Transforms and Partial Differential Equations
 9789352601677, 935260167X, 9789352601905, 9352601904

Table of contents :
Title
Contents
1 Partial Differential Equations
2 Fourier Series
3 Applications of Partial Differential Equations
4 Fourier Transforms
5 Z-Transforms and Difference Equations
Appendix A
Appendix B
Appendix C
Appendix D
Question Papers
Index

Citation preview

Transforms and Partial Differential Equations Third Edition

[For Semester III]

About the Author T Veerarajan is Dean (Retd), Department of Mathematics, Velammal College of Engineering and Technology, Viraganoor, Madurai, Tamil Nadu. A Gold Medalist from Madras University, he has had a brilliant academic career all through. He has 53 years of teaching experience at undergraduate and postgraduate levels in various established engineering colleges in Tamil Nadu including Anna University, Chennai.

Transforms and Partial Differential Equations Third Edition

[For Semester III]

T Veerarajan (Retd) Dean Department of Mathematics Velammal College of Engineering and Technology Viraganoor, Madurai Tamil Nadu

McGraw Hill Education (India) Private Limited New Delhi McGraw Hill Education Offices New Delhi New York St louis San Francisco Auckland Bogotá Caracas Kuala lumpur lisbon london Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016 Transforms and Partial Differential Equations (Third Edition) Copyright © 2016, 2014, 2013, 2012 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listing (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. Print Edition ISBN 13: 978-93-5260-167-7 ISBN 10: 93-5260-167-X E-Book Edition ISBN 13: 978-93-5260-190-5 ISBN 10: 93-5260-190-4 Managing Director: Kaushik Bellani Director—Products (Higher Education & Professional): Vibha Mahajan Manager—Product Development: Koyel Ghosh Specialist—Product Development: Sachin Kumar Head—Production (Higher Education & Professional): Satinder S Baveja Assistant Manager—Production: Anuj K Shriwastava Senior Production Executive: Jagriti Kundu Assistant General Manager—Product Management (Higher Education & Professional): Shalini Jha Product Manager—Product Management: Ritwick Dutta General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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Contents Preface Roadmap to the Syllabus

ix xi

1. Partial Differential Equations

1-1–1-108

1.1 Introduction 1-1 1.2 Formation of Partial Differential Equations 1-1 1.3 Elimination of Arbitrary Constants 1-2 1.4 Elimination of Arbitrary Functions 1-2 Worked Example 1(a) 1-3 Exercise 1(a) 1-19 1.5 Solutions of Partial Differential Equations 1-21 1.6 Procedure to Find General Solution 1-22 1.7 Procedure to Find Singular Solution 1-23 1.8 Complete Solutions of First Order Non-linear P.D.E.S 1-23 1.9 Equations Reducible to Standard Types—Transformation 1-26 Worked Example 1(b) 1-28 Exercise 1(b) 1-48 1.10 General Solutions of Partial Differential Equations 1-50 1.11 Lagrange’s Linear Equation 1-51 d x d y dz 1.12 Solution of the Simultaneous Equations = = 1-52 P Q R Worked Example 1(c) 1-53 Exercise 1(c) 1-68 1.13 Linear P.D.E.S. of Higher Order with Constant Coefficients 1-71 1.14 Complementary Function for a Non-Homogeneous Linear Equation 1-76 1.15 Solution of P.D.E.S. by the Method of Separation of Variables 1-76 Worked Example 1(d) 1-77 Exercise 1(d) 1-96 Answers 1-99

2. Fourier Series 2.1 2.2

Introduction 2-1 Dirichlet’s Conditions

2-1–2-96 2-2

Contents

vi

2.3 Euler’s Formulas 2-2 2.4 Definition of Fourier Series 2-5 2.5 Important Concepts 2-5 2.6 Fourier Series of Even and Odd Functions 2-8 2.7 Theorem 2-9 2.8 Convergence of Fourier Series at Specific Points 2-11 Worked Example 2(a) 2-12 Exercise 2(a) 2-39 2.9 Half-Range Fourier Series and Parseval’s Theorm 2-42 2.10 Root-Mean Square Value of a Function 2-45 Worked Example 2(b) 2-47 Exercise 2(b) 2-70 2.11 Harmonic Analysis 2-73 2.12 Complex Form of Fourier Series 2-75 Worked Example 2(c) 2-76 Exercise 2(c) 2-89 Answers 2-91

3A. Applications of Partial Differential Equations— Vibration of Strings 3.A1 3.A2 3.A3 3.A4

3-1–3-62

Introduction 3-1 Transverse Vibrations of a Stretched String—One Dimensional Wave Equation 3-4 Transmission Line Equations 3-6 Variable Separable Solutions of the Wave ∂2 y ∂2 y Equation 2 = a 2 2 3-8 ∂t ∂x

3.A5 Choice of Proper Solution 3-10 3.A6 Solution of a Damped Vibrating String Equation Worked Example 3(a) 3-12 Exercise 3(a) 3-57 Answers 3-61

3B. One Dimensional Heat Flow

3-10

3-63–3-115

3.B1 Introduction 3-63 3.B2 Equation of Variable Heat Flow in One Dimension 3-63 3.B3 Variable Separable Solutions of the Heat Equation 3-65 Worked Example 3(b) 3-67 Exercise 3(b) 3-109 Answers 3-113

3C. Steady State Heat Flow in Two Dimensions [Cartesian Coordinates] 3.C1

Introduction

3-116

3-116–3-167

Contents

vii

3.C2

Equation of Variable Heat Flow in Two Dimensions in Cartesian Coordinates 3-116 3.C3 Variable Separable Solutions of Laplace Equation 3-119 3.C4 Choice of Proper Solution 3-120 Worked Example 3(c) 3-121 Exercise 3(c) 3-162 Answers 3-165

4. Fourier Transforms 4.1 4.2 4.3 4.4 4.5

4-1–4-71

Introduction 4-1 Fourier Integral Theorem 4-1 Fourier Transforms 4-4 Alternative Form of Fourier Complex Integral Formula Relationship Between Fourier Transform and Laplace Transform 4-7 Worked Example 4(a) 4-8 Exercise 4(a) 4-23 4.6 Properties of Fourier Transforms 4-26 Worked Example 4(b) 4-34 Exercise 4(b) 4-53 4.7 Finite Fourier Transforms 4-55 Worked Example 4(c) 4-59 Exercise 4(c) 4-66 Answers 4-67

4-6

5. Z-Transforms and Difference Equations

5-1–5-50

5.1 Introduction 5-1 5.2 Properties of Z-Transforms 5-2 5.3 Z-Transforms of Some Basic Functions 5-7 Worked Example 5(a) 5-12 Exercise 5(a) 5-23 5.4 Inverse Z-Transforms 5-26 5.5 Use of Z-Transforms to Solve Finite Difference Equations Worked Example 5(b) 5-27 Exercise 5(b) 5-41

5-27

Appendix A

A.1–A.13

Appendix B

B.1–B.11

Appendix C

C.1–C.14

Appendix D

D.1–D.11

Solved Solved Solved Solved

Question Question Question Question

Papers-1 Papers-2 Papers-3 Papers-4

SQP1.1–SQP1.11 SQP2.1–SQP2.10 SQP3.1–SQP3.11 SQP4.1–SQP4.14

Contents

viii

Solved Solved Solved Solved Solved Solved Solved Index

Question Question Question Question Question Question Question

Papers-5 Papers-6 Papers-7 Papers-8 Papers-9 Papers-10 Papers-11

SQP5.1–SQP5.11 SQP6.1–SQP6.9 SQP7.1–SQP7.7 SQP8.1–SQP8.9 SQP9.1–SQP9.9 SQP10.1–SQP10.9 SQP11.1–SQP11.9 I-1–I-2

Preface Transforms and Partial Differential Equations, 3e, has been designed specifically to cater to the needs of third semester B Tech students of Anna University. The current edition aims at preparing the students for examination alongside strengthening the fundamental concepts related to Transforms and Partial Differential Equations. Lucidity of the text, ample worked examples and notes highlighted within the text help students navigate through complex topics seamlessly. Stepwise explanation, use of multiple methods of problem solving, and additional information presented by the means of appendices are few other notable features of the content. In addition, solved question papers of 2009–2015 and plentiful practice exercises are the highlighting feature of this edition.

Salient Features • Strict adherence to the latest AU syllabus • In-depth coverage of topics like Fourier Series, Fourier Transform, Complex Analysis and Solution of PDE • Addition of topics such as classification of PDE of the second order and inclusion of solved and unsolved examples • Stepwise solutions of solved problems which will enable students to score marks • Solved university questions papers from 2009 to 2015 (R-08, R-13) • Rich pedagogy:  284 examples within the chapter  611 short and long answer type questions

Chapter Organization The book is organised into 5 chapters. Chapter 1 deals with Partial Differential Equations. Chapter 2 explains in detail about Fourier Series. Chapter 3 discusses the Applications of Partial Differential Equations and has been divided into three major parts—Vibrations of Strings; One Dimensional Heat Flow; and Steady State Heat Flow in Two Dimension. Chapter 4 focuses on Fourier Transforms while Chapter 5 elaborates on Z-Transforms and Difference Equations.

x

Preface

Acknowledgements I am deeply grateful to all the students and teachers of India who have enthusiastically responded to the previous editions of Transforms and Partial Differential Equations (III Semester BE/B Tech courses). I hope that both the faculty and the students will receive the present edition as willingly as the earlier editions and my other books. A number of reviewers took pains to provide valuable feedback for the book. We are grateful to all of them and their names are mentioned as follows: A. Manoharan Ganadipathy Tulsi’s Jain Engineering College, Vellore, Tamil Nadu K. Thanalakshmi Kamaraj College of Engineering and Technology, Chennai, Tamil Nadu R. Ranganathan Dhirajlal Gandhi College of Technology, Salem, Tamil Nadu K. Basari Kodi Ramco Institute of Technology, Virudhunagar, Tamil Nadu

T Veerarajan Publisher’s Note Critical evaluation and suggestions for the improvement of the book will be highly appreciated and acknowledged. Hence, we look forward to receiving your feedback, comments and ideas to enhance the quality of this book. You can reach us at [email protected]. Kindly mention the title and the author’s name as the subject. In case you spot piracy of this book, please do let us know.

Roadmap to the Syllabus Transforms and Partial Differential Equations Regulation – 2013 Module I: PARTIAL DIFFERENTIAL EQUATIONS Formation of partial differential equations—Singular integrals—Solutions of standard types of first order partial differential equations—Lagrange’s linear equation—Linear partial differential equations of second and higher order with constant coefficients of both homogeneous and non-homogeneous types.

Go to

CHAPTER 1. PARTIAL DIFFERENTIAL EQUATIONS

Module II: FOURIER SERIES Dirichlet’s conditions—General Fourier series—Odd and even functions—Half range sine series—Half range cosine series—Complex form of Fourier series— Parseval’s identity—Harmonic analysis.

Go to

CHAPTER 2. FOURIER SERIES

Module III: APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS Classification of PDE—Method of separation of variables—Solutions of one dimensional wave equation—One dimensional equation of heat conduction— Steady state solution of two dimensional equation of heat conduction (excluding insulated edges).

CHAPTER 3. APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS CHAPTER 3A. VIBRATIONS OF STRINGS

Go to

CHAPTER 3B. ONE DIMENSIONAL HEAT FLOW CHAPTER 3C. STEADY STATE HEAT FLOW IN TWO DIMENSION

xii

Roadmap to the Syllabus

Module IV: FOURIER TRANSFORMS Statement of Fourier integral theorem—Fourier transform pair—Fourier sine and cosine transforms—Properties—Transforms of simple function—Convolution theorem—Parseval’s identity.

Go to

CHAPTER 4. FOURIER TRANSFORMS

Module V: Z-TRANSFORMS AND DIFFERENCE EQUATIONS Z-transforms—Elementary properties—Inverse Z-transform (using partial fraction and residues)—Convolution theorem—Formation of difference equations—Solution of difference equations using Z-transform.

Go to

CHAPTER 5. Z-TRANSFORMS AND DIFFERENCE EQUATIONS

Chapter

1

Partial Differential Equations 1.1

1.2

1-2

Transforms and Partial Differential Equations

1.3 1

3

1

1.4 1

w

Partial Differential Equations

1

1

Worked Examples 1(a)

1-3

1-4

Transforms and Partial Differential Equations

Partial Differential Equations

1-5

1-6

Transforms and Partial Differential Equations

with respect to y, we would have got

Partial Differential Equations

px

Example 5

1-7

1-8

Transforms and Partial Differential Equations

Partial Differential Equations

1-9

1-10

Transforms and Partial Differential Equations

Partial Differential Equations

1-11

1-12

Transforms and Partial Differential Equations

Partial Differential Equations

1-13

1-14

Transforms and Partial Differential Equations

Partial Differential Equations

1-15

1-16

Transforms and Partial Differential Equations

Partial Differential Equations

1-17

1-18

Transforms and Partial Differential Equations

Partial Differential Equations

Exercise 1(a) -

1-19

1-20

Transforms and Partial Differential Equations

f

Partial Differential Equations

1.5

1

1-21

Transforms and Partial Differential Equations

1-22

1

1.6

Partial Differential Equations

1-23

1.7

1

1.8

1-24

Transforms and Partial Differential Equations

Partial Differential Equations

1-25

1-26

1.9

Transforms and Partial Differential Equations

Partial Differential Equations

1-27

1-28

Transforms and Partial Differential Equations

Worked Examples

1(b)

Partial Differential Equations

1-29

1-30

Transforms and Partial Differential Equations

Partial Differential Equations

obtained from (3), satisfy Eq. 2.

1-31

1-32

Transforms and Partial Differential Equations

Partial Differential Equations

1-33

1-34

Transforms and Partial Differential Equations

Partial Differential Equations

1-35

1-36

Transforms and Partial Differential Equations

Partial Differential Equations

1-37

1-38

Transforms and Partial Differential Equations

Partial Differential Equations

1-39

1-40

Transforms and Partial Differential Equations

Partial Differential Equations

Solving (4), we get

1-41

1-42

Transforms and Partial Differential Equations

Partial Differential Equations

P

Q

1-43

1-44

Transforms and Partial Differential Equations

Partial Differential Equations

1-45

1-46

Transforms and Partial Differential Equations

Partial Differential Equations

1-47

Transforms and Partial Differential Equations

1-48

1 -

C

Partial Differential Equations

1-49

Transforms and Partial Differential Equations

1-50

1.10

1 which will be discussed in Section 1.12.

Partial Differential Equations

1.11

1-51

1-52

Transforms and Partial Differential Equations

1.12

R are, in general, functions

Partial Differential Equations

Worked Examples

1(c)

1-53

1-54

Transforms and Partial Differential Equations

Partial Differential Equations

1-55

1-56

Transforms and Partial Differential Equations

Partial Differential Equations

1-57

1-58

Transforms and Partial Differential Equations

Partial Differential Equations

1-59

(2)

(3)

1-60

Transforms and Partial Differential Equations

Partial Differential Equations

1-61

1-62

Transforms and Partial Differential Equations

Partial Differential Equations

n

1-63

1-64

Transforms and Partial Differential Equations

Partial Differential Equations

1-65

1-66

Transforms and Partial Differential Equations

Partial Differential Equations

1-67

Transforms and Partial Differential Equations

1-68

1

-

Partial Differential Equations

1-69

1-70

Transforms and Partial Differential Equations

Partial Differential Equations

1.13

1-71

1-72

Transforms and Partial Differential Equations

n

D¢z Dn–r D¢yr

Partial Differential Equations

1-73

1-74

Transforms and Partial Differential Equations

The Particular Integral of the solution of f (D, D¢)z = R(x, y).

Partial Differential Equations

1-75

1-76

1.14

1.15

Transforms and Partial Differential Equations

Partial Differential Equations

1-77

1

Worked Examples

1(d)

1-78

Transforms and Partial Differential Equations

Partial Differential Equations

1-79

1-80

Transforms and Partial Differential Equations

Partial Differential Equations

1-81

1-82

Transforms and Partial Differential Equations

Partial Differential Equations

1-83

1-84

Transforms and Partial Differential Equations

Partial Differential Equations

1-85

1-86

Transforms and Partial Differential Equations

(D2 – 5 DD¢ + 6 D¢2) z = y sin x

Partial Differential Equations

1-87

1-88

Transforms and Partial Differential Equations

RSx + 2 F a - 1 xI UV T H 2 KW

Partial Differential Equations

1-89

1-90

Transforms and Partial Differential Equations

- 2e z = f 1 ( y + x) + e–2x f 2 ( y + 2 x) +

1 2 1 1 1 1 x + x y - y - - cos (2 x + y) 2 2 4 4 2

Partial Differential Equations

1-91

1-92

Transforms and Partial Differential Equations

Partial Differential Equations

Y

1-93

1-94

Transforms and Partial Differential Equations

c

Partial Differential Equations

1-95

Transforms and Partial Differential Equations

1-96

C1

C2

C1

1

-

C2

Partial Differential Equations

1-97

1-98

Transforms and Partial Differential Equations

Partial Differential Equations

1 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

1-99

1-100

20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33.

34.

35.

36. 37.

Transforms and Partial Differential Equations

Partial Differential Equations

38. 39. 40. 41. 42. 43. 44. 45.

1 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

1-101

1-102

17.

18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

Transforms and Partial Differential Equations

Partial Differential Equations

35.

36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50.

1 1. 2.

1-103

1-104

3. 4. 5. 6. 7. 8. 9. 10. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

Transforms and Partial Differential Equations

Partial Differential Equations

23. 24. 25. 26. 27. 28. 29. 30.

31. 32.

33. 34. 35. 36. 37.

38.

1-105

Transforms and Partial Differential Equations

1-106

39. 40. 41.

42. 43. 44.

1 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Partial Differential Equations

13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

1-107

1-108

31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.

42.

43. 44. 45.

Transforms and Partial Differential Equations

Chapter

2

Fourier Series 2.1

2-2

2.2

2.3

Transforms and Partial Differential Equations

Fourier Series

2-3

2-4

Transforms and Partial Differential Equations

Fourier Series

2.4

2.5

2-5

2-6

Transforms and Partial Differential Equations

2.1, 2.2 and 2.3.

2

2

2

2.1

2.3 2 2.1 2.1

Fourier Series

2.3

2-7

Transforms and Partial Differential Equations

2-8

Fourier

2.6

2

2

2

Fourier Series

2

2-9

2

2

2

2

2

2

2.7 1.7

2-10

Transforms and Partial Differential Equations

Fourier Series

2.8

2-11

2-12

Transforms and Partial Differential Equations

Worked Examples

2(a)

Fourier Series

2-13

2-14

Transforms and Partial Differential Equations

Fourier Series

2-15

and

2-16

Transforms and Partial Differential Equations

Fourier Series

2-17

2-18

Transforms and Partial Differential Equations

Fourier Series

2-19

2-20

Transforms and Partial Differential Equations

Fourier Series

2-21

Since the function f(x) (1) is defined in a range of length 2, it can be expanded as a Fourier series of period 2.

2-22

Transforms and Partial Differential Equations

Fourier Series

2-23

Transforms and Partial Differential Equations

2-24

(2) -

4 2 (- p )

(1),

(2)

Fourier Series

p

ò

2-25

2-26

Transforms and Partial Differential Equations

Fourier Series

2-27

2-28

Transforms and Partial Differential Equations

Fourier Series

2-29

2-30

Transforms and Partial Differential Equations

Fourier Series

LMRS - cos (4n - 1)x + cos (4n - 1)x UVOP 4n - 1 WQ NT 4 n + 1

2-31

p 4

p 4

2-32

Transforms and Partial Differential Equations

Fourier Series

2-33

2-34

Transforms and Partial Differential Equations

Fourier Series

2-35

2-36

Transforms and Partial Differential Equations

8 p2

2

Fourier Series

2

2-37

2-38

Transforms and Partial Differential Equations

2

2

(3)

Fourier Series 2-39

2

-

2-40

Transforms and Partial Differential Equations

Fourier Series

2-41

2-42

2.9

Transforms and Partial Differential Equations

Fourier Series

2

2-43

2-44

Transforms and Partial Differential Equations

Fourier Series 2-45

2.10 Definition

Parseval's theorem

2-46

Transforms and Partial Differential Equations

Fourier Series

Worked Examples

2-47

2(b)

Transforms and Partial Differential Equations

2-48

R| S|T 2 p

p2 - 43 n n 2p , n

, if n is odd if n is even

Fourier Series

2-49

2-50

Transforms and Partial Differential Equations

Fourier Series

[Sum of the Fourier series of f (x)]x = 0 = f (0)

2-51

2-52

Transforms and Partial Differential Equations

Fourier Series

2-53

2-54

Transforms and Partial Differential Equations

Fourier Series

2-55

2-56

Transforms and Partial Differential Equations

Fourier Series

2-57

2-58

Transforms and Partial Differential Equations

Fourier Series

2-59

2-60

Transforms and Partial Differential Equations

Fourier Series

2-61

Transforms and Partial Differential Equations

2-62

ò

Fourier Series

2-63

2-64

Transforms and Partial Differential Equations

Fourier Series

2-65

2-66

Transforms and Partial Differential Equations

Fourier Series

2-67

2-68

Transforms and Partial Differential Equations

0

Fourier Series

2-69

Transforms and Partial Differential Equations

2-70

2

-

Fourier Series

2-71

2-72

Transforms and Partial Differential Equations

Fourier Series

2.11

2-73

2-74

Transforms and Partial Differential Equations

Fourier Series

2.12

2-75

2-76

Transforms and Partial Differential Equations

Worked Examples

2(c)

Fourier Series

2-77

2-78

Transforms and Partial Differential Equations

Fourier Series

2

2

2

a

2

2-79

2-80

Transforms and Partial Differential Equations

2

Fourier Series

2-81

2

Transforms and Partial Differential Equations

2-82

2

2

2

2

2

¥

å =

p

¥



2

p

=

2

Fourier Series

Fig. 2.17

2

2-83

Fig. 2.18

2-84

Transforms and Partial Differential Equations

2

2

,

2

,

2

Fourier Series

2-85

,

2

2

2-86

Transforms and Partial Differential Equations

, Fig. 2.22)

2

2

2

, Fig.2 2.23)

, Fig.222.24)

Fourier Series

2-87

2-88

Example 9

Transforms and Partial Differential Equations

Fourier Series

2 -

2-89

2-90

Transforms and Partial Differential Equations

p

4p / 3

5p / 3

2p

1.30 - 0.88 -0.25 1.98

Fourier Series

2 19. 20.

21.

22.

23.

24.

25.

26.

2-91

2-92

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

Transforms and Partial Differential Equations

Fourier Series

41. 42. 43. 44. 45. 46.

47.

48.

49. 50.

2 7. 8. 11.

12.

13.

2-93

2-94

14. 15. 16.

17.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

Transforms and Partial Differential Equations

Fourier Series

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

2 6.

7.

2-95

2-96

8.

9.

10.

11.

12. 13. 14. 15. 16.

17. 18.

19.

20.

Transforms and Partial Differential Equations

Chapter

3

Applications of Partial Differential Equations— Part

A

Vibration of Strings 3A.1

Transforms and Partial Differential Equations

3-2

Classification of Partial Differential Equations of the Second Order The most general form of linear partial differential equation of the second order in two independent variables is A

�2 u �x 2

�B

�2 u �2 u �u �u �C 2 �D �E � Fu � G �x �y �x �y �y

(1)

where A, B, C, D, E, F, and G are functions of x and y only. An equation of the from A

�2 u �x 2

�B

�2 u �2 u �u �u � C 2 � f ( x, y, u, , ) � 0 �x �y �x �y �y

(2)

in which the terms involving the second order partial derivatives alone are linear, is known as a quasi-linear P.D.E. of the second order. The equation (1) or (2) is said to be of 1. The elliptic type, if B2 – 4 AC < 0 2. The parabolic type, if B2 – 4 AC = 0 3. The hyperbolic type, if B2 – 4 AC > 0 For example, let us consider the following equations: (i)

�2 u �x Or

2

� 2x

�2 u �2 u �u � x2 2 � 2 �0 �x �y �y �y

uxx � 2 xuxy � x 2 uyy � 2uy � 0

Here A = 1, B = –2x, C = x2 B2 – 4 AC = (–2x)2 – 4x2 = 0 � �� The equation is parabolic at all points.

� (ii)

y 2 uxx � x 2 uyy � 0 Here A = y2, B = 0, C = x2



B2 – 4 AC = 0 – 4x2y2 < 0, for all x and y (� 0) � ��The equation is elliptic at all points. (iii) x2uxx – y2uyy = 0 Here A = x2, B = 0, C = –y2 B2 – 4 AC = 0 + 4x2y2 > 0, for all x and y (� 0)



� �� The equation is hyperbolic at all points. (iv) yxx – x uyy = 0 Here A = 1, B = 0, C = –x

Applications of Partial Differential Equations—Vibration of Strings

3-3

B2 – 4 AC = 0 + 4x � �� The equation is elliptic, parabolic and hyperbolic, according as x < 0, x = 0 and x > 0 respectively.



(v) uxx + 4uxy + (x2 + 4y2)uyy = sin(x + y) Here A = 1, B = 4, C = x2 + 4y2 B2 – 4 AC = 16 – 4 (x2 + 4y2) = 4(4 – x2 – 4y2) � �� The equation is of the elliptic type if 4 – x2 – 4y2 < 0



x2 + 4y2 > 4

i.e.,

x 2 y2 � �1 4 1

i.e.,

i.e., outside the ellipse

x 2 y2 � � 1. 4 1

The given P.D.E. is of the parabolic type on the ellipse the hyperbolic type inside the ellipse

x 2 y2 � � 1 and of 4 1

x 2 y2 � �1. 4 1

A liner or quasi-linear P.D.E. of the second order can be converted into simple forms, canonical forms or normal forms by effecting suitable transformations of independent variables, the discussion of which is beyond the scope of this book. The frequently occurring canonical forms of P.D.E’s are (i)

(ii)

(iii)

(iv)

�2 u �x 2 �2 u �x 2





�2 u �y 2 �2 u �y 2

� 0 (Laplace equation)

� � f ( x, y) (Poisson equation)

�u �2 u � � 2 2 (One dimensional heat flow equation) �t �x �2 u �t 2

� a2

�2 u �x 2

(One dimensional wave equation)

It can be easily verified that equations (i) and (ii) are of the elliptic type, equation (iii) is of the parabolic type and equation (iv) is of the hyperbolic type.

Transforms and Partial Differential Equations

3-4

3A.2 -

3

3A.1

Applications of Partial Differential Equations—Vibration of Strings

3-5

3-6

Transforms and Partial Differential Equations

3A.3

3A.2

3A.2)

3A.3

By Kirchhoff’s voltage law, the potential drop across the segment PQ 3A.3)

By Kirchhoff’s current law, loss of current while crosssing the segment PQ = loss

Applications of Partial Differential Equations—Vibration of Strings

3(B)

3-7

3-8

3A.4

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-9

3-10

3A.5

3A.6

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-11

3-12

Transforms and Partial Differential Equations

Worked Examples

3A.4

3A.4)

3A

Applications of Partial Differential Equations—Vibration of Strings

3-13

3-14

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

Example 2

3-15

3-16

Transforms and Partial Differential Equations

3A.5

3

Applications of Partial Differential Equations—Vibration of Strings

3-17

Transforms and Partial Differential Equations

3-18

-

Applications of Partial Differential Equations—Vibration of Strings

3-19

Transforms and Partial Differential Equations

3-20

r2 Pa

r

3A.6

3A.6).

r

r

Applications of Partial Differential Equations—Vibration of Strings

3-21

3-22

Transforms and Partial Differential Equations

n2

3A.7).

Applications of Partial Differential Equations—Vibration of Strings

3A.7

3-23

3-24

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-25

3-26

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-27

3-28

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

which is Fourier half-range sine series of k(2lx – x2) in (0, 2l),

3-29

3-30

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-31

3-32

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-33

3-34

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-35

3-36

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-37

3-38

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-39

3-40

Transforms and Partial Differential Equations

3A.6]

Applications of Partial Differential Equations—Vibration of Strings

3-41

3-42

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

PROBLEMS ON TRANSMISSION LINE EQUATION

3A.3]

3-43

3-44

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-45

Transforms and Partial Differential Equations

3-46

(1)

(2) (2). (2) (2) (2) (2)

(3)

(2)

(3)

(3),

Applications of Partial Differential Equations—Vibration of Strings

3-47

3-48

Transforms and Partial Differential Equations

Equation (8) is readily solved, as the end conditions, (13) and (14) for e2(x, t) contain only zero values

Applications of Partial Differential Equations—Vibration of Strings

3-49

3-50

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

e

3-51

3-52

Transforms and Partial Differential Equations

3

Applications of Partial Differential Equations—Vibration of Strings

PROBLEMS ON VIBRATION OF MEMBRANES

is

3-53

3-54

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Vibration of Strings

3-55

Transforms and Partial Differential Equations

3-56

a

Applications of Partial Differential Equations—Vibration of Strings

3-57

Exercise 3(a) Part A (Short-Answer Questions) 1. When do you say that a quasi-linear P.D.E. is (i) elliptic, (ii) parabolic, (iii) hyperbolic? 2. Classify the following P.D.E.s as to whether they are elliptic, parabolic or hyperbolic. (i) uxx + 2uxy + uyy = 0 (ii) uxy – ux = 0 (iii) x2uxx + (1 – y2) uyy = 0; – � < x < �; – 1< y 0; y >0. (v) fxx – 2fxy = 0

3-58

3. 4. 5. 6. 7. 8. 9.

Transforms and Partial Differential Equations

(vi) fxx + 2fxy + 4fyy = 0 (vii) (x + 1)fxx + 2(x + 2) fxy + (x + 3)fyy = 0 State the assumptions made while deriving the equation of vibration of a string in the simple form. 2 �2 y 2 � y ? What does a2 represent in the equation 2 � a �t �x 2 Write down the two transmission line equations. Write down the telephone equations for potential and current. Write down the radio equations for potential and current. Write down the telegraph equations for potential and current. When do the telephone equations reduces to the form of one dimensional wave equation?

10. When do the telephone equations reduce to the form of one dimensional heat flow equation? 11. Write down the three mathematically possible solutions of one dimensional wave equation. 12. Write down the appropriate solution of the vibration of string equation. How is it chosen? 13. Write down the differential equation that represents damped vibration of a string. 14. Write down the suitable variable separable solution of the equation that represents damped vibration of a string. 15. Write down the differential equation that represents the vibration of a membrane. 16. Write down the appropriate solution of the equation representing the vibration of a membrane. 17. Write down the form of the general solution of the vibration of string equation, if the string fixed at the ends, is given a non zero initial displacement and zero initial velocity. 18. Write down the form of the general solution of the vibration of string equation, if the string, fixed at the ends, is given a non zero initial velocity from its equilibrium position. 19. Write down the form of the general solution of the vibration of string equation, if the string is fixed at its ends. Part B 20. A string is stretched and fastened to two points l apart. Motion is started by 2� x 3� x displacing the string into the form of the curve (a) y � 2 sin � 3sin l l 3� x 2� x cos and then releasing it from this position at time and (b) y � 2 sin l l t = 0. Find the displacement function y(x, t).

Applications of Partial Differential Equations—Vibration of Strings

3-59

21. The ends of a uniform string of length 2l are fixed. The initial displacement is y(x, 0) = kx(2l – x), 0 < x < 2l, while the initial velocity is zero. Find the displacement at any distance x from the end x = 0 at any time t. 22. A tightly stretched string of length � is fastened at both ends. The mid-point of the string is displaced by a distance d transversely and the string is released from rest in this position. Find the displacement of any point of the string at any subsequent time. 23. Solve Problem 22, if length of the string is 50 cm and d = 1 cm. 24. A tightly stretched string of length l has its ends x = 0 and x = l fixed. The point x = l/3 is drawn aside by a small distance h and released from rest at time t = 0. Find y(x, t) at any subsequent time t. 25. An elastic string is stretched between two points at a distance l. One end is 2l taken as the origin and at a distance from this end, the string is displaced 3 a distance d transversely and is released from rest when it is in this position. Find the displacement function y(x, t). 26. The points of trisection of a string of length � are pulled aside through a distance b on opposite sides of the position of equilibrium and the string is released from rest. Find the subsequent displacement of the string. 27. A tightly stretched string with fixed end x = 0 and x = � is initially at rest in its equilibrium position. If it is set in motion by giving each point a velocity v = 16 sin5 x, find the displacement of any point of the string at any time t. 28. A taut string of length 50 cm fastened at both ends, is disturbed from its position of equilibrium by imparting to each of its points an initial velocity of magnitude kx for 0 < x < 50. Find the displacement function y(x, t). 29. A string is stretched between two fixed points at a distance of l cm and the points of the string are given initial velocity v = �(lx – x2), for 0 < x < l. Find the displacement function y(x, t). 30. A string is stretched between two fixed points at a distance of 2l cm and the points of the string are given initial velocities v, where v=

kx in 0 � x � l l

k (2l � x ) in l � x � 2l l x being the distance from an end point. Find the displacement of the string at any time. 31. A string is stretched and fastened to two point l apart. Motion is started by n� and also by displacing the string into the form of the curve y � y0 sin l imparting a constant velocity v0 to every point of the string in this position at time t = 0. Determine the displacement function y(x, t). 32. A string is stretched between two fixed points at a distance of � cm and the points of the string are given initial velocities v, where =

Transforms and Partial Differential Equations

3-60

v = x, in 0 � x �

� 2

� � x�� 2 after displacing it to the position y = x(� – x) at t = 0. Find the displacement of the string at any time. 33. A tightly stretched string of length l is fastened at both ends. The mid-point of the string is taken to a distance h and the various points of the string are given (different) velocities v, where v = kx(l – x), in this position at t = 0. Find the subsequent displacement of the string. = � � x, in

�2 u �2 u 34. Solve the wave equation 2 � a 2 2 , if u(0, t) = 0, u(l, t) = A sin �t and �t �x u(x, 0) = 0. 35. The differential equation of a vibrating string that is viscously damped is �2 y �t

2

= a2

�2 y �x

2

� 2b

�y �t

If the string is fastened at both ends and given zero initial velocity, find the displacement function y(x, t) when the length of the string is l and initial � �x� displacement is sin 3 � � . � l � 36. Solve the viscously damped vibrating string problem, if the mid-point of the string of length l is taken to a distance h and the string is released from rest from this position at time t = 0. Assume that the ends of the string are fisec. 37. Neglecting the resistance R and leakage G, find the e.m.f. v(x, t) in a line a km long t seconds after the ends were suddenly grounded, if initially �x 7� x �v � E7 sin i(x, 0) = i0 so that (x, 0) = 0 and v(x, 0) = v( x, 0) � E1 sin . a a �t 38. Neglecting R and G, find e.m.f. e(x, t) in a line of length 1 km, t seconds after � �e � � 0 and the ends were suddenly grounded, if initially i(x0) = i0 and � � � �t � t � 0 e(x, 0) = 2E sin 3�x(1 + cos �x). 0

39. Show that the solution of the equation

�2 e

� LC

�2 e

appropriate to the case �x 2 �t 2 when a periodic e.m.f. v0 cos pt is applied at the end x = 0 of a line is given by e � v0 cos( pt � p LC x ) .

