Control Systems Engineering Solution Manual (Nise-2004)

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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 = 1.59 5 x 2π rad

2 Chapter 1: Introduction

2.

Desired temperature

Voltage difference

Temperature difference

Actual temperature

Fuel flow

+

Amplifier and valves

Thermostat

Heater

-

3.

-

Roll angle

Roll rate

Aileron position control

+

Pilot controls

Aileron position

Error voltage

Input voltage

Desired roll angle

Aircraft dynamics

Integrate

Gyro

Gyro voltage

4. Input voltage

Desired speed

Speed Error voltage

+

transducer

Motor and drive system

Amplifier

Voltage proportional to actual speed

Dancer position sensor

Actual speed

Dancer dynamics

5. Input voltage

Desired power Transducer

Rod position

Power Error voltage

+

Amplifier

-

Voltage proportional to actual power

Motor and drive system

Sensor & transducer

Actual power Reactor

Solutions to Problems 3

6.

Desired student population

Desired student rate

Population error

+

Graduating and drop-out rate

Administration

Actual student rate

+

Actual student population

Net rate of influx Integrate

Admissions

-

7. Voltage proportional to desired volume

Desired volume

+

Transducer

Volume error

Voltage representing actual volume Volume control circuit

Radio

Effective volume

+ -

Transducer

Speed

Voltage proportional to speed

Actual volume

4 Chapter 1: Introduction

8. a. Fluid input

Valve Actuator

Power amplifier +V Differential amplifier + -

R

Desired level

-V

+V

R Float

-V

Tank

Drain

b. Desired level Potentiometer

voltage in + Amplifiers -

Actuator and valve

Actual level

Flow rate in + Integrate -

Drain Flow rate out

voltage out

Displacement Potentiometer

Float

Solutions to Problems 5

9.

Desired force

Current

+ Transducer

Displacement

Amplifier

Actual force

Displacement Actuator and load

Valve

Tire

-

Load cell

10.

Commanded blood pressure +

Actual blood pressure

Isoflurane concentration Vaporizer

Patient

-

11.

Desired depth +

Controller & motor

-

Force

Feed rate Grinder

Depth Integrator

12.

Coil voltage +

Desired position

Coil circuit

Transducer

Coil current

Solenoid coil & actuator

-

LVDT

13. a. L

di + Ri = u(t) dt

Force

Armature & spool dynamics

Depth

6 Chapter 1: Introduction

b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB = 1,

1 R . The characteristic equation is LM + R = 0, from which M = - . Thus, the total R L 1 1 solution is i(t) = Ae-(R/L)t + . Solving for the arbitrary constants, i(0) = A + = 0. Thus, A = R R 1 1 1 -(R/L)t 1 −( R / L)t . The final solution is i(t) = -= (1 − e e ). R R R R from which B =

c.

14.

di 1 idt + vC (0) = v(t) + dt C ∫ d 2i di b. Differentiating and substituting values, + 30i = 0 2 +2 dt dt

a. Writing the loop equation, Ri + L

Writing the characteristic equation and factoring, 2

M + 2 M + 30 = M + 1 + 29 i M + 1 -

29 i .

The general form of the solution and its derivative is -t

i = e 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 dt v (0) 1 di Using i(0) = 0; (0) = L = =2 L L dt i 0 = A =0

di (0) = − A + 29 B =2 dt 2 . Thus, A = 0 and B = 29 The solution is

29 t

Solutions to Problems 7

i=

2 29

-t

29 t

29 e sin

c.

i

t 15. a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

35 From which, C = 53

10 and D = 53 .

The characteristic polynomial is

Thus, the total solution is

35 35 Solving for the arbitrary constants, x(0) = A +53 = 0. Therefore, A = - 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 2

M +6 M+ 8 = M+ 4 M+ 2 Thus, the total solution is

18 1 cos 3 t sin 3 t 65 65 18 Solving for the arbitrary constants, x(0) = C + D − =0. 65 x =C e

-4t

+ De

-2t

+ -

Also, the derivative of the solution is

dx = - 3 cos 3 t + 54 sin 3 t -4t -2t - 4 C e -2 D e dt 65 65 . 15 3 3 . Solving for the arbitrary constants, x(0) − − 4C − 2D = 0 , or C = − and D =

10

65

26

The final solution is

x =-

18 3 - 4 t 15 - 2 t 1 e e + sin 3 t cos 3 t 26 65 10 65

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 2

M + 8 M + 25 = M + 4 + 3 i M + 4 - 3 i Thus, the total solution is

x=

2 -4t +e B sin 3 t + C cos 3 t 5

Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of the solution is

dx -4t dt = 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e

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 2 ⎛8 − e −4t sin(3 t ) + cos(3t )⎞ ⎝ 15 ⎠ 5 5

16. a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

1 From which, C = - 5

1 and D = - 10 .

The characteristic polynomial is

Thus, the total solution is

11 1 Solving for the arbitrary constants, x(0) = A - 5 = 2. Therefore, A = . Also, the derivative of the 5

solution is

dx dt . 3 Solving for the arbitrary constants, x(0) = - A + B - 0.2 = -3. Therefore, B = − . The final solution

5

is

1 1 3 −t ⎛ 11 x (t ) = − cos(2t ) − sin(2 t ) + e cos( t ) − sin(t )⎞ ⎝5 ⎠ 5 10 5 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 = ( − A + B )e − t − Bte −t − 10e −2 t + 1 dt . 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

1 Equating like coefficients, C = 4 , D = 0, and 2C + 4E = 0. 1 1 From which, C = 4 , D = 0, and E = - 8 . The characteristic polynomial is

Thus, the total solution is

9 1 Solving for the arbitrary constants, x(0) = A - 8 = 1 Therefore, A = 8 . 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. Spring displacement Desired force

Input voltage + Input transducer

F up Controller

Actuator

Pantograph dynamics

F out Spring

-

Sensor

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