About the Author
A Nagoor Kani is a multifaceted personality with efficient technical expertise and management skills. He
obtained his BE degree in Electrical and Electronics Engineering from Thiagarajar College of Engineering,
Madurai, and MS (Electronics and Control) through Distance Learning program of BITS, Pilani. He is a
life member of ISTE and IETE.
He started his career as a self-employed industrialist (1986 –1989) and then changed over to teaching in
1989. He has worked as Lecturer in Dr M G R Engineering College (1989–1990) and as Asst. Professor
in Satyabhama Engineering College (1990–1997). In 1993, he started a teaching centre for BE students
named Institute of Electrical Engineering, which was renamed RBA Tutorials in 2005.
A Nagoor Kani launched his own organisation in 1997. The ventures currently run by him are RBA
engineering (involved in manufacturing of lab equipments, microprocessor trainer kits and undertake
Electrical contracts and provide electrical consultancy), RBA Innovations (involved in developing projects
for engineering students and industries), RBA Tutorials (conducting tutorial classes for engineering students
and coaching for GATE, IES, IAS) and RBA Publications (publishing of engineering books), RBA Software
(involved in web-design and maintenance). His optimistic and innovative ideas have made the RBA Group
a very successful venture.
A Nagoor Kani is a well-known name in major engineering colleges in India. He is an eminent writer and
till now he has authored several engineering books (published by RBA Publications) which are very
popular among engineering students. He has written books in the areas of Control Systems,
Microprocessors, Microcontrollers, Digital Signal Processing, Electric Circuits, Electrical Machines and
Power Systems.
A Nagoor Kani
Founder
RBA Group
Chennai
Tata McGraw Hill Education Private Limited
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iv
v
Contents
Dedicated to my father
Contents
vii
Contents
Preface
xv
Acknowledgements
xviii
List of Symbols and Abbreviations
xx
Chapter 1 : Introduction to Signals and Systems
1.1
1.2
Signal
1. 1
1.1.1
Continuous Time Signal
1.1. 2
Discrete Time Signal
1.1. 3 Digital Signal
System
1. 2
1. 2
1. 3
1.2.1
Continuous Time System 1. 3
1.2.2
Discrete Time System 1. 4
1. 3
1.3
Frequency Domain Analysis of Continuous Time Signals and Systems
1. 5
1.4
Frequency Domain Analysis of Discrete Time Signals and Systems
1. 6
1.5
Importance of Signals and Systems
1. 7
1.6
Use of MATLAB in Signals and Systems
1. 8
Chapter 2 : Continuous Time Signals and Systems
2.1
Introduction
2. 1
2.2
Standard Continuous Time Signals
2. 1
2.3
Classification of Continuous Time Signals
2. 6
2.4
2.5
2.3.1
Deterministic and Nondeterministic Signals 2. 6
2.3.2
Periodic and Nonperiodic Signals
2.3.3
Symmetric (Even) and Antisymmetric (Odd) Signals 2. 11
2.3.4
Energy and Power Signals 2. 13
2.3.5 Causal, Noncausal and Anticausal Signals 2. 15
Mathematical Operations on Continuous Time Signals
2. 16
2.4.1
Scaling of Continuous Time Signals 2. 16
2.4.2
Folding (Reflection or Transpose) of Continuous Time Signals
2.4.3
Time Shifting of Continuous Time Signals
2.4.4
Addition of Continuous Time Signals
2.4.5
Multiplication of Continuous Time Signals
2.4.6
Differentiation and Integration of Continuous Time Signals 2. 21
2. 18
2. 18
2. 20
2. 20
Impulse Signal
2.5.1 Properties of Impulse Signal 2. 25
2.5.2
2.6
2. 6
Representation of Continuous Time Signal as Integral of Impulses
2. 24
2. 27
Continuous Time System
2. 29
2.6.1
Mathematical Equation Governing LTI Continuous Time System 2. 29
2.6.2
Block Diagram and Signal Flow Graph Representation of LTI Continuous Time System 2. 31
Contents
viii
2.7
2.8
2.9
2.10
2.11
Response of LTI Continuous Time System in Time Domain
2.7.1
Homogeneous Solution
2.7.2
2.7.3
Particular Solution 2. 36
Zero-input and Zero-state Response
2.7.4
Total Response
2. 34
2. 35
2. 36
2. 37
Classification of Continuous Time Systems
2. 43
2.8.1
Static and Dynamic Systems
2. 43
2.8.2
Time Invariant and Time Variant Systems
2.8.3
Linear and Nonlinear Systems
2.8.4
Causal and Noncausal Systems
2.8.5
2.8.6
Stable and Unstable Systems 2. 54
Feedback and Nonfeedback Systems
2. 43
2. 47
2. 51
2. 59
Convolution of Continuous Time Signals
2. 59
2.9.1
Response of LTI Continuous Time System using Convolution 2. 59
2.9.2
Properties of Convolution
2. 60
2.9.3
Interconnections of Continuous Time Systems
2.9.4
Procedure to Perfom Convolution
2.9.5
Unit Step Response using Convolution 2. 65
2. 62
2. 64
Inverse System and Deconvolution
Summary of Important Concepts
2. 76
2. 77
2.12
Short Questions and Answers
2. 78
2.13
MATLAB Programs
2. 84
2.14
Exercises
2. 94
Chapter 3 : Laplace Transform
3.1
Introduction
3. 1
3.2
Region of Convergence
3. 4
3.3
3.4
Properties and Theorems of Laplace Transform
Poles and Zeros of Rational Function of s
3. 18
3. 33
3.5
3.6
3.7
3.4.1
Representation of Poles and Zeros in s-plane 3. 34
3.4.2
ROC of Rational Function of s
3.4.3
Properties of ROC 3. 37
3. 35
Inverse Laplace Transform
3. 38
3.5.1
Inverse Laplace Transform by Partial Fraction Expansion Method
3.5.2
Inverse Laplace Transform using Convolution Theorem
3. 39
3. 44
Analysis of LTI Continuous Time System using Laplace Transform
3.6.1 Transfer Function of LTI Continuous Time System 3. 48
3.6.2
Impulse Response and Transfer Function 3. 48
3.6.3
Response of LTI Continuous Time System using Laplace Transform 3. 49
3.6.4
Convolution and Deconvolution using Laplace Transform 3. 50
3.6.5
Stability in s-Domain
3. 51
Structures for Realization of LTI Continuous Time Systems in s-domain
3.7.1
Direct Form-I Structure
3. 48
3. 80
3. 79
Contents
3.7.2
Direct Form-II Structure
3.7.3
Cascade Structure
ix
3. 81
3. 84
3.8
3.7.4 Parallel Structure 3. 85
Summary of Important Concepts
3. 96
3.9
Short Questions and Answers
3. 97
3.10
MATLAB Programs
3.101
3.11
Exercises
3.107
Chapter 4 : Fourier Series and Fourier Transform of Continuous Time Signals
4.1
Introduction
4.2
Trigonometric Form of Fourier Series
4.3
4.4
4. 1
4.2.1
Definition of Trigonometric Form of Fourier Series
4.2.2
4.2.3
Conditions for Existence of Fourier Series 4. 2
Derivation of Equations for a0, an and bn 4. 2
4. 1
4. 1
Exponential Form of Fourier Series
4. 5
4.3.1
Definition of Exponential Form of Fourier Series
4.3.2
Negative Frequency
4. 5
4.3.3
Derivation of Equation for cn 4. 6
4.3.4
Relation Between Fourier Coefficients of Trigonometric and Exponential Form 4. 6
4. 5
4.3.5 Frequency Spectrum (or Line Spectrum) of Periodic Continuous Time Signals 4. 7
Fourier Coefficients of Signals with Symmetry
4.4.1
Even Symmetry
4. 9
4.4.2
Odd Symmetry
4. 11
4.4.3
4.4.4
Half Wave Symmetry (or Alternation Symmetry)
Quarter Wave Symmetry 4. 15
4. 9
4. 14
4.5
Properties of Fourier Series
4. 15
4.6
Diminishing of Fourier Coefficients
4. 17
4.7
Gibbs Phenomenon
4. 18
4.8
4.9
Solved Problems in Fourier Series
Fourier Transform
4. 19
4. 41
4.9.1
Development of Fourier Transform from Fourier Series 4. 41
4.9.2
Frequency Spectrum using Fourier Transform 4. 43
4.10
Properties of Fourier Transform
4. 44
4.11
Fourier Transform of Some Important Signals
4. 53
4.12
Fourier Transform of a Periodic Signal
4. 65
4.13
Analysis of LTI Continuous Time System Using Fourier Transform
4. 65
4.13.1
4.13.2
Transfer Function of LTI Continuous Time System in Frequency Domain 4. 65
Response of LTI Continuous Time System Using Fourier Transform 4. 67
4.13.3
Frequency Response of LTI Continuous Time System
4. 67
4.14
Relation Between Fourier and Laplace Transform
4. 69
4.15
Solved Problems in Fourier Transform
4. 71
4.16
Summary of Important Concepts
4. 80
4.17
Short Questions and Answers
4. 81
Contents
x
4.18
MATLAB Programs
4. 88
4.19
Exercises
4. 92
Chapter 5 : State Space Analysis of Continuous Time Systems
5.1
5.2
Introduction
State Model of a Continuous Time System
5. 1
5. 1
5.3
State Model of a Continuous Time System from Direct Form-II Structure
5. 4
5.4
Transfer Function of a Continuous Time System from State Model
5. 6
5.5
Solution of State Equations and Response of Continuous Time System
5. 6
5.6
Solved Problems in State Space Analysis
5. 8
5.7
Summary of Important Concepts
5. 16
5.8
Short Questions and Answers
5. 16
MATLAB Programs
Exercises
5. 19
5. 20
5.9
5.10
Chapter 6 : Discrete Time Signals and Systems
6.1
Discrete and Digital Signals
6. 1
6.1.1
Generation of Discrete Signals
6. 1
6.1.2
Representation of Discrete Time Signals
6. 2
6.2
Standard Discrete Time Signals
6. 3
6.3
Sampling of Continuous Time (Analog) Signals
6. 6
6.4
6.3.1 Sampling and Aliasing 6. 7
Classifications of Discrete Time Signals
6. 10
6.5
6.6
6.7
6.4.1
Deterministic and Nondeterministic Signals
6.4.2
Periodic and Aperiodic Signals
6. 10
6.4.3
Symmetric (Even) and Antisymmetric (Odd) Signals
6.4.4
Energy and Power Signals
6.4.5
Causal, Noncausal and Anticausal Signals 6. 16
6. 10
6. 12
6. 14
Mathematical Operations on Discrete Time Signals
6. 17
6.5.1
6.5.2
Scaling of Discrete Time Signals 6. 17
Folding (Reflection or Transpose) of Discrete Time Signals
6.5.3
Time Shifting of Discrete Time Signals
6.5.4
Addition of Discrete Time Signals
6.5.5
Multiplication of Discrete Time Signals
6. 18
6. 18
6. 19
6. 19
Discrete Time System
6. 20
6.6.1
Mathematical Equation Governing Discrete Time System
6.6.2
Block Diagram and Signal Flow Graph Representation of Discrete Time System 6. 22
Response of LTI Discrete Time System in Time Domain
6.7.1 Zero-Input Response or Homogeneous Solution 6. 25
6.7.2
Particular Solution
6. 26
6.7.3
Zero-State Response 6. 27
6.7.4
Total Response 6. 27
6. 20
6. 25
Contents
6.8
6.9
6.10
6.11
6.12
Classifications of Discrete Time Systems
Static and Dynamic Systems
6.8.2
6.8.3
Time Invariant and Time Variant Systems 6. 32
Linear and Nonlinear Systems 6. 35
6.8.4
Causal and Noncausal Systems
6. 32
6. 41
6.8.5
Stable and Unstable Systems
6.8.6
FIR and IIR Systems
6. 43
6.8.7
Recursive and Nonrecursive Systems
6. 45
6. 45
Discrete or Linear Convolution
6. 46
6.9.1
Representation of Discrete Time Signal as Summation of Impulses
6.9.2
6.9.3
Response of LTI Discrete Time System using Discrete Convolution 6. 48
Properties of Linear Convolution 6. 49
6.9.4
Interconnections of Discrete Time Systems 6. 51
6.9.5
Methods of Performing Linear Convolution
6. 47
6. 56
Circular Convolution
6. 63
6.10.1
Circular Representation and Circular Shift of Discrete Time Signal
6.10.2
Circular Symmetrics of Discrete Time Signal
6. 63
6. 65
6.10.3
Definition of Circular Convolution
6.10.4
6.10.5
Procedure for Evaluating Circular Convolution 6. 67
Linear Convolution via Circular Convolution 6. 68
6.10.6
6. 66
Methods of Computing Circular Convolution
6. 68
Sectioned Convolution
6. 78
6.11.1
Overlap Add Method 6. 78
6.11.2
Overlap Save Method 6. 79
Inverse System and Deconvolution
6. 92
Inverse System 6. 92
Deconvolution 6. 93
Correlation, Crosscorrelation and Autocorrelation
6.13.1
6.14
6. 32
6.8.1
6.12.1
6.12.2
6.13
xi
Procedure for Evaluating Correlation
6. 95
6. 96
Circular Correlation
6.104
6.14.1
Procedure for Evaluating Circular Correlation 6.105
6.14.2
Methods of Computing Circular Correlation
6.105
6.15
Summary of Important Concepts
6.110
6.16
6.17
Short Questions and Answers
MATLAB Programs
6.111
6.115
6.18
Exercises
6.119
Chapter 7 : Z - Transform
7.1
Introduction
7. 1
7.2
Region of Convergence
7. 3
7.3
Properties of Z-Transform
7. 11
7.4
Poles and Zeros of Rational Function of z
7. 27
7.4.1
Representation of Poles and Zeros in z-plane
7. 28
Contents
xii
7.5
7.6
7.7
7.4.2
ROC of Rational Function of z
7.4.3
Properties of ROC 7. 30
7. 29
Inverse Z-Transform
7.5.1 Inverse Z-Transform by Contour Integration or Residue Method 7. 31
7.5.2
Inverse Z-Transform by Partial Fraction Expansion Method
7.5.3
Inverse Z-Transform by Power Series Expansion Method 7. 35
7. 31
7. 32
Analysis of LTI Discrete Time System Using Z-Transform
7. 47
7.6.1
Transfer Function of LTI Discrete Time System 7. 47
7.6.2
Impulse Response and Transfer Function 7. 48
7.6.3
Response of LTI Discrete Time System Using Z-Transform 7. 49
7.6.4
7.6.5
Convolution and Deconvolution Using Z-Transform 7. 50
Stability in z-Domain 7. 51
Relation Between Laplace Transform and Z-Transform
7. 56
7.7.1
Impulse Train Sampling of Continuous Time Signal
7. 56
7.7.2
Transformation from Laplace Transform to Z-Transform 7. 56
7.7.3
Relation Between s-Plane and z-Plane 7. 57
7.8
Structures for Realization of LTI Discrete Time Systems in z-Domain
7. 59
7.9
Summary of Important Concepts
7. 72
7.10
7.11
Short Questions and Answers
MATLAB Programs
7. 73
7. 79
7.12
Exercises
7. 84
Chapter 8 : Fourier Series and Fourier Transform of Discrete Time Signals
8.1
8.2
8.3
Introduction
Fourier Series of Discrete Time Signals (Discrete Time Fourier Series)
8.2.1
Frequency Spectrum of Periodic Discrete Time Signals
8.2.2
Properties of Discrete Time Fourier Series 8. 4
8. 1
8. 1
8. 3
Fourier Transform of Discrete Time Signals (Discrete Time Fourier Transform)
8.3.1
8.3.2
8. 9
Development of Discrete Time Fourier Transform from
Discrete Time Fourier Series 8. 9
Definition of Discrete Time Fourier Transform 8. 10
8.3.3
Frequency Spectrum of Discrete Time Signal 8. 11
8.3.4
Inverse Discrete Time Fourier Transform
8.3.5
Comparison of Fourier Transform of Discrete and Continuous Time Signals 8. 12
8. 11
8.4
Properties of Discrete Time Fourier Transform
8. 12
8.5
Discrete Time Fourier Transform of Periodic Discrete Time Signals
8. 20
8.6
Analysis of LTI Discrete Time System Using Discrete Time Fourier Transform
8. 22
8.6.1
8.6.2
Transfer Function of LTI Discrete Time System in Frequency Domain 8. 22
Response of LTI Discrete Time System Using Discrete Time Fourier Transform 8. 23
8.6.3
Frequency Response of LTI Discrete Time System
8.6.4
Frequency Response of First Order Discrete Time System 8. 25
8.6.5
Frequency Response of Second Order Discrete Time System
8. 23
8. 31
Contents
8.7
xiii
Aliasing in Frequency Spectrum Due to Sampling
8.7.1
8. 36
Signal Reconstruction (Recovery of Continuous Time Signal) 8. 38
8.8
8.7.2 Sampling of Bandpass Signal 8. 39
Relation Between Z-Transform and Discrete Time Fourier Transform
8. 40
8.9
Summary of Important Concepts
8. 62
8.10
Short Questions and Answers
8. 63
8.11
MATLAB Programs
8. 68
8.12
Exercises
8. 73
Chapter 9 : Discrete Fourier Transform (DFT) and Fast Fourier Transform (FFT)
9.1
Introduction
9. 1
9.2
Discrete Fourier Transform (DFT) of Discrete Time Signal
9. 1
9.2.1
9.2.2
Development of DFT from DTFT 9. 1
Definition of Discrete Fourier Transform (DFT)
9.2.3
Frequency Spectrum Using DFT
9.2.4
Inverse DFT
9. 2
9. 2
9. 3
9.3
Properties of DFT
9. 4
9.4
Relation Between DFT and Z-Transform
9. 10
9.5
Analysis of LTI Discrete Time Systems Using DFT
9. 10
9.6
9.7
Fast Fourier Transform (FFT)
Decimation In Time (DIT) Radix-2 FFT
9. 19
9. 21
9.8
9.7.1
8-Point DFT Using Radix-2 DIT FFT
9. 23
9.7.2
Flow Graph for 8-Point DIT Radix-2 FFT
9. 27
Decimation In Frequency (DIF) Radix-2 FFT
9. 29
9.8.1
8-Point DFT Using Radix-2 DIF FFT
9. 32
9.8.2
Flow Graph for 8-Point DIF Radix-2 FFT
9. 34
9.8.3
Comparison of DIT and DIF Radix-2 FFT
9. 37
9.9
9.10
Computation of Inverse DFT Using FFT
Summary of Important Concepts
9. 37
9. 56
9.11
Short Questions and Answers
9. 57
9.12
MATLAB Programs
9. 60
9.13
Exercises
9. 65
Chapter 10 : Structures for Realization of IIR and FIR Systems
10.1
10.2
Introduction
Discrete Time IIR and FIR Systems
10.2.1
10.3
10. 1
Discrete Time IIR System
10. 1
10. 1
10.2.2 Discrete Time FIR System 10. 2
Structures for Realization of IIR Systems
10.3.1
Direct Form-I Structure of IIR System
10.3.2
Direct Form-II Structure of IIR System
10. 3
10. 4
10. 5
Contents
xiv
10.4
10.3.3
Cascade Form Realization of IIR System
10.3.4
Parallel Form Realization of IIR System
10. 8
10. 8
Structures for Realization of FIR Systems
10.4.1 Direct Form Realization of FIR System 10. 30
10.4.2
Cascade Form Realization of FIR System
10.4.3
Linear Phase Realization of FIR System 10. 31
10. 29
10. 30
10.5
Summary of Important Concepts
10. 37
10.6
Short Questions and Answers
10. 38
10.7
Exercises
10. 40
Chapter 11 : State Space Analysis of Discrete Time Systems
11.1
Introduction
11. 1
11.2
11.3
State Model of Discrete Time Systems
State Model of a Discrete Time System from Direct Form-II Structure
11. 1
11. 4
11.4
Transfer Function of a Discrete Time System from State Model
11. 5
11.5
Solution of State Equations and Response of Discrete Time System
11. 6
11.6
Solved Problems in State Space Analysis of Discrete Time System
11. 8
11.7
Summary of Important Concepts
11. 14
11.8
11.9
11.10
Short Questions and Answers
11. 14
MATLAB Programs
Exercises
11. 17
11. 19
Appendix 1
Important Mathematical Relations
A. 1
Appendix 2
MATLAB Commands and Functions
A. 5
Appendix 3
Summary of Various Standard Transform Pairs
A. 11
Appendix 4
Summary of Properties of Various Transforms
A. 17
Index
I. 1
Contents
xv
Preface
The main objective of this book is to explore the basic concepts of signals and systems in a simple and
easy-to-understand manner.
This text on signals and system has been crafted and designed to meet students’ requirements. Considering
the highly mathematical nature of this subject, more emphasis has been given on the problem-solving
methodology. Considerable effort has been made to elucidate mathematical derivations in a step-by-step
manner. Exercise problems with varied difficulty levels are given in the text to help students get an
intuitive grasp on the subject.
This book with its lucid writing style and germane pedagogical features will prove to be a master text for
engineering students and practitioners.
Salient Features
The salient features of this book are
• Separate discussions on continuous time and discrete time signals for thorough understanding of the
concepts
• Proof of properties of transforms clearly highlighted by shaded boxes for quick review
• Additional explanations for solutions and proofs provided in separate boxes
• Different types of fonts used for text, proof and solved problems providing better clarity and user-friendliness
Organization
In this book, the concepts of continuous time signals and systems are organized in four chapters. The
concepts of discrete time signals and systems are organized in six chapters, and one chapter is devoted to
a general discussion on signals and systems. Each chapter provides the foundations and practical implications
with a large number of solved numerical examples for better understanding.
The important concepts are summarized at the end of each chapter which help in quick reference. Another
significant aspect of this book is MATLAB based computer exercises with complete explanations given in
each chapter. This will be of great assistance to both instructors and students.
Chapter 1 deals with a general introduction to various types of signals, systems and their importance in
real life. Basic definitions of signals, their mathematical representation, significance of their frequency
domain analysis and usage of MATLAB in this course are presented in a brief manner.
Chapter 2 introduces analysis of continuous time signals and systems in depth. It explores various
classifications of continuous time signals, systems, and possible mathematical operations such as
scaling, folding, time shifting, addition, multiplication, differentiation and integration on them. Then, it
discusses block diagrams and signal flow graph representation of continuous time systems, LTI systems
characterized by linear differential equations and methods to solve those equations.
Another vital aspect of Chapter 2 is the discussion on graphical convolution operation of continuous-time
signals by clearly separating the shift index and the time index. This will aid in clear understanding of the
concepts.
Chapter 3 discusses the analysis of continuous time systems using Laplace transform. The rational
functions of 's' and their representation in terms of poles and zeros, region of convergence of Laplace
transform and its properties are presented in a crisp and easy manner. The stability of the LTI systems and
their response via Laplace transforms are dealt with lucidly.
xvi
Preface
Further, the inverse Laplace transform using partial-fraction method and convolution theorem are
discussed. The convolution and deconvolution operations are explained with simple numerical examples.
The realization structures for continuous-time systems characterized by differential equations are also
presented in this chapter.
Chapter 4 is concerned with Fourier analysis of continuous-time systems which forms the basis for
frequency domain analysis. The first half of this chapter is dedicated to Fourier series in both
trigonometric and exponential forms, Fourier coefficients of various signals with symmetry, properties
of Fourier series and the Gibbs phenomenon.
The second half of the chapter explains the development of Fourier transform from Fourier series, frequency
spectrum, various properties of Fourier transform, and Fourier transform of some standard signals. It
also covers the computation of frequency responses of LTI systems using Fourier transform explained
with examples. The chapter also talks about the relation between Fourier transform and Laplace transform
of continuous-time signals.
Chapter 5 deals with the concepts of state space analysis of continuous-time systems. In this chapter,
the development of state model, solutions to state equations and response of continuous-time systems to
state models are discussed.
Chapter 6 is devoted to concepts of discrete-time signals and systems and is more concerned with the
generation, representation, classification, mathematical operations of discrete time signals and systems,
block diagrams and signal-flow graph notations.
The chapter also presents the methods of obtaining responses of LTI discrete-time systems and various
convolution methods. The deconvolution, correlation techniques and the inverse systems are clearly
explained with solved numerical examples. In addition, the concept of sampling and its importance are
dealt with briefly.
Chapter 7 explains Z-transform and its application to signals and systems. The concepts are similar to
Laplace transform except as applied to discrete-time signals and systems. All the important properties of
Z-transform are presented explicitly. Inverse Z-transform and solution of difference equations describing
the discrete-time systems are demonstrated with numerical examples. Also given are the systems
interconnections and standard system realization using structures.
Chapter 8 is dedicated to discrete-time Fourier series and Fourier transform which forms the basis for
frequency domain analysis of discrete-time signals and systems. In the first half of this chapter, the
discrete-time Fourier series and the frequency spectrum using discrete-time Fourier series are discussed
with relevant examples.
The second half of the chapter details the development of discrete-time Fourier transform from discretetime Fourier series, frequency spectrum, various properties of Fourier transform, and Fourier transform
of some standard discrete-time signals. In addition, the computation of frequency responses of LTI
discrete-time systems using Fourier transform are also explained with numerous examples. The relation
between Fourier transform and Z-transform of discrete-time signals is also discussed in the chapter.
Chapter 9 extends the understanding of the concepts of Discrete-Time Fourier Transform (DTFT) to
DFT (Discrete Fourier Transform) and FFT (Fast Fourier Transform). Development of DFT from DTFT,
properties of DFT, relation between DFT and Z-transform, analysis of the LTI systems using DFT and
FFT are extensively discussed.
Chapter 10 focuses on structures for realization of discrete-time systems with special attention to IIR
and FIR systems.
Preface
xvii
Chapter 11 presents the concepts of state space analysis of discrete-time systems. In this chapter, the
development of state model of discrete-time systems, solutions to state equations and response of discretetime systems from state models are discussed in an easy manner.
Web Supplements
This book is accompanied by a comprehensive website which can be accessed at http://www.mhhe.com/
nagoorkani/signals1e. It has been designed to provide valuable resources for students, instructors and
professionals.
• Students can access Interactive Quiz, Objective-Type Questions and Short-Answer-Type Questions
on the website.
• Supplementary teaching material for Instructors includes chapterwise PowerPoint slides for effective
lecture presentation and an on-request Solution Manual.
Feedback
I have taken care to present the concepts of signals and systems in a user-friendly manner and hope that
the teaching and student community will welcome the book. The readers may feel free to convey their
criticism and suggestions for further improvement of the book. The feedback is welcome at my email
address: kani@vsnl.com
A Nagoor Kani
Publisher's Note
Tata McGraw Hill Education looks forward to receiving views, comments and suggestions for improvement
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Contents
xviii
Acknowledgements
I express my heartfelt thanks to my wife, Ms C Gnanaparanjothi Nagoor Kani, and my sons N Bharath
Raj, alias Chandrakani Allaudeen, and N Vikram Raj for the support, encouragement and cooperation they
have extended to me throughout my career.
It's my pleasure to acknowledge the contribution of our technical editors Ms K Jayashree, Ms B
Srimathi, Ms B Hemavathy for editing and proofreading of the manuscript, and Mrs A Selvi towards
typesetting and preparing the layout of the book.
I am thankful to all the reviewers for their valuable suggestions and comments which helped me explore
the subject to a greater depth.
Sarika Sachdeva
Galgotia College of Engineering
Greater Noida, Uttar Pradesh
D K Sharma
Meerut Institute of Engineering and Technology
Meerut, Uttar Pradesh
Prateek Kumar
Maharana Pratap Engineering College
Kanpur, Uttar Pradesh
Asish Suri
Sri Mata Vaishno Devi University
Jammu, Jammu and Kashmir
Abhay Vidyarti
NRI Institute of Technology and Management
Gwalior, Madhya Pradesh
Smriti Agrawal
Ramswaroop College of Engineering
Lucknow, Uttar Pradesh
D Sriramkumar
National Institute of Technology (NIT)
Tiruchirapalli, Tamil Nadu
S Anand
M S Engineering College
Madurai, Tamil Nadu
Y V Ramana Rao
College of Engineering
Guindy, Tamil Nadu
Harpal Thethi
Kalinga Institute of Industrial Technology
Bhubaneswar, Orissa
U M Chaskar
College of Engineering
Pune, Maharashtra
Yogesh Angal
D Y Patil Institute of Engineering and Technology,
Pune, Maharashtra
Kishor Kinage
D J Sanghvi Engineering College, Mumbai, Maharashtra
B Lokeswara Rao
Geetanjali College of Engineering,
Hyderabad, Andhra Pradesh
Dayananda Nayak
Manipal Institute of Technology
Manipal, Karnataka
Acknowledgements
xix
I am also grateful to Mr Michael Hays, Mr Raghu Srinivasan, Ms Vibha Mahajan, Mr Ebi John,
Mr Manish Choudhary, Mr H R Nagaraja, Mr Suman Sen, Mr P L Pandita and Ms Sohini Mukherjee of
Tata McGraw Hill Education for their concern and care in publishing this work.
I thank all my office staff for their cooperation in carrying out my day-to-day activities.
Finally, a special note of appreciation is due to my sisters, brothers, relatives, friends, students
and the entire teaching community for their overwhelming support and encouragement during the process
of book writing.
A Nagoor Kani
Contents
xx
List of Symbols and Abbreviations
Symbols
a o, a n , b n
B
Cn
ck
E
f
F
Fo
Fm
Fs
H
j
L
nWo
P
p
R
s
t
T
W
z
z
z–1
W
Wo
W max
w
wk
s
*
*
-
Fourier coefficients of trigonometric form of Fourier series of x(t)
Bandwidth in Hz
Fourier coefficients of exponential form of Fourier series of x(t)
Fourier coefficients of discrete time signal x(n)
Energy of a signal
Frequency of discrete time signal in Hz/sample
Frequency of continuous time signal in Hz
Fundamental frequency of continuous time signal in Hz
Maximum frequency of continuous time signal
Sampling frequency of continuous time signal in Hz
System operator
Complex operator, -1
Inductance
Harmonic angular frequency, where n = 1,2,3.......
Power of a signal
Pole
Resistor
Complex frequency (s = s + jW)
Time in seconds
Time period in seconds
Phase factor or Twiddle factor
Complex variable (z = u + jv)
Unit advance operator or zero
Unit delay operator
Angular frequency of continuous time signal in rad/sec
Fundamental angular frequency
Maximum angular frequency in rad/sec
Angular frequency of discrete time signal
Sampling frequency point
Neper frequency (Real part of s)
Convolution operator
Circular convolution operator
z
-
Integration operator
d
dt
-
Differentiation operator
List of Symbols and Abbreviations
Standard/Input/Output Signals
h(n)
h(t)
r xy(m)
rxx(m)
rxy ( m)
-
Impulse response of discrete time system
Impulse response of continuous time system
Cross-correlation sequence of x(n) and y(n)
Auto-correlation sequence of x(n)
-
Circular cross-correlation sequence of x(n) and y(n)
rxx ( m)
sgn(t)
sinc (t)
u(n)
u(t)
x(n)
x(n)
xo(n)
x e(n)
x(n–m)
x((n–m)) N
x(t)
xo(t)
x e(t)
x(t–m)
y(n)
yzs(n)
yzi(n)
y(t)
yzs(t)
yzi(t)
d(t)
d(n)
P(t)
-
Circular auto-correlation sequence of x(n)
Signum signal
Sinc signal
Discrete time unit step signal
Continuous time unit step signal
Discrete time signal
Input of discrete time system
Odd part of discrete time signal x(n)
Even part of discrete-time signal x(n)
Delayed or linearly shifted x(n) by m units
Circularly shifted x(n) by m units, where N is period
Continuous time signal or Input of continuous time system
Odd part of continuous time signal x(t)
Even part of continuous time signal x(t)
Delayed or linearly shifted x(t) by m units
Output / Response of discrete time system
Zero state response of discrete time system
Zero input response of discrete time time system
Output / Response of continuous time system
Zero state response of continuous time system
Zero input response of continuous time system
Continuous time impulse signal
Discrete time impulse signal
Unit pulse signal
Transform Operators and Functions
DFT
DFT–1
F
F–1
H(s)
L
L–1
X(e jw)
X r(e jw)
Xi(e jw)
-
Discrete Fourier Transform (DFT)
Inverse DFT
Fourier Transform
Inverse Fourier Transform
Laplace Transform of h(t)
Laplace Transform
Inverse Laplace Transform
Discrete Time Fourier Transform of x(n)
Real part of X(ejw)
Imaginary part of X(ejw)
xxi
List of Symbols and Abbreviations
xxii
X(jW)
X(k)
Xr (k)
Xi(k)
X(s)
X(z)
Z
Z–1
-
Fourier Transform of x(t)
Discrete Fourier Transform of x(k)
Real part of X(k)
Imaginary part of X(k)
Laplace Transform of x(t)
Z-transform of x(n)
Z-transform
Inverse Z-transform
Matrices and Vectors
A
An
B
C
D
eAt
I
Q(t)
Q(n)
& ( t)
Q
& ( n)
Q
X(t)
X(n)
Y(t)
Y(n)
-
System matrix
State transition matrix of discrete time state model
Input matrix
Output matrix
Transmission matrix
State transition matrix of continuous time state model
Identity / Unit matrix
State vector of continuous time state model
State vector of discrete time state model
First derivative of continuous time state vector
First derivative of discrete time state vector
Input vector of continuous time state model
Input vector of discrete time state model
Output vector of continuous time state model
Output vector of discrete time state model
-
Bounded Input Bounded Output
Continuous Time
Continuous Time Fourier Series
Continuous Time Fourier Transform
Discrete Fourier Transform
Decimation In Frequency
Decimation In Time
Discrete Time
Discrete Time Fourier Series
Discrete Time Fourier Transform
Fast Fourier Transform
Finite Impulse Response
Infinite Impulse Response
Left Half Plane
Linear Time Invariant
Right Half Plane
Region Of Convergence
Abbreviations
BIBO
CT
CTFS
CTFT
DFT
DIF
DIT
DT
DTFS
DTFT
FFT
FIR
IIR
LHP
LTI
RHP
ROC
1. 1
Signals & Systems
CHAPTER 1
Introduction to Signals and Systems
1.1 Signal
Any physical phenomenon that conveys or carries some information can be called a signal. The
music, speech, motion pictures, still photos, heart beat, etc., are examples of signals that we normally
encounter in day to day life.
Usually, the information carried by a signal will be a function of an independent variable. The
independent variable can be time, spatial coordinates, intensity of colours, pressure, temperature, etc.,.
The most popular independent variable in signals is time and it is represented by the letter “t”.
The value of a signal at any specified value of the independent variable is called its amplitude. The
sketch or plot of the amplitude of a signal as a function of independent variable is called its waveform.
Mathematically, any signal can be represented as a function of one or more independent
variables.Therefore, a signal is defined as any physical quantity that varies with one or more independent
variables.
For example, the functions x1(t) and x2(t) as defined by the equations (1.1) and (1.2) represents
two signals: one that varies linearly with time “t” and the other varies quadratically with time “t”. The
equation (1.3) represents a signal which is a function of two independent variables “p” and “q”.
x1(t) = 0.7t
x2(t) = 1.8t
.....(1.1)
2
x(p,q) = 0.6p + 0.5q + 1.1q
.....(1.2)
2
.....(1.3)
The signals can be classified in number of ways. Some way of classifying the signals are,
I. Depending on the number of sources for the signals.
1.One-channel signals
2. Multichannel signals
II. Depending on the number of dependent variables.
1.One-dimensional signals
2. Multidimensional signals
III. Depending on whether the dependent variable is continuous or discrete.
1.Analog or Continuous signals
2. Discrete signals
1. One-channel signals
Signals that are generated by a single source or sensor are called one-channel signals.
The record of room temperature with respect to time, the audio output of a mono speaker, etc.,
are examples of one-channel signals.
Chapter 1 - Introduction to Signals and Systems
1. 2
2. Multichannel signals
Signals that are generated by multiple sources or sensors are called multichannel signals.
The audio output of two stereo speakers is an example of two-channel signal.The record of ECG
(Electro Cardio Graph) at eight different places in a human body is an example of eight-channel signal.
3. One-dimensional signals
A signal which is a function of single independent variable is called one-dimensional signal.
The signals represented by equation (1.1) and (1.2) are examples of one-dimensional signals.
The music, speech, heart beat, etc., are examples of one-dimensional signals where the single
independent variable is time.
4. Multidimensional signals
A signal which is a function of two or more independent variables is called multidimensional signal.
The equation (1.3) represents a two dimensional signal.
A photograph is an example of a two-dimensional signal. The intensity or brightness at each point
of a photograph is a function of two spatial coordinates “x” and “y”, (and so the spatial coordinates are
independent variables). Hence, the intensity or brightness of a photograph can be denoted by b(x, y).
The motion picture of a black and white TV is an example of a three-dimensional signal. The
intensity or brightness at each point of a black and white motion picture is a function of two spatial
coordinates “x” and “y”, and time “t”. Hence, the intensity or brightness of a black and white motion
picture can be denoted by b(x, y, t).
5. Analog or Continuous signals
When a signal is defined continuously for any value of independent variable, it is called analog or
continuous signal. Most of the signals encountered in science and engineering are analog in nature. When
the dependent variable of an analog signal is time, it is called continuous time signal.
6. Discrete signals
When a signal is defined for discrete intervals of independent variable, it is called discrete signal .
When the dependent variable of a discrete signal is time, it is called discrete time signal. Most of the
discrete signals are either sampled version of analog signals for processing by digital systems or output
of digital systems.
1.1.1 Continuous Time Signal
In a signal with time as independent variable, if the signal is defined continuously for any value of
the independent variable time “t”, then the signal is called continuous time signal. The continuous time
signal is denoted as “x(t)”.
The continuous time signal is defined for every instant of the independent variable time and so the
magnitude (or the value) of continuous time signal is continuous in the specified range of time. Here both
the magnitude of the signal and the independent variable are continuous.
1.1.2 Discrete Time Signal
In a signal with time as independent variable, if the signal is defined only for discrete instants of the
independent variable time, then the signal is called discrete time signal.
1. 3
Signals & Systems
In discrete time signal the independent variable time “t” is uniformely divided into discrete
intervals of time and each interval of time is denoted by an integer “n”, where “n” stands for discrete
interval of time and “n” can take any integer value in the range -¥ to +¥. Therefore, for a discrete time
signal the independent variable is “n” and the magnitude of the discrete time signal is defined only for
integer values of independent variable “n”. The discrete time signal is denoted by “x(n)”.
1.1.3 Digital Signal
The quantized and coded version of the discrete time signals are called digital signals. In digital
signals the value of the signal for every discrete time “n” is represented in binary codes. The process of
conversion of a discrete time signal to digital signal involves quantization and coding.
Normally, for binary representation, a standard size of binary is chosen. In m-bit binary representation
we can have 2m binary codes. The possible range of values of the discrete time signals are usually divided
into 2m steps called quantization levels, and a binary code is attached to each quantization level. The
values of the discrete time signals are approximated by rounding or truncation in order to match the
nearest quantization level.
1.2 System
Any process that exhibits cause and effect relation can be called a system. A system will have an
input signal and an output signal. The output signal will be a processed version of the input signal. A
system is either interconnection of hardware devices or software / algorithm.
A system is denoted by letter H. The diagrammatic representation of a system is shown in fig 1.1.
System
Input signal
or Excitation
® H
Output signal
® or Response
Fig 1.1 : Representation of a system.
The operation performed by a system on input signal to produce output signal can be expressed as,
Output = H{Input}
where H denotes the system operation (also called system operator).
The systems can be classified in many ways.
Depending on type of energy used to operate the systems, the systems can be classified into
Electrical systems, Mechanical systems, Thermal systems, Hydraulic systems, etc.
Depending on the type of input and output signals, the systems can be classified into Continuous
time systems and Discrete time systems.
1.2.1 Continuous Time System
A system which can process continuous time signal is called continuous time system, and so the
input and output signals of a continuous time system are continuous time signals.
A continuous time system is denoted by letter H. The input of continuous time system is denoted
as x(t) and the output of continuous time system is denoted as y(t). The diagrammatic representation of
a continuous time system is shown in fig 1.2
Chapter 1 - Introduction to Signals and Systems
1. 4
Continuous
time system
x(t)
Input signal
or Excitation
® H
y(t)
®
Output signal
or Response
Fig 1.2 : Representation of continuous time system.
The operation performed by a continuous time system on input to produce output or response can
be expressed as,
Response, y(t) = H{x(t)}
where, H denotes the system operation (also called system operator).
When a continuous time system satisfies the properties of linearity and time invariance then it is
called LTI (Linear Time Invariant) continuous time system . Most of the practical systems that we
encounter in science and engineering are LTI systems.
The input-output relation of an LTI continuous time system is represented by constant coefficient
differential equation shown below(equation (1.4)).
a0
dN
dN 1
dN 2
d
dM
y(t)
+
a
y(t)
+
a
y(t)
+.......+
a
y(t)
+
a
y(t)
=
b
x( t )
1
N
1
2
N
0
dt
dt N
dt N 1
dt N 2
dt M
dM 1
dM 2
d
b1 M 1 x( t ) + b 2 M 2 x( t ) +.......+b M 1 x( t ) b M x( t )
dt
dt
dt
.....(1.4)
where, N = Order of the system, M £ N, and a0 = 1.
The solution of the above differential equation is the response y(t) of the continuous time system,
for the input x(t).
1.2.2 Discrete Time System
A system which can process discrete time signal is called discrete time system, and so the input and
output signals of a discrete time system are discrete time signals.
A discrete time system is denoted by the letter H. The input of discrete time system is denoted as
“x(n)” and the output of discrete time system is denoted as “y(n)”. The diagrammatic representation of a
discrete time system is shown in fig 1.3.
x(n)
Input signal
or Excitation
Discrete
time system
® H
y(n)
®
Output signal
or Response
Fig 1.3 : Representation of discrete time system.
The operation performed by a discrete time system on input to produce output or response can be
expressed as,
Response, y(n) = H{x(n)}
where, H denotes the system operation (also called system operator).
1. 5
Signals & Systems
When a discrete time system satisfies the properties of linearity and time invariance then it is called
LTI (Linear Time Invariant) discrete time system .
The input-output relation of an LTI discrete time system is represented by constant coefficient
difference equation shown below(equation (1.5)).