[Hint: Assume the solution in the form e(x, t) = k cos (ax + bt) and find the values of k, a, b, such that the assumed solution satisfies the given equation and the given boundary condition.] 40. A tightly stretched unit square membrane with sides x = 0, x = 1, y = 0 and y = 1 starts vibrating from rest and its initial displacement is (a) k sin 2�x sin �y, (b) k sin �x sin 2�y. Find the deflection of any point (x, y) of the membrane at any time t.

Applications of Partial Differential Equations—Vibration of Strings

3-61

41. Find the deflection z(x, y, t) of a rectangular membrane (0 < x < 2, 0 < y < 1) whose boundary is fixed, given that it starts from rest and z(x, y, 0) = �xy(2 – x)(1 – y). 42. Find the deflection z(x, y, t) of a rectangular membrane (0 < x < a, 0 < y < b) whose boundary is fixed, given that its starts from rest and z(x, y, 0) = xy(a2 – x2)(b2 – y2).

Answers Exercise 3(a) 2. (i) Parabolic (v) Hyperbolic

(ii) Hyperbolic (iii) Elliptic (iv) Elliptic (vi) Elliptic (vii) Hyperbolic 2 n� 2� at 3� x 3� at 20. (a) y( x, t ) � 2 sin cos � 3sin � cos l l l l n� � at 5� x 5� at cos � sin cos . l l l l

(b) y( x, t ) � sin 21.

22.

23.

24.

25.

26.

y( x, t ) � y( x, t ) � y( x, t ) � y( x, t ) � y( x, t ) � y( x, t ) �



32 kl 2 �

� (2n � 1)2 sin

2

n �1 �

n� n� x n� at . sin cos 3 l l

1

2 n� n� x n� at . sin cos 3 l l

1

2 n� sin 2 nx cos 2 nat . 3

n �1 �

� n2 sin

2

n �1 �

9b

� n2 sin

2

(2 n � 1)� x (2 n � 1)� at . � cos 60 60

1

� n2 sin

2

9d



(�1)n �1



9h



(�1)n �1

n �1

8



(2 n � 1)� x (2 n � 1)� at . cos 2l 2l

� (2n � 1)2 sin(2n � 1) x � cos(2n � 1) at

�2



n �1



8d

1

� (2n � 1)3 sin

3

n �1

27.

y( x, t ) �

10 5 1 sin x sin at � sin 3 x sin 3 at � sin 5 x sin 5 at . a 3a a

28.

y( x, t ) �

5000 k

29.

30.

y( x, t ) � y( x, t ) �

2

� a 8� l �

4



1

� n2 (�1)n�1 sin n �1

3 �

1

� (2n � 1)4 sin a

16 kl

n �1 �

(�1)n �1

� (2n � 1)3 sin � 3a n �1

n� x n� at . cos 50 50

(2 n � 1)� x (2 n � 1)� at . � sin 2l l (2 n � 1)� x (2 n � 1)� at sin 2l 2l

Transforms and Partial Differential Equations

3-62

31.

y( x, t ) � y0 sin

32.

y( x, t ) � �

33.

8 �

�x � at 4 v0 l � 1 n� x n� at � 2 � 2 sin cos sin l l l l � a n �1 n



1

� (2n � 1)3 sin(2n � 1) x � cos(2n � 1) at n �1

4 (�1)n �1 sin(2 n � 1) x � sin(2 n � 1) at . � � n �1 (2 n � 1)2 �

y( x, t ) �

8h



(�1)n �1

� (2n � 1)2 sin �2 n �1

8kl � 34. 35.

4

3 �

(2 n � 1)� x (2 n � 1)� at cos l l

(2 n � 1)� x (2 n � 1)� at . � sin l l

1

� (2n � 1)4 sin a n �1

�l �x � sin � sin � t . a a 3 �x b 1 3� x y( x, t ) � e � bt sin (cos c1t � sin c1t ) � e � bt sin (cos c3 t � 4 l c1 4 l

u( x, t ) � A cosec

b � 2 a2 sin c3 t ), where c1 � � b2 and c3 � c3 l2 36.

y( x, t ) �

8h �

38.

1

n �1 n

2

sin

l2

� b2 .

n� n� x b sin � (cos cn t � sin cn t ) , 2 l cn

n2� 2 a 2

� b2 . l2 �x �t 7� x 7� t . v( x, t ) � E1 sin cos � E7 sin cos a a a LC a LC

where cn = 37.

2



e � bt �

9� 2 a 2

� 2� t 3� t 4� t � � 2 sin 3� x cos � sin 4� x cos e( x, t ) � E0 �sin 2� x cos �. LC LC LC � �

40. (a) z(x, y, t) = k sin 2� x sin � y cos 5� at ; (b) z(x, y, t) = k sin � x sin 2� y cos 5� at . 41.

z( x, y, t ) �

256 � �

6



z( x, y, t ) �

144 a 3 b3 �



1

m �1 n �1 �

sin(2 n � 1)� y � cos

42.



� � �� (2m � 1)3 (2n � 1)3 sin

6

(2 m � 1)� x 2

�� (2 m � 1)2 � (2 n � 1)2 � at � . 4 �� �



�� 1

� � � m3 n3 sin

m �1 n �1 � �

�� m� x n� y m 2 n2 sin cos � 2 � ct � . 2 a b a b ��

Part

B

One-Dimensional Heat Flow

3B.1

3B.2

3B.1

r,

3-64

Transforms and Partial Differential Equations

3

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3B.3

3-65

3-66

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

Worked Examples

3B

3B.2

3B.2)

3-67

3-68

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-69

3-70

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-71

Transforms and Partial Differential Equations

3-72

3

A

3B.3

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-73

3-74

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-75

3-76

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-77

3-78

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-79

3-80

Transforms and Partial Differential Equations

3B.4]

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3B.4

3-81

3-82

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-83

B at 100°C until steady state conditions prevail. At time t = 0, the end B is suddenly

When steady state conditions prevail in the rod, the temperature distribution is

3-84

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-85

3-86

Transforms and Partial Differential Equations

3B.5

3B.5

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-87

3-88

Transforms and Partial Differential Equations

3B.6)

3B.6

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-89

3-90

Transforms and Partial Differential Equations

3(A).

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-91

3-92

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-93

3-94

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-95

3-96

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-97

3-98

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-99

3-100

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-101

3-102

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3(A)

3-103

3-104

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-105

3-106

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-107

3-108

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

Exercise3B -

Part B

find the temperature distribution in the bar

3-109

3-110

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3-111

3-112

Transforms and Partial Differential Equations

x = 0, while the end x = l is maintained at 0°C temperature. Find the temperature in the bar at a subsequent time.

Applications of Partial Differential Equations—One-Dimentional Heat Flow

3B 10.

11.

12.

13. 14.

15.

17.

18.

19.

20.

3-113

3-114

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—One-Dimentional Heat Flow

33.

35.

3-115

Part

C

Steady State Heat Flow in Two Dimensions [Cartesian Coordinates]

3C.1

6

3C.2

3C.1.

Applications of Partial Differential Equations—Steady state Heat Flow ...

Fig. 3C.1

3(B)

3-117

3-118

Transforms and Partial Differential Equations

Deduction

3(B).

Applications of Partial Differential Equations—Steady state Heat Flow ...

3C.3

3-119

3-120

Transforms and Partial Differential Equations

3C.4 a

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-121

according as non-zero boundary value is prescribed on the side x = k or y = k.

Worked Examples

3C

PROBLEMS ON TEMPERTURE DISTRIBUTION IN VERY LONG PLATES

3C.2)

3C.2

3-122

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3C.3)

3-123

3-124

Transforms and Partial Differential Equations

3C.3

(6¢)

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-125

3-126

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

in

3-127

3-128

Transforms and Partial Differential Equations

3C.4).

Applications of Partial Differential Equations—Steady state Heat Flow ...

3C.4

3-129

3-130

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3C.5).

3C.5

3-131

3-132

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-133

PROBLEMS ON TEMPERTURE DISTRIBUTION IN FINITE PLATES

3C.6).

3-134

Transforms and Partial Differential Equations

3C.6

Applications of Partial Differential Equations—Steady state Heat Flow ...

(1)

3-135

3-136

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3C.7).

3C.7

3-137

3-138

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-139

3C.8).

3C.8

3C.9).

3C.9

3-140

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

nxY 2a

3-141

Transforms and Partial Differential Equations

3-142

3

3

Fig. 3C.10

Fig. 3C.11

Fig. 3C.12

3

Applications of Partial Differential Equations—Steady state Heat Flow ...

100,

3-143

3-144

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-145

3-146

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-147

3-148

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-149

3-150

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-151

3-152

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-153

3-154

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-155

3-156

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-157

3-158

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-159

3-160

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-161

Transforms and Partial Differential Equations

3-162

3C -

Applications of Partial Differential Equations—Steady state Heat Flow ...

3-163

3-164

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

3(A)]

Exercise 3(C) 6.

3-165

3-166

7.

8.

9.

10.

11. 12.

13.

14.

15.

Transforms and Partial Differential Equations

Applications of Partial Differential Equations—Steady state Heat Flow ...

16.

17.

18. 19.

20.

21.

22.

23.

24.

25.

26.

3-167

Chapter

4

Fourier Transforms 4.1

Section 4.2.

4.2

4-2

Transforms and Partial Differential Equations

Fourier Transforms

–¥

4-3

–¥

4-4

4.3

Transforms and Partial Differential Equations

Fourier Transforms

4-5

4-6

4.4

Transforms and Partial Differential Equations

Fourier Transforms

Equation (1) can be re-written as

4.5

4-7

4-8

Transforms and Partial Differential Equations

Worked Examples

4(a)

Fourier Transforms

4-9

4-10

Transforms and Partial Differential Equations

Fourier Transforms

Example 5

x

4-11

4-12

Transforms and Partial Differential Equations

Fourier Transforms

4-13

4-14

Transforms and Partial Differential Equations

Fourier Transforms

4-15

4-16

Transforms and Partial Differential Equations

Fourier Transforms

4-17

4-18

Transforms and Partial Differential Equations

Fourier Transforms

4-19

4-20

Transforms and Partial Differential Equations

Fourier Transforms

4-21

(n ) s

n

cos

nπ 2

i sin

nπ 2

Transforms and Partial Differential Equations

4-22

s

0

Fourier Transforms

s

4

4-23

4-24

Transforms and Partial Differential Equations

Fourier Transforms

R|p sin x , S|02, T

4-25

when 0 £ x £ p when x > p

Transforms and Partial Differential Equations

4-26

4.6 1. Linearity property ,

Fourier Transforms

2. Change of scale property

x 3. Shifting property (Shifting in x)

s

4-27

4-28

Transforms and Partial Differential Equations

5. Modulation theorem

6. Conjugate symmetry property

Fourier Transforms

5. Transform of derivatives

4-29

4-30

Transforms and Partial Differential Equations

s

8. Derivatives of the transform

Fourier Transforms

9. Convolution theorem

4-31

4-32

Transforms and Partial Differential Equations

10. Parseval's identity (or) energy theorem

Fourier Transforms

4-33

4-34

Transforms and Partial Differential Equations

Worked Examples

4(b)

Fourier Transforms

4-35

4-36

Transforms and Partial Differential Equations

Fourier Transforms

4-37

4-38

Transforms and Partial Differential Equations

Fourier Transforms

4-39

4-40

Transforms and Partial Differential Equations

Fourier Transforms

4(a),

4-41

4-42

Transforms and Partial Differential Equations

Fourier Transforms

4-43

4-44

Transforms and Partial Differential Equations

Fourier Transforms

4-45

4-46

Transforms and Partial Differential Equations

Fourier Transforms

4-47

4-48

Transforms and Partial Differential Equations

(5)

6

(7)

Fourier Transforms

4-49

4-50

Transforms and Partial Differential Equations

Fourier Transforms

4-51

4-52

Transforms and Partial Differential Equations

Fourier Transforms

4-53

Exercise 4(b) Part A (Short-Answer Questions)

4

4-54

Transforms and Partial Differential Equations

4(a).]

4(a).]

Fourier Transforms

=a2

4.7

¶2u ¶x2

4-55

4-56

Transforms and Partial Differential Equations

Inversion formulas

Fourier Transforms

4-57

(2)

4-58

Transforms and Partial Differential Equations

Finite Fourier transforms of derivatives

Fourier Transforms

Worked Examples Example 1

Example 2

4-59

4(c)

4-60

Transforms and Partial Differential Equations

in

Fourier Transforms

4-61

4-62

Transforms and Partial Differential Equations

Example 6

Example 7

1

Example 8

Fourier Transforms

4-63

4-64

Transforms and Partial Differential Equations

Fourier Transforms

4-65

4-66

Transforms and Partial Differential Equations

4

Fourier Transforms

4

4-67

4-68

Transforms and Partial Differential Equations

Fourier Transforms

4

4-69

Transforms and Partial Differential Equations

4-70

4

Fourier Transforms

4-71

Chapter

5

Z-Transforms and Difference Equations 5.1

IntroductIon

In Communication Engineering, two basic types of signals (that are represented mathematically as functions of one or more independent variables) are encountered. They are continuous time signals and discrete time signals. Continuous time signals are defined for a continuum of values of the independent variable, namely time and are denoted by a function {f(t)}. On the other hand, discrete time signals are defined only at discrete set of values of the independent variable and are denoted by a sequence {f(n)}. A discrete time signal {f(n)} may represent a phenomenon for which the independent variable is inherently discrete or successive samples of the underlying phenomenon (for which the independent variable is continuous) at the sampling instants 0, T, 2T, .... Here T is called the sampling period. Laplace transform and Fourier transform play important roles in the study of continuous time signals. Z-transform which is the discrete time counterpart of the Laplace transform plays an important role in discrete time signal analysis. Signal Analysis is an engineering discipline of broad scope. It is used not only in communication technology, but also in the fields of astronomy, oceanography, crystallography, bio-engineering, antenna design, system theory, computer sciences and in many other fields.

Definition of the Z-transform •

If {f(n)} is a sequence defined for n = 0, ±1, ±2, ±3, ..., then

Â

f(n)z–n is

n =- •

called the two-sided or bilateral Z-transform of {f(n)} and denoted by Z{f(n)} or f (z), where z is a complex variable in general. If {f(n)} is a causal sequence, i.e. if f(n) = 0 for n < 0, then the Z-transform is called one-sided or unilateral Z-transform of {f(n)} and is defined as •

Z {f(n)} = f (z) =

Â

n=0

f(n)z–n

We shall mostly deal with one sided Z-transform which will be hereafter referred to as Z-transform.

Transforms and Partial Differential Equations

5-2

note



The series

Â

f(n)z-n will be convergent only for certain values of z i.e. only

n=0

in a certain region of the z-plane (called the region of convergence of the z-transform) depending on the sequence { f(n)}. The inverse Z-transform of f (z) = Z{f(n)} is defined as Z–1 { f (z)} = {f(n)}. If the function f(t) is defined by means of a sequence of its sampled values at 0, T, 2T, ..., then the Z-transform of f(t) is defined as Z{ f(t)} =



Â

f(nT)z–n

n=0

We shall denote both Z{f(n)} and Z{f(t)} by f (z).

5.2 1.

ProPertIes of Z-transforms Linearity

The Z-transform is linear i.e. Z{af(n) + bg(n)} = aZ{f(n)} + bZ{g(n)}.

Proof •

Z{af(n) + bg(n)} =

Â

{af(n) + bg(n)}z–n

n=0 •

=a

Â

f(n)z–n + b

n=0

Similarly,



Â

g(n)z–n

n=0

= aZ{f(n)} + bZ{g(n)} Z{af(t) + bg(t)} = aZ{f(t)} + bZ{g(t)}

2. time shifting (i) Z{f(n – n0)} = z–n0 ◊ Z{f(n)} or z–n0 f (z) (ii) Z{f(t + T)} = z{ f (z) – f(0)}

Proof •

(i)

Z{f(n – n0)} =

Â

f(n – n0)z–n

n=0 •

=

Â

f(m) z–(m + n0)

m = - n0

= z–n0



Â

f(m) z–m {∵ f(n) is causal}

m=0

= z–n0 f (z) if n ≥ n0

Z-Transforms and Difference Equations •

(ii)

Z{f(t + T)} =

Â

5-3

f (nT + T)z–n [∵ Z{f(t) =

n=0

 f (nT)z–n]

n=0



=



 f {(n + 1)T}z–n

n=0



=

 f (mT)z–(m–1)

m =1

È • ˘ = z Í Â f (mT )z - m - f (0)˙ ÍÎ m = 0 ˙˚ = z{ f (z) – f (0)} Extending this result, we get È f (T ) f (2T ) f {(k - 1)T } ˘ Z{f(t + kT)} = zk Í f ( z ) - f (0) - 2 -º˙ z z zk - 1 ˚ Î

3.

frequency shifting Ê zˆ (i) Z{an f(n)} = f Á ˜ Ë a¯

(ii) Z{an f(t)} = f (z/a)

Proof (i)

Z{an f(n)} =



 a n f(n)z–n

n= 0 •

=

 f (n) ÊÁË z ˆ˜¯ a

n= 0

-n

= f (z/a)

Similarly (ii) can be proved.

Corollary If Z{f(t)} = f (z), then Z{e–at f(t)} = f (zeaT) This result follows, if we replace an by e–anT or (e–aT)n, i.e. ‘a’ by e–aT in (ii).

4. time reversal for bilateral Z-transform If Z{f(n)} = f (z), then Z{f(– n)} = f

Proof

() 1 z



Z{f(– n)} =

Â

f (–n)z–n

n=-• -•

=

Â

m=•

f (m)zm

Transforms and Partial Differential Equations

5-4



Â

=

m=-•

Ê 1ˆ f (m) Á ˜ Ë z¯

-m

Ê 1ˆ = f Á ˜ Ë z¯

5.

differentiation in the Z-domain (i) Z{nf(n)} = – z

d f (z) dz

(ii) Z{nf(t)} = – z

d f (z) dz

Proof (i)

f (z) = Z{f(n) =



 f (n)z

–n

n=0

d f (z) = dz

\



 - n f(n)z–n–1

n=0

= \

1 • Â {n f (n)}z - n z n=0 d f (z) dz

Z{nf(n)} = – z

Similarly, (ii) can be proved.

6.

Initial value theorem (i) If Z{f(n)} = f (z), then f(0) = lim f ( z ) . zƕ

(ii) If Z{f(t)} = f (z), then f(0) = lim f ( z ) . zƕ

Proof •

(i)

f (z) =

 f (n)z–n

n=0

= f(0) + \

lim ÈÎ f ( z )˘˚ = f(0)

f (1) f (2) + 2 +º• z z

zƕ

Similarly, (ii) can be proved.

7.

final value theorem (i) If Z{f(n)} = f (z), then lim [f(n)] = lim {(z – 1) f (z)}. nÆ•

z Æ1

(ii) If Z{f(t)} = f (z), then lim [f(t)] = lim {(z – 1) f (z)}. t Æ•

z Æ1

Z-Transforms and Difference Equations

5-5

Proof •

(i)

 f (n + 1)z–n

Z{f(n + 1) =

n=0 •

 f (m)z–m + 1 = z [ f (z) – f(0)]

=

m =1

\

z f (z) – zf(0) – f (z) = Z{f(n + 1)} – Z{f(n)} •

i.e.

Â

(z – 1) f (z) – zf(0) =

{f(n + 1) – f(n)} z–n

n=0

Taking limits as z Æ 1,



Â

lim [(z – 1) f (z)] – f(0) = z Æ1

{f(n + 1) – f(n)}

n=0

= lim {f(1) – f(0)} + {f(2) – f(1)} nÆ•

+ {f(3) – f(2)} +  + {f(n + 1) – f(n)}] = lim [f(n + 1) – f(0)] nÆ•

= lim [f(n) – f(0)] nÆ•

\

lim [f(n)] = lim [(z – 1) f (z)] nÆ•

z Æ1

Similarly, (ii) can be proved, starting with Property 2(ii).

8. convolution theorem Definitions The convolution of the two sequences {f(n)} and {g(n)} is defined as (i) {f(n)* g(n)} •

Â

=

f (r) · g(n – r) if the sequences are non-causal and

r =-• n

(ii) {f(n)*gn)} =

 f (r) g(n – r), if the sequences are causal.

r =0

The convolution of two functions f(t) and g(t) is defined as f(t)*g(t) = n

 f (rT) · g(n - r)T, where T is the sampling period.

r =0

Statement of the theorem (i) If Z{f(n)} = f (z) and Z{g(n)} = g (z), then Z{f(n)*g(n)} = f (z) ◊ g (z). (ii) If Z{f(t)} = f (z) and Z{g(t)} = g (z), then Z{f(t)*g(t)} = f (z) ◊ g (z).

5-6

Transforms and Partial Differential Equations

Proof (for the bilateral Z-transform) È • ˘ (i) Z{f(n)*g(n)} = Z Í Â f (r ) ◊ g(n - r )˙ ÍÎr = - • ˙˚ •

È



 ÍÍ Â

=

n = - • Îr = - • •

Â

=

r =-•

˘ f (r )g(n - r )˙ z - n ˙˚

È • ˘ f (r ) Í Â g(n - r ) z - n ˙ , ÍÎ n = - • ˙˚

by changing the order of summation. •

Â

=

r =-• •

Â

=

r =-• •

Â

=

È • ˘ f (r ) Í Â g(m) z - ( m + r ) ˙ , by putting n – r = m ÍÎ m = - • ˙˚ È • ˘ f (r )z - r Í Â g(m) z - m ˙ ÍÎ m = - • ˙˚ f (r )z - r ◊ g ( z )

r =-•

= g (z) ◊



Â

f (r )z - r = f ( z ) ◊ g ( z )

r =-•

(ii)

È• ˘È • ˘ f ( z ) ◊ g ( z ) = Í Â f (rT )z - r ˙ Í Â g(sT )z - s ˙ ÍÎr = 0 ˙˚ ÍÎ s = 0 ˙˚ •

=

 h(nT )z - n

(1)

n=0

say, where h (nT) = f(0 ◊ T) g(nT) + f(1T) ◊ g{(n – 1)T} + f(2T) ◊ g{(n – 2)T} +  + f(nT) ◊ g(0 ◊ T) n

=

 f (rT ) ·g{(n – r)T}

r =0

Using (2) in (1), we get •

f (z) ◊ g (z) =

n=0

=

È

n

˘

ÎÍr = 0

˚˙

Â Í Â f (rT ) g {(n - r )T}˙ z - n •

 [ f (t )* g(t )] z - n

n=0

= Z [f(t)*g(t)]

(2)

Z-Transforms and Difference Equations

5-7

note Proof of (i) in the unilateral Z-transform case can be given as in the proof of (ii).

5.3

Z-transforms of some BasIc functIons

1. Z{d(n)}, where d(n) is the unit impulse sequence defined by Ï1, for n=0 d(n) = Ì Ó0, for n π 0 •

Z{d(n)} =

 d (n)z–n = 1

n=0

2. Z(k) and Z{U(n)}, where k is a constant and U(n) is the unit step sequence defined by Ï1, for n ≥ 0, i.e. for n = 0, 1, 2, º U(n) = Ì Ó0, for n < 0 •

(i)

Z(k) =

Ê

1

1

ˆ

 kz - n = k ÁË1 + z + z2 +º•˜¯

n=0

= k◊

1 1 1z

=

kz z -1

1 where the region of convergence (ROC) is < 1 or |z| > 1 z (ii) In particular, z Z{U(n)} = Z(1) = , if |z| > 1 and z -1 z Z{U(t)} = z -1 3. Z{an}, Z{(–1)n}, Z{eat}, Z{e–at} and Z(an–1) (i)

Z{an} =



 an ◊ z-n =

n=0

=

1 a 1z

or



n

Ê aˆ  ÁË z ˜¯ , if n ≥ 0 n=0

z , where the ROC is z-a

a < 1 or |z| > |a| z (ii) In particular, Z{(–1)n} =

z , where the ROC is |z| > 1. z +1

Transforms and Partial Differential Equations

5-8

Z{eat} = Z{eanT} = Z{(eaT)n}

(iii)

z

=

, where the ROC is |z| > e aT

z-e Alternatively, by Property 3, Z{eat} = Z{eat U(t)} = Z{U(t)}zÆze – aT aT

ze - aT z Ê z ˆ = - aT = = Á ˜ Ë z - 1¯ z Æ ze- aT ze - 1 z - e aT z

(iv) Similarly, Z{e–at} =

z - e - aT

(v) Z{an–1) = z–1 Z(an), by Property 2, if n ≥ 1 = z –1 ◊ =

z z-a

1 , if n ≥ 1 z-a

4. Z(n), Z(nan), Z(n2), Z{n(n – 1)}, z{t} •

(i)

Z(n) =

1

2

3

 nz - n = z + z2 + z3 +º

n=0

1 = z = =

2 ÏÔ ¸Ô Ê 1ˆ Ê 1ˆ 1 + 2 + 3 Ì ÁË z ˜¯ ÁË z ˜¯ +º˝ ÓÔ ˛Ô

1 Ê 1ˆ 1- ˜ z ÁË z¯ z ( z - 1)2

-2

, if

1 < 1 or |z| > 1 z

, where the ROC is |z| > 1

Alternatively Z(n) = Z{nU(n)}

(ii)

= -z

d U ( z ) , by Property 5 dz

= -z

d Ê z ˆ z = Á ˜ d z Ë z - 1¯ ( z - 1)2

Z{nan} = - z = -z =

d {Z(an)}, by Property 5 dz d Ê z ˆ d z ÁË z - a ˜¯ az

( z - a )2

, where the ROC is |z| > |a|

Z-Transforms and Difference Equations Z{n2} = z{n ◊ n} = - z

(iii)

5-9

d {Z(n)} dz

d ÔÏ z ¸Ô Ì ˝ d z ÔÓ ( z - 1)2 Ô˛

= -z

È ( z - 1)2 - 2 z ( z - 1) ˘ = -z Í ˙ ( z - 1)4 ÍÎ ˙˚ = (iv)

z( z + 1) ( z - 1)3

Z{n(n – 1)} = Z(n2) – Z(n) [by Property 1] z( z + 1) z = 3 ( z - 1) ( z - 1)2 =

(v)

2z ( z - 1)3

Z(t) = Z(nT) = T ◊ Z(n) =

Tz ( z - 1)2

5. Z(nk) and Z(tk) (i) Z(nk) = Z{n ◊ nk – 1} = -z In particular, Z(n) = - z

d {Z(nk – 1)}, which is a recurrence formula. dz d z {Z (1)} = dz ( z - 1)2

d d Ï z ¸ {Z (n)} = - z Ì ˝ dz d z Ó ( z - 1)2 ˛ z( z + 1)

Z(n2) = - z = Similarly, (ii)

Z(n3) =

( z - 1)3 z( z 2 + 4 z + 1) ( z - 1)4

and so on.

Z(t k) = Z{(nT)k} = T k Z(nk) d = –Tk ◊ z {Z(nk – 1)} from (i) above dz = –Tz

d {Z(nT)k – 1} dz

= –Tz

d {Z(tk – 1)}, which is a recurrence formula. dz

Transforms and Partial Differential Equations

5-10

In particular, Z(t) = –Tz = –Tz Similarly,

Z(t2) = Z(t3) =

d {Z(1)} dz d Ê z ˆ Tz = d z ÁË z - 1˜¯ ( z - 1)2

T 2 z( z + 1) ( z - 1)3

and

T 3 z( z 2 + 4 z + 1) ( z - 1)4

and so on.

Ê 1ˆ Ê 1 ˆ 6. Z Á ˜ and Z Á Ë n¯ Ë n + 1 ˜¯ (i)

Ê 1ˆ Z Á ˜ = Ë n¯



Â

n =1

1 –n z n 2

=

3

1 1 Ê 1ˆ 1 Ê 1ˆ + + Á ˜ +º• z 2 ÁË z ˜¯ 3 Ë z¯

1 Ê 1ˆ = –log Á 1 - ˜ , where the ROC is < 1 or |z| > 1 Ë z¯ z Ê z ˆ = log Á Ë z - 1˜¯ (ii)

Ê 1 ˆ Z Á = Ë n + 1˜¯



Â

n=0

1 z–n n +1 2

1 Ê 1ˆ 1 Ê 1ˆ = 1 + Á ˜ + Á ˜ +º• 2 Ë z¯ 3 Ë z¯ È1 1 Ê 1ˆ 2 1 Ê 1ˆ 3 ˘ = z Í + Á ˜ + Á ˜ +º• ˙ 3 Ë z¯ ÍÎ z 2 Ë z ¯ ˙˚ Ê 1ˆ Ê z ˆ = –z log Á 1 - ˜ = z log Á Ë z¯ Ë z - 1 ˜¯ ÏÔ a n ¸Ô Ï1¸ 7. Z Ì ˝ and Z Ì ˝ n! Ó n! ˛ ÓÔ ˛Ô (i)

ÏÔ a n ¸Ô ZÌ ˝ = ÓÔ n ! ˛Ô



Â

n=0

an -n • 1 Ê a ˆ z = ÁË ˜¯ n! n = 0 n! z 2

= 1+

n

1 Ê aˆ 1 Ê aˆ a/z + +º• = e 1! ÁË z ˜¯ 2! ÁË z ˜¯

Ï1¸ (ii) Putting a = 1, we get Z Ì ˝ = e1/z. Ó n! ˛

Z-Transforms and Difference Equations n

5-11

n

8. Z(r cos nq), Z(r sin nq), Z(cos nq), Z(sin nq) z , if |z| > a z-a iq Putting a = re , we have

(i) We know that Z(an) =

Z{(reiq)n} =

z z - reiq

, if |z| > |r|

i.e. Z{rn (cos nq + i sin nq)} = = =

z z - r (cos q + i sin q )

z{( z - r cos q ) + ir sin q} ( z - r cos q )2 + r 2 sin 2 q z( z - r cos q ) + izr sin q} z 2 - 2 zr cos q )2 + r 2

Equating the real parts on both sides, we get Z(rn cos nq) =

z( z - r cos q )

, if |z| > |r|

z - 2 zr cos q + r 2 2

(1)

(ii) Equating the imaginary parts on both sides, we get Z(rn sin nq) =

zr sin q

, if |z| > r

z - 2 zr cos q + r 2 2

(2)

(iii) Putting r = 1 in Eq. (1), we have Z(cos nq) =

z( z - cos q ) z 2 - 2 zr cos q + 1

, if |z| > 1

(3)

np ˆ z2 Ê In particular, Z Á cos = Ë 2 ˜¯ z 2 + 1 (iv) Putting r = 1 in Eq. (2), we have Z(sin nq) =

z sin q z - 2 z cos q + 1 2

, if |z| > 1

(4)

np ˆ z Ê In particular, Z Á sin = 2 ˜ Ë ¯ 2 z +1 9. Z(cos wt), Z(sin wt), Z(e–at cos bt), Z(e–at sin bt) (i) Z(cos wt) = Z(cos wnT) or Z{cos n(wT)} =

z(z - cos wT ) z2 - 2 z cos wT + 1

,

if |z| > 1 [on using Eq. (3) of result 8] (ii)

Z(sin wt) =

z sin wT z - 2 z cos wT + 1 2

, if |z| > 1 [on using Eq. (4) of Function 8],

Transforms and Partial Differential Equations

5-12

Z(cos wt) =

1 Z(eiwt + e–iwt) 2

˘ 1È z z [by steps (iii) and (iv) + 2 ÍÎ z - eiwT z - e-iwT ˙˚ of result (3)] z È 2 z - (eiwT + e-iwT ) ˘ = Í 2 ˙ 2 ÍÎ z - (eiwT + e-iwT )z + 1 ˙˚ =

= Z(sin wt) =

(iii) Z{e

z - 2 z cos wT + 1 1 Z(eiwT – e–iwT) 2i

=

˘ z È 1 1 2i ÍÎ z - eiwT z - e-iwT ˙˚

=

˘ z È eiwT - e-iwT Í 2 ˙ w w i T T 2i ÍÎ z - (e + e )z + 1 ˙˚

= –at

z(z - cos wT ) 2

z sin wT z - 2 z cos wT + 1 2

cos bt} = Z(cos bt)z Æ zeaT, by corollary under Property 3 È z( z - cos bT ) ˘ = Í 2 ˙ Î z - 2 z cos bT + 1 ˚ z Æ zeaT =

(iv) Z{e

–at

ze aT ( zeaT - cos bT )

z 2 e2 aT - 2 ze aT cos bT + 1 sin bt} = Z(sin bt)z Æ zeaT, by the same rule Ê ˆ z sin bT = Á Ë z 2 - 2 z cos bT + 1˜¯ z Æ zeaT =

ze aT sin bT z 2 e2 aT - 2 zeaT cos bT + 1

Worked examples

5(a)

Example 1 Find the bilateral Z-transforms of (i) an d(n – k), (ii) – an U(– n – 1); and (iii) – nan U(– n – 1) (i)

Ï1, for n = k d (n – k) = Ì Ó0, for n π k

Z-Transforms and Difference Equations •

Z{d(n – k)} =

Â

5-13

d(n – k)z–n = z–k

n =-•

Ê zˆ By Property 3, Z{an d(n – k)} = Á ˜ Ë a¯

-k

Ï 1, if - n - 1 ≥ 0, ie. if n £ -1 U(– n – 1) = Ì Ó0, if - n - 1 < 0, i.e. if n > -1 •

Z{U(– n – 1)} =

Â

U(– n – 1)z–n

n =-• -1

=

Â

z–n or

n =-•



Â

zn

n =1

= z(1 + z + z2 + ...•) z 1- z (ii) Z{an f(n)} = f (z/a), by Property 3, which is true for bilateral Z-transform also. =

\

Z{an U(– n – 1)} =

\

Z{–an U(– n – 1)} =

(iii) Z{nf(n)} = –z \

z /a z or 1 - z /a a-z z z -a

d { f (z)}, by Property 5. dz

Z{– nan U(– n – 1)} = –z

d Ê z ˆ az = d z ÁË z - a ˜¯ ( z - a )2

Example 2 Find the bilateral Z-transform of f(n), if ÏÔa n , for 0 £ n £ N –1 (i) f(n) = Ì ÔÓ0, otherwise (ii) f(n) = b|n|, b > 0 •

(i)

Z{f(n)} =

Â

f(n)z–n

n =-• N -1

=

Â

an z–n

n=0 2

= 1+

a Ê aˆ Ê aˆ + Á ˜ + ... + Á ˜ Ë z¯ z Ë z¯

N -1

5-14

Transforms and Partial Differential Equations N

Ê aˆ 1- Á ˜ Ë z¯ = Ê aˆ 1- Á ˜ Ë z¯ (ii)

f(n) = b|n|

i.e.

