Control Systems Engineering By Norman S. Nise 6th Edition – Test Bank

Complete Test Bank With Answers

 

 

Sample Questions Posted Below

 

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

50 volts

1. Five turns yields 50 v. Therefore K =

= 1.59

5 x 2πrad

Copyright © 2011 by John Wiley & Sons, Inc.1-2 Chapter 1: Introduction

2.

Desired

temperature

Temperature

difference

Voltage

difference

Fuel

flow

+

Thermostat Amplifier and

valves

Heater

–

3.

Desired

roll

angle

Pilot

controls

4.

Desired

speed

transducer Input

voltage

+

–

Input

voltage

+

–

Error

voltage

Aileron

position

Roll

rate

Aileron

position

control

Aircraft

dynamics

Integrate

Gyro

Gyro voltage

Speed

Error

voltage

Amplifier

Motor

and

drive

system

Dancer

position

sensor

Dancer

dynamics

Voltage

proportional

to actual speed

Copyright © 2011 by John Wiley & Sons, Inc.

Actual

temperature

Roll

angle

Actual

speed1-3 Solutions to Problems

5.

Desired

power

Transducer Input

voltage

+

–

Power

Error

voltage

Rod

position

Amplifier

Motor

and

drive

system

Actual

power

Reactor

Sensor &

transducer

Voltage

proportional

to actual power

6.

Desired

student

population +

Population

error

Desired

student

rate

Actual

student

rate +

Graduating

and

drop-out

rate

–

Net rate

of influx

Actual

student

population

Administration

Admissions

Integrate

–

7.

Voltage

proportional

to desired

Desired

volume

volume +

Transducer

–

Volume

error

Volume

control circuit

Effective

volume

Voltage

representing

actual volume Actual

volume

Radio

Voltage

proportional

to speed

+

–

Transducer

–

Speed

Copyright © 2011 by John Wiley & Sons, Inc.1-4 Chapter 1: Introduction

8.

a.

Fluid input

Valve

Actuator

Power

amplifier

Differential

amplifier

+V

R

-V

Desired

level

–

+

+V

R

-V

Float

Tank

Drain

b.

Desired

level

Integrate

Drain

Actual

level

voltage

in

+

Flow

rate in

+

Potentiometer

Amplifiers Actuator

and valve

–

–

Flow

rate out

Displacement

voltage

out

Potentiometer

Float

Copyright © 2011 by John Wiley & Sons, Inc.9.

Desired

force

Transducer +

Current Displacement Displacement

1-5 Solutions to Problems

Actual

force

Amplifier Valve Actuator

and load

Tire

–

Load cell

10.

Commanded

blood pressure

+

Isoflurane

concentration

Actual

blood

pressure

Vaporizer Patient

–

11.

Desired

depth +

–

12.

Desired

position Controller

&

motor

Force Feed rate

Depth

Grinder

Integrator

Coil

voltage

+

Coil

current Transducer

Coil

circuit

Force Solenoid coil

& actuator

Armature

&

spool dynamics

Depth

–

LVDT

Copyright © 2011 by John Wiley & Sons, Inc.1-6 Chapter 1: Introduction

13.

a.

Nervous

system

electrical

impulses

Brain Desired

Light

Intensity

+

Internal eye

muscles

Retina’s

Light

Intensity

Retina + Optical

b.

Nervous

system

electrical

impulses

Desired

Light

Intensity

Brain +

Internal eye

muscles

Retina’s

Light

Intensity

Retina + Optical

Nerves

External

Light

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.

Copyright © 2011 by John Wiley & Sons, Inc.1-7 Solutions to Problems

14.

Desired HT’s

+

Amplifier

Actual

Gyroscopic

15.

16.

17.

a. L

di

dt + Ri = u(t)

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

Copyright © 2011 by John Wiley & Sons, Inc.1-8 Chapter 1: Introduction

1

1

solution is i(t) = Ae-(R/L)t +

1

–

R . The final solution is i(t) =

R . Solving for the arbitrary constants, i(0) = A +

1

1

e-(R/L)t =

R

1

R (1− e−( R/ L)t ).

R = 0. Thus, A =

R —

c.

18.

a. Writing the loop equation, Ri + L di

1

+

dt

C

∫= v(t)

idt + vC (0)

2

b. Differentiating and substituting values,

2 2 25 0 d i di i

+ + =

dt dt

Writing the characteristic equation and factoring,

M M M i M i + + = + + + − .