N
M
bg
b
yn =
am y n
m
g
b
m +
1
bm x n
m
g
m
.....(1.5)
0
where, N = Order of the system, and M £ N.
The solution of the above difference equation is the response y(n) of the discrete time system, for
the input x(n).
1.3 Frequency Domain Analysis of Continuous Time Signals and Systems
Physically, we realize any signal or system in time domain. In time domain, the continuous time
systems are governed by differential equations. The analysis of continuous time signals and systems in
time domain involves solution of differential equations. The solution of differential equations are difficult
due to assumption of a solution and then solving the constants using initial conditions.
In order to simplify the task of analysis, the signals can be transformed to some other domain,
where the analysis is easier. One such transform exists for continuous time signals is Laplace transform.
The Laplace transform, will transform a function of time “t” into a function of complex frequency “s”
where s = s + jW. Therefore, Laplace transform of a continuous time signal will transform the time
domain signal into s-domain signal.
On taking Laplace transform of the differential equation governing the system, it becomes algebraic
equation in “s” and the solution of algebraic equation will give the response of the system as a function
of “s” and it is called s-domain response. The inverse Laplace transform of the s-domain response, will
give the time domain response of the continuous time system. Also, the stability analysis of the continuous
time systems are much easier in s-domain.
Another important characteristics of any signal is frequency, and for most of the applications the
frequency content of the signal is an important criteria. The frequency contents of a signal can be
Table 1.1 : Frequency Range of Some Electromagnetic Signals
Type of signal
Wavelength (m)
Frequency range (Hz)
Radio broadcast
104
to 102
3 ´ 104
to 3 ´ 106
Shortwave radio signals
102
to 10–2
3 ´ 106
to
3 ´ 1010
Radar / Space communications
1
to
10–2
3 ´ 108
to
3 ´ 1010
Common-carrier microwave
1
to
10–2
3 ´ 108
to
3 ´ 1010
Infrared
10–3 to
10–6
3 ´ 1011
to
3 ´ 1014
Visible light
3.9´10–7 to 8.1´10–7
3.7 ´ 1014 to
7.7 ´ 1014
Ultraviolet
10–7 to
10–8
3 ´ 1015
to
3 ´ 1016
Gamma rays and x-rays
10–9 to 10–10
3 ´ 1017
to
3 ´ 1018
Chapter 1 - Introduction to Signals and Systems
1. 6
Table 1.2 : Frequency Range of Some Biological and Seismic Signals
Type of Signal
Frequency Range (Hz)
Electroretinogram
0
to
20
Electronystagmogram
0
to
20
Pneumogram
0
to
40
Electrocardiogram (ECG)
0
to
100
Electroencephalogram (EEG)
0
to
100
Electromyogram
10
to
200
Sphygmomanogram
0
to
200
Speech
100
to
4000
Wind noise
100
to
1000
Seismic exploration signals
10
to
100
Earthquake and nuclear explosion signals
0.01 to
10
Seismic noise
0.1
1
to
studied by taking Fourier transform of a signal. The Fourier transform of a signal is a particular class of
Laplace transform in which s = jW, where “W” is real frequency.
The Fourier transform, will transform a function of time “t” into a function of real frequency
“W”. Therefore, Fourier transform of a continuous time signal will transform the time domain signal
into frequency domain signal. From the Fourier transform of a continuous time signal, the frequency
spectrum of the signal can be obtained which is used to study the frequency content of a signal. The
frequency range of some of the signals are listed in table 1.1 and 1.2.
1.4 Frequency Domain Analysis of Discrete Time Signals and Systems
Mostly, the discrete time systems are designed for analysis of discrete time signals. Physically, the
discrete time systems are also realized in time domain. In time domain, the discrete time systems are
governed by difference equations. The analysis of discrete time signals and systems in time domain
involves solution of difference equations. The solution of difference equations are difficult due to assumption
of a solution and then solving the constants using initial conditions.
In order to simplify the task of analysis, the discrete time signals can be transformed to some
other domain, where the analysis may be easier. One such transform exists for discrete time signals is
Z-transform.The Z-transform, will transform a function of discrete time “n” into a function of complex
variable “z” and it is expressed as, z = rejw. Therefore, Z-transform of a discrete time signal will transform
the time domain signal into z-domain signal.
On taking Z -transform of the difference equation governing the discrete time system, it becomes
algebraic equation in “z” and the solution of algebraic equation will give the response of the system as a
function of “z” and it is called z-domain response. The inverse Z -transform of the z-domain response,
will give the time domain response of the discrete time system. Also, the stability analysis of the discrete
systems are much easier in z-domain.
1. 7
Signals & Systems
The frequency contents of a discrete time signal can be studied by taking Fourier transform of the
discrete time signal. The Fourier transform of discrete time signal is a particular class of Z-transform in
which z = ejw, where “w” is the frequency of the discrete time signals.
The Fourier transform, will transform a function of discrete time “n” into a function of frequency
“w”. Therefore, Fourier transform of a discrete time signal will transform the discrete time signal into
frequency domain signal. From the Fourier transform of the discrete time signal, the frequency spectrum
of the discrete time signal can be obtained which is used to study the frequency content of the discrete
time signal.
1.5 Importance of Signals and Systems
Every part of the universe is a system which generates or processes some type of signal [Of
course the universe itself is a system and said to be controlled by signals (or commands) issued by God].
The signals and systems play a vital role in almost every field of Science and Engineering. Some of
the applications of signals and systems in various field of Science and Engineering are listed here.
1. Biomedical
v ECG is used to predict heart diseases.
v EEG is used to study normal and abnormal behaviour of the brain.
v EMG is used to study the condition of muscles.
v X-ray images are used to predict the bone fractures and tuberclosis.
v Ultrasonic scan images of kidney and gall bladder is used to predict stones.
v. Ultrasonic scan images of foetus is used to predict abnormalities in a baby.
v MRI scan is used to study minute inner details of any part of the human body.
2. Speech Processing
v
v
v
v
v
Speech compression and decompression to reduce memory requirement of storage systems.
Speech compression and decompression for effective use of transmission channels.
Speech recognization for voice operated systems and voice based security systems.
Speech recognization for conversion of voice to text.
Speech synthesis for various voice based warnings or annoucements.
3. Audio and Video Equipments
v The analysis of audio signals will be useful to design systems for special effects in audio
systems like stereo, woofer, karoke, equalizer, attenuator, etc.
v Music synthesis and composing using music keyboards.
v Audio and video compression for storage in DVDs.
4. Communication
v The spectrum analysis of modulated signals helps to identify the information bearing frequency
component that can be used for transmission.
Chapter 1 - Introduction to Signals and Systems
1. 8
v The analysis of signals received from radars are used to detect flying objects and thier velocity.
v Generation and detection of DTMF signals in telephones.
v Echo and noise cancellation in transmission channels.
5. Power electronics
v The spectrum analysis of the output of coverters and inverters will reveal the harmonics
present in the output, which in turn helps to design suitable filter to eliminate the harmonics.
v The analysis of switching currents and voltages in power devices will help to reduce losses.
6. Image processing
v
v
v
v
Image compression and decompression to reduce memory requirement of storage systems.
Image compression and decompression for effective use of transmission channels.
Image recognization for security systems.
Filtering operations on images to extract the features or hidden information.
7. Geology
v The seismic signals are used to determine the magnitude of earthquake and valconic eruptions.
v The seismic signals are also used to predict nuclear explosions.
v The seismic noise are also used to predict the movement of earth layers (tectonic plates).
8. Astronomy
v The analysis of light received from a star is used to determine the condition of the star.
v The analysis of images of various celestial bodies gives vital information about them.
1.6 Use of MATLAB in Signals and Systems
The MATLAB (MATrix LABoratory) is a software developed by The MathWork Inc, USA, which
can run on any windows platform in a PC(Personal Computer). This software has number of tools for the
study of various engineering subjects. It includes a tool for signal processing also. Using this tool a wide
variety of studies can be made on signals and systems. Some of the analysis that is relevant to this
particular text book are given below.
v
v
v
v
v
Sketch or plot of signals as a function of independent variable.
Spectrum analysis of signals.
Solution of LTI systems.
Perform convolution and deconvolution operations on signals.
Perform various transforms on signals like Laplace transform, Fourier transform, Z-transform,
Fast Fourier Transform (FFT), etc.
v Determination of state model from transfer function and viceversa.
v Stability analysis of signals and systems in various domains.
Signals & Systems
2. 1
CHAPTER 2
Continuous Time Signals and Systems
2.1 Introduction
Time is an important independent variable required to measure/monitor any activity. Hence, whatever
phenomena we observe in nature are always measured as a function of time.
Time is a continuous independent variable represented by the letter ‘t’. Any physical observation or
a measure is a continuous function of time represented by x(t) and called signal. The signal x(t) is called
the continuous time (CT) signal which is a function of the independent variable, time. In x(t),the unit of
time and the unit of the value of x(t) at any time is not considered but only the numerical values are
considered.
The signal x(t) can be used to represent any physical quantity, and the start of any observation or
a measure of the physical quantity is taken as time t = 0. The time after the start of observation is taken as
positive time and the time before the start of observation is taken as negative time.
The physical devices which are all sources of continuous time signals are called continuous time
systems. The standard continuous time signals, mathematical operation on continuous time signals and
classification of continuous time signals are discussed in this chapter. The mathematical representation of
continuous time systems and their analyses are also presented in this chapter. Wherever required, the
discussion on LTI systems are presented separately.
2.2 Standard Continuous Time Signals
d(t)
1. Impulse signal
¥
The impulse signal is a signal with infinite magnitude and zero duration,
but with an area of A. Mathematically, impulse signal is defined as,
+¥
Impulse Signal, d ( t ) = ¥ ; t = 0
z
and
d ( t ) dt = A
-¥
= 0; t ¹ 0
0
t
The unit impulse signal is a signal with infinite magnitude and zero duration, Fig 2.1 : Impulse signal
but with unit area. Mathematically, unit impulse signal is defined as,
(or Unit Impulse signal).
+¥
Unit Impulse Signal, d ( t ) = ¥ ; t = 0
z
and
d ( t ) dt = 1
-¥
= 0; t ¹ 0
2. Step signal
The step signal is defined as,
x(t) = A ; t ³ 0
=0;t<0
The unit step signal is defined as,
x(t)
u(t)
A
1
0
x(t) = u(t) = 1 ; t ³ 0
=0;t<0
t
Fig 2.2 : Step signal.
0
t
Fig 2.3 : Unit step signal.
Chapter 2 - Continuous Time Signals and Systems
2. 2
3. Ramp signal
The ramp signal is defined as,
x(t) = At ; t ³ 0
=0 ; t<0
x(t)
x(t)
2A
2
A
1
The unit ramp signal is defined as,
0
x(t) = t ; t ³ 0
1
2
0
t
2
t
Fig 2.5 : Unit ramp signal.
Fig 2.4 : Ramp signal.
=0 ; t<0
1
4. Parabolic signal
x(t)
x(t)
The parabolic signal is defined as,
At 2
x( t ) =
; for t ³ 0
2
=0
;
4.5A
4.5
2A
2
t < 0
0.5
0.5A
The unit parabolic signal is defined as,
x( t ) =
;
1
2
3
0
t
Fig 2.6 : Parabolic signal.
t2
; for t ³ 0
2
=0
0
1
2
3
t
Fig 2.7 :Unit parabolic signal.
t< 0
P(t)
5. Unit pulse signal
1
The unit pulse signal is defined as,
FG
H
x ( t ) = P( t ) = u t + 1
2
IJ
K
FG
H
- u t- 1
2
IJ
K
–0.5
6. Sinusoidal signal
Case i : Cosinusoidal signal
The cosinusoidal signal is defined as,
x(t) = A cos(W0t + f)
2p
= Angular frequency in rad/sec
T
F0 = Frequency in cycles/sec or Hz
where, W0 = 2pF0 =
T = Time period in sec
When
f = 0,
x(t) = A cosW 0 t
When
f = Positive,
x(t) = A cos(W0t + f)
When
f = Negative, x(t) = A cos(W0t – f)
0.5
t
Fig 2.8 : Unit pulse signal.
Signals & Systems
2. 3
Case ii : Sinusoidal signal
The sinusoidal signal is defined as,
x(t) = A sin(W0t + f)
2p
= Angular frequency in rad/sec
T
where, W0 = 2pF0 =
F0 = Frequency in cycles/sec or Hz
T = Time period in sec
x(t) = A sin W 0 t
When
f = 0,
When
f = Positive, x(t) = A sin(W0t + f)
When
f = Negative, x(t) = A sin(W0t – f)
x(t)
x(t)
f =0
A
f =0
A
0
0
t
–A
t
–A
x(t)
x(t)
f = Positive
­
f = Positive
t
f
t
-W
f
-W
0
0
x(t)
f = Negative
f
W0
­
x(t)
f = Negative
t
Fig 2.9 : Cosinusoidal signal.
t
­
f
W0
Fig 2.10 : Sinusoidal signal.
7. Exponential signal
Case i : Real exponential signal
The real exponential signal is defined as,
x(t) = A ebt
where, A and b are real
Here, when b is positive, the signal x(t) will be an exponentially rising signal; and when b is negative
the signal x(t) will be an exponentially decaying signal.
Chapter 2 - Continuous Time Signals and Systems
x(t)
2. 4
x(t)
b = Negative
b = Positive
A
A
0
0
t
t
Fig 2.11 : Real exponential signal.
Case ii : Complex exponential signal
The complex exponential signal is defined as,
x ( t ) = A e jW 0 t
where, W0 = 2pF0 =
2p
= Angular frequency in rad/sec
T
F0 = Frequency in cycles/sec or Hz
T = Time period in sec
Im
The complex exponential signal can be represented in a
complex plane by a rotating vector, which rotates with a
constant angular velocity of W0 rad/sec.
Complex plane
A
W0t
The complex exponential signal can be resolved into real
and imaginary parts as shown below,
Real
x( t ) = A e jW0 t = A ( cosW 0 t + j sin W 0 t )
= A cos W 0 t + jA sin W 0 t
\ A cosW 0 t = Real part of x(t)
Fig 2.12 : Complex exponential signal.
A sin W 0 t = Imaginary part of x(t)
From the above equation, we can say that a complex exponential signal is the vector sum of two
sinusoidal signals of the form cosW 0 t and sin W 0 t .
8. Exponentially rising/decaying sinusoidal signal
The exponential rising/decaying sinusoidal signal is defined as,
x(t) = A ebt sin W 0 t
2p
= Angular frequency in rad/sec
T
F0 = Frequency in cycles/sec or Hz
where, W0 = 2pF0 =
T = Time period in sec
Here, A and b are real constants. When b is positive, the signal x(t) will be an exponentially rising
sinusoidal signal; and when b is negative, the signal x(t) will be an exponentially decaying sinusoidal signal.
Signals & Systems
2. 5
x(t)
x(t) = A ebt sin W0t,
x(t)
where, b = Positive
x(t) = A ebt sin W0t,
where, b = Negative
A
A
0
t
Fig 2.13 : Exponentially rising sinusoid.
0
t
Fig 2.14 : Exponentially decaying sinusoid.
9. Triangular pulse signal
x(t)
The Triangular pulse signal is defined as
x( t ) = D a (t) = 1 -
t
a
= 0
;
;
1
t £ a
t > a
–a
a
0
t
Fig 2.15 : Triangular pulse signal.
10. Signum signal
x(t)
The Signum signal is defined as the sign of the independent variable t.
Therefore, the Signum signal is expressed as,
1
0
t
–1
x(t) = sgn(t) = 1 ; t > 0
= 0
; t=0
= -1
; t<0
11. Sinc signal
x(t)
1
The Sinc signal is defined as,
x ( t ) = sinc(t) =
Fig 2.16 : Signum signal.
sin t
; -¥ < t < ¥
t
t
Fig 2.17 : Sinc signal.
ga(t)
12. Gaussian signal
1
The Gaussian signal is defined as,
x(t) = g a (t) = e - a
2t2
; -¥ < t < ¥
t
Fig 2.18 : Gaussian signal.
Chapter 2 - Continuous Time Signals and Systems
2. 6
2.3 Classification of Continuous Time Signals
The continuous time signals are classified depending on their characteristics. Some ways of classifying
continuous time signals are,
1. Deterministic and Nondeterministic signals
2. Periodic and Nonperiodic signals
3. Symmetric and Antisymmetric signals (Even and Odd signals)
4. Energy and Power signals
5. Causal and Noncausal signals
2.3.1 Deterministic and Nondeterministic Signals
The signal that can be completely specified by a mathematical equation is called a deterministic
signal. The step, ramp, exponential and sinusoidal signals are examples of deterministic signals.
Examples of deterministic signals: x1(t) = At
x2(t) = Xm sin W 0 t
The signal whose characteristics are random in nature is called a nondeterministic signal. The
noise signals from various sources like electronic amplifiers, oscillators, radio receivers, etc., are best
examples of nondeterministic signals.
2.3.2 Periodic and Nonperiodic Signals
A periodic signal will have a definite pattern that repeats again and again over a certain period of
time. Therefore the signal which satisfies the condition,
x(t + T ) = x(t) is called a periodic signal.
A signal which does not satisfy the condition, x(t + T) = x(t) is called an aperiodic or nonperiodic
signal. In periodic signals, the term T is called the fundamental time period of the signal. Hence, inverse
of T is called the fundamental frequency, F0 in cycles/sec or Hz, and 2pF0 = W0 is called the fundamental
angular frequency in rad/sec.
The sinusoidal signals and complex exponential signals are always periodic with a periodicity of T,
1
2p
where, T =
=
W 0 . The proof of this concept is given below.
F0
Proof :
a) Cosinusoidal signal
Let,
x(t) = A cosW0 t
b g
b
\ x(t + T) = A cos W 0 t + T = A cos W 0t + W 0 T
2p I
FG
TJ
H
T K
= A cos bW t + 2pg = A cos W t = xbtg
= A cos W0 t +
0
0
g
W 0 = 2 pF0 =
2p
T
cos( q + 2 p) = cosq
Signals & Systems
2. 7
b) Sinusoidal signal
Let, x(t) = A sinW0t
b g
b
g
2p I
F
TJ
= A sin G W t +
H
T K
= A sin bW t + 2p g = A sin W t = xbt g
\ x(t + T) = A sin W 0 t + T = A sin W 0 t + W 0 T
W 0 = 2 p F0 =
0
0
2p
T
sin(q + 2 p ) = sinq
0
c) Complex exponential signal
Let ,
x(t) = A e jW 0t
\ x( t + T ) = A ejW0 ( t + T ) = A ejW0 t e jW 0T = A e jW 0t e
j
2p
T
T
= A e jW0t ej2 p
cos 2 p = 1, sin 2 p = 0
= A e jW 0t (cos 2p + j sin 2p ) = A e jW0t ( 1 + j 0) = x(t)
When a continuous time signal is a mixture of two periodic signals with fundamental time periods
T1 and T2 , then the continuous time signal will be periodic, if the ratio of T1 and T2 (i.e., T1/T2) is a rational
number. Now the periodicity of the continuous time signal will be the LCM (Least Common Multiple) of
T1 and T2 .
Note : 1. The ratio of two integers is called a rational number.
5
7
8
Example of rational number : 2 , 9 , 11 .
Example of non-rational number :
2 7
4
,
,
5 2p
7
.
2. When T1 /T2 is a rational number, then F01 /F02 and W01 /W02 are also rational numbers.
Example 2.1
Verify whether the following continuous time signals are periodic. If periodic, find the fundamental period.
g bg
a x t = 2 cos
t
4
g bg
b x t = e at ; a > 1
g
c x(t) = e
- j2 pt
7
g
Solution
a) Given that, x(t) = 2 cos
t
4
The given signal is a cosinusoidal signal, which is always periodic.
On comparing x(t) with the standard form “A cos 2pF0t” we get,
1
1
2pF0 =
Þ
F0 =
4
8p
Period, T =
1
= 8p
F0
\ x(t) is periodic with period, T = 8p.
b) Given that, x(t) = eat ; a > 1
\ x(t + T) = e a(t + T)
= eat eaT
For any value of a, eaT ¹ 1 and so x(t + T) ¹ x(t)
Since x(t + T) ¹ x(t), the signal x(t) is non-periodic.
FG
H
d x(t) = 3cos 5t +
p
6
IJ
K
g
FG
H
e x(t) = cos 2 2t -
p
4
IJ
K
Chapter 2 - Continuous Time Signals and Systems
c) Given that, x(t) = e
2. 8
- j2 pt
7
The given signal is a complex exponential signal, which is always periodic.
On comparing x(t) with the standard form "A e - j2 pF0 t "
1
We get, F0 =
7
\ Peroid, T =
1
= 7
F0
\ x(t) is periodic with period, T = 7.
FG
H
p
6
d) Given that, x(t) = 3 cos 5t +
IJ
K
The given signal is a cosinusoidal signal, which is always periodic.
FG
H
\ x(t + T) = 3 cos 5(t + T) +
Let 5T = 2p,
\T =
p
6
IJ
K
FG
H
2p
5
FG FG p IJ + 5 ´ 2p IJ
5K
HH 6K
F pI
= 3 cosG 5t + J = x(t)
H 6K
\ x(t + T) = 3 cos 5t +
FG
H
p
3
FG FG
HH
FG
H
1 + cos 2 2 t =
FG
H
2p
3
IJ
K
2
FG
H
2p
1 + cos 4t + 4T
3
=
2
2p
p
Let 4T = 2 p , \ T =
=
4
2
FG
H
1 + cos 4t \ x(t + T) =
FG
H
1 + cos 4t =
2
p
3
2
1 + cos 4(t + T) \ x(t + T) =
FG FG
HH
p
6
= 3 cos 5t +
p
6
IJ
K
IJ
K
IJ
K
+ 5T
IJ
K
+ 2p
For integer values of M,
cos(q + 2pM) = cosq
2p
5
IJ
K
F p IJ
x(t ) = cos G 2t H 3K
2
IJ
K
= 3 cos 5t +
Since x(t + T) = x(t), the signal x(t) is periodic with period, T =
e) Given that, x(t) = cos2 2t -
p
6
= 3 cos 5t + 5T +
IJ
K
FG
H
1 + cos 4 t =
IJ
K
cos 2 q =
FG
H
=
2p
3
IJ
K
1+ cos 2q
2
2
IJ
K
IJ
K
FG FG
HH
1 + cos 4t =
2p
3
IJ
K
+ 2p
IJ
K
2
FG
H
1 + cos 2 2t =
2p
3
2
1 + cos 4t + 4T -
2p
p
+ 4 ´
3
2
2
2p
3
IJ
K
p
3
2
Since x(t + T) = x(t), the signal x(t) is periodic with period, T =
p
2
IJ
K
FG
H
= cos 2 2t -
p
3
IJ
K
= x(t)
For integer values of M,
cos (q + 2pM) = cos q
Chapter 2 - Continuous Time Signals and Systems
2. 10
Since x1(t) and x2(t) are periodic and the ratio of T1 and T2 is a rational number, the signal x(t) is also periodic. Let
T be the periodicity of x(t). Now the periodicity of x(t) is the LCM (Least Common Multiple) of T1 and T2 , which is
calculated as shown below.
2p
=
3
2p
T2 =
=
7
Now LCM of 7 and 3
2p
´
3
2p
´
7
is 21.
\ Period, T = 21 ¸
21
2p
= 21 ´
= 2p
2p
21
T1 =
21
= 7
2p
21
= 3
2p
Note : To find LCM, first convert T1 and T2 to integers
by multiplying by a common number. Find LCM
of integer values of T1 and T2 . Then divide this
LCM by the common number.
Proof : x(t + T) = 2 cos 3(t + T) + 3 sin 7(t + T)
= 2 cos(3t + 3T) + 3 sin(7t + 7T)
Put, T = 2p
= 2 cos(3t + 3 ´ 2p) + 3 sin (7t + 7 ´ 2p)
= 2 cos(3t + 6p) + 3 sin(7t + 14p)
For integer values of M,
cos(q + 2pM) = cosq
= 2 cos 3t + 3 sin 7t = x(t)
sin(q + 2pM) = sinq
(c) Given that, x(t) = 5 cos 4pt + 3 sin 8pt
Let, x1(t) = 5 cos 4pt
Let T1 be the periodicity of x1(t). On comparing x1(t) with the standard form “A cos 2pF01t”, we get,
F01 = 2 ;
\ Period, T1 =
1
1
=
F01
2
Let, x2(t) = 3 sin 8pt
Let T2 be the periodicity of x2(t). On comparing x2(t) with the standard form “A sin 2pF02t”, we get,
F02 = 4 ;
Now,
1
1
=
F02
4
\ Period, T2 =
1 4
T1
1
= ´
= T1 ´
= 2
2 1
T2
T2
Since x1(t) and x2(t) are periodic and the ratio of T1 and T2 is a rational number, the signal x(t) is also periodic. Let
T be the periodicity of x(t). Now, the periodicity of x(t) is the LCM (Least Common Multiple) of T1 and T2, which is calculated
as shown below.
1
1
=
´ 4 = 2
2
2
1
1
T2 =
=
´ 4 = 1
4
4
Now LCM of 2 and 1 is 2.
Note : To find LCM, first convert T1 and T2 to integers
by multiplying by a common number. Find LCM
of integer values of T1 and T2 . Then divide this
LCM by the common number.
T1 =
\ Period, T = 2 ¸ 4 = 2 ´
1
1
=
4
2
Proof : x(t + T) = 5 cos 4p(t + T) + 3 sin 8p(t + T)
= 5 cos(4pt + 4pT) + 3 sin(8pt + 8pT)
FG
H
= 5 cos 4pt + 4p ´
1
2
IJ
K
FG
H
+ 3 sin 8pt + 8p ´
1
2
IJ
K
Put, T =
1
2
= 5 cos (4pt + 2p) + 3 sin (8pt + 2p)
For integer values of M,
cos(q+ 2pM) = cosq
= 5 cos 4pt + 3 sin 8pt = x(t)
sin(q + 2pM) = sinq
Signals & Systems
2. 11
2.3.3 Symmetric (Even) and Antisymmetric (Odd) Signals
The signals may exhibit symmetry or antisymmetry with respect to t = 0.
When a signal exhibits symmetry with respect to t = 0 then it is called an even signal. Therefore,
the even signal satisfies the condition, x(-t) = x(t).
When a signal exhibits antisymmetry with respect to t = 0, then it is called an odd signal. Therefore,
the odd signal satisfies the condition, x(-t) = -x(t).
Since cos(-q) = cosq, the cosinusoidal signals are even signals and since sin(-q) = -sinq, the
sinusoidal signals are odd signals.
x(t)
x(t)
x(t) = A cosW 0 t
A
x(t) = A sin W 0 t
A
0
0
t
–A
t
–A
Fig 2.19a : Symmetric or Even signal.
Fig 2.19b : Antisymmetric or Odd signal.
Fig 2.19 : Symmetric and antisymmetric continuous time signals.
A continuous time signal x(t) which is neither even nor odd can be expressed as a sum of even and
odd signal.
Let, x(t) = x e (t) + x o (t)
where, x e (t) = Even part of x(t) and x o (t) = Odd part of x(t)
Now, it can be proved that,
1
x e (t) =
x(t) + x( - t)
2
1
x o (t) =
x(t) - x( - t)
2
Proof :
Let, x(t) = xe(t) + xo(t)
On replacing t by –t in equation (2.1) we get,
x(–t) = xe(–t) + xo(–t)
.....(2.1)
.....(2.2)
Since xe(t) is even, xe(–t) = xe(t)
Since xo(t) is odd, xo(–t) = – xo(t)
Hence the equation (2.2) can be written as,
x(–t) = xe(t) – xo(t)
On adding equations (2.1) & (2.3) we get,
x(t) + x(–t) = 2 xe(t)
1
\ x e (t) =
x(t) + x( - t)
2
On subtracting equation (2.3) from equation (2.1) we get,
x(t) – x(–t) = 2 xo(t)
\ x o (t) =
1
x(t) - x( - t)
2
.....(2.3)
Chapter 2 - Continuous Time Signals and Systems
The properties of signals with symmetry are given below without proof.
1. When a signal is even, then its odd part will be zero.
2. When a signal is odd, then its even part will be zero.
3. The product of two odd signals will be an even signal.
4. The product of two even signals will be an even signal.
5. The product of an even and odd signal will be an odd signal.
Example 2.3
Determine the even and odd part of the following continuous time signals.
a) x(t) = et
b) x(t) = 3 + 2t + 5t2
c) x(t) = sin 2t + cos t + sin t cos 2t
Solution
a) Given that, x(t) = et
\ x(–t) = e–t
1
1 t
[ x( t) + x(- t)] =
[e + e - t ]
2
2
1
1 t
Odd part, x o (t ) =
[ x(t ) - x(- t)] =
[e - e - t ]
2
2
Even part, x e ( t) =
b) Given that, x(t) = 3 + 2t + 5t2
\ x(–t) = 3 + 2(–t) + 5(–t)2
= 3 – 2t + 5t2
Even part, x e ( t) =
=
Odd part, x o ( t) =
=
1
1
[ x( t) + x(- t)] =
[3 + 2t + 5t 2 + 3 - 2t + 5t 2 ]
2
2
1
[6 + 10t 2 ] = 3 + 5t 2
2
1
1
[ x( t) - x(- t)] =
[ 3 + 2t + 5t 2 - 3 + 2t - 5t 2 ]
2
2
1
[4t ] = 2t
2
c) Given that, x(t) = sin 2t + cos t + sin t cos 2t
\ x(–t) = sin 2(–t) + cos(–t) + sin(–t) cos 2(–t)
= –sin 2t + cos t – sin t cos 2t
Even part, x e (t ) =
=
1
1
[ x(t ) + x( - t)] =
[sin 2t + cos t + sin t cos 2t - sin 2t + cos t - sin t cos 2t ]
2
2
1
[2 cos t ] = cos t
2
1
1
[ x( t) - x(- t)] =
[ sin 2t + cos t + sin t cos 2t + sin 2t - cos t + sin t cos 2t ]
2
2
1
=
[2 sin 2t + 2 sin t cos 2t ] = sin 2t + sin t cos 2t
2
Odd part, x o ( t) =
2. 12
Signals & Systems
2. 13
2.3.4 Energy and Power Signals
The signals which have finite energy are called energy signals. The nonperiodic signals like
exponential signals will have constant energy and so nonperiodic signals are energy signals.
The signals which have finite average power are called power signals. The periodic signals like
sinusoidal and complex exponential signals will have constant power and so periodic signals are power
signals.
The energy E of a continuous time signal x(t) is defined as,
z
T
Energy, E = Lt
T®¥
x(t)
2
dt in joules
-T
The average power of a continuous time signal x(t) is defined as,
Power, P = Lt
T®¥
1
2T
z
T
x(t)
2
dt in watts
-T
For periodic signals, the average power over one period will be same as average power over an
infinite interval.
\ For periodic signals, power, P =
1
T
z
T
x(t)
2
dt
0
For energy signals, the energy will be finite (or constant) and average power will be zero.
For power signals the average power is finite (or constant) and energy will be infinite.
i.e., For energy signal, E is constant (i.e., 0 < E < ¥) and P = 0.
For power signal, P is constant (i.e., 0 < P < ¥) and E = ¥ .
Proof :
The energy of a signal x(t) is defined as ,
T
E =
Lt
T®¥
z
x(t)
2
dt
.....(2.4)
-T
The power of a signal is defined as ,
P =
Lt
T®¥
1
2T
T
z
2
x(t) dt =
-T
Lt
T®¥
1
2T
T
Lt
T®¥
z
2
x(t) dt
.....(2.5)
-T
Using equation (2.4), the equation (2.5) can be written as,
P =
Lt
T®¥
1
´ E
2T
.....(2.6)
In equation (2.6), When E = constant,
P = E ´
= E ´
Lt
T®¥
1
2T
1
= E ´ 0= 0
2´¥
From the above analysis, we can say that when a signal has finite energy the power will be zero. Also, from the above
analysis we can say that the power is finite only when energy is infinite.
Chapter 2 - Continuous Time Signals and Systems
2. 14
Example 2.4
Determine the power and energy for the following continuous time signals.
a) x(t) = e–2t u(t)
FG
H
j 2t +
b) x( t ) = e
p
4
IJ
K
c) x(t) = 3 cos 5 W0 t
Solution
a) Given that, x(t) = e –2t u(t)
x(t) = e–2t u(t) ; for all t
Here,
; for t ³ 0
\ x(t) = e–2t
z
z
T
\
T
2
x(t) dt =
-T
ee j
-2t
dt =
0
-4T
z
T
Energy, E =
z
dt =
Lt
T® ¥
2
x(t) dt =
-4t
e -4t dt =
0
OP = LM 1
PQ N 4
e0
-4
-
LM e OP
N -4 Q
T
-2t 2
de i
0
LM e
MN -4
=
z
T
2
Lt
T® ¥
-T
-
e -4T
4
LM 1
N4
-
T
0
OP
Q
e -4T
4
OP
Q
1
e -¥
1
0
1
=
=
joules
4
4
4
4
4
=
Power, P =
z
T
1
2T
Lt
T ®¥
Lt
T®¥
-T
LM
N
1 1
e -¥
¥ 4
4
=
2
x(t) dt =
OP = 0 ´ L 1
MN 4
Q
- 0
LM
N
1 1
e -4T
2T 4
4
OP =
Q
OP
Q
0
Since energy is constant and power is zero, the given signal is an energy signal.
af
b) Given that, x t
FG
H
j 2t+
= e
FG
H
j 2t +
Here,
x(t) = e
p
4
IJ
K
p
4
IJ
K
FG
H
= 1 Ð 2t +
p
4
IJ
K
\ x( t) = 1
z
T
z
T
2
x(t) dt =
-T
1 ´ dt = t
T
-T
= T + T = 2T
-T
z
T
Energy, E =
Lt
T® ¥
2
x(t) dt =
Lt
T® ¥
2T = ¥
-T
Power, P =
Lt
T ®¥
1
2T
z
T
-T
2
x(t) dt =
Lt
T ®¥
1
´ 2T = 1 watt
2T
Since power is constant and energy is infinite, the given signal is a power signal.
Signals & Systems
2. 15
c) Given that, x(t) = 3 cos 5 W 0 t
z
T
\
z
z
zb
z
b3cos 5W tg
z FGH
IJ
K
T
2
x(t) dt =
-T
2
dt =
0
-T
T
2
0
dt =
-T
T
=
zb
T
d 3cos 5W t i
T
9 cos2 5W 0t dt = 9
-T
-T
T
g
1+ cos10W 0t dt =
-T
2
-T
1+ cos10W 0 t
dt
2
cos 2 q =
LM t + sin10W t OP
N 10W Q
9 L
sin10W T F
sin10W ( - T ) I O
T +
=
- G-T +
M
JK PPQ
2 MN
10W
10W
H
2p O
L
T P
sin 10
9 L
9 M
sin10W T O
T
2T + 2
2T + 2
=
=
M
P
M
PP
2
p
2 N
10W
Q 2 MN
10
T
Q
9 L
T
9 L
T
O
O
2T +
sin 20p P =
=
2 MN
10p
Q 2 MN 2T + 10p ´ 0 PQ =
9
=
2
g
3cos 5W 0t dt
1+ cos2q
2
T
9
2
0
0
0
-T
0
0
0
sin(- q ) = – sin q
0
W 0 =2pF0 =
0
z
9T
For integer M,
sin pM = 0
T
Energy, E =
Lt
T® ¥
2
x(t) dt =
Lt 9T = ¥
T® ¥
-T
Power, P =
Lt
T ®¥
1
2T
z
T
2
x(t) dt =
Lt
T®¥
-T
1
´ 9T =
2T
Lt
T ®¥
9
9
=
= 4.5 watts
2
2
Since energy is infinite and power is constant, the given signal is a power signal.
2.3.5 Causal, Noncausal and Anticausal Signals
A signal is said to be causal, if it is defined for t ³ 0.
Therefore if x(t) is causal, then x(t) = 0, for t < 0.
A signal is said to be noncausal, if it is defined for either t £ 0, or for both t £ 0 and t > 0.
Therefore if x(t) is noncausal, then x(t) ¹ 0, for t < 0.
When a noncausal signal is defined only for t £ 0, it is called anticausal signal.
x(t) = A
Unit step signal,
x(t) = u(t) = 1 ; t ³ 0
Exponential signal,
x(t) = A ebt u(t)
Complex exponential signal, x( t ) = A e jW 0 t u(t)
Exponential signal,
x(t) = A ebt ; for all t
Complex exponential signal, x( t ) = A e jW0 t ; for all t
123
;t³0
Step signal,
Causal signals
123
Examples of causal and noncausal signals
Noncausal signals
Note : On multiplying a noncausal signal by u(t), it becomes causal.
2p
T
Chapter 2 - Continuous Time Signals and Systems
2. 16
2.4 Mathematical Operations on Continuous Time Signals
2.4.1 Scaling of Continuous Time Signals
The two types of scaling continuous time signals are,
1. Amplitude Scaling
2. Time Scaling
1. Amplitude Scaling
The amplitude scaling is performed by multiplying the amplitude of the signal by a constant.
Let x(t) be a continuous time signal. Now Ax(t) is the amplitude scaled version of x(t), where A is
a constant.
When |A| > 1, then Ax(t) is the amplitude magnified version of x(t) and when |A| < 1, then Ax(t) is
the amplitude attenuated version of x(t).
Example : 1
Let, x(t) = at + be–ct
Let x 1(t) and x 2 (t) be the amplitude scaled versions of x(t), scaled by constants 4 and
1
4
FG 1 = 0.25IJ respectively.
H4
K
Now, x1(t) = 4x(t) = 4(at + be–ct) = 4at + 4be–ct
x2(t) = 0.25x(t) = 0.25(at + be–ct) = 0.25at + 0.25be–ct
Example : 2
A continuous time signal and its amplitude scaled version are shown in fig 2.20.
x1(t) = 2x(t)
x1(t)
4
x1(t) = 2x(t) = 2 ; 0 < t < t1
x(t)
3
= 4 ; t1< t < t2
2
= 2 ; t2 < t < t3
1
n
(t)
2x
tio
ica
f
i
pl
Am
2
1
t1
= 2 ; t1 < t < t2
= 1 ; t2 < t < t3
t
t3
t
0.
5x
(t)
n
io
= 1 ; 0 < t < t1
t3
t2
at
nu
X(t)
t2
t1
tte
A
0
0
x2(t) = 0.5x(t)
x2(t)
2
x2(t) = 0.5x(t) = 0.5 ; 0 < t < t1
1
0.5
=1
; t1 < t < t2
= 0.5 ; t2 < t <t3
0
t1
t2
t3
t
Fig 2.20 : A continuous time signal and its amplitude scaled version.
2. 17
Signals & Systems
2. Time Scaling
The time scaling is performed by multiplying the variable time by a constant.
If x(t) is a continuous time signal, then x(At) is the time scaled version of x(t), where A is a
constant.
When |A| > 1, then x(At) is the time compressed version of x(t) and when |A| < 1, then x(At) is the
time expanded version of x(t).
Example : 1
Let, x(t) = at + be–ct
Let x1(t) and x2(t) be the time scaled versions of x(t), scaled by constants 4 and 1/4 (0.25)respectively.
Now, x1(t) = x(4t) = a ´ 4t + be–c ´ 4t = 4at + be–4ct
x2(t) = x(0.25t) = a ´ 0.25t + be–c ´ 0.25t = 0.25at + be–0.25ct
Example : 2
A continuous time signal and its time scaled version are shown in fig 2.21.
x1(t) = x(2t)
x1(t)
2
1
0
t)
ion
x(2 press
om
ec
m
i
T
x(t)
2
t1 t 2 t 3
2 2 2
t
1
0
t1
t2
t1
t
t
; x 1 1 = x 2 1 = x( t 1 )
2
2
2
When t =
t2
;
2
When t =
t3
;
2
2
2
1
2
3
3
3
1
t
n
sio
an
xp )
t
ee
m (0.5
x
Ti
x(t) = 1 ; 0 < t < t1
= 2 ; t1 < t < t2
= 1 ; t2 < t < t3
t3
FG IJ FG IJ
H K H K
Ft I F t I
x G J = xG 2 J = x(t )
H 2K H 2K
Ft I F t I
x G J = xG 2 J = x(t )
H 2K H 2K
When t =
x2(t)
x2(t) = x(0.5t)
2
1
0
2t 1
2t 2
2t 3
t
When t = 2t1 ; x1(2t1) = x(0.5 ´ 2t1) = x(t1)
When t = 2t2; x1(2t2) = x(0.5 ´ 2t2) = x(t2)
When t = 2t3; x1(2t3) = x(0.5 ´ 2t3) = x(t3)
Fig 2.21 : A continuous time signal and its time scaled version.
Chapter 2 - Continuous Time Signals and Systems
2. 18
2.4.2 Folding (Reflection or Transpose) of Continuous Time Signals
The folding of a continuous time signal x(t) is performed by changing the sign of time base t in
the signal x(t).
The folding operation produces a signal x(–t) which is a mirror image of the original signal x(t)
with respect to the time origin t = 0.
Example : 1
Let, x(t) = at + be–ct
Let x1(t) be folded version of x(t).
Now, x1(t) = x(– t) = a (– t) + be–c(– t) = – at + bect
Example : 2
A continuous time signal and its folded version is shown in fig 2.22.
x(t)
x1(t) = x(–t)
x1(t)
x(–t)
2
2
Folding
1
1
0
t1
t3
t2
t
x(t) = 1 ; 0 < t < t1
–t3
–t 2
–t 1
0
t
When t = –t1; x1(–t1) = x(–(–t1)) = x(t1)
=2;
t1 < t < t2
When t = –t2; x1(–t2) = x(–(–t2)) = x(t2)
=1;
t2 < t < t3
When t = –t3; x1(–t3) = x(–(–t3)) = x(t3)
Fig 2.22 : A continuous time signal and its folded version.
2.4.3 Time Shifting of Continuous Time Signals
The time shifting of a continuous time signal x(t) is performed by replacing the independent
variable t by t – m, to get the time shifted signal x(t – m), where m represents the time shift in seconds.
In x(t – m), if m is positive, then the time shift results in a delay by m seconds. The delay results
in shifting the original signal x(t) to right, to generate the time shifted signal x(t – m).
In x(t – m), if m is negative, then the time shift results in an advance of the signal by |m| seconds.