ÏÔb n , if n ≥ 0 f(n) = Ì -n ÔÓb , if n < 0 -1

\

Z{b|n|} =

Â



b–n z–n +

n =-• •

=

Â

n =1

=

(bz)n +

Â

bn z–n

n=0 •

Ê bˆ

 ÁË z ˜¯

n

n=0

bz 1 bz z + + = b 1 - bz 1 - bz z - b 1z

Example 3 (In what follows, Z-transform refers to one sided Z-transform, unless otherwise specified.) Find the Z-transforms of (i) an cosh an; and (ii) 2n sinh 3n ÏÔ Ê e an + e - an ˆ ¸Ô (i) Z{an cosh an} = Z Ìa n Á ˜˝ 2 ¯ ˛Ô ÓÔ Ë =

1 [Z{aea)n} + Z{(ae–a)n}] 2

=

˘È 1È z z z ˘ + (∵ z(b n )) = 2 ÍÎ z - ae a z - ae - a ˙˚ ÍÎ z - b ˙˚

=

z È 2 z - a (e a + e - a ) ˘ Í ˙ 2 ÍÎ z 2 - a(e a + e - a )z + a 2 ˙˚

=

(ii)

z( z - a cosh a ) z - 2 a(cosh a )z + a 2 2

ÏÔ Ê e3n - e -3n ˆ ¸Ô Z{2n sinh 3n} = Z Ì2 n Á ˜˝ 2 ¯ ˛Ô ÓÔ Ë =

1 [Z{(2e3)n} – Z{(2e–3)n}] 2

=

˘ 1È z z 2 ÍÎ z - 2e3 z - 2e -3 ˙˚

Z-Transforms and Difference Equations

= =

5-15

˘ zÈ 2(e3 - e-3 ) Í 2 ˙ 3 3 2 ÎÍ z - 2(e + e )z + 4 ˚˙ 2 z sinh 3 z 2 - 4 z cosh 3 + 4

Example 4 Find the Z-transforms of (i) t2 e–t; and (ii) (t + T)e–(t + T) (i)

Z(t2) =

T 2 z( z + 1) ( z - 1)3

[Refer to basic transforms (5)]

Z{e–at f(t)} = f (zeaT), by corollary under Property 3. \

È T 2 z( z + 1) ˘ Z{t2 ¥ e–t} = Í 3 ˙ ÍÎ ( z - 1) ˙˚ z Æ zeT =

(ii)

Z(t) =

T 2 zeT ( zeT + 1) ( zeT - 1)3 Tz ( z - 1)2

[Refer to basic transform (4)]

È Tz ˘ TzeT Z{te–t} = Í = ˙ 2 ( zeT - 1)2 Î ( z - 1) ˚ z Æ zeT Now \

Z{f(t + T)} = z f (z) – f(0), by Property 2 Z{(t + T)e–(t + T)} =

Tz 2 eT ( zeT - 1)2

–0=

TeT z 2 ( zeT - 1)2

Example 5 1 Find the Z-transforms of (i) f(n) = an2 + bn + c; and (ii) f(n) = (n + 1)(n + 2). 2 (i) f(n) = an2 + bn + c \ Z{ f(n)} = aZ(n2) + bZ(n) + cZ(1) z( z + 1) z z +b +c = a◊ , 3 2 z -1 ( z - 1) ( z - 1) by Z-transforms of basic functions z = [a(z + 1) + b(z – 1) + c(z – 1)2] ( z - 1)3 z = [cz2 + (a + b – 2c)z + (a – b + c)] ( z - 1)3 (ii) \

1 1 2 (n + 1)(n + 2) = (n + 3n + 2) 2 2 1 Z{ f(n)} = {Z(n2) + 3Z(n) + 2Z(1)} 2 f(n) =

Transforms and Partial Differential Equations

5-16

=

1 È z( z + 1) 3z 2z ˘ + + Í ˙ 3 2 2 Î ( z - 1) z - 1˚ ( z - 1) z

2 (z - 1)

3

=

Example 6

È(z + 1) + 3 (z - 1) + 2 (z - 1)2 ˘ ÍÎ ˙˚

z3 ( z - 1)3

Find Z-transforms of (i)

f(n) =

1 ; and n(n - 1)

(ii)

f(n) =

2n + 3 (n + 1)(n + 2)

(i)

f(n) =

1 = 1 -1 n(n - 1) n -1 n

Ï 1 ¸ Z Ì ˝ = Ó n - 1˛



Ê 1 ˆ

 ÁË n - 1˜¯

z–n

n=2

2

3

4

1 Ê 1ˆ 1 Ê 1ˆ 1 Ê 1ˆ = ◊ Á ˜ + ◊ Á ˜ + ◊ Á ˜ + ◊◊◊ 1 Ë z¯ 2 Ë z¯ 3 Ë z¯ 1È Ê 1ˆ ˘ 1 Ê z ˆ Í- log ÁË 1 - ˜¯ ˙ = log ÁË zÎ z ˚ z z - 1˜¯ Ï1 ¸ Ê z ˆ Z Ì ˝ = log Á [Refer to basic transform (6)] Ë z - 1˜¯ n Ó ˛ =

Ï 1 ¸ Ï1 ¸ ˝- Z Ì ˝ n 1 Ó ˛ Ón˛

\

Z{ f(n)} = Z Ì

Ê1 ˆ Ê z ˆ Ê z - 1ˆ Ê z - 1ˆ = Á - 1˜ log Á = Á log Á Ë z - 1˜¯ Ëz ¯ Ë z ˜¯ Ë z ˜¯ (ii)

f(n) =

2n + 3 1 1 + = , by partial fractions. (n + 1)(n + 2) n +1 n + 2 Ê z ˆ Ï 1 ¸ Z Ì [Refer to basic transform (6)] ˝ = z log Á Ë z - 1˜¯ n + 1 Ó ˛ Ï 1 ¸ Z Ì ˝ = Ón + 2˛



Â

n=0

1 ◊ z–n n+2 2

=

1 1 1 1 Ê 1ˆ + ◊ + ◊ + ◊◊◊ 2 3 z 4 ÁË z ˜¯

È 1 Ê 1ˆ 2 1 Ê 1ˆ 3 ˘ = z Í 2 ◊ ÁË z ˜¯ + 3 ◊ ÁË z ˜¯ + ◊◊◊˙ ÍÎ ˙˚ 2

Z-Transforms and Difference Equations

5-17

È Ê 1ˆ 1 ˘ = z2 Í- log Á 1 - ˜ - ˙ Ë z¯ z˚ Î Ê z ˆ = z2 log Á –z Ë z - 1˜¯ Ê 1 ˆ Ê 1 ˆ Z{ f(n)} = Z Á +Z Á Ë n + 1˜¯ Ë n + 2 ˜¯

\

Ê z ˆ = z(z + 1) log Á –z Ë z - 1˜¯

Example 7 Given that f (z) = log (1 + az–1), for |z| > |a|, find f(n) and also Z{nf(n)}. f (z) = log (1 + az–1) 1 1 (az–1)2 + (az–1)3 –  + 3 2 1 (– 1)n–1 ¥ (az–1)n +  n n -1 n (-1) a –n z , for |z| > |a| n

= az–1 – •

=

Â

n =1

\

f(n) = coefficient of z–n in the R.H.S. ( - a )n = n d Z{nf(n)} = –z f (z), by Property 5. dz d = –z log (1 + az–1) dz 1 = –z (– az–2) 1 + az -1

Now

=

az -1 1 + az -1

Example 8 Find the Z-transforms of Ê np ˆ (i) sin2 Á ˜ Ë 4 ¯ Ê np ˆ (ii) sin3 Á ˜ ; and Ë 6 ¯ (iii) cos (np/2 + p/4) (i)

Let

1Ê np ˆ Ê np ˆ f(n) = sin2 Á ˜ = Á 1 - cos ˜ Ë 4 ¯ 2Ë 2 ¯

Transforms and Partial Differential Equations

5-18

\

Z{f(n)} = Z(1/2) –

1 Ê np ˆ Z Á cos ˜ 2 Ë 2 ¯

pˆ Ê z Á z - cos ˜ Ë z 1 2¯ = p 2( z - 1) 2 2 z - 2 z cos + 1 2 [Refer to basic transform (8)] =

1È z z2 ˘ - 2 Í ˙ 2 ÎÍ z - 1 z + 1 ˚˙

np 6 3 np 1 np - sin = sin 4 6 4 2 3 Ê np ˆ 1 Ê np ˆ Z{ f(n)} = Z Á sin ˜ - Z Á sin ˜ 4 Ë 6 ¯ 4 Ë 2 ¯ Let f(n) = sin3

(ii)

\

=

3 z sin p /6 1 z sin p /2 - ◊ 2 2 4 z - 2 z cos p /6 + 1 4 z - 2 z cos p /2 + 1

by basic transform (8)

(iii)

z z 3 1 ◊ - ◊ 8 z2 - z 3 + 1 4 z2 + 1 Let f(n) = cos (np/2 + p/4) np p np p = cos cos – sin sin 2 4 2 4

\

Z{ f(n)} =

=

= =

1 Ï Ê np ˆ np ˆ ¸ Ê Ì Z Á cos ˜ - Z Á sin ˜ ˝ Ë 2 ¯ 2 ¯˛ 2Ó Ë z ¸Ô 1 ÏÔ z 2 - 2 Ì 2 ˝ 2 ÔÓ z + 1 z + 1 Ô˛ 1 z( z - 1) 2 z2 + 1

Example 9 Find the Z-transforms of (i) an sin a n; and (ii) n cos nq. z sin a (i) Z(sin a n) = 2 z - 2 z cos a + 1 \

Z{an sin a n} = {Z(sin a n)}z Æ z/a, by Property 3 z sin a a z sin a = 2 a = 2 z - 2a z cos a + a 2 z z 2 cos a + 1 a a2

Z-Transforms and Difference Equations

(ii)

Z (cos nq) =

\

5-19

z( z - cos q ) z - 2 z cos q + 1 2

Z (n cos nq) = –z ◊ = -z

d {Z(cos nq)} dz d È z 2 - z cos q ˘ Í ˙ dz ÎÍ z 2 - 2 z cos q + 1 ˚˙

È 2 ˘ 2 Í ( z - 2 z cos q + 1)(2 z - cos q ) - ( z - z cos q ) ˙ (2 z - 2 cos q ) ˙ = –z Í 2 2 Í ˙ + ( z 2 z cos q 1) Î ˚ =

z( z 2 cos q - 2 z + cos q ) ( z 2 - 2 z cos q + 1)2

Example 10 Find the Z-transforms of (i) cos2 t (ii) cos3 t; and (iii) cosh at sin bt 1 1 Z(1) + Z (cos 2t) 2 2 z 1 z( z - cos 2T ) + = 2 2( z - 1) 2 ( z - 2 z cos 2T + 1)

(i)

Z (cos2 t) =

(ii)

1 Ê3 ˆ Z (cos3 t) = Z Á cos t + cos3t ˜ Ë4 ¯ 4 =

(iii)

3 z( z - cos T ) 1 z( z - cos3T ) + 4 ( z 2 - 2 z cos T + 1) 4 z 2 - 2 z cos 2T + 1

1 Z {(eat + e–at) sin bt} 2 1 = [Z (eat sin bt) + Z (e–at sin bt)] 2 Z (cosh at sin bt) =

=

˘ 1È ze - aT sin bT ze aT sin bT + Í 2 -2 aT ˙ 2 ÎÍ z e - 2 ze- aT cos bT + 1 z 2 e2 aT - 2 zeaT cos bT + 1 ˚˙

Example 11 (i) use initial value theorem to find f(0) when f (z) =

ze aT ( zeaT - cos bT )

z 2 e2 aT - 2 ze aT cos bT + 1 (ii) Use final value theorem to find f (•), when f (z) =

Tze aT ( zeaT - 1)2

5-20

Transforms and Partial Differential Equations

(i) By initial value theorem, f(0) = lim ÈÎ f ( z )˘˚ z Æ• È aT Ê aT 1 ˆ Í e ÁË e - z cos bT ˜¯ = lim Í 2 1 z Æ• Í 2 aT - e aT cos bT + 2 Íe z z Î =1

˘ ˙ ˙ ˙ ˙ ˚

(ii) By final value theorem, f(•) = lim {(z – 1) f (z)} zÆ1

È ( z - 1)Tze aT ˘ = lim Í ˙ =0 z Æ1 Í ( ze aT - 1)2 ˙ Î ˚

Example 12 Find the Z-transform of f(n)* g(n), where n

Ê 1ˆ (i) f(n) = U(n) and g(n) = d(n) + Á ˜ U(n) Ë 2¯ (ii) f(n) = nU(n) and g(n) = 2n U(– n – 1) (i) By convolution theorem, Z{ f(n)*g(n)} = f (z) ◊ g (z) z f (z) = Z{U(n)} = z -1 ÏÔÊ 1 ˆ n ÔÓ 2

¸Ô

g (z) = Z{d(n)} + Z ÌÁ ˜ U (n) ˝ Ë ¯

= 1+ \

Z{ f(n)* g(n)} =

˛Ô

z 2z 4z - 1 = 1+ or z - 1/2 2z - 1 2z - 1

z(4 z - 1) ( z - 1)(2 z - 1)

f (z) = Z{nU(n)} =

z ( z - 1)2

g (z) = Z{2n U(– n – 1)} =

(bilateral Z-transform taken)



Â

2n U(– n – 1)z–n

n =-• -1

=

Â

n =-• •

=

Â

n =1

È Ï1, if n £ -1˘ 2n z–n Í∵ U (- n - 1) = Ì ˙ Ó0, if n > -1˚ Î

2–n zn

Z-Transforms and Difference Equations 2

5-21

3

= (z/2) + (z/2) + (z/2) + ⋅⋅⋅ ∞ z /2 z = = 1 - z /2 2 - z \

Z{ f (n)}* g(n)} =

z2 ( z - 1)2 (2 - z )

Example 13 Find the Z-transform of f (n)* g(n), where Ê 1ˆ (i) f (n) = Á ˜ Ë 2¯

n

and g(n) = cos nπ

Ï(1 / 3)n , for n ≥ 0 Ô n (ii) f(n) = Ì Ê 1ˆ -n Ô(1 / 2) , for n < 0; and g(n) = Á ˜ U (n) Ë 2¯ Ó n z 2z ÔÏÊ 1 ˆ ¸Ô f (z) = Z ÌÁ ˜ ˝ = = Ë 2¯ Ô z - 1/2 2z - 1 ÓÔ ˛ z g (z) = Z{cos nπ} = Z {(– 1)n} = z +1

(i)

By convolution theorem Z{ f (n)* g(n)} = f (z) g (z) =

2z z 2z2 ◊ = 2z - 1 z + 1 2z2 + z - 1 -1

(ii)

f (z) =

Â

n =-• •

=

Â

n =1

Ê 1ˆ ÁË 2 ˜¯

-n

z–n +

n

Z{ f (n)* g(n)} =

n

z 3z ◊ 2 - z 3z - 1 2z 2z - 1

6 z3 (2 - z )(2 z - 1)(3z - 1)

n

Ê 1 ˆ –n ÁË 3 ˜¯ z

1ˆ Ê ÁË 1 - 3z ˜¯

g ( z ) = Z {(1/2)n U(n)} =

\

Â

n=0

• Ê zˆ Ê 1ˆ ÁË 2 ˜¯ +  ÁË 3z ˜¯ n=0

z = (1 – z/2)–1 + 2 =



-1

Transforms and Partial Differential Equations

5-22

Example 14 Use convolution theorem to find the sum of the first n natural numbers. n

1 + 2 + 3 + ⋅⋅⋅ + n =

Âk

k =0 n

=

Â

rU(r) U(n – r) [ U(r) = 1, as r ≥ 0

r =0

and U(n – r) = 1, as r ≤ n] = {nU(n)}*U(n) \

By convolution theorem,

ÔÏ n Ô¸ Z Ì Â k ˝ = Z{nU(n)} Z{U(n)} ÔÓ k = 0 Ô˛ z2 z z ◊ = = ( z - 1)3 ( z - 1)2 z - 1 Taking Inverse Z-transforms, n

Âk

k =0

ÏÔ z 2 ¸Ô = Z –1 Ì 3˝ ÓÔ ( z - 1) ˛Ô È 1 ÔÏ z( z + 1) + z( z - 1) Ô¸˘ = Z–1 Í Ì ˝˙ ( z - 1)3 Ô˛˚˙ ÎÍ 2 ÔÓ =

1 È -1 ÏÔ z( z + 1) ¸Ô Ô z ¸Ô˘ -1 Ï ÍZ Ì ˙ ˝+ Z Ì 3 2˝ 2 ÎÍ ÓÔ ( z - 1) ˛Ô ÓÔ ( z - 1) ˛Ô˚˙

=

1 2 (n + n) 2

=

1 n(n + 1) 2

Example 15 Use convolution theorem to find the inverse Z-transform of (i)

(i)

z2 8z 2 and (ii) (2 z - 1)(4 z + 1) ( z + a )2 8z 2 ÔÏ Ô¸ z z ¸ –1 Ï Z –1 Ì ˝ =Z Ì ◊ ˝ z 1/2 z + 1/4 ˛ ÔÓ (2 z - 1)(4 z + 1) Ô˛ Ó Ï z ¸ Ï z ¸ –1 = Z –1 Ì ˝ ˝* Z Ì Ó z + 1/ 4 ˛ Ó z - 1/2 ˛ n

Ê 1ˆ = Á ˜ * (– 1/4)n Ë 2¯

Z-Transforms and Difference Equations n

=

Â

r =0

Ê 1ˆ ÁË 2 ˜¯

Ê 1ˆ = Á ˜ Ë 2¯

n n

Ê 1ˆ = Á ˜ Ë 2¯

n n

5-23

n -r

Â

r =0

Â

r =0

(– 1/4)r (1/2)–r (– 1/4)r Ï Ê 1 ˆ n +1 ¸ Ô1 - Á ˜ Ô Ê 1ˆ Ì Ë 2¯ ˝ r (–1/2) = Á ˜ Ô Ë 2 ¯ Ó 1 - (-1/2) Ô˛

=

n ¸ 2 ÏÔÊ 1 ˆ 1 nÔ ÌÁ ˜ + ◊ (-1/4) ˝ 3 ÔË 2 ¯ 2 Ó ˛Ô

=

2 1 (1/2)n + (– 1/4)n 3 3

n

2 z ¸ Ï z ÔÏ z Ô¸ ◊ Z –1 Ì = Z–1 Ì ˝ 2˝ Óz + a z + a˛ ÔÓ ( z + a ) Ô˛ Ê z ˆ Ê z ˆ = Z –1 Á *Z –1 Á ˜ Ë z + a¯ Ë z + a ˜¯

(ii)

= (– a)n * (– a)n n

=

Â

(– a)r ⋅ (– a)n – r

r =0 n

=

Â

(– a)n = (n + 1) (– a)n

r =0

Exercise 5(a) Part A (Short-Answer Questions) 1. Define unilateral and bilateral Z-transforms of { f (n)}. 2. Define one sided Z-transform of f (t). 3. Prove that Z{ f (n – m)} = z–m f (z). 4. 5. 6. 7. 8. 9. 10. 11.

Prove that Z{ f(t + T)} = z { f (z) – f(0)}. Express Z{an f (n)} in terms of Z{ f(n)}. Prove that Z{ f(t) e–at} = Z{ f(t)}z Æ zeaT. Derive the time-reversal property of a bilateral Z-transform. How are Z{ f(n)} and Z{nf (n)} related? State initial value theorem in Z-transforms. State final value theorem in Z-transforms. Define convolution of two sequences with respect to unilateral and bilateral Z-transforms. 12. State convolution theorem in Z-transforms.

Transforms and Partial Differential Equations

5-24

Find the Z-transforms of the following sequences/functions: 13. 2n δ(n – 1) 14. an U(n) + bn U(– n – 1) 15. abn (a, b ≠ 0) 16. a(– a)n U(n) n

18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

n

Ê 1ˆ Ê 1ˆ ÁË 2 ˜¯ U(n) + ÁË 3 ˜¯ U(n)

17.

3 ⋅ 2n + 4 ⋅ (– 1)n eat + b e2(t + T) (n – 1)an – 1 nC2 1 n(n + 1) np sin 3 np cos 4 sin (t + T ) cos (2t + T ) cosh t sinh 2t sinh (t + T)

Part B Find the Z-transforms of the following sequences/functions: 1 31. 2n – 1 + ⋅ 4 n – 3 n. 2 1 n 1 1 32. 6 – ⋅ 3n + ⋅ 2n 12 3 4 n n¸ Ï 1 ÔÊ 5 + 1ˆ Ê 1 - 5 ˆ Ô 33. ÌÁ ˜ -Á ˜ ˝ 5 ÔË 2 ¯ Ë 2 ¯ Ô Ó ˛ 34. 35. 36. 37. 38. 39. 40.

Ê 1 - 3i ˆ Ê 1 + 3i ˆ n n ÁË 4 ˜¯ (1 + i) + ÁË 4 ˜¯ (1 – i) 1 1 1 - (– 2)n + (– 2i)n + (2i)n 4 4(1 - i ) 4(1 + i ) 1 n 1 2 – ⋅ (– 3)n – 5n(– 3)n – 1 5 5 2(– 1)n + 2n (n + 2) an sinh an 2n cosh 5n t3 e–2t

Z-Transforms and Difference Equations

5-25

n

41. n(n – 1)2 42. nC3 1 43. n(n + 1)(n + 2) n-2 44. n(n - 1) 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61.

1 n (a + a–n) n! Ê np ˆ + a˜ cos Á Ë 8 ¯ np cos2 6 np cos3 4 a n cos an n sin nq nan cos nq nan sin nq r n cos (nq + f) r t sin (wt + f) sin2 2t sin3 t t sin t t cos t et sin 2t e–2t cos 3t Find the initial and final values of f(n), if f (z) =

0.4 z 2 ( z - 1)( z 2 - 0.736 z + 0.136)

62. Find the Z-transform of f (n)* g(n), when (i) f(n) = a n U(n) and g(n) = a n U(n) (ii) f(n) = U(n) and g(n) = 2n U(n) 63. Find the Z-transform of f(n)* g(n), if (i) f (n) = a n U(n) and g(n) = bn U(n) n

Ê 1ˆ Ê 1ˆ (ii) f(n) = Á ˜ U(n – 1) and g(n) = 1 + Á ˜ Ë 4¯ Ë 2¯ n

n

64. Express (i) (n + 1) and (ii)  n2 as convolution products and hence find k =0 their Z-transforms. 65. Use convolution theorem to find the inverse Z-transform of z2 1 (i) (ii) ( z + a )( z + b) Ê 1 -1 ˆ Ê 1 -1 ˆ ÁË 1 - 2 z ˜¯ ÁË 1 - 4 z ˜¯

Transforms and Partial Differential Equations

5-26

5.4

Inverse Z-transforms

The inverse Z-transform of f (z) has been already defined as Z –1{ f (z)} = f(n), when Z{f (n)} = f (z). Z –1 { f (z)} can be found out by any one of the following methods:

Method 1 (Expansion Method) If f (z) can be expanded in a series of ascending powers of z–1, i.e. in the form •

Â

f(n)z–n, by binomial, exponential and logarithmic theorems, the coefficient of

n=0

z–n in the expansion gives Z–1 { f (z)}.

Method 2 (Long Division Method) When the usual methods of expansion of f (z) fail and if f (z) =

g( z -1 ) h( z -1 )

g(z–1) is divided by h(z–1) in the classical manner and hence the expansion

, then



 f(n)z–n

n=0

is obtained in the quotient.

Method 3 (Partial Fractions Method) When f (z) is a rational function in which the denominator can be factorised, f (z) is resolved into partial fractions and then Z –1 { f (z)} is derived as the sum of the inverse Z-transforms of the partial fractions.

Method 4 (By Cauchy’s Residue Theorem) By using the relation between the Z-transform and Fourier transform of a sequence, it can be proved that 1 n–1 f(n) = Ú f (z) z dz, 2p i  C where C is a circle whose centre is the origin and radius is sufficiently large to include all the isolated singularities of f (z). C may also be a closed contour including the origin and all the isolated singularities of f (z). By Cauchy’s residue theorem,

Ú f (z) z

n–1

dz = 2π i × sum of the residues of f (z)zn–1 at the isolated singularities.

C

\ f (n) = Sum of the residues of f (z)z n–1 at the isolated singularities.

Z-Transforms and Difference Equations

5.5

5-27

formatIon of dIfference eQuatIons and use of Z-transform to soLve tHem

Z-transforms can be used to solve finite difference equations of the form ay(n + 2) + by(n + 1) + cy(n) = φ(n) with given values of y(0) and y(1). Taking Z-transforms on both sides of the given difference equation and using the values of y(0) and y(1), we will get y (z). Then Z –1 { y (z)} will give y(n) To express Z{y(n + 1)} and Z{y(n + 2)} in terms of y (z), the following results may be noted. •

(i)

Z{y(n + 1)} =

Â

y(n + 1)z–n = z



Â

y(n + 1)z–(n + 1)

n=0

n=0 –1

–2

= z[ y(1)z + y(2)z + ...] = z[ y(0) + y(1)z–1 + y(2)z–2 + ...] – zy(0) = z y (z) – zy(0) (ii) Similarly, Z{ y(n + 2)} = z2 y (z) – z2y (0) – zy(1) Z{y(n + 3) = z3 y (z) – z3 y(0) – z2 y(1) – zy(2) and so on.

and

Worked examples

5(b)

Example 1 Find the inverse Z-transform of

1 + 2 z -1 , by the long division method. 1 - z -1

(1 – z–1) 1 + 2z–1 (1 + 3z–1 + 3z–2 1 – z–1 3z–1 3z–1 – 3z–2. 3z–2 3z–2 – 3z–3 3z–3. -1

Thus

1 + 2z = 1 - z -1



Â

n=0

f(n)z–n = 1 + 3z–1 + 3z–2 + ... + 3z–n + ...

\

Ï1, for n = 0 f(n) = Ì Ó3, for n ≥ 1

or

f (n) = 1 + 2U(n – 1)

Transforms and Partial Differential Equations

5-28

Example 2 Ï 4z ¸ Find Z –1 Ì by the long division method. 3˝ Ó ( z - 1) ˛ 4 z -2 4z 4 z -2 f (z) = = = (1 - z -1 )3 ( z - 1)3 1 - 3z -1 + 3z -2 - z -3 1 – 3z–1 + 3z–2 – z–3) 4z–2 (4z–2 + 12z–3 + 24z–4 4z–2 – 12z–3 + 12z–4 – 4z–5 12z–3 – 12z–4 + 4z–5 12z–3 – 36z–4 + 36z–5 – 12z–6 24z–4 – 32z–5 + 12z–6 24z–4 – 72z–5 + 72z–6 – 24z–7 40z–5 – 60z–6 + 24z–7

4z = ( z - 1)3

Thus



Â

f(n)z–n = 4z–2 + 12z–3 + 24z–4 + ...

n=0

= 2[1.2z–2 + 2.3z–3 + 3.4z–4 + ... + (n – 1) ⋅ nz–n + ...]

Ï 4 z ¸ Ï0, for n = 0, 1 \ Z –1 Ì = Ì 3˝ 2( n 1) n , for n ≥ 2 Ó ( z - 1) ˛ Ó = 2(n – 1)nU(n)

Example 3 2 ÔÏ z + z Ô¸ Find Z –1 Ì by the long division method. 3˝ ÔÓ ( z - 1) Ô˛ z2 + z z -1 + z -2 f (z) = = ( z - 1)3 (1 - z -1 )3

1 – 3z–1 + 3z–2 – z–3) z–1 + z–2 (z–1 + 4z–2 + 9z–3 + 16z–4 z–1 – 3z–2 + 3z–3 – z–4 4z–2 – 3z–3 + z–4 4z–2 – 12z–3 + 12z–4 – 4z–5 9z–3 – 11z–4 + 4z–5 9z–3 – 27z–4 + 27z–5 – 9z–6 16z–4 – 23z–5 + 9z–6 16z–4 – 48z–5 + 48z–6 – 16z–7 25z–5 – 39z–6 + 16z–7

ÏÔ z + z ¸Ô = z–1 + 4z–2 + 9z–3 + 16z–4 + ..... ∞ Ì 3˝ ( z 1) ÓÔ ˛Ô 2

Thus



i.e.

Â

n=0

f(n)z–n =



 n2 z–n n =1

Z-Transforms and Difference Equations

\

ÏÔ0, for n = 0 f(n) = Ì 2 ÔÓn , for n ≥ 1

or

f (n) = n2 U(n – 1) or n2 U(n)

5-29

Example 4 Ï 1 ¸ Find Z –1 Ì by the long division method. -2 ˝ Ó1 + 4z ˛ 1 + 4z–2) 1 (1 – 4z–2 + 16z–4 – 64z–6 1 + 4z–2 – 4z–2 – 4z–2 – 16z–4 16z–4 16z–4 + 64z–6 – 64z–6 – 64z–6 – 256z–8 256z–8

1 = 1 – 4z–2 + 16z–4 – 64z–6 + ... 1 + 4z -2 • np –n = Â 2n cos z 2 n=0 np f (n) = 2n cos 2

Thus

\

Example 5 Find Z Let \

–1

2 ÔÏ 3z - 18z + 26 Ô¸ Ì ˝ by the partial fractions method. ÓÔ ( z - 2)( z - 3)( z - 4) ˛Ô

A B C 3z 2 - 18z + 26 + + = z-2 z-3 z-4 ( z - 2)( z - 3)( z - 4) 3z2 – 18z + 26 = A(z – 3)(z – 4) + B(z – 2)(z – 4) + C(z – 2)(z – 3) A=1=B=C

2 ÔÏ 3z - 18z + 26 Ô¸ –1 Ï 1 ¸ –1 Ï 1 ¸ –1 Ï 1 ¸ \ Z –1 Ì ˝ +Z Ì ˝ +Z Ì ˝ ˝ =Z Ì Óz - 2˛ Óz - 3˛ Óz - 4˛ ÔÓ ( z - 2)( z - 3)( z - 4) Ô˛

= 2n – 1 + 3n – 1 + 4n – 1 1 n 1 n 1 n = ◊ 2 + 3 + 4 , where n ≥ 1 2 3 4

Transforms and Partial Differential Equations

5-30

Example 6 ÏÔ ¸Ô 4 z3 Find Z –1 Ì ˝ , by the method of partial fractions. 2 ÓÔ (2 z - 1) ( z - 1) ˛Ô Let \

=

f (z) A B C 4z2 + + = = 2 2 z (2 z - 1) (2 z - 1) z -1 (2 z - 1) ( z - 1) 4z2 = A(2z – 1)(z – 1) + B(z – 1) + C(2z – 1)2

\

f (z) 6 2 4 = + 2 z 2 z - 1 (2 z - 1) z -1

\

f (z) = –

\

3z 1 z 4z - ◊ + 2 z - 1/2 2 ( z - 1/2) z -1

Ï z ¸ Ô Z –1 { f (z)} = – 3Z –1 Ô 1˝ Ì z ÔÓ 2 Ô˛

Ï ¸ Ô 1z Ô Ï z ¸ Ô Ô – Z –1 Ì 2 2 ˝ + 4 Z -1 Ì ˝ Ó z - 1˛ ÔÊ z - 1ˆ Ô Ô ÁË 2 ˜¯ Ô˛ Ó n

Ê 1ˆ = – 3(1/2)n – n Á ˜ + 4 Ë 2¯ È z az ˘ n and Z (na n ) = Í∵ Z (a ) = ˙ z-a ( z - a )2 ˚ Î

Example 7

Ê 1ˆ = 4 – (n + 3) Á ˜ Ë 2¯

n

ÏÔ 4 - 8z -1 + 6 z -2 ¸Ô Find Z –1 Ì by the method of partial fractions. -1 -1 2 ˝ ÓÔ (1 + z )(1 - 2 z ) ˛Ô 4 - 8z -1 + 6 z -2 4 z3 - 8z 2 + 6 z Let = f (z) = (1 + z -1 )(1 - 2 z -1 )2 ( z + 1)( z - 2)2 Let \

\ \

A B C f (z) 4 z 2 - 8z + 6 + + = = 2 z + 1 z 2 z ( z - 2)2 ( z + 1)( z - 2) 2 2 4z – 8z + 6 = A(z – 2) + B(z + 1)(z – 2) + C(z + 1) A=B=C=2 z z 2z +2◊ + f (z) = 2 z +1 z - 2 ( z - 2)2 Ï 2z ¸ Ï z ¸ –1 Ï z ¸ –1 Z –1 { f ( z ) } = 2Z –1 Ì ˝ + 2Z Ì ˝ +Z Ì 2˝ Ó z + 1˛ Óz - 2˛ Ó ( z - 2) ˛ = 2(–1)n + 2 ⋅ 2n + n ⋅ 2n = 2(–1)n + (n + 2) ⋅ 2n

Z-Transforms and Difference Equations

5-31

Example 8 2 ÔÏ z + 2 z Ô¸ Find Z –1 Ì 2 ˝ by the method of partial fractions. ÔÓ z + 2 z + 4 Ô˛ f (z) z+2 z+2 Let = 2 = z z + 2z + 4 ( z + 1 - i 3)( z + 1 + i 3)

=

( z + 1) - i 3

+

B z +1+ i 3

z + 2 = A(z + 1 + i 3 ) + B(z + 1 – i 3 )

\

A= \

A

f (z) =

\

1 2 3

2 3

z

( 3 - i) ◊

Z–1

1

( 3 - i ) and B = 1

2 3 z

( 3 + i)

( 3 + i) z +1- i 3 2 3 z +1+ i 3 1 { f ( z )} = [( 3 - i )(-1 + i 3)n + ( 3 + i )(-1 - i 3)n ] 2 3 =

+

1

È 2p 2p nÊ n + i sin Í( 3 - i )2 ÁË cos 3 3 2 3 Î 2p 2p ˆ ˘ Ê n - i sin n ˙ ( 3 + i )2 n Á cos Ë 3 3 ˜¯ ˚ 1

ˆ n˜ + ¯

[ the modulus-amplitude form of (-1 ± i 3)

2p 2p ˆ ˘ Ê ± i sin ˜ ˙ is 2 Á cos Ë 3 3 ¯˚ =

2p n 2p n ˘ È ◊ 2 n ◊ 2 Í 3 cos + sin 3 3 ˙˚ 2 3 Î 1

È 2p n 1 2p n ˘ + sin = 2n Ícos ˙ 3 3 ˚ 3 Î

Example 9 z2 ÔÏ Ô¸ Find Z –1 Ì ˝ by the method of partial fractions. 2 ÔÓ ( z + 2)( z + 4) Ô˛ Let

f (z) z A Bz + C = + 2 = 2 z z + 2 ( z + 2)( z + 4) z +4 1 1 1 z+ 4 4 2 + = z + 2 z2 + 4 -

\

f (z) = –

1 z 1 z2 1 2z + ◊ 2 + 4 z + 2 4 z + 4 4 z2 + 4

Transforms and Partial Differential Equations

5-32

np np 1 1 1 (–2)n + ⋅ 2n cos + ⋅ 2n sin 2 2 4 4 4 2 È np ˆ z np ˆ az ˘ Ê n Ê and Z Á a n sin ˜ = 2 Í∵ Z Á a cos ˜ = 2 ˙ 2 Ë Ë 2 ¯ z +a 2 ¯ z + a2 ˚ Î

Z –1 { f ( z )} = –

\

Example 10 z2 ÔÏ Ô¸ Find Z –1 Ì ˝ , by using Residue theorem. ÓÔ ( z - a)( z - b) ˛Ô ÏÔ ¸Ô z2 1 z n -1 ◊ z 2 Z –1 Ì = ˝ Ú (z - a)(z - b) dz, where C is the circle whose 2p i  ÓÔ ( z - a)( z - b) ˛Ô C centre is the origin and which includes the singularities z = a and z = b. = [(Res.)z = a + (Res.)z = b] of Cauchy’s residue theorem z = a and z = b are simple poles.

