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Control Systems Engineering Nise 6th edition solutions CH01


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. Five turns yields 50 v. Therefore K =

50 volts = 1.59 5 x 2π rad

Copyright ? 2011 by John Wiley & Sons, Inc.

1-2 Chapter 1: Introduction

2.

Desired temperature

Temperature difference

Voltage difference Amplifier and valves

Fuel flow Heater

Actual temperature

+
Thermostat

-

3.

Desired roll angle Pilot controls

Input voltage

Error voltage Aileron position control

Aileron position Aircraft dynamics

Roll rate Integrate

Roll angle

+ -

Gyro voltage

Gyro

4.
Desired speed transducer Input voltage Speed Error voltage Amplifier

+ -

Motor and drive system Dancer dynamics

Actual speed

Voltage proportional to actual speed

Dancer position sensor

Copyright ? 2011 by John Wiley & Sons, Inc.

1-3 Solutions to Problems

5.
Input voltage Desired power Transducer Power Error voltage Amplifier Rod position Motor and drive system Actual power Reactor

+ -

Voltage proportional to actual power

Sensor & transducer

6.
Graduating and drop-out rate Actual student rate Admissions

Desired student population

Population error

Desired student rate Administration

+

Net rate of influx Integrate

Actual student population

+ -

7.
Voltage proportional to desired volume Voltage representing actual volume Volume control circuit Radio

Desired volume Transducer

+

Volume error

Actual volume

Effective volume

+ 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 +V Differential amplifier + +V -V

R

Desired level

R
Float Tank -V Drain

b.
Desired level Potentiometer voltage in + Amplifiers Flow rate in + Integrate Actual level

Actuator and valve

Drain Flow rate out Displacement Potentiometer Float

voltage out

Copyright ? 2011 by John Wiley & Sons, Inc.

1-5 Solutions to Problems

9.

Desired force Transducer

+ Amplifier -

Current Valve

Displacement Actuator and load

Displacement Tire

Actual force

Load cell

10.

Commanded blood pressure + Vaporizer -

Isoflurane concentration Patient

Actual blood pressure

11.

Desired depth + -

Controller & motor

Force Grinder

Feed rate Integrator

Depth

12.

Desired position

Coil voltage + Transducer -

Coil circuit

Coil current

Solenoid coil & actuator

Force

Armature & spool dynamics

Depth

LVDT

Copyright ? 2011 by John Wiley & Sons, Inc.

1-6 Chapter 1: Introduction

13.

a.

Nervous system electrical impulses

Desired Light Intensity

Brain
+

Internal eye muscles

Retina’s Light Intensity

Retina + Optical

b.
Desired Light Intensity

Nervous system electrical impulses

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 + 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 R . The characteristic equation is LM + R = 0, from which M = - . Thus, the total R L

Copyright ? 2011 by John Wiley & Sons, Inc.

1-8 Chapter 1: Introduction

1 1 . Solving for the arbitrary constants, i(0) = A + = 0. Thus, A = R R 1 1 1 -(R/L)t 1 ?( R/ L) t ). = (1 ? e . The final solution is i(t) = -e R R R R
solution is i(t) = Ae-(R/L)t +

c.

18. a. Writing the loop equation,

di 1 + idt + vC (0) = v(t) dt C ∫ d 2i di + 2 + 25i = 0 b. Differentiating and substituting values, 2 dt dt Ri + L
Writing the characteristic equation and factoring,

M 2 + 2M + 25 = ( M + 1 + 24i)( M + 1 ? 24i) .
The general form of the solution and its derivative is

i = Ae?t cos( 24t ) + Be? t sin( 24t ) di = (? A + 24 B)e ?t cos( 24t ) ? ( 24 A + B)e? t sin( 24t ) dt di vL (0) 1 = =1 Using i (0) = 0; (0) = dt L L
i 0 = A =0

di (0) = ? A + 24 B =1 dt 1 Thus, A = 0 and B = . 24
The solution is

i=

1 ?t e sin( 24t ) 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,

10 35 From which, C = 53 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

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

M2 + 6 M + 8 = M + 4 M + 2
Thus, the total solution is

18 1 cos 3 t sin 3 t 65 65 18 =0. Solving for the arbitrary constants, x(0) = C + D ? 65 x =C e
-4t

+ De

-2t

+ -

Also, the derivative of the solution is

dx = - 3 cos 3 t + 54 sin 3 t - 4 C e - 4 t - 2 D e - 2 t dt 65 65
. Solving for the arbitrary constants, x(0) The final solution is

?

15 3 3 ? 4C ? 2D = 0 , or C = ? and D = . 26 65 10

x =-

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

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 + 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

Copyright ? 2011 by John Wiley & Sons, Inc.

1-11 Solutions to Problems

dx -4t dt = 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) =
20.

2 2 ?8 ? e ?4t ? sin(3t) + cos(3t )? ? 5 5 15

a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

1 1 From which, C = - 5 and D = - 10 . The characteristic polynomial is

Thus, the total solution is

1 Solving for the arbitrary constants, x(0) = A - 5 = 2. Therefore, A = solution is

11 . Also, the derivative of the 5

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

3 ? . The final solution 5

1 1 3 ? 11 x(t) = ? cos(2t) ? sin(2t) + e ?t ? cos(t) ? sin(t)? ? 5 10 5 5
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 = (? 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

Copyright ? 2011 by John Wiley & Sons, Inc.

1-13 Solutions to Problems

21.
Spring displacement Desired force Input voltage + Input transducer F up Controller Actuator Pantograph dynamics F out

Spring

-

Sensor

22.

Desired Amount of HIV viruses

RTI
Controller

Amount of HIV viruses

Patient

PI

Copyright ? 2011 by John Wiley & Sons, Inc.

1-14 Chapter 1: Introduction

23.

a.
Climbing & Rolling Resistances
Motive Force

Inverter Control Command Desired Speed
ECU +

Controlled Voltage

Inverter

Electric Motor

Actual

Vehicle
+

Aerodynamic
Aerodynamic

Speed

Copyright ? 2011 by John Wiley & Sons, Inc.

1-15 Solutions to Problems

b.
Climbing & Rolling Resistances
Desired

Speed _

Accelerator Displacement ECU

Motive

Actual Vehicle

Accelerator,
+

+

Aerodynamic
Aerodynamic

Speed

Copyright ? 2011 by John Wiley & Sons, Inc.

1-16 Chapter 1: Introduction

c.

Accelerator Accelerator Speed Error Desired ECU
+

ICE

Climbing & Rolling Resistances Actual
+

Power

Planetary Gear Control

Vehicle

Inverter Control Command

Inverter & Electric Motor

Total Motive Force Aerodynamic Motor Aerodynamic

Speed

Copyright ? 2011 by John Wiley & Sons, Inc.

1-17 Solutions to Problems

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Copyright ? 2011 by John Wiley & Sons, Inc.


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