The advance results in shifting the original signal x(t) to left, to generate the time shifted signal x(t – m).
Example : 1
Let, x(t) = at + be–ct
Let x1(t) and x 2(t) be time shifted version of x(t), shifted by m units of time.
Let x1(t) be delayed version of x(t) and x2(t) be advanced version of x(t).
Now, x1(t) = a(t – m) + be–c(t – m)
x2(t) = a(t + m) + be
–c(t + m)
2. 19
Signals & Systems
Example : 2
A signal and its shifted version are shown in fig 2.23.
x1(t)
2
x(t
1
0
1
2
3
When t = 4; x1(4) = x(4 – 2) = x(2) = 2
1
When t = 5; x1(5) = x(5 – 2) = x(3) = 1
2
1
t
5
4
3
)
–2
De
l ay
t
Ad
va
nc
e
x(t) = 1 ; 0 < t < 1
t+
x(
=2 ;1<t<2
When t = 3; x1(3) = x(3 – 2) = x(1) = 1
2
0
x(t)
When t = 2; x1(2) = x(2 – 2) = x(0) = 1
x1(t) = x(t – 2)
x2(t) = x(t + 2)
x2(t)
2)
=1 ;2<t<3
When t = –2 ; x1(–2) = x(–2 + 2) = x(0) = 1
2
When t = –1 ; x1(–1) = x(–1 + 2) = x(1) = 1
1
–2
0
–1
1
2
When t = 0 ;
x1(0) = x(0 + 2) = x(2) = 2
When t = 1
x1(1) = x(1 + 2) = x(3) = 1
;
t
Fig 2.23 : A continuous time signal and its shifted version.
Delayed Unit Impulse Signal
d(t)
d(t-m)
¥
The unit impulse signal is defined as,
¥
+¥
d( t) = ¥ ; t = 0
and
= 0 ; t ¹0
z
d ( t ) dt = 1
-¥
0
m
0
t
t
The unit impulse signal delayed by m units of Fig 2.24a : Impulse. Fig 2.24b : Delayed impulse.
Fig 2.24 : Impulse and delayed impulse signal.
time is denoted as d(t – m), and it is defined as,
+¥
d ( t - m) = ¥ ; t = m and
= 0; t ¹m
z
d ( t - m) dt = 1
-¥
Delayed Unit Step Signal
u(t)
u(t-m)
1
1
The unit step signal is defined as,
u(t) = 1 ; for t ³ 0
= 0 ; for t < 0
0
t
0
m
t
The unit step signal delayed by
Fig 2.25a : Unit step signal. Fig 2.25b : Delayed unit step signal.
m units of time is denoted as u(t – m),
Fig 2.25 : Unit step and delayed unit step signal.
and it is defined as,
u(t – m) = 1 ; t ³ m
=0;t<m
Chapter 2 - Continuous Time Signals and Systems
2. 20
2.4.4 Addition of Continuous Time Signals
The addition of two continuous time signals is performed by adding the value of the two signals
corresponding to the same instant of time.
The sum of two signals x1(t) and x2(t) is a signal y(t), whose value at any instant is equal to the
sum of the value of these two signals at that instant.
i.e., y(t) = x1(t) + x2(t)
Example :
Graphical addition of two continuous time signals is shown in fig 2.26.
y(t)
x2(t)
x1(t)
2
2
+
1
0
2
1
1
0
1
2
1
t
3
x1(t) = 1 ; 0 < t < 1
y(t) = x1(t) + x2(t)
3
2
x2(t) = t
;0<t<1
=2 ;1<t<2
=1
;1<t<2
=1 ;2<t<3
=3–t;2<t<3
t
3
0
1
2
3
t
For t = 0 to 1 ; y(t) = x1(t) + x2(t) = 1 + t
For t = 1 to 2 ; y(t) = x1(t) + x2(t) = 2 + 1 = 3
For t = 2 to 3 ; y(t) = x 1(t) + x2(t) = 1 + 3 – t = 4 – t
Fig 2.26 : Addition of two continuous time signals.
2.4.5 Multiplication of Continuous Time Signals
The multiplication of two continuous time signals is performed by multiplying the value of the
two signals corresponding to the same instant of time.
The product of two signals x1(t) and x2(t) is a signal y(t), whose value at any instant is equal to the
product of the values of these two signals at that instant.
i.e., y(t) = x1(t) ´ x2(t)
Example :
Graphical multiplication of two continuous time signals is shown in fig 2.27.
y(t)
x1(t)
x2(t)
2
2
2
1
1
´
1
0
1
2
3
x1(t) = 1 ; 0 < t < 1
t
0
1
2
3
t
0
y(t) = x1(t) ´ x2(t)
1
2
x 2(t) = t ; 0 < t < 1
For t = 0 to 1;
y(t) = x1(t) ´ x2(t) = 1 ´ t = t
=2;1<t<2
=1;1<t<2
For t = 1 to 2;
y(t) = x 1(t) ´ x2(t) = 2 ´ 1 = 2
=1;2<t<3
=3–t;2<t<3
For t = 2 to 3; y(t) = x1(t) ´ x2(t) = 1 ´ (3 – t) = 3 – t
Fig 2.27 : Multiplication of two continuous time signals.
3
t
2. 21
Signals & Systems
2.4.6 Differentiation and Integration of Continuous Time Signals
Differentiation is a mathematical operation used to estimate the rate of change of a continuous
time signal at any instant of time.
Differentiation is denoted by the operator d .
dt
Therefore, the differentiation of a continuous time signal x(t) is denoted by
FG
H
IJ
K
d
dx( t )
.
x( t ) or
dt
dt
The differentiation of a continuous time signal x(t) is defined as,
dx( t )
=
dt
Lt
Dt ® 0
x(t) - x(t - Dt)
Dt
Integration, is the inverse process of differentiation. More appropriately. Integration is the process
of identifying the signal from its differentiation.
The integration is denoted by the operaor
z ....
dt
z
Therefore, the integration of a continuous time signal x(t) is denoted by x(t) dt .
Differentiation and integration of standard continuous time signals are listed in table 2.1
Table 2.1 : Differentiation and Integration of Standard Continuous Time Signals
Signal, x(t)
Differentiation of x(t),
dx(t)
dt
Integration of x(t),
d(t)
––
1
u(t)
d(t)
t
t
u(t)
t2
2
t2
2t
t3
3
sin t
cost
- cos t
cost
- sin t
sin t
e–at
– ae–at
e - at
-a
eat
aeat
e at
a
sin W 0 t
W 0 cos W 0 t
cosW 0 t
- W 0 sin W 0 t
-cos W 0 t
W0
sin W 0 t
W0
z
x(t) dt
Chapter 2 - Continuous Time Signals and Systems
2. 22
The standard signals such as impulse, step, ramp and parabolic signals are related through integration
and differentiation as shown below.
¾integration
¾¾¾®
d( t )
bImpulseg
bImpulseg
d
u(t)
dt
bUnit rampg
bUnit parabolicg
differentiation
¬¾¾¾
¾¾
u(t)
t
b Unit rampg
b Unit stepg
=
t =0
Du
Dt
=
t =0
t2
2
¾integration
¾¾¾®
t
bUnit stepg
differentiation
d( t ) ¬¾¾¾
¾¾
Note :
¾integration
¾¾¾®
u(t)
t2
2
differentiation
¬¾¾¾
¾¾
b Unit parabolicg
u(0 + ) - u(0 - )
Dt
=
t =0
1 - 0
0
= ¥
= Impulse
t =0
t =0
Example 2.5
A continuous time signal is defined as,
x(t) = t ; 0 £ t £ 3
=0 ;t>3
Sketch the waveform of x(–t) and x(2 – t).
Solution
The given signal is shown in fig 1.
The signal x(–t) is the folded version of x(t). The signal x(–t) is shown in fig 2.
The signal x(2 – t) = x(–t + 2) is the advanced version of the folded signal. The signal x(–t + 2) is shown in fig 3.
x(t)
x(t)
x(2 – t)
3
3
3
2
2
2
1
1
1
0
1
2
t
3
–3
Fig 1 : x(t).
–2
–1
0
1
2
3
t
–5
–4
–3
–1
–2
0
1
2
3
t
Fig 3 : x(2 – t).
Fig 2 : x(–t).
Example 2.6
Sketch the even and odd parts of the following signals.
a)
b)
x(t)
1
1
0
Solution
a)
x(t)
a
t
1
[x( t ) + x(- t)] =
2
1
[x( t ) - x( - t)] =
The odd part of the signal is given by, x o (t ) =
2
The even part of the signal is given by, x e (t ) =
0
1
x( t) +
2
1
x(t) 2
a
t
1
x( - t)
2
1
x(- t)
2
.....(1)
.....(2)
2. 23
Signals & Systems
From equations (1) and (2), it is observed that the even and odd parts of the signal can be obtained from the folded
and scaled versions of the signal. Hence the given signal is folded, scaled and then graphically added and subtracted to
get the even and odd parts as shown below.
x(t)
x(–t)
Fold
1
0
a
1
–a
t
0
1
Scale by
2
Scale by
1
x(t)
2
1
x( - t)
1
2
a
0
–a
t
1
1
1
[x(t ) + x( - t)] =
x( t) +
x( - t)
2
2
2
x o (t ) =
0
t
1
1
1
[x( t ) - x( - t)] =
x(t) x(- t)
2
2
2
xe(t)
xo(t)
1
2
1
2
–a
0
–a
b)
1
2
2
1
2
x e (t ) =
t
a
0
t
The even part of the signal is given by, x e (t ) =
a
- 1
2
1
1
1
[ x(t ) + x( - t)] =
x( t) +
x( - t)
2
2
2
t
.....(1)
1
1
1
[x( t ) - x( - t)] =
x(t) x(- t)
.....(2)
2
2
2
From equations (1) and (2), it is observed that the even and odd parts of the signal can be obtained from the
folded and scaled versions of the signal. Hence the given signal is folded, scaled and then graphically added and
subtracted to get the even and odd parts as shown below.
x(t)
x(–t)
Fold
1
1
The odd part of the signal is given by, x o (t ) =
0
a
0
–a
t
1
2
Scale by
t
Scale by
1
x( - t)
2
1
x(t)
2
1
2
1
2
0
a
–a
t
0
t
1
2
Chapter 2 - Continuous Time Signals and Systems
x e (t ) =
2. 24
1
1
1
[x( t ) + x(- t)] =
x( t) +
x( - t)
2
2
2
xe(t)
x o (t ) =
1
1
1
[x(t ) - x(- t)] =
x( t) x(- t)
2
2
2
xe(t)
1
2
1
2
–a
–a
0
a
a
0
t
- 1
2
t
2.5 Impulse Signal
The impulse signal is a special signal which can be derived as follows.
Consider a pulse signal, PD(t) with height A/D and width D as shown in fig 2.28. Now, the
PD(t)
pulse signal, PD(t) can be defined as,
A
A
; 0£t£D
D
= 0
; t>D
The area of the pulse signal for any value of t is given by,
A
Area = Height ´ Width =
´D =A
D
D
PD ( t ) =
0
t
D
Fig 2.28 : Pulse signal.
In the signal, PD(t) if the width D is reduced, then the height A /D increases, but the area of the
pulse remains same as A. When the width D tends to zero, the height A /D tends to infinity. This limiting
value of the pulse signal is called impulse signal, d(t). Even when the width D tends to zero, the area of
d(t)
the pulse remains as A.
¥
A
; t=0
\ Impulse Signal, d( t ) = Lt PD ( t ) = Lt
D®0
D®0 D
; t ¹0
= 0
+¥
\ Impulse Signal, d( t ) = ¥ ; t = 0
z
and
0
d( t ) dt = A
-¥
= 0 ; t ¹0
t
Fig 2.29 : Impulse signal
(or Unit impulse signal).
In the above impulse signal if A=1, then the impulse signal is called a unit impulse signal.
The impulse signal or unit impulse signal can be represented graphically as shown in fig 2.29. An impulse
with infinite magnitude and zero duration is a mathematical fiction and does not exist in reality. However
a signal with large magnitude and short duration (when compared to time constant of a system) can be
considered as an impulse signal. Practically, the magnitude of the impulse is measured by its area.
Definition of impulse signal : The impulse signal is a signal with infinite magnitude and zero duration,
but with an area of A. Mathematically, an impulse signal is defined as,
+¥
Impulse Signal, d ( t ) = ¥ ; t = 0
= 0 ; t ¹0
and
z
d ( t ) dt = A
-¥
Definition of unit impulse signal : The unit impulse signal is a signal with infinite magnitude and zero
duration, but with unit area. Mathematically, a unit impulse signal is defined as,
+¥
Unit Impulse Signal, d ( t ) = ¥ ; t = 0
= 0 ; t ¹0
and
z
-¥
d ( t ) dt = 1
2. 25
Signals & Systems
2.5.1 Properties of Impulse Signal
+¥
z
Property -1:
d(t) dt = 1
-¥
Proof :
Consider a narrow pulse signal, P D(t) of width D l and height 1/Dl as shown in fig 2.30.
Now the pulse signal is defined as,
PD (t ) =
1
Dl
1
Dl
; 0 £ t £ Dl
= 0
PD(t)
; t > Dl
Now the impulse signal can be represented as,
d(t) =
Lt
0
PD ( t)
Dl ® 0
Dl
On integrating the above equation we get,
+¥
z
Fig 2.30.
+¥
d(t)dt =
-¥
z
=
+¥
Lt
-¥
Dl ® 0
Lt
Dl ® 0
t
PD (t) dt
1
t
Dl
Dl
=
=
0
Lt
Dl ® 0
Lt
Dl ® 0
z
Dl
PD (t) dt =
-¥
1
Dl - 0 =
Dl
Lt
Dl ® 0
Lt
Dl ® 0
z
0
1
dt
Dl
1
=
Lt
Dl ® 0
1
Dl
Dl
z
dt
0
1
=
+¥
Property - 2:
z
x(t) d( t ) dt = x(0)
-¥
Proof :
+¥
z
Since d(t) is nonzero only at t=0,
x(t) is replaced by x(0).
+¥
x( t) d(t) dt =
-¥
z
x( 0) d(t) dt
Since x(0) is constant, it is
taken outside integration.
-¥
+¥
=
x( 0)
z
d(t) dt = x(0) ´ 1 =
x(0)
Using property-1
-¥
+¥
Property - 3:
z
x(t) d( t - t 0 ) dt = x( t 0 )
-¥
Proof :
+¥
z
Since d(t0) is nonzero only at t=t0,
x(t) is replaced by x(t0).
+¥
x(t) d(t - t 0 ) dt =
-¥
z
x(t 0 ) d(t - t 0 ) dt
-¥
+¥
=
x(t 0 )
z
d(t - t 0 ) dt =
x(t 0 ) ´ 1 = x(t 0 )
Since x(t0) is constant, it is
taken outside integration.
Using property-1
-¥
+¥
Property - 4:
z
x(l ) d( t - l ) dl = x(t)
-¥
Proof :
+¥
Consider the property-3 of impulse signal.
z
x(t) d(t - t 0 ) dt =
x(t 0 )
-¥
+¥
On substituting t = l in the above equation we get,
z
-¥
x( l ) d( l - t 0 ) dl =
x(t 0 )
Chapter 2 - Continuous Time Signals and Systems
2. 26
+¥
On substituting t0 = t in the above equation we get,
z
x( l ) d( l - t) dl =
x(t)
-¥
Since impulse signal is even, d(l-t) = d(t-l), Therefore the above equation is written as shown below.
+¥
z
x( l ) d( t - l) dl =
x(t)
-¥
d(at ) =
Property - 5 :
1
dt
a
bg
Proof :
Consider a narrow pulse signal, P D(t) of width D l and height 1/Dl as shown in fig 2.31(a).
P D (t)
P D (at)
1
Dl
1
Dl
Time scale
0
0
t
Fig 2.31 (a) .
Dl
t
Fig 2.31 (b).
Dl a
Fig 2.31 : A pulse signal and its time scaled version.
Now the signal, PD(at) will be time scaled version of signal PD(t) as shown in fig 2.31(b).Now the pulse signal and
time scaled pulse signal can be mathematically defined as,
1
Dl
= 0
z
z
PD (at) =
; t > Dl
+¥
\
1
Dl
= 0
; 0 £ t £ Dl
PD (t ) =
Dl
PD (t) dt =
Dl a
+¥
z
PD (at) dt =
0
-¥
; t>
1
dt
Dl
z
0
-¥
; 0 £ t £ Dl
a
1
1
dt =
a
Dl
Dl
a
....(2.7)
Dl
z
0
1
1
dt =
a
Dl
+¥
z
PD (t) dt
....(2.8 )
Using equation (2.7)
-¥
Now the impulse signal and time scaled impulse signal can be represented as,
d(t) =
Lt
Dl ® 0
PD (t )
and
d(at) =
Lt
Dl ® 0
PD (at)
On integrating the time scaled impulse signal we get,
+¥
z
+¥
d(at) dt =
-¥
z
-¥
1
=
a
+¥
\
z
-¥
1
d(at) dt =
a
+¥
Lt
Dl ® 0
PD ( at) dt =
+¥
Lt
Dl ® 0
1
Lt PD (t ) dt =
Dl ® 0
a
-¥
z
z
z
PD (at) dt =
-¥
1
d(t)
a
Dl ® 0
1
a
+¥
z
-¥
PD (t ) dt
Using equation (2.8)
+¥
z
d(t) dt
-¥
+¥
d(t) dt
-¥
On differentiating the above equation we get ,
d(at) =
Lt
Using definition of
impulse signal
2. 27
Signals & Systems
2.5.2 Representation of Continuous Time Signal as Integral of Impulses
Let x(t) be a continuous time signal as shown in fig 2.32.
t
Fig 2.32.
0
Dt
2Dt
3Dt
4Dt
5Dt
6Dt
7Dt
0
–3Dt
–2Dt
–Dt
~x( t )
x(t)
t
Fig 2.33.
Let us divide x(t) as narrow pulses of width Dt as shown in fig 2.33.
Now the signal x(t) can be expressed as,
x( t ) = Lt ~x(t)
Dt ® 0
Each narrow pulse of fig 2.33 can be interpretted as shown below,
x( -2Dt)
x( - Dt )
x(0)
x( Dt )
x(2Dt)
\ x( t ) =
=
M
M
=
=
=
=
=
M
M
~
x(t)
~x(t)
~x(t)
~
x(t)
~
x(t)
;
;
;
;
;
for - 2Dt < t < - Dt
for - Dt < t < 0
for
0 < t < Dt
for
Dt < t < 2Dt
for 2Dt < t < 3Dt
Lt ~x(t)
Dt ® 0
Lt [..... x(-2 Dt ) + x(- Dt ) + x(0) + x(Dt ) + x(2Dt ) + .....]
.....(2.9)
Consider the pulse signal of width Dt and height 1/ Dt as shown in fig 2.34. This pulse signal can
be expressed as,
Dt ® 0
PD(t)
1
PD ( t ) =
; 0 £ t £ Dt
Dt
= 0
; otherwise
1
Dt
Now, PD(t) ´ Dt = A pulse of unit amplitude.
\ On multiplying PD(t) ´ Dt with the signal ~
x( t ) , the signal ~x(0) is selected.
\ ~x(0) = ~x(t) PD ( t ) Dt
0
Dt
Consider the shifted version of the pulse signal of fig 2.34, as shown in fig 2.35.
Fig 2.34.
t
Chapter 2 - Continuous Time Signals and Systems
2. 28
right shift
(or delay)
t
0
0
t
Dt
0
t
PD(t – 2Dt)
PD(t – Dt)
PD(t)
Dt
2Dt
PD(t + Dt)
–Dt
0
–2Dt
–Dt
PD(t + 2Dt)
t
0
Dt
2Dt
3Dt
left shift
(or advance)
t
Fig 2.35.
If we multiply ~x ( t ) with shifted pulse signals shown in fig 2.35, then each product will select one
pulse of the signal ~x ( t ) as shown below.
M
M
x( -2Dt) = ~
x(t) PD (t + 2Dt ) Dt
~
x(- Dt) = x(t) PD (t + Dt ) Dt
x(0) = ~x(t) PD (t) Dt
x( Dt) = ~x(t) PD (t - Dt ) Dt
x(2 Dt) = ~
x(t) PD (t - 2Dt) Dt
M
M
In the above equation ~x ( t ) can be replaced by respective selected pulses itself as shown below,
M
M
x( -2Dt) PD (t + 2Dt ) Dt
\ x( -2 Dt) = ~
x( - Dt) = ~
x(- Dt ) PD (t + Dt ) Dt
~
x(0) = x(0) PD (t) Dt
x( Dt) = ~
x( Dt ) PD (t - Dt) Dt
x(2Dt) = ~x(2Dt ) PD (t - 2Dt ) Dt
M
M
On substituting the above equations in equation (2.9) we get,
~x(t) =
Lt
Dt ® 0
LM..... ~x(~-2Dt) P (t + 2Dt) Dt ~+ ~x(- Dt) P (t + Dt) Dt + ~x(0) P (t) Dt OP
N + x(Dt) P (t - Dt) Dt + x(2Dt) P (t - 2Dt) Dt + ........................Q
D
D
D
D
D
+¥
=
Lt
Dt ® 0
å ~x(nDt)
PD ( t - nDt) Dt
n = -¥
On applying limit Dt ® 0 the signal ~x( nDt ) becomes continuous, the signal PD(t - nDt) becomes
an impulse and so the summation becomes integration.
Hence the above equation can be expressed as,
+¥
x(t) =
z
x( t) d (t - t ) dt
....(2.10)
-¥
The equation (2.10) is used to represent any continuous time signal x(t) as an integral of impulses.
Signals & Systems
2. 29
2.6 Continuous Time System
A continuous time system (or Analog system) is a physical device that operates on a continuous
time signal (or an analog signal) called input or excitation, according to some well defined rule, to produce
another continuous time signal (or an analog signal) called output or response. We can say that the input
signal x(t) is transformed by the system into a signal y(t), and the transformation can be expressed
mathematically as shown in equation (2.11).The diagrammatic representation of continuous time system
is shown in fig 2.36.
Response, y(t) = H{x(t)}
.....(2.11)
where, H denotes the transformation (also called an operator).
Continuous
time system
x(t)
y(t)
H
Input signal
or Excitation
Output signal
or Response
Fig 2.36 : Representation of continuous time system.
LTI System
A continuous time system is linear if it obeys the principle of superposition and it is time invariant
if its input-output relationship does not change with time. When a continuous time system satisfies the
properties of linearity and time invariance then it is called an LTI system (Linear Time Invariant system).
Impulse Response
When the input to a continuous time system is a unit impulse signal d(t) then the output is called
an impulse response of the system and it is denoted by h(t).
\ Impulse Response, h(t) = H{d(t)}
d(t)
Impulse input
.....(2.12)
Continuous
time system
H
h(t)
Impulse response
Fig 2.37 : Continuous time system with impulse input.
2.6.1 Mathematical Equation Governing LTI Continuous Time System
The electric heaters, motors, generators, etc., are examples of electrical continuous time systems.
The continuous time systems that operate on electrical energy can be modelled by three basic elements
Resistor(R), Inductor(L) and Capacitor(C). The models constructed using these fundamental elements
are called electric circuits.
In electric circuits the inputs and outputs are either voltage signals or current signals. The continuous
time voltage signal is denoted by v(t) and current signal by i(t).
The basic RL, RC, and RLC circuits and their time domain KVL (Kirchoff’s Voltage Law) equations
are shown in fig 2.38, fig 2.39 and fig 2.40 respectively. From these circuits it can be observed that the
equations governing the continuous time systems are differential equations.
Also, it can be shown that all continuous time systems like Mechanical systems, Thermal systems,
Hydraulic systems, etc., are all governed by differential equations.
Chapter 2 - Continuous Time Signals and Systems
+
R i(t)
di(t)
= n(t)
dt
R i(t ) + L
L
R
2. 30
–
– +
L
di( t )
dt
¯
Replace i(t) by y(t) and n(t) by x(t)
dy( t )
= x(t)
dt
dy( t )
R
1
\
+
y( t ) =
x(t)
dt
L
L
i(t) = y(t)
Ry( t ) + L
+n(t) = x(t)
Fig 2.38 : RL circuit and the mathematical equation governing RL circuit.
R i( t) +
C
R
–
+
R i(t)
1
C
–
+
z
z
1
C
i(t) dt = n(t)
differentiate
¯
i ( t ) dt
R
i(t) = y(t)
dn(t)
1
di(t)
i(t) =
+
dt
C
dt
Replace i(t) by y(t) and n(t) by x(t)
+-
¯
n(t) = x(t)
dy(t)
1
dx(t)
+
y(t) =
dt
C
dt
dy(t)
1
1 dx(t)
\
+
y(t) =
dt
RC
R dt
R
Fig 2.39 : RC circuit and the mathematical equation governing RC circuit.
R i(t ) + L
L
R
+
R i(t)
–
+
L
di( t )
dt
C
–
+
1
C
z
–
i ( t ) dt
z
i(t) dt = n(t)
differentiate
¯
R
i(t) = y(t)
di(t)
d 2i( t )
1
dn(t)
+ L
+
i(t) =
dt
C
dt
dt 2
Replace i(t) by y(t) and n(t) by x(t)
¯
+n(t) = x(t)
1
di(t)
+
C
dt
R
dy(t)
d 2 y( t )
1
dx(t)
+ L
+
y(t) =
dt
C
dt
dt 2
\
d 2 y( t )
R dy(t)
1
1 dx(t)
+
+
y(t) =
L dt
LC
L dt
dt 2
Fig 2.40 : RLC circuit and the mathematical equation governing RLC circuit.
Signals & Systems
2. 31
In general, the input-output relation of an LTI (Linear Time Invariant) continuous time system is
represented by a constant coefficient differential equation shown below (equation (2.18)).
a0
dN
d N -1
d N -2
d
dM
y(t)
+
a
y(t)
+
a
y(t)
+
.......
+
a
y(t)
+
a
y(t)
=
b
x( t )
1
N
1
2
N
0
dt
dt N
dt N - 1
dt N - 2
dt M
d M -1
dM-2
d
+ b1 M - 1 x( t ) + b 2 M - 2 x( t ) + .......+ b M -1 x( t ) + b M x( t ) .....(2.13)
dt
dt
dt
where, N = Order of the system, M £ N, and a0 = 1.
The solution of the above differential equation is the response y(t) of the system, for the input x(t).
Note : A system is linear if it obeys the principle of superposition and it is time invariant if its
input-output relationship do not change with time.
2.6.2 Block Diagram and Signal Flow Graph Representation of LTI Continuous Time System
Block Diagram
A block diagram of a system is a pictorial representation of the functions performed by the system.
The block diagram of a system is constructed using the mathematical equation governing the system.
The basic elements of a block diagram are Differentiator, Integrator, Constant Multiplier and Signal
Adder. The symbols used for the basic elements and their input-ouput relation are listed in table 2.2.
Table 2.2 : Basic Elements of Block Diagram and Signal Flow Graph
Description
Elements of
block diagram
Elements of
signal flow graph
Differentiator
x(t)
d
dt
d
x( t )
dt
x(t)
Integrator
(with zero initial
condition)
x(t)
z
z x(t) dt
x(t)
a
a x(t)
x(t)
Constant Multiplier
Signal Adder
x(t)
x1(t)
z
a
x1(t)
+
x2(t)
d
dt
x1(t) + x2(t)
d
x( t )
dt
z x(t) dt
a x(t)
1
x1(t) + x2(t)
x2(t)
1
Chapter 2 - Continuous Time Signals and Systems
2. 32
Signal Flow Graph
A signal flow graph of a system is a graphical representation of the functions performed by the
system. The signal flow graph shows the flow of signals from one point of a system to another and gives
the relationship among the signals. The signal flow graph of a system is constructed using the mathemetical
equation governing the system.
The basic elements of a signal flow graph are nodes and directed branches. Each node represents
a signal. The signal at a node is given by the sum of all incoming signals. Each branch has an input node
and an output node. The direction of signal flow is marked by an arrow on the branch and the operation
performed by the signal is indicated by an operator like integrator/differentiator. When the signal passes
from the input node to the output node, it is operated by the operation specified by the branch. The basic
operations performed by the branches of a signal flow graph are listed in table 2.2.
Example 2.7
Construct the block diagram and signal flow graph of the system described by the equation,
d2 y(t )
dy(t )
dx( t)
+ 2
+ 3 y(t) = 4
+ 5 x(t)
dt 2
dt
dt
Solution
Case i : Block diagram and signal flow graph using differentiators
Given that,
d2 y(t)
dy(t)
dx(t)
+ 2
+ 3 y(t) = 4
+ 5 x(t)
dt
dt
dt 2
\ y(t) = -
1 d2 y(t )
2 dy( t)
4 dx(t )
5
+
+
x(t)
3 dt 2
3 dt
3 dt
3
.....(1)
The equation (1) is used to construct the block diagram and signal flow graph using differentiators as shown in
fig 1 and fig 2 respectively.
5
x(t)
d
dx(t )
dt
5
x( t)
3
3
+
+
y(t)
d
dt
4
4 dx( t)
3 dt
3
-
+
2 dy(t )
3 dt -2
dy( t)
dt
3
d
2
-
1 d y(t )
3 dt 2
dt
-1
3
Fig 1 : Block diagram using differentiators.
dt
d 2 y ( t)
dt 2
Signals & Systems
5
x(t)
d
5
4 dx(t)
x(t ) +
3
3 dt
2. 33
3
y(t)
y(t)
d
dt
4
3
-
dx(t )
dt
-2
2 dy(t) 1 d2 y(t)
3 dt
3 dt 2
dt
dy( t)
dt
3
d
-1
3
dt
d 2 y ( t)
dt 2
Fig 2 : Signal flow graph using differentiators.
Case ii : Block diagram and signal flow graph using integrators
Given that,
d2 y(t)
dy(t)
dx(t)
+ 2
+ 3 y(t) = 4
+ 5 x(t)
dt
dt
dt 2
¯
Integrate with zero
initial conditions
z
dy( t)
+ 2 y(t) + 3 y(t) dt = 4 x(t) + 5
dt
Integrate with zero
initial conditions
y( t) + 2
z
¯
y(t) dt + 3
\ y( t) = - 2
z
zz
y(t) dt dt = 4
y(t) dt - 3
zz
z
z
x(t) dt
x(t) dt + 5
y(t) dt dt + 4
z
zz
x(t) dt dt
x(t) dt + 5
zz
x(t) dt dt
.....(2)
The equation (2) is used to construct the block diagram and signal flow graph using integrators as shown in fig 3
and fig 4 respectively.
x(t)
z
z x(t) dt
z
4 x( t ) dt
4
+
+
y(t)
z
z
zz x(t) dt dt
zz
5 x(t ) dtdt
5
+
z y(t) dt
z
-2 y(t) dt
–2
z
zz
-3 y( t) dt dt
Fig 3 : Block diagram using integrators.
–3
zz y(t) dt dt
Chapter 2 - Continuous Time Signals and Systems
2. 34
dt
dt
dt
z x(t)
z
+
4
y(t)
4
x(t)
z x(t) dt
z
5
t)
x(
y(t)
z
5
z
zz x(t) dt dt
zz
-2 y(t) dt - 3 y(t ) dt dt
–2
–3
z
z y(t) dt
z
zz y(t) dt dt
Fig 4 : Signal flow graph using integrators.
2.7 Response of LTI Continuous Time System in Time Domain
The general equation governing an LTI continuous time system is,
a0
dN
d N -1
d N-2
d
dM
y( t ) + a1 N - 1 y( t ) + a 2 N - 2 y( t ) + ....... + a N - 1 y( t ) + a N y( t ) = b 0 M x(t )
N
dt
dt
dt
dt
dt
+ b1
d M -1
dM-2
d
(
)
+
x
t
b
x( t ) + ....... + b M - 1 x( t ) + b M x( t ) .....(2.14)
2
M -1
M-2
dt
dt
dt
where, N = Order of the system, M £ N, and a0 = 1.
The solution of the differential equation (2.14) is the response y(t) of the LTI system, which
consists of two parts. In mathematics, the two parts of the solution y(t) are the homogeneous solution
yh(t) and the particular solution yp(t).
\ Response, y(t) = yh(t) + yp(t)
.....(2.15)
The homogeneous solution is the response of the system when there is no input.The particular
solution yp(t) is the solution of difference equation for specific input signal x(t) for t ³ 0.
In signals and systems the two parts of the solution y(t) are called zero-input response yzi(t) and
zero-state response yzs(t).
\ Response, y(t) = yzi(t) + yzs(t)
.....(2.16)
The zero input response is given by the homogeneous solution with constants evaluated using
initial values of output (or initial conditions). The zero-input response is mainly due to initial output
conditions (or initial stored energy) in the system. Hence the zero-input response is also called free
response or natural response.
The zero-state response is given by the sum of homogeneous solution and particular solution
with zero initial conditions.The zero-state response is the response of the system due to input signal and
with zero initial output condition. Hence the zero-state response is also called forced response.
Signals & Systems
2. 35
2.7.1 Homogeneous Solution
The homogeneous solution is obtained when x(t) = 0. On substituting x(t) = 0 in the system
equation (2.14) we get,
dN
d N -1
d N -2
d
y(t) + a1 N -1 y(t) + a 2 N - 2 y(t) + .......+ a N -1 y(t) + a N y(t) = 0
N
dt
dt
dt
dt
.....(2.17)
Now, the solution of equation (2.17) is the homogeneous solution.
Let us assume that the solution of equation (2.17) is in the form of an exponential.
i.e., y(t) = Celt
d
y( t ) = C l e lt
dt
d2
y( t ) = C l2 e lt
dt 2
............
\
dN
y( t ) = C lN e lt
N
dt
On substituting the above assumed solution in equation (2.17) we get,
C lN e lt + a1 C lN -1 e lt + a 2 C lN - 2 e lt +......+ a N - 1C l e lt + a N C e lt = 0
\ lN + a 1 lN - 1 + a 2 lN - 2 + ......+ a N -1l + a N
d
i Ce
lt
=0
Now the characteristic polynomial of the system is given by,
lN + a 1 lN -1 + a 2 lN - 2 +......+ a N - 1l + a N = 0
The characteristic polynomial has N roots, which are denoted as l1, l2,...lN.
The roots of the characteristic polynomial may be distinct real roots, repeated real roots or complex
roots. The assumed solutions for various types of roots are given below.
Distinct Real Roots
Let the roots l1, l2, l3, ... lN be distinct real roots. Now the homogeneous solution will be in the
form,
y h ( t ) = C1 e l1t + C2 e l 2 t +......+C N -1 e l N -1t + C N e l N t
where, C1, C2, .....CN are constants that can be evaluated using initial conditions.
Repeated Real Roots
Let one of the real roots l1 repeats p times and the remaining (N – p) roots are distinct real roots.
Chapter 2 - Continuous Time Signals and Systems
2. 36
Now the homogeneous solution is in the form,
y h ( t ) = (C1 + C2 t + C3 t 2 + ..... + C p t p - 1 ) e l1t + C p + 1 e
l p +1t
+ .... + C N e l N t
where, C1 , C 2 , C3 ,...... C N are constants that can be evaluated using initial conditions.
Complex Roots
Let the characteristic polynomial have a pair of complex roots l and l* and the remaining (N – 2) roots
be distinct real roots. Let, l = a + jb and l* = a - jb. Now, the homogeneous solution will be in the form,
bg
b
g
y h t = e at C1cosbt + jC2 sinbt + C3 e l 3t +......+ C N -1 e
l N - 1t
+ CN el N t
where, C1, C2, C3 ... CN are constants that can be evaluated using initial conditions.
2.7.2 Particular Solution
The particular solution, yp(t) is the solution of Nth order differential equation governing the system
for specific input signal x(t) for t ³ 0. Since the input signal may have a different form, the particular
solution depends on the form or type of the input signal x(t).
If x(t) is constant, then yp(t) is also a constant.
Example :
Let x(t) = u(t), now yp (t) = K u(t)
If x(t) is exponential, then yp(t) is also an exponential.
Example :
Let x(t) = eat u(t), now yp(t) = K eat u(t)
If x(t) is sinusoid, then yp(t) is also a sinusoid.
Example :
Let x(t) = A cos W0t, now yp(t) = K1 cos W0t + K2 sin W0t
The general form of a particular solution for various types of inputs are listed in table 2.3.
Table 2.3 : Particular Solution
Input
x(t)
x(t)
x(t)
signal, x(t)
= A
= Au(t)
= A eat (if a ¹ li)
x(t) = A cos W 0 t
x(t) = A sin W0 t
123
x(t) = A eat (if a= li)
Particular solution, yp(t)
yp(t) = K
yp(t) = K u(t)
yp(t) = K eat
yp(t) = K t eat
yp(t) = K1 cos W 0 t + K2 sin W 0 t
Note : li is one of the root
of characteristic
polynomial.
2.7.3 Zero-Input and Zero-State Response
The zero-input response yzi(t) (or free response or natural response) is obtained from the
homogeneous solution yh(t) with constants evaluated using initial output (or initial conditions).
Signals & Systems
2. 37
\ Zero - input response, y zi ( t ) = yh ( t ) with constants evaluated using initial output conditions
The zero-state response yzs(t) (or forced response) is obtained from the sum of homogeneous
solution and particular solution and evaluating the constants with zero initial conditions.
\ Zero - state response, y zs ( t ) = y h ( t ) + yp ( t )
with const ants C 1 , C2 , ....CN evaluated with zero initial output cond itions
The sum of zero-input response and zero-state response will be the total response or complete
response of a system.
2.7.4 Total Response
The total response y(t) of a continuous time system can be obtained by the following two methods.
Method -1
The total response y(t) is given by the sum of homogeneous solution and particular solution.
\ Total response, y(t) = yh(t) + yp(t)
Procedure to Determine Total Response
1. Determine homogeneous solution yh(t) with constants C1, C2, .....CN.
2. Determine particular solution yp(t) and estimate the constant K by evaluating the given system
equation using yp(t) for any value of t ³ 1 so that no term of the system equation vanishes.
3. Now the total response y(t) is given by, the sum of yh(t) and yp(t)
\ Total response, y(t) = yh (t) + yp(t)
4. The total response y(t) will have N number of constants C1, C2, .....CN. Evaluate the constants
C1, C2, .....CN using initial outputs (or initial conditions).
Method -2
The total response y(t) is given by the sum of zero-input response and zero-state response.
\ Total response, y(t) = yzi(t) + yzs(t)
Procedure to Determine Total Response
1. Determine the homogeneous solution yh(t) with constants C1, C2, ....CN.
2. Determine the zero-input response yzi(t) which is obtained from the homogeneous solution yh(t)
by evaluating the constants C1, C2, ....CN using initial outputs (or initial conditions).
3. Determine particular solution yp(t) and estimate the constant K by evaluating the given system
equation using yp(t) for any value of t ³ 1 so that no term of the system equation vanishes.
4. Determine the zero-state response, yzs(t) which is given by sum of homogeneous solution and
particular solution and evaluating the constants C1, C2, ....CN with zero initial conditions.
5. Now the total response y(t) is given by the sum of zero-input response and zero-state response.
\ Total response, y(t) = yzi(t) + yzs(t)
Chapter 2 - Continuous Time Signals and Systems
2. 38
Example 2.8
Determine the natural response of the system described by the equation,
d2 y(t)
dy(t)
dx(t)
+ 6
+ 5y(t) =
+ 4x(t) ; y(0) = 1;
dt 2
dt
dt
dy(t)
dt
= -2
t =0
Solution
The natural response is response of the system due to initial conditions and so it is given by zero-input response.
Zero - input response , y zi (t) = yh (t) with constants evaluated using initial conditions
where, y h (t) = Homogeneous solution
The given system equation is,
d2 y(t)
dy(t)
dx(t)
+ 6
+ 5y(t) =
+ 4x(t)
dt
dt
dt2
.....(1)
Homogeneous Solution
The homogeneous solution is the solution of the system equation when x(t) = 0.
On substituting x(t) = 0 in system equation (equation (1)) we get,
d2 y(t)
dy(t)
+ 6
+ 5y(t) = 0
dt 2
dt
d
d2
Let, y( t) = C e lt ; \
y( t) = C l elt ;
y(t ) = C l2 elt
dt
dt 2
On substituting the above terms in equation (2) we get,
.....(2)
C l2 e lt + 6 C l e lt + 5 C e lt = 0
\ l2 + 6 l + 5 C e lt = 0
d
i
The characteristic polynomial of the above equation is,
l2 + 6l+ 5 = 0
Þ
(l + 1) (l + 5) = 0
Þ
l = –1, –5
Now the homogeneous solution is given by,
Homogeneous solution, yh (t) = C1 el 1t + C2 el 2 t = C1 e - t + C2 e -5 t
Natural Response (or Zero-input Response)
Zero - input response , y zi (t) = yh (t) with const ants evalu ated using initial c onditions
= C1 e - t + C 2 e -5 t
with C and C evaluated using ini tial condi tions
1
2
dy zi (t)
\
= - C1 e - t - 5C 2 e -5 t
dt
At t = 0, y zi (0) = C1 e 0 + C 2 e0 = C1 + C 2
Given that, y zi (0) = 1,
At t = 0,
Given that,
dy zi (t)
dt
\ C1 + C2 = 1
.....(3)
dy zi (t)
= - C1 e0 - 5C 2 e0 = - C1 - 5C 2
dt
= - 2,
t=0
\ - C1 - 5C 2 = - 2
.....(4)
Signals & Systems
2. 39
On adding equations (3) and (4) we get,
-4C2 = - 1 Þ
C2 =
1
4
From equation (3), C1 = 1- C2 = 1 \ Natural response, y zi (t) =
1
3
=
4
4
3 -t
1 -5 t
1
e +
e ; t³0 =
3e- t + e -5 t u(t)
4
4
4
d
i
Example 2.9
Determine the forced response of the system described by the equation,
5
dy(t)
+ 10 y(t) = 2 x(t) , for the input, x(t) = 2 u(t).
dt
Solution
The forced response is the response of the system due to input signal with zero initial conditions and so it is given
by zero-state response.
Zero state response , y zs (t) = yh (t) + y p (t)
with const ants evalu ated with zero initi al conditions
where, y h (t) = Homogeneous solution and y p (t) =Particular solution
The given system equation is,
5
dy(t)
+ 10 y(t) = 2 x(t)
dt
.....(1)
Homogeneous Solution
The homogeneous solution is the solution of the system equation when x(t) = 0.