Ê z n +1 ˆ a n +1 b n +1 (Res.)z = a = Á = and (Res.) = z=b ˜ a-b b-a Ë z - b ¯ z=a

\

\

z n +1 , by ( z - a)( z - b)

ÏÔ ¸Ô z2 1 a b Z –1 Ì (an + 1 – bn + 1) or ◊ an ◊ bn ˝ = ( z a )( z b ) a b a b a b ÓÔ ˛Ô

Example 11 ÏÔ z( z 2 - z + 2) ¸Ô Find Z –1 Ì , by using Residue theorem. 2˝ ÓÔ ( z + 1)( z - 1) ˛Ô By residue theorem,

ÏÔ z( z 2 - z + 2) ¸Ô È n -1 z n ( z 2 - z + 2) ˘ Z –1 Ì = Sum of the residue of z f ( z ) = ˝ Í ˙ 2 ( z + 1)( z - 1)2 ˚ ÓÔ ( z + 1)( z - 1) ˛Ô Î at the poles. z = – 1 is a simple pole of zn–1 f (z) \

R(z = – 1) =

(-1)n ◊ (1 + 1 + 2) = (– 1)n (-1 - 1)2

z = 1 is a double pole of zn – 1 f ( z ) . \

È d ÏÔ z n ( z 2 - z + 2) ¸Ô˘ R(z = 1) = Í Ì ˝˙ z +1 ˛Ô˚˙ z =1 ÎÍ dz ÓÔ È ( z + 1){(n + 2)z n +1 - (n + 1)z n + 2 nz n -1} - ( z n + 2 - z n +1 + 2 z n ) ˘ = Í ˙ ( z + 1)2 Î ˚ z =1

Z-Transforms and Difference Equations

=

5-33

1 {n + 2 – (n + 1) + 2n – 1} = n 2

ÏÔ z( z 2 - z + 2) ¸Ô Z –1 Ì = (–1)n + n 2˝ ÓÔ ( z + 1)( z - 1) ˛Ô

Example 12 2 ÔÏ 2 z + 4 z Ô¸ Find Z –1 Ì , by using Residue theorem. 3 ˝ ÔÓ ( z - 2) Ô˛ By residue theorem, 2 ÔÏ 2 z + 4 z Ô¸ Z –1 Ì = the residue of 3 ˝ ÔÓ ( z - 2) Ô˛

Rz = 3 =

n +1 n ÔÏ 2 z + 4 z Ô¸ Ì ˝ at the only triple pole (z = 2). 3 ÔÓ ( z - 2) Ô˛

1 È d2 Í 2! ÎÍ dz 2

ÏÔ 2 z n +1 + 4 z n ¸Ô˘ ( z - 2)3 ˝˙ Ì 3 ÓÔ ( z - 2) ˛Ô˚˙

1 {2(n + 1)nzn – 1 + 4n(n – 1)zn – 1}z = 2 2 1 = [2(n + 1) n ⋅ 2n – 1 + 4n(n – 1)2n – 2} 2 =

= n2 2n \

ÏÔ 2 z 2 + 4 z ¸Ô Z –1 Ì = n2 2n 3 ˝ ÓÔ ( z - 2) ˛Ô

Example 13 z2 ÔÏ Ô¸ Find Z –1 Ì ˝ , by the method of residues. 2 ÔÓ ( z + 2)( z + 4) Ô˛ z2 z n +1 ÔÏ Ô¸ By residue theorem, Z –1 Ì = Sum of the residues of ˝ 2 ( z + 2)( z 2 + 4) ÔÓ ( z + 2)( z + 4) Ô˛ at the singularities. z = – 2, ± 2i are simple poles. Ê z n +1 ˆ 1 R(z = – 2) = Á 2 = (– 2)n + 1 ˜ 8 Ë z + 4 ¯ z =-2 Ê (2i )n +1 ˆ ÏÔ ¸Ô z n +1 R(z = 2i) = Ì = ˝ Á ˜ ÓÔ ( z + 2)( z + 2i ) ˛Ôz = 2i Ë 4i(2 + 2i ) ¯ 2 n (1 - i )i n 2n i n = = 8 4(1 + i ) Similarly

R(z = – 2i) =

2 n (1 + i )(-i )n 8

Transforms and Partial Differential Equations

5-34

\

ÏÔ ¸Ô z2 1 2n n +1 Z –1 Ì = (– 2) + (1 – i) ˝ 2 8 8 ÓÔ ( z + 2)( z + 4) ˛Ô

np np ˆ Ê ÁË cos 2 + i sin 2 ˜¯

+

np np ˆ 2n Ê (1 + i) Á cos - i sin ˜ Ë 2 2 ¯ 8

=

1 2 n +1 Ê np np ˆ + sin ˜ (– 2)n + 1 + cos 8 8 ÁË 2 2 ¯

= -

1 1 np np ˆ Ê (– 2)n + ◊ 2 n Á cos + sin ˜ 4 Ë 4 2 2 ¯

Example 14 Ï ¸ z Find Z –1 Ì 2 ˝ , by the method of residues. Ó z + 2z + 2 ˛ z Ï ¸ zn By residue theorem, Z –1 Ì 2 at its ˝ = sum of the residues of 2 z + 2z + 2 Ó z + 2z + 2 ˛ singularities. zn The singularities of 2 are given by z2 + 2z + 2 = 0, i.e. z = – 1 + i and z + 2z + 2 – 1 – i which are simple poles.

Ê zn ˆ 1 R(z = – 1 + i) = Á = (– 1 + i)n ˜ 2i Ë z + 1 + i ¯ z =-1+i Ê zn ˆ 1 R(z = – 1 – i) = Á z + 1 - i ˜ =– (– 1 – i)n Ë ¯ z =-1-i 2i \

z 1 1 Ï ¸ Z –1 Ì 2 (– 1 + i)n – (– 1 – i)n ˝ = 2i 2i Ó z + 2z + 2 ˛ ( 2 )n = 2i

n n ÏÔÊ 3p 3p ˆ 3p 3p ˆ ¸Ô Ê + i sin ˜ ÌÁ cos ÁË cos 4 i sin 4 ˜¯ ˝ 4 4¯ ÔÓË Ô˛

( 2 )n 3np ⋅ 2i sin 2i 4 3np = ( 2 )n sin 4 =

Example 15 Form the difference equation by eliminating the arbitrary constants A and B from the relations (i) yx = A ⋅ 3x + B ⋅ 4x and (ii) yx = (Ax + B) (–2)x (i) yx = A⋅3x + 1 + B⋅4x (1) x+1 x+1 ∴ yx + 1 = A ⋅ 3 + B⋅4 (2) x+ 2 x+2 and yx + 2 = A ⋅ 3 + B⋅4 (3)

Z-Transforms and Difference Equations

5-35

Eliminating A and B from equations (1), (2) and (3), we get

i.e.

yx

3x

yx + 1

3 x +1

4 x +1 = 0

yx + 2

3x + 2

4x +2

yx 3 ¥ 4 yx +1 x

x

4x

1 3

yx + 2

1 yx 4 = 0 (or) y x +1

9 16

yx + 2

i.e.

y x (21 - 9) - (7 y x + 1 - y x + 2 ) = 0

i.e.

y x + 2 - 7 y x + 1 + 12 y x = 0 y x = ( Ax + B)(- 2) x

(ii)

y x + 1 = ( Ax + A + B)(- 2) x + 1

or

yx + 1

and

y x + 2 = ( Ax + 2 A + B)(- 2) x + 2

= ( Ax + A + B)(- 2) x

yx + 2

= ( Ax + 2 A + B)(- 2) x 4 Now, (1) + (3) – 2 × (2) gives or

yx + i.e.,

yx + 2 4

-2 ¥

yx + 1 -2

9 7

(1)



-2

1 0 3 1 =0

(2)

(3)

=0

yx + 2 + 4 yx + 1 + 4 yx = 0

Example 16 From the difference equation by eliminating the arbitrary constants a and b from the relations (i) yn = a cos nq + b sin n and (ii) yn = an2 + bn (i) yn = a cos nq + b sin nq (1) ∴ yn + 1 = a cos (n + 1)q + b sin (n + 1)q (2) and yn + 2 = a cos (n + 2)q + b sin (n + 2)q (3) yn+2 + yn = a [cos (n + 2) q + cos nq] + b[sin (n + 2) q + sin n q] = 2 a cos (n + 1) q cos q + 2b sin (n + 1) q cos q = 2 cos q ◊ yn + 1 ∴ The required D.E, is yn + 2 – 2yn +1 cos q + yn = 0 (ii) yn = an2 + bn (1) Dyn = yn + 1 – yn = a{(n + 1)2– n2} + b = a(2n + 1) + b (2)

Transforms and Partial Differential Equations

5-36

D2yn = 2a From (3),

a=

(3)

1 2 D yn 2

(4)

(2 n + 1) 2 D yn 2 Using (4) and (5) in (1), we get Using (4) in (2), b = Dyn -

yn =

n2 2 (2 n + 1) 2 ¸ Ï D yn + n Ì Dyn D yn ˝ 2 2 Ó ˛

2 2 2yn = {n - n(2 n + 1}D yn + 2 n Dyn

i.e. i.e.

(5)

n(n + 1)D 2 yn - 2 n Dyn + 2 yn = 0

Example 17 Show that n straight lines, no two of which are parallel and no three of which meet in 1 a point, divide a plane into (n2 + n + 2) parts. 2 Let yn denote the number of sub-regions formed by n straight lines. When the (n + 1)th lines is drawn, it will intersect each of the previous n lines at n points and hence generate (n + 1) more sub-regions in addition to the previous yn sub-regions. \ yn + 1 = yn + (n + 1) i.e. Dyn = n + 1 = n[1] + 1 yn = D - 1 n[1] + D - 1 (1)

\

n[2] + n[1] + c 2 1 i.e. yn = n (n - 1) + n + c 2 Clearly, when n = 1, yn = 2 Using this in (1), we get 2 = 1 + c ∴ c = 1 1 ∴ yn = (n2 + n + 2) 2 =

(1)

Example 18 Form the difference equation satisfied by the nth order determinant

Dn =

1 + x2

x

0

x

1 + x2

x

0

x

1+ x







2

0



0

0



0

x

 0   

Z-Transforms and Difference Equations

5-37

Expanding Dn in terms of elements of the first row, 1 + x2 Dn = (1 + x 2 )

x 1+ x

x 0 

0

0

x

0

 

x

0 0

-x

1 + x2 x  0    

x  1

= (1 + x 2 )Dn -1 - x 2

0 2

1

0

1+ x

2

x

0

x

1+ x







2



0



0

0

x 1+ x

0 

0 2

x 1+ x 

x 

2



0



0

 0  

 0  

= (1 + x 2 )Dn - 1 - x 2 Dn - 2 ; viz., Dn - (1 + x 2 )Dn - 1 + x 2 Dn - 2 = 0 Changing n into (n + 2), the required difference equation becomes Dn + 2 - (1 + x 2 )Dn + 1 + x 2 Dn = 0

Example 19 p

cos nq dq , where n is an integer and 0 < a < π, form the difference cos q - cos a 0

If In = Ú

equation satisfied by In. p

In =

cos n q

Ú cos q - cos a dq 0

p

In + 2 =

cos (n + 2)q

Ú cos q - cos a dq 0

p

Now

In + 2 + In =

cos (n + 2)q + cos n q dq cos q - cos a 0

Ú

p

=

2 cos (n + 1)q cos n q dq cos q - cos a 0

Ú

p

=

2 cos (n + 1)q {(cos q - cos a ) + cos a} dq cos q - cos a 0

Ú

p

p

0

0

= 2 Ú cos (n + 1) q dq + 2 cos a

cos (n + 1)q

Ú cos q - cos a dq

p

Ï sin (n + 1)q ¸ = 2Ì ˝ + 2 cos a ◊ I n +1 Ó n + 1 ˛0 \ The required difference equation is In + 2 – 2 cos a ◊ In + 1 + In = 0.

Transforms and Partial Differential Equations

5-38

Example 20 Solve the difference equation y(n + 3) – 3y(n + 1) + 2y(n) = 0, given that y(0) = 4, y(1) = 0 and y(2) = 8 Taking Z-transforms on both sides of the given equation, we have Z{y(n + 3)} – 3Z{y(n + 1) + 2Z{y(n)} = 0 i.e. {z3 y (z) – z3 y(0) – z2 y(1) – zy(2)] – 3 {z y (z) – zy(0)] + 2 y (z) = 0 (z3 – 3z + 2) y (z) = 4z3 – 4z

i.e. \

\

A B C y (z) 4z2 - 4 + + = = 2 2 z - 1 ( z - 1) z+2 z ( z - 1) ( z + 2) 8/3 4/3 = + z -1 z + 2 8 z 4 z y (z) = + 3 z -1 3 z + 2 8 4 Inverting, we get y(n) = + (– 2)n 3 3

Example 21 Solve the equation f (n) + 3f (n – 1) – 4f (n – 2) = 0, n ≥ 2, given that f(0) = 3 and f(1) = – 2. Changing n into (n + 2) in the given equation, it becomes f (n + 2) + 3f (n + 1) – 4 f(n) = 0, n ≥ 0. Taking Z-transforms of this equation, we have [z2 f ( z ) – z2 f (0) – z f (1)] + 3[z f ( z ) – z f (0)] – 4 f ( z ) = 0 i.e.

(z2 + 3z – 4) f ( z ) = 3z2 + 7z

\ \ \

3z + 7 1 2 f (z) = = + ( z + 4)( z - 1) z + 4 z - 1 z z 2z f (z) = + z + 4 z -1 f (n) = (– 4)n + 2

Example 22 Solve the equation yn + 2 – 7yn + 1 + 12yn = 2n, given that y0 = y1 = 0.

note yn means y(n). Taking Z-transforms of the given equation, [z2 y (z) – z2 y(0) – zy(1)] – 7[z y (z) – zy(0)] + 12 y (z) = i.e.

z z-2

(z2 – 7z + 12) y (z) =

z z-2

Z-Transforms and Difference Equations

5-39

1/ 2 1 1/ 2 y (z) 1 + = = z-2 z-3 z-4 z ( z - 2)( z - 3)( z - 4) 1 z z 1 z + \ y (z) = 2 z-2 z-3 2 z-4 Inverting, we get yn = 1/2 ⋅ 2n – 3n + 1/2 ⋅ 4n \

Example 23 Solve the equation xn + 2 – 5xn + 1 + 6xn = 36, given that x0 = x1 = 0. Taking Z-transforms of the given equation, z [z2 x (z) – z2 x(0) – z x(1)] – 5[z x (z) – zx(0)] + 6 x (z) = 36 z -1 36 z i.e. (z2 – 5z + 6) x (z) = z -1 i.e. \ Inverting, we get

36 x (z) 18 36 18 = = + ( z - 1)( z - 2)( z - 3) z z -1 z - 2 z - 3 z z z x ( z ) = 18 - 36 + 18 z -1 z-2 z-3 xn = 18 – 36 ⋅ (2)n + 18 ⋅ (3)n

Example 24 Solve the equation y(x + 2) + 4y(x + 1) + 4y(x) = x, given that y(0) = 0 and y(1) = 1.

note The letter ‘x’ is used instead of ‘n’ Taking Z-transforms of the given equation, [z2 y (z) – z2 y(0) – zy(1)] + 4[z y (z) – zy(0)] + 4 y (z) =

i.e.

È z ˘ Í∵ Z ( x ) ∫ Z (n) = ˙ ( z - 1)2 ˚ Î z (z2 + 4z + 4) y (z) = z + ( z - 1)2

z ( z - 1)2

y (z) 1 1 = + 2 2 z ( z + 2) ( z - 1) ( z + 2)2

\

=

1 A B C D + + + + 2 2 z - 1 ( z - 1) z + 2 ( z + 2)2 ( z + 2)

y (z) = –

\

2 z 1 z 2 z 10 z + ◊ + ◊ + 2 27 z - 1 9 ( z - 1) 27 z + 2 9 ( z + 2)2

Inverting, we get y(n) = –

2 2 1 5 (– 2)n – n (– 2)n + n+ 27 27 9 9

Transforms and Partial Differential Equations

5-40

Example 25 Solve the equation yn + 2 + yn = 2n ⋅ n Taking Z-transforms of the given equation, [z2 y (z) – z2 y(0) – zy(1)] + y (z) = Z{n ⋅ 2n} È az ˘ = Í∵ Z (na n ) = ˙ ( z - 2) ( z - a )2 ˚ Î Since y(0) and y(1) are not given, we assume that y(0) = A and y(1) = B. 2z \ (z2 + 1) y (z) = Az2 + Bz + ( z - 2)2 =

y (z) = A

\ Taking inverses

y(n) = A cos

z2 z +1 2

2z

2

+B

z z +1 2

+

2z ( z + 1) ( z - 2)2 2

ÏÔ 2z Ô¸ np np + B sin + Z –1 Ì 2 2˝ 2 2 ÓÔ ( z + 1)( z - 2) ˛Ô

2z ÔÏ Ô¸ Z –1 Ì 2 = Sum of the residues of 2˝ ÓÔ ( z + 1)( z - 2) ˛Ô

(1)

È ˘ 2zn Í 2 2 ˙ at its poles. ÎÍ ( z + 1)( z - 2) ˚˙

È ˘ 2zn i n -1 (3 + 4i )i n -1 = = R(z = i) = Í ˙ 2 25 ÍÎ ( z + i )( z - 2) ˙˚ z = i 3 - 4i = Similarly

3 È p p˘ 4 È np np ˘ cos(n - 1) + i sin (n - 1) ˙ + cos + i sin Í Í 25 Î 2 2 ˚ 25 Î 2 2 ˙˚ R(z =− i) =

3 4 [cos (n − 1) π/2 − i sin (n − 1) π/2] + 25 25

È d ÏÔ 2 z n ¸Ô˘ R(z = 2) = Í Ì 2 ˝˙ ÎÍ d z ÓÔ z + 1 ˛Ô˚˙ z = 2

np np ˘ È ÍÎcos 2 - i sin 2 ˙˚ ( z = 2 is a double pole)

È ( z 2 + 1)2 nz n -1 - 2 z n .2 z ˘ = Í ˙ ( z 2 + 1)2 ÍÎ ˙˚ z = 2 5.2 n .2 n -1 - 4.2 n +1 5n - 8 n = .2 25 25 Using these values in (1), we get np 6 np p 8 np (5n - 8) n + y(n) = A cos + B sin cos (n − 1) + cos 2 + 2 25 2 2 25 2 25 =

= A cos

np np 6 np 8 np (5n - 8) n + B sin sin cos 2 + + + 2 2 25 2 25 2 25

Z-Transforms and Difference Equations = C1 cos

5-41

np np (5n - 8) n + C2 sin 2 + 2 2 25

Example 26 Solve the simultaneous difference equations xn + 1 = 7xn + 10yn; yn + 1 = xn + 4yn given that x0 = 3 and y0 = 2. Taking Z-transforms of both the equations, z · x (z) – z · x(0) = 7 x (z) + 10 y (z) and z y (z) – zy(0) = x (z) + 4 y (z) i.e. (z – 7) x − 10 y = 3z and x − (z − 4) y = 2z

(1) (2)

Solving (1) and (2), we get x (z) 3z + 8 y (z) 2 z - 11 and = = z ( z - 2)( z - 9) z ( z - 2)( z - 9) x (z) =

\

-2 z 5z z z + and y (z )= + z-2 z-9 z-2 z-9

Inverting, we get x(n) = −2n + 1 + 5 · 9n and y(n) = 2n + 9n

Exercise 5(b) Part A (Short-Answer Questions) 1. Write down the formula for finding Z−1 { f (z)} using Cauchy’s residue theorem. 2. Express Z{ f (n + 2)} and Z{ f (n + 3) in terms of f (z). Ï z ¸ 3. Find Z−1 Ì ˝ by expansion method. Óz - a˛ Ï 1 ¸ 4. Find Z−1 Ì ˝ by expansion method. Óz +2˛ 5. Find Z−1 (ea/z) by expansion method. 6. Find Z−1 {e−2z} by expansion method. 7. Find Z−1 {log (1 – z–1 – 6z–2)} by expansion method. z-a¸ Ï 8. Find Z−1 Ìlog ˝ by expansion method. z+b˛ Ó 9. Find f (z) from the equation f (n + 2) – 3 f (n + 1) + 2 f (n) = 0, if f(0) = 0 and f(1) = 1. 10. Find f (z) from the equation f (n + 1) − 2f (n) = U(n), if f (0) = 0.

Transforms and Partial Differential Equations

5-42

Part B Find the inverse Z-transforms of the following functions by the long division method. 4z 11. ( z - 1)2 12. 13. 14.

z2 + z ( z - 1)2 z ( z - 1)( z - 2) 2z2 + 4z ( z - 2)3

Find the inverse Z-transforms of the following functions by the partial fractions method. 8z 2 15. 8z 2 - 6 z + 1 16.

17. 18.

19. 20. 21. 22.

23. 24.

25.

z+2 z - 5z + 6 2

z 2 - 3z z - 3z + 10 2

z2 + 2z ( z - 1)( z - 2)( z - 3) z -2 -1

(1 - z )(1 - 2 z -1 )(1 - 3z -1 ) 2z2 - z ( z - 1)( z - 2)2 z ( z + 1)( z - 1)2 z -2 (1 + z -1 )2 (1 - z -1 ) z z - 2z + 2 2

z2 + z ( z - 1)( z 2 + 1) z2 - 3 ( z + 2)( z 2 + 1)

Z-Transforms and Difference Equations

5-43

Find the inverse Z-transforms of the following functions by the method of residues. z 26. 2 z + 7 z + 10 27. 28. 29. 30.

31.

32. 33. 34. 35. 36.

37.

38.

39. 40.

z-4 z + 5z + 6 2

z2 - 4z ( z - 2)2 4z2 - 2z ( z - 1)( z - 2)2 2z ( z - 1)( z 2 + 1) z2 + 1 z - 2z + 2 2

z( z 2 - 1) ( z 2 + 1)2 Form the difference equation by eliminating the constants a and b from the relation yx = a ◊ 2x + b ◊ (–2)x. Form the difference equation by eliminating the constants a and b from the relation yn = (a – bn) ◊ 3n Form the difference equation by eliminating a and b from yn = an2 – bn. Form the difference equation by eliminating A and B from np np ˆ Ê yn = ( 2 )n Á A cos + B sin . Ë 4 4 ˜¯ Form the difference equation by eliminating a and b from px pxˆ Ê y x = 2 x Á a cos + b sin Ë 3 3 ˜¯ n circles are drawn in a plane so that each circle intersects all others and no three meet in a point. Prove that the plane is divided into (n2 – n + 2) parts, by forming the difference equation satisfied by the number yn of sub-regions obtained. A solid is so constructed that every face is a triangle. Form the difference equation satisfied by yn, the number of faces for n vertices. Form the difference equation satisfied by Dn, where Dn is the nth order determinant given by 2 cos q 1 0 0  0 1 2 cos q 1 0  0 Dn = 0 1 2 cos q 1  0  0

 0

and hence find the value of Dn.

 0

   0  2 cos q

Transforms and Partial Differential Equations

5-44

41. Find the difference equation satisfied by p

In =

cos n x

Ú 5 - 3 cos x dx 0

and hence find In. 42. Find the difference equation satisfied by p

In =

cos n q - cos n a dq cos q - cos a 0

Ú

and hence find In. Solve the following difference equation, using Z-transforms. 43. y(n + 1) – y(n) = 3n, given that y(0) = 0. 44. y(n + 2) + y(n) = 0. 45. y(n + 3) – 6y(n + 2) + 12y(n + 1) – 8y(n) = 0, given that y(0) = −1, y(1) = 0 and y(2) = 1. 46. y(n) + y(n – 1) – 8y(n – 2) – 12y(n – 3) = 0, n ≥ 3, given that y(0) = 1, y(1) = –6, y(2) = 12. 47. y(n + 2) – 4y(n + 1) + 4y(n) = 0, given that y(0) = 1 and y(1) = 0. 48. y(n + 2) – y(n + 1) – 2y(n) = 4, given that y(0) = – 1 and y(1) = 3. 49. y(n + 2) – 4y(n + 1) + 4y(n) = π, given that y(0) = 0 and y(1) = 0. 50. y(n + 2) – 3y(n + 1) + 2y(n) = 2n, given that y(0) = 3 and y(1) = 6. 51. y(n + 2) + 6y(n + 1) + 9y(n) = 2n, given that y(0) = 0 and y(1) = 0. 52. y(n + 2) – y(n) = 2n, given that y(0) = 0 and y(1) = 1. 53. y(n + 2) – 5y(n + 1) + 6y(n) = 4n, given that y(0) = 0 and y(1) = 1. 54. y(n + 2) – 4y(n) = 2n, given that y(0) = 0 and y(1) = 0. 55. y(n + 2) – 5y(n + 1) + 6y(n) = 5n, given that y(0) = 0 and y(1) = 0. 56. y(n + 2) + 2y(n + 1) + y(n) = n, given that y(0) = 0 and y(1) = 0. 57. y(n + 2) – 5y(n + 1) + 6y(n) = n, given that y(0) = 0 and y(1) = 0. 58. y(n + 2) – 5y(n + 1) + 6y(n) = n(n − 1), given that y(0) = 0 and y(1) = 0. 59. y(n + 2) – 4y(n + 1) + 3y(n) = 2n · n2, given that y(0) = 0 and y(1) = 0. 60. x(n + 2) – y(n) = 1 and y(n + 2) – x(n) = 1, given that x(0) = 0 and y(0) = –1.

Answers Exercise 5(a) 13.

2 ; z

14.

z z + ; z-a z-b

15.

az ; z-b

Z-Transforms and Difference Equations

16.

az ; z+a

3 ¸ Ï 2 17. z Ì + ˝; Ó 2 z - 1 3z - 1 ˛ 4 ¸ Ï 3 + 18. z Ì ˝; z 2 z +1˛ Ó z

19. eb . 20. e2T . 21. 22.

;

z - e aT z

;

z - e2T

a ( z - a )2 z ( z - 1)3

; ;

Ê z ˆ 23. (1 – z) log Á ; Ë z - 1˜¯ 24.

25. 26. 27.

3z 2 z - z +1 2

(

z z -1

; 2

);

z - z 2 +1 2

z 2 sin T z 2 - 2 z cos T + 1

;

z 2 ( z - cos 2T ) z 2 - 2 z ◊ cos 2T + 1

– 1;

28.

zÊ 1 1 ˆ ; + 2 ÁË z - eT z - e -T ˜¯

29.

zÊ 1 1 ˆ ; 2 ÁË z - e2T z - e -2T ˜¯

30.

z2 Ê 1 1 ˆ ; 2 ÁË z - eT z - e -T ˜¯

31.

z( z 2 - 6 z + 6) ; 2(2 - z )(3 - z )(4 - z )

32.

z ; ( z - 2)( z - 3)( z - 6)

5-45

Transforms and Partial Differential Equations

5-46

33. 34.

35. 36. 37. 38. 39.

40.

41. 42.

z z - z -1 2

;

z( z + 2) 2( z 2 - 2 z + 2) z2 ( z + 2)( z 2 + 4) 5z ( z - 2)( z + 3)2 4 z3 - 8z 2 + 6 z ( z + 1)( z - 2)2

;

; ; ;

az sinh a

;

z - 2 az cosh a + a 2 2

z( z - 2 cosh 5) z - 2 z cosh 5 + 4 2

;

T 3 ze2T (z 2 e 4T + 4 ze2T + 1)

(ze2T - 1)4 8z ( z - 2)3 z ( z - 1)4

;

; ;

43.

1È Ê z ˆ ˘ 2 Í(1 - z ) log ÁË ˜ - z˙ ; 2Î z - 1¯ ˚

44.

1ˆ z Ê ; ÁË 2 - ˜¯ log z z -1

45. ea/z + e1/az; 46.

z 2 cos a - z cos(a 8 - a ) ; p z 2 - 2 z cos + 1 8

47.

zÈ 1 z -1 2 ˘ + 2 ; Í 2 Î z - 1 z - z + 1 ˚˙

48.

3 È z( z - 1 2 ) ˘ Í ˙ +1 4 Î z2 - z 2 + 1 ˚

49.

z( z - a cos a ) z - 2 za cos a + a 2 2

;

È z( z + 1 2 ) ˘ 4Í 2 ˙; Î z + z 2 + 1˚

Z-Transforms and Difference Equations

50.

51.

52.

53. 54.

(z3 - z )sin q

;

(z2 - 2 z cos q + 1)2 a (z 3 cos q - 2 az 2 + a 2 z cos q ) ; (z2 - 2az cos q + a2 )2 az (z 2 - a 2 )sin q

(z2 - 2az cosq + a2 )

2

;

z 2 cos f - zr cos (q - f ) z 2 - 2 zr cos q + r 2

;

z 2 sin f + r T z sin(w T - f ) z 2 - 2r T z cos w T + r 2T

;

55.

z 1 z( z - cos 4T ) + . 2 ; 2( z - 1) 2 z - 2 z cos 4T + 1

56.

3 z sin T 1 z sin 3T . 2 - . 2 ; 4 z - 2 z cos T + 1 4 z - 2 z cos3T + 1

57.

58.

59.

60.

Tz (z 2 - 1)sin T

(z2 - 2 z cos T + 1)2

;

T (z 3 cos T - 2 z 2 + z cos T )

(z2 - 2 z cos T + 1)2 ze -T sin 2T z 2 e -2T - 2 ze -T cos 2T + 1 ze2T (ze2T - cos3T ) z 2 e 4T - 2 ze2T cos3T + 1

;

61. 0; 1 62. (a)

z2 ( z - a )2

(b)

z2 ; ( z - 1)( z - 2)

63. (a)

z2 ; ( z - a )( z - b)

(b)

4 z 2 - 3z ; ( z - 1)(2 z - 1)(4 z - 1)

;

;

5-47

Transforms and Partial Differential Equations

5-48

64. (i)

(ii)

65. (i)

z2 ( z - 1)2 z3 + z2 ( z - 1)4

;

;

(-1)n {bn + 1 – an + 1; b-a

Ê 1ˆ (ii) Á ˜ Ë 2¯

n -1

n

Ê 1ˆ -Á ˜ . Ë 4¯

Exercise 5(b) 3. an; 4. (–2)n – 1; 5. 6. 7. 8. 9. 10.

an ; n! (-2)n ; n! 1 [(–1)n2n – 3n]; n 1 [(–b)n – an]; n z f (z) = 2 ; z - 3z + 2 f ( z) =

z ; ( z - 1)( z - 2)

11. 12. 13. 14. 15. 16.