2 2 25 ( 1 24 )( 1 24 )

The general form of the solution and its derivative is

cos( 24 ) sin( 24 )t t i Ae t Be t− −

= +

di A B e t A B e t

− −

= − + − +

( 24 ) cos( 24 ) ( 24 ) sin( 24 )t t

dt

di v

Using (0) 1 (0) 0; (0) 1 L

i

dt L L= = = =

i 0 A= =0

di A B

(0) 24

= − + =1

dt

Thus, A= 0 and

B=

.

The solution is

t i e t−

=

1 sin( 24 )

24

1

24

Copyright © 2011 by John Wiley & Sons, Inc.1-9 Solutions to Problems

c.

19.

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

Copyright © 2011 by John Wiley & Sons, Inc.1-10 Chapter 1: Introduction

xp = Asin3t + Bcos3t

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

M + 2

1

65 sin 3 t

M2 + 6 M + 8 = M + 4 Thus, the total solution is

x = C e– 4 t + D e– 2 t + –

Solving for the arbitrary constants, x(0)= C + D−

18

65 cos 3 t –

18

65= 0 .

Also, the derivative of the solution is

dx

dt

3

= –

65 cos 3 t +

54

65 sin 3 t – 4 C e– 4 t – 2 D e– 2 t

.

Solving for the arbitrary constants, x

(0) −

3

65− 4C− 2D= 0 , or C = −

3

and D =

10

The final solution is

18

65 cos 3 t –

15

26.

x = –

c. Assume a particular solution of

1

65 sin 3 t –

3

– 4 t +

10 e

15

26 e

– 2 t

xp = A

Substitute into the differential equation and obtain 25A = 10, or A = 2/5.

The characteristic polynomial is

M2 + 8 M + 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

Copyright © 2011 by John Wiley & Sons, Inc.1-11 Solutions to Problems

dx

dt

– 4 t

= 3 B -4 C cos 3 t – 4 B + 3 C sin 3 t e

.

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

2

⎝ ⎞ ⎠

15 sin(3t) +

5 cos(3t )

20.

a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

From which, C = – 1

5 and D = – 1

10 .

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A – 1

5 = 2. Therefore, A =

11

5 . Also, the derivative of the

solution is

dx

dt

3

.

Solving for the arbitrary constants, x

(0) = – A + B – 0.2 = -3. Therefore, B = −

5

. The final solution

is

x(t)= −

1

5 cos(2t)−

1

⎛

10 sin(2t) + e−t 11

3

⎝ ⎞ ⎠

5 cos(t)−

5 sin(t)

b. Assume a particular solution of

xp = Ce-2t + Dt + E

Substitute into the differential equation and obtain

Copyright © 2011 by John Wiley & Sons, Inc.1-12 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 = 1

4 , D = 0, and 2C + 4E = 0.

From which, C = 1

4 , D = 0, and E = – 1

8 .

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A – 1

8 = 1 Therefore, A = 9

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

Copyright © 2011 by John Wiley & Sons, Inc.1-13 Solutions to Problems

21.

Spring

displacement

Desired

force

Input

transducer

Input

voltage +

Controller Actuator Fup

Pantograph

dynamics Fout

Spring

–

Sensor

22.

Desired

Amount of

HIV viruses

RTI

Amount of

HIV viruses

Controller Patient

PI

Copyright © 2011 by John Wiley & Sons, Inc.1-14 Chapter 1: Introduction

23.

a.

Inverter

Control

Command

Controlled

Voltage

Climbing &

Rolling

Resistances

Desired

Speed

Motive

Force

Actual

ECU Inverter

Electric

Motor

+ +

Vehicle

Aerodynamic

Aerodynamic

Speed

Copyright © 2011 by John Wiley & Sons, Inc.b.

Desired Speed +

_

1-15 Solutions to Problems

Climbing &

Rolling

Resistances

Accelerator

Displacement

Motive

Actual

ECU

Accelerator,

Vehicle

+

Aerodynamic

Aerodynamic

Speed

Copyright © 2011 by John Wiley & Sons, Inc.1-16 Chapter 1: Introduction

c.

Accelerator

ICE

Accelerator

Climbing &

Rolling

Resistances

Speed

Error

Desired

ECU Power Planetary

Gear

Control

Actual

+

Vehicle

+

Total

Motive

Force

Inverter

Control

Command

Inverter

&

Electric

Aerodynamic

Motor

Aerodynamic

Motor

Speed