On substituting x(t) = 0 in system equation (equation (1)) we get,
dy(t)
5
+ 10 y(t) = 0
dt
d
Let, y( t) = C elt ; \
y( t) = C l e lt
dt
On substituting the above terms in equation (2) we get,
5 C l e lt + 10 C elt = 0
b
g
\ 5 l + 10 C elt = 0
The characteristic polynomial of the above equation is,
5l+ 10 = 0
Þ
l+2=0
Þ
l = –2
Now the homogeneous solution is given by,
Homogeneous solution, y h (t) = Ce lt = C e -2 t
Particular Solution
The particular solution is the solution of the system equation (equation (1)) for specific input.
Here input,
x(t) = 2 u(t)
Let the particular solution, yp(t) is of the form,
yp(t) = K x(t)
.....(2)
Chapter 2 - Continuous Time Signals and Systems
\ y p (t) = 2K u(t) ;
dy p (t)
2. 40
d
u(t) = d(t)
dt
= 2K d(t)
dt
On substituting the above terms and the input in system equation (equation (1)) we get,
5
dy(t)
+ 10 y(t) = 2 x(t)
dt
ß
10 K d(t) + 20 K u(t) = 4 u(t)
Þ
At t = 1, 10 K d(1) + 20 K u(1) = 4 u(1)
Þ
20 K = 4
K = 1/5
d(1) = 0, u(1) = 1
2
u(t)
5
\Particular solution, y p (t) =
Forced Response (or Zero-State Response)
Zero state response , y zs (t) = yh (t) + y p (t)
= Ce
At t = o,
y zs (t) = C e0 +
Since,
y zs (0) = 0,
-2t
2
2
u(0) = C +
5
5
C+
\ Forced Response, y zs (t ) = -
with constants evaluated with zero initial conditions
2
u(t)
+
5
2
=0
5
Þ
C=-
2
5
2 -2 t
2
2
e
+
u(t); for t ³ 0 =
1 - e -2 t
5
5
5
d
i u(t)
Example 2.10
Determine the complete response of the system described by the equation,
d2 y(t)
dy(t)
dx(t)
+ 5
+ 4 y(t) =
;
dt 2
dt
dt
y(0) = 0 ;
dy(t)
dt
t=0
= 1 , for the input, x(t) = e–2t u(t)
Solution
The given system equation is,
d2 y(t)
dy(t)
dx(t)
+ 5
+ 4 y(t) =
dt
dt
dt2
.....(1)
Homogeneous Solution
The homogeneous solution is the solution of the system equation when x(t) = 0.
On substituting x(t) = 0 in system equation (equation (1)) we get,
d2 y(t)
dy(t)
+ 5
+ 4 y(t) = 0
dt 2
dt
Let, y( t) = C e lt ;
d
y( t) = C l e lt and
dt
\
.....(2)
d2
y( t) = C l2 e lt
dt
On substituting the above terms in equation (2) we get,
C l2 e lt + 5 C l e lt + 4 C e lt = 0
\ l2 + 5 l + 4 C e lt = 0
d
i
Signals & Systems
2. 41
The characteristic polynomial of the above equation is,
l2 + 5l + 4 = 0
Þ
Þ
(l + 4) (l + 1) = 0
l = –4, –1
Now the homogeneous solution is given by,
Homogeneous solution, yh (t) = C1 el 1t + C2 el 2 t = C1 e -4 t + C2 e - t
Particular Solution
The particular solution is the solution of the system equation (equation (1)) for specific input.
Here input,
x(t) = e–2t u(t)
\ x(t) = e–2t ; for t ³ 0
dx(t)
= - 2e -2t
dt
\
Let the particular solution, yp(t) is of the form,
yp(t) = K x(t)
\ y p (t) = K e -2t ;
dyp (t)
dt
d2 y( t)
= 4K e -2t
dt 2
= - 2K e -2t ;
On substituting the above terms and the input in system equation (equation (1)) we get,
d2 y(t)
dy(t)
dx(t)
+ 5
+ 4 y(t) =
dt 2
dt
dt
ß
4K e–2t – 10K e–2t + 4K e–2t = –2 e–2t
On dividing throughout by e–2t we get,
Þ
4K – 10K + 4K = –2
– 2K = –2
Þ
K=1
\ Particular solution, yp(t) = e
–2t
Total (or Complete) Response
Method - 1
Total response, y(t) = yh(t) + yp(t)
\ y(t) = C1 e–4t + C2 e–t + e–2t
When t = 0, y(t) = y(0) = C1 e0 + C2 e0 + e0 = C1 + C2 + 1
Given that y(0) = 0, \ C1 + C2 + 1 = 0
.....(3)
Here,
dy(t)
= - 4C1 e -4t - C 2 e - t - 2 e -2t
dt
Now,
dy(t)
dt
Given that,
= - 4C1 e0 - C 2 e0 - 2 e 0 = - 4C1 - C 2 - 2
t=0
dy(t)
dt
= 1;
.....(4)
\ -4C1 - C 2 - 2 = 1
t=0
On adding equation (3) and (4) we get,
-3C1 - 1 = 1 Þ
- 3C1 = 2
Þ
C1 = -
2
3
Chapter 2 - Continuous Time Signals and Systems
2. 42
From equation (3), C 2 = - 1 - C1 = - 1 +
\ Total Response, y(t) = -
FG e
H
(or) y(t) =
2
1
= 3
3
2 -4 t
1 -t
e
e + e -2 t ; t ³ 0
3
3
-2 t
IJ
K
2 -4 t
1 -t
e
e
u(t)
3
3
-
Method - 2
y(t) = yzi(t) + yzs(t)
Total response,
Zero - input response, y zi (t) = yh (t)
with const ants evalu ated using initial c ondition
yh(t) = C1 e–4t + C2 e–t
\
dyh (t)
= - 4C1 e -4t - C 2 e - t
dt
At t = 0, y h (0) = C1 e0 + C2 e0 = C1 + C2
\ C1 + C2 = 0
Given that, yh(0) = 0,
At t = 0,
Given that,
dyh (t)
dt
.....(5)
dyh (t)
= - 4C1 e 0 - C 2 e 0 = - 4C1 - C2
dt
\ - 4C1 - C2 = 1
= 1,
.....(6)
t=0
On adding equations (5) and (6) we get,
-3C1 = 1 Þ
C1 = -
From equation (5), C2 = - C1 =
\ y zi (t) = -
1
3
1
3
1 -4t
1 -t
e
+
e
3
3
Zero - state response , y zs (t) = yh (t) + y p (t)
\ yzs(t) = C1 e
–4t
\
–t
with const ants evalu ated with zero initi al conditions
–2t
+ C2 e + e
dy zs (t)
= - 4C1 e -4t - C2 e - t - 2 e -2 t
dt
At t = 0,
y zs (0) = C1 e0 + C 2 e0 + e0 = C1 + C 2 + 1
At t = 0,
dy zs (t)
= - 4C1 e 0 - C 2 e0 - 2 e0 = - 4C1 - C 2 - 2
dt
Since initial conditions are zero,
C1 + C2 + 1 = 0
.....(7)
–4C1 – C2 – 2 = 0
.....(8)
On adding equations (7) and (8) we get,
-3C1 - 1 = 0
Þ
C1 = -
1
3
Signals & Systems
2. 43
From the equation (7), C2 = - 1 - C1 = -1 +
\ y zs (t) = -
1
2
= 3
3
1 -4t
2 -t
e e + e-2 t
3
3
\ Total Response, y(t) = y zi (t) + yzs (t)
= -
1 -4 t
1 -t
1 -4 t
2 -t
e +
e e e + e-2 t
3
3
3
3
= -
2 -4 t
1 -t
e e + e-2t ; t ³ 0
3
3
=
FG e
H
-2 t
-
IJ
K
2 -4 t
1 -t
e
e u(t)
3
3
2.8 Classification of Continuous Time Systems
The continuous time systems are classified based on their characteristics. Some of the classifications
of continuous time systems are,
1. Static and dynamic systems
2. Time invariant and time variant systems
3. Linear and nonlinear systems
4. Causal and noncausal systems
5. Stable and unstable systems
6. Feedback and nonfeedback systems
2.8.1 Static and Dynamic Systems
A continuous time system is called static or memoryless if its output at any instant of time t
depends at most on the input signal at the same time but not on the past or future input. In any other case,
the system is said to be dynamic or to have memory.
y(t) = a x(t)
y(t) = t x(t) + 6 x3(t)
y(t) = t x(t) + 3 x(t2)
y(t) = x(t) + 3 x(t – 2)
123 123
Example :
Static systems
Dynamic systems
2.8.2 Time Invariant and Time Variant Systems
A system is said to be time invariant if its input-output characteristics does not change with time.
Definition : A relaxed system H is time invariant or shift invariant if and only if
H
H
x(t) ¾¾
® y(t) implies that, x(t - m) ¾¾
® y(t - m)
for every input signal x(t) and every time shift m.
i.e., in time invariant systems, if y(t) = H{x(t)} then y(t – m) = H{x(t – m)}.
Chapter 2 - Continuous Time Signals and Systems
2. 44
Alternative Definition for Time Invariance
A system H is time invariant if the response to a shifted (or delayed) version of the input is
identical to a shifted (or delayed) version of the response based on the unshifted (or undelayed) input.
The diagrammatic explanation of the above definition of time invariance is shown in fig 2.41.
x(t)
Delay
Input signal
Delayed input
x(t)
Input signal
x(t - m)
H
System
y 1(t)
H
Response for
delayed input
System
y(t)
y 2(t)
Delay
Response for
undelayed input
Delayed
response
If y1(t) = y2(t) then the system is time invariant
Fig 2.41 : Diagrammatic explanation of time invariance.
Procedure to test for time invariance
1. Delay the input signal by m units of time and determine the response of the system for this
delayed input signal. Let this response be y1(t).
2. Delay the response of the system for unshifted input by m units of time. Let this delayed
response be y2(t).
3. Check whether y 1(t) = y2(t). If they are equal then the system is time invariant.
Otherwise the system is time variant.
Example 2.11
State whether the following systems are time invariant or not.
a) y(t) = 2t x(t)
b) y(t) = x(t) sin20pt
c) y(t) = 3x(t2)
d) y(t) = x(-t)
Solution
a) Given that, y(t) = 2t x(t)
Test 1 : Response for delayed input
x(t)
Input signal
Delay
x(t - m)
Delayed input
H
System
y1(t) = 2t x(t - m)
Response for
delayed input
Test 2 : Delayed response
x(t)
Input signal
H
System
y(t) = 2t x(t)
Response for
undelayed input
Delay
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
y2(t) = 2 (t - m) x(t - m)
Delayed response
Signals & Systems
2. 45
b) Given that, y(t) = x(t) sin20pt
Test 1 : Response for delayed input
x(t)
Input signal
Delay
x(t - m)
Delayed input
y1(t) = x(t - m) sin20pt
H
Response for
delayed input
System
Test 2 : Delayed response
x(t)
Input signal
H
System
y(t) =x(t) sin20pt
Response for
undelayed input
Delay
y2(t) = x(t - m) sin(20p(t - m))
Delayed response
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
c) Given that, y(t) = 3x(t2)
Test 1 : Response for delayed input
x(t)
x(t - m)
Delay
Input signal
Delayed input
y1(t) = 3 x(t2 - m)
H
Response for
delayed input
System
Test 2 : Delayed response
x(t)
Input signal
H
System
y(t) = 3x(t2)
Response for
undelayed input
y2(t) = 3 x((t - m)2)
Delay
Delayed response
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
d) Given that, y(t) = x(-t)
Test 1 : Response for delayed input
x(t)
Input signal
Delay
x(t - m)
Delayed input
y1(t) = x(- t - m)
H
Response for
delayed input
System
Test 2 : Delayed response
x(t)
Input signal
H
y(t) = x(- t)
Response for
System
undelayed input
Delay
y2(t) = x(- (t - m))=x(- t + m)
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
Delayed response
Chapter 2 - Continuous Time Signals and Systems
2. 46
Example 2.12
State whether the following systems are time invariant or not.
a) y(t) = 2 ex(t)
c) y(t) = 3x2(t)
b) y(t) = x(t) + C
d) y(t) = x(t) +
dx(t)
dt
e) y(t) = x(t) +
Solution
a) Given that, y(t) = 2 ex(t)
Test 1 : Response for delayed input
x(t)
Input signal
x(t - m)
Delay
Delayed input
y1(t) = 2 ex(t - m)
H
System
Response for
delayed input
Test 2 : Delayed response
x(t)
Input signal
H
System
y(t) = 2 ex(t)
Response for
undelayed input
Delay
y2(t) = 2 ex(t - m)
Delayed response
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
b) Given that, y(t) = x(t) + C
Test 1 : Response for delayed input
x(t)
Input signal
x(t - m)
Delay
Delayed input
H
System
y1(t) = x(t - m) + C
Response for
delayed input
Test 2 : Delayed response
x(t)
Input signal
H
System
y(t) = x(t) + C
Response for
undelayed input
Delay
y2(t) = x(t - m) + C
Delayed response
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
c) Given that, y(t) = 3x2(t)
Test 1 : Response for delayed input
x(t)
Input signal
Delay
x(t - m)
Delayed input
H
System
y1(t) = 3x2(t - m)
Response for
delayed input
Test 2 : Delayed response
x(t)
Input signal
H
System
y(t) = 3x2(t)
Response for
undelayed input
Delay
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
y2(t) = 3x2(t - m)
Delayed response
z
x(t) dt
Signals & Systems
2. 47
d) Given that, y( t ) = x( t ) +
dx ( t)
dt
Test 1 : Response for delayed input
x(t)
Delay
Input signal
x(t - m)
Delayed input
H
System
dx(t - m)
dt
Response for
delayed input
y1(t) = x(t - m) +
Test 2 : Delayed response
dx(t)
dt
H
Response for
System
undelayed input
y(t) = x(t) +
x(t)
Input signal
Delay
y2 (t) = x(t - m) +
dx(t - m)
dt
Delayed response
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
e) Given that, y ( t) = x( t ) +
z
x ( t) dt
Test 1 : Response for delayed input
x(t)
Input signal
x(t - m)
Delay
Delayed input
H
System
z
y1(t) = x(t - m) + x(t - m) dt
Response for
delayed input
Test 2 : Delayed response
x(t)
Input signal
y(t ) = x( t) +
H
z
x( t) dt
Response for
System undelayed input
Delay
y2 (t) = x(t - m) +
z
x(t - m) dt
Delayed response
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
2.8.3 Linear and Nonlinear Systems
A linear system is the one that satisfies the superposition principle.
The principle of superposition requires that the response of a system to a weighted sum of the
signals is equal to the corresponding weighted sum of the responses to each of the individual input signals.
Definition : A relaxed system H is linear if
H{a1 x1(t) + a2 x2(t)} = a1 H{x1(t)} + a2 H{x2(t)}
for any arbitrary input signal x1(t) and x2(t) and for any arbitrary constants a1 and a2.
If a relaxed system does not satisfy the superposition principle as given by the above definition, the
system is nonlinear. The diagrammatic explanation of linearity is shown in fig. 2.42.
Chapter 2 - Continuous Time Signals and Systems
x1(t)
a1
2. 48
a1x1(t)
+
x2(t)
a2
x1(t)
H
a1x1(t) + a2x2(t)
H
H{a1x1(t) + a2x2(t)}
a2x2(t)
H {x1(t)}
a1
a1H{x1(t)}
+
x2(t)
H
H{x2(t)}
a2
a1H{x1(t)} + a2H{x2(t)}
a2H {x2(t)}
The system, H is linear if and only if, H{a1 x1(t) + a2 x2(t)} = a1 H{x1(t)} + a2 H{x2(t)}
Fig 2.42 : Diagrammatic explanation of linearity.
Procedure to test for linearity
1. Let x1(t) and x2(t) be two inputs to the system H, and y1(t) and y2(t) be the corresponding responses.
2. Consider a signal, x3(t) = a1 x1(t) + a2 x2(t) which is a weighed sum of x1(t) and x2(t).
3. Let y3(t) be the response for x3(t).
4. Check whether y3(t) = a1 y1(t) + a2 y2(t). If equal then the system is linear, otherwise it is nonlinear.
Example 2.13
Test the following systems for linearity.
b) y(t) = x(t2),
a) y(t) = t x(t),
c) y(t) = x2(t),
e) y(t) = ex(t).
d) y(t) = A x(t) + B,
Solution
a) Given that, y(t) = t x(t)
Let H be the system operating on x(t) to produce, y(t) = H{x(t)} = t x(t).
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
\ y1(t) = H{x1(t)} = t x1(t)
....(1)
y2(t) = H{x2(t)} = t x2(t)
.....(2)
Let x3(t) = a1 x1(t) + a2 x2(t).
A linear combination of inputs x1(t) and x2(t)
Let y3(t) be the response of the system H for input x3(t).
\ y3(t) = H{ x3(t)} = H{ a1 x1(t) + a2 x2(t)}
= t(a1 x1(t) + a2 x2(t)) = a1 t x1(t) + a2 t x2(t)
= a1 y1(t) + a2 y2(t))
Since, y3(t) = a1 y1(t) + a2 y2(t), the given system is linear.
Using equations (1) and (2)
Signals & Systems
2. 49
b) Given that, y(t) = x(t2)
Let H be the system operating on x(t) to produce, y(t) = H{x(t)} = x(t2).
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
\ y1(t) = H{x1(t)} = x1(t2)
....(1)
y2(t) = H{x2(t)} = x2(t2)
.....(2)
A linear combination of inputs x1(t) and x2(t)
Let x3(t) = a1 x1(t) + a2 x2(t).
Let y3(t) be the response of the system H for input x3(t).
\ y3(t) = H{x3(t)} = H{a1 x1(t) + a2 x2(t)}
= (a1 x1(t2) + a2 x2(t2))
Using equations (1) and (2)
= a1 y1(t) + a2 y2(t))
Since, y3(t) = a1 y1(t) + a2 y2(t), the given system is linear.
c) Given that, y(t) = x2(t)
Let H be the system operating on x(t) to produce, y(t) = H{x(t)} = x2(t).
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
\ y1(t) = H{x1(t)} = x12 (t)
y2(t) = H{x2(t)} =
.....(1)
x 22 (t)
Let x3(t) = a1 x1(t) + a2 x2(t).
.....(2)
A linear combination of inputs x1(t) and x2(t)
Let y3(t) be the response of the system H for input x3(t).
\ y3(t) = H{x3(t)} = H{ a1 x1(t) + a2 x2(t)} = (a1 x1(t) + a2 x2(t))2
= a12 x12 ( t) + a 22 x 22 ( t) + 2 a1 a 2 x1( t) x1( t)
= a12 y1( t) + a 22 y 2 ( t) + 2 a1 a 2 x1( t) x1( t)
Using equations (1) and (2)
Here, y3(t) ¹ a1 y1(t) + a2 y2(t). Hence the given system is nonlinear.
d) Given that, y(t) = A x(t) + B
Let H be the system operating on x(t) to produce, y(t) = H{x(t)} = A x(t) + B.
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
\ y1(t) = H{x1(t)} = A x1(t) + B
y2(t) = H{x2(t)} = A x2(t) + B
Let x3(t) = a1 x1(t) + a2 x2(t).
.....(1)
.....(2)
A linear combination of inputs x1(t) and x2(t)
Let y3(t) be the response of the system H for input x3(t).
\ y3(t) = H{x3(t)}= H{ a1 x1(t) + a2 x2(t)}
= A ( a1 x1(t) + a2 x2(t) ) + B = A a1 x1(t) + A a2 x2(t) + B
= a1 A x1(t) + a2 A x2(t) + B
= a1 ( y1(t) - B ) + a2 ( y2(t) - B ) + B
Here, y3(t) ¹ a1 y1(t) + a2 y2(t). Hence the given system is nonlinear.
Using equations (1) and (2)
Chapter 2 - Continuous Time Signals and Systems
2. 50
x(t)
e) Given that, y(t) = e
Let H be the system operating on x(t) to produce, y(t) = H{x(t)} = ex(t).
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
\ y1(t) = H{x1(t)} = e x 1( t )
y2(t) = H{x2(t)} = e
.....(1)
x 2 (t)
.....(2)
Let x3(t) = a1 x1(t) + a2 x2(t).
A linear combination of inputs x1(t) and x2(t)
Let y3(t) be the response of the system H for input x3(t).
\ y3(t) = H{x3(t)} = H{ a1 x1(t) + a2 x2(t)}
= e( a1 x1( t ) + a 2 x 2 ( t )) = e a1x1( t) ea 2 x 2 ( t )
e
= e x1( t )
a1
a2
b g by (t)g
j ee j
x 2 (t)
= y1(t)
a1
a2
Using equations (1) and (2)
2
Here, y3(t) ¹ a1 y1(t) + a2 y2(t). Hence the given system is nonlinear.
Example 2.14
Test the following systems for linearity.
a) y(t) = 4 x(t) + 2
dx(t)
dt
d2y(t)
dy(t)
+ 2
+ 3 y(t) = x(t)
dt 2
dt
b)
Solution
a) Given that, y(t) = 4 x(t) + 2
dx(t)
dt
Let H be the system operating on x(t) to produce, y(t) = H{x(t)} = 4 x(t) + 2
dx(t)
dt
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
dx1(t)
dt
dx2 (t)
y2 (t) = H{ x 2 (t)} = 4 x 2 (t) + 2
dt
Let x3(t) = a1 x1(t) + a2 x2(t).
\ y1(t) = H{ x1(t)} = 4 x1(t) + 2
.....(1)
.....(2)
A linear combination of inputs x1(t) and x2(t)
Let y3(t) be the response of the system H for input x3(t).
\ y 3 (t) = H{ x 3 (t)}
= 4 x 3 (t) + 2
dx 3 (t)
dt
d
(a1 x1(t) + a 2 x 2 ( t))
dt
dx1( t)
dx 2 ( t)
= 4a1 x1(t) + 4a 2 x 2 (t) + 2a1
+ 2a 2
dt
dt
= 4(a1 x1(t) + a2 x 2 (t)) + 2
FG
H
= a1 4 x1( t) + 2
dx1( t)
dt
IJ
K
FG
H
+ a 2 4 x 2 (t ) + 2
= a1 y1( t) + a2 y2 ( t)
Since, y3(t) = a1 y1(t) + a2 y2(t), the given system is linear.
dx 2 ( t)
dt
IJ
K
Using equations (1) and (2)
Signals & Systems
2. 51
b) Given that,
d2 y(t)
dy(t)
+ 2
+ 3 y(t) = x(t)
dt
dt2
Let H be the system operating on x(t) to produce, y(t).
Consider two signals x1(t) and x2(t).
Let y1(t) and y2(t) be the response of the system H for inputs x1(t) and x2(t) respectively.
When the input is x1(t), the response is y1(t). Hence the system equation for the input x1(t) can be written as,
d2 y1(t)
dy1(t)
+ 2
+ 3 y1(t) = x1(t)
dt 2
dt
.....(1)
When the input is x2(t), the response is y2(t). Hence the system equation for the input x2(t) can be written as,
d2y 2 (t)
dy 2 (t)
+ 2
+ 3 y 2 (t) = x 2 (t)
dt 2
dt
Let, x3(t) = a1 x1(t) + a2 x2(t).
.....(2)
A linear combination of inputs x1(t) and x2(t)
Let, y3(t) be the response of the system H for input x3(t).
When the input is x3(t), the response is y3(t). Hence the system equation for the input x3(t) is given by,
d2y 3 (t)
dy 3 (t)
+ 2
+ 3 y 3 (t) = x 3 (t)
dt 2
dt
.....(3)
Let us multiply equation (1) by a1.
\ a1
d2y1(t)
dy1(t)
+ 2a 1
+ 3a1 y1(t) = a1 x1(t)
dt 2
dt
.....(4)
Let us multiply equation (2) by a2.
\ a2
d2 y 2 (t)
dy 2 (t)
+ 2a 2
+ 3a 2 y 2 (t) = a 2 x 2 (t)
dt 2
dt
.....(5)
On adding equation (4) and (5) we get,
a1
d2 y1(t)
dy1(t)
d2 y 2 (t)
dy 2 (t)
+ 2a1
+ 3a 1 y1(t) + a2
+ 2a 2
+ 3a 2 y 2 (t) = a1 x1(t) + a2 x 2 (t)
dt 2
dt
dt 2
dt
d2
d
[ a1 y1(t ) + a 2 y 2 ( t)] + 2
[a1 y1(t ) + a 2 y 2 (t )] + 3[a1 y1(t ) + a 2 y 2 (t )] = a1 x1( t) + a2 x 2 (t ) .....(6)
dt
dt 2
On comparing equations (3) and (6) we can say that,
if, x3(t) = a1 x1(t) + a2 x2(t), then y3(t) = a1 y1(t) + a2 y2(t)
Hence the sytem is linear.
2.8.4 Causal and Noncausal Systems
Definition : A system is said to be causal if the output of the system at any time t depends only on the
present input, past inputs and past outputs but does not depend on the future inputs and outputs.
If the system output at any time t depends on future inputs or outputs then the system is called a
noncausal system.
The causality refers to a system that is realizable in real time. It can be shown that an LTI system
is causal if and only if the impulse response is zero for t < 0, (i.e., h(t) = 0 for t < 0).
Chapter 2 - Continuous Time Signals and Systems
2. 52
Example 2.15
Test the casuality of the following systems.
a) y(t) = x(t) – x(t – 1)
b) y(t) = x(t) +2 x(3 – t)
d) y(t) = x(t) +
z
c) y(t) = t x(t)
3t
t
x(l) dl
e) y(t) = x(t) +
0
z
x(l) dl
f) y(t) = 2x(t) +
0
dx(t)
dt
Solution
a) Given that, y(t) = x(t) – x(t – 1)
When t = 0, y(0) = x(0) – x(–1)
Þ
The response at t = 0, i.e., y(0) depends on the
present input x(0) and past input x(–1).
When t = 1, y(1) = x(1) – x(0)
Þ
The response at t = 1, i.e., y(1) depends on the
present input x(1) and past input x(0).
From the above analysis we can say that for any value of t, the system output depends on present and past inputs.
Hence the system is causal.
b) Given that, y(t) = x(t) + 2 x(3 – t)
When t = – 1, y(– 1) = x(– 1) + 2 x(4)
Þ
The response at t = – 1, i.e., y(– 1) depends on the
present input x(– 1) and future input x(4).
When t = 0, y(0)
= x(0) + 2 x(3)
Þ
The response at t = 0, i.e., y(0) depends on the
present input x(0) and future input x(3).
When t = 1, y(1)
= x(1) + 2 x(2)
Þ
The response at t = 1, i.e., y(1) depends on the
present input x(1) and future input x(2).
When t = 2, y(2)
= x(2) + 2 x(1)
Þ
The response at t = 2, i.e., y(2) depends on the
present input x(2) and past input x(1).
From the above analysis we can say that for t< 2, the system output depends on present and future inputs. Hence
the system is noncausal.
c) Given that, y(t) = t x(t)
When t = 0, y(0) = 0 ´ x(0)
Þ
The response at t = 0, i.e., y(0) depends on the present input x(0).
When t = 1, y(1) = 1 ´ x(1)
Þ
The response at t = 1, i.e., y(1) depends on the present input x(1).
When t = 2, y(2) = 2 ´ x(2)
The response at t = 2, i.e., y(2) depends on the present input x(2).
Þ
From the above analysis we can say that the response for any value of t depends on the present input. Hence the
system is causal.
t
d) Given that, y(t) = x(t) +
z
z
x( l ) dl
0
t
y(t) = x(t) +
t
x(l ) dl = x(t) + z( l ) 0 = x( t) + z( t) - z(0),
0
z
where, z(l ) = x(l ) dl
When t = 0, y(0) = x(0) + z(0) - z(0)
Þ
The response at t = 0, i.e., y(0) depends on present input.
When t = 1, y(1) = x(1) + z(1) - z(0)
Þ
The response at t = 1, i.e., y(1) depends on present and past input.
When t = 2, y(2) = x(2) + z(1) - z(0)
Þ
The response at t = 2, i.e., y(2) depends on present and past input.
From the above analysis we can say that the response for any value of t depends on the present and past input.
Hence the system is causal.
Signals & Systems
2. 53
3t
z
e) Given that, y(t) = x(t) +
x ( l ) dl
0
3t
y(t) = x(t) +
z
3t
x(l) dl = x(t) + z(l) 0 = x(t) + z(3t) - z(0),
When t = 0, y(0) = x(0)+z(0)- z(0)
z
where z(l) = x(l) dl
0
Þ The response at t = 0, i.e., y(0) depends on present input.
When t = 1, y(1) = x(1) + z(3) - z(0) Þ The response at t =1, i.e., y(1) depends on present, past and future inputs.
When t = 2, y(2) = x(2) + z(6) - z(0) Þ The response at t = 2, i.e., y(2) depends on present, past and future inputs.
From the above analysis we can say that the response for t > 0 depends on the present, past and future inputs.
Hence the system is noncausal.
dx(t)
dt
f) Given that, y(t) = 2x(t) +
y(t) = 2 x(t) +
= 2 x(t) +
dx(t)
dt
Lt
Dt ® 0
x( t) - x( t - Dt)
Dt
(Using definition of differentiation, refer section 2.4.6)
In the above equation, for any value of t, the x(t) is present input and x(t-Dt) is the past input.
Therefore we can say that the response for any value of t depends on present and past input. Hence the system
is causal.
Example 2.16
Test the causality of the following systems.
a) y(t) = x(t) + 3 x(t + 4)
b) y(t) = x(t2)
c) y(t) = x(2t)
d) y(t) = x(–t)
Solution
a) Given that, y(t) = x(t) + 3 x(t + 4)
W hen t = 0, y(0) = x(0) + 3 x(4)
Þ
The response at t = 0, i.e., y(0) depends on the
present input x(0) and future input x(4).
When t = 1, y(1) = x(1) + 3 x(5)
Þ
The response at t = 1, i.e., y(1) depends on the
present input x(1) and future input x(5).
From the above analysis we can say that the response for any value of t depends on present and future inputs.
Hence the system is noncausal.
b) Given that, y(t) = x(t2)
When t = –1 ; y(–1) = x(1)
Þ
The response at t = –1, depends on the future input x(1).
When t = 0 ; y(0) = x(0)
Þ
The response at t = 0, depends on the present input x(0).
When t = 1 ; y(1) = x(1)
Þ
The response at t = 1, depends on the present input x(1).
When t = 2 ; y(2) = x(4)
Þ
The response at t = 2, depends on the future input x(4).
From the above analysis we can say that the response for any value of t (except t = 0 & t = 1) depends on future input.
Hence the system is noncausal.
Chapter 2 - Continuous Time Signals and Systems
2. 54
c) Given that, y(t) = x(2t)
When t = –1 ; y(–1) = x(–2)
Þ
The response at t = –1, depends on the past input x(–2).
When t = 0
; y(0) = x(0)
Þ
The response at t = 0, depends on the present input x(0).
When t = 1
; y(1) = x(2)
Þ
The response at t = 1, depends on the future input x(2).
From the above analysis we can say that the response of the system for t > 0, depends on future input. Hence the
system is noncausal.
d) Given that, y(t) = x(–t)
When t = –2 ; y(–2) = x(2)
Þ
The response at t = –2, depends on the future input x(2).
When t = –1 ; y(–1) = x(1)
Þ
The response at t = –1, depends on the future input x(1).
When t = 0
; y(0)
= x(0)
Þ
The response at t = 0, depends on the present input x(0).
When t = 1
; y(1)
= x(–1)
Þ
The response at t = 1, depends on the past input x(–1).
From the above analysis we can say that the response of the system for t < 0 depends on future input. Hence the
system is noncausal.
2.8.5 Stable and Unstable Systems
Definition : An arbitrary relaxed system is said to be BIBO stable (Bounded Input-Bounded Output
stable) if and only if every bounded input produces a bounded output.
Let x(t) be the input of continuous time system and y(t) be the response or output for x(t).
The term bounded input refers to finite value of the input signal x(t) for any value of t. Hence if
input x(t) is bounded then there exists a constant Mx such that |x(t)| £ Mx and Mx< ¥, for all t.
Examples of bounded input signal are step signal, decaying exponential signal and impulse signal.
Examples of unbounded input signal are ramp signal and increasing exponential signal.
The term bounded output refers to finite and predictable output for any value of t. Hence if output
y(t) is bounded then there exists a constant My such that |y(t)| £ My and My< ¥, for all t.
In general, the test for stability of the system is performed by applying specific input. On applying
a bounded input to a system if the output is bounded then the system is said to be BIBO stable.
Condition for Stability of an LTI System
For an LTI (Linear Time Invariant) system, the condition for BIBO stability can be transformed to
a condition on impulse response, h(t). For BIBO stability of an LTI continuous time system, the integral
of impulse response should be finite.
+¥
\
z
| h( t )| dt < ¥ , for stability of an LTI system.
-¥
Proof :
The response of a system y(t) for any input x(t) is given by convolution of the input and impulse response.
+¥
y( t) =
z
-¥
h( t) x(t - t) dt
.....(2.18)
Signals & Systems
2. 55
On taking the absolute value on both sides of equation (2.18), we get,
+¥
y( t ) =
z
z
+¥
h( t) x(t - t ) dt
=
-¥
+¥
=
z
h( t) x(t - t ) dt
-¥
h( t ) x(t - t) dt
.....(2.19)
-¥
If the input x(t) is bounded then there exists a constant MX, such that | x(t-t) | £ MX < ¥. Hence equation(2.19) can be written as,
+¥
y( t ) = M x
z
h( t) dt
.....(2.20)
-¥
From equation (2.20) we can say that the output y(t) is bounded, if the impulse response satisfies the condition,
z
+¥
h( t) dt < ¥
-¥
Since t is a dummy variable in the above condition we can replace t by t.
+¥
\
z
h(t ) dt < ¥
-¥
Example 2.17
Test the stability of the following systems.
a) y(t) = cos (x(t))
b) y(t) = x(–t – 2)
c) y(t) = t x(t)
Solution
a) Given that, y(t) = cos (x(t))
The given system is a nonlinear system, and so the test for stability should be performed for specific inputs.
The value of cos q lies between –1 to +1 for any value of q. Therefore the output y(t) is bounded for any value of input
x(t). Hence the given system is stable.
b) Given that, y(t) = x(–t – 2)
The given system is a time variant system, and so the test for stability should be performed for specific inputs.
The operations performed by the system on the input signal are folding and shifting. A bounded input signal will remain
bounded even after folding and shifting. Therefore in the given system, the output will be bounded as long as input is bounded.
Hence the given system is BIBO stable.
c) Given that, y(t) = t x(t)
The given system is a time variant system, and so the test for stability should be performed for specific inputs.
Case i : Let x(t) tends to ¥ or constant, as t tends to infinity. In this case, y(t) = t x(t) will be infinity as t tends to inifnity
and so the system is unstable.
Case ii : Let x(t) tends to 0, as t tends to infinity. In this case y(t) = t x(t) will be zero as t tends to infinity and so the
system is stable.
Chapter 2 - Continuous Time Signals and Systems
2. 56
Example 2.18
Test the stability of the LTI systems, whose impulse responses are given below.
a) h(t) = e–5|t|
b) h(t) = e4t u(t)
c) h(t) = e–4t u(t)
d) h(t) = t e–3t u(t)
e) h(t) = t cost u(t)
f) h(t) = e–t sint u(t)
Solution
a) Given that, h(t) = e–5|t|
+¥
For stability,
z
h(t) dt < ¥
-¥
+¥
\
z
+¥
h(t) dt =
-¥
z
z
+¥
e -5|t| dt =
-¥
0
=
+¥
z
h(t) dt =
-¥
z
e-5t dt =
0
0
Here,
e -5|t| dt
+¥
e5t dt +
-¥
=
z
-¥
-¥
e
5
-
e
5
5t
0
-¥
LM e OP
N -5 Q
- 5t ¥
+
0
0
-¥
+
LM e OP
N5Q
e
-5
-
e
1
1
2
- 0 + 0 +
=
=
-5
5
5
5
2
= constant. Hence the system is stable.
5
b) Given that, h(t) = e4t u(t)
+¥
For stability,
z
h(t) dt < ¥
-¥
+¥
\
z
+¥
h(t) dt =
-¥
z
z
+¥
e4t u(t) dt =
-¥
e4t u(t) dt
-¥
+¥
=
z
e4t dt =
LM e OP
N4Q
4t
¥
e¥
e0
1
= ¥
= ¥ 4
4
4
=
0
0
+¥
Here,
z
h(t) dt = ¥. Hence the system is unstable.
-¥
c) Given that, h(t) = e–4t u(t)
+¥
For stability,
z
h(t) dt < ¥
-¥
+¥
\
z
+¥
h(t) dt =
-¥
z
z
+¥
e-4t u(t) dt =
-¥
0
+¥
Here,
z
-¥
h(t) dt =
e-4t u(t) dt
-¥
+¥
=
z
e-4t dt =
LM e OP
N -4 Q
-4t ¥
0
=
e-¥
e0
1
1
= 0 +
=
4
4
-4
-4
1
= constant. Hence the system is stable.
4
Signals & Systems
2. 57
d) Given that, h(t) = t e
–3t
u(t)
+¥
For stability,
z
h(t) dt < ¥
-¥
+¥
\
z
+¥
z
h(t) dt =
-¥
+¥
=
+¥
z
h(t) dt =
-¥
t e-3t dt
0
-¥
LMt e
N -3
z
-3t
Since,
z
t e -3t u(t) dt =
-
-3t
1 ´
e
dt
-3
OP
Q
LM- t e
N 3
¥
-3t
=
0
=-
¥ ´ e -¥
e -¥
0 ´ e0
e0
+
+
3
9
3
9
=-
¥ ´ 0
1
1
- 0 + 0 +
=
9
3
9
e -3t
9
-
OP
Q
¥
0
z
uv = u
z z LNM z OQP
v -
du v
1
= constant, the system is stable.
9
e) Given that, h(t) = t cost u(t)
+¥
For stability,
z
h(t) dt < ¥
-¥
+¥
\
z
+¥
h(t) dt =
-¥
z
+¥
t cost u(t) dt =
=
z
t cost dt
0
-¥
LMt sint N
z
1 ´ sint dt
OP
Q
¥
= t sint + cost
0
z
¥
0
z z LMN z OPQ
uv = u v -
du
v
= ¥ ´ sin ¥ + cos ¥ - 0 ´ sin 0 - cos 0
= ¥ + cos ¥ - 0 - 1 = ¥
+¥
Since,
z
h(t) dt = ¥, the system is unstable.
-¥
f) Given that, h(t) = e–t sint u(t)
+¥
For stability,
z
h(t) dt < ¥
-¥
+¥
\
z
-¥
+¥
h(t) dt =
z
+¥
e- t sint u(t) dt =
¥
z
¥
e - t sint dt = e - t ( - cost)
z
.....(1)
e- t sint dt
0
-¥
¥
0
0
z
z
0
z
e- t
( -cost ) dt
-1
uv = u
z z LNM z OQP
v -
du v
¥
= -e- t cost
¥
0
e - t cost dt
0
= -e - t cost
¥
0
LM e
MN
¥
-t
¥
sint
0
z
z
0
OP
PQ
e- t
sint dt
-1
z
z z LNM z OQP
uv = u v -
du
v
¥
= -e - t cost
¥
0
- e- t sint
¥
0
0
e- t sint dt
.....(2)
Chapter 2 - Continuous Time Signals and Systems
2. 58
From equation (2) we can write,
¥
z
z
2
e - t sint dt = -e- t cost
¥
0
- e- t sint
¥
0
0
¥
\
1
-e - t cost
2
e - t sint dt =
0
1 -t
e sint
2
¥
0
¥
.....(3)
0
Using equation (3), the equation (1) can be written as,
+¥
¥
z
h(t) dt
z
=
0
-¥
+¥
z
h(t) dt =
-¥
1
- e - t cost
2
1 -t
e sint
2
¥
0
¥
0
1
1 -¥
-e -¥ cos ¥ + e0 cos 0 e sin ¥ - e0 sin 0
2
2
=
Since,
e- t sint dt =
=
1
1
-0 ´ cos ¥ + 1 0 ´ sin ¥ - 0
2
2
=
1
1
1
0 + 1 0 - 0 =
2
2
2
1
= constant, the system is stable.
2
Example 2.19
Determine the range of values of "a" and "b" for the stability of LTI system with impulse response.
h(t) = eat u(t) + e-bt u(t)
Solution
Given that, h(t) = eat u(t) + e-bt u(t).
+¥
z
+¥
|h(t )| dt =
-¥
z
z
+¥
z
eat u(t) + e - bt u( t) dt =
-¥
¥
=
LM e OP
NaQ
¥
e at dt +
0
=
( e at u( t) + e -bt u(t)) dt
-¥
e
z
=
0
a´¥
a
e
0
LM e OP
N -b Q
-b ´ 0
a´¥
at
e - bt dt
a ´0
+
e
a
-b ´ ¥
-b
e
¥
- bt
+
=
-b
e
¥
0
a
1 e -b ´ ¥
1
+
+
-b
a
b
In the above equation if " a" is negative and " b" is positive then it converges to finite value.
Therefore, when " a" is negative and " b" is positive,
¥
z
|h(t )| dt =
0
=
0
1
0
1
+
+
-b
a
a
b
-
1
1
+
a
b
=
a-b
ab
= Cons tan t
Here the integral of impulse response is a constant when "a" is negative and "b" is positive.
Therefore the range of values of "a" and "b" for stability of LTI system are, a < 0 and b > 0.
2. 59
Signals & Systems
2.8.6 Feedback and Nonfeedback Systems
The system in which the output y(t) at any time t depends on past output, past input and present
input is called a feedback system. The integration and differentiation of a signal at any time depends on
past value and so the equations governing feedback systems will have terms involving differentiations
and integrations of output and input.
The equations governing feedback systems will be in the form,
a0
dN
d N -1
dN-2
d
dM
y(t)
+
a
y(t)
+
a
y(t) + ....... + a N -1 y(t) + a N y(t) = b0 M x( t )
1
2
N
N -1
N-2
dt
dt
dt
dt
dt
M -1
M -2
d
d
d
+ b1 M - 1 x( t ) + b 2 M - 2 x( t ) +.......+ b M - 1 x( t ) + b M x( t )
dt
dt
dt
The system in which the output depends only on the present and past input is called a nonfeedback
system. The equations governing nonfeedback systems will not have terms involving differentiations and
integrations of output.