4n; (2n + 1); 2n – 1; n 2 2 n; 21 – n – 4–n; 5 · 3n – 1 – 4.2n – 1; 2 n 5 17. ·5 + · (–2)n; 7 7 3 5 18. - 4 ◊ 2 n + 3n ; 2 2 1 1 - 2 n + 3n ; 2 2 3 20. 1 – 2n + n · 2 n; 2 19.

Z-Transforms and Difference Equations

21. 22. 23.

1 1 n (–1)n – + ; 4 4 2 n 1 1 - (–1)n + (–1)n; 2 4 4 ( 2 )n sin np ; 4

24. 1 – cos 25. 26. 27. 28. 29. 30.

-

np ; 2

1 1 np 1 np sin (–2)n + cos ; 10 10 2 20 2

1 [(–2)n – (–5)n]; 3 7 3 · (–2)n – · (–3)n 3 (1 – n) · 2n; 2 + (3n – 2) · 2n; np np ˆ Ê 1 – Á cos + sin ˜ Ë 2 2 ¯

Ê 1 - 3i ˆ n Ê 1 + 3i ˆ n ÁË ˜ (1 + i ) + ÁË ˜ (1 - i ) ; 4 ¯ 4 ¯ n n–1 32. [i + (–i)n – 1]; 2 33. yx + 2 – 4yx = 0 31.

34. yn + 2 –6yn + 1+9yn = 0 35. (n2 + n) D2yn – 2nDyn + 2yn= 0 36. yn + 2 – 2yn +1 + 2yn = 0 37. yx + 2 – 2yx +1 + 4yx = 0 38. yn + 1 = yn + 2n 39. yn + 1 = yn + 2 40. Dn + 2 – 2 cos q Dn + 1 + Dn = 0; Dn = n

41. In =

p Ê 1ˆ ◊ 4 ÁË 3 ˜¯

42. In =

p sin na sin a

1 n (3 – 1); 2 np np 44. y(n) = A cos + B sin ; 2 2 43. y(n) =

sin(n + 1)q sin q

5-49

5-50

Transforms and Partial Differential Equations

11 3 2¸ Ï 45. y(n) = 2n Ì-1 + n - n ˝ ; 8 8 ˛ Ó 46. y(n) = -

8 n 33 6 .3 + .(-2)n + n(-2)n ; 25 25 5

y(n) = 2n (1 – n); y(n) = – 2 – (–1)n + 2 · 2n; y(n) = π [1 + (n – 2) · 2n – 1]; y(n) = 1 + 2n + 1 + n · 2n – 1; 1 n 1 1 .2 - .(-3)n + n · (–3)n; 51. y(n) = 25 25 5 47. 48. 49. 50.

1 n [2 – (–1)n] 3 1 n 53. y(n) = (4 – 3n); 2

52. y(n) =

1 {(–2)n – 2n + n · 2n + 1} 16 1 n 1 n 55. y(n) = .2 - .3 ; 3 2 54. y(n) =

1 (n – 1) {1 + (–1)n – 1}; 4 n 1 3 Ê 1ˆ n n 57. y(n) = ·3 –2 + Á ˜ + Ë 2¯ 4 4 1 58. y(n) = . 3n – 2n + 1 + 3/2 + n; 2 59. y(n) = 3 + 5 · 3n – 2n (n2 + 8); 56. y(n) =

60. x(n) = 1 ÊÁ 1 - cos np - sin np ˆ˜ and y(n) = 1 ÊÁ sin np - cos np - 1ˆ˜ ¯ 2Ë 2 2 ¯ 2Ë 2 2

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Appendix A

A-3

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D-6

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D-8

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Appendix D

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Transforms and Partial Differential Equations

D-10

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53185 B.E./B.Tech. Degree Examinations, November/December 2010 Regulations 2008 Third Semester Common to all branches MA 2211 Transforms and Partial Differential Equations Time: Three Hours

Maximum: 100 Marks Answer ALL Questions Part A – (10

2 = 20 Marks)

1. Find the constant term in the expansion of cos2 x as a Fourier series in the interval (– , ). 2. Find the root mean square value of f (x) = x2 in (0, l). 3. Write the Fourier transform pair. 4. Find the Fourier sine transform of f (x) = e–ax, a > 0. 5. Form the partial differential equation by eliminating the arbitrary function from z2 – xy = f

.

6. Find the particular integral of (D2 – 2DD + D 2) z = ex – y. 7. Write down the three possible solutions of one dimensional heat equation. equation. 10. Form a difference equation by eliminating the arbitrary constant A from yn = A 3 n. Part B – (5

16 = 80 Marks)

11. (a) (i) Find the Fourier series expansion of f (x) = Also, deduce that =

(10)

(ii) Find the Fourier series expansion of f (x) = 1 – x2 in the interval (–1, 1). (6) OR

SQP1.2

Transforms and Partial Differential Equations

11. (b) (i) Obtain the half range cosine series for f (x) = x in (0, ). (8) (ii) Find the Fourier series as far as the second harmonic to represent the function f (x) with period 6, given in the following table. x 0 1 2 3 4 5 f (x) 9 18 24 28 26 20 (8) 12. (a) (i) Derive the Parseval’s identity for Fourier Transforms. (8) (ii) Find the Fourier integral representation of f (x

f (x) =

(8)

OR 12. (b) (i) Find the Fourier sine transform of f (x) =

(ii) Evaluate

(8)

using Fourier cosine transforms of

e–ax and e–bx. (8) 13. (a) (i) Form the PDE by eliminating the arbitrary function from (x2 + y2 + z2, ax + by + cz) = 0. (8) 2 2 (ii) Solve the partial differential equation x (y – z) p + y (z – x) q = z2 (x – y). (8) OR 13. (b) (i) Solve the equation [D3 + D2D – 4DD 2 – 4D 3]z = cos (2x + y). (8) (8) (ii) Solve [2D2 – DD – D 2 + 6D + 3D ]z = xey. 14. (a) A tightly stretched string of length 2l is fastened at both ends. The midpoint of the string is displaced by a distance ‘b’ transversely and the string is released from rest in this position. Find an expression for the transverse displacement of the string at any time during the subsequent motion. (16) OR 14. (b) A square plate is bounded by the lines x = 0, y = 0, x = 20 and y = 20. Its faces are insulated. The temperature along the upper horizontal edge is given by u(x, 20) = x(20 – x), 0 < x < 20 while the other two edges are kept at 0°C. Find the steady state tem perature distribution in the plate. (16)

Solved Question Papers-1

15. (a) (i) Find the Z Z

SQP1.3

n and sin n . Hence deduce the n + 1) and an sin n . (10)

(ii) Find the inverse Z

by residue method. (6)

OR 15. (b) (i) Form the difference equation from the relation yn = a + b 3n. (8) (ii) Solve yn + 2 + 4yn + 1 + 3yn = 2n with y0 = 0 and y1 = 1, using Z transform. (8)

Solutions Part – A 1. a0 = constant term in the F.S. expansion =

.

2. e–isx dx and

3. F{f (x)} = F –1

eixs ds

= f (x) =

4. Fs(e–ax) =

e–ax sin sx dx = =

5. z2 – xy =

(1)

Differentiating (1) partially w.r.t. x, 2zp – y = f

(2)

Differentiating (1) partially w.r.t. y; 2zq – x = f (2)

(3) gives

(3)

SQP1.4

Transforms and Partial Differential Equations

i.e., – 2pqxz + xyq = 2z2q – 2pqxz – xz + x2p i.e., x2p + (2z2 – xy)q = xz ex – y =

6. P.I.

ex–y

7. Three possible solutions of the equations ut = 2uxx are 2 2 (i) u(x, t) = (Ae px + Be –px) e p t 2 2 (ii) u(x, t) = (A cos px + B sin px) e– p t and (iii) u(x, t) = Ax + B, where A, B, p are arbitrary constants 8. Three possible solutions of the equation uxx + uyy = 0 are (i) u(x, y) = (Aepx + Be–px) (C cos py + D sin py) (ii) u(x, y) = (A cos px + B sin px) (Cepy + De–py) (iii) u(x, y) = (Ax + B) (Cy + D), where A, B, C, D, p are arbitrary constants 9. The unit step sequence un Z{un} =

un = (or)

10. yn = A 3n ...(1); yn + 1 = 3A 3n ...(2) (2)

(1) gives

= 3 (or) yn + 1 – 3yn = 0. Part – B

11. (a) (i) Let the F.S. of f (x) be (0, 2 ) an = = =

=

a0 = =

=

cos nx +

sin nx, in

Solved Question Papers-1

SQP1.5

bn = =

=

=0

F.S. of f (x) ~

cos nx in (0, 2 )

Putting x = 0, that is a point of continuity of f (x), we get =0

11. (a) (ii) f (x) = 1 – x2 is an even function in (– 1, 1). F.S. of f (x) will be of period 2 and will not contain sine terms. Let the F.S. of f (x) be an = 2

cos n x dx =

(– 1)n or

= 2

a0 = 2

(1 – x2) dx = 2

Required F.S. is

.

cos n x in (–1, 1)

11. (b) (i) Given an even extension for f (x) in (– , 0). i.e. put f (x) = – x in (– , 0) Then f (x) is even in (– , ) F.S. of f (x) ~

cos nx in (– , )

SQP1.6

Transforms and Partial Differential Equations

an =

= a0 =

x is

cos nx in (0, )

11. (b) (ii) x : x0 = 0, x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5 y = f (x) : y0 = 9, y1 = 18, y2 = 24, y3 = 28, y4 = 26, y5 = 20 2l = 6 l = 3

Let the required F.S. be y =

a0 = 2

+

125 = 41.67

a1 = 2 = =

(– 19 – 12

0.5) = – 8.33

b1 = 2 =

–4

0.866 = – 1.15

Solved Question Papers-1

SQP1.7

a2 = = =

[37 – 88

0.5] = – 2.33

b2 = =

=0

The required F.S. is 20.84 – cos

– 2.33

.

12. (a) (i) (The derivation is available in Q.10 in page 4.32 of the book.) 12. (a) (ii) (This problem is the worked example (1) in page 4.8 of the book.)

=

=

(2 sin s – sin 2s) (or)

12. (b) (ii) Let f (x) = e–ax and g(x) = e–bx Then

=

and

SQP1.8

Transforms and Partial Differential Equations

f (x) g(x) dx =

e–(a + b) x dx =

=

1 1.84 of the book]

14. (a) This is the worked example (2) given in page 3.13 of the book.]

We have to solve (1) satisfying the following boundary conditions: u(0, y) = 0, 0

y

20

(2)

u(20, y) = 0, 0

y

20

(3)

u(x, 0) = 0, 0

x

20

(4)

u(x, 20) = x(20 – x), 0

x

20

(5) y = 20, the

proper solution of (1) is u(x, y) = (A cos px + B sin px) (C cosh py + D sinh py) Using (2) in (6); A = 0

(6)

Solved Question Papers-1

SQP1.9

Using (3) in (6); B sin 20 p = 0 B

p=

0

(n = 0, 1, 2, ... )

Using (4) in (6); C = 0 The required solution becomes u(x, y) =

, n = 0, 1, 2, ...

sin

The general solution is u(x, y) =

(7)

Using (5) in (7), we have (

n

sinh n ) sin

= x(20 – x) in 0 =

n

bn sin

x

(Fourier H.R. series)

sinh n = bn

=

=

=

=

n

=

u(x, y) =

cosech n sin

20

sinh

SQP1.10

Transforms and Partial Differential Equations

15. (a) (i) Consider Z(ein ) = Z{(ei )n} = i.e., Z(cos n + i sin n ) = = Equating the R.P.’s; Z (cos n ) = Equating the I.P.’s; Z (sin n ) = Now Z {cos (n + 1) ] = cos

Z (cos n ) – sin

Z (sin n )

= Z {an f (n)} = Z {an sin n } =

=

=

15. (a) (ii)

f(n) = Sum of the residues of larities. = Res. of

at z = 1, which is a triple pole

= = = = n2

at the isolated singu

{(n + 1)n + n(n – 1)}

Solved Question Papers-1

SQP1.11

15. (b) (i) yn = a + b 3n (1); yn + 1 = a + 3b 3n (2); yn + 2 = a + 9b 3n (3); yn + 1 – yn = 2b 3n (4); yn + 2 – yn + 1 = 6b 3n (5); = 3 i.e., yn + 2 – 4yn + 1 + 3yn = 0 is

From (4) and (5); the required difference equation. 15. (b) (i) Taking Z [z2

(z) – z2 y(0) – z

y(1)] + 4 [z

= i.e., (z2 + 4z + 3)

(z) =

+ z or

(z) =

Now (z) = Inverting, the required solution is yn =

(z) – z

y (0)] + 3 (z)

K 406 B.E./B.Tech. Degree Examinations, November/December 2010 Third Semester Civil Engineering MA 31—Transforms and Partial Differential Equations Common to all branches Regulations 2008 Time: Three Hours

Maximum: 100 Marks Answer ALL Questions Part A – (10

2 = 20 Marks)

1. State Dirichlet’s conditions for a function f (x) to be represented as a Fourier series in an interval (0, 2 ). 2. Write down the form of the Fourier series of an odd function in (– l, l) and 3. State the Fourier integral theorem. 4. Let F (s) denote the Fourier transform of f (x). Show that F [f (ax)] =

.

5. Form the partial differential equation by eliminating the arbitrary function f from z = f (xy). . 6. Find the complete integral of z = px + qy + 2 7. What does a ut = a2 uxx? 8. The ends A and B of a rod of length 10 cm long have their temperature kept 20°C and 80°C. Find the steady state temperature distribution on the rod. 9. Find the Z-transform of n2. 10. Form a difference equation by eliminating arbitrary constants from yn = A + B 2 n. Part B – (5

16 = 80 marks)

11. (a) (i) If f (x) = hence deduce the value of

f (x) and (8)

SQP2.2

Transforms and Partial Differential Equations

sine terms in the Fourier series of y = f (x following data in (0, 2 ): (8) x: 0 /3 2 /3 4 /3 5 /3 2 f (x): 1.98 1.30 1.05 1.30 –0.88 –0.25 1.98 Or (b) (i) Find the Fourier series for the function f (x) = | x |, – l < x < l. Hence –2 + 3–2 + 5–2 + ... (8) (ii) Find the half-range sine series for a function f (x) = x ( – x), 0 < x < . Hence deduce

=

3

/32.

12. (a) Find the Fourier transform of f (x) =

(8)

and also

the inverse transform. Hence deduce that (16) Or (b) Find the Fourier sine and cosine transform of e–x. Hence evaluate (i)

(ii)

and

.

(16)

13. (a) (i) Find the general solution of x (y – z) p + y (z – x) q – z (x – y) = 0. (8) (ii) Solve (D2 + DD – 6D 2) z = cos (x + 2y). (8) Or (b) (i) Find the singular solution of z = px + qy + p2 – q2. (8) (ii) Solve (D2 + 2DD + D 2) z = – x sin y. (8) 14. (a) A string is stretched and fastened to two points l apart. Motion is started by displacing the string into the form y = k (lx – x2) and then released it from this position at time t = 0. Find the displacement of the point of the string at a distance x from one end at any time t. (16) Or (b) A long rectangular plate with insulated surface is l cm wide. If the temperature along one short edge y = 0 is u (x, 0) = k for 0 < x < l, while the two long edges

Solved Question Papers-2

x = 0 and x = l steady state temperature function u(x, y). 15. (a) (i) Find the Z-transform of (n + 1)2 and sin (3n + 5). (ii) Find the inverse Z-transform of

SQP2.3

(16) (4 + 4)

.

(8)

Or Z-transform of .

(6)

Z-transform, yn + 2 + 4yn + 1 + 3yn = 3n, given that y0 = 0 and y1 = 1. (10)

Solutions Part – A 1. [Answer is available in page 2.2 of the book] c x c + 2l starting with (ii)] 3. [Answer is available in 4.2 given in page 4.1 of the book.] 4. [Answer is available in page 4.27 of the book under (2). Change of Scale 5. [This is the same as the problem (13) in page 1.20 of the book.] z = f (xy);

. 7. In the equation ut = the bar, equal to heat and

2

uxx,

2

represents the diffusivity of the material of

, where k is the thermal conductivity, c

is the density of the material of the bar.

8. The steady-state distribution is given by = 0 subject to u(0) = 20 and u(10) = 80. The general solution of the equation is u(x) = Ax + B. Using the given boundary conditions, we get A = 6 and B = 20. The required temperature distribution is given by u(x) = 6x + 20.

SQP2.4

Transforms and Partial Differential Equations

9. Z(n2) = Z(n n) =

=

= 10. yn = A + B 2n (1); yn + 1 = A + 2B 2n (2); yn + 2 = A + 4B 2n (3) Eliminating A and (B 2n) from (1), (2), (3) the required equation is = 0. i.e., yn + 2 – 3yn + 1 + 2yn = 0. Part – B 11. (a) (i) f (x) = (– x) = 1 + x = 2(x) f (x) is even in (– , ) and so the Fourier series of f (x) will not contain sine terms and will be of period 2 . Let F.S. of f (x) be an cos nx in (– , ) 1

an =

(1 + x) cos nx dx

=

=

a0 =

(1 + x) dx =

=2+ cos nx in (– , )

The required F.S. of f (x) is Putting x = 0, that is a point of continuity of f (x), =1

11. (a) (ii) The length of the interval = 2 and that of the sub-interval = .

The 6 values of y = f (x) should be considered at the left/right

end points of the sub-intervals.

Solved Question Papers-2

Accordingly, x : x0 = 0, x1 =

, x2 =

, x3 = , x4 =

SQP2.5

, x5 =

y : y0 = 1.98, y1 = 1.30, y2 = 1.05, y3 = 1.30, y4 = – 0.88, y5 = – 0.25

Let the required F.S. be y = a0 = 2

=

a1 = 2

cos xr

+ (a1 cos x + b1 sin x) + ...

4.50 = 1.50

= =

= 0.373

b1 = 2 =

yr sin xr = 3.48

0.866 = 1.005

The required F.S. of y = f (x) is 0.75 + 0.373 cos x + 1.005 sin x 11. (b) (i) f (x) = | x | is even in – l < x < l. The F.S. of f (x) will not contain sine terms and will be of period 2l. in (– l, l) Let the F.S. of f (x) be an =

=

=

(or)

SQP2.6

Transforms and Partial Differential Equations

a0 =

=l

|x|~

in (– l, l)

(1)

Putting x = 0, that is a point of continuity of | x |, in (1), 1–2 + 3–2 + 5–2 +

=

11. (b) (ii) Give an odd extension for f (x) in (– , 0) i.e., put f (x) = x ( + x) in (– , 0). Then f (x) has become an odd function in (– , ) F.S. of f (x) will contain only sine terms and will be of period 2 . bn sin nx in Let the required half-range F.S.S of f (x) be (0, ). bn =

x ( – x) sin nx dx

=

=

x ( – x) ~ Putting x =

sin nx in (0, )

(1)

, that is a point of continuity for x ( – x) in (1),

12. (a) [Major part of this problem is the Worked example (5) in page 4.37 of the book.] It is proved in the W.E. that F{f (x)} =

Solved Question Papers-2

F–1

SQP2.7

e ixs ds

=

= f (x) = Putting x = 0 (or) f (0) = 1, we get

ds = 1 i.e.,

= , on putting

The deduction

=t

is a part of the W.E.

12. (b) [This problem is the Worked example (6) in page 4.38 of the book, in which we have to put a = 1.] 13. (a) (i) x (y – z) p + y (z – x) q = z(x – y), which is a Lagrange Linear equation with P = x(y – z), Q = y(z – x) and R = z(x – y). The subsidiary equations are

(1)

Using the multipliers 1, 1, 1, each ratio in (1) = d (x + y + z) = 0 i.e., x + y + z = a is one independent solution of Lagrange’s subsidiary equations Using the multipliers

; each ratio in (1) = =0

Integrating, we get log (xyz) = log b or xyz = b is the second independent solution of Lagrange’s subsidiary equations General solution of the given equation is f (x + y + z, xyz) = 0, where f is an arbitrary function. 13. (a) (ii) (D2 + DD – 6D 2) Z = cos (x + 2y). This is a homogeneous linear equation of the second order. A.E. is m2 + m – 6 = 0 i.e., (m + 3) (m – 2) = 0 m = – 3, 2.

SQP2.8

Transforms and Partial Differential Equations

C.F. =

1

(y – 3x) +

2

P.I. = or

(y + 2x), where

1

,

2

are arbitrary functions.

cos (x + 2y) =

cos (x + 2y)

cos (x + 2y)

G.S. of the given equation is Z = C.F. + P.I. 13. (b) (i) z = px + qy + p2 – q2. This is a Clairaut’s type equation. C.S. is z = ax + by + a2 – b2 Differentiating (1) partially w.r.t. ‘a’, x + 2a = 0 Differentiating (1) partially w.r.t. ‘b’, y – 2b = 0 From (2), a =

(1) (2) (3)

; From (3), b =

Using these values of a and b in (1), we get , i.e., 4z = y2 – x2. This is the required

z=

singular solution. 13. (b) (ii) (D2 + 2DD + D 2) Z = – x sin y. This is a homogeneous linear equation of second order. A.E. is m2 + 2m + 1 = 0 or (m + 1)2 = 0 \ m = – 1, – 1. C.F. = 1(y – x) + x 2(y – x), where 1 and 2 are arbitrary functions. P.I. =

I.P. of (– xeiy) = I.P. of (– eiy)

(x)

= – I.P. of = I.P. of [e iy {1 + 2i (D + D ) + ...} (x)] = I.P. of [(cos y + i sin y) (x + 2i)] = x sin y + 2 cos y G.S. of the given equation is Z = C.F. + P.I. 14. (a) [This problem is the W.E. (1) in page 3.10 of the book. Only part (ii) of this W.E. provides the required solution.] 14. (b) This problem is the W.E. (1) in page 3.119 of the book. We have to change a into l and put f (x) = k

.

Steps (1) to (10) are the same as in the W.E. Using the boundary condition (5), namely, u(x, 0) = k

in step (10), we have (11)

Solved Question Papers-2

Comparing like terms in (11), we have = k; 3 = 3k; 2 = 4 = 5 = 6 = ... = 0. 1 Using these values of n in (10), the required solution is e–

u(x, y) = k sin

y/l

e–3 y/l.

+ 3k sin

15. (a) (i) Z{(n + 1)2} = Z(n2 + 2n + 1) = Z{sin (3n + 5)} = Z {sin 3n cos 5 + cos 3n sin 5} = (cos 5)

;

15. (a) (ii) Let

Z–1 i.e., f (n) = 15. (b) (i) Z–1

= Z–1

= (4n) * (5n) =

= 5n

= 5n + 1 – 4n + 1

15. (b) (ii) yn + 2 + 4yn + 1 + 3yn = 3n Taking Z-transforms of the given equation Z(yn + 2) + 4Z(yn + 1) + 3Z(yn) = Z(3n) i.e., [z2

(z) – z2y(0) – z y(1)] + 4[z (z) – zy(0)] + 3 (z) =

i.e., (z2 + 4z + 3)

(z) =

+ z or

SQP2.9

SQP2.10

Transforms and Partial Differential Equations

(z) =

or

Now (z) = Inverting, we get y(n) = solution.

, which is the required

Anna University of Technology, Coimbatore B.E./B.Tech. Degree Examinations : Nov/Dec 2010 Regulations : 2008 Third Semester 080100008-Transforms and Partial Differential Equations (Common to Civil/EEE/EIE/ICE/ECE/Biomedical/Biotech/ AERO/AUTO/CSE/IT/Mechanical/Chemical/FT/II/TC) Time 3 Hours

Max. Marks: 100 Part-A Part A – (20

2 = 40 Marks)

Answer All Questions 1. 2. 3. 4.

State the conditions for f(x) to have Fourier series expansion. Write a0, an in the expansion of x + x3 as Fourier Series in (– , ). Expand f(x) = 1 in a sine series in 0 < x < Find Root Mean Square value of the function f(x) = x in the interval (0, l). e–2x. f (x of eiax f (x) is F(S – a). a and b

z = (x2 + a) (y2 + b). z = f (x + t) + g (x – t).

q = 2px. 12. Solve (D3 – 3DD 2 + 2D 3)z = 0 uxx + 4uxy + uyy + 2ux – uy = 0. I y = y0 sin t

17. Find Z[e –an].

-

SQP3.2

Transforms and Partial Differential Equations

Z[n] = Z[ f (n + 1)] = zF(z) – zf (0)

Part – B (5

12 = 60 Marks)

Answer any Five Questions 21 (a). If f(x) =

in the interval

(0, 2 ). (8)

Hence deduce that f (x) = x in the interval (– , )

(4)

f (x) =

is

. Hence deduce that

23. (a) Solve (mz – ny)p + (nx – lz)q = ly – mx D3 + D2 D – DD 2 – D 3)z = e x cos 2y

y(x, t). 25. (a) Evaluate Z–1

(6) u(n + 2) – 5u(n + 1) + 6u(n) = 4n

u(0) = 0, u(1) = 1

(6) y = f (x

x f (x)

0 9

1 18

2 24 f (x) =

3 28

4 26

5 20

6 9

Solved Question Papers-3

SQP3.3

dx

u(x, t 28. (a) Solve p(1 + q) = qz

(6)

x = 0 at A. (6) Z–1

(6)

Solutions Part – A f (x) = x + x 3 is an

, odd function in (– , ). f (x) in (– , 0 and an = 0. 3. Give an odd extension for f(x) in – x 0) s + a2 2



4.

È e- ax ˘ cos sx dx = Í 2 (- a cos sx + s sin sx )˙ 2 Îs + a ˚0

F{ f ( x - a )} =

Ú

f ( x - a ) e - isx dx =

-•

= e- ias



Ú

f (t ) e - is( t + a ) dt , on putting t = x – a

-• •

Ú

f (t ) e- ist dt = e- ias F (s)

-•

[Note: If the definition is taken as F(s) = F{f(x) =



Úe

- isx

f ( x ) dx , the re-

-•

sulting given in the problem would have been obtained.] 5. z = (x2 + a2) (y2 + b2) .. (1) Differing (1) partially w.r.t. x and then w.r.t. y, we get p = 2x(y2 + b2) ... (2) and q = 2y(x2 + a2) ... (3) Substituting for (x2 + a2) and (y2 + b2) obtained from (3) and (2) in (1), we q p get the required P.D.E. as z = or pq = 4xyz ◊ 2y 2x 1 p 1 6. pq = x; i.e., = = a, say, \ p = ax and q = a x q

SQP5.4

Transforms and Partial Differential Equations

dz = p dx + q dy = ax dx + solution as z =

1 dy; Integrating both sides, we get the complete a

ax 2 1 + y + b  (1) 2 a

Putting b = f(a), (1) becomes = z =

a 2 1 x + y + f (a )  (2) 2 a

x2 1 - 2 y + f ¢(a)  (3) 2 a Eliminant of a from (2) and (3) is the G.S. of the given equation. Differentiating (2) w.r.t. a partially, 0 =

∂y 7. Boundary condition are (i) y(0, t) = 0, t ≥ 0; (ii) y(l, t) = 0, t ≥ 0; (iii) ∂t 3 Ê pxˆ y sin , 0 < x < l . (x, 0) = 0, 0 < x < l and (iv) y(x, 0) = 0 ÁË l ˜¯ ∂2 u ∂2 u + = 0 are ∂x 2 ∂y 2 (i) u(x, y) = (Aepx + Be–px) (C cos py + D sin py)

8. The three possible solutions of

(ii) u(x, y) = (A cos px + B sin px) (Cepy + De–py)

9.

(iii) u(x, y) = (Ax + B) (Cy + D), where A, B, C, D are arbitrary constants. z z z z( z + i ) Z (a n ) = , if |z| > a; Z (ein p /2 ) = Z (ei p /2 )n = = = 2 i p /2 z-a z i z-e z +1 np np ˆ z2 z np ˆ z Ê Ê i.e. Z Á cos + i sin = +i 2 \ Z Á sin = 2 ˜ ˜ 2 Ë Ë 2 2 ¯ z +1 2 ¯ z +1 z +1

10. yn = a + bn ◊ 2n (1) [Note the question is wrong. It has been corrected] yn+1 = a + 2(b ◊ 2n) ... (2); yn + 2 = a + 4(b ◊ 2n) ...(3) (2) – (1) gives; yn+1 – yn = b ◊ 2n ... (4); (3) – (2) gives; yn+2 – yn+1 = 2b ◊ 2n .. (5) From (4) and (5); yn+2 – yn+1 = 2(yn+1 – yn); i.e., yn+2 – 3yn+1 + 2yn = 0 is the difference equation required. Part - B 11. (a)(i) We assume that the Fourier series of period 2 is required for f(x) = 2x – x2 in 0 < x < 2. The problem in which F.S. of period 2l for f(x) = 2lx – x2 in (0, 2l) is a worked example in the book. We simply put l = 1. 2 4 • 1 Ans. 2 x - x 2 = - 2 Â 2 cos (n p x ) in (0,2) 3 p n =1 n (ii) This problem is a worked example in the book.

Solved Question Papers-5

SQP5.5

11. (b)(i) Let the complex form of the Fourier series of f(x) = eax in –p < x < p •

Â

be f(x) =

n = -•

cn ein x . p

p

1 Then cn = 2p

Úe

ax

◊e

- in x

-p

1 dx = 2p

È e( a - in ) x ˘ Í ˙ Î a - in ˚ - p

=

1 {e( a - in )p 2p (a - in)

=

1 sinh ap (-1)n {eap (-1)n - e- ap (-1)n } = 2p (a - in) p a - in

=

sinh a p (-1)n (a + in) p n2 + a 2

- ( a - in )p

\ Required F.S. is f(x) =

}

sinh a p p



(-1)n (a + in) in x e in (-p , p ) n2 + a 2 n = -•

Â

(ii) The internal (0, 6) is divided into 6 sub-intervals each of length 1. 2l = 6 \ l = 3; y0 = 9, y1 = 18, y2 = 24, y3 = 28, y4 = 26, y5 = 20. Let the required F.S. be y = a0 = 2 ¥

1 6

5

 yr =

r =0

a0 np x np x  (1) + Â an cos + bn sin 2 3 3

125 1 p = 41.67; a1 = Â yr cos 3 3 3

1 p ÈÎ( y0 - y3 ) + ( y1 + y5 - y2 - y4 ˘˚ ¥ cos 3 3 1 25 = [(9 - 28) + (18 + 20 - 24 - 26) ¥ 0.5] = = - 8.33 3 3 p xr 1 1 p b1 = Â yr sin = ( y1 + y2 - y4 - y5 ) sin 3 3 3 3 4 = - ¥ 0.866 = - 1.15 3 2p xr 1 1 a2 = Â yr cos = [( y0 + y3 ) - ( y1 + y2 + y4 + y5 ) cos 60∞] 3 3 3 7 = - = - 2.33 3 =

b2 =

1 2p 1 yr sin xr = ( y1 + y4 - y2 - y5 ) sin 60∞ = 0. Â 3 3 3

Required F.S. is

SQP5.6

Transforms and Partial Differential Equations

px pxˆ 2p x Ê y = 20.84 + Á - 8.33 cos - 1.15 sin - 2.33 cos . ˜ Ë 3 3 ¯ 3 12. (a)(i) The problem is a worked example in the book. (ii) The problem of finding F (e- a -a It is derived that F (e

2 2

x

)=

2 2

x

) is a worked example in the book.

1 a 2

e- s

2

/4 a 2

.

1 1 or taking a = in this result, we get 2 2

Putting a 2 = F (e - x

2

/2

) = e- s

2

/2



12. (b)(i) FS{f(x)} =



Ú f ( x) sin sx dx and FC { f ( x )} = Ú f ( x ) cos sx dx 0

0



\ FS{sin x} =

1

a

Ú sin x ◊ sin sx dx = 2 Ú [cos (s - 1) x - cos (s + 1) x] dx 0

0

a

1 È sin (s - 1) x sin (s + 1) x ˘ = Í ˙ 2 Î s -1 s +1 ˚0

a

1 È sin (s - 1)a sin (s + 1)a ˘ = Í ˙ 2 Î s -1 s +1 ˚0 a

FC{sin x} =

a

Ú cos sx ◊ sin x dx = 0

1 [sin (s + 1) x - sin (s - 1) x ] dx 2 Ú0 a

È cos(s + 1) x cos(s - 1) x ˘ + Í˙ s +1 s -1 ˚0 Î

=

1 2

=

˘ 1È 1 1 {1 - cos(s + 1) a} {1 - cos (s - 1)a}˙ Í 2 Îs + 1 s -1 ˚

(ii) Let f(t) = e–at and g(t) = e–bt

\

fC (s) =

a b ; gC (s) = 2 2 s +a s + b2 2

By the property of F¢ cosine transforms, •

Ú 0



\

2 f (t ) g(t ) dt = p

-(a + b) t dt = Úe 0

2 p





Ú fC (s) ◊ gC (s) ds 0

ab

Ú (s2 + a2 )(s2 + b2 ) ds 0

Solved Question Papers-5



ds p Ú (s2 + a2 )(s2 | b2 = 2ab 0

\

SQP5.7



-(a + b) t ¸ p ÔÏ e Ô Ì ˝ = ÓÔ -(a + b) ˛Ô 0 2 ab (a + b)

Changing s as t, the required result follows 13. (a)(i) (x2 – yz)p + (y2 – zx)q = z2 – xy This is a Lagrange’s liner equation Pp + Qq = R. dx dy dz = = This subsidiary equation are 2 x - yz y 2 - zx z 2 - xy Each ratio in (1) is equal to d ( x - y) d ( y - z) d(z - x) = = ( x - y)( x + y + z ) ( y - z )( x + y + z ) ( z - x )( x + y + z )

(1)

(2)

From the first two ratios in (2), we get log (x – y) = log (y – z) + log a viz.,

x-y = a (3) y-z

Also each ratio is (1) =

d ( x + y + z ) x dx + y dy + z dx = ( Sx 2 - Syz ) Sx 3 - 3 xyz

1 d Sx 2 d ( Sx ) 2 i.e., ;i.e., 2Sxd(Sx) = d(Sx2) = Sx 2 - Syz ( Sx ) ( Sx 2 - Syz ) Solving, we get (Sx)2 – (Sx2) = 2b; i.e. xy + yz + zx = b

(4)

Ïx-y ¸ \ G.S of the given L.L.E. is f Ì , xy + yz + zx ˝ = 0. Ó y - z) ˛ 13. (a)(ii) p(1 + q) = qz... (1). Let a solution of (1) be z = z(x + ay) (2) dz dz From (2), p = and q = a . du du dz Ï dz ¸ dz Ì1 + a ˝ = a ◊ z du Ó du ˛ du dz dz adz π 0, we get 1 + a = az or Ú = du + b Since du du az - 1 Ú Since (2) is a solution of (1),

\ C.S. of(1) is log (az – 1) = x + ay + b...(3) To get G.S., we put b = f(a) in (3), where f is an arbitrary function i.e., (3) becomes log (az – 1) = x + ay + f(a) (4) z Differentiating (4) partially w.r.t. a; = y + f ¢(a )  (5) az - 1 Eliminating a between (4) and (5), the G.S. of (1) is obtained. 13. (a)(iii) p2 + q2 = x2 + y2 ... (1); i.e. p2 – x2 = y2 – q2 = a2...(2) where a is an arbitrary constant.

SQP5.8

Transforms and Partial Differential Equations

From (2); p = x 2 + a 2 and q = y 2 - a 2 Now dz = pdx + q dy 2 2 2 2 \ C.S. of (1) is z = Ú x + a dx + Ú y - a dy + b y Ê xˆ i.e., 2z = x x 2 + a 2 + a 2 sinh -1 Á ˜ + y y 2 - a 2 - a 2 cosh -1 + b Ë a¯ a (3) is the C.S. Putting b = f(a) and differentiating. (3)w.r.t. a, we get a result (4) The eliminant of a between (3) and (4) is the G.S. of (1) 13. (b) (i) This problem is a worked example in the book. 13. (b)(ii) (D2 + 2DD¢ + D¢2 – 2D – 2D¢) z = sin (x + 2y) (1) i.e., (D + D¢)(D + D¢ – 2)z = sin (x + 2y) Since the part of the C.F. Corresponding to (D – aD¢ – b)z = 0 is ebx f(y + ax), the C.F. of the solution of (1) = f1(y – x) + e2x f2(y – x) (2) 1 sin ( x + 2 y) P.I. = 2 D + 2 DD ¢ + D ¢ 2 - 2 D - 2 D ¢ 1 sin ( x + 2 y) = -1 - 4 - 4 - 2D - 2D¢ (2 D + 2 D ¢ - 9) sin ( x + 2 y) = 81 - (2 D + 2 D ¢ )2

1 [2 cos ( x + 2 y) + 4 cos ( x + 2 y) - 9 sin ( x + 2 y)] 117 1 = (3) [2 cos ( x + 2 y) - 3 sin ( x + 2 y)] 39 \ G.S. of (1) is z = C.F. + P.I, which are given by (2) and (3) 14. (a) This problem is the same as a worked example in the book. In the W.E., the length of the string is taken as (2l) instead of l. \ In the W.E., wherever we get 2l, it is to be replaced by b or l should be replaced by l/2. Finally the required solution will become =

8b • 1 np np x n p at  sin 2 sin l cos l . p 2 n =1 n2 14. (b) In steady-state temperature at any point (x, y) in the plate is given by uxx + uyy = 0... (1). We have to solve (1) satisfying the following boundary conditions. u(0, y) = 0 ... (2); u(10, y) = 0 ... (3); u(x, •) = 0... (4) and u(x, 0) in 0 £ x £ 5 Ï20 x, u (x, 0) = Ì (5) Ó20(10 - x ), in 5 £ x £ 10 y(x, t) =

Solved Question Papers-5

SQP5.9

The 3 possible solutions of (1) are u(x, y) = (A epx + Be–px) (C cos py + D sin py) ... (6); u(x, y) = (A cos px + B sin px) (C epy + De–py) ... (7) and u(x, y) = Ax + B) (Cy + D) ... (8). Since u Æ 0 as y Æ • by B.C.(4), (7) is the proper solution with C = 0. Using B.C. (2) in (7) (with C deleted), we get A = 0 np Using B.C. (3) in (7), we get sin 10p = 0 \ p = , n = 0, 1, 2, ... • 10 n p x - n p y /10 \ Required solution is u(x, y) = l sin , where l = B.D. e 10 The general form of the required solution is • n p x - n p y /10 , e u(x, y) = Â ln sin (9) 10 n =1 •

Using B.C. (5) in (9), we have

 ln sin

n =1

np x = u( x, 0) as given. 10

Let the Fourier half-range sine series of u(x, 0) in (0, 10) be

 bn sin i.e.,

np x 2 ... (10), where bn = ln = 10 10

ln =

È5

1 np x dx + Í Ú 20 x sin 5 ÍÎ 0 10

10

Ú u( x, 0) sin 0

10

Ú 20(10 - x) sin 5

ÈÏ -1 np x ˆ Ê ÍÔ Ê - cos n p x ˆ - sin Á ˜ Á 10 10 ˜ = 4 ÍÌ x ◊ Á ÍÔ Ë n p / 10 ˜¯ - ÁË n2p 2 /102 ˜¯ ÍÓ ÍÎ

np x dx 10

np x ˘ dx ˙ 10 ˙˚ 5

¸ Ô ˝ Ô ˛0 10 ˘

Ï np x ˆ Ê np x ˆ ¸ Ê - cos - sin Ô Ô Á Á ˜ Ô 10 + Á 10 ˜˜ Ô + Ì(10 - x ) Á ˝ ˜ 2 2 np ˜Ô Ô Á ˜ ÁÁ n p ˜¯ Ô Ë ¯ Ë 102 10 ÔÓ ˛5

˙ ˙ ˙ ˙ ˙˚

ÈÏ 50 np 100 np ¸ + 2 2 sin = 4 ÍÌcos ˝ n p 2 2 ˛ n p ÎÓ np 100 np ¸˘ 800 np Ï 50 +Ì + 2 2 sin cos ˝˙ = 2 2 sin n p 2 2 2 np Ó ˛˚ n p Using this value of ln in (9), the required solution is u(x, y) =

800 p2



1

 n2 sin

n =1

np n p x - n p y /10 sin e 2 10

SQP5.10

Transforms and Partial Differential Equations

15. (a)(i) un+2 – 2un+1 + un = 2n; u0 = 2 and u1 = 1 Taking z-transforms of the given equation, z z-2

[ z 2 u ( z ) - z 2 u(0) - z u(1)] - 2[ z u ( z ) - zu(0)] + u ( z ) = i.e., ( z 2 - 2 z + 1) u ( z ) - 2 z + 4 z = i.e.,

z z-2

u (z) 1 2z 3 1 1 2 = + = + 2 2 2 z z - 2 z - 1 ( z - 1)2 ( z - 1) ( z - 2) ( z - 1) ( z - 1)

z z 2z + z - 2 z - 1 ( z - 1)2 Inverting, we get un = 2n + 1 – 2n. 2 ÔÏ z Ô¸ ÔÏÊ z ˆ Ê z ˆ Ô¸ = Z -1 ÌÁ (ii) Z -1 Ì ˜ *Á ˜ ˝ by convolution theorem 2˝ ÔÓË z - 4) ¯ Ë z - 4 ¯ Ô˛ ÔÓ ( z - 4) Ô˛ \ u (z) =

n

= 4n * 4n =

 4r ◊ 4n - r = (n + 1) 4n

(1)

r =0

Ï z 3 Ô¸ Ï z 2 Ô¸ -1 -1 Ô -1 Ô Now Z Ì = Z *Z ˝ Ì 3 2˝ ÔÓ ( z - 4) Ô˛ ÔÓ ( z - 4) Ô˛ = (n + 1)4n * 4n by (1)

Ï z ¸ Ì ˝ Óz - 4˛

n

=

 (r + 1)4r ¥ 4n - r , by convolution theorem

r =0

=

1 n 1 4 (n + 1)(n + 2) or (n2 + 3n + 2) ◊ 4 n 2 2

2 ÔÏ z ( z - z + 2 Ô¸ 1 15. (b) (i) (1) Z -1 Ì = z { f ( z )}, say 2˝ ÔÓ ( z + 1)( z - 1) Ô˛

Then

f (z) z2 - z + 2 A B C = = + + 2 z z + 1 z - 1 ( z - 1)2 ( z + 1)( z - 1) =

\

1 1 + z + 1 ( z - 1)2

z z + ; Inverting, f(n) = (–1)n + n z + 1 ( z - 1)2 (Note: This problem is worked out in the book, by Residue theorem method) ¸ z -1 Ï -1 (2) Z Ì ˝ = Z { f ( z )}, say ( z 1)( z 2) Ó ˛ f (z) =

Solved Question Papers-5

Then

SQP5.11

f (z) 1 1 1 z z = = \ f (z) = z ( z - 1)( z - 2) z - 2 z - 1 z - 2 z -1

Inverting, we get, f(n) = 2n – 1. (ii) Z(nan sin n q). Z(sin n q) =

z sin q z / a sin q ; \ z(a n sin q ) = 2 2 z - 2 z cos q + 1 z /a - 2 z /a cos q + 1 2

i.e., Z (a n sin n q ) =

a z sin q z 2 - 2az cos q + a 2

z (n a n sin n q ) = - z

\

d dy

ÏÔ ¸Ô a z sin q Ì 2 2˝ ÓÔ z - 2az cos q + a ˛Ô

È ( z 2 - 2az cos q + a 2 ) ◊1 - z(2 z - 2a cos q ) ˘ = - (a sin q ) z Í ˙ ( z 2 - 2az cos q + a 2 )2 Î ˚ =

(a sin q ) z ( z 2 - a 2 ) ( z 2 - 2az cos q + a 2 )2

B.E./B.Tech. Degree Examinations, April/May 2011 Regulations 2008 Third Semester Common to all branches MA2211 Transforms and Partial Differential Equations Time: Three Hours

Maximum: 100 Marks Answer ALL Questions Part A – (10 ¥ 2 = 20 Marks)

1. Give the expression for the Fourier Series co-efficient bn for the function f(x) defined in (–2, 2). 2. Without finding the values of a0, an and bn, the Fourier coefficients of Fourier series, for the function F(x) = x2 in the interval (0, p) find the value of • È a2 ˘ Í 0 + Â (an2 + bn2 ˙ ÍÎ 2 n =1 ˙˚ 3. State and prove the change of scale property of Fourier Transform. 4. If Fc(s) is the Fourier cosine transform of f(x), prove that the Fourier cosine 1 Ê sˆ transform of f (ax ) is Fc Á ˜ . Ë a¯ a 5. From the partial differential equation by eliminating the arbitrary constants a and b from z = (x2 + a) (y2 + b). 6. Solve the equation (D – D¢)3 z = 0. 7. A rod 40 cm long with insulated sides has its ends A and B kept at 20°C and 60°C respectively. Find the steady state temperature at a location 15 cm from A. 8. Write down the three possible solutions of Laplace equation in two dimensions. 9. Find the Z-transform of an. 10. What advantage is gained when Z-transform is used to solve difference equation? Part B – (5 ¥ 16 = 80 marks) 11. (a) (i) Expand f(x) = x(2p – x) as Fourier series in (0, 2p) and hence deduce 1 1 1 1 that the sum of 2 + 2 + 2 + 2 +  1 2 3 4 (ii) Obtain the Fourier series for the function f(x) given by

SQP6.2

Transforms and Partial Differential Equations

Ï1 - x, - p < x < 0 f ( x) = Ì . Hence deduce that Ó1 + x, 0 < x < p 1

+

12

1 32

+

1 52

+=

p2 . 8 Or

l Ï in 0 £ x £ ÔÔ x 2 . (b)(i) Obtain the sine series for f ( x ) = Ì Ôl - x in l £ x £ l ÔÓ 2 (ii) Find the Fourier series up to second harmonic for y = f(x) from the following values.

x:

0

p/3 2p/3 p

4p/3 5p/3

2p

y:

1.0

1.4

1.5

1.0

1.9

1.7

1.2

ÏÔ1 - x 2 12. (a)(i) Find the Fourier transform of f ( x ) = Ì ÔÓ0 •

evaluate

if x < 1 if x > 1

. Hence

x Ê x cos x - sin x ˆ ˜¯ cos 2 dx. 3 x

Ú ÁË 0

ÏÔ1 for x < a (ii) Find the Fourier transform of f(x) given by f ( x ) = Ì ÔÓ0 for x > a > 0 •

and using Parseval’s identity prove that

Ú 0

Or

2

p Ê sin t ˆ ÁË t ˜¯ dt = 2 .

0 < x 2 Ó • dx (ii) Evaluate Ú 2 using Fourier cosine transforms of e–ax 2 2 2 0 ( x + a )( x + b ) and e–bx. 13. (a)(i) Form the partial differential equation by eliminating arbitrary functions f and f from z = f(x + ct) + f(x – ct). (ii) Solve the partial differential equation (mz – ny)p + (nx – lz)q = ly – mx. Or (b)(i) Solve (D2 – D¢2) z = ex–y sin (2x + 3y). (ii) Solve (D2 – 3DD¢ + 2D¢2 – 2D¢) z = x + y + sin (2x + y).

Solved Question Papers-6

SQP6.3

14. (a) A uniform string is stretched and fastened to two points l apart. Motion is started by displacing the string into the form of the curve y = kx(l – x) and then released from this position at time t = 0. Derive the expression for the displacement of any point of the string at a distance x from one end at time t. Or (b) A rectangular plate with insulated surface is 20 cm wide and so long compared to its width that it may be considered infinite in length without introducing an appreciable error. If the temperature of the short edge for 0 £ y £ 10 Ï10 y x = 0 is given by u = Ì and the two long edges Ó10(20 - y) for 10 £ y £ 20 as well as the other short edge are kept at 0°C. Find the steady state temperature distribution in the plate. 15. (a)(i) Using convolution theorem, find inverse Z-transform of z2 . ( z - 1)( z - 3)

(ii) Find the Z-transforms of an cos nq and e–at sin bt. (b)(i) Solve the difference equation y(n + 3) – 3y(n + 1) + 2y(n) = 0, given that y(0) = 4, y(1) = 0 and y(2) = 8. (ii) Derive the difference equation from yn = (A + Bn)(–3)n.

(6)

Solutions Part A 1. If the function f(x) is an odd function in (–2, 2), then the Fourier series will np x be of the form  bn sin , where 2 bn =

2 2

2

Ú

2

f ( x ) sin

0

np x np x dx or Ú f ( x ) sin dx 2 2 0

If the function f(x) is an even function in (–2, 2), then the F.S. will be a np x of the form 0 + Â an cos and hence bn = 0. 2 2 2.

• a02 1 + Â (an2 + bn2 ) = 2 y 2 = 2 ¥ 2 n =1 p

p

2 Ú y dx = 0

p

2 4 x dx, since the given function p Ú0

2

is y = f(x) = x in (0, p) 2 p5 2 4 ¥ = p . p 5 5 This is a standard result, worked out in the book.

\ Required value =

3.

SQP6.4

Transforms and Partial Differential Equations



4.

Ú f ( x) cos sx dx = Fc (s) 0



\

Ú



f (ax ) cos sx dx =

0

i.e.,

Ê sˆ

1

Ú f (t ) cos ÁË a ˜¯ t ◊ a dt, on putting ax = t(a > 0) 0

Fc { f (ax )} =

1 Ê sˆ Fc Á ˜ Ë a¯ a

5. z = (x2 + a) (y2 + b) ... (1); Differentiating (1) partially w.r.t. x and then w.r.t. y, we get p = 2x(y2 + b) ... (2) and q = 2y(x2 + a) ... (3). Using the values of (x2 + a) and (y2 + b) obtained from (3) and (2) in (1), we q p ◊ get, z = or pq = 4xyz, which is the P.D.E. required. 2y 2x 6. (D – D¢)3 z = 0; The A.E. is (m – 1)3 = 0 \ m = 1, 1, 1. \ the G.S. of the given PDE is z = f1(y + x) + x f2(y + x) + x2 f3(y + x), where f1, f2, f3 are arbitrary functions. d 2u 7. The steady-state temperature in the rod is given by = 0 ... (1) with the dx 2 B.C¢s u(0) = 20 ... (2) and u(40) = 60... (3). Solving (1), we get u(x) = Ax + B ... (4); Using (2) in (4), B = 20; Using (3) in (4); 40 A + 20 = 60 \ A = 1. \

Solution of (1) is u(x) = x + 20 ... (5)

\

[u(x)]x = 15 = 35°C

8. The three possible solutions of the Laplace equation

∂2 u ∂x 2

+

∂2 u ∂y 2

= 0 are

(i) u(x, y) = (Aepx + Be–px) (C cos py + D sin py) (ii) u(x, y) = (A cos px + B sin px) (Cepy + De–py) and (iii) u(x, y) = (Ax + B) (Cy + D), where A, B, C, D and p are arbitary constants 9.

Z (a n ) =



 an z-n =

n=0



n

Ê aˆ  ÁË z ˜¯ = n=0

1 1-

a z

=

z , if z > a z-a

10. When we solve a difference equation, by other methods, we first get the G.S. involving arbitrary constants. Then the arbitrary constants are evaluated by using the given boundary conditions. Thus we get the required P.S. When the Z-transform method is used, the given boundary conditions are absorbed in the beginning of the problem and the required P.S. is directly obtained.

Solved Question Papers-6

SQP6.5

PART - B 11. (a)(i) This problem is worked out in the book taking the interval as (0, 2l). We have to simply change l as p. 4 4 Then an = - 2 ; a0 = p 2 ; bn = 0 3 n \ The required F.S. of

• 2 2 1 p - 4 Â 2 cos nx ... (1) in (0, 2p) 3 n =1 n

f(x) = x(2p – x) = 1

+

1

2

2

+

1

+  can be obtained by putting 1 2 32 x = 0 or x = 2p in (1), that are points of continuity for f(x). \ [sum of the F.S] • • 2 1 1 p2 i.e., p 2 - 4 Â 2 = 0 \ Â 2 = 3 6 n =1 n n =1 n

The required sum of

11. (a)(ii) The given function is an even function in (–p, p) a0 + Â an cos n x in (-p , p ) Let the F.S. of f(x) be 2 p

Then an =

2 ( x + 1) cos n x dx p Ú0 p

È Ê sin nx ˆ Ê - cos nx ˆ ˘ Í( x + 1) ÁË ˙ n ˜¯ ÁË n2 ˜¯ ˚ 0 Î

=

2 p

=

Ï 4 , if n is odd Ôn 2 {( 1) 1} = Ì pn 2 pn Ô0, if n is even Ó 2

p

= a0 =

2 1 1 p ( x + 1) dx = {( x + 1)}0 = {(p + 1)2 - 1} = p + 2. Ú p 0 p p

\ Required F.S. of f(x) =

4 p +12 p



1

 n2

cos nx in(-p , p )

n =1

Putting x = 0, which is a point of continuity of f(x), in (1), we get •

Ê 1ˆ ÁË 2 ˜¯ = f (0) = 1 n = 1, 3, 5, ... n

4 p +12 p •

\

Â

n = 1, 3, 5, ...

Â

1 n

2

=

p2 . 8

(1)

SQP6.6

Transforms and Partial Differential Equations •

11. (b)(i) Let the F.H.R. sine series of f(x) be

 bn sin

n =1

np x in (0, l) l

l Èl /2 ˘ Then bn = 2 Í x sin np x dx + (l - x ) sin np x dx ˙ Ú Ú l ÍÎ 0 l l ˙˚ l /2 l /2 2 ÈÏ Ê cos np x /l ˆ Ê - sin np x /l ˆ ¸ Í = Ìx ˝ l ÍÓ ÁË np /l ˜¯ ÁË n2p 2 / l 2 ˜¯ ˛0 Î l ˘ ÏÔ Ê - cos np x / l ˆ Ê sin np x / l ˆ ¸Ô ˙ + Ì(l - x ) Á ˜¯ + Á - n2p 2 / l 2 ˜ ˝ ˙ np / l Ë Ë ¯ ˛Ôl /2 ÓÔ ˚˙

=

ÈÏÔ l 2 np l2 np ¸Ô cos + 2 2 sin ÍÌ˝ 2 2 Ô˛ n p ÎÍÔÓ 2 np

2 l

ÏÔ l 2 np l2 np ¸Ô˘ +Ì cos + 2 2 sin ˝˙ 2 2 Ô˛˚˙ n p ÓÔ 2 np

=

4l 2

2

sin

n p \ Required F.S. of f(x) •

np 2

np np x sin in (0, l) 2 l p n =1 a 11. (b)(ii) Let the required F.S. be 0 + Â an cos nx + Â bn sin nx in (0, 2p) 2 y0 = y(0) = 1.0; y1 = y(p/3) = 1.4, y2 = 1.9, y3 = 1.7, y4 = 1.5, y5 = 1.2 1 5 1 a0 = 2 ¥ Â yr = ¥ 8.7 = 2.9 6 r =0 3 4l

2

1

 n2 sin

p xr px 1 and b1 = 1 ¥ Â yr sin r 3 6 3 1È p˘ 1 p = Í( y0 - y3 ) + ( y1 + y5 - y2 - y4 ) cos ˙ b1 = [ y1 + y2 - y4 - y5 )sin 3Î 3˚ 3 3

a1 = 2 ¥

1 6

 yr cos

1 = [ -0.7 + (-0.8) ¥ 0.5] = - 0.367 3 a2 = 2 ¥

1 6

 yr cos

2p xr 3

1È p˘ = Í( y0 + y3 ) - ( y1 + y2 - y4 - y5 ) cos ˙ 3Î 3˚ 1 = [2.7 - 7.0 ¥ 0.5] = - 0.1 3

=

1 [0.6 ¥ 0.866] = 0.173 3

b2 = 2 ¥ Â yr sin

2p xr 3

1È p˘ = Í y1 + y4 - y2 - y5 )sin ˙ 3Î 3˚ = - 0.058

Solved Question Papers-6

SQP6.7

\ Required F.S. is f(x) ~ 1.45 – 0.367 cos x + 0.173 sin x – 0.1 cos 2 x - 0.058 sin 2 x. 12. (a)(i) This problem is a worked example in the book. a

Úe

12. (a)(ii) F{f(x)} =

a

- isx

Ú (cos sx - i sin sx) dx

dx =

-a

-a

a

= 2 Ú cos sx dx = 0

2 2 (sin sx )0a = sin as. s s



Ú

By Parseval’s identity,

2

f ( x ) dx =

-•

1 2p



Ú

2

f (s) ds

-•

2



2 Ê sin as ˆ \ p ¥ 2 Ú ÁË s ˜¯ ds = 2a 0 2



Ê sin t ˆ dt 4 i.e., ¥ 2 ÚÁ = 2 a, p Ë t / a ˜¯ a 0 •

i.e., 4 ÊÁ sin t ˆ˜ p Ú0 Ë t ¯ •

\

Ú 0

2

a dt = 2 a, on putting as = t

2

p Ê sin t ˆ ÁË t ˜¯ dt = 2 . 1

2

12. (b)(i) FS{f(x)} = Ú x sin sx dx + Ú (2 - x ) sin sx dx 0

\

1

ÈÏ Ê - cos sx ˆ Ê - sin sx ˆ ¸1 + FS{sin x} = ÍÌ x Á ˝ s ˜¯ ÁË s2 ˜¯ ˛0 ÍÓ Ë Î

2 ÏÔ Ê - cos sx ˆ Ê - sin sx ˆ ¸Ô˘˙ + Ì(2 - x ) Á ˝ Ë s ˜¯ ÁË s2 ˜¯ 1 Ô˙ ÓÔ ˛˚

ÏÊ cos s ˆ Ê sin s ˆ ¸ Ê sin 2 s cos s sin s ˆ + + + 2 ˜ = ÌÁ ˝+ s ˜¯ ÁË s2 ˜¯ ˛ ÁË s2 s s ¯ ÓË

=

2 sin s s

2

-

sin 2 s s

2

=

2 sin s s2

(1 - cos s)

- ax - bx 12. (b)(ii) Let f(x) = e and g(x ) = e \ fc (s) =

a s +a

By the property of F¢ cosine transforms. •

Ú 0

f ( x ) g( x ) dx =

2 p



Ú fc (s) gc (s) ds 0

2

2

and gc (s) =

b s + b2 2

SQP6.8

Transforms and Partial Differential Equations •

i.e.,

-(a + b) x dx = Úe 0



2 p



ab

Ú (s2 + a2 )(s2 + b2 ) ds 0

p = \ Ú 2 2 2 2 2 ab 0 ( s + a )( s + b ) ds



-(a + b) x ¸ p ÔÏ e Ô Ì ˝ ÔÓ - (a + b) Ô˛0 2 ab (a + b)

Changing s into x, we get the required result. 13. (a)(i) z = f(x + cy) + f(x – cy) ... (1) changing t as y. p = f ¢ + f¢ ... (2); q = C f ¢ – C f¢ ... (3) Differentiating (2); w.r.t. x; r = f ¢¢ + f¢¢ ... (4) Differentiating (3) w.r.t. y; t = C2 f ¢¢ + C2 f ¢¢ .... (5) From (4) and (5), we get, t = C2r; viz.,

∂2 z ∂y

2

- c2

Again changing y into t, the required P.D.E. is

∂2 z ∂x 2

∂2 z ∂x 2

=0

-

1 ∂2 z C 2 ∂t 2

= 0.

13. (a)(ii) This problem is a worked example in the book. 13. (b) (i) This problem is a worked example in the book. 13. (b)(ii) This problem is a worked example in the book. 14. (a) This problem is a worked example in the book. 14. (b) This problem is a worked example in the book. Ï ¸Ô z2 -1 15. (a)(i) Z -1 ÔÌ ˝=Z ÔÓ ( z - 1)( z - 3) Ô˛

ÔÏÊ z ˆ Ê z ˆ ¸Ô ÌÁ ˜Á ˜˝ ÔÓË z - 1¯ Ë z - 3 ¯ Ô˛ Ï z ¸ -1 = Z -1 Ì ˝* Z z 1 Ó ˛

Ï z ¸ Ì ˝ Óz - 3˛

n

= 1* 3n =

 1n - r ◊ 3r = 1 + 3 + 32 +  + 3n

r =0

=

3n + 1 - 1 1 n + 1 = (3 - 1) 3-1 2

Ê zˆ 15. (a)(ii) (1) z {a n f (n) } = f Á ˜ ; \ Z{a n cos n q} = z(cos n q )z Æ z / a Ë a¯ ÏÔ z ( z - cos q ¸Ô z ( z - a cos q ) = 2 = Ì 2 ˝ 2 ÓÔ z - 2 z cos q + 1 ˛Ôz Æ z / a z - 2 az cos q + a

Solved Question Papers-6

SQP6.9

(2) Z{e- at sin bt} = Z (sin bt )z Æ z eaT ÏÔ ¸Ô z sin bT z e aT sin bT = = Ì 2 ˝ 2 2 aT - 2 ze aT cos bT + 1 ÓÔ z - 2 z cos bT + 1 Ô˛z Æ z eaT z e

15. (b) (i) This problem is a worked example in the book (ii) yn = (A + Bn) (–3)n ... (1); yn + 1) = {A + B(n + 1)} (–3)n+1 ... (2) yn+2 = {A + B(n + 2)} (–3)n+2 ... (3). yn Putting = zn ; we get zn = A + Bn ...(1)¢; Zn + 1 = A + Bn + B... (-3)n (2)¢ and Zn+2 = A + Bn + 2B... (3)¢ From (2)¢ and (3)¢; Zn+2 – 2Zn+1 = –(A + Bn) = –Zn, using (1)¢ i.e.,

yn + 2 (-3)

n+2

-2

yn +1 (-3)

n +1

+

yn ((-3)n

=0

\ The required D.E. is yn+2 + 6yn+1 + 9yn = 0

B.E./B.Tech. Degree Examinations, May/June 2012 Regulations 2008 Third Semester Common to all branches MA2211 Transforms and Partial Differential Equations Time: Three Hours

Maximum: 100 Marks Answer ALL Questions Part A – (10 ¥ 2 = 20 Marks)

1. Find the constant term in the expansion of cos2 x as a Fourier series in the interval (–p, p). 2. Define Root Mean square value of a function f(x) over the interval (a, b). 3. What is the Fourier transform of f(x – a), if the Fourier transform of f(x) is F(s)? 4. Find the Fourier since transform of f(x) = e–ax, a > 0. 5. Form the partial differential equation by eliminating the arbitrary function Ê xˆ from z 2 - xy = f Á ˜ . Ë z¯ 6. Solve (D2 – 7DD¢ + 6D¢2) z = 0. 7. What is the basic difference between the solution of one dimensional wave equation and one dimensional heat equation with respect to the time? 8. Write down the partial differential equation that represents steady state heat flow in two dimensions and name variables involved. 9.

Ï an Ô Find the Z-transform of x(n) = Ì n ! Ô0 Ó

for n ≥ 0 otherwise

10. Solve yn+1 – 2yn = 0 given y0 = 3. Part B – (5 ¥ 16 = 80 marks) 11. (a) (i) Find the Fourier series of f(x) = (p – x)2 in (0, 2p) of periodicity 2p. (ii) Obtain the Fourier series to represent the function f(x) = |x|, –p < x < •

p and deduce

Â

n = 1 (2 n

1 - 1)2

=

p2 . 8

Or

SQP7.2

Transforms and Partial Differential Equations

(b)(i) Find the half-range Fourier cosine series of f(x) = (p – x)2 in the interval (0, p). Hence find the sum of the series

1

x:

0

1

2

3

4

5

y:

9

18

24

28

26

20

+

1

4

+

1

++ • 1 2 34 (ii) Find the Fourier series up to second harmonic for y = f(x) from the following values. 4

12. (a)(i) Derive the Parseval’s identity for Fourier Transforms.

(ii) Find the Fourier integral representation of f(x) defined as Ï 0 for x < a Ô 1 f(x) = ÔÌ for x = 0 Ô 2 ÔÓ- e - x for x > 0 Or (b) (i) State and prove convolution theorem on Fourier transform. (ii) Find Fourier sine and cosine transform of xn–1 and hence prove 1/ x is self reciprocal under Fourier sine and cosine transforms. 13. (a)(i) From the PDE by eliminating the arbirary functions f from f(x2 + y2 + z2, ax + by + cz) = 0. (ii) Solve the partial differential equation x2(y – z)p + y2(z – x)q = z2(x – y). Or (b)(i) Solve the equation (D3 + D2D¢ – 4DD¢2 – 4D¢3)z = cos (2x + y). (ii) Solve (2D2 – DD¢ – D¢2 + 6D = 3D¢) z = xey. 14. (a) The ends A and B of a rod 40 cm long have their temperatures kept at 0°C and 80°C respectively, until steady state condition prevails. The temperature of the end B is then suddenly reduced to 40°C and kept so, while that of the end A is kept at 0°C. Find the subsequent temperature distribution u(x, t) in the rod. Or (b) A long rectangular plate with insulated surfaceis l cm wide. If the temperature along one short edge (y = 0) is u(x, 0) = k(lx – x2) degrees, for 0 < x < l, while the other two long edges x = 0 and x = l as well as the other short edge are kept at 0°C, find the steady state temperature function u(x, y).

Solved Question Papers-7

SQP7.3

15. (a)(i) Find Z[n(n – 1)(n – 2)]. (ii) Using Convolution theorem, find the inverse Z-transform of 8z 2 (2 z - 1)(4 z - 1)

(b)(i) Solve the difference equation y(k + 2) + y(k) = 1, y(0) = y(1) = 0, using Z-transform. (ii) Solve yn+2 + yn = 2n◊ n, using Z-transform

Solutions Part A 1. a0 =

2 p

p

Ú cos

p

2

x dx =

0

2 1 1 ¥ (1 + cos 2 x ) dx = 1\ Constant Term in F.S. = p 2 Ú0 2 b

2. If y is the RMS value of y = f(x) in (a, b), then y = •

3. F(s) = F({f(x)} =

Ú

1 [ f ( x )]2 dx. b - a Úa

f ( x )e - i s x dx

-• •

Ú

\ F{f(x – a)}=

f ( x - a ) e - i s x dx =

-•

=e • –ax

4. Fs(e ) =

–ias

Úe



Ú

f (t )e - i s(t + a ) dt , on putting x – a = t

-•

F(s). •

- ax

0

- ax s ÔÏ e Ô¸ sin sx dx = Ì 2 (- a sin sx - s cos sx )}˝ = 2 2 s + a2 ÔÓ s + a Ô˛0

Ê xˆ 5. z2 – xy = f Á ˜  (1); Differentiating (1) partially w.r.t. x; we get Ë z¯ Ï z ◊1 - xp ¸ x 2zp – y = f ¢(u) ◊ Ì ˝  (2), where u = . 2 z z Ó ˛ Differentiating (1) partially w.r.t. y, we get Ï x ¸ 2z q – x = f ¢(u) ◊ Ì- q ˝  (3) 2 Ó z ˛

(2) ∏ (3) gives;

2 z p - y z - xp = ; viz., qxy = 2z2q – xz + x2p 2zq - x - xq

i.e., the required P.D.E. is x2p + (2z2 – xy) q = xz. 6. (D2 – 7DD¢ + 6D¢2)z = 0; The A.E. is (m2 = 7m + 6 = 0; \ m = 1, 6 \ Solution of the P.D.E. is z = f1(y + x) + f2(y + 6x), where f1 and f2 are arbitrary functions.

Transforms and Partial Differential Equations

SQP7.4

7. The proper solution of one dimensional wave equation will have cos pat or sin pat in the value of y(x, t), depending on the initial condition. The proper - p2 a 2 t solution of one dimensional heat flow equation will have e in the value of u(x, t), irrespective of the end conditions. 8. The P.D.E. that represents steady-state heat flow in two dimensions in ∂2 u ∂2 u + = 0 ; x, y that are the distances of the typical point (x, y) in the ∂x 2 ∂y 2 plate from the y- and x-axes are independent variables and u representing the S.S. temperature at the point (x, y) of the plate is the dependent variable. 9.

n • Ê an ˆ 1 Ê aˆ Z Á ˜ = Â ◊ Á ˜ = ea / z Ë n ¯ n=0 n Ë z ¯

10. yn+1 – 2yn = 0; y0 = 3 Taking Z-transforms of the given equation, we have [ z y ( z ) - z y (0)] - 2 y ( z ) = 0 i.e. ( z - 2) y ( z ) = 3z or y ( z ) = 3 ◊ z  (1) z -2 Ê z ˆ Inverting (1); y(n) = 3Z -1 Á = 3 ¥ 2 n or yn = 3 ¥ 2 n is the required ˜ z 2) Ë ¯ solution.

PART - B 11. (a)(i) Let the F.S. of f(x) = (p – x)2 in (0, 2p) be a0 + Â an cos nx + Â bn sin nx. 2 an = =

1 p

a0 = bn = =

1 p

1 p

2p

Ú (p - x)

2

cos n x dx

0

2p

È 4 Ê cos nx ˆ Ê sin nx ˆ ˘ 2 sin nx - {- 2(p - x )} Á + 2Á- 3 ˜˙ = 2 Í(p - x ) 2 ˜ Ë ¯ Ë ¯ n n n n Î ˚0 1 p 1 p

2p

Ú

(p - x )2 dx = -

0

1 2p 2 {(p - x )3}20p = 3p 3

2p

Ú (p - x)

2

sin nx dx

0

È 2 Í(p - x ) Î

2p

Ê cos nx ˆ Ê sin nx ˆ Ê cos nx ˆ ˘ ÁË - n ˜¯ - {-2 (p - x )} ÁË - 2 ˜¯ + 2 ÁË ˜˙ = 0 n n3 ¯ ˚ 0

\ Required F.S. of f(x) is

• 1 p2 + 4 Â 2 cos nx in (0, 2p) 3 n =1 n

Solved Question Papers-7

SQP7.5

11. (a)(ii) f(x) = |x| is an even function in = p < x < p. a \ Let the F.S. of f(x) be 0 + Â an cos nx in (–p, p) 2 an =

2 p

p

Ú

x cos nx dx =

0

2 p

p

Ú x cos nx dx 0

p

=

2 Ï Ê sin nx ˆ 2 Ê cos nx ˆ ¸ n -1ÁÌx Á ˝ = 2 {(-1) - 1} ˜ ˜ 2 Ë ¯ Ë ¯ p Ó n n ˛0 n p

If n even Ï 0, Ô = Ì 4 Ô- 2 , If n odd Ó n p

a0 =

2 p

p

Úx 0

= x =

\

dx =

p 4 2 p

1 2 p ( x )0 = p . p •

Â

n = 1, 3...

1 n2

cos nx in (-p , p )  (1)

Putting x = 0 in the R.S. of (1), \

p 4 2 p



1

 (2n - 1)2

= f (0) = 0;

n =1



1 p2 = Â 8 n = 1 (2 n - 1)

11. (b)(i) This problem is a worked example in the book. 11. (b)(ii) 2l = 6 \ l = 3; The required F.S. of y is of the form a0 np x np x in (0,6) + Â an cos + Â bn sin 2 3 3 Given: y0 = 9, y1 = 18, y2 = 24, y3 = 28, y4 = 26 and y5 = 20. a0 = 2 ¥ 1 3

i.e., a1 =

1 6

 yr =

125 1 = 41.67; a1 = 3 3

 yr cos

p xr 3

p˘ 25 È Í( y0 - y3 ) + ( y1 + y5 - y2 - y4 ) cos 3 ˙ = - 3 = - 8.33 Î ˚

p xr 1 p 4 = ( y1 + y2 - y4 - y5 ) sin = - ¥ 0.866 = - 1.15 3 3 3 3 2p xr 1 È 1 p˘ = Í( y0 + y3 ) - ( y1 + y2 + y4 + y5 ) cos ˙ a2 = Â yr cos 3 3 3Î 3˚

b1 =

1 3

=-

 yr sin

7 = - 2.33 3

SQP7.6

Transforms and Partial Differential Equations

1 2p 1 p  yr sin 3 xr = 3 ( y1 + y4 - y2 - y5 ) sin 3 = 0. 3 \Required F.S. is px pxˆ 2p x Ê - 1.15 sin - 2.33 cos . y = 20.84 + Á -8.33 cos Ë 3 3 ˜¯ 3

b2 =

12. (a)(i) The derivation of Parseval’s identity for Fourier transforms is a standard book-work, available in the book. 12. (a)(ii) This problem is a worked example in the book 12. (b)(i) Statement and proof of convolution theorem in Fourier transforms is a standard result, that is available in the book 12. (b)(ii) This problem is a worked example in the book 13. (a)(i) This problem is a worked example in the book. In the book, the function is taken as f instead of f. 13. (a)(ii) This problem is a worked example in the book. 13. (b) (i) This problem is a worked example in the book. 13. (b)(ii) This problem is a worked example in the book. In the worked example, there are 2 terms in the R.S. of the P.D.E., namely xey + yex. As only one term, namely xey, is given in the question paper problem in the R.S. So it is enough we find (P.I.)1. The general solution of the given P.D.E. is z = f1(2y – x) + e–3x f2(y + x) + 1/4 (2x – 5)ey. 14. (a) This problem is a worked example in the book. 14. (b) This problem is a worked example in the book. 15. (a)(i) Z{n(n - 1)(n - 2)} = Z{n3 - 3n2 + 2n} = =

z( z 2 + 4 z + 1) ( z - 1)

4

-3

z( z + 1) ( z - 1)

3

+2

z ( z - 1)2

z

È( z 2 + 4 z + 1) - 3( z 2 - 1) + 2( z 2 - 2 z + 1)˘ ˚ ( z - 1)4 Î

= 6

z ( z - 1)4

8z 2 ÔÏ Ô¸ Z -1 ÏÔ z ◊ z ¸Ô = Z -1 ÌÏ z ˝¸ * Z -1 ÏÔ z ¸Ô 15. (a)(ii) Z -1 Ì ˝ = 1 1˝ Ì Ì 1˝ Ó z - 1/ 2 ˛ ÔÓ (2 z - 1)(4 z - 1) Ô˛ ÔÓ z - 2 z - 4 Ô˛ ÔÓ z - 4 Ô˛ n

n

n Ê 1ˆ Ê 1ˆ Ê 1ˆ = Á ˜ *Á ˜ = ÂÁ ˜ Ë 2¯ Ë 4¯ Ë ¯ r =0 2

n-r

Ê 1ˆ ◊Á ˜ Ë 4¯

r

Solved Question Papers-7

SQP7.7

Ï Ê 1 ˆ n +1 ¸ Ô n Ô1 - Á ˜ n -1 2n Ê 1ˆ Ô Ë 2¯ Ê 1ˆ Ô Ê 1ˆ =Á ˜ Ì -Á ˜ ˝=Á ˜ 1 Ô Ë 2¯ Ë 2¯ Ô Ë 2¯ 12 Ô Ô Ó ˛ 15. (b) (i) yk + 2 + yk = 1; y0 = 0 and y1 = 0 Taking Z-transforms of the equation, z [ z 2 y ( z ) - z 2 y(0) - z y(1)] + y ( z ) = z -1

i.e., ( z 2 + 1) y ( z ) = \ y ( z) = \

z ( z - 1)

z ( z - 1)( z + 1) 2

;\

y ( z) 1 A Bz + C = = + 2 2 z ( z - 1)( z + 1) z - 1 z + 1

y (z) 1 È 1 z 1 ˘ = Í - 2 - 2 ˙ z 2 ÍÎ z - 1 z + 1 z + 1 ˚˙

\ y (z) =

1 2

È z z2 z ˘ - 2 - 2 Í ˙ ÍÎ z - 1 z + 1 z + 1 ˙˚

Inverting, y(k) =

1 2

kp kp ˘ È Í1 - cos 2 - sin 2 ˙ Î ˚

(ii) This problem is a worked example in the book.

Question Paper Code 21522 B.E./B.Tech. Degree Examinations, May/June 2013 Regulations 2008 Third Semester Common to all branches MA2211/MA 31/MA 1201 A/CK 201/10177 MA 301/09010008/080210001 Transforms and Partial Differential Equations Time: Three Hours

Maximum: 100 Marks Answer ALL Questions Part A – (10 ¥ 2 = 20 Marks)

1. State the Dirichlet’s conditions for Fourier series. 2. What is meant by Harmonic Analysis? 3. Find the Fourier Sine Transform of e–ax 4. If F{f(x)} = F(s), prove that F{f(ax)} = 5. 6. 7. 8.

1 Ê sˆ ◊FÁ ˜ Ëa¯ 2

Form the PDE from (x – a)2 + (y – b)2 + z2 = r2. Find the complete integral of p + q = pq. In the one dimensional heat equation ut = c2 ◊ uxx, what is c2? Write down the two dimensional heat equation both in transient and steady states.

9. Find Z(n). È ˘ z 10. Obtain Z -1 Í ˙ ( z 1)( z 2) + + Î ˚

Part B – (5 ¥ 16 = 80 marks) 11. (a) (i) Find the Fourier series of x2 in (–p, p) and hence deduce that p4 90 14 2 4 34 (ii) Obtain the Fourier cosine series of 1

+

1

+

1

+

0 < x < l/2 Ï kx, f(x) = Ì Ók (l - x ) l / 2 < x < 1

Or

(8)

(8)

SQP8.2

Transforms and Partial Differential Equations

11. (b)(i) Find the complex form of Foureir series of cos ax in (–p, p), where a is not an integer. (ii) Obtain the Fourier cosine series of (x – 1)2, 0 < x < 1 and hence show that

1 12

1

+

22

+

1 32

+=

p2 . 6

ÏÔ1, 12. (a)(i) Find the Fourier transform of f ( x ) = Ì ÔÓ0, • sin x Ú x dx 0

x a

(ii) Verify the convolution theorem under Fourier transform, for f(x) = g(x) = e–x

2

Or 2 (b)(i) Obtain the fourier transform of e- x /2 . •

(ii) Evaluate

dx

Ú ( x 2 + a 2 )2

(8)

using Parseval’s identity

0

13. (a)(i) Solve: x(y2 – z2)p + y(z2 – x2)q = z(x2 – y2) (ii) Solve: (D2 + DD¢ – 6D¢2)z = y ◊ cos x. Or (b)(i) Solve: z = px + qy +

p2 + q 2 + 1

(ii) Solve: (D3 – 7DD¢2 – 6D¢3) z = sin (2x + y). 14. (a) A tightly stretched string between the fixed end points x = 0 and x = l is initially at rest in its equilibrium position. If each of its points is given a velocity kx(l – x), find the displacement y(x, t) of the string. Or (b) An infinitely long rectangular plate is of width 10 cm. The temperature along the short edge y = 0 is given by 0< x 1. 2 z Ë z - 1˜¯ z ( z - 1)

Ï ¸ z -1 10. Z -1 Ì ˝ ∫ Z { f ( z )}, say ( z + 1)( z + 2) Ó ˛ f (z) 1 1 1 z z = = ; f (z) = \ z ( z + 1)( x + 2) z + 1 z + 2 z +1 z + 2

\ Z -1 { f ( z )} = (-1)n - (-2)n PART - B 11. (a)(i) Since f(x) = x is an even function in (–p, p), let the F.S. of f(x) = x2 a be 0 + Â an cos nx 2 2

4

=

2 È 2 sin nx Ê cos nx ˆ Ê sin nx ˆ ˘ Ú x cos nx dx = p ÍÎ x n - 2 x ÁË - n2 ˜¯ + 2 ÁË - n3 ˜¯ ˙˚ 0 0 2

(-1)n

n2

a0 =

p

p

2 an = p



2 p

Úx

2

dx =

0

2 2 ¥ p 3 = p 2. 3p 3

• (-1)n p + 4 Â 2 cos nx in (–p, p). 3 n =1 n 2

2 \ x ~

By Parseval’s theorem, 1 2 1 a0 + 4 2

i.e.,

 an2 =

1 4 1 ¥ 4p + 4 9 2 •



Â

1 2p 1

n=1 n

4

p

Úx

4

dx

-p

p

=

1 Ê x5 ˆ 1 ◊ = p4 5 p ÁË 5 ˜¯

4 4 Ê 1 1ˆ = Á - ˜ p4 = p Ë ¯ 5 9 45 n =1 n

\ 8Â •

\

1

 n4

n =1

1

4

=

4 1 4 p4 = p 45 ¥ 8 90

0

Solved Question Papers-8

SQP8.5

11. (a)(ii) Give an even extension for f(x) in (–l, 0), so that f(x) is made an even function in (–l, b). Let the F.C. series of f(x) in (0, l) be an =

=

2 l

l

Ú f ( x) cos 0

a0 np x . + Â an cos 2 l

np x dx l

l /2 l 2k È np x np x ˘ dx + Ú (l - x ) cos dx ˙ Í Ú x cos l ÎÍ 0 l l ˙˚ 0

l È 2 Ï ¸ Ê ˆ Í Ê np x ˆ np x Ô sin Á cos ˜ ÔÔ ˜ 2 k ÍÍÔ Á l - 1 Á - 2 l2 ˜ ˝ = Ìx Á ˜ l ÍÔ Á np ˜ n p ˜Ô Á ÍÔ Ë ÁË ˜¯ Ô ¯ l l2 ˛0 ÍÓ Î

˘ ˙ ˙ ˙ ˙ ˙ ˙ 2˚

l

Ï np x ˆ Ê np x ˆ ¸ Ê Ô sin cos Á ˜Ô Á Ô l ˜ + Ál ˜Ô + Ì(l - x ) Á ˝ ˜ np n2 p 2 ˜ Ô Ô Á ˜ Á ˜¯ Ô l Ë ¯ ÁË l ÔÓ l2 ˛

=

2k l

ÈÔÏ l 2 np l2 Ê np ˆ Ô¸ sin + 2 Á cos - 1˜ ˝ ÍÌ Ë ¯ Ô˛ 2 2 n p ÍÔ ÎÓ 2 np

\

-

l2 np l2 - 2 2 sin 2 np 2 n p

np ¸˘ Ï n Ì(-1) - cos ˝˙ 2 ˛˙˚ Ó

0, If n is odd Ï Ô = Ì 4 kl n /2 Ô 2 2 {(-1) - 1, If n is even Ón p

a0 =

2 l

l Èl /2 ˘ Í Ú kx dx + Ú k (l - x ) dx ˙ ÍÎ 0 ˙˚ l /2

l ˘ ÈÊ 2 ˆ l /2 Ï (l - x ) ¸ ˙ 2 k È l 2 l 2 ˘ kl Í x + = Í + ˙= Ì ˝ ÍÁË 2 ˜¯ l ÎÍ 8 8 ˙˚ 2 -2 ˛l /2 ˙ Ó 0 Î ˚ \ Required F.H.R. Cosine series of

=

f(x) is

2k l

kl 4 kl + 4 p2

Â

1

2 n even n

{(-1)n /2 - 1} cos

np x l

SQP8.6

Transforms and Partial Differential Equations

11. (b)(i) This problem is a worked example in the book. 11. (b)(ii) Giving an even extension for f(x) in (–1, 0), the function f(x) is made an even function in (–1, 1) • a Let the Fourier series of f(x) be 0 + Â an cos n p x in (–1, 1) 2 n =1 1

an =

2 ( x - 1)2 cos n p x dx 1 Ú0 1

È Ê cos n p x ˆ Ê sin np x ˆ ˘ sin n p x ˆ = 2 Í( x - 1)2 ÊÁ - 2( x - 1) Á - 2 2 ˜ + 2 Á - 3 3 ˜ ˙ ˜ Ë np ¯ Ë n p ¯ Ë n p ¯ ˚˙ 0 ÎÍ

=

4 2

n p2 1

a0 =

1 2 2 2 ( x - 1)2 dx = ÈÎ( x - 1)3 ˘˚ = Ú 0 1 0 3 3

1 4 + 3 p2

\ F.H.R.C. series of f(x) ~



1

 n2

cos np x in (0, 1)

(1)

n =1

Putting x = 0 in (1), which is a point of continuity of f(x) = (x – 1)2, we get



1 4 + 3 p2

1

 n2

=1 \

n =1



1

 n2

=

n =1

1ˆ p 2 p2 Ê 1- ˜ = . Á 4 Ë 3¯ 6

12. (a)(i) This problem is a worked example in the book 12. (a)(ii) Convolution theorem in Fourier transform is F [ f ( x ) * g( x )] = f (s) g (s) ... (1) 2

If we assume that f(x) = g(x) = e- x , then f (s) = g (s) = p e- s L.S. of (1) =

È• ˘ -i s x Ú ÍÍ Ú f ( x - u) ◊ g(u) du˙˙ e dx -• Î -• ˚

=

Ê • - ( x - u )2 - u 2 ˆ - i s x e du˜ e dx ÚÁÚe ¯ -• Ë -•

2

/4







=

Ú

2

e - u du

-•



Ú

2

e - ( x - u ) e - i s x dx =

-•



Úe

- u2

2

◊ e - i u s F (e - x ),

-•

by shifting property 2

= F (e - x ) ◊



Úe

-•

2

- (u + i u s )

2

du = F (e - x ) ◊



Ú

-•

2 ÈÊ i sˆ i 2 s2 ˘ ˙ - ÍÁ u + ˜ 2¯ 4 ˙ ÍÎË ˚ e

du

Solved Question Papers-8

2

= F (e - x ) ¥ e - s

2

• /4

Úe

- v2

is 2

dv, where v = u +

-• 2

= F (e - x ) ◊

SQP8.7

( p e )= f (s) ◊ g (s) = R.S. of (1) - s2 /4

Hence convolution theorem is verified 12. (b)(i) F (e- a

2 2

x

)=

p - s2 /4 a2 e . This problem is a worked example in the a

book. Putting a =

1 2

F (e - x

in that worked example, we get 2

/2

) = 2p e - s

2

/2

12. (b)(ii) This problem is a worked example in the book. We have to apply Parseval’s identity to FC(e–ax). 13. (a)(i) x(y2 – z2) p + y(z2 – x2) q = z(x2 – y2) This is a Lagrange’s equation with P = x(y2 – z2) etc. The LSSE’s are

dx x( y - z ) 2

2

=

dy y( z - x ) 2

2

=

dz z( x - y 2 ) 2

(1)

Using the multiplies 1/x, 1/y, 1/z in (1), Each ratio in (1) = \

1/ x dx + 1/ y dy + 1/ x dz 0

1 1 1 dx + dy + dz = 0; Integrating, log xyz = log a or xyz = a ...(2) x y z

Using the multipliers x, y, z in (1), x dx + y dy + z dz ; \ x dx + y dy + z dz = 0 0 Integrating, x2 + y2 + z2 = b ... (3).

Each ratio =

\

(

)

G.S. of the given equation is f xyz, Â x 2 = 0

13. (a)(ii) (D2 + DD¢ – 6D¢2) z = y cos x The A.E. is m2 + m – 6 = 0, i.e. (m + 3)(m – 2) = 0 \ m = –3, 2. \ C.F. = f1(y – 3x) + f2 (y + 2x) P.I. =

1 y ¥ R.P. of eix ( D + 3D ¢ )( D - 2 D ¢ )

È 1 = R.P. of eix Í Î {( D + i ) + 3D ¢}{( D + i ) - 2 D ¢}

˘ y˙ ˚

SQP8.8

Transforms and Partial Differential Equations

Ê ˆ 1 = R.P. of eix ◊ Á y = R.P. of [ -eix (1 + iD ¢ ) y] Ë -1 + iD ¢ ˜¯ = R.P of [–(cos x + i sin x) (y + i] = –y cos x + sin x

\ G.S. of the given equation is z = C.F. + P.I. 13. (b) (i) z = px + qy +

p2 + q 2 + 1

This is a worked example in the book. We have to put c = 1 in the worked example 13. (b)(ii) (D3 – 7DD¢2 – 6D¢3) z = sin (2x + y) A.E. is m2 – 7m – 6 = 0, viz., (m + 1) (m + 2) (m – 3) = 0 \ m = –1, –2, 3 \ C.F. f1(y – x) + f2(y – 2x) + f3(y + 3x), where f1, f2, f3 are arbitrary functions. 1 -1 sin (2 x + y) = sin (2 x + y) P.I. = 3 2 3 4( D - 5D ¢ ) D - 7 DD ¢ - 6 D ¢ 1 ( D + 5D ¢ ) = - ◊ sin (2 x + y) 4 D 2 - 25D ¢ 2 1 1 ¥ 7 cos (2 x + y) = - cos (2 x + y) 84 12 \ G.S is z = C.F + P.I. =-

14. (a) This problem is a worked example in the book. In the worked example, the length of the string is taken as 2 l. If we replace 2 l by L or l by (L/2) and work out the problem, the required solution will be obtained as y( x, t ) =

8kL3 p 4a



1

(2 n - 1) p x 2 n p at ◊ sin L L

1

(2 n - p ) p x (2 n - 1)p at sin l l

 (2n - 1)4 sin

n =1

\ The required solution is y( x, t ) =

8kl 3 4

p a



 (2n - 1)4 sin

n =1

14. (b) This problem is the same as the problem given in question 14(b) of the April/May 2010 question paper. The solution is given under April/May 2010 question paper. z z 15. (a)(i) Z (a n ) = ; \ Z (ei nq ) = Z{(ei q )n } = ( z - a) z - eiq i.e., Z (cos n q + i sin nq) =

z z{( z - cos q ) + i sin q} = ( z - cos q ) - i sin q ( z - cos q )2 + sin 2 q

Solved Question Papers-8

SQP8.9

Equating the real parts, z( z - cos q ) , if z > 1 Z(cos nq) = = 2 z - 2 z cos q + 1 np ˆ z2 \ Z ÊÁ cos = Ë 2 ˜¯ z 2 + 1

15. (a)(ii) yn+2 – 3yn+1 – 10yn = 0; y0 = 1 and y1 = 0 Taking z-transform of the given equation, [ z 2 y ( z ) - z 2 y(0) - z y(1)] - 3[ z y ( z ) - z y(0)] - 10 y ( z ) = 0

i.e., [ z 2 - 3z - 10) y ( z ) = z 2 - 3z i.e., y ( z ) =

z 2 - 3z y (z) z-3 2 1 5 1 \ = = ◊ + ◊ ( z - 5)( z + 2) z ( z - 5)( z + 2) 7 z - 5 7 z + 2

2 z 5 z + ◊ \ y (z) = ◊ 7 z-5 7 z+2 2 5 ¥ 5n + ¥ (-2)n 7 7 15. (b) (i) Time shifting property:

Inverting; we get yn =

If Zf (n) = f ( z ), then Z{f (n - n0 )} = z - n0 ◊ f ( z ) z Frequency shifting property: Z{a n f (n)} = f ÊÁ ˆ˜ . Ë a¯ The proofs are available in the book ÏÔ ¸Ô z2 z ¸ -1 Ï -1 15. (b)(ii) Z -1 Ì ˝ = Z Ì ˝* Z ( z a )( z b ) ( z a ) Ó ˛ ÓÔ ˛Ô by convolution theorem = an * bn

Ï z ¸ Ì ˝, Óz - b˛

n

=

 ar ◊ bn - r

r =0

Ï Ê a ˆ n +1 ¸ - 1Ô ÔÁ ˜ n Ê aˆ ÔË b¯ Ô = bn  Á ˜ = bn ◊ Ì ˝ a Ë ¯ b r=Ô -1 Ô Ô b Ô Ó ˛ n

È a n +1 - b n +1 ˘ = bn Í n ˙ ÍÎ (a - b) b ˙˚ 1 [ a n +1 - b n +1 ] = a-b

B.E./B.Tech. Degree Examinations, November/December 2014 Question paper code: 97107 Third Semester/ Civil Engineering Common to all branches MA6351 Transforms and Partial Differential Equations (Regulation - 2013) Time: Three hours

Maximum: 100 marks Answer ALL Questions Part A – (10 ¥ 2 = 20 marks)

1. Form the partial differential equation by eliminating the arbitrary function Ê yˆ f from z = f Á ˜ . Ë x¯ 2. Find the complete solution of p + q = 1 3. State the sufficient conditions for existence of Fourier series. • p2 cos nx +4 4. If (p - x) 2 = in 0 < x < 2p, then deduce that value of 2 3 n =1 n

Â



1

Ân n =1

5

2

Classify the partial differential equation (1 - x 2 ) z xx - 2 xy z xy + (1 - y 2 ) z yy + xz x + 3x 2 yz y - 2 z = 0

6. Write down the various possible solutions of one dimensional heat flow equation. 7. State and prove modulation theorem on Fourier transforms. 8. If F {f (x)} = F(s), then find F {eiaxf (x)}. 9. Find the z-transform of n. 10. State initial value theorem on z-transforms. Part B – (5 × 16 = 80 marks) 11. (a) (i) Find the singular solution of z = px + qy + p2 – q2 (ii) Solve (D2 – 2DD¢) z – x3y + e2x – y

(8) (8)

Or 11. (b) (i) Solve x(y – z)p + y (z – x) q = z (x – y). 11. (b) (ii) Solve (D3 – 7DD¢2 – 6D¢3) z – sin (x + 2y)

(8) (8)

SQP9.2

Transforms and Partial Differential Equations

12. (a) (i) Find the Fourier series of f (x) = x2 in – p < x < p. Hence, deduce the •

value of

1

Ân n =1

2

.

(8)

(ii) Find the half range cosine series expansion of (x – 1)2 in 0 < x 0. Hence show that e- x /2 is self reciprocal under the Fourier transform. (8) x dx (ii) Evaluate (8) , using Fourier transforms 0 ( x 2 + a 2 )( x 2 + b 2 ) 15. (a) (i) Find Z(cos n q) and Z (sin n q). (8) (ii) Using z-transforms, solve un + 2 - 3un +1 + 2un = 0 given that u0 = 0, u1 = 1. (8) Or 1 , for n ≥ 1 (b) (i) Find the z-transforms of (8) n(n + 1)

Ú

Solved Question Papers-9

(ii) Find the inverse Z-transforms of

SQP9.3

z2 + z , using partial fraction ( z - 1)( z 2 + 1) (8)

Solutions Part A Ê yˆ 1. z = f ÁË ˜¯ x

(1)

Ê yˆ Differentiating (1) partially w.r.t. x, we get p = f ¢(u). Á - 2 ˜ Ë x ¯ y where u = . x Similarly, differentiating (1) partially w.r.t. y, we get q = f ¢(u) ÊÁ 1 ˆ˜ Ë x¯

(2),

(3)

Ê pˆ y (2) ∏ (3) gives Á ˜ = - ÊÁ ˆ˜ ; i.e. px + qy = 0 is the required P.D.E. Ë x¯ Ë q¯ 2. p + q = 1 (1) Let the C.S. be z = ax + by + c (2) From (2), p = a and q = b. Since (2) is a solution of (1), a + b = 1 ∴ b = 1 – a. ∴ The required C.S. is z = ax + (1 – a) y + c 3. The conditions are available in Chapter 2, page 2-2 (Dirichlet’s Conditions). •

4.

(p - x) 2 =

p2 cos nx +4 in 0 < x < 2p 2 3 n =1 n

Â

(1) •

Putting x = 0 in the R.H.S. of (1), we get

1

Ân n =1

2

But x = 0 is a point of discontinuity at the left extremity. ∴ (Sum of the R.H.S. series)x = 0 = •

i.e.

p2 1 1 +4 = lim 2 3 2 hÆ0 n =1 n

Â

= ∴

Â

1 lim [f (0 – h) + f (0 + h)] 2 hÆ0

[ f (2p + h) +

f (h) ]

1 lim ÈÎ( - p - h)2 + (p - h)2 ˘˚ = p 2 2

1 p2 = 6 n2

SQP9.4

Transforms and Partial Differential Equations

5. A = 1 –x2, B = –2xy, C = 1 –y2 B2 –4AC = 4[x2y2 – (1 – x2) (1 – y2)] = 4 (x2 + y2 – 1) ∴ The given P.D.E. is of the elliptic type inside the circle x2 + y2 = 1, parabolic type on the circle x2 + y2 = 1and hyperbolic type outside the circle x2 + y2 = 1. 6. The possible solutions of the equation ut = a2uxx are 2 2 (i) u (x, t) = (Aepx + Be – px)ep a t 2 2 (ii) u (x, t) = (A cos px + B sin px)e–p a t and (iii) u (x, t) = (A x + B) 7. This is a standard property of Fourier transforms, explained in Chapter 4 on page 4-28. 8.

F{eiax f ( x)} =

Ú



=

Ú



-•

-•

eiax f ( x)e - isx dx f ( x)e - i ( s - a ) x dx = f ( s - a )

9. This is a standard result given Chapter 5 on page 5-8. 10. This is a standard property of z-transform given in Chapter 5 on page 5-4. PART - B 11. (a)(i) z = px + qy + p2 – q2 The C.S. of (1) is z = ax + by + (a2 – b2)

(1) (2)

Differentiating (2) w.r.t. a, we get x + 2a = 0

(3)

Differentiating (2) w.r.t. b, we get y –2 b = 0 y x From (3) and (4), we get a = – and b = 2 2 Using (5) in (2), the required singular solution is

(4)

z=

- x2 y 2 x2 y 2 + + 2 2 4 4

i.e. z =

y 2 x2 or 4 z = y 2 - x 2 4 4

(ii) (D2 – 2DD¢) z = x3y + e2x – y A.E. is m2 – 2m = 0; i.e. m(m – 2) = 0 or m = 0, 2 ∴

C.F. = f1(y) + f2 (y + 2x)

P.I.1 =

1 1 Ê 2D¢ ˆ x3 y = 2 Á1 2 D ˜¯ D - 2 DD ¢ D Ë

-1

x3 y

(5)

Solved Question Papers-9

=

1 Ê 2D¢ ˆ 3 Á1 + D ˜¯ x y D2 Ë

2D ¢ ˆ 1 5 1 6 Ê 1 = Á 2 1 + 3 ˜ x3 y = x y+ x ËD 20 60 D ¯ P.I.2 =

1 1 1 e2 x - y = e2 x - y = e2 x - y 4 - 2 ¥ 2 ¥ ( - 1) 8 D - 2 DD ¢ 2

∴ G.S. is z = C.F. + P.I.1 + P.I.2 11. (b) (i) x(y–z) p + y (z – x) q = z(x – y) dx dy dz LSSE’S are = =  (1) x( y - z ) y ( z - x) z ( x - y ) Using the multipliers 1, 1, 1, each ratio in (1) = dx + dy + dz 0 ∴ dx + dy dz = 0, hence, one solution is x + y + z = a 1 1 1 Using the multipliers , , , x y z 1 1 1 dx + dy + dz x y z Each ratio in (1) = 0 1 1 1 dx + dy + dz = 0 \ x y z Integrating, we get log x + log y + log z = log b ∴ Second solution is xy z = b ∴ The G.S. of the given equation is f (x + y + z, xyz) = 0 (ii) (D3 – 7DD¢2 – 6D¢3) z = sin (x + 2y) A.E. is m3 – 7m – 6 = 0; i.e. (m + 1)(m + 2)(m – 3) = 0 ∴ m = –1, – 2, 3 ∴ C.F. = f1 (y – x) + f2 (y – 2x) + f3 (y + 3x) P.I. = =

1 sin (x + 2y) D - 7 DD ¢ 2 - 6 D ¢3 3

1 38D ¢ + D sin( x + 2 y ) = sin( x + 2 y ) 38D ¢ - D 1444 D ¢ 2 - D 2

1 74 (76 + 1) cos ( x + 2 y ) = cos( x + 2 y ) - 5775 5775 ∴ G.S. is z = CF + P.I. =

SQP9.5

SQP9.6

Transforms and Partial Differential Equations

12. (a) (i) f (x) = x2 is an even function of x in – p 0.

(10)

Solved Question Papers-10

SQP10.3

(ii) Find the Fourier cosine transform of xn–1

(6)

-1 -1 -2 15. (a)(i) Find Z(rn cos nq) and Z ÈÎ(1 - az ) ˘˚ .

(8)

È ˘ z2 -1 (ii) Using convolution theorem, find Z Í ˙. ÍÎ ( z - 1 / 2)( z - 1 / 4) ˙˚

(8)

(b)(i) Using Z-transform, solve the difference equation x(n + 2) – 3x(n + 1) + 2x(n) = 0 given that x(0) = 0, x(1) = 1. È ˘ z -1 (ii) Using residue method, find Z Í 2 ˙. Î z - 2z + 2 ˚

(8) (8)

Solutions Part A 1. log (az – 1) = x + ay + b

(1)

ap Differentiating (1) p.w.r.t x, =1 az - 1

Differentiating (1) p.w.r. t. y, (3) ∏ (2) gives

q =a p

q pq p = 1 or = 1 or z = p + is the required qz qz - p q -1 p

P.D.E. 2. q = 2px = a \

p=

a and q = a 2x

dz = p dx + q dy =

3.

I=

C.S. is z = •

Â

(3)

(4)

Using (4) in (2), we get

\

aq =a az - 1

(2)

a dx + a dy 2x

a log x + ay + b 2

I n sin (nw t + a n ) =

n = 1,3,5



Â

I n (cos a n )sin nw t + ( I n sin a n ) cos nw t

n = 1,3,

Effective value of I = =

1 2

n odd

1 2

 I 22n -1

 I n2 (cos2 a n + sin2 a n )



n =1

SQP10.4

Transforms and Partial Differential Equations

sin 2 x sin 3 x 4x Ê ˆ + - sin + ˜ in –p < x < p 4. x ~ 2 Á sin x Ë ¯ 2 3 4

Put x = 1-

(1)

p 1 1 ˆ p Ê in (1); 2 Á 1 - + ˜ = Ë 2 3 5 ¯ 2

1 1 1 p + - +• = 3 5 7 4

5. A = 1 – x2, B = –2xy, C = 1 – y2 B2 – 4 AC = 4 (x2 + y2 – 1) The given P.D.E. is of the elliptic or parabolic or hyperbolic type, according as x2 + y2 – 1 < 0, = 0 or > 0, viz., inside, on or outside the circle x2 + y2 = 1. d 2u 6. The SST is given by 2 = 0 or u(x) = Ax + B dx Using u(0) = 20, B = 20; using u(30) = 80, A = 2. \

SST is u(x) = 2x + 20

7. This is a standard property of F¢ transforms, covered in Chapter 4, page 4.27. 8. The solution to this problem is provided in Chapter 4. It is proved in the • sin sx p dx = (s > 0) . The L.H.S is the F¢ sine transform of example that Ú x 2 0 1 . x 9. This is a standard property of z-transform, given in 3(i) Chapter 5, page 5-3. 10. This is a standard theorem, given in Chapter 5, page 5-5. PART - B 11. (a)(i) LSSE’s are

i.e.,

dx x - yz 2

=

dy y - zx 2

=

dz z - xy 2

 (1)

(1)

viz.,

d ( x - y) d ( y - z) d(z - x) = = ( x - y) ( x + y + z ) ( y - z ) ( x + y + z ) ( z - x )( x + y + z )

viz.,

d ( x - y) d ( y - z ) ; Solving, log (x – y) – log (y – z) = log a = x-y y-z

x-y =a y-z

Also each ratio in (1) is equal to d ( x + y + z) x dx + y dy + z dz = 2 Sx - Syz ( x + y + x ) ( Sx 2 - Syz )

Solved Question Papers-10

i.e.

( x + y + z )2 1 2 = ( x + y2 + z2 ) + b 2 2

or

xy + yz + zx = b

SQP10.5

Êx-y ˆ \ Reqd. G.S. is f Á , xy + yz + zx ˜ = 0 y z Ë ¯

\

11. (a)(ii) A.E. is m2 – 3m + 2 = 0

m = 1, 2

C.F. = f1 (y + x) + f2(y + 2x) P.I. =

1 (2 + 4 x ) e x + 2 y ( D - D ¢ )( D - 2 D ¢ )

x + 2y = e

1 (2 + 4 x ) {D + 1 - ( D ¢ + 2)}{D + 1 - 2 ( D ¢ + 2)

x + 2y = e

1 (2 + 4 x ) ( D - D ¢ - 1)( D - 2 D ¢ - 3)

=

=

1 x+2y e 3

1 1 (2 + 4 x ) = e x + 2 y Dˆ 3 Ê (1 - D) Á 1 - ˜ Ë 3¯

1 x+ 2y e 3

4 ˆ Ê ÁË 1 + 3 D˜¯ (2 + 4 x )

22 ˆ Ê ÁË 4 x + 3 ˜¯

\ G.S is z = C.F. + P.I 11. (b)(i) This problem has been worked out in Chapter 1, page 1-41, worked example 16. 11. (b)(ii) The C.S. of the given PDE is z = ax + by + a2b2 Differentiating (1) p.w.r.t, a, we get x + 2ab2 = 0 Differentiating (1) p.w.r.t b, we get y + 2a2 b = 0 a y a b From (2) and (3), we get = ; = = k b x y x Using this in (2), k= -

¸ \ a = - y 2/3 /21/3 x1/3 Ô (2 xy) ˝ 2/3 1/3 1/3 Ô and b = - x /2 y ˛ 1

1/3

(1) (2) (3)

(4)

Using (3) and (4) in (1), we get 16z3 + 27x2y2 = 0, which is the required S.S. 12. (a)(i) Given an odd extension for f(x) in (–p, 0) \ f(x) is odd in (–p, p) Let the F.H.R.S.S. of f(x) be

 bn sin nx is (0, p) p

SQP10.6

Transforms and Partial Differential Equations p Èp /2 ˘ Í Ú x sin nx dx + Ú (p - x ) sin nx dx ˙ ÍÎ 0 ˙˚ p /2

2 bn = p

=

2 p

p È Ï Ê cos nx ˆ sin nx ¸p /2 Ï Ê cos nx ˆ sin nx ¸ ˘ ÍÌ x Á + + ( x ) p ˝ Ì ˝ ˙ ÁË n ˜¯ n ˜¯ ÍÓ Ë n2 ˛0 n2 ˛p /2 ˙ Ó Î ˚

Ï 0, np Ô = 2 sin =Ì 4 2 Ô± 2 n p p¢ n Ó

if n is even

4

\

f(x) ~

if n is odd

4 Ï1 1 1 ¸ Ì 2 sin x - 2 sin 3 x + 2 sin 5 x -  •˝ in (0, p ) p Ó1 3 5 ˛

p 1 1 1 p p p2 , we get 2 + 2 + 2 +  • = ¥ = 2 4 2 8 1 3 5 12. (a)(ii) Refer to the worked example (8) in Chapter 2 on page 2-87.

Putting x =

In the example, put a = 1 and l = 1 Required F.S. is –x

e = (sinh 1)



( -1)n (1 - i n p )

n = -•

1 + n2 p 2

Â

- ei n p x in ( - 1, 1)

12. (b)(i) F(x) = |sin x|; f(–x) = |-sin x| = f(x) \ f(x) is an even function in (–p, p) Let the F.S. be f ( x ) ~ an =

2 p

a0 + San cos nx in( -p , p ) 2

p

Ú

p

sin x cos nx dx =

0

1 2 sin x cos nx dx p Ú0

p

=

1 [sin (n + 1) x - sin (n - 1) x ] dx p Ú0

4 Ï , if n even 1 2 Ôn -1 2 {(-1) = ◊ 2 - 1} = Ì p (n - 1) p n -1 Ô 0 , if n odd Ó

a0 = \

|sin x| =

2 p



Ú sin x

dx =

0

2 4 p p

2 4 (- cos x )p0 = p p



Â

n = 2, 4, 6

1 n -1 2

cos nx in ( - p , p )

Solved Question Papers-10

SQP10.7

12. (b)(ii) Refer to question 12 (b) (i) of Solved Question Paper 7 (Nov-Dec 2014) for the solution. The same problem has been worked out in 12 (b) (i)

a3 = (y0 + y2 + y4) – (y1 + y3 + y5) = (1 + 1.9 + 1.5) – (1.4 + 1.7 + 1.2) = 4.4 – 4.3 = 0.1 and b3 = 0 \ Regd. F.S. is y = 1.45 – 0.367 cos x + 0.173 sin x – 0.1 cos 2x –0.058 sin 2x + 0.1 cos 3x Ï x, in 0 £ x £ l / 2 13. (a) B.C.’s (2) u(0, t) = 0; (3) u(l, t) = 0; (4) u(x, 0) = Ì Ól - x in l / 2 £ x £ l Proper solution of u(t) = a2 uxx (1) is u(x, t) = ( A cos px + B sin px ) e - p

Using (2) in (5), A = 0; Using (3) in (5), p = •

np x e \ Given solution is u(x, t) = Â Bn sin l n =1 Using (4) in (6);

Bn = =

2 l

2

a 2t

(5)

np ; n = 0, 1, 2, ... • l

n 2 p 2a 2 l2

l

(6)

/ Èl /2 np x np x ˘ dx + Ú (l - x ) sin dx ˙ Í Ú x sin l l ÍÎ 0 ˙˚ l /2

4l 2

n p

\ u(x, t) =

2

4l p

2

sin

¸Ô np ÏÔ 4l ( -1)n + 1 =Ì 2 ; n = 1, 2, 3, ˝ 2 2 ÓÔ p (2n - 1) ˛Ô



( -1)n + 1

 (2n - 1)2 sin

n =1

(2 n - 1) p x - (2 n - 1)2 p 2a 2 t / l 2 e l

13. (b) This problem has been worked out in Chapter 3, page 3-12 as worked example 1 (ii). 14. (a) This problem has been worked out in Chapter 4, page 4-11 in worked example 5. • • By Parseval’s identity, | f ( x )|2 dx = 1 | f (s)|2 ds Ú Ú 2p -• -• a

i.e.

2 Ú 1 dx =

-a

1 2p



Ú

-•

4 s

2

sin 2 as ds

SQP10.8

Transforms and Partial Differential Equations

i.e.

a= •

Ú

\

1 p



Ú

t

s2

-•

sin 2 t

0

sin 2 as

2

dt =



ds ; Putting a = 1,

Ú

sin 2 s

-•

s2

ds = p ;

p 2

14. (b)(i) This problem has been worked out in Chapter 4, page 4-14 in worked example 8. 14. (b)(ii) This problem has been worked out in Chapter 4, page 4.21 as worked example 16. 15. (a)(i) Z(rn cos n q) – This problem has been worked out in Chapter 5, page 5-11. -2 ÈÊ aˆ ˘ 2 a 3a 2 ÔÏ Ô¸ Z–1 [(1 – az–1)–2] = Z -1 ÍÁ 1 - ˜ ˙ = z -1 Ì1 + + 2 + ˝ Ë ¯ z a z ÍÎ ˙˚ ÔÓ Ô˛ =



 (n + 1) a n z - n

n=0

= Coefficient of z–n in S = (n + 1) an -1 15. (a)(ii) Z

È ˘ Í ˙ 2 z Í ˙ = Z -1 ÍÊ 1ˆ Ê 1ˆ ˙ Í ÁË z - 2 ˜¯ ÁË z - 4 ˜¯ ˙ Î ˚ n

1 1 = ÊÁ ˆ˜ * ÊÁ ˆ˜ Ë 2¯ Ë 4¯ •

=

Ê 1ˆ  ÁË 2 ˜¯ r =0

Ê 1ˆ =Á ˜ Ë 2¯

n

n -r

ÈÏ ¸ Ï ¸˘ z z ÍÌ ˝* Ì ˝˙ ÍÎÓ ( z - 1 / 2 ˛ Ó ( z - 1 / 4 ˛˙˚

n

r

Ê 1ˆ Ê 1ˆ ÁË 4 ˜¯ = ÁË 2 ˜¯

n

n

r

Ê 1ˆ  ÁË 2 ˜¯ by Convolution r =0

n +1 ¸ ÏÔÊ 1 ˆ n 1 ÔÏ 1 - (1/2) Ô Ì ˝ = 2 ÌÁ ˜ 2 ÔÓ 1 - 1 / 2 Ô˛ ÔÓË 2 ¯ n

Ê 1ˆ Ê 1ˆ = 2Á ˜ - Á ˜ Ë 2¯ Ë 4¯

Ê 1ˆ ÁË 4 ˜¯



Ô ˝ Ô˛

n

15. (b) (i) Taking z-transform on both sides of the difference equation, we have [ z 2 x ( z ) - z 2 x (0) - z x(1)] - 3 [ z x ( z ) - z x(0)] + 2 x ( z ) = 0 i.e., ( z 2 - 3z + 2) x ( z ) = z \

x (z) 1 1 1 = 2 = z z - 3z + 2 z - 2 z - 1

Solved Question Papers-10

\ \

SQP10.9

z z z - 2 z -1 x(n) = 2n – 1 x (z) =

Ï zn z Ô¸ = sum of the residues of 15. (b)(ii) Z -1 ÔÌ at its ˝ 2 z2 - 2z + 2 ÔÓ z - 2 z + 2 Ô˛ singularities. zn

are given by z2 – 2z + 2 = 0, z2 - 2z + 2 i.e. z = 1 + i and z = 1 – i which are simple poles The singularities of

Ê zn ˆ 1 = (1 + i )n Res z = 1 + i = Á ˜ Ë z - 1 + i ¯ z = 1 + i 2i Ê zn ˆ 1 = - (1 - i )n Res z = 1 – i = Á ˜ z 1 i 2 i Ë ¯ z =1- i Ï z Ô¸ 1 -1 Ô n n \ z Ì 2 ˝ = {(1 + i ) - (1 - i ) } ÔÓ z - 2 z + 2 Ô˛ 2i =

( 2 )n 2i

ÏÊ np np ˆ Ê np np ˆ ¸ + i sin ÌÁ cos ˜ - ËÁ cos 4 - i sin 4 ¯˜ ˝ Ë ¯ 4 4 Ó ˛

( 2 )n np ¥ 2i sin 2i 4 np = ( 2 )n ◊ sin 4

=

B.E./B.Tech. Degree Examinations, November/December 2015 Question paper code: 27327 Third Semester/ Civil Engineering Common to all branches MA6351 Transforms and Partial Differential Equations (Regulation - 2013) Time: Three hours

Maximum: 100 marks Answer ALL Questions Part A – (10 ¥ 2 = 20 marks)

1. Construct the partial differential equation of all spheres whose centers lie on the Z – axis, by the elimination of arbitrary constants. 2. Solve (D + D¢– 1) (D – 2D¢ + 3)z = 0. 3. Find the root mean square value of f(x) = x(l – x) in 0 £ x £ l. 4. Find the sine series of function f(x) = 1, 0 £ x £ p. ∂u ∂u 5. Solve 3 x - 2 y = 0 by method of separation of variables. ∂x ∂y 6. Write all possible solutions of two dimensional heat equation ∂2 u ∂2 u + =0 ∂x 2 ∂y 2 7. If

is the Fourier 1 Ê sˆ F{ f (ax )} = F ,aπ0 | a | ÁË a ˜¯ F(s)

Transform

of

f(x),

prove

that



8. Evaluate

s2 Ú (s2 + a2 )(s2 + b2 ) using Fourier Transforms. 0

1 . n +1 10. State the final value theorem in Z-transform. 11. (a)(i) Find complete solution of z 2 ( p2 + q 2 ) = ( x 2 + y 2 ) 9. Find the Z-transform of

(8)

(ii) Find the general solution of ( D 2 + 2 DD ¢ + D ¢ 2 ) z = 2 cos y - x sin y. (8) Or (b)(i) Find the general solution of ( z 2 - y 2 - 2 yz ) p + ( xy + zx )q = ( xy - zx ). (8) 2 2 2 2 (ii) Find the general solution of (D + D¢ ) z = x y . (8)

SQP11.2

Transforms and Partial Differential Equations

12. (a)(i) Find the Fourier series expansion the following periodic function of period 4. Ï2 + x - 2 £ x £ 0 ¸ 1 1 1 p2 f ( x) = Ì (8) ˝ Hence deduce that 2 + 2 + 2 + = 8 1 3 5 Ó2- x 0 £ x £ 2 ˛ (ii) Find the complex form of Fourier series of f(x) = eax in the interval (–p, p) where a is a real constant. Hence, deduce that •

p ( - 1)n = 2 2 a sinh a p n = -• a + n

Â

Or (b)(i) Find the half range cosine series of f(x) = (p – x)2, 0 < x < p. Hence 1 1 1 find the sum of series 4 + 4 + 4 +  (8) 1 2 3 (ii) Determine the first two harmonics of Fourier series for the following data. p 2p 4p 5p x:0 p 3 3 3 3 y0 y1 y2 y3 y4 y5 f(x): 1.98 1.80 1.05 1.30 – 0.88 – 0.25 13. (a) A tightly stretched string with fixed end points x = 0 and x = 1 is initially at rest in its equilibrium position. If it is vibrating by giving to each of its points a velocity 1 Ï 2kx in 0 < x < ÔÔ l 2 v=Ì Ô 2k (l - x ) in 1 < x < 1 ÔÓ 2 l Find the displacement of the string at any distance x from one end at any time t. Or (b) A bar 10 cm long with insulated sides has its ends A and B maintained at temperature 50°C and 100°C, respectively, until steady state conditions prevails. The temperature at A is suddenly raised to 90°C and at the same time lowered to 60°C at B. Find the temperature distributed in the bar at time t. (16) 14. (a) (i) Find the Fourier sine integral representation of the function f(x) = e–x sin x. (8) e- ax - e- bx (ii) Find the Fourier cosine transform of the function f ( x ) = . x Or

Solved Question Papers-11

SQP11.3

Ï1 - | x |, x £ 1 ¸ (b) (ii) Find the Fourier transform of the function f ( x ) Ì ˝. | x | > 1˛ Ó0, •

4

p Ê sin t ˆ Hence deduce that Ú Á dt = . ˜ Ë ¯ t 3 0

(8)

(ii) Verify the convolution theorem for Fourier transform if f(x) = g (x) = 2 e–x . z3 + z 15. (a)(i) If u ( z ) = find the value of u0, u1, and u2. (8) ( z - 1)3 Ï z2 Ô¸ (ii) Use convolution theorem to evaluate Z -1 ÔÌ ˝ ÓÔ ( z - 3)( z - 4) ˛Ô (b) (i) Using the inversion integral method (Residue Theorem), find the z2 ( z + 2)( z 2 + 4) (ii) Using the z-transform solve the difference un + 2 + 4un +1 + 3un = 3n , given u0 = 0, u1 = 1. inverse Z-transform of u( z ) =

equations

Solutions PART - A 1. The equation of any sphere whose centre is (0, 0, a) [viz. a point on the z-axis] and whose radius is b is x2+ y2+ (z – a)2 = b2 (1) If a and b are treated as arbitrary constants, (1) represents the family of the given spheres Differentiating (1) p.w.r. t. x; 2x + 2(z – a) p = 0 (2) Differentiating (1) p.w r t x; 2y + 2 (z – a) q = 0 (3) x p From (2) and (3), we get = or qx - py = 0 , which is the required y q PDE. 2. Since the solutions of (D – a D¢– b) z = 0 is z = ebx f (y + ax), the solution of the given equation is z = ex f1 (y – x) + e–3xf2 (y + 2x) l

3. R.M.S. Value =

=

l

1 2 1 2 2 x (l - x )2 dx = (l x - 2lx 3 + x 4 ) dx Ú l 0 l Ú0 1 Ê l5 l5 l5 ˆ 1 4 l2 l + = l ÁË 3 2 5 ˜¯ 30 30

4. Give an odd extension for f(x) in – p £ x £ 0; i.e. put f (x) = –1 in (–p, 0) f (x) is odd in (–p, p) Let the F.S.S. be  bn sin nx.

SQP11.4

Transforms and Partial Differential Equations

if n is even Ï0, p 2 Ô bn = Ú 1.sin nx dx = Ì 4 p 0 ÔÓ np , if n is odd \

Required sine series is 1 =

4 p



1 sin nx in (0, p ) n = 1,3,5,... n

Â

5. Let u = X(x), Y(y) be a solution of the given equation X¢ Y¢ Then 3 x = 2 y = k , say \ X 3 = Ax k or X = ax k /3 and Y = by k /2 X Y k /3 k /2 k /3 k /2 \ Required solution is y = ab x y or y = c x y .

6. All possible solution are given in Chapter 3, page 3-120. 7. This is the standard change of scale property of F¢ transforms that is given in Chapter 4, page 4-27. 8. This problem is worked out in Chapter 4, page 4-39 as worked example 7 (ii). 9. This problem is worked out in Chapter 5, page 5-10 as worked example (Z-transform of basic functions) 6 (ii). 10. Final value theorem in Z-transform is available as Property (7) in in Chapter 5, page 5-4. PART - B ∂Z ∂Z 2 11. (a) (i) Put Z = y \ P = = 2 yp and Q = = 2 yq ∂x ∂y Using these in the given equation, it becomes P2 + Q2 = 4 (x2 + y2) i.e. P2 – 4x2 = 4y2 – Q2 = 4a2, say \

P = 2 x 2 + a 2 and Q = 2 y 2 - a 2

dZ = Pdx + Qdy = 2 x 2 + a 2 dx + 2 y 2 - a 2 dy Integrating, x y + y y 2 - a 2 - a 2 cosh -1 + b a a (ii) A.E is m2 + 2m + 1, i.e. m = –1, –1 \ C.F = x f1(y – x ) + f2(y – x) 1 2 2 cos y = cos y = - 2 cos y P.I.1 = 2 0 + 0 -1 D + 2 DD ¢ + D ¢ 2 ( Z = ) y 2 = x x 2 + a 2 + a 2 sinh -1

1 1 x I.P. of x eiy = I.P. of eiy 2 (D + D ¢) ( D + D ¢ + i )2 = – I.P. of eiy (x + 2i) = –x sin y – 2 cos y \ G.S is Z = C.F + P.I1 – P.I2 P.I.2 =

Solved Question Papers-11

SQP11.5

11. (b) (i) This problem is given in Chapter 1, page 1-60 as worked example 11. (ii) (D2 + D¢2)z = x2y2. A.E is m2 + 1 = 0 \ m = ± i, \ C.F = f1 (y + ix) + f2(y – ix) 1 1 Ê D ¢2 ˆ 2 2 2 2 PI = x y 1 = Á ˜x y D2 Ë D2 ¯ D ¢2 ˆ 2Ê D Á1 + 2 ˜ D ¯ Ë 1 2 2 1 x4 Ê 2 x2 ˆ 2 ( x y ) 2 x y + ˜ = 12 ÁË 15 ¯ D2 D4 \ G.S. is z = C.F + P.I =

Ïf1 ( x ) = 2 + x, in ( - 2 £ x £ 0) 12. (a) (i) f ( x ) = Ì (0 £ x £ 2) Óf2 ( x ) = 2 - x, in

2l = 4 \ l = 2

Since f1 (–x) = f2 (x), the function f(x) is even in (–2 £ x £ 2) a0 np x Let the F.S. of f(x) be + Â an cos 2 2 2

an =

2

2 np x np x f ( x )cos dx = Ú (2 - x )cos dx Ú 20 2 2 0 2

Ï Ê np x ˆ Ê np x ˆ ¸Ô Ô sin Á cos ÁË 2 ˜¯ Ë 2 ˜¯ Ô = ÔÌ(2 - x ) ˝ 2 2 n p Ê ˆ Ên p ˆ Ô Ô ÁË 2 ˜¯ Á 4 ˜ Ô Ô Ë ¯ ˛0 Ó

if n is even ¸ Ï 0, 4 Ô Ô n = - 2 2 {(- 1) - 1} = Ì 8 ˝ np Ô n2p 2 , if n is odd Ô Ó ˛ 2

2 Ê 2 x2 ˆ a0 = Ú (2 - x )dx = Á 2 x - ˜ = 2 20 2 ¯0 Ë • 8 1 Ê np x ˆ cos Á in - 2 £ x £ 2 (1)  2 Ë 2 ˜¯ p n = 1,3,5,... n2 Putting x = 0 in (1), which is a point of continuity of f (x), 8 Ï1 1 1 ¸ We get 1 + 2 Ì 2 + 2 + 2 +  • ˝ = f (0) = 2 p Ó1 3 5 ˛

\ Required F.S. is f(x) = 1 +

SQP11.6

Transforms and Partial Differential Equations

1 1 1 p2 + + +  • = 8 12 32 52

\

(ii) The complex form of the F.S. of f(x) = e–ax in (– l, l) is worked out in Chapter 2 on page 2.87. Simply change a to –a and l to p in that W.E. and obtain eax = sinh (- ap )

i.e.



(- 1)n (- ap - in p ) inx  a2p 2 + n2p 2 e in (- p , p ) n = -•

Ê sinh a p ˆ • ( - 1)n (a + in ) inx eax = Á e in (- p , p ) Â Ë p ˜¯ n = - • a 2 + n2

(1)

Putting x = 0 in (1); we get • Ê p ˆ ( - 1)n (a + in ) =  a 2 + n2 ÁË sinh a p ˜¯ n = -• \ Equating the R.P.’s; we get



p ( - 1)n  a2 + n2 = a sinh ap n = -•

12. (b)(i) This problem has been worked out in Chapter 2, page 2-67 in worked example 18. 5p /3 Ï0 p /3 2p /3 p 4p /3 12. (b)(ii) x : Ì x3 x5 x1 x2 x4 Ó x0 Ï1.98 1.80 1.05 1.30 - 0.88 - 0.25 y:Ì y1 y2 y3 y4 y5 Ó y0 Since f (x) is defined in (0, 2p), the F.S. is of the form a0 + Â an cos n x + Â bn sin n x 2 a0 = 2 ¥

1 5 1 + Â yr = ¥ 5 = 1.67 6 r =0 3

1 1 yr cos xr = ÎÈ( y0 - y3 ) + ( y1 + y5 - y2 - y4 )cos60∞˚˘ Â 3 3 1 = [0.68 - (1.80 - 0.25 - 1.05 + 0.88) ¥ 0.5]= - 0.033 3 1 b1 = ( y1 + y2 - y4 - y5 )sin 60∞ 3 1 = [(1.80 - 1.05 + 0.88 - 0.25 ) ¥ 0.866 ]= 1.149 3

a1 =

Solved Question Papers-11

SQP11.7

1 È( y0 + y3 ) - ( y1 + y4 + y2 + y5 )cos60∞˚˘ 3Î 1 = [(1.98 + 1.30) - (1.80 - 0.88 + 1.05 - 0.25) ¥ 0.5] 3

a2 =

=

1 [3.28 - 0.86]= 0.807 3

1 b2 = ( y1 + y4 - y2 - y5 )sin 60∞ 3 1 = {1.80 - 0.88 - 1.05 + 0.25}¥ 0.866 = 0.035 3 \ Required F.S. is y = 0.835 + (– 0.033cos x + 1.149 sin x) + (0.807 cos 2x + 0.035 sin 2x) 13. (a) This problem has been worked out in Chapter 3, page 3-30 as worked example 8. We have to replace l by k and 60 is given instead of l. The required solution will then be y( x, t ) =

8kl • (-1)n +1 (2n - 1)p x (2n - 1)p at sin ¥ sin 3 Â 3 l l p a n = 1 (2n - 1)

13. (b) This problem has been worked out in Chapter 3, page 3-94 in worked example 13. 14. (a) (i) Fourier sine integral of f(x) = e–x sin is given by f ( x) = \ e - x sin x =

2 p

••

Ú Ú f (t )sin st.sin sx dt ds 0 0

• È• - t ˘ 2 x s ds sin Í Ú e sin t sin st dt ˙ Ú p 0 ÍÎ 0 ˙˚ •

=

=



2 1 sin x s ds ◊ Ú e - t {cos(s - 1)t - cos(s + 1)t }dt Ú p 0 20 • È e -t 1 sin x s ds Í - cos(s - 1)t + (s - 1)sin(s - 1)t } Ú 2 { p 0 Î1 + (s - 1) •

˘ e- t - cos(s + 1)t + (s + 1)sin(s + 1)t }˙ 2 { 1 + (s + 1) ˚0 i.e., e - x sin x =



4 s sin x s ds p Ú0 s 4 + 4

SQP11.8

Transforms and Partial Differential Equations •

(ii) fc (e - ax ) = Ú e - ax cos sx dx = 0

a s2 + a 2



fc (e - l x ) = Ú e - l x cos sx dx =

\

0

l s + l2 2

Integrating both sides w.r.t. l between a and b, we have b



b Ê e -lx ˆ È1 2 2 ˘ = + l cos sx dx log ( s ) Ú ÁË - x ˜¯ Í2 ˙ Î ˚a 0 a

(1)



i.e.,

Ê e - ax - e - bx ˆ Ê s2 + b2 ˆ 1 cos sx dx = log Á 2 ˜ ˜ x 2 ¯ Ë s + a2 ¯ 0

Ú ÁË

(2)

Ê e - ax - e - bx ˆ 1 Ê s2 + b2 ˆ i.e., Fc Á = log Á 2 ˜ ˜ x Ë ¯ 2 Ë s + a2 ¯ 14. (b) (i) This problem has been worked out in Chapter 4, example 5, page 4.37. (ii) F (e - a \ F{e

2 2

x

)=

p - s2 /4 a2 (Refer to Worked Example 8 in Chapter 4. page e a 4-14)

2

F (e - x ) = p e - s - x2

*e

- x2

2

/4

È• ˘ 2 2 } = Ú Í Ú e - ( x - u ) e - u du ˙ e - i sx dx ˙˚ -• Í Î -• •

È• ˘ - u2 - ( x - u )2 - i sx Í ˙ du e e e dx Ú Ú ÍÎ - • ˙˚ -• •

=



=

Ú

2

2

e - u e - i us F (e - x ) dx, by shifting property

-•

=

p e -s

2

2

/4

p e -s

2

/4

2

= F (e - x ) F (e - x ) Thus F [f ( x )* g( x )]= f (s).g (s) is verified 15. (a) (i) u =

z3 + z 1 + z -2 1 + z -2 = = ( z - 1)3 (1 - z -1 )3 1 - 3z -1 + 3z - 2 - z - 3

Solved Question Papers-11

SQP11.9

We apply the long division method to obtain the quotient 1 – 3z–1 + 3z–2 – z – 3 1 + z –2 1 + 3 z –1 + 7z–2 1 – 3z–1 + 3z–2 – z–3 ————————— 3z–1 – 2z –2 + z–3 3z–1 – 9z –2 + 9z–3 – 3z–4 —————————— 7z–2 – 8z –3 + 3z–4 7z–2 – 21z –3 + 21z–4 – 7z–5 ——————————— 13z –3 – 18z–4 + 7z–5 \



 u(n)z - n = 1 + 3z -1 + 7z - 2 + 

n=0

\ u(0) = 1, u(1) = 3 and u2 = (7) ÏÔ ¸Ô Ê z ˆ z2 z ¸ -1 Ï z -1 Ê z ˆ ◊ Z -1 Ì * Z -1 Á (ii) ˝=Z Ì ˝=Z Á ˜ Ë z - 3¯ Ë z - 4 ˜¯ Óz -3 z -4˛ ÓÔ ( z - 3)( z - 4) ˛Ô n n = 3 *4 =

n

 3r.4n - r

r=0

È Ê 3ˆ n +1 ˘ Í1 - Á ˜ ˙ r n Ë 4¯ Ê 3ˆ ˙ = 4n  Á ˜ = 4n Í Í ˙ 3 Ë ¯ 4 r=0 Í 1˙ 4 ˙ ÎÍ ˚ = 4n + 1 – 3n + 1 15.(b) (i) This problem has been worked out in Chapter 5, page 5-33 in worked example 13. (ii) un + 2 + 4un + 1 + 3un = 3n ; u(0) = 0 and u(1) = 1 Taking Z-Transform z È z 2 u ( z ) - z 2 u(0) - zu(1)˘ + 4 [z u ( z ) - zu(0)]+ 3u ( z ) = Î ˚ z -3 i.e. ( z 2 + 4 z + 3)u ( z ) =

z +z z-3

z2 - 2z \ u (z) = ( z + 1)( z + 3)( z - 3)

3 5 1 u (z) 8 12 24 \ = + z z + 1 z +3 z -3

3 z 5 z 1 z \ u (z) = ◊ - ◊ + ◊ 8 z + 1 12 z + 3 24 z - 3 3 5 1 n Inverting, un = ( - 1)n - ◊ ( - 3)n + ◊3 8 12 24

Index Auxiliary equations,

1-72

Boundary conditions, 3(a)-11, 3(b)-5 values, 3(b)-5, 3(b)-27 Complementary function, 1-72, 2-76 Convolution product, 4-31 theorem, 4-31, 7.5 Derivative of Fourier transform, 4-30 Diffusion equation, 3(b)-3 Diffusivity, 3(b)-3, 3(c)-3 Dirichlet’s conditions, 4-2, 4-5, 4-39 Effective value, 4-70 Electrostatic potential, 3(c)-4, 3(c)-48 Energy theorem, 4-32 Euler’s formulas, 4-2, 4-5, 4-39, 4-43, 446, 4-73, 4-76 modified, 4-43 Even extension, 4-47, 4-48, 4-50, 4-54, 4-56 function, 4-8, 4-19, 4-20, 4-23, 4-24 Final value theorem, 5-4 Finite difference equation, 5-27 Fourier series, 4-1, 4-2, 4-5 coefficients, 4-2, 4-39, 4-42, 4-45, 4-46 complex form of, 4-76 convergence of, 4-11 exponential form of, 4-75 half-range, 4-42, 4-50, 4-70, 4-89 of an even function, 4-10 of an odd function, 4-10 Fourier complex integral, 4-3 cosine integral, 4-3 cosine transform, 4-5 inverse, 4-5 pair, 4-5

integral representation, 4-3 theorem, 4-1 sine integral, 4-4 sine transform, 4-5 inverse, 4-5 pair, 4-6 Fourier transform, 4-4 complex pair of, 4-4 finite, 4-55 cosine, 4-56 sine, 4-55 of derivatives, 4-29 inverse, 4-56 properties of, 4-26 Fundamental term, 4-73 Harmonic analysis, 4-75, 4-81 nth, 4-75, 4-81 Harrison circle method, 4-79 Heat flow one-dimensional, 3(b)-1 steady-state, 3(c)-1 transient, 3(b)-4 two-dimensional, 3(c)-1 variable, 3(b)-4 Initial value theorem, 5-4 Lagrange’s linear equation, 1-51 multipliers, 1-53 subsidiary simultaneous equations, 1-52 Laplace equation, 3(c)-3 Mean square value, 4-45, 4-70, 4-74 Modulation theorem, 4-28 Octave, 4-73 Odd extension, 4-43, 4-45, 4-47 function, 4-8

Index

I-2

Parseval’s identity, 4-32 Partial differential equations, 1-1 applications of, 3(a)-1 clairat’s type, 1-24 first order, non-linear, 1-23 formation of, 1-1 higher order, 1-71 integral of, 1-21 complete, 1-21, 1-23 D’Alembert’s, 3(a)-47, 3(a)-49 general, 1-22 particular, 1-22 singular, 1-22 linear, 1-71 homogeneous, 1-71 non-homogeneous, 1-71 order of, 1-1 primitive of, 1-21 solution of, 1-21 by separation of variables, 1-76 Particular integral, 1-74 Root mean-square value, Steady-state part,

4-45

3(b)-28

Telegraph equations, 3(a)-5 Telephone equations, 3(a)-5, 3(a)-49 Temperature gradient, 3(b)-5 Transient part, 3(b)-27

Transmission line equation, 3(a)-41 first, 3(a)-4 second, 3(a)-5

3(a)-4,

Unit impulse sequence, 5-7 step sequence, 5-7 Vibration of strings, 3(a)-1 damped, 3(a)-8 of membrane, 3(a)-51 transverse, 3(a)-2 Wave equation, one-dimensional, 3(a)-2 proper solution of, 3(a)-8 variable separable solution of, 3(a)-6 Z-transform, 5-1 bilateral (two-sided), 5-1 inverse, 5-2, 5-26 expansion method, 5-26 long division method, 5-26 partial fraction method, 5-26 residue theorem method, 5-26 properties of, 5-2 region of convergence of, 5-2 unilateral (one-sided), 5-1