Vb -∆Vpv Irradiance, r(t) Voltage,Vpv DC/DC Converte r Solar Array Battery + -∆Vpv MPPT Controller b. Time of Day Battery SOC Actual AC Voltage, Vac Desired AC Voltage Voltage Error Va Power Manager & Controller DC/AC Inverter Current Controller + + _ _ Output Inductor Ia + Iu IL Load with Plug-in HEV _ Voltage Sensor @solutionmanual1

8. 1-8 Chapter 1: Introduction https://gioumeh.com/product/control-systems-engineering-solution/ cl i ck h ere t o d ow nl oad 12. @solutionmanual1

9. https://gioumeh.com/product/control-systems-engineering-solution/https://gioumeh.com/product/control-systems-engineering-solution/ Solutions to Problems 1-9 cl i ck h ere t o d ow nl oad 13. a. Ldi + Ri = u(t) dt b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB = 1, from which B = 1 . The characteristic equation is LM + R = 0, from which M = -R . Thus, the total R L solution is i(t) = Ae-(R/L)t +1 . Solving for the arbitrary constants, i(0) = A + 1 = 0. Thus, A = R R - 1 . The final solution is i(t) = 1 -- 1 Re-(R/L)t = 1 R(1−e−( R/L)t). R R c. @solutionmanual1

10. 1-10 Chapter 1: Introduction https://gioumeh.com/product/control-systems-engineering-solution/ cl i ck h ere t o d ow nl oad 14. 1 C d di  + + + = ) 0 ( c ( ) Ri L idt v v t a. Writing the loop equation, dt 2 i di + + = 2 16 0 i b. Differentiating and substituting values, 2 dt dt Writing the characteristic equation and factoring, 2 + + = + + + − 2 16 ( 1 15 )( 1 15 ) M M M i M i The general form of the solution and its derivative is − + = − t t cos( 15 ) sin( 15 ) i Ae t Be t Using i(0) = 0; di − − t t = − + − + ( 15 ) cos( 15 ) ( 15 ) sin( 15 ) A B e t A B e t dt ) 0 ( L 1 L di v L = = = ) 0 ( 2 dt 2 di = − + =  = ) 0 ( 15 2 A B B i(0) = A = 0 and dt 15 2 t − = ( ) 15 sin( 15 ) i t e t The solution is: 15 c. 0.5 0.4 0.3 Current, i(t), A 0.2 0.1 0 -0.1 -0.2 0 0.5 1 1.5 2 2.5 3 3.5 4 Time, sec @solutionmanual1