The equations governing nonfeedback systems will be in the form,
y(t) = b 0
dM
d M -1
dM-2
d
(
)
+
(
)
+
x
t
b
x
t
b
x( t ) +.......+ b M -1 x( t ) + b M x( t )
1
2
dt
dt M
dt M - 1
dt M - 2
2.9 Convolution of Continuous Time Signals
The convolution of two continuous time signals x1(t) and x2(t) is defined as,
+¥
x3 (t) =
z
x1 (l ) x 2 ( t - l ) dl
....(2.21)
-¥
where, x3(t) is the signal obtained by convolving x1(t) and x2(t),
and l is a dummy variable used for integration.
The convolution relation of equation (2.21) can be symbolically expressed as,
x3(t) = x1(t) * x2(t)
..... (2.22)
where the symbol * indicates convolution operation.
2.9.1 Response of LTI Continuous Time System Using Convolution
In an LTI continuous time system, the response y(t) of the system for an arbitrary input x(t) is
given by convolution of input x(t) with impulse response h(t) of the system. It is expressed as,
+¥
y( t ) = x( t ) * h( t ) =
z
x(l ) h( t - l ) dl
....(2.23)
-¥
where the symbol * represents convolution operation.
In an LTI system, if the input x(t) is a unit step signal, then the response is called a unit step
response.
Chapter 2 - Continuous Time Signals and Systems
2. 60
Proof :
Let y(t) be the response of system H for an input x(t)
\ y(t) = H[x(t)]
.....(2.24)
From equation (2.10) we know that the signal x(t) can be expressed as an integral of impulses,
+¥
z
i. e., x( t) =
x( l ) d( t - l ) dl
.....(2.25)
-¥
where, d(t – l) is the delayed unit impulse signal
From equation (2.24) and (2.25) we get,
y( t ) = H
LM
MN
+¥
z
x(l ) d( t - l ) dl
-¥
OP
PQ
+¥
=
z
z
H x( l ) d( t - l ) dl
In linear system, integration and system
operation H can be interchanged
-¥
+¥
=
x( l ) H d(t - l ) dl
.....(2.26)
-¥
The system H is a function of
t and not a function of l.
Let the response of the LTI system to the unit impulse input d(t) be denoted by h(t).
\ h(t) = H[d(t)]
Then by time invariance property, the response of the system to delayed unit impulse input d(t – l) is given by,
H[d(t – l)] = h(t – l)
.....(2.27)
Using equation (2.27), the equation (2.26) can be expressed as,
+¥
y( t ) =
z
x( l ) h(t - l ) dl
.....(2.28)
-¥
The equation (2.28) represents the convolution of input x(t) with the impulse response h(t) to yield the output
y(t). Hence it is proved that the response y(t) of LTI continuous time system for an arbitrary input x(t) is given
by convolution of input x(t) with impulse response h(t) of the system.
2.9.2 Properties of Convolution
The convolution of continuous time signals will satisfy the following properties.
Commutative property
: x1(t) * x2(t) = x2(t) * x1(t)
Associative property
: [ x1(t) * x2(t) ] * x3(t) = x1(t) * [ x2(t) * x3(t) ]
Distributive property
: x1(t) * [ x2(t) + x3(t) ] = [ x1(t) * x2(t) ] + [ x1(t) * x3(t) ]
Proof of Commutative Property :
Consider two continuous time signals, x1(t) and x2(t).
By Commutative property we can write,
x1(t) * x2(t) = x2(t) * x1(t)
(LHS)
(RHS)
2. 61
Signals & Systems
LHS = x1(t) * x2(t)
+¥
z
=
x 1( m) x 2(t - m) dm
.....(2.29)
m = -¥
where m is a dummy variable used for convolution operation.
Let,
t–m=p
when m = –¥, p = t – m = t + ¥ = +¥
\m = t – p
when m = +¥, p = t – m = t – ¥ = –¥
dm = –dp
On replacing m by (t – p) and (t – m) by p in equation (2.29) we get,
-¥
z
LHS = -
+¥
x 1( t - p) x 2( p) dp =
p = +¥
z
x 2( p) x 1( t - p) dp
p = -¥
Here p is a dummy variable used
for convolution operation
= x2(t) * x1(t)
= RHS
Proof of Associative Property :
Consider three continuous time signals x1(t), x2(t) and x3(t). By Associative property we can write,
[ x1(t) * x2(t) ] * x3(t) = x1(t) * [ x2(t) * x3(t) ]
LHS
RHS
y1(t) = x1(t) * x2(t)
Let,
.....(2.30)
Let us replace t by p.
\ y1(p) = x1(p) * x2(p)
+¥
z
=
x 1( m) x 2(p - m) dm
.....(2.31)
m = -¥
y2(t) = x2(t) * x3(t)
Let,
.....(2.32)
+¥
z
\ y 2 (t ) =
x 2( q) x 3(t - q) dq
q = -¥
+¥
z
\ y 2 (t - m) =
x 2 ( q) x 3(t - q - m) dq
.....(2.33)
q = -¥
where p, m and q are dummy variables used for convolution operation.
LHS = [ x1 (t) * x2(t) ] * x3(t)
= y1(t) * x3(t)
Using equation (2.30)
+¥
=
z
z z
z
z
y 1( p) x 3(t - p) dp
p = -¥
+¥
=
+¥
x 1 ( m) x 2 (p - m) x 3 (t - p) dm dp
p = - ¥ m = -¥
+¥
=
Using equation (2.31)
+¥
x1 (m) dm
m = -¥
x 2 (p - m) x 3 (t - p) dp
p = -¥
.....(2.34)
Chapter 2 - Continuous Time Signals and Systems
Let,
p–m=q
\p = q + m
2. 62
when p = –¥,
q = p – m = –¥ – m = –¥
when p = +¥,
q = p – m = +¥ – m = +¥
dp = dq
On replacing (p – m) by q and p by (q + m) in the equation (2.34) we get,
+¥
+¥
z
z
LHS =
z
x 1(m) dm
m = -¥
x2 ( q) x3 (t - q - m) dq
q = -¥
+¥
=
x 1( m) y 2 (t - m) dm
Using equation (2.33)
m = -¥
= x1(t) * y2(t)
= x1(t) * [x2(t) * x3(t)]
Using equation (2.32)
= RHS
Proof of Distributive Property :
Consider three continuous time signals x1(t), x2(t) and x3(t). By distributive property we can write,
x1(t) * [ x2(t) + x3(t) ] = [ x1(t) * x2(t) ] + [ x1(t) * x3 (t) ]
LHS
RHS
LHS = x1(t) * [ x2(t) + x3(t) ]
x4(t) = x2(t) + x3(t)
= x1(t) * x4(t)
m is dummy variable
used for integration
+¥
=
z
z
z
x1( m) x 4 (t - m) dm
m = -¥
+¥
=
m = -¥
+¥
=
if, x4(t) = x 2(t) + x3(t), then
x4(t-m) = x2(t-m) + x3(t-m)
x1( m) [ x 2 ( t - m) + x 3 (t - m) ] dm
+¥
z
x1( m) x2 (t - m) dm +
m = -¥
x 1( m) x 3(t - m) dm
m = -¥
= [ x1(t) * x2(t) ] + [ x1(t) * x3(t) ]
= RHS
2.9.3 Interconnections of Continuous Time Systems
® h (t)
1
x(t)
®h (t)
1
y 1(t)
® h (t)
2
y(t)
®
x(t)
®
® ®
Smaller continuous time systems may be interconnected to form larger systems. Two possible
basic ways of interconnection are cascade connection and parallel connection. The cascade and parallel
connections of two continuous time systems with impulse responses h1(t) and h2(t) are shown in
fig 2.43.
+
® h (t)
2
Fig 2.43a : Cascade connection.
Fig 2.43b : Parallel connection.
Fig 2.43 : Interconnection of continuous time systems.
y(t)
®
2. 63
Signals & Systems
Cascade Connected Continuous Time Systems
Two cascade connected continuous time systems with impulse response h1(t) and h2(t) can be
replaced by a single equivalent continuous time system whose impulse response is given by convolution
of individual impulse responses.
x(t)
® h (t)
y 1(t)
1
® h (t)
2
y(t)
®
Þ
x(t)
® h (t)*h (t)
1
2
y(t)
®
Fig 2.44 : Cascade connected continuous time systems and their equivalent.
Proof :
With reference to fig 2.44 we can write,
y1(t) = x(t) * h1(t)
.....(2.35)
y(t) = y1(t) * h2(t)
.....(2.36)
Using equation (2.35) the equation (2.36) can be written as,
y(t) =[ x(t) * h1(t) ] * h2(t)
Using associative property
= x(t) * [ h1(t) * h2(t) ]
= x(t) * h(t)
.....(2.37)
where, h(t) = h1(t) * h2(t)
From equation (2.37) we can say that the overall impulse response of two cascaded continuous time systems
is given by convolution of individual impulse responses.
Parallel Connected Continuous Time Systems
Two parallel connected continuous time systems with impulse responses h1(t) and h2(t) can be
replaced by a single equivalent continuous time system whose impulse response is given by the sum of
individual impulse responses.
® h (t)
y 1(t)
®
1
y(t)
x(t)
®
® h (t)
y2(t)
®
®
+
Þ
x(t)
® h (t)+h (t)
1
2
y(t)
®
2
Fig 2.45 : Parallel connected continuous time systems and their equivalent.
Proof:
With reference to fig 2.45 we can write,
y1(t) = x(t) * h1(t)
.....(2.38)
y2(t) = x(t) * h2(t)
.....(2.39)
y(t) = y1(t) + y2(t)
.....(2.40)
On substituting for y1(t) and y 2(t) from equations (2.38) and (2.39) in equation (2.40) we get,
y(t) = [ x(t) * h1(t) ] + [ x(t) * h2(t) ]
.....(2.41)
By using distributive property of convolution, the equation (2.41) can be written as shown below.
y(t) = x(t) * [ h1(t) + h2(t) ]
= x(t) * h(t)
.....(2.42)
where, h(t) = h1(t) + h2(t)
From equation (2.42) we can say that the overall impulse response of two parallel connected continuous time
systems is given by the sum of individual impulse responses.
Chapter 2 - Continuous Time Signals and Systems
2. 64
2.9.4 Procedure to Perform Convolution
The convolution of two continuous time signals x1(t) and x2(t) is defined as,
+¥
x 3 ( t ) = x1 ( t ) * x 2 ( t ) =
z
x1 (l ) x2 ( t - l ) dl
-¥
where,
x3(t) is the signal obtained by convolving x1(t) and x2(t),
l is a dummy variable used for integration,
* indicates convolution operation.
The computation of x3(t) using the above convolution equation for any value of t involves the
following operations,
1. Change of time index : The time index t in signals x1(t) and x2(t) is changed to l to get x1(l)
and x2(l).
2. Folding
: The signal x2(l) is folded to get x2(-l).
3. Shifting
: The signal x2(-l) is shifted by t units of time to get x2(t-l).
4. Multiplication
: The signals x1(l) and x2(t-l) are multiplied to get a product signal.
5. Integration
: The product signal is integrated to get x3(t). Let the product signal is
nonzero in the interval l=l1 to l=l2 , Now the signal x3(t) is given by,
l =l2
x 3 ( t) =
z
x1 (l ) x 2 ( t - l ) dl
l = l1
If both x1(t) and x2(t) are defined for t > 0, (i.e., both x1(t) and x2(t)
are causal) then the product signal is nonzero in the interval l=0 to l=t ,
Now the signal x3(t) is given by,
l=t
x3 (t) =
z
x1 (l ) x 2 ( t - l ) dl
l=0
In order to determine the range of values of l for product signal, graphical representation of
signals will be very useful. The operations like folding, shifting and multiplication can be performed
graphically to ascertain the range of values of l over which the product signal is nonzero.
In the above convolution, if the signals x1(t) and x2(t) are defined by a single mathematical equation
for t = -¥ to +¥, then time shift is valid for any value of t in the range t = -¥ to +¥. Therefore the time
shift, multiplication and integration are performed only once by taking general time shift t.
In the above convolution, if the signals x1(t) and x2(t) are defined by different mathematical equations
in various intervals of time, then the time shift, multiplication and integration are performed in each
interval of time by considering a time shift t in each interval.
2. 65
Signals & Systems
2.9.5 Unit Step Response Using Convolution
In general the response y(t) of a system is given by convolution of input x(t) and impulse response
h(t) of the system.
l = +¥
z
y( t ) = x( t ) * h( t ) =
x(l ) h( t - l ) dl
l = -¥
Let the input x(t) be unit step input u(t), and the corresponding response be s(t). Now the unit step
response s(t) is given by,
Unit Step Response, s( t ) = u( t ) * h( t )
= h( t ) * u( t )
Using Commutative property
l =+¥
z
=
h(l ) u( t - l ) dl
l =-¥
In the above convolution operation, u(l) = 1 for l > 0,
u(-l) = 1 for l < 0,
u(t-l) = 1 for l < t, and u(t-l) = 0 for l > t.
Therefore the unit step response s(t) is given by,
l=t
Unit Step Response, s( t ) =
z
h(l ) dl
l = -¥
Example 2.20
Perform convolution of the following causal signals.
a) x1(t) = 2 u(t), x2(t) = u(t)
b) x1(t) = e–2t u(t) ,
x2(t) = e–5t u(t)
c) x1(t) = t u(t), x2(t) = e–5t u(t)
d) x1(t) = cos t u(t) ,
x2(t) = t u(t)
Solution
a) Given that, x1(t) = 2 u(t) = 2 ; t ³ 0
x 2(t) = u(t)
Let,
=1 ; t³0
x3(t) = x1(t) * x2(t)
Since x1(t) and x2(t) are causal,
the limits of integration is 0 to t.
By definition of convolution,
l=t
x 3 (t) = x1(t) * x 2 (t) =
z
l=t
x1(l ) x 2 (t - l ) dl =
l=0
–2t
b) Given that, x1(t) = e
x 2(t) = e
Let,
–5t
u(t) = e
l=0
t
= 2l
0
= 2 t-0
–2t
; t³0
–5t
; t³0
u(t) = e
z
l=t
2 ´ 1 dl = 2
= 2t
dl
; for t ³ 0 = 2 t u( t)
x3(t) = x1(t) * x2(t)
Since x1(t) and x2(t) are causal,
the limits of integration is 0 to t.
By definition of convolution,
l=t
x 3 (t) = x1(t) * x 2 (t) =
z
l=0
z
l=t
x1(l ) x 2 ( t - l ) dl =
l=0
z
l=t
e -2 l e -5( t - l ) dl =
l=0
z
l=0
e - 2 l e - 5 t e5 l dl
Chapter 2 - Continuous Time Signals and Systems
2. 66
l=t
= e -5 t
LM e OP
N3Q
l=t
z
l=0
3l
z
e -2 l + 5 l dl = e -5 t
e3 l dl
= e -5 t
l=0
t
LM e
N3
3t
= e -5 t
0
-
e0
3
OP
Q
1 -2 t
1 -2 t
e -5 t 3 t
e -1 =
e - e -5 t ; for t ³ 0 =
e - e -5 t u( t)
=
3
3
3
d
i
d
i
d
i
c) Given that, x1(t) = t u(t) = t ; t ³ 0
x 2(t) = e–5t u(t) = e–5t ; t ³ 0
Let,
x3(t) = x1(t) * x2(t)
By definition of convolution,
Since x1(t) and x2(t) are causal,
the limits of integration is 0 to t.
l=t
x 3 (t) = x1(t) * x 2 (t) =
z
z
x1(l ) x2 ( t - l) dl
l=0
l=t
=
l=t
z
l e -5( t - l ) dl =
l=0
= e
z z z z
l=0
l=t
-5 t
l e -5 t e 5 l d l
z
l e
5l
dl = e
-5 t
l=0
LMl
MN
e5 l
5
z
LM1 ´ e OP dl OP
N 5 Q PQ
5l
uv = u v - du v
t
u = l
v = e5l
0
LM e - e OP = e LMt e - e - 0 ´ e + e OP
5
25 Q
N 5 25 Q
N 5 25
e
1
d5te - e + 1i ; for t ³ 0 = 25 de + 5t - 1i u(t)
25
5l
5l
t
5t
= e -5 t l
5t
0
0
-5 t
0
-5 t
=
5t
d) Given that, x1(t) = cos t u(t) = cos t
x 2(t) = t u(t)
Let,
=t
5t
-5 t
; t³0
;t³0
x3(t) = x1(t) * x2(t)
Since x1(t) and x2(t) are causal,
the limits of integration is 0 to t.
By definition of convolution,
z z z z
l=t
x 3 (t) = x1(t) * x 2 (t) =
z
z
uv = u v - du v
x1(l ) x 2 ( t - l ) dl
u = l
l=0
l=t
=
l=t
cos l ´ (t - l ) dl = t
l=0
= t sin l
z
cos l dl -
l=0
t
0
- l sin l -
z
v = cos l
l=t
z
l cos l dl
l=0
t
1 ´ sin l dl
= t sin l
0
t
0
- l sin l + cos l
t
0
= t sin t - sin 0 - t sin t + cos t - 0 ´ sin 0 - cos 0
= t sin t - 0 - t sin t - cos t + 0 + 1 = 1 - cos t
; for t ³ 0 = ( 1 - cos t ) u( t )
Example 2.21
Determine the unit step response of the following systems whose impulse responses are given below.
a) h(t) = 3t u(t)
b) h(t) = e–5t u(t)
d) h(t) = u(t - 2)
e) h(t) = u(t + 2) + u(t - 2)
c) h(t) = u(t + 2)
2. 67
Signals & Systems
Solution
a) Given that, h(t) = 3t u(t) = 3t
; t³0
l=t
z
Unit Step Response, s(t) =
l=t
h(l ) dl
l = -¥
2 t
LM l OP
N2Q
=3
b) Given that, h(t) = e–5t u(t) = e–5t
LM t
N2
2
=3
0
0
2
-
z
-5 t
z
=
e0
-5
OP
Q
-5 l
t
0
; t ³ -2
z
l=t
h( l ) dl
z
=
dl =
l
t
-2
=
t +2
l = -2
= t+2;
for t ³ -2 = ( t + 2) u( t + 2)
; t ³2
l=t
Unit Step Response, s(t) =
LM e OP
N -5 Q
e -5 l dl =
1
1
=
( 1 - e -5 t ); for t ³ 0 =
( 1 - e -5 t ) u( t)
5
5
l = -¥
d) Given that, h(t) = u(t - 2) = 1
3 2
3 2
t ; for t ³ 0 =
t u(t )
2
2
l=0
l=t
Unit Step Response, s(t) =
OP =
Q
l=t
h(l ) dl
LM e
N -5
c) Given that, h(t) = u(t + 2) = 1
l dl
l=0
; t³0
l = -¥
=
z
3l dl = 3
l=0
l=t
Unit Step Response, s(t) =
l=t
z
=
z
l=t
h(l ) dl
l = -¥
z
=
dl =
l
t
2
= t -2
l= 2
= t - 2 ; for t ³ 2 = (t - 2) u(t - 2)
e) Given that, h(t) = u(t + 2) + u(t - 2)
Let, h(t) = h1(t) + h2(t)
where, h1(t) = u(t+2) = 1
; t ³ -2
h2(t) = u(t-2) = 1
; t ³2
Unit step response, s(t) = h(t) * u(t) = [ h1(t) + h2(t) ] * u(t)
= [ h1(t) * u(t) ] + [ h2(t) * u(t) ] = s1(t) + s2(t)
where, s1(t) = h1(t) * u(t)
s2(t) = h2(t) * u(t)
l=t
s1(t) =
z
z
l=t
h1(l ) dl =
dl = l
l=t
l=t
h( l ) dl
l = -¥
t
-2
= t+2 = t+2
; for t > - 2
l = -2
l = -¥
s 2 (t) =
z
z
=
l= 2
l=t
dl =
z
dl
= l
t
2
= t - 2 = t - 2 ; for t > 2
l= 2
Now, Unit step response, s(t) = s1(t)
; for t = - 2 to 2
= s1(t) + s2(t) ; for t > 2
\ Unit step response, s(t) = t + 2
= t + 2 + t - 2 = 2t
; for t = - 2 to 2
; for t > 2
Chapter 2 - Continuous Time Signals and Systems
2. 68
Example 2.22
Perform convolution of the following signals, by graphical method.
b) x1(t) = e–at ; 0 £ t £ T,
a) x1(t) = e–3t u(t), x2(t) = t u(t)
x2(t) = 1 ; 0 £ t £ 2T
Solution
a) Given that, x1(t) = e–3t u(t) = e–3t ; t ³ 0
x 2(t) = t u(t)
=t
;t³0
x3(t) = x1(t) * x2(t)
Let,
By definition of convolution,
l = +¥
z
x 3 (t) =
x1(l) x 2 (t - l) dl
l = -¥
Let us change the time index t, in x1(t) and x2(t) to l, to get x1(l) and x2(l), and then fold x2(l) to get x2(–l) graphically
as shown below.
x1(l) = e–3l ­
x1(t) = e–3t ­
t ®l
®
Change of
index
1
1
®
0
®t
0
t®l
®
Change of
index
x2(–l) = –l
­
0
l ® –l
®
Fold
®
l
®
l
–l
Fig 5 : x2(–l).
Fig 4 : x2(l).
Fig 3 : x2(t).
­
­
x2(l) = l
­
l
Fig 2 : x1(l).
Fig 1 : x1(t).
x2(t) = t
®
0
t
Let us shift x2(–l) by t units of time to get x2(t – l) and then multiply with x1(l) as shown below.
x1(l) = e–3l
x1(l) = e–3l ­
­
1
1
0
–l
t
0
No overlap
region
®l
®
shift index
Fig 6 : x1(l) and x2(t – l) when time shift, t < 0.
­
®
l
x2(t – l) = t – l
­
®
®l
­
­
x2(t – l) = t – l ­
l
–l
–l
­
–l
­
®
0
overlapping region
­
®
0
0
t
®
shift index
Fig 7 : x1(l) and x2(t – l) when time shift, t > 0.
Signals & Systems
2. 69
From fig 6 and fig 7, it is observed that the product of x1(l) and x2(t – l) is non-zero only for time shift t > 0.
For any time shift t > 0, the non-zero product exists in the overlapping region shown in fig 7. Here the overlapping
region is l = 0 to l = t. Hence integration of x1(l) and x2(t – l) is performed from l = 0 to l = t.
l=t
z
\ x 3 ( l) =
=
x1(l) x 2 ( t - l) dl ; t ³ 0
l= 0
t
-3 l
z
e
t
( t - l) dl =
0
z
0
LM e
N -3
-3t
= t
-3 l
=
e -3t
1
t
+
9
3
9
=
Ft
GH 3
+
z
-3t
0
-
z
uv = u v - du v
l e -3 l dl
v = e -3 l
u = l
0
LMl e - 1 ´
N -3
e O
P - LMNt e-3 - e9
-3 Q
e -3 l d l -
z z z z
t
t e -3 l dl -
0
t
= t
z
t
e -3 l
dl
-3
OP
Q
- 0 +
e0
9
-3t
LM e OP
N -3 Q
-3 l
= t
0
OP =
Q
-
t
LMl e
N -3
-3 l
0
-
e -3l
9
OP
Q
t
0
t e -3t
e -3t
1
t e -3 t
t
+
+
+
9
3
3
3
9
; t ³ 0
I
JK
e -3t
1
1 -3t
u(t) =
e + 3t - 1 u(t)
9
9
9
d
i
b) Given that, x1(t) = e–at ; 0 £ t £ T
x2(t) = 1
; 0 £ t £ 2T
x3(t) = x1(t) * x2(t)
Let,
By definition of convolution,
l = +¥
x 3 (t) =
z
x1(l) x 2 (t - l) dl
l = -¥
Let us change the time index t in x1(t) and x2(t) to l, to get x1(l) and x2(l), and then fold x2(l) to get x2(–l) graphically
as shown below.
x1(t) = e–at ; t = 0 to T
­
1
1
t®l
®
Change of
index
0
®
T
x1(l) = e–al ; l= 0 to T
0
t
x2(t) = 1 ; t = 0 to 2T
1
­
t®l
®
Change of
index
0
T
®t
2T
Fig 3 : x2(t).
l
Fig 2 : x1(l).
Fig 1 : x1(t).
­
®
T
x2(l) = 1 ; l = 0 to 2T
1
0
x2(–l) ­
1
l ® –l
®
Fold
T
®l
2T
Fig 4 : x2(l).
­
­
–l
0
–2T
–T
®l
Fig 5 : x2(–l).
Chapter 2 - Continuous Time Signals and Systems
2. 70
Let us shift x2(–l) by t units of time to get x2(t – l) and then multiply with x1(l) as shown below.
x1(l)
x1(l)­
­
1
1
–T
–2T
x2(t – l)­
­
l
2T
T
–l
No overlap
region
1
–2T
Overlapping
region
­
­
–l
®
0
–T
®
0
T
2T
l
x2(t – l)
­1
t – 2T
t
2T
0
®l
–l
®
shift index
T
l
2T
x2(t – l)
1­
®
0
T
–T
t – 2T
–l
0
2T
t
®l
®
shift index
Fig 8 : x1(l) and x2(t – l)
when time shift,
t = T to 2T.
®
0
T
l
3T
2T
®
t
shift index
2T
T
4T
3T
No overlap
region
­
­
®l
0
x2(t – l)
x2(t – l)
1
1
®
®
0
0
T
t – 2T
­
0
–T
­
Overlapping region
–l
Overlapping region
®
®
l
1
1
®
0
2T
x1(l)­
x1(l)­
1
T
–T
Fig 7 : x1(l) and x2(t – l) when time shift, t = 0 to T.
Fig 6 : x1(l) and x2(t – l) when time shift, t < 0.
x1(l)­
®
0
–2T
3T
2T
t
l
®
shift index
Fig 9 : x1(l) and x2(t –l)
when time shift,
t = 2T to 3T.
0
0
T
2T
t – 2T
T
–T
t – 2T
–2T
­
­
­
0
–l
®l
3T
4T
t
®
shift index
Fig 10 : x1(l) and x2(t – l)
when time shift,
t > 3T.
From fig 6 to fig 10, it is observed that the product of x1(l) and x2(t – l) is non-zero only for time shift t = 0 to 3T. For
any time shift in the range t = 0 to 3T, the non-zero product exists in the overlapping regions shown in fig 7, fig 8 and fig 9.
Signals & Systems
2. 71
For t = 0 to T
In this interval, the overlapping region is l = 0 to l = t. Hence the integration of the product of
x1(l) and x2(t – l) is performed from l = 0 to l = t.
:
l=t
z
\ x 3 (t) =
t
x1(l) x 2 ( t - l) dl =
l= 0
- at
e -al ´ 1 dl =
0
LM e OP
N -a Q
- al
t
0
e0
1
1
1
= - e -at +
=
(1 - e -at )
a
a
a
-a
e
=
-a
For t = T to 2T
z
In this interval, the overlapping region is l = 0 to l = T. Hence the integration of the product of
x1(l) and x2(t – l) is performed from l = 0 to l = T.
:
l=T
z
\ x 3 (t) =
T
x1(l) x 2 ( t - l) dl =
l= 0
- aT
z
e -al ´ 1 dl =
0
LM e OP
N -a Q
- al
T
0
1
1
e0
1
(1 - e -aT )
=
= - e -aT +
a
a
a
-a
e
=
-a
In this interval, the overlapping region is l = t – 2T to l = T. Hence the integration of the product
of x1(l) and x2(t – l) is performed from l = t – 2T to l = T.
For t = 2T to 3T :
l=T
z
\ x 3 ( t) =
T
l = t - 2T
- aT
e
=
-a
z
x1(l) x 2 ( t - l) dl =
e - a l ´ 1 dl =
t - 2T
LM e OP
N -a Q
- al
T
t - 2T
e -a(t - 2T)
1 -a ( t - 2T )
1 -a( t - 2T)
1
= - e - aT +
e
=
(e
- e -aT )
a
a
a
-a
\ x 3 ( t) = 0
;
t < 0
;
0 < t < T
=
1
(1 - e - at )
a
=
1
(1 - e - aT )
a
=
1 - a(t - 2T)
(e
- e - aT ) ; 2T < t < 3T
a
; T < t < 2T
= 0
;
t > 3T
Example 2.23
Perform convolution of the following signals, by graphical method and sketch the resultant signal.
a)
x2(t) ­
x1(t) ­
1
0
1
1
2
®t
0
1
2
®t
–1
b)
x1(t)
x2(t)
­
0
­
1
1
®
1
2
3
t
®
0
1
2
3
t
Chapter 2 - Continuous Time Signals and Systems
2. 72
Solution
a)
x3(t) = x1(t) * x2(t)
Let,
By definition of convolution,
l = +¥
z
x 3 (t) =
x1(l) x 2 (t - l) dl
l = -¥
Let us change the time index t in x1(t) and x2(t) to l, to get x1(l) and x2(l), and then fold x2(l) to get x2(–l) graphically
as shown below.
x1(t) ­
t®l
®
Change of
index
1
x1(l)
®
0
1
®
0
t
2
­
1
1
Fig 2 : x1(l).
Fig 1 : x1(t).
x2(l) ­
x2(t) ­
t ®l
®
Change of
index
1
0
l
2
1
®
2
t
x2(–l) ­
l ® –l
®
Fold
1
0
1
2
®
l
1
®
–l
®
–2
–1
–1
–1
l
–1
Fig 5 : x2(–l).
Fig 4 : x2(l).
Fig 3 : x2(t).
0
Let us shift x2(–l) by t units of time to get x2(t – l) and then multiply with x1(l) as shown below.
x1(l)
­
–2
–1
®
0
1
x2(t – l)
l
2
®
®
–l –3
®
–l
No overlap
region
­
0
–1
–2
1
–3
®
–2
–1
0
1
2
l
®
®
–l
–2
0
–1
1
Fig 6 : x1(l) and x2(t – l) when time shift, t < 0.
t–1
shift index
t–2
t–1
2
®
l
®
®
t–2
l
2
–1
–1
t 0
®
1
x2(t – l) ­
1
®
–l
­
1
1
Overlapping
region
x1(l)
0 t
shift index
Fig 7 : x1(l) and x2(t – l) when time shift, t = 0 to 1.
Signals & Systems
2. 73
x1(l) ­
x1(l) ­
1
1
®
®
0
x2(t – l)
­
1
1
1
0
l
2
1
2
3
2
3
®
®
®1
®
t
0
shift index
x1(l)
x1(l)
­
Overlapping
region
t
shift index
Fig 9 : x1(l) and x2(t – l) when time shift,
t = 2 to 3.
Fig 8 : x1(l) and x2(t – l) when time shift,
t = 1 to 2.
­
1
®
1
t–2
t–1
t–2
0
t–1
®
0
l
–1
–1
1
l
x2(t – l) ­
® ®
0
–1
®
0
l
2
Overlapping
region
1
Overlapping
region
–1
®
2
®l
3
0
4
x2(t – l) ­
®
1
3
2
5
4
No overlap
region
x2(t – l) ­
1
l
1
®
®
0
1
2
4
3
®
l
0
–1
1
2
4
3
5
l
–1
®
0
shift index
Fig 10 : x1(l) and x2(t – l) when time shift,
t = 3 to 4.
t–1
t
t–2
t–2
t–1
®
0
t
shift index
Fig 11 : x1(l) and x2(t – l) when time shift,
t > 4.
From fig 6 to fig 11, it is observed that the product of x1(l) and x2(t – l) is non-zero only for time shift t = 0 to 4. For
any time shift in the range t = 0 to 4, the non-zero product exists in the overlapping regions shown in fig 7 to fig 10.
For t = 0 to 1
: In this interval, the overlapping region is l = 0 to l = t. Hence the integration of the product of x1(l)
and x2(t – l) is performed from l = 0 to l = t.
l=t
\ x 3 (t) =
z
t
x1( l) x 2 ( t - l) dl =
l= 0
z
0
1 ´ 1 dl = l
t
0
= t
Chapter 2 - Continuous Time Signals and Systems
For t = 1 to 2
2. 74
: In this interval, the overlapping region is l = 0 to l = t. Hence the integration of the product of x1(l)
and x2(t – l) is performed from l = 0 to l = t.
l=t
z
\ x 3 (t) =
t -1
x1( l) x 2 ( t - l ) dl =
l= 0
z
t
1 ´ ( -1) d l +
z
1 ´ 1 dl = - l
t -1
+ l
0
t
t -1
t-1
0
= - ( t - 1) + 0 + t - (t - 1) = - t + 1 + t - t + 1 = 2 - t
For t = 2 to 3
: In this interval, the overlapping region is l = t – 2 to l = 2. Hence the integration of the product of
x1(l) and x2(t – l) is performed from l = t – 2 to l = 2.
l=2
z
\ x 3 (t) =
t-1
x1(l ) x 2 ( t - l ) dl =
l=t -2
z
2
z
1 ´ ( -1) d l +
t -2
1 ´ 1 dl = -l
t -1
t -2
2
+ l
t -1
t -1
= - ( t - 1) + (t - 2) + 2 - (t - 1) = - t + 1 + t - 2 + 2 - t + 1 = 2 - t
For t = 3 to 4
: In this interval, the overlapping region is l = t – 2 to l = 2. Hence the integration of the product of
x1(l) and x2(t – l) is performed from l = t – 2 to l = 2.
l=2
z
\ x 3 (t) =
2
x1(l ) x 2 ( t - l ) dl =
l=t -2
z
1 ´ ( -1) d l = - l
2
t -2
t -2
= - 2 + ( t - 2) = - 2 + t - 2 = t - 4
=t
At t = 0, x3(0) = 0
1
At t = 1, x3(1) = t = 1
x3(t)
;0<t<1
0
=2–t;1<t<3
=t–4;3<t<4
=0
b)
®
The result of the convolution of x1(t) with
x2(t) is given below and the resultant waveform is
shown in fig 12.
x1(t) * x 2(t) = x3(t) = 0 ; t < 0
1
2
®
At t = 2, x3(2) = 2 – t = 2 – 2 = 0
t
At t = 3, x3(3) = 2 – t = 2 – 3 = –1
4
3
At t = 4, x3(4) = t – 4 = 4 – 4 = 0
–1
Fig 12 : x3(t).
;t>4
x 3(t) = x1(t) * x2(t)
Let,
By definition of convolution,
l = +¥
z
x 3 (t) =
x1(l) x 2 (t - l) dl
l = -¥
Let us change the time index t in x1(t) and x2(t) to l, to get x1(l) and x2(l), and then fold x2(l) to get x2(–l) graphically
as shown below.
x1(t)
­
t®l
®
Change of
index
1
x1(l)­
1
®
0
1
2
1
­
x2(l)
1
®
0
1
2
Fig 3 : x2(t).
3
t
2
l
3
Fig 2 : x1(l).
Fig 1 : x1(t).
x2(t)
®
0
t
3
t ®l
®
Change of
index
x2(–l)
­
0
­
1
l ® –l
1
®
1
2
Fig 4 : x2(l) .
3
l
®
Fold
®
–l
®
–3
–2
–1
Fig 5 : x2(–l).
0
l
Signals & Systems
2. 75
Let us shift x2(–l) by t units of time to get x2(t – l) and then multiply with x1(l) as shown below.
x1(l)
x1(l)
1
1
­
­
0
1
2
0
–1
–2
t–2
t
3
2
1
®l
–l
–3
0
t
­
®
¬
–2
0
–1
2
1
3
4
®l
Overlapping region
1
x1(l ) x 2 (t - l ) d l
1 ´ 1 dl = l
shift index
1
x2(t – l) ­
l=1
t-2
l
x1(l)
–l
z
1
3
2
Fig 7 : x1(l) and x2(t – l)
when time shift, t > 0.
l=t -2
=
0
–1
–2
shift index
For t > 3 :
z
No overlap
for 0 < t < 3
®
From fig 6, fig 7 and fig 8, it is observed that the product
of x1(l) and x2(t – l) is non-zero only for time shift t ³ 3. For any
time shift t > 3, the non-zero product exists in the overlapping
region shown in fig 8.
=
®l
®
®
Fig 6 : x1(l) and x2(t – l)
when time shift, t < 0.
x 3 (t)
3
2
t>0
1
®
0
1
x2(t – l) ­
t–2
–3
0
No overlap
for t < 0
1
®
®l
t<0
x2(t – l) ­
–l
3
t>3
®
¬
t -2
–l
1
–2
0
–1
1
2
0
t–2
1
= t - 2 - 1
= t - 3 ; t ³ 3
3
4
®l
®
t shift index
Fig 8 : x1(l) and x2(t – l) when time shift, t >3.
The result of the convolution of x1(t) with x2(t) is given
below and the resultant waveform is shown in fig 9.
x1(t) * x 2 (t) = x 3 (t) = t - 3
;t ³ 3
x3(t) ­
4
At t = 3, x3(t) = t – 3 = 3 – 3 = 0
3
At t = 4, x3(t) = t – 3 = 4 – 3 = 1
2
At t = 5, x3(t) = t – 3 = 5 – 3 = 2
1
At t = 6, x3(t) = t – 3 = 6 – 3 = 3
0
1
2
3
4
5
Fig 9 : x3( t ).
6
7
®
t
At t = 7, x3(t) = t – 3 = 7 – 3 = 4
Chapter 2 - Continuous Time Signals and Systems
2. 76
2.10 Inverse System and Deconvolution
Inverse System
The inverse system is used to recover the input from the response of a system. For a given system,
the inverse system exists, if distinct inputs to a system leads to distinct outputs. The inverse systems exist
for all LTI systems. Consider an amplifier with gain “A”. When we pass the output of the amplifier
through an attenuator with gain “1/A” then the output of the attenuator will be same as that of input of the
amplifier. Therefore the attenuator with gain “1/A” is an inverse system for the amplifier with gain “A”.
The inverse system is denoted by H–1. If x(t) is the input and y(t) is the output of a system, then
y(t) is the input and x(t) is the output of its inverse system.
x(t)
®
H
y(t)
y(t)
®
Fig 2.46a : System.
®H
-1
w(t) = x(t)
®
Fig 2.46b : Inverse system.
Fig 2.46 : A system and its inverse system.
Let h(t) be the impulse response of a system and h'(t) be the impulse response of inverse system.
Let us connect the system and its inverse in cascade as shown in fig 2.47.
Identity system
H
x(t)
® h(t)
y(t)
H –1
® h'(t)
® w(t) = x(t)
Fig 2.47 : Cascade connection of a system and its inverse.
Now it can be proved that,
h(t) * h'(t) = d(t)
.....(2.43)
Therefore the cascade of a system and its inverse is the identity system.
Proof :
With reference to fig 2.47 we can write,
y(t) = x(t) * h(t)
.....(2.44)
w(t) = y(t) * h'(t)
.....(2.45)
On substituting for y(t) from equation (2.44) in equation (2.45) we get,
w(t) = x(t) * h(t) * h'(t)
.....(2.46)
In equation (2.46), if h(t) * h'(t) = d(t) then x(t) * d(t) = x(t)
In a inverse system, w(t) = x(t), and so,
h(t) * h'(t) = d(t). Hence proved.
Deconvolution
In an LTI system the response y(t) is given by convolution of input x(t) and impulse response h(t).
i.e., y(t) = x(t) * h(t)
The process of recovering the input signal from the response of a system is called deconvolution.
Alternatively, the process of recovering x(t) from y(t), where y(t) = x(t)*h(t), is called deconvolution.
The deconvolution can be more conveniently handled in the Laplace domain (or s-domain), and so
solved problems on deconvolution are presented in Chapter-3. [Refer example 3.29]
2. 77
Signals & Systems
2.11 Summary of Important Concepts
1.
An analog signal is a continuous function of an independent variable.
2.
When the independent variable of an analog signal is time 't' then the analog signal is called continuous
time (CT) signal.
3.
In a continuous time signal the magnitude and the independent variable are continuous.
4.
The sinusoidal and complex exponential signals are always periodic.
5.
The sum of two periodic signals is also periodic if the ratio of their fundamental periods is a rational
number.
6.
A signal which is neither even nor odd can be expressed as a sum of even and odd signals.
7.
Periodic signals are power signals and nonperiodic signals are energy signals.
8.
The power of an energy signal is zero and the energy of a power signal is infinite.
9.
The causal signals are defined only for t ³ 0.
10. The noncausal signals are defined for either t £ 0 or all t.
11. Ideally, an impulse signal is a signal with infinite magnitude and zero duration.
12. Practically, an impulse signal is a signal with large magnitude and short duration.
13. The homogenous solution is the response of a system when there is no input, whereas the particular
solution is the response for specific input.
14. The free or natural response is the response of the system due to initial stored energy, whereas the forced
response is the response due to a particular input when there is no initial energy.
15. The response (or total response) of an LTI system is the sum of natural and forced response.
16. The response of a static system depends on present input whereas, the response of a dynamic system
depends on present, past and future inputs.
17. The dynamic systems require memory whereas the static systems do not require memory.
18. In time invariant systems the input-output characteristics do not change with time.
19. In time invariant systems, if a delay is introduced either at input or at output, the response remains same.
20. Linear systems will satisfy the principle of superposition.
21. In linear systems, the response for weighted sum of inputs is equal to similar weighted sum of individual
responses.
22. In causal systems, the response depends only on past and present values of inputs/outputs.
23. In noncausal systems, the response depends on future inputs/outputs.
24. In stable systems, for any bounded input, the output will be bounded.
25. The present output of a feedback system depends on past outputs.
26. The response of an LTI system is given by convolution of input and impulse response.
27. The unit step response of an LTI system is given by the integral of its impulse response.
28. The convolution operation satisfies the commutative, associative and distributive properties.
29. When LTI systems are connected in cascade, the overall impulse response is given by the convolution of
individual impulse responses.
30. When LTI systems are connected in parallel, the overall impulse response is given by the sum of its
individual impulse responses.
31. The inverse system is used to recover the input from the response of a system.
32. The cascade of a system and its inverse is the identity system.
33. The deconvolution is the process of recovering input from the response of a system.
Chapter 2 - Continuous Time Signals and Systems
2.78
2.12 Short Questions and Answers
Q2.1
Prove that sinc (0) = 1.
Proof
sin c ( 0) = Lt sin c ( t) = Lt
t® 0
Q2.2
t ®0
sin t
=1
t
bUsing L'Hospital' s ruleg
bg
Consider the complex valued exponential signal x t = Ae
a t + jW t
, a > 0 . Evaluate the real and
imaginary components of x(t) for the following cases.
i) a real , a = a
Solution
ii) a imaginary , a = j W
1
iii) a complex, a = a 1 + jW
1
1
case i : a = a1
b
x(t ) = Ae at + jWt = Aea 1t + jWt = Ae a1t e jWt = Aea 1t cos Wt + j sin Wt
\ Real part of x( t) = Ae
a1t
g
Imaginary part of x( t ) = Aea 1t sin Wt
cos Wt ;
case ii : a + jW1
x(t ) = Ae at + jWt = Ae jW1t + jWt = Ae
b
b
g
j W1 + W t
d b
g
b
gi
= A cos W1 + W t + j sin W1 + W t
g
b
\ Real part of x( t) = A cos W1 + W t ;
g
Imaginary part of x(t ) = A sin W1 + W t
case iii : a = a 1 + jW1
x(t ) = Aeat + jWt = Ae( a1 + jW1)t + jWt = Ae a1t + j( W1+ W) t = Ae a1t e b
g
j W1 + W t
d b
g
b
gi
= Aea 1t cos W1 + W t + j sin W1 + W t
\ Real part of x( t) = Ae
Q2.3
a1t
b
g
b
g
Imaginary part of x( t) = Ae a1t sin W1 + W t
cos W1 + W t ;
Determine whether the signal, x( t ) = 3 cos 2 t + 7 cos 5p t is periodic.
Solution
Given that x(t ) = 3 cos 2 t + 7 cos 5p t
bg
Let, x1 t = 3 cos 2 t . Let T1 be period of x1(t). On comparing x1(t) with standard form A cos 2pF01t we get,
2 pF01 = 2
Þ
F01 =
2
,
2p
\ T1 =
1
2p
=
F01
2
bg
Let, x 2 t = 7 cos 5pt . Let T2 be period of x2(t). On comparing x2(t) with standard form A cos 2pF02t we get,
2 pF02 = 5p
Now,
Q2.4
Þ
F02 =
5
5p
=
2
2p
,
T1
5p
1
2p 5
= T1 ´
=
´ =
.
T2
T2
2
2 2
\ T2 =
-j
2p
t
3 .
bg
Here T1 T2 is not a rational number and so x t is nonperiodic.
Determine the period of the signal x( t ) = 0 .1 e
Solution
1
2
=
F02 5
-j
2p
t
3
+ 0 .3 sin p t .
Let T1 be period of x1(t). On comparing x1(t) with standard form A e- j2pF01t we get,
2p
1
1
Þ
F01 =
\ T1 =
=3
2pF01 =
F01
3
3
Let, x2(t) = 0.3 sinpt. Let T2 be period of x2(t). On comparing x2(t) with standard form A sin2pF02 t we get,
1
1
Þ
F02 =
\ T2 =
=2
2 pF02 = p
F02
2
Let, x1( t) = 0.1 e
2. 79
Signals & Systems
Here,
T1 3
= = Rational number.
T2 2
\x(t) is periodic.
\T = LCM of 2 and 3 = 2 ´ 3 = 6.
Let T be the period of x(t), which is given by LCM of T1 and T2.
Q2.5
Determine the even and odd parts of a complex exponential signal.
Solution
bg
b
g
Complex exponential signal, x t = Ae jW0 t = A cos W0 t + j sin W0 t = A cos W0 t + jA sin W0 t
b g
b g
bg b g
bg b g
b g
Now, x - t = A cos W0 - t + jA sin W0 - t = A cos W0 t - jA sin W0 t
1
Even part, x e t = x t + x - t = A cos W0 t
2
1
Odd part, x o t = x t - x - t = jA sin W0 t
2
bg
bg
Q2.6
Sketch the even parts and odd parts of a unit step signal.
Solution
–u(–t)
u(– t)
u(t)
1
0
1
t
–1
0
0
t
bg
bg b g
bg
Even part , ue t = 0.5 u t + u - t
b g
t
bg
= 0.5 u t + 0.5 u - t
0.5
0.5
0
0
t
bg
Unit step signal,
T
z bg
=1; t³ 0
=0; t<0
ut
T
2
x t dt =
-T
T
z bg z
z bg
z bg
2
u t dt =
0
T
T ®¥
Power, P = Lt
T ®¥
T
12 dt
=
0
Energy, E = Lt
Q2.8
–0.5
Determine the energy and power of a unit step signal.
Solution
\
bg
uo(t)
ue(t)
Q2.7
bg b g
z
dt =
t
T
0
= T-0 = T
0
2
x t dt = Lt T = ¥
-T
1
2T
T
T ®¥
2
x t dt =
-T
Lt
T ®¥
b g
Odd part , uo t = 0.5 u t - u - t = 0.5u t - 0.5u - t
1
1
watts
´T =
2T
2
Compare energy and power signals.
Energy Signal
1. Energy of the signal is constant.
Power Signal
1. Energy of the signal is infinite.
2. Power of the signal is zero.
3. Nonperiodic signals.
2. Power of the signal is constant.
3. Periodic signals.
t
Chapter 2 - Continuous Time Signals and Systems
Q2.9
2.80
Differentiate between causal and noncausal signals.
The causal signals are defined only for t ³ 0, whereas the noncausal signals are defined for either t £ 0
or for all t (i.e, for both t £ 0 and t > 0).
Q2.10
Sketch the signal, x(t) = 2 u(t) + t u(t) – (t – 1) u(t – 1) –3 u(t – 2)
2u(t)
Solution
2
1
+
0
3
2
1
t
t u(t)
2
+
x(t)
1
3
0
3
2
1
t
(t–1)u(t–1)
2
2
1
–
1
0
2
1
3
3u(t–2)
0
3
2
1
t
t
3
2
–
1
0
Q2.11
1
3
2
b g FGH
t
IJ b g
K
t -1
+P t-1
Sketch the signal, x t = P
2
Solution
x(t)
2
P(t)
P(t–1)
1
1
c h
P
1
Q2.12
0.5
1
Time scale
(Time expansion)
delay
–0.5
t -1
2
0
t
0.5
1.5
0 0.5 1.0 1.5 2.0 2.5 3.0
t
t
0
0.5 1.0 1.5 2.0 2.5 3.0
x(t)
A continuous time signal is shown in fig Q 2.12.
Sketch the following versions of the signal.
g b g
g
a x t-3
bg
b -2x t
g b g
bg
c x t-3 -2x t
1
d
g d xdtbt g
0
–1
Solution
1
Fig Q : 2.12.
b)
a)
–2 x(t)
–1
x (t–3)
1
0
1
0
1
2
3
4
t
t
–2
t
t
2. 81
Signals & Systems
b g b g
b g b g
b g b g
c)
d) x(t ) = u t + 1 - u t - 1
d
d
d
\ x( t) =
u t +1 - u t -1
dt
dt
dt
= d t +1 - d t -1
x(t–3) – 2 x(t)
1
–1
0
2
1
3
4
dx( t )
¥ dt
t
–1
d
u( t) = d( t)
dt
Using time
invariant property
d
u t ±k =d t ±k
dt
b g b g
–2
–1
1
0
t
¥
Q2.13
A continuous time signal is shown in fig Q 2.13.
Find the following versions of the signal.
x (t)
1
a) x(– t)
b) – x(t)
0
–1
2
1
Solution
–1
a)
b)
x (–t)
–x (t)
1
–2
1
0
–1
1
0
–1
t
–1
Q2.14
t
Fig Q : 2.13.
2
1
t
–1
x(t)
A continuous time signal is shown in fig Q 2.14 .
Find the following versions of the signal.
Comment on the result.
a) x(t–2)
b) x(–t)
c) x(–t +2)
2
1
Fig Q : 2.14.
Solution
a)
–3
b)
x(t–2)
2
0
c)
x(–t)
x(–t+2)
2
1
1
1
2
t
t
0
2
1
–1
–1
–2
0
1
2
3
t
–2
–1
0
1
t
Comment : The folded version of a delayed signal will be same as advance of its folded version
(i.e., delay and fold is equal to fold and advance).
Chapter 2 - Continuous Time Signals and Systems
Q2.15
2.82
x(t)
A continuous time signal is shown in fig Q 2.15.
Find the following versions of the signal.
Comment on the result.
a) x(t +2)
b) x(–t)
c) x(–t – 2)
2
1
Fig Q : 2.15.
Solution
0
x(t+2)
a)
x(–t)
b)
2
1
t
x(–t–2)
c)
2
2
2
3
1
1
1
–2
0
–1
1
–3
t
–2
0
–1
0
–1
t
2
1
t
Comment : The folded version of an advanced signal will be same as delay of its folded version
(i.e., advance and fold is equal to fold and delay).
Q2.16
Evaluate the following integrals,
¥
¥
i
g z d b t + 3g e
-t
dt
ii
z
g
bg
5
b g
d t cos t + d t - 1 sin t dt
iii
-¥
-¥
g z dbt g sin 2pt dt
0
Solution
¥
Using Property of impulse
¥
z
z bg
z
b g
d t + 3 e-t dt =
i)
-¥
c b gh
e - t d t - -3 dt = e - t
= 20.0855
-¥
¥
ii)
t = -3
e3
=
¥
b g
d t cos t + d t - 1 sin t dt =
-¥
z
-¥
z
b g
sin t d t - 1 dt = cos t t =0 + sin t t =1
Using Property of impulse
z bg bg
bg
x t d t dt = x 0
d(t) sin2pt dt = sin2pt
= sin 0 = 0
0
Q2.17
b g
-¥
5
z
g
x t d t - t 0 dt = x t 0
¥
bg
cos t d t dt +
= cos 0 + sin1 = 1+ 0.8415 = 1.8415
iii)
z bg b
t=0
Determine natural response of the first order system governed by the equation,
dy t
+ 3y t = x t ; y 0 = 2
dt
Solution
bg
bg bg bg
Homogeneous Solution, yh(t)
Put, x(t) = 0 in the given equation.
\
bg
dy t
dt
bg
Let, y h t
bg
+ 3y t = 0
= Ce lt ;
\
..... (1)
bg
dyh t
dt
= C l e lt
bg
C b l + 3ge
On substituting the above terms in equation 1 we get,
Cle
lt
lt
+ 3C e = 0
Þ
lt
=0
2. 83
Signals & Systems
The characteristic equation is, l + 3 = 0
bg
lt
\ yh t = C e = C e
l = -3
Þ
-3 t
Natural Response
b
g bg
bg
Natural response or zero input response , y zi t
= yh t
with constants evaluated using initial conditions
= C e-3t
At t = 0, y zi (0) = C e0 = C
Given that,
y(0) = 2, \ C = 2
bg
\ Natural response, y zi t = 2 e -3t ; t ³ 0
Q2.18
bg
bg
y zi t = 2e -3tu t
Þ
Determine the unit step response of the first order system governed by the equation,
dy( t )
+ 0 .5 y( t ) = x( t ), with zero intial conditions.
dt
Solution
Homogeneous Solution, yh(t)
Put, x(t) = 0 in the given equation.
dy( t )
+ 0.5y( t) = 0
dt
dy t
Let, y h t = C e lt ;
\ h
= C l e lt
dt
On substituting the above terms in equation 1 we get,
bg
..... 1
\
bg
bg
bg
Cle
lt
lt
+ 0.5C e = 0
Þ
b
g
C l + 0.5 e lt = 0
The characteristic polynomial is, l + 0.5 = 0
bg
\ yh t = C e -0.5 t ; t ³ 0
Þ
\ l = -0.5
bg
bg
y h t = C e -0.5 t u t
Particular Solution, yp (t)
Here, x(t) = u(t)
dyp (t )
= k d( t)
dt
On substituting the above terms in the given equation we get,
Let,
yp (t) = k u(t),
\
K d( t) + 0.5 Ku( t) = u( t)
At t = 1,
K d(1) + 0.5 K u(1) = u(1)
Þ
K ´ 0 + 0.5 K ´ 1 = 1
Þ
\K =
1
=2
0.5
\ yp (t ) = 2 u( t)
Unit Step Response (or Total Response)
Unit step response (or Total Response),
At t = 0,
Here, y(0) = 0,
y(t) = yh ( t) + y p ( t) = Ce -0.5tu(t ) + 2 u(t )
y(0) = Ce0 + 2 = C + 2
\C + 2 = 0
Þ
C = -2
\ Unit step response, y(t) = Ce -0.5tu(t ) + 2 u(t ) = - 2e-0.5tu( t) + 2 u(t ) = 2 (1 - e -0.5t ) u(t )
Q2.19
-
If two LTI systems with impulse responses, h1 ( t ) = e - at u( t ) and h2 ( t ) = e bt u( t ) are
connected in cascade, what will be the overall impulse response of cascaded system?
Solution
When LTI systems are in cascade, the overall impulse response is given by convolution of individual impulse
responses.
Chapter 2 - Continuous Time Signals and Systems
Let,
Now,
bg
h b tg = h b t g * h b tg
2.84
ho t = Overall impulse response of cascaded system.
o
1
2
t
=
z
t
bg b g
h1 l h2 t - l dl =
l =0
z
t
e - al e - bte bl dl = e - bt
l =0
=
Q2.20
e - al e
b g dl
-b t - l
l =0
t
=
z
z
e
dl = e - bt
l =0
1
- e-at + e-bt
a-b
;t ³ 0
LM e b g OP
MN -ba - bg PQ
- a-b l
b g
- a -b l
t
LM e b g - e OP
MN -ba - bg -ba - bg PQ
- a -b t
= e - bt
0
0
1
- e - at + e -bt u(t)
a-b
=
® h (t)
®
Find the overall impulse response of the system shown in fig Q2.20.
Take, h1(t) = t u(t); h2(t) = 3 u(t); h3(t) = 2 u(t).
1
®
+
®
Solution
® h2(t)
t
t
z
z
bg b g
h1 l h3 t - l d l +
l =0
z
bg b g
h2 l h 3 t - l d l
l =0
t
=
3
Fig Q : 2.20.
Overall impulse response,ho (t) = h1(t ) + h2 ( t) * h3 (t ) = h1(t ) * h3 (t ) + h 2 ( t) * h3 ( t)
=
® h (t)®
t
l ´ 2 dl +
l =0
z
t
z
3 ´ 2 d l = 2 l dl + 6
l =0
Lt O
2 M - 0P + 6
N2 Q
l =0
t
z
LM l OP
N2Q
2
dl = 2
l =0
t
+6 l
t
0
0
2
=
Q2.21
2
t - 0 = t + 6t ; t ³ 0 =
2
t + 6t u(t)
What is the importance of convolution?
The convolution operation can be used to determine the response of an LTI system for any input
from the knowledge of its impulse response. [The response of an LTI system is given by convolution
of input and impulse response of the LTI system].
2.13 MATLAB Programs
Program
2.1
Write a MATLAB program to generate standard signals like unit impulse,
unit step, unit ramp, parabolic, sinusoidal, triangular pulse, signum, sinc
and Gaussian signals.
%*******************
tmin=-5; dt=0.1;
t=tmin:dt:tmax;
program
to
plot
some
standard
signals
tmax=5;
%set
a
tim e
vecto r
%******************* unit impulse signal
x1=1;
x2=0;
x=x1.*(t==0)+x2.*(t~=0);
%generate unit impulse signal
subplot(3,3,1);plot(t,x);
%plot the generated unit impulse
xlabel(‘t’);ylabel(‘x(t)’);title(‘unit impulse signal’);
%******************* unit step signal
x1=1;
x2=0;
x=x1.*(t>=0)+x2.*(t<0);
%generate unit step signal
subplot(3,3,2);plot(t,x);
%plot the generated unit step
xlabel(‘t’);ylabel(‘x(t)’);title(‘unit step signal’);
%*******************
unit
ramp
signal
signal
signal
2. 85
Signals & Systems
x1=t;
x2=0;
x=x1.*(t>=0)+x2.*(t<0);
%generate unit ramp signal
subplot(3,3,3);plot(t,x);
%plot the generated unit ramp
xlabel(‘t’);ylabel(‘x(t)’);title(‘unit ramp signal’);
%******************* parabolic signal
A=0.4;
x1=(A*(t.^2))/2;
x2=0;
x=x1.*(t>=0)+x2.*(t<0);
%generate parabolic signal
subplot(3,3,4);plot(t,x);
%plot the generated parabolic
xlabel(‘t’);ylabel(‘x(t)’);title(‘parabolic signal’);
%******************* sinusoidal signal
T=2;
%declare time period
F=1/T;
%compute frequency
x=sin(2*pi*F*t);
%generate sinusoidal signal
subplot(3,3,5);plot(t,x);
%plot the generated sinusoidal
xlabel(‘t’);ylabel(‘x(t)’);title(‘sinusoidal signal’);
signal
signal
signal
%******************* triangular pulse signal
a=2;
x1=1-abs(t)/a;
x2=0;
x=x1.*(abs(t)<=a)+x2.*(abs(t)>a); %generate triangular pulse signal
subplot(3,3,6);plot(t,x);
%plo t the tr ia ng ula r pulse sig na l
xlabel(‘t’);ylabel(‘x(t)’);title(‘triangular pulse signal’);
%*******************
signum
signal
Fig P2.1 : Output waveforms of program 2.1.
Chapter 2 - Continuous Time Signals and Systems
2.86
x1=1;
x2=0;
x3=-1;
x=x1.*(t>0)+x2.*(t==0)+x3.*(t<0); %generate signum signal
subplot(3,3,7);plot(t,x);
%plot the generated signum
xlabel(‘t’);ylabel(‘x(t)’);title(‘signum signal’);
%******************* sinc Pulse
x=sinc(t);
%generate sinc Pulse
subplot(3,3,8);plot(t,x);
%plot the generated sinc
xlabel(‘t’);ylabel(‘x(t)’);title(‘sinc signal’);
signal
Pulse
%******************* gaussian signal
a=2;
x=exp(-a.*(t.^2));
%generate gaussian signal
subplot(3,3,9);plot(t,x);
%plo t the g ener a ted g a ussia n
xlabel(‘t’);ylabel(‘x(t)’);title(‘gaussian signal’);
sig na l
OUTPUT
The output waveforms of program 2.1 are shown in fig P2.1.
Program
2.2
Write a MATLAB program to find the even and odd parts of the signal
x(t)=e2t.
%To
find
the
even
tmin=-3; tmax=3;
t=tmin:dt:tmax;
a nd
o dd
dt=.1;
%set
x1=exp(2*t);
x2=exp(-2*t);
ymin=min([min(x1),
ymax=max([max(x1),
tim e
%generate
%generate
if(x2==x1)
disp(‘“The given
else if (x2==(-x1))
disp(‘“The given
else
disp(‘“The given
end
end
xe=(x1+x2)/2;
xo=(x1-x2)/2;
a
pa r ts
of
the
the
the
given signal
time folded signal
is
even
signal
is
odd
signal
is
neither
min(x2),
max(x2),
x(t)=exp(2 *t)
vecto r
signal
%compute
%compute
sig na l,
signal”’);
signal”’);
even
nor
odd
signal”’);
even part
odd part
min(xe),
max(xe),
min(xo)]);
max(xo)]);
subplot(2,2,1);plot(t,x1);axis([tmin tmax
xlabel(‘t’);ylabel(‘x1(t)’);title(‘signal
ymin ymax]);
x(t)’);
subplot(2,2,2);plot(t,x2);axis([tmin tmax
xlabel(‘t’);ylabel(‘x2(t)’);title(‘signal
ymin ymax]);
x(-t)’);
subplot(2,2,3);plot(t,xe);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘xe(t)’);title(‘even part of x(t)’);
subplot(2,2,4);plot(t,xo);axis([tmin tmax ymin
xlabel(‘t’);ylabel(‘xo(t)’);title(‘odd part of
ymax]);
x(t)’);
2. 87
Signals & Systems
OUTPUT
"The given signal is neither even nor odd signal"
The input and output waveforms of the program 2.2 are shown in fig P2.2.
Fig P2.2 : Input and Output waveforms of program 2.2.
Program
2.3
Write a MATLAB program to find the energy and power of the signal
x(t)=10sin(10 Pt).
%Program to compute the
%x(t)=10sin(10*pi*t)
signal
tmax=10; dt=.01; T0=10;
t=-tmax:dt:tmax;
%set up
a
energy
tim e
a nd
power
of
the
signal
vecto r
x = 1 0 * s i n ( 1 0 * p i * t ) ; %generate the given signal x(t)
xsq=x.^2;
%co m pute the squa r e o f the g iven sig na l x(t)
E = trapz(t,xsq);
%use trapezoidal-rule numerical integration to
%find energy
P = E /(2*T0);
%divide the energy by period to compute power
disp([‘
disp([‘
Energy, E
=’,num2str(E),’ Joules’]);
Power, P =’,num2str(P),’ Watts’]);
OUTPUT
Energy, E
Power,
P
= 1000 Joules
= 50 Watts
Chapter 2 - Continuous Time Signals and Systems
Program
2.88
2.4
Write a MATLAB program to perform Amplitude scaling, Time scaling, and
Time shift on the signal x(t) = 1+t; for t = 0 to 2.
Program to declare the given signal as function y(t)
% decla r e the g iven sig na l a s
functio n x = y(t)
x=(1.0 + t).*(t>=0 & t<=2);
functio n
y(t)
Note: The above program should be stored as a separate file in the
current working directory
Program to perform amplitude & time scaling and time shift on y(t)
%To perform Amplitude scaling, Time scaling and Time shift
%on the signal x(t)=1.0+t; for t= 0 to 2
%include y. m file in wo r k dir ecto r y which decla r e the g iven
%as function y(t)
tmin=-3; tmax=5;
t=tm in:dt:tm a x;
y0 =y(t);
y1 =1.5*y(t);
y2 =0. 5*y(t);
y3=y(2*t);
y4=y(0.5*t);
y5=y(t-2);
y6=y(t+2);
dt=0.2;
%set a
tim e
sig na l
vecto r
%a ssig n the g iven sig na l a s y0
%compute the amplified version of x(t)
%co mpute the a ttenua ted ver sio n o f x(t)
%compute the time compressed version of x(t)
%compute the time expanded version of x(t)
%compute the delayed version of x(t)
%compute the advanced version of x(t)
%co m pute the m in a nd m a x va lue fo r y-a xis
ymin=min([min(y0), min(y1), min(y2), min(y3),
min(y5),min(y6)]);
ymax=max([max(y0), max(y1), max(y2), max(y3),
max(y5),max(y6)]);
min(y4),
max(y4),
%plot the given signal and amplitude scaled signal
subplot(3,3,1);plot(t,y0);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x(t)’);title(‘Signal x(t)’);
subplot(3,3,2);plot(t,y1);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x1(t)’);title(‘Amplified signal 1.5x(t)’);
subplot(3,3,3);plot(t,y2);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x2(t)’);title(‘Attenuated signal 0.5x(t)’);
%plo t the g iven sig na l a nd tim e sca led sig na l
subplot(3,3,4);plot(t,y0);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x(t)’);title(‘Signal x(t)’);
subplot(3,3,5);plot(t,y3);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x3(t)’);title(‘Time comp. signal x(2t)’);
subplot(3,3,6);plot(t,y4);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x4(t)’);title(‘Time expan. signal x(0.5t)’);
%plot the given signal and time shifted signal
subplot(3,3,7);plot(t,y0);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x(t)’);title(‘Signal x(t)’);
subplot(3,3,8);plot(t,y5);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x5(t)’);title(‘Delayed signal x(t-2)’);
subplot(3,3,9);plot(t,y6);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x6(t)’);title(‘Advanced signal x(t+2)’);
OUTPUT
The input and output waveforms of program 2.4 are shown in fig P2.4.
2. 89
Signals & Systems
Fig P2.4 : Iuput and Output waveforms of program 2.4.
Program
2.5
Write a MATLAB program to perform addition and multiplication on the
following two signals.
xa(t)=1; 0<t<1
xb(t)=t;
0<t<1
=2; 1<t<2
=1;
1<t<2
=1; 2<t<3
=3–t; 2<t<3
%To perform addition
%1) xa(t)=1;0<t<1
%
=2;1<t<2
%
=1;2<t<3
tmin=-1; tmax=5;
t=tmin:dt:tmax;
and
dt=0.1;
%Set a
multiplication of the
2) xb(t)=t;0<t<1
=1;1<t<2
=3-t;2<t<3
tim e
following
vecto r
x1=1;
x2=2;
x3=3-t;
xa=x1.*(t>0&t<1)+x2.*(t>=1&t<=2)+x1.*(t>2&t<3);
xb=t.*(t>0&t<1)+x1.*(t>=1&t<=2)+x3.*(t>2&t<3);
xadd=xa+xb;
%Add the two signals
xmul=xa.*xb;
%Multiply two signals
xmin=min([min(xa), min(xb), min(xadd), min(xmul)]);
xmax=max([max(xa), max(xb), max(xadd), max(xmul)]);
subplot(2,3,1);plot(t,xa);axis([tmin tmax xmin xmax]);
xlabel(‘t’);ylabel(‘xa(t)’);title(‘Signal xa(t)’);
subplot(2,3,2);plot(t,xb);axis([tmin tmax xmin xmax]);
xlabel(‘t’);ylabel(‘xb(t)’);title(‘Signal xb(t)’);
two
signals
Chapter 2 - Continuous Time Signals and Systems
2.90
subplot(2,3,3);plot(t,xadd);axis([tmin tmax xmin xmax]);
xlabel(‘t’);ylabel(‘xadd(t)’);title(‘Sum of xa(t) and xb(t)’);
subplot(2,3,4);plot(t,xa);axis([tmin tmax xmin xmax]);
xlabel(‘t’);ylabel(‘xa(t)’);title(‘Signal xa(t)’);
subplot(2,3,5);plot(t,xb);axis([tmin tmax xmin xmax]);
xlabel(‘t’);ylabel(‘xb(t)’);title(‘Signal xb(t)’);
subplot(2,3,6);plot(t,xmul);axis([tmin tmax xmin xmax]);
xlabel(‘t’);ylabel(‘xmul(t)’);title(‘Product of xa(t) and
xb(t)’);
OUTPUT
The input and output waveforms of program 2.5 are shown in fig P2.5.
Fig P2.5 : Input and Output waveforms of program 2.5.
Program
2.6
Write a MATLAB program to perform convolution of the following two
signals.
x1(t)=1; 1<t<10
x2(t)=1;
2<t<10
%******************Program to
%******************x1 (t)=1 ; t=
tmin=0; tmax=10;
t=tmin:dt:tmax;
x1=1.*(t>=1
x2=1.*(t>=2
&
&
perform convolution of two signals
1 to 1 0 a nd x2 (t)=1 ; t= 2 to 1 0
dt=0.01;
t<=10);
t<=10);
%set
time
vector
for
given
%generate
%generate
signal
signal
x1(t)
x2(t)
signal
2. 91
Signals & Systems
x3=conv(x1,x2);
n3=length(x3);
t1=0:1:n3-1;
%perform
%set
convolution
time
vector
for
of
signals
x3(t)
x1(t)
signal
subplot(3,1,1);plot(t,x1);
xlabel(‘t’);ylabel(‘x1(t)’);
title(‘signal x1(t)’);
subplot(3,1,2);plot(t,x2);
xlabel(‘t’);ylabel(‘x2(t)’);
title(‘signal x2(t)’);
subplot(3,1,3);plot(t1,x3); xlim ([0 600]);
xlabel(‘t / dt’);ylabel(‘x3(t) / dt’);
title(‘signal, x3(t) = x1(t)* x2(t)’);
OUTPUT
The input and output waveforms of program 2.6 are shown in fig P2.6.
Fig P2.6 : Input and Output waveforms of program 2.6.
and
x2(t)
Chapter 2 - Continuous Time Signals and Systems
Program
2.92
2.7
Write a MATLAB program to perform convolution of the following two
signals.
x1(t)=1; 0<t<2
x2(t)= 1;
0<t<1
=-1;
1<t<2
%******************Program to perform convolution of two signals
%******************x1(t)=1; t= 0 to 2 and x2(t)= 1; for t= 0 to 1
%
= -1; for t= 1 to 2
tmin=0; tmax=4; dt=0.01;
t=tmin:dt:tmax;
%set time vector for given signal
x1=1.*(t>=0 & t<=2);
xa=1;
xb=-1;
x2=xa.*(t>=0 & t<=1)+
x3=conv(x1,x2);
n3=length(x3);
t1=0:1:n3-1;
%generate
xb.*(t>=1
%set
time
signal
&
x1(t)
t<=2); %
%
%
vector
generate signal x2(t)
perform convolution of
x1(t) & x2(t)
for
signal
subplot(3,1,1);plot(t,x1);
xlabel(‘t’);ylabel(‘x1(t)’);title(‘signal
x1(t)’);
subplot(3,1,2);plot(t,x2);
xlabel(‘t’);ylabel(‘x2(t)’);title(‘signal
x2(t)’);
x3(t)
Fig P2.7 : Input and Output waveforms of program 2.7.
2. 93
Signals & Systems
subplot(3,1,3);plot(t1,x3);
xlabel(‘t / dt’);ylabel(‘x3(t) / dt’);
title(‘signal, x3(t) = x1(t)* x2(t)’);
OUTPUT
The input and output waveforms of program 2.7 are shown in fig P2.7.
Program
2.8
Given that, x1(t) = e-0.7t; 0<t<1
and
x2(t)=1;
0<t<2
Write a MATLAB program to perform following operations.
1. Convolution of x1(t) and x2(t).
2. Deconvolution of output of convolution with x2(t) to extract x1(t).
3. Deconvolution of output of convolution with x1(t) to extract x2(t).
%****************** Program to perform convolution and deconvolution
%****************** x1(t)=exp(-0.7t);0<=t<=T and x2(t)=1;0<=t<=T
T=2;
tmin = 0; tmax=2*T; dt=0.01;
t=tmin:dt:tmax;
%set time vector for given signals
x1=exp(-0.7*t).*(t>=0
x2=1.*(t>=0 & t<T);
x3=conv(x1,x2);
&
t<T);
%generate signal x1(t)
%generate signal x2(t)
%perform convolution of x1(t)
n3=length(x3);
Fig P2.8 : Input and Output waveforms of program 2.8.
and
x2(t)
Chapter 2 - Continuous Time Signals and Systems
2.94
t1=0:1:n3-1;
%set time vector for x3(t) signal
%plot the given signals and the signal obtained by convolution
subplot(3,2,1);plot(t,x1);title(‘signal x1(t)’);
xlabel(‘t’);ylabel(‘x1(t)’);
subplot(3,2,2);plot(t,x2);title(‘signal x2(t)’);
xlabel(‘t’);ylabel(‘x2(t)’);
subplot(3,2,3:4);plot(t1,x3);title(‘signal, x3(t)=x1(t)*x2(t)’);
xlabel(‘t / dt’);ylabel(‘x3(t) / dt’);
x1d=deconv(x3,x2);
x2d=deconv(x3,x1);
ymin=min([min(x1d),
ymax=max([max(x1d),
%perform
%perform
deconvolution
deconvolution
to
to
extract
extract
x1(t)
x2(t)
min(x2d)]);
max(x2d)]);
%plot the signals obtained by deconvolution
subplot(3,2,5);plot(t,x1d);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x1(t)’);title(‘extracted signal, x1(t)’);
subplot(3,2,6);plot(t,x2d);axis([tmin tmax ymin ymax]);
xlabel(‘t’);ylabel(‘x2(t)’);title(‘extracted signal, x2(t)’);
OUTPUT
The input and output waveforms of program 2.8 are shown in fig P2.8.
2.14 Exercises
I. Fill in the blanks with appropriate words
1. The ––––––––– signal is continuous function of an independent variable.
2. The ––––––––– signal can be represented by a rotating vector in a complex plane.
3. The sum of two periodic signals will also be periodic if the ratio of their fundamental periods is a
––––––––– number.
4. When a periodic signal is a sum of two or more periodic signals then the period of the signal is given by
––––––––– of the period of its components.
5. When a signal exhibits ––––––––– with respect to t ³ 0 it is called even signal.
6. When a signal exhibits antisymmetry with respect to t = 0 it is called ––––––––– signal.
7. The periodic signals will have constant –––––––––.
8. The nonperiodic signals will have constant –––––––––.
9. The ––––––––– signals are defined for t ³ 0.
10. When the sign of t is changed in x(t) the resultant signal is called ––––––––– signal.
11. The ––––––––– is the reverse process of differentiation.
12. A signal with large magnitude and short duration is called –––––––––.
13 The response of a system for impulse input is called –––––––––.
14. The ––––––––– system is governed by constant coefficient differential equation.
15. The ––––––––– response is the response of the system due to initial conditions alone.
16. The ––––––––– response is the response of the system due to input alone.
17. The systems that do not require memory are called ––––––––– systems.
18. A system is said to be ––––––––– if its input-output characteristics do not change with time.
19. A ––––––––– system is one that satisfies the superposition principle.
20. A ––––––––– system is one in which the response does not depend on future inputs/outputs.
21. When the output of a system is finite for any finite input then the system is called ––––––––– stable.
22. For ––––––––– of LTI system the integral of impulse response should be finite.
23. In ––––––––– system the present output depends on past outputs.
2. 95
Signals & Systems
24. The response of an LTI system is given by ––––––––––––– of input and impulse response.
25. In ––––––––––––––––– connection the output of a system becomes input for another system.
Answers
1.
2.
3.
4.
5.
6.
7.
8.
analog
complex exponential
rational
LCM (Least Common Multiple)
symmetry
odd
power
energy
9. causal
10. folded
11. integration
12. impulse
13. impulse response
14. LTI
15. natural or free or zero input
16. forced or zero state
17. static
18. time invariant
19. linear
20. causal
21. BIBO
22. stability
23. feedback
24. convolution
25. cascade
II. State whether the following statements are True/False
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
In an analog signal, both the magnitude of the signal and the independent variable are continuous.
A complex exponential signal is a three dimensional signal.
The sinc signal is a periodic signal.
The sinusoidal and complex exponential signals are always periodic.
The product of two odd signals will be an odd signal.
When the energy of a signal is finite then the power will be infinite.
When the power of a signal is finite then the energy will be zero.
The causal signals are defined for all t.
The time shift of a signal x(t) is obtained by replacing t by t ± m .
The rate of change of a signal is called differentiation.
An impulse signal has infinite magnitude but constant area.
A system which satisfies the property of linearity and time invariance is called an LTI system.
For a constant coefficient differential equations, the homogeneous solution will be always exponential.
For a constant coefficient differential equation, the particular solution and input will be the same type of signal.
The systems that require memory are called dynamic systems.
In a time invariant system, if a delay is introduced either at input or output, then the overall response will
remain same.
A nonlinear system is one which satisfies the principle of superposition.
The noncausal systems can be realised in real time.
The response of a noncausal system depends only on past inputs.
All LTI systems are BIBO stable.
The stability of non-LTI systems can be tested by using its impulse response.
The nonfeedback systems depend on past inputs.
Convolution is a linear operation.
The convolution can be used to determine the response of non-LTI systems.
25. The convolution operation satisfies commutative property.
Answers
1. True
2. True
3. False
4. True
5. False
6.
7.
8.
9.
10.
False
False
False
True
True
11.
12.
13.
14.
15.
True
True
True
True
True
16.
17.
18.
19.
20.
True
False
False
False
False
21.
22.
23.
24.
25.
False
True
True
False
True
Chapter 2 - Continuous Time Signals and Systems
2.96
III. Choose the right answer for the following questions
1. The unit impulse is defined as,
a) d( t ) = ¥;t = 0
b) d ( t ) = ¥;t = 0
= 0;t ¹ 0
bg
c) d (t) = ¥; t = 0
d) d t = ¥; t = 0
=0; t ¹0
+¥
and
z bg
d t dt = A
+¥
and
-¥
z bg
d t dt = 1
-¥
2.
Which of the following is a periodic signal?
b) x(t) = A e - jbt
a) x(t) = A u(t)
3.
d) x(t) = A t
c) x( t) = A e bt
d) x( t) = A e - jbpt
Which of the following is a nonperiodic signal?
a) x(t) = A e- j
4.
c) x(t) = A e bt
bt
b) x(t) = A e
-j
pt
b
Which of the following statements is false?
i) the product of two odd signals is an odd signal.
ii) the product of two even signals is an even signal.
iii) the product of even and odd signals is an even signal. iv) the product of even and odd signals is an odd signal.
a) i and iii
b) i only
c) iii only
d) iv only
5.
Which of the following is a purely even signal?
a) x(t) = eat
6.
b) x(t) = Ae bt
b) x(t) = A sin W 0 t
Which of the following statements are true?
i) the periodic signals are power signals.
iii) for energy signals, the power is zero.
a) i only
b) i and ii only
9.
d) x(t) = u( t )
c) x(t) = A sin nW 0 t
d) x(t) = t 2
c) x(t) = B cos W 0 t
d) x(t) = e - at u( t )
Which of the following is an energy signal?
a) x(t) = Ae jW 0 t
8.
c) x(t) = e jW 0 t
Which of the following is a purely odd signal?
a ) x(t) = e jbt
7.
b) x(t) = cosW 0 t
ii) the nonperiodic signals are energy signals.
iv) for power signals, the energy is zero.
c) i, ii and iii only
d) all of the above
Which of the following signal is a causal signal?
a) x(t) = A
b) x(t) = t
d) x(t) = e- at
c) x(t) = u( t )
x(t)
10. Which of the following represents the pulse signal shown in fig 10.?
1
a) u (t + 0.5) – u (t – 0.5)
b) u (t – 0.5) + u (t + 0.5)
–0.5
c) u (t – 0.5) – u (t + 0.5)
d) u (t + 0.5) + u (t – 0.5)
0.5
Fig 10.
11. The differentiation of a unit step signal is,
a) ramp signal
b) impulse signal
c) exponential signal
d) parabolic signal
12. The integral of a unit impulse signal is,
a) infinity
b) zero
c) one
13. If x(t) is a continuous time signal and d (t) is a unit impulse signal then,
d) constant
z
¥
-¥
a) x(t)
b) d(t)
c) x( t 0 )
bg b
x t d t - t0
d) d(t0)
g is equal to,
t
2. 97
Signals & Systems
14. If x(t) is a continuous time signal and d(t) is unit impulse signal then,
z
¥
-¥
a) x(t)
b) d(t)
bg b
g
x t d t - T dt is equal to,
d) d(t – t)
c) x(t) d( t )
15. Which of the following responses of an LTI system does not depend on initial conditions?
a) natural response
b) free response
c) forced response
16. The homogeneous solution of the differential equation
a) Ce
At
b) C1 + C2 e
At
17. The total solution of the differential equation
a) eAt x(t)
bg
dy t
dt
b) CeAt + K x(t)
bg
dy t
dt
c) eAt
d) total response
bg bg b g
+ a y t = x t ; y 0 = b , will be in the form,
d) C1 + C2
bg bg b g
+ a y t = x t ; y 0 = b , will be in the form,
c) K1 + K2 x(t)
d) eAt + x(t)
18. Which of the following is not a static system?
a) y(t) = x( t 2 )
b) y(t) = x 2 (t)
c) y(t) = t x(t)
d) y(t) = e x (t)
c) y(t) = x( - t)
d) y(t) = x (t 2 )
19. Which of the following is a time invariant system?
a) y(t) = t x(t)
b) y(t) = ex bt g
20. Which of the following is a linear system?
a) y(t) = x(t 2 )
b) y(t) = x2 (t)
c) y(t) = x(t) + C
d) y(t) = e x(t)
21. Which of the following is a nonlinear system?
a) y(t) = t x(t)
z
b) y(t) = x(t) dt
dx(t)
dt
d) y(t) = x(t)
c) y(t) = x( - t)
d) y(t) = x(2 t )
c) y(t) = ex( t )
d) y(t) = x(t 2 )
c) y(t) =
22. Which of the following is a causal system?
a) y(t) = x(t 2 )
b) y(t) = x2 (t)
23. Which of the following is a noncausal system?
dx(t)
t
a) y(t) =
b) y(t) = x(l ) dl
0
dt
z
24. Which of the following statements are true ?
i) an LTI system is always stable.
ii) an LTI system is stable only if the integral of its impulse response is finite.
iii) in a system, if the input is bounded then the output is always bounded.
iv) in a system, even if the input is unbounded the output can be bounded
a) ii only
b) ii and iv only
c) iii only
d) i and iv only
c) y(t) = e t x( t )
d) y(t) = e - t x ( t )
25. Which of the following is a stable system?
a) y(t) = t x(t)
b) y(t) = t 2 x(t)
26. Which of the following impulse responses of LTI systems represents an unstable system?
a) h(t) = eat u(t)
b) h(t) = e -at u(t)
c) h(t) = t e - t u(t)
d) h(t) = e - t sin t u(t)
27. Which of the following impulse responses of LTI systems represents a stable system?
a) h(t) = e t cost u(t)
b) h(t) = e t sin t u(t)
c) h(t) = e - t cost u(t)
d) h(t) = t sin t u(t)
Chapter 2 - Continuous Time Signals and Systems
2.98
28. Which of the following represents the convolution of two causal signals x1(t) and x2(t)?
t
i)
z
t
x1 ( l ) x 2 ( t - l ) dl
l
z
l =0
l
z
ii ) x 2 ( l ) x1 ( t - l ) dl
iii ) x1 ( t ) x2 (l - t ) dt
l=0
z
iv) x2 ( t ) x1 (l - t ) dt
t=0
t =0
a) i only
b) ii only
c) i and ii only
d) all of the above
29. If h1(t) and h2(t) are impulse responses of two stable LTI systems in cascade, then overall impulse response
of the cascaded system is,
a) h1 (t) * h2 (t)
b) h1 (t) h2 (t)
c) h1 (t) + h2 (t)
d) h1 (t) - h2 (t)
–at
30. If the impulse response of an LTI causal system is in the form e then its response for step input of value
A will be in the form,
A
c)
1- e-at u( t )
a) A 1- e -at u( t )
d) A - e -at u( t )
b) Ae -at u( t )
a
d
Answers
1. d
2. b
3. c
4. a
e
i
5.
6.
7.
8.
b
c
d
c
9.
10.
11.
12.
c
a
b
c
13. c
14. a
15. c
16. a
17. b
18. a
19. b
20. a
j
21.
22.
23.
24.
d
b
d
b
e
25.
26.
27.
28.
d
a
c
d
j
29. a
30. c
IV. Answer the following questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Define and sketch impulse and unit impulse signal.
Define and sketch step and unit step signal.
Define and sketch ramp signal.
Define and sketch sinc signal.
Define and sketch signum signal.
What are the various ways of classifying the continuous time signals?
Define deterministic and nondeterministic signals. Give examples.
Define periodic and nonperiodic signals. Give examples.
Define even and odd signals. Give examples.
1
Prove that the even part of a signal is given by xe t = x t + x - t and the odd part of a signal is given by
2
1
xo t = x t - x - t .
2
Define energy of a continuous time signal. Give examples.
Define power of a continuous time signal. Give examples.
Prove that power of an energy signal is zero.
Prove that energy of a power signal is infinite.
Define causal and noncausal signals. Give examples.
List the properties of impulse signal.
Write any two properties of impulse signal and prove.
Show that any continuous time signal x(t) can be expressed as an integral of impulses.
What is homogenous and particular solution?
What is free and forced response?
What is zero-input and zero-state response?
Define static and dynamic systems. Give examples.
Explain the time invariant property of a system.
Define linear systems.
Define causal and noncausal systems. Give examples.
Define stable and unstable systems.
bg
bg
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
bg b g
bg b g
2. 99
27.
28.
29.
30.
31.
32.
33.
34.
35.
Signals & Systems
What is the condition for stability of an LTI system?
Define feedback and nonfeedback systems.
Define convolution of two continuous time signals.
Show that the response of an LTI system can be obtained by convolution of input and impluse response?
List the properties of convolution.
State and prove the associative and commutative properties of convolution.
State and prove the distributive property of convolution.
What are the two ways of interconnection of LTI systems?
What is the relation between impulse response and unit step response of an LTI system?
V. Solve the following problems
E2.1
Verify whether the following signals are periodic. If periodic find the fundamental period.
a) x(t) = 4 sin 7t
b) x(t) = 2e0.7t
a) x(t) = e j2 t
FG
H
g bg
e x t = sin 2 3t -
p
5
IJ
K
g
c x(t) = 6 sin 5t + 3 cos 4 pt
b) x(t) = 4 + e3t
c) x(t) = t + 3t 2 + cos2 t
g
d x(t) = sin2 t + e j5 pt
Determine the energy and power of the following continuous time signals.
E2.4
a) x(t) = 0.9 e -3 t u(t)
b) x(t) = 3 e- j0.5pt
c) x(t) = 1.2 sin7W o t
d) x(t) = t u(t)
Sketch the even and odd parts of the signals shown in fig E2.5.1, E2.5.2, E2.5.3, E2.5.4.
x(t)
a)
b)
c) x(t)
d)
E2.5
x(t)
A
A
0
2
t
Fig E2.5.1.
2
2
x(t)
1
1
0
2
t
–1
0
Fig E2.5.2.
–1
a) x (–t)
b) – x (t)
c) x (t +3)
d) x (–t + 3)
e) x(t – 3)
f) x (–t – 3)
2
t
Fig E2.5.3.
1
–1
0
2
1
Fig E2.5.4.
x(t)
A continuous time signal is shown in fig E2.6
Determine the following versions of the signal.
2
Fig E2.6.
1
0
2
1
3
t
Evaluate the following integrals.
+¥
z
a) e -0.5 t d ( t - 3)dt
-¥
E2.8
IJ
K
Determine the even part and odd part of the following continuous time signals.
E2.3
E2.7
p
3
Determine the periodicity of the following continuous time signals.
2 pt
-j
2 pt
2 pt
a) x(t) = 4 sin
+ 5 sin
b) x(t) = 0.5 e 5 + 0.7 cos 8pt
9
7
E2.2
E2.6
FG
H
c ) x(t) = 3e-j0.1pt d) x(t) = 9 sin 6t +
+¥
z
b) e j0.1t d( t - 10)dt
-¥
+¥
z
c) t 3d ( t + 1)dt
-¥
Determine the natural response of the following systems.
d 3y(t)
d 2 y(t)
dy(t)
dx(t)
dy(t)
a)
+
8
+ 17
+ 10y(t) =
+ 0.5x(t) ; y(0) = 2,
3
2
dt
dt
dt
dt
dt
d 2 y(t)
dy(t)
dy(t)
+ 1.6
+ 0.63y(t) = 0 ;
= - 0.05, y(0) = 0.02
dt 2
dt
dt t = 0
d 2 y(t)
dy(t)
dy(t)
c)
+2
+ 0.75y(t) = 0 ; y(0) = 0.25,
= -1
dt 2
dt
dt t = 0
b)
= -4 ;
t=0
d 2 y(t)
dt 2
= 10
t=0
t
Chapter 2 - Continuous Time Signals and Systems
E2.9
2.100
Determine the forced response of the following systems.
d 3y ( t )
d 2 y( t )
dy( t )
dx( t )
a)
+7
+ 14
+ 8 y( t ) =
+ 0.5x( t ) ; x( t ) = e -3t u( t )
3
dt
dt 2
dt
dt
d 2 y( t )
dy( t )
dx( t )
b)
+ 18
.
+ 0.45y( t ) = 0.5
+ 0.18x( t ) ; x( t ) = u( t )
dt 2
dt
dt
d 2 y( t )
dy( t )
c)
+ 0.7
+ 01
. y( t ) = 0.4 x( t ) ; x( t ) = 0.2e-0.3t u( t )
dt 2
dt
E2.10 Determine the total response of the following systems.
d 2 y( t ) dy( t )
dx( t )
dy( t )
a)
+
+ 0.21y( t ) =
+ x( t ) ;x(t) = 0.2e -0.5t u( t ) ;
dt 2
dt
dt
dt
b)
c)
E2.11
d 2 y( t )
dy( t )
dy( t )
+9
+ 14 y( t ) = 2x( t ) ;x(t) = 0.7 u( t ) ;
2
dt
dt
dt
= -1 ; y(0) = 0.3
t =0
= -0.5 ; y(0) = 0.2
t =0
d 2 y( t )
dy( t )
dy( t )
+ 31
.
+ 2.2 y( t ) = 0.1x( t ) ;x(t) = 0.7e -2.5t u( t ) ;
= -0.5 ; y(0) = 0
dt 2
dt
dt t = 0
Verify whether the following systems are time invariant or time variant.
a) y(t) = e t x(t)
b) y(t) = cost x(t)
c) y(t) = x - t + 2
b
g
z
dx(t)
d) y(t) = 3 x(t) + 5
e) y(t) = 3
+ 7 x(t)
f) y(t) = ex (t) + x(t) dt
dt
E2.12 Verify whether the following systems are linear or nonlinear.
a) y(t) = t 2 x(t) b) y(t) = x (t)
z
g
c) y(t) = e t x (t) d) y(t) = x (t) dt e) y(t) = t + x (t)
f y(t) = cos t x(t)
E2.13 Determine the linearity of the LTI systems governed by the following differential equations.
a)
d 2 y( t )
dy(t)
+ 0.3
+ 0.5y(t) = 2 x(t)
dt
dt 2
b) y(t) = 4
d 2 x(t )
dx(t)
+2
+ 5 x(t)
dt
dt 2
E2.14 Verify whether the following systems are causal or noncausal.
a) y(t) = e t x(t)
b) y(t) = t - 2 u t + 2
c) y(t) = t + 2 x(t)
b gb g
b g
dx(t)
d) y(t) = e +
e) y(t) = cost x(t)
f) y(t) = sin t x(t + 2)
dt
E2.15 Verify whether the following systems are stable or unstable.
a) y(t) = e t x(t)
b) y(t) = e- t x(t)
c) y(t) = t 2 x(t)
x (t)
d) y(t) = sin t x(t)
e) y(t) = e- t sin t x(t)
f) y(t) = x(t + 2)
E2.16 Verify the stability of LTI systems whose impulse responses are given below.
a) h(t) = t 2 u(t)
b) h(t) = e-7 t u(t)
c) h(t) = e5 t u(t)
d
i
e) h(t) = A + Be- Ct u(t)
f) h(t) = 2e -3 t cost u(t)
®
® h (t)
3
®
®
+
Fig E2.17a
h1 (t) = e -6 t u(t); h2 (t) = e2 t u(t); h3 (t) = t u(t)
®
®
2
®
E2.17 Determine the overall impulse response of the following system.
a)
b)
® h1(t)
® h (t)
® h1(t)
+
®h (t)
2
®
d) h(t) = t sin t u(t)
® h (t)®
3
Fig E2.17b
h1 (t) = cos t u(t); h2 (t) = sin t u(t); h3 (t) = u(t)
E2.18 Perform the convolution of the following signals.
a) x1 ( t ) = t 2 u(t); x2 ( t ) = u(t - 1)
b) x1 ( t ) = te -4 t u(t);
c) x1 ( t ) = t u(t); x2 ( t ) = sin t u(t)
x2 ( t ) = u(t)
2. 101
Signals & Systems
E2.19 Determine the unit step response of the following systems whose impulse responses are given below.
E2.20
a) h(t) = te-9 t u(t)
b) h(t) = e- t cost u(t)
c) h(t) = e-5t u(t - 2)
d) h(t) = u(t - 3) + u(t + 2)
Perform convolution of the following signals by graphical method and sketch the resultant waveform.
x2(t)
x1(t)
a)
2
b)
x1(t)
2
2
x2(t)
1
1
–1
1
0
t
0
1
2
1
–1
t
1
0
0
t
1
t
Answers
2p
p
p
b) nonperiodic c) T = 20 d) T =
e) T =
7
3
3
E2.1
a) T =
E2.3
a) xe ( t ) = cos 2t ; xo ( t ) = j sin 2t
1
1
b) xe ( t ) = 4 + (e3t + e-3t ) ; xo ( t ) = (e 3t - e -3 t )
2
2
d) xe ( t ) = sin2 t + cos5pt ; xo ( t ) = jsin 5pt
c) xe ( t ) = 3t 2 + cos2 t ; xo ( t ) = t
E2.4 a) E = 0.135 joules, P = 0
E2.5
a)
–2
E2.2) a) T = 63 b) T = 5 c) nonperiodic
b) E = ¥, P = 9 watts c) E = ¥ , P = 0.72 watts
xe(t)
xo(t)
A/2
A/2
–1
t
2
1
0
–2
–1
0
xe(t)
–1
1
2
–2
t
–1
0
Fig E2.5b.1
1.5
0
xe(t)
1
2
t
–2
0
–1
1
2
t
–0.5
Fig E2.5c.2
–1
1.5
xo(t)
0.5
–2
–1
Fig E2.5b.2
1
0.5
–2
t
0.5
Fig E2.5c.1
d)
2
xo(t)
0.5
–1
1
–A/2
xe(t)
–2
Fig E2.5a.2
A/2
0
c)
t
xo(t)
A/2
–2
2
1
–A/2
Fig E2.5a.1
b)
d) E = ¥ , P = ¥
0
1
0
–1
2
t
Fig E2.5d.1
–0.5
1
2
t
Fig E2.5d.2
Chapter 2 - Continuous Time Signals and Systems
2.102
x(t+3)
–x(t)
x(–t)
2
0
1
E2.6 a)
b)
-3
-2
2
t
1
c)
–3
Fig E2.6b
Fig E2.6a
x(t–3)
x(–t–3)
2
2
1
1
d)
–5 –4 –3
–1
–2
f)
0
t
0
1
a) e -1.5 = 0.2231
E2.8
a) y zi ( t ) =
FG 1 e
H2
2
3
4
5
IJ
K
4 -2t 1 -5t
e + e
u(t)
3
6
c) y zi ( t ) = ( -0.625e -0.5t + 0.875e -1.5t ) u(t)
-5 - t 5 -2 t 5 -4 t 5 -3t
a) y zs ( t ) =
e + e + e - e
u(t)
12
4
12
4
+
FG
H
F 8 + 4 e - 4e IJ u(t)
c) y ( t ) = G e
H3
K
3
a) y ( t ) = d -0.725e
+ 3.525e
- 2.5e
-0 . 2 t
-0.5t
IJ
K
2
3
c) t 3
t = -1
= -1
b) y zi ( t ) = ( -0.16e -0.7t + 0.18e -0.9t ) u(t)
b) yzs ( t ) = ( -0.5e-0.3t + 0.1e -1.5t + 0.4) u(t)
-0.3t
zs
E2.10
-0.2t
c) y zs ( t ) = ( -0.5e
-1.1t
a)
d)
b)
e)
E2.13
a)
b) – Linear
E2.15
b)
d)
a)
c) – Unstable
a) ho ( t ) =
b) ho ( t ) =
+ 0.4e
-2 t
+ 01
.e
-2.5 t
-0.5 t
i u(t)
f) – Stable
e)
d
E2.12) a)
b)
c)
d) f) – Linear
e) – NonLinear
E2.14) a)
b)
E2.16) b)
c) d) e) – Causal
f) – Noncausal
f) – stable
a)
2t
1
2
E2.19
a) s(t) =
1
2
e) – Unstable
-6 t
3
b)
1
1 - e -9 t - 9 te -9 t u(t)
81
e -10
-5 t - 2
c) s(t) =
1 - e b g u(t - 2)
5
d
i
e
E2.20
d)
3
3
a)
c)
F1 1 I
h b t g * h b t g + h b t g = G e - e + tJ u(t)
H8 8 K
h b t g + h b t g * h b t g = bsin t - cost + 1g u(t)
FG t - 1IJ u( t - 1)
H 3 K
c) b t + sin t g u(t)
E2.18
i
b) y ( t ) = 0.04e -2 t + 0.06e -7 t + 01
. u(t)
) u(t)
c) – Time variant
f) – Time invariant
E2.11
E2.17
-0 . 7 t
j
d
i
a) x1 ( t ) * x2 ( t ) = t + 1) 2 u( t + 1
(or) x1 (t) * x2 ( t ) = ( t + 1)2 ; t ³ -1
t
Fig E2.6f
b) e j = cos1 + jsin1 = 0.5403 + j0.8415
(Note : Calculate in radians mode)
-t
1
0
t
6
Fig E2.6e
Fig E2.6d
E2.7
t
0
1
e)
–6
–2 –1
Fig E2.6c
2
x(-t+3)
E2.9
3
2
-2
t
0
-1
1
-1
1
1 - e -4t - 4te -4t u(t)
16
d
i
1
1 + e - t sin t - e - t cos t u(t)
2
i
d) s(t) = b t - 3g ub t - 3g + b t + 2g ub t + 2g
b) x ( t ) * x ( t ) = 2t - 2b t - 2) u ( t - 2g
b) s(t) =
1
d
2
(or) x1 (t) * x2 ( t ) = 2 t ; t = 0 to 2
= 4 ; t³ 2
CHAPTER 3
Laplace Transform
3.1 Introduction
The Laplace transform is used to transform a time signal to complex frequency domain.
(The complex frequency domain is also known as Laplace domain or s-domain). This transformation
was first proposed by Laplace (in the year 1780) and later adopted for various engineering applications
for solving differential equations. Hence this transformation is called Laplace transform.
In signals and systems the Laplace transform is used to transform a time domain system to
s-domain. In time domain the equations governing a system will be in the form of differential equations.
While transforming the system to s-domain, the differential equations are transformed to simple
algebraic equations and so the analysis of systems will be much easier in s-domain.
In this chapter a brief discussion about Laplace transform and its applications for analysis of
signals and systems are presented.
Complex Frequency
The complex frequency is defined as,
Complex frequency, s = s + jW
where, s = Neper frequency in neper per second
W = Radian (or Real) frequency in radian per second
The complex frequency is involved in the time domain signal of the form Kest. The signal Kest can
be thought of as an universal signal which represents all types of signals and takes a particular form for
various choices of s and W as shown below.
Let, x(t) = A est = A e(s + jW)t
.....(3.1)
Let us analyse the signal of equation (3.1) for various choice of s and W.
Case i : When s = 0, W =W0
On substituting s = 0 in equation (3.1) we get,
x( t ) = A e jW 0t
= A (cos W 0 t + j sin W 0 t)
= A cos W 0 t + j A sin W 0 t
.....(3.2)
The real part of equation (3.2) represents a cosinusoidal signal and the imaginary part represents a
sinusoidal signal.
Chapter 3 - Laplace Transform
3. 2
i. e., Re[x(t)] = A cos W 0 t
Im[ x( t )] = A sin W 0 t
..... Cosinusoidal signal
..... Sinusoidal signal
Note : Re - stands for ‘real part of’
Im - stands for ‘imaginary part of’
Case ii : When W = 0
On substituting W = 0 in equation (3.1) we get,
x(t) = A est
....(3.3)
In equation (3.3) if s is positive then the signal will be an exponentially increasing signal.
In equation (3.3) if s is negative then the signal will be an exponentially decreasing signal.
i.e.,
x(t) = A est
..... Exponentially increasing signal
x(t) = A e-st ..... Exponentially decreasing signal
Case iii : When s = 0 and W = 0
On substituting s = 0 and W = 0 in equation (3.1) we get,
x(t) = A e0 = A
.....(3.4)
The equation (3.4) represents a step signal.
Case iv : When s ¹ 0 ,W ¹ 0 and W =W0 .
When both s and W are non-zero and when W =W0 , the equation (3.1) can be expressed as shown
below.
x( t ) = A e( s + jW 0 ) t = A est A e jW 0t
= A est (cos W 0 t + j sin W 0 t )
= A est cos W 0 t + j A est sin W 0 t )
.....(3.5)
The real part of equation (3.5) represents an exponentially increasing/decreasing cosinusoidal signal.
The imaginary part of equation (3.5) represents an exponentially increasing/decreasing sinusoidal
signal.
i. e., Re[x(t)] = A e st cos W 0 t
st
® A e cos W 0 t
® A e - st cos W 0 t
..... Exponentially increasing cosinusoidal signal
..... Exponentially decreasing cosinusoidal signal
Im[x(t)] = A est sin W 0 t
st
® A e sin W 0 t
® A e- st sin W 0 t
..... Exponentially increasing sinusoidal signal
..... Exponentially decreasing sinusoidal signal
3. 3
Signals & Systems
Complex Frequency Plane or s-Plane
jW
­
The complex frequency is defined as,
where, s = Real part of s
W = Imaginary part of s
- s®
­
Complex frequency, s = s + jW
s-plane
s
­
-jW
The s and W can take values from - ¥ to + ¥ . A two dimensional Fig 3.1: Complex frequency
complex plane with values of s on horizontal axis and values W on vertical
plane or s-plane.
axis as shown in fig 3.1 is called complex frequency plane or
s-plane.
The s-plane is used to represent various critical frequencies (poles and zeros) of signals which are
functions of s and to study the path taken by these critical frequencies when some parameters of the
signals are varied. This study will be useful to design systems for a desired response.
Definition of Laplace Transform
In order to transform a time domain signal x(t) to s-domain, multiply the signal by e–st and then
integrate from –¥ to ¥. The transformed signal is represented as X(s) and the transformation is denoted
by the script letter L.
Symbolically the Laplace transform of x(t) is denoted as,
X(s) = L{x(t)}
Let x(t) be a continuous time signal defined for all values of t. Let X(s) be Laplace transform of
x(t). Now the Laplace transform of x(t) is defined as,
+¥
L{x( t )} = X(s) =
z
z
x( t ) e - st dt
.....(3.6)
-¥
If x(t) is defined for t ³ 0, (i.e., if x(t) is causal) then,
+¥
L{x( t )} = X(s) =
x( t ) e - st dt
.....(3.7)
0
The definition of Laplace transform as given by equation (3.6) is called Two sided Laplace transform
or Bilateral Laplace Transform and the definition of Laplace transform as given by equation (3.7) is
called One sided Laplace transform or Unilateral Laplace transform.
Definition of Inverse Laplace Transform
The s-domain signal X(s) can be transformed to time domain signal x(t) by using inverse Laplace
transform.
The Inverse Laplace transform of X(s) is defined as,
L-1 X(s) = x(t) =
l q
1
2 pj
s = s + jW
z
X(s) e st ds
s = s - jW
Chapter 3 - Laplace Transform
3. 4
The signal x(t) and X(s) are called Laplace transform pair and can be expressed as,
x(t)
L
® X(s)
L–1
¬
Existence of Laplace Transform
The computation of Laplace transform involves integral of x(t) from t = -¥ to +¥.
Therefore Laplace transform of a signal exists if the integral,
z
+¥
-¥
x( t ) e - st dt converges (i.e., finite).
The integral will converge if the signal x(t) is sectionally continuous in every finite interval of t and if it is
of exponential order as t approaches infinity.
A causal signal x(t) is said to be exponential order if a real, positive constant s (where s is real
part of s) exists such that the function, e - st x( t ) approaches zero as t approaches infinity.
i.e., if, Lt e - st x( t ) = 0 , then x(t) is of exponential order.
t®¥
For a causal signal, if Lt e -st x( t ) = 0 for s > sc , and if Lt e - st x( t ) = ¥ for s < sc, then
t®¥
t®¥
sc is called abscissa of convergence, (where sc is a point on real axis in s-plane).
The integral
z
+¥
-¥
x( t ) e - st dt converges only if the real part of s is greater than the abscissa of
convergence sc. The values of s for which the integral
z
+¥
-¥
x( t ) e - st dt converges is called Region Of
Convergence (ROC). Therefore for a causal signal the ROC includes all points on the s-plane to the right
of abscissa of convergence.
3.2 Region of Convergence
+¥
The Laplace transform of a signal is given by
+¥
z x( t ) e
- st
z
x( t ) e - st dt . The values of s for which the integral
-¥
dt converges is called Region Of Convergence (ROC). The ROC for the following three types
-¥
of signals are discussed here.
Case i : Right sided (causal) signal
Case ii : Left sided (anticausal) signal
Case iii : Two sided signal.
Case i : Right sided (causal) signal
Let, x(t) = e-at u(t), where a > 0
= e-at for t ³ 0
Now, the Laplace transform of x(t) is given by,
+¥
L{x( t )} = X(s) =
z
+¥
x( t ) e - st dt =
-¥
+¥
=
z
0
=
z
e - at u(t) e- st dt
-¥
+¥
e - at e - st dt =
z
0
e- ( s + a ) t dt =
LM e
OP
N - (s + a) Q
-(s + a) t
¥
0
e0
1
e - ( s + jW + a ) ¥
e - ( s + a ) ´ ¥ e - jW ´ ¥
= +
s + a
s + a
-( s + a )
- (s + a)
Put,
s=s+jW
3. 5
Signals & Systems
1
e - k ´ ¥ e - jW ´ ¥
+
s + a
s + a
where, k = s + a = s -(-a)
When s > -a, k = s -(-a) = Positive, \ e-k¥ = e-¥ = 0
When s < –a, k = s -(-a) = Negative, \ e-k¥ = e+¥ = ¥
\ L x(t) = -
l q
jW
s-plane ­
Hence we can say that, X(s) converges, when s > -a, and does
not converge for s < -a.
\ Abscissa of convergence, sc = -a.
sc = –a
®s
Abscissa of
®
convergence
Fig 3.2 : ROC of x(t) = e–at u(t).
When s > -a, the X(s) is given by,
L{x( t )} = X(s) = -
ROC
X
1
0 ´ e - jW ´ ¥
1
1
e - k ´ ¥ e - jW ´ ¥
+
= +
=
s + a
s + a
s + a
s + a
s + a
Therefore for a causal signal the ROC includes all points on the s-plane to the right of abscissa of
convergence, sc= -a, as shown in fig 3.2.
Case ii : Left sided (anticausal) signal
Let, x(t) = e-bt u(-t) = e-bt for t £ 0, where b > 0
Now, the Laplace transform of x(t) is given by,
+¥
L{x( t )} = X(s) =
z
+¥
x( t ) e - st dt =
-¥
-¥
0
=
z
e - ( s + b) t dt =
-¥
= -
1
e
+
s + b
z
0
e - bt u(- t) e- st dt =
LM e
OP
N -(s + b) Q
- ( s + b) t
( s + b) ´ ¥
e- bt e- st dt
-¥
0
e
e ( s + jW + b) ¥
-(s + b)
-(s + b)
0
=
-¥
jW ´ ¥
e
s + b
z
k´¥
= -
Put,
s=s+jW
jW ´ ¥
1
e
e
+
s + b
s + b
where, k = s + b = s -(-b)
When s > -b, k = s -(-b) = Positive, \ ek¥ = e¥ = ¥
When s < -b, k = s -(-b) = Negative, \ ek¥ = e-¥ = 0
Hence we can say that, X(s) converges, when s < -b, and
does not converge for s > -b.
\ Abscissa of convergence, sc = -b.
jW
ROC
­
s-plane
®
scX= -b s
Abscissa of
¬convergence
Fig 3.3 : ROC of x(t) = e–btu(–t).
When s < -b, the X(s) is given by,
L{x( t )} = X(s) = -
1
1
0 ´ e jW ´ ¥
1
e k ´ ¥ e jW ´ ¥
+
= +
= s + b
s + b
s + b
s + b
s + b
Therefore for an anticausal signal the ROC includes all points on the s-plane to the left of abscissa of
convergence, sc = -b, as shown in fig 3.3.
Chapter 3 - Laplace Transform
3. 6
Case iii: Two sided signal
Let, x(t) = e-at u(t) + e-bt u(-t) , where a > 0, b > 0, and a > b (i.e., - a < -b)
Now, the Laplace transform of x(t) is given by,
+¥
z
L{x( t )} = X(s) =
+¥
x( t ) e - st dt =
-¥
z
e - at u(t) + e - bt u(- t) e - st dt
-¥
+¥
=
z
+¥
0
e
- at
e
- st
dt +
z
e
- bt
e
- st
dt =
-¥
0
Le
OP
= M
N -( s + a ) Q
-(s + a) t
¥
0
z
0
e
-(s + a) t
z
dt +
-¥
0
Le
OP
+ M
N -(s + b) Q
- ( s + b) t
0
-¥
e- ( s + b) t dt
Le
OP
= M
N -(s + a) Q
- ( s + jW + a ) t
¥
Le
OP
+ M
N - ( s + b) Q
- ( s + jW + b ) t
0
e0
e0
e ( s + jW + b ) ¥
e - ( s + jW + a ) ¥
+
=
-(s + a )
-(s + a )
- (s + b)
-(s + b)
= -
0
-¥
Put,
s=s +jW
1
1
e - p ´ ¥ e - jW ´ ¥
eq ´ ¥ e jW ´ ¥
+
+
s + a
s + a
s + b
s + b
where, p = s + a = s - (-a) and q = s + b = s - (-b)
When s > -a, p = s - (-a) = Positive, \ e-p¥ = e-¥ = 0
When s < -a, p = s - (-a) = Negative, \ e-p¥ = e+¥ = ¥
When s > -b, q = s - (-b) = Positive, \ eq¥ = e¥ = ¥
When s < -b, q = s - (-b) = Negative, \ eq¥ = e-¥ = 0
Hence we can say that, X(s) converges, when s lies between -a and -b (i.e., -a < s < -b ), and
does not converge for s < -a and s > -b.
\ Abscissa of convergences, sc1 = -a and sc2 = -b.
When -a < s < -b, the X(s) is given by,
1
1
e - p ´ ¥ e - jW ´ ¥
e q ´ ¥ e jW ´ ¥
+
+
s + a
s + a
s + b
s + b
0 ´ e - jW ´ ¥
1
1
0 ´ e jW ´ ¥
= +
+
s + a
s + a
s + b
s + b
1
1
=
jW
­ s-plane
s + a
s + b
L{x( t )} = X(s) = -
Therefore for a two sided signal the ROC
includes all points on the s-plane in the region in
between two abscissa of convergences, sc1= -a and
sc2= -b, as shown in fig 3.4.
ROC
X
X
®
sc1 = –a sc2 = –b s
of
Abscissa of®
¬Abscissa
convergence
convergence
Fig 3.4 : ROC of x(t) = e–at u(t) + e–bt u(–t).
3. 7
Signals & Systems
Example 3.1
Determine the Laplace transform of the following continuous time signals and their ROC.
a) x(t) = A u(t)
c) x(t) = e–3t u(t)
b) x(t) = t u(t)
e) x(t) = e-4½t½
d) x(t) = e–3t u(-t)
Solution
a) Given that, x(t) = A u(t) = A ; t ³ 0
By definition of Laplace transform,
+¥
z
L{x(t )} = X(s) =
¥
z
x(t) e - st dt =
-¥
LM e OP
N -s Q
- st
= A
A e - st dt = A
0
¥
= A
0
LM e
N
- (s + jW )t
-s
OP
Q
¥
z
Put,
s=s+jW
e- st dt
0
¥
= A
0
LM e
N
- (s + jW )¥
e0
-s
-
-s
OP = A LM e
Q N
-s ´ ¥
e - jW ´ ¥
1
+
s
-s
jW
When, s > 0, (i.e., when s is positive), e-s¥ = e-¥ = 0
When, s < 0, (i.e., when s is negative), e-s¥ = e¥ = ¥
s-plane
­
ROC
Therefore we can say that, X(s) converges when s > 0.
OP
Q
-s ´ ¥
LM
N
OP
Q
1
0 ´ e - jW ´ ¥
1
A
e- jW ´ ¥
+
= A
+
=
-s
s
-s
s
s
Fig 1 : ROC of x(t) = A u(t).
A
; with ROC as all points is s-plane to the right of line passing through s = 0.
s
(or ROC is right half of s-plane).
X(s) = A
\
®s
X
s =0
When s > 0, the X(s) is given by,
LM e
N
L lA u(t)q =
OP
Q
b) Given that, x(t) = t u(t) = t ; t ³ 0
By definition of Laplace transform,
¥
z
L{x(t )} = X(s) =
¥
x(t) e - st dt =
-¥
LMt e OP
N -s Q
LM¥ ´ e
N
LM¥ ´ e
N
- st
=
=
t e - st dt
0
¥
¥
z
-
0
=
z
0
- ( s + jW ) ¥
-s
1 ´
LMt e OP - LM e OP
N -s Q N s Q
e
e O
+
P
s
s Q
e - st
dt =
-s
- 0 ´
e0
-s
- st
¥
- st
¥
=
2
0
- (s + jW )¥
0
2
2
-s ´ ¥
0
1
e - jW ´ ¥
e - s ´ ¥ e - jW ´ ¥
- 0 + 2
-s
s2
s
When, s > 0, (i.e., when s is positive), e-s¥ = e-¥ = 0
When, s < 0, (i.e., when s is negative), e-s¥ = e¥ = ¥
LMt e
N
z uv = u z v OP - LM e
-s Q
N s
- ( s + j W) t
¥
-( s + j W) t
2
0
du
OP
Q
zv
¥
0
Put,
s=s+jW
OP
Q
jW
s-plane
­
ROC
X
s =0
Therefore we can say that, X(s) converges when s > 0.
z
®s
When s > 0, the X(s) is given by,
LM¥ ´
N
L
= M¥ ´
N
X(s) =
1
e - s ´ ¥ e - jW ´ ¥
e - s ´ ¥ e - jW ´ ¥
+ 2
-s
s2
s
0 ´ e - jW ´ ¥
0 ´ e - jW ´ ¥
1
+ 2
-s
s2
s
OP =
Q
OP
Q
Fig 2 : ROC of x(t) = t u(t).
1
s2
Chapter 3 - Laplace Transform
3. 8
1
; with ROC as all points is s-plane to the right of line passing through s = 0.
s2
\ L{t u(t)} =
(or ROC is right half of s-plane).
c) Given that, x(t) = e–3t u(t) = e–3t ; t ³ 0
By definition of Laplace transform,
+¥
L{x(t )} = X(s) =
z
¥
x(t) e - st dt =
-¥
0
LM e
OP
(s
+
3)
N
Q
-( s + 3 ) t
=
=-
z
¥
e -3t e - st dt =
¥
z
e - (s +
3 )t
dt
0
e -( s + 3 ) ¥
e0
e -( s + j W + 3 ) ¥
1
= +
-(s + 3)
-(s + 3)
s + 3
s + 3
=
0
e - (s + 3 ) ´ ¥ e - jW ´ ¥
1
e - k ´ ¥ e - jW ´ ¥
1
+
= +
s + 3
s + 3
s + 3
s + 3
Put,
s=s +jW
jW
­s-plane
where, k = s + 3 = s - (- 3)
ROC
When, s > -3, k = s - (- 3) = Positive. \ e -k¥ = e-¥ = 0
When, s < -3, k = s - (- 3) = Negative. \ e -k¥ = e¥ = ¥
®s
X
s = –3
Therefore we can say that, X(s) converges when s > -3.
When s > -3, the X(s) is given by,
X(s) = -
n
e
-k ´ ¥
- jW ´ ¥
e
s + 3
s
\ L e -3 t u(t) =
Fig 3 : ROC of x(t) = e–3t u(t).
- jW ´ ¥
0 ´ e
1
= s + 3
s + 3
+
+
1
1
=
s + 3
s + 3
1
; with ROC as all points in s-plane to the right of line passing through s = - 3.
s + 3
d) Given that, x(t) = e–3t u(-t) = e–3t ; t £ 0
By definition of Laplace transform,
z
-¥
0
=
-¥
z
0
e -3t e - st dt =
-¥
LM e
OP
(s
+
3)
N
Q
-( s + 3 ) t
=
z
0
x(t) e - st dt =
e -( s +
3)t
dt
-¥
Put,
s=s +jW
e0
e( s + 3 )¥
1
e( s + j W + 3 )¥
= +
s + 3
s + 3
-(s + 3)
-(s + 3)
1
e(s + 3 ) ´ ¥ e jW ´ ¥
1
ek ´ ¥ e jW ´ ¥
= +
= +
s + 3
s + 3
s + 3
s + 3
jW
where, k = s + 3 = s - (- 3)
ROC
®
When, s > -3, k = s - (- 3) = Positive. \ ek¥ = e¥ = ¥
When, s < -3, k = s - (- 3) = Negative. \ ek¥ = e-¥ = 0
­
s-plane
–s
X
+¥
L{x(t )} = X(s) =
s = –3
®s
Therefore we can say that, X(s) converges when s < -3.
When s < -3, the X(s) is given by,
X(s) = -
n
Fig 4 : ROC of x(t) = e–3t u(–t).
0 ´ e jW ´ ¥
1
ek ´ ¥ e jW ´ ¥
1
1
+
= +
= s + 3
s + 3
s + 3
s + 3
s + 3
s
\ L e -3 t u( - t) = -
1
; with ROC as all points in s-plane to the left of line passing through s = - 3.
s + 3
3. 9
Signals & Systems
e) Given that, x(t) = e-4½t½ = e4t ; t £ 0
= e–4t ; t ³ 0
By definition of Laplace transform,
+¥
L{x(t )} = X(s) =
z
e 4t e - st dt +
-¥
OP
LM e
(s
4)
Q
N
=
z
0
-¥
-( s - 4 ) t
¥
z
x(t) e - st dt =
0
OP
LM e
(s
+
4)
Q
N
-¥
¥
0
e -( s -
4) t
dt +
-¥
LM e
N -(s - 4)
0
=
¥
0
0
-( s + 4 ) t
+
z
e -4 t e - st dt =
-
( s - 4 )¥
e
-(s - 4)
z
e -( s + 4 )t dt
0
OP + LM e
Q N -(s + 4)
-( s + 4 ) ¥
-
e0
-(s + 4)
1
e ( s + jW - 4 ) ¥
e - ( s + jW + 4 ) ¥
1
= +
+
s - 4
s - 4
s + 4
s + 4
Put,
s=s+jW
1
e( s - 4 ) ´ ¥ e jW ´ ¥
e - ( s + 4 ) ´ ¥ e - jW ´ ¥
1
= +
+
s - 4
s - 4
s + 4
s + 4
When, s < 4, s - 4 = Negative.
When, s > 4, s - 4 = Positive.
OP
Q
\ e(s - 4)¥ = e-¥ = 0
\ e(s - 4)¥ = e¥ = ¥
\ e - (s + 4)¥ = e¥ = ¥
\ e - (s + 4)¥ = e-¥ = 0
When, s < - 4, s + 4 = Negative.
When, s > -4, s + 4 = Positive.
Therefore we can say that, X(s) converges when s lies between -4 and +4.
When s lies between - 4 and + 4, the X(s) is given by,
1
e( s - 4 ) ´ ¥ e jW ´ ¥
e - (s + 4 ) ´ ¥ e - jW ´ ¥
1
+
+
s - 4
s - 4
s + 4
s + 4
jW
­
1
e -¥ e jW ´ ¥
e -¥ e - jW ´ ¥
1
= +
+
s - 4
s - 4
s + 4
s + 4
= -
1
1
+ 0 - 0 +
s + 4
s - 4
\L e
-
0 ´ e
s + 4
1
s + 4
+
1
1
s - 4 - (s + 4)
8
=
= - 2
s + 4
s - 4
(s + 4) (s - 4)
s - 16
-4 t
}
= -
®
0 ´ e
1
+
s - 4
s - 4
ROC
s-plane
X
- jW ´ ¥
= -
=
{
jW ´ ¥
X
X(s) = -
s =4
–s s = –4
Fig 5 : ROC of x(t) = e
®s
-4 t
.
(a+b)(a-b) = a2-b2
8
; with ROC as all points in s-plane in between the lines passing through s = - 4
s 2 - 16
and s = 4 .
Example 3.2
Determine the Laplace transform of the following signals.
a) x(t) = sin W0t u(t)
b) x( t) = cos W0 t u(t)
c) x( t) = cos h W0t u(t)
d) x( t) = e - a sin W0 t u(t)
e)x( t) = e - at cos W0 t u(t)
Solution
a) Given that, x(t) =sin W0 t u(t) =sin W0 t ; t ³ 0
sinq =
By definition of Laplace transform,
¥
L{x(t )} = X(s) =
z
0
=
1
2j
¥
x(t) e - st dt =
z
¥
sin W0 t e - st dt =
0
0
¥
ze
0
z
j
e -( s - jW0 )t - e -( s + jW0 ) t dt =
LM
MN
e jW0 t - e - jW0 t - st
e dt
2j
1 e - ( s - jW 0 ) t
e - ( s + jW 0 ) t
2 j -(s - jW0 )
-(s + jW0 )
OP
PQ
¥
0
e jq - e- jq
2j
Chapter 3 - Laplace Transform
=
=
3. 10
LM
N
1L
M0 - 0 + s -1jW - s +1jW OPQ
2j N
1 L s + j W - s + jW O
P = 21j LMMN s 2+jWW OPPQ =
M
2 j N (s - jW ) (s + jW ) Q
e -¥
e -¥
e0
e0
1
+
-(s + jW0 )
-(s - jW0 )
-(s + jW0 )
2 j -(s - jW0 )
0
=
0
0
0
0
2
0
\ L{sinW0 t u(t)} =
OP
Q
2
0
0
W0
s 2 + W0 2
(a+b)(a-b) = a2-b2
j2 = -1
W0
s 2 + W0 2
b) Given that, x(t) =cos W0 t u(t) =cos W0 t ; t ³ 0
By definition of Laplace transform,
¥
L{x(t )} = X(s) =
z
¥
x(t) e- st dt =
0
=
1
2
z
¥
cos W0 t e - st dt =
0
z
-( s - j W 0 ) t
W0 t
ej
0
¥
ee
z
j
W
+ e -( s + j 0 ) t dt =
0
+ e
2
- jW 0 t
LM
MN
1 e -( s - jW 0 ) t
e -( s + j W 0 ) t
+
2 -(s - jW0 )
-(s + jW0 )
LM
N
OP
1L
1
1
+
0 + 0 +
M
2N
s - jW
s + jW Q
1 L s + jW + s - jW O
P = 21 LMMN s +2sW OPPQ =
M
2 N (s - jW ) (s + jW ) Q
1
e -¥
e -¥
e0
e0
+
=
-(s + jW0 )
-(s - jW 0 )
-(s + jW 0 )
2 -(s - jW0 )
=
0
=
0
OP
Q
OP
PQ
cos q =
e jq + e - jq
2
¥
0
0
s
s 2 + W0 2
0
2
0
e - st dt
2
0
0
(a+b)(a-b) = a2-b2
j2 = -1
s
s 2 + W0 2
\ L{cosW0 t u(t)} =
c) Given that, x(t) =cosh W0 t u(t) =cosh W0 t ; t ³ 0
By definition of Laplace transform,
¥
L{x(t )} = X(s) =
z
¥
x(t) e
- st
dt =
0
=
1
2
z
¥
cosh W0 t e
z
dt =
0
-( s - W 0 ) t
z
0
¥
ee
- st
j
+ e - (s + W0 )t dt =
0
eW 0 t + e - W0 t - st
e dt
2
LM
MN
1 e -( s - W 0 ) t
e -( s + W 0 ) t
+
2 -(s - W0 )
-(s + W0 )
LM
N
1L
M0 + 0 + s -1 W + s +1 W OPQ
2N
1 Ls + W + s - W O
P = 21 LMMN s -2sW OPPQ =
M
2 N (s - W ) (s + W ) Q
1
e -¥
e -¥
e0
e0
=
+
-(s + W 0 )
-(s - W0 )
-(s + W0 )
2 -(s - W 0 )
=
0
=
0
\ L{coshW0 t u(t)} =
0
0
s
s 2 - W0 2
eq + e - q
2
¥
0
0
2
0
OP
Q
OP
PQ
cosh q =
2
0
s
s 2 - W0 2
(a+b)(a-b) = a2-b2
3. 11
Signals & Systems
d) Given that, x(t) = e - at sin W0 t u(t) = e- at sin W0 t ; t ³ 0
By definition of Laplace transform,
¥
L{x(t )} = X(s) =
z
¥
x(t) e- st dt =
0
=
z
¥
e - at sin W0 t e - st dt =
0
z
ee
0
-¥
- j W0 t
e - st dt
e - (s + a + jW 0 )t
-(s + a + jW0 )
-
0
0
0
0
0
0
0
2
2
2
2
2
0
cos q =
By definition of Laplace transform,
¥
L{x(t )} = X(s) =
z
¥
x(t) e- st dt =
0
1
2
2
2
W0
(s + a)2 + W0 2
e) Given that, x(t) = e - at cos W0 t u(t) = e - at cos W0 t ; t ³ 0
z
¥
e- at cos W0 t e - st dt =
0
z
e- at
0
¥
ze
e jq + e- jq
2
e jW0 t + e - jW 0 t - st
e dt
2
LM
MN
OP
PQ
OP
e
-(s + a + jW ) Q
e -( s + a + jW0 )t
1 e - (s + a - j W 0 ) t
+
2 -(s + a - jW0 )
-(s + a + jW0 )
j
e -( s + a - jW0 )t + e - (s + a + jW0 )t dt =
0
¥
0
LM
N
1L
M0 + 0 + s + a1- jW + s + a1+ jW OPQ = 21 LMN (ss ++ aa +- jWjW )+(ss ++ aa +- jWjW ) OPQ
2N
1L
M 2(s + a) OP = (s + sa)+ a+ W
2 MN (s + a) + W PQ
j = -1
(a + b)(a - b) = a - b
e -¥
e -¥
e0
1
=
+
-(s + a + jW0 )
-(s + a - jW0 )
2 -(s + a - jW0 )
0
0
0
0
=
0
0
\ L{e - at sin W0 t u(t)} =
=
¥
0
0
0
=
OP
PQ
0
0
0
=
- e
2j
OP
e
e
e
e
1L
+
M
-(s + a + jW )
-(s + a - jW )
-(s + a + jW ) Q
2 j N -(s + a - jW )
L
O
L
1
M0 - 0 + s + a1- jW - s + a1+ jW PQ = 21j MN (s(s+ +a a+ -jWjW) )-(s(s++a a+ -jWjW) ) OPQ
2j N
OP =
W
1L
M 2 jW
2 j MN (s + a) + W PQ
(s + a) + W
j = -1
(a + b)(a - b) = a - b
0
=
W 0t
LM
MN
- e (s + a + jW 0 )t dt =
-¥
=
ej
e - at
e -( s + a - jW0 )t
1
2 j -(s + a - jW0 )
j
- (s + a - j W 0 ) t
z
0
¥
1
2j
e jq - e- jq
2j
sinq =
2
0
2
2
0
0
0
2
0
2
0
2
2
s + a
(s + a)2 + W0 2
\ L{e - at cos W0 t u(t)} =
Example 3.3
Determine the Laplace transform of the signals shown below.
a)
b)
c)
d)
x(t)
x(t)
A
0
x(t)
x(t)
a
1
1
T
2
T
t
0
–A
Fig 3.3.1.
a
t
Fig 3.3.2.
0
a
Fig 3.3.3.
t
0
a
t
Fig 3.3.4.
Chapter 3 - Laplace Transform
3. 12
Solution
a)
The mathematical equation of the signal shown in fig 3.3.1 is,
x (t) = A
;for 0 < t < T/2
= –A ;for T/2 < t < T
By definition of Laplace transform,
¥
L{x(t)} = X(s) =
z
z
T
x(t ) e - st dt
=
-¥
z
0
T/2
=
LM A e
MM -s
N
= -
T/2
OP
Q
0
- sT
2
z
A e - st dt +
- sT
2
=
b-Ag e dt = LM A -es
N
O LM A e
Ae P
Ae
+ M
- s PP
s
s
Q MN
T
0
=
x( t) e - st dt
-
- sT
2
- st
- st
- sT
T/2
LM -A e OP
N -s Q
- st
+
0
T
T/2
OP
PP
Q
- sT
2
Ae
s
+
A
A e - sT
Ae
+
s
s
s
LM
MN
A
1 + e - sT - 2 e
s
- sT
2
OP =
PQ
LM
MN
A
1 - e
s
- sT
2
OP
PQ
2
(a - b)2 = a2 + b2 - 2ab
b)
The mathematical equation of the signal shown in fig 3.3.2 is,
x(t) = 1 ;for 0 £ t £ a
= 0 ;for t > a
By definition of Laplace transform,
¥
L{x(t)} = X(s) =
z
-¥
- as
=
e
-s
-
z
1 ´ e- st dt =
+
1
1
1 - e - as
=
s
s
a
x(t ) e - st dt =
a
0
0
- as
e
e
= -s
s
z
e- st dt =
0
d
LM e OP
N -s Q
- st
a
0
i
c)
x(t)
To Find Mathematical Equation for x(t)
Consider the equation of straight line,
Here, y = x(t),
y - y1
x - x1
=
y1 - y 2
x1 - x 2
x = t.
\ The equation of straight line can be written as,
Consider points P and Q, as shown in fig 1.
Coordinates of point - P = [t1, x(t1)] = [0, 0]
Coordinates of point - Q = [t2, x(t2)] = [a, a]
Q
a
P
0
a
t
Fig 1.
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
.....(1)
3. 13
Signals & Systems
On substituting the coordinates of points - P and Q in equation - (1) we get,
x( t) - 0
t - 0
=
0 - a
0 - a
\ x(t) = t
x( t)
t
=
-a
-a
Þ
Þ
x(t) = t
; for t = 0 to a
= 0
;for t > a
To Evaluate Laplace transform of x(t)
+¥
z
L{x(t)} = X(s) =
z
a
x(t) e - st dt =
-¥
LMt ´ e
N -s
L ae
= MN s
- st
z
-
=
- sa
t e - st dt
0
-
e - st
dt
-s
1´
- sa
OP
Q
a
LM- t e
N s
- st
=
0
e
e0
+ 0 + 2
s2
s
OP =
Q
-
e - st
s2
- as
OP
Q
a
z uv = uz v - z duz v
u=t
0
v = e -st
- as
1
e
ae
s2
s2
s
1
1 - e - as (1 + as)
s2
=
d)
To Find Mathematical Equation for x(t)
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Consider the equation of straight line,
Here, y = x(t),
x = t.
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
\ The equation of straight line can be written as,
.....(1)
Consider points P and Q, as shown in fig 1.
Coordinates of point - P = [t1, x(t1)] = [0, 1]
Coordinates of point - Q = [t2, x(t2)] = [a, 0]
x(t)
On substituting the coordinates of points - P and Q in equation - (1) we get,
x( t) - 1
t - 0
=
0 - a
1 - 0
t
\ x(t) = 1 a
Þ
x(t ) - 1 = -
t
a
Þ
x(t) = 1 -
1
P
t
a
Q
0
; for t = 0 to a
a
t
Fig 1.
= 0
; for t > a
To Evaluate Laplace transform of x(t)
+¥
L{x(t )} = X(s) =
z
x(t ) e - st dt
-¥
FG1- t IJ e dt =
H aK
L e OP - 1 LMt ´ e
=M
N -s Q a N -s
z
a
=
- st
0
- st
a
0
- st
z
a
e - st dt -
0
-
z
1
a
z
a
t e - st dt
0
e - st
1´
dt
-s
OP
Q
a
z uv = uz v - z duz v
u=t
0
v = e - st
Chapter 3 - Laplace Transform
3. 14
LM e OP - 1 LM- t e - e OP
s Q
N s Q aN s
L e + e OP - 1 LM- a e = MN s s Q aN s
a
- st
= -
- st
a
- st
2
0
0
- as
e - as
e0
+ 0 + 2
2
s
s
OP
Q
1
e - as
1
e - as
e - as
+
+
+
s
s
s
as2
as2
= =
- as
0
1
e - as
1
1
e - as + a s - 1
+
=
s
as2
as2
as2
Example 3.4
x(t)
Determine the Laplace transform of the sine pulse shown in fig 3.4.1.
A
Solution
p
The mathematical equation of the sine pulse shown in fig 3.4.1. is,
x(t) = A sin t ;for 0 < t < p
t
Fig 3.4.1.
;for t > p
=0
¥
L{x(t)} = X(s) =
p
z
x(t) e - st dt =
0
z
p
sin t e
- st
dt = A
0
=
=
A
2j
=
=
=
0
z
e -( s - j)t - e - (s + j) t dt =
0
LM e
N (s + j)
A L (s - j) e
M
2j N
-( s + j)t
A
2j
=
=
A
d
i
2j s2 + 1
A
d
i
2j s 2 + 1
A
d
2
i
2j s + 1
A
d
i
2j s 2 + 1
A
d
z
e jt - e - jt - st
e dt
2j
p
- st
=
A sin t e- st dt
0
p
= A
z
e
s2
-( s - j)t
p
-
e -( s + j)t
-(s + j)
=
0
- (s + j)t
A
2j
- st
OP
Q
e jq - e - jq
2j
p
0
-( s - j)t
OP
Q
p
0
p
jt
2
0
(s - j) e - st e - jt - (s + j) e - st e jt
p
0
(s - j) e - sp e - jp - (s + j) e- sp e jp - (s - j) e0 + (s + j) e0
-(s - j) e - sp + (s + j) e - sp - (s - j) + (s + j)
e ± jp = cos p ± j sin p
= -1 ± j0 = -1
e - ps ( -s + j + s + j) - s + j + s + j
2j e
id
2j s 2 + 1
e
(s - j)
- jt
LM e
N -(s - j)
LM (s - j)e - (s + j) e
(s + j)(s - j)
N
- (s + j) e e O
PQ
- j
-( s - j)t
OP
Q
A
2j
sinq =
- ps
i
+ 2j =
A
e - ps + 1
s2 + 1
d
i
3. 15
Signals & Systems
Table 3.1 : Laplace Transform of Some Standard Signals
Waveform
x(t)
X(s) = L{x(t)}
x(t)
x(t) = A ; 0 < t < T
=0 ; t>T
A
0
T
X(s) =
A
(1 - e-sT)
s
X(s) =
A
1 - e -sT (1 + sT)
Ts2
X(s) =
2A
1 - e
Ts2
t
x(t)
At
; 0<t<T
T
= 0 ; t>T
A
x(t) =
0
t
T
x(t)
A
x(t) =
0
T
2
; 0< t<
T
2
2At T
;
<t<T
T
2
= 2A -
t
T
2At
T
F
GH
- sT
2
I
JK
2
x(t)
A
x(t) = A ; 0 < t <
0
T
2
T
t
= –A ;
–A
T
2
T
<t<T
2
F
GH
sT
A
1 - e 2
X(s) =
s
I
JK
2
x(t)
x(t) = A sin t ; 0 < t < T
=0
;
t>T
A
T
0
t
X(s) =
A
e - sT + 1
s + 1
2
d
i
x(t)
2
1
0
-1
-2
1
2
3
4
5
x(t) = 1
= –2
= 2
t
= –2
= 0
;
;
;
;
;
0<t<1
1<t<2
2<t<4
4<t<5
t>5
X(s) =
1
(1 - 3e-s + 4e-2s - 4e-4s + 2e-5s)
s
Chapter 3 - Laplace Transform
3. 16
Table 3.1 Continued.....
Waveform
x(t)
X(s) = L{x(t)}
x(t)
x(t) = A sin t ; 0 < t < T
A
0
2T
T
t
3T
and x(t + nT) = x(t)
x(t)
A
0
T
T
2
t
3T
2
x(t)
T
x(t) = A sin t ; 0 < t <
2
T
=0
;
<t<T
2
and x(t + nT) = x(t)
x(t) =
A
T
2
2At
T
; 0< t <
=A -
–A
X(s) =
F
GH
1 + e - sT
A
s + 1 1 - e - sT
2
I
JK
A
ds + 1i FGH1 - e IJK
-
2
T
2
2At T
;
<t<T
T
2
and x(t + nT) = x(t)
t
2T
3T
2
T
X(s) =
sT
2
LM FG TsIJ e OP
MN H 2 K PQ
F
I
Ts G 1 + e J
H
K
-
2A 1 - 1 +
X(s) =
-
2
sT
2
sT
2
x(t)
A
x(t) = A 2T
T
3T
t
2At
;0<t<T
T
F
GH
I
JK
X(s) =
2A T 1 + e - sT
1
- sT
Ts 2 1 - e
s
X(s) =
A 1 - e - as
s 1 - e - sT
and x(t + nT) = x(t)
–A
x(t)
x(t) = A ; 0 < t < a
=0 ; a<t<T
A
0
a
T+a
T
2T
t
2T+a
and x(t + nT) = x(t)
x(t)
x(t) = A
A
0
T
T
2
3T
2
–A
x(t)
t
T
2
T
= –A ;
<t<T
2
and x(t + nT) = x(t)
At
; 0 < t < T
T
and x(t + nT) = x(t)
A
X(s) = s
x(t) =
A
0
2T
; 0<t<
T
2T
3T
t
X(s) =
F1 - e
GG
H1 + e
A
Ts2
-
sT
2
-
sT
2
I
JJ
K
LM1 - e (1 + sT) OP
N 1-e
Q
- sT
- sT
3. 17
Signals & Systems
Table 3.2 : Some Standard Laplace Transform Pairs
Note : s = Real part of s
x(t)
X(s)
ROC
d(t)
1
Entire s-plane
u(t)
1
s
1
s2
s>0
t u(t)
tn - 1
u(t)
(n - 1)!
s>0
1
sn
s>0
where, n = 1, 2, 3, ......
e -at u(t)
–e -at u(–t)
1
s + a
1
s + a
s > –a
s < –a
n
t u(t)
n!
where, n = 1, 2, 3 .....
t e -at u(t)
1
t n - 1 e - at u(t)
(n - 1)!
where, n = 1, 2, 3, .....
tn e -at u(t)
s
n+1
s>0
1
(s + a) 2
s > –a
1
(s + a) n
s > –a
n!
(s + a) n + 1
s > –a
where, n = 1, 2, 3, .....
bg
cosW t ub t g
sin h W t ub t g
sin W 0 t u t
0
0
W0
s + W 02
s
s>0
2
s2 + W 0
s>0
2
W0
s > W0
s - W 02
s
2
bg
cos h W 0 t u t
e - at sin W 0 t u t
bg
e - at cosW 0 t u t
bg
s2 - W 0
s >W0
2
W0
(s + a) 2 + W 0 2
s > –a
s+a
s > –a
(s + a) 2 + W 0
2
Chapter 3 - Laplace Transform
3. 18
3.3 Properties and Theorems of Laplace Transform
The properties and theorems of Laplace transform are listed in table 3.3. The proof of properties
and theorems are presented in this section.
1. Amplitude Scaling
In amplitude scaling, if the amplitude (or magnitude) of a time domain signal is multiplied by a
constant A, then its Laplace transform is also multiplied by the same constant.
i.e., if L{x(t)} = X(s), then
L{A x(t)} = A X(s)
Proof :
By definition of Laplace transform,
z
+¥
X ( s ) = L {x( t ) } =
x ( t ) e - st dt
.....(3.8)
-¥
z
+¥
m
r
L A x (t ) =
A x(t ) e - st dt
-¥
z
+¥
= A
x(t ) e - st dt
-¥
Using equation (3.8)
= A X(s)
2. Linearity
The linearity property states that, Laplace transform of weighted sum of the two or more signals is
equal to similar weighted sum of Laplace transforms of the individual signals.
i.e., if L{x1(t)} = X1(s) and L{x2(t)} = X2(s), then
L{a1 x1(t) + a2 x2(t)} = a1 X1(s) + a2 X2(s)
Proof :
By definition of Laplace transform,
z
z
+¥
X 1( s) = L {x 1(t ) } =
x 1( t) e - st dt
.....(3.9)
x 2 ( t) e - st dt
.....(3.10)
-¥
+¥
X 2 ( s ) = L {x 2 ( t } =
-¥
z
+¥
m
r
L a1 x 1( t) + a2 x2 (t) =
a1 x 1( t) + a2 x 2 ( t) e - st dt
-¥
z
+¥
= a1
z
+¥
x 1( t) e - st dt + a2
-¥
= a1 X 1 (s) + a2 X 2 (s)
-¥
x 2 ( t) e - st dt
Using equations
(3.9) and (3.10)
3. 19
Signals & Systems
3. Time Differentiation
The time differentiation property states that if a causal signal x(t) is piecewise continuous, and
d
x( t ) is given by sX(s) - x(0).
Laplace transform of x(t) is X(s) then, Laplace transform of
dt
i.e., If L{x(t)} = X(s), then
L
RS d x(t )UV = sX(s) - x(0)
T dt W
; where, x(0) is value of x(t) at t = 0.
Proof :
By definition of Laplace transform, the Laplace transform of a causal signal is given by,
z
z
¥
X ( s) = L {x(t)} =
x(t ) e - st dt
.....(3.11)
0
\L
RS d x(t)UV =
T dt W
¥
0
= e
z
¥
dx( t) - st
e dt =
dt
- st
x( t)
z
0
dx(t )
dt
dt
¥
¥
0
e - st
-se
- st
zu v = u z v - z
x(t ) dt
u = e - st
v=
0
z
du
zv
dx(t )
dt
¥
= e -¥ x(¥) - e 0 x( 0) + s
x( t) e - st dt
e -¥ = 0 and e 0 = 1
0
z
¥
= s
x( t) e - st dt - x(0) = s X(s) - x( 0)
Using equation (3.11)
0
4. Time Integration
The time integration property states that, if a causal signal x(t) is continuous and Laplace transform
z
X(s)
of x(t) is X(s), then the Laplace transform of x( t ) dt is given by,
+
s
i.e., If L{x(t)} = X(s), then
L
z
{
X(s)
x( t ) dt =
+
s
}
z
z x(t) dt
t= 0
s
x( t ) dt
t=0
s
Proof :
By definition of Laplace transform, the Laplace transform of a causal signal is given by,
z
z LMNz
¥
m r
X(s) = L x(t ) =
x(t ) e - st dt
.....(3.12)
0
\L
{
z
¥
}
x( t) dt =
OP
Q
LML x(t) dtO
PQ
MNMN
x(t ) dt e - st dt
0
=
z
e - st
-s
OP
PQ
¥
0
z
¥
0
e - st
x(t )
dt
-s
z
uv = u
z
z z z
v -
u = x(t) dt
du
v
v = e - st
Chapter 3 - Laplace Transform
3. 20
z
x( t) dt
1
s
z
LM
N
=
=
LM
N
t=¥
x(t ) dt
X(s)
+
s
=
OP
Q
LM
N
OP
Q
e -¥
-s
LM
N
1
s
z
+
t= 0
z
x(t ) dt
OP
Q
z
x(t ) dt
OP
Q
e0
1
+
s
-s
t= 0
z
¥
x(t ) e - st dt
0
¥
x( t) e - st dt
e -¥ = 0 and e 0 = 1
0
Using equation (3.12)
t= 0
s
5. Frequency shifting
The frequency shifting property of Laplace transform says that,
If, L x(t) = X( s), then
l q
L ne x( t )s = X(s m a )
n
± at
s
n
s
i. e., L e at x( t ) = X(s - a ) and L e - at x( t ) = X(s + a )
Proof :
By definition of Laplace transform,
z
z
z
+¥
X ( s ) = L {x( t ) } =
x ( t ) e - st dt
.....(3.13)
-¥
+¥
The term
+¥
{
\L e
± at
}=
x(t)
e
± at
x(t) e
- st
dt
- (s m a )t
dt is similar to the form
-¥
of definition of Laplace transform (equation(3.13))
except that s is replaced by (s m a ) .
-¥
+¥
+¥
x(t ) e -( s m a) t dt
=
z x(t) e
\
-¥
z x(t) e
b
-(s m a ) t
g
dt = X s m a
-¥
= X(s m a)
6. Time shifting
The time shifting property of Laplace transform says that,
l q
L lx( t ± a )q = e
If, L x(t) = X(s), then
± as
l
q
l
q
i.e., L x( t + a ) = e as X(s) and L x( t - a ) = e - as X(s)
X(s)
Proof :
By definition of Laplace transform,
z
z
z
+¥
X ( s ) = L {x( t ) } =
x ( t ) e - st dt
.....(3.14)
-¥
+¥
m
r
\ L x(t ± a) =
z
+¥
x(t ± a) e
-¥
- st
dt =
x( t) e
- s( t m a )
+¥
=
dt
-¥
z
Let, t ± a = t
\t = t m a
On differentiating
dt = dt
+¥
x( t ) e - st ´ e ± as dt = e ± as
-¥
-¥
z
+¥
= e ± as
-¥
x(t ) e - st dt = e ± as X ( s)
x( t ) e - st dt
Since t is a dummy
variable for integration
we can change t to t.
Using equation (3.14)
3. 21
Signals & Systems
7. Frequency Differentiation
The frequency differentiation property of Laplace transform says that,
i.e., If L{x(t)} = X(s), then
d
L t x( t ) = - X(s)
ds
l
q
Proof :
By definition of Laplace transform,
z
+¥
X ( s) = L {x(t)} =
x( t) e - st dt
-¥
On differentiating the above equation with respect to s we get,
d
d
X( s ) =
ds
ds
F
GG
H
z
+¥
x(t ) e - st dt
-¥
z FGH
zb
+¥
=
x( t)
-¥
I
JJ
K
IJ
K
d - st
e
dt =
ds
+¥
g
e
Interchanging the order of
integration and differentiation
j
x(t) -t e - st dt
-¥
m
-t x( t) e - st dt
=
z
+¥
r
m r
= L -t x( t) = - L t x( t)
-¥
d
X( s )
ds
m r
\ L t x(t ) = -
8. Frequency Integration
The frequency integration property of Laplace transform says that,
i.e., If L{x(t)} = X(s), then
L
RS1 x(t)UV =
Tt W
z
¥
X(s) ds
s
Proof :
By definition of Laplace transform,
z
+¥
X ( s) = L {x(t)} =
x( t) e - st dt
-¥
On integrating the above equation with respect to s between limits s to ¥ we get,
z
¥
z LMMN z
z LMMNz
z LMMN
z LMMN
¥
X ( s) ds =
s
+¥
s
-¥
¥
+¥
=
x(t)
+¥
x(t )
-¥
+¥
=
-¥
e - st
s
-¥
=
OP
PQ
O
ds P dt
PQ
x( t) e - st dt ds
e - st
-t
OP
PQ
¥
Interchanging the
order of integrations.
z
+¥
dt =
s
-¥
OP
PQ
e - st
x(t) 0 +
dt =
t
LM e - e OP dt
MN -t -t PQ
LM 1 x(t)OP e dt = L RS 1 x(t)UV
Tt W
Nt Q
-¥
x(t)
z
+¥
-¥
- st
- st
Chapter 3 - Laplace Transform
3. 22
9. Time scaling
The time scaling property of Laplace transform says that,
If L{x(t)} = X(s), then
l q
L x(at) =
1
X as
a
ej
Proof :
By definition of Laplace transform,
z
+¥
X ( s) = L {x(t)} =
x(t ) e - st dt
.....(3.15)
-¥
z
+¥
\ L { x(at) } =
-¥
-s
x( t ) e
FG t IJ
H aK
z
+¥
-
x( t) e
FG s IJ t
H aK
dt =
-¥
Put, at =
dt
a
-¥
1
a
=
z
+¥
x(at) e - st dt =
t
t
a
On differentiating
\t =
FH IK
1
X s
a
a
dt
a
dt =
The above transform is applicable for positive values of "a".
z
+¥
If "a" happens to be negative it can be proved that,
The term
-
x( t) e
FG s IJ t
H a K dt
is similar
-¥
L{x(at)} = -
1
X
a
to the form of definition of Laplace
transform(equation(3.15)) except
di
s
a
that s is replaced by
Hence in general,
1
X
a
di
L{x(at)} =
z
+¥
s
a
\
for both positive and negative values of "a"
-
x( t) e
FG s IJ t
H a K dt
-¥
FG s IJ .
H aK
FH IK
= X s
a
10. Periodicity
The periodicity property of Laplace transform says that,
T
l q z x (t) e
If x(t) = x(t+nT), and x1(t) be one period of x(t), and L x1 (t) =
l
1
1 - e - sT
q
L x(t + nT) =
z
1
- st
dt , then
0
T
x1 (t) e - st dt
0
Proof :
By definition of Laplace transform,
z
z
¥
L { x(t + nT) } =
x(t + nT) e - st dt
0
=
0
z
x 1 (t) e - st dt +
z
z
3T
2T
T
x 1( t - T ) e - s( t + T ) dt +
x1 (t - 2 T ) e - s( t + 2 T ) dt + ..............
2T
T
(p + 1) T
............ +
pT
x 1( t - pT ) e - s( t + pT ) dt + ..........
3. 23
Signals & Systems
z
( p + 1) T
¥
å
=
p=0
å
p=0
z
z
z
pT
z
T
¥
=
x1 (t - pT ) e - s( t + pT ) dt
0
T
=
T
x 1(t ) e - st
0
1
1 - e - sT
=
- sT
p=0
0
- sT
z
Interchanging the order of
integration and summation
p
¥
x 1(t ) e - st
I dt
JK
I dt
JK
IJ dt
K
- psT
p=0
T
=
F e
GH å
F e
GH å
FG 1
H1 - e
¥
x 1(t ) e - st
0
=
The periodic signal will be
identical in every period
and so, x1(t+pT) = x1(t).
x 1( t) e - st e -psT dt
Using infinite geometric
series
sum formula
¥
1
Cn =
1 - C
n=0
å
1
1 - e- sT
is independent of t
T
The term
x 1 (t) e - st dt
0
11. Initial Value Theorem
The initial value theorem states that, if x(t) and its derivative are Laplace transformable then,
Lt x( t ) = Lt s X(s)
t® 0
s®¥
i. e., Initial value of signal, x(0) =
Lt x(t) =
t®0
Lt
s®¥
s X(s)
Proof :
We know that, L
RS dx(t) UV = s X(s) - x(0)
T dt W
On taking limit s ® ¥ on both sides of the above equation we get,
Lt
L
Lt
z
s®¥
RS dx(t) UV =
T dt W
dx(t ) - st
e dt =
dt
0
z
¥
0
dx(t)
dt
0 =
s X( s) - x(0 )
By definition of Laplace transform,
¥
s®¥
Lt
s®¥
FH
Lt
s® ¥
s® ¥
\ x( 0) =
Lt
IK
Lt
s® ¥
e - st dt =
FH
s X( s) - x( 0)
Lt
s X ( s) - x(0 )
s® ¥
IK
s X ( s) - x(0 )
L
RS dx(t) UV =
T dt W
z
¥
0
dx(t) - st
e dt
dt
dx(t)
and x(0) are
dt
not functions of s
Here
Lt
s®¥
Lt
s® ¥
Lt x( t) =
\ t®
0
e - st = 0
s X ( s)
Lt
s® ¥
s X ( s)
x(0 ) = Lt x(t )
t®0
Chapter 3 - Laplace Transform
3. 24
12. Final Value Theorem
The final value theorem states that if x(t) and its derivative are Laplace transformable then
Lt x( t ) = Lt s X(s)
t®¥
s®0
i.e., Final value of signal, x(¥ ) =
Lt x(t) = Lt s X(s)
t®¥
s®0
Proof :
We know that, L
RS dx(t) UV = s X( s) - x(0)
T dt W
On taking limit s ® 0 on both sides of the above equation we get,
Lt L
s®0
RS dx(t) UV =
T dt W
Lt
s® 0
By definition of Laplace transform
s X( s) - x(0)
L
z
¥
Lt
s®0
0
z
z
¥
0
¥
0
dx(t)
dt
dx(t) - st
e dt =
dt
FH Lt
s®0
IK
e - st dt =
FH
s X(s) - x(0)
IK
Lt s X(s) - x(0)
s®0
z
¥
0
dx( t) - st
e dt
dt
dx(t)
and x(0) are
dt
not functions of s
Here
Lt e- st = 1
dx(t)
dt =
dt
x(t)
¥
0
=
x( ¥ ) - x(0) =
\ x( ¥) =
\
Lt
s®0
RS dx(t) UV =
T dt W
Lt x( t) =
t®¥
s®0
Lt s X ( s) - x(0)
s®0
Lt s X ( s) - x(0)
s®0
Lt s X ( s ) - x(0)
s®0
Lt s X ( s)
s®0
Lt s X( s)
x(¥ ) =
s®0
Lt
t® ¥
x(t)
13. Convolution Theorem
The convolution theorem of Laplace transform says that, Laplace transform of convolution of
two signals is given by the product of the Laplace transform of the individual signals.
i.e., if L{x1(t)} = X1(s) and L{x2(t)} = X2(s) then,
L{x1(t) * x2(t)} = X1(s) X2(s)
.....(3.16)
The equation (3.16) is also known as convolution property of Laplace transform.
With reference to chapter-2, section 2.9 we get,
z
+¥
x1 ( t ) * x 2 ( t ) =
x1 (l ) x2 ( t - l ) dl
-¥
where, l is a dummy variable used for integration.
....(3.17)
3. 25
Signals & Systems
Proof :
Let x1(t) and x2(t) be two time domain signals.
By definition of Laplace transform,
z
z
+¥
X 1( s) = L {x 1(t ) } =
x 1( t) e - st dt
.....(3.18)
x 2 ( t) e - st dt
.....(3.19)
-¥
+¥
X 2 ( s ) = L {x 2 ( t } =
-¥
Let x3(t) be the signal obtained by convolution of x1(t) and x2(t).Now from equation (3.17) we get,
z
+¥
x 3 (t) = x1(t) * x2 (t) =
x1(l ) x2 (t - l ) dl
.....(3.20)
-¥
Let, L{x3(t)} = X3(s). Now by definition of Laplace transform we can write,
z
+¥
m r
X 3 (s) = L x 3 (t =
x 3 ( t) e - st dt
..... (3.21)
-¥
On substituing for x3(t) from equation (3.20) in equation (3.21) we get,
z LMMN z
+¥
X 3 ( s) =
-¥
OP
PQ
+¥
x 1 (l ) x2 ( t - l ) dl e - st dt
-¥
Let, e–st = esl ´ e–sl ´ e–st = e–sl ´ e–s(t – l) = e–sl ´ e–sM
where, M = t – l and so, dM = dt
..... (3.22)
..... (3.23)
.....(3.24)
Using equations (3.23) and (3.24), the equation (3.22) can be written as,
zz
z
+¥ +¥
X 3 ( s) =
x 1( l ) x 2 (M) e - sl e - sM dl dM
-¥ -¥
+¥
=
z
+¥
x 1( l ) e - sl dl ´
-¥
x 2 (M ) e - sM dM
.....(3.25)
-¥
In equation (3.25), l and M are dummy variables used for integration, and so they can be changed to t.
Therefore equation (3.25) can be written as,
z
+¥
X 3 ( s) =
z
+¥
x 1 (t) e - st dt ´
-¥
x2 ( t) e - st dt
-¥
= X 1 ( s) X 2 ( s )
m
r
\ L x 1 ( t ) * x 2 ( t ) = X 1 ( s) X 2 ( s)
Using equations
(3.18) and (3.19)
Chapter 3 - Laplace Transform
3. 26
Table 3.3 : Properties of Laplace Transform
Note : L{x(t)} = X(s); L{x1(t)} = X1(s); L{x2(t)} = X2(s)
Property
Time domain signal
s-domain signal
Amplitude scaling
A x(t)
A X(s)
Linearity
a1x1(t) ± a2 x2(t)
a1 X1(s) ± a2 X2(s)
Time differentiation
d
x(t)
dt
s X(s) - x (0)
dn
x(t)
dt n
sn X(s) - å sn - K
d (K - 1) x(t)
dt K - 1
n
K=1
where n = 1, 2, 3 ......
Time integration
z
z z
X(s)
+
s
x(t) dt
.....
x(t) (dt) n
where n = 1, 2, 3 ......
e ± at x(t)
X(s m a)
Time shifting
x( t ± a )
e ±as X(s)
tn x(t)
t=0
s
n
X(s)
1
+ å n-K+1
n
K=1 s
s
Frequency shifting
Frequency differentiation t x(t)
z x(t) dt
t=0
LM
N
z z
.... x( t ) (dt) k
dX(s)
ds
dn
( -1) n
X(s)
ds n
-
where n = 1, 2, 3 ......
1
x(t)
t
z
x(at)
1
X s
a
a
x(t + nT)
1
1 - e - sT
¥
Frequency integration
Time scaling
Periodicity
X(s) ds
s
ej
z
T
0
x1 (t) e - st dt
where, x1(t) is one period of x(t).
Initial value theorem
Final value theorem
Lt x(t) = x(0)
t® 0
Lt s X(s)
s®¥
Lt x(t) = x(¥)
t®¥
Lt s X(s)
s® 0
x1 (t) * x2 (t)
Convolution theorem
z
+¥
=
-¥
x1 (l ) x 2 (t - l ) dl
X1 (s) X 2 ( s)
OP
Q
t=0
3. 27
Signals & Systems
Example 3.5
x(t)
Determine Laplace transform of periodic square wave shown in fig 3.5.1.
A
T
2T
t
3T
–A
Solution
Fig 3.5.1.
The given waveform satisfy the condition, x(t + nT) = x(t), and so it is periodic.
Let x1(t) be one period of x(t). The equation for one period of the periodic waveform of fig 3.5.1 is,
x1( t) = A ; for t = 0 to
T
2
T
to T
2
= - A ; for t =
From periodicity property of Laplace transform,
l q
If X(s) = L x( t) , and if x( t) = x(t + nT ) then, X(s) =
1
\ L x(t) = X(s) =
1 - e - sT
l q
z
z
T
1
1 - e - sT
x1(t) e - st dt, where x1( t) is one period of x( t).
0
T
x1(t) e - st dt
0
Using the result of example 3.3(a), the above equation can be written as,
X(s) =
sT
2
2
-
=
sT
2
-
-
sT
2
-
sT
2
sT
2
F1 - e I
GH
JK
-
sT
2
2
2
a 2 - b 2 = (a + b) (a - b)
sT
2
Example 3.6
A
s
x1(t) e - st dt =
0
- sT
-
=
z
T
LM A F
IO
1
1 - e J P
G
M
1 - e
K PQ
Ns H
LM A F
IO
1
1 - e J P
G
F1 + e I F1 - e I MN s H
K PQ
JK
GH
JK GH
F
I
A G1 - e
JJ
s GG
H 1 + e JK
-
From example 3.3(a) we get,
x(t)
Determine the Laplace transform of the periodic
signal whose waveform is shown in fig 3.6.1.
p
2p
3p
0
4p
5p
6p
t
–1
Fig 3.6.1.
Solution
The given waveform satisfies the condition, x(t + nT) = x(t), and so it is periodic. Here, T = 2p.
Let x1(t) be one period of x(t). The equation for one period of the periodic waveform of fig 3.6.1 is,
x1(t) = 0
; for t = 0 to p
= sin t ; for t = p to 2p
From periodicity property of Laplace transform,
l q
If X(s) = L x( t) , and if x( t) = x(t + nT ) then, X(s) =
1
1 - e - sT
z
T
0
x1(t) e - st dt, where x1( t) is one period of x( t).
Chapter 3 - Laplace Transform
3. 28
1
1 - e - sT
l q
\ X(s) = L x( t) =
1
=
1 - e -2 ps
LM
MN
z
z
=
=
=
=
=
=
1
1 - e -2 ps
0
z
2p
0´e
- st
dt +
0
e
- e
2j
p
- jt
d
2j 1- e
x1( t) e - st dt
0
O
dt P =
PQ
Put, T = 2p
1
1 - e -2 ps
iz
2p
1
e - st dt =
d
2 j 1 - e -2 ps
e
b g
z
2p
- s- j t
sin t e
- st
dt
sinq =
p
- e
b g
- s+j t
dt
p
- s+ j t
2p
jt
-2 ps
- st
- jt
2p
- st
-2 ps
p
p
j2 p -2 ps
1
d
e jq - e - j q
2j
LM e b g - e b g OP =
Lee e e O
1
Mi N -(s - j) -bs + jg PQ 2jd1 - e i MN- s - j + s + j PQ
LM- e e + e e + e e - e e OP
s+ j
s-j
s+ j Q
iN s-j
LM- e + e - e + e OP
e = -1 and e
=1
+
j
+
j
s
j
s
s
j
s
iN
Q
a - b = (a + b) (a - b)
LM -(s + j) e + (s - j) e - (s + j) e + (s - j) e OP
j = -1
(s
j
)
(s
+
j)
iN
Q
LM -se - je + se - j e - se - je + se - je OP
s +1
iN
Q
LM -2je - 2je OP = -2je d1+ e i =
e
i N s + 1 Q 2j ds + 1id1 + e id1- e i ds + 1ide - 1i
- s- j t
1
sin t ´ e
- st
p
jt
z
2p
1
1 - e -2 ps
x1(t ) e - st dt =
p
2p
=
z
T
- j2 p
-2 ps
jp
- ps
- jp - ps
2 j 1 - e -2 ps
-2 ps
1
d
2j 1- e
-2 ps
- ps
- ps
± jp
-2 ps
2
-2 ps
1
d
-2 ps
± j2 p
2
- ps
- ps
2
2 j 1 - e -2 ps
-2 ps
1
d
2j 1- e
d
-2 ps
-2 ps
-2 ps
- ps
- ps
- ps
- ps
2
- ps
1
2 j 1- e
-2 ps
-2 ps
- ps
-2 ps
2
2
- ps
- ps
- ps
2
- ps
- ps
Example 3.7
Determine the Laplace transform of the following signals using properties of Laplace transform.
a) x(t) = (t2 – 2t) u(t – 1)
c) x(t)
b) Unit ramp signal starting at t = a.
4
Solution
a) Given that, x(t) = (t2 – 2t) u(t – 1) = t2 u(t – 1) – 2t u(t – 1)
0
From table 3.2 we get,
2
L t u(t ) = 3
s
1
L t u( t) = 2 ;
s
n s
l q
2
t
Fig 3.7.1.
2
.....(1)
l
q
\ L 2 t u( t) =
2
s2
.....(2)
From time delay property of Laplace transform,
If L{x(t) u(t)} = X(s), then L{x(t) u(t – a)} = e–as X(s)
i
{d
} n
Lnt u(t)s - e Ll2t u(t)q
2
2
F 1 - s IJ = 2e
= 2e G
- e
H s K
s
s
s
l
q
\ X(s) = L{x( t)} = L t 2 - 2t u(t - 1) = L t 2 u(t - 1) - L 2t u(t - 1)
= e- s
= e- s
-s
2
-s
3
-s
2
3
Using time delay property
-s
(1 - s)
s3
Using equations (1) and (2)
3. 29
Signals & Systems
b) Given that, Unit ramp signal starting at t = a.
x(t)
x(t – a)
The unit ramp starting at t = a, is unit ramp delayed by
“a” units of time. The unit ramp waveform and the ramp
waveform starting at t = a are shown in fig 1 and fig 2 respectively.
The equation of unit ramp and delayed ramp are given below.
Unit ramp,
t
0
x(t) = t u(t)
Fig 1 : Ramp.
t
a
Fig 2 : Ramp starting at t = a.
Delayed unit ramp, x(t - a) = (t - a) u(t - a)
From table 3.2 we get,
1
s2
l q
L t u(t ) =
.....(3)
From time delay property of Laplace transform,
If L{x(t) u(t)} = X(s), then L{x(t) u(t – a)} = e–as X(s)
Therefore Laplace transform of delayed ramp signal is,
l
q
l q
L lt u(t)q
L x( t - a) = e - as L x(t)
= e
- as
= e - as
Using time delay property
1
e - as
=
2
s
s2
Using equation (3)
c)
The given signal can be decomposed into two signals as shown in fig 4 and fig 5.
x(t)
x1(t)
0
8
–2
6
4
2
4
Þ
2
0
+
2
3
4
-t
–4
3
0
t
1
1
Fig 3.
2
4
t
x2(t)
Fig 5.
Fig 4.
The mathematical equations of signals, x1 (t) and x2(t) are given below.
x1(t) = 2t u(t)
x2(t) = -2(t - 2) u(t - 2)
\
x(t) = x1(t) + x2(t) = 2t u(t) - 2(t - 2) u(t - 2)
From table 3.2 we get,
1
s2
l q
L t u(t ) =
l
;
q
\ L 2t u( t) =
l
2
s2
q
.....(4)
l
q
\ X(s) = L {x(t )} = L x1(t ) + x 2 ( t) = L 2t u(t) - 2(t - 2) u(t - 2)
l q l
q
= L l2t u(t)q - e L l2 t u(t)q
2d1 - e i
2
2
- e
=
=
= L 2t u(t) - L 2(t - 2) u(t - 2)
-2 s
Using time delay property
-2 s
-2 s
s
2
s
2
s
2
Using equation (4)
Chapter 3 - Laplace Transform
3. 30
Example 3.8
Let, X(s) = L{x(t)}. Determine the initial value, x(0) and the final value, x(¥), for the following signals using initial value
and final value theorems.
a) X(s) =
1
s(s - 1)
b) X(s) =
s + 1
s 2 + 2s + 2
d) X(s) =
s2 + 1
s 2 + 6s + 5
e) X(s) =
s + 5
s2 (s + 9)
7s + 6
s(3s + 5)
c) X(s) =
Solution
a) Given that, X(s) =
1
s(s - 1)
Initial value, x(0) =
Lt s X(s) =
s®¥
=
1
s
Lt
s®¥
Final value, x(¥) =
b) Given that, X(s) =
Lt
1
=
s(s - 1)
s
s®¥
1
FG1 - 1IJ
H sK
Lt sX(s) =
s® 0
1
¥
=
1
= 0 ´
FG1 - 1 IJ
H ¥K
1
=
s(s - 1)
Lt s
s® 0
Lt
s®¥
1
F 1I
sG 1 - J
H sK
1
= 0
1 - 0
1
1
=
= -1
0 - 1
(s - 1)
Lt
s® 0
s + 1
s 2 + 2s + 2
Initial value, x(0) =
Lt
s®¥
sX(s) =
1 +
=
Lt
Lt
s®¥
1
s
1 +
= 0´
2
2
1 +
+ 2
s
s
=
1
¥
Lt
s®¥
=
Lt
sX(s) =
s®¥
Lt
s®¥
7s + 6
s
=
s(3s + 5)
=
Lt
Lt
s® 0
FG
H
s2 1 +
Lt
s®¥
7s + 6
0 + 6
6
=
=
3s + 5
0 + 5
5
1
s
FG 6 IJ
H sK
s
F 5I
s G3 + J
H sK
2
IJ
K
2
2
+ 2
s
s
1+ 0
= 1
1+ 0 + 0
s 7 +
6
6
7 +
s =
¥ = 7 + 0 = 7
5
5
s®¥
3 + 0
3
3 +
3 +
s
¥
7s + 6
Final value, x(¥) = Lt sX(s) = Lt s
s® 0
s® 0
s(3s + 5)
7 +
=
s
0 + 1
= 0
0 + 0 + 2
7s + 6
s(3s + 5)
Initial value, x(0) =
FG
H
s 1 +
s + 1
s 2
=
s + 2s + 2
2
2
1 +
+
¥
¥
s + 1
Final value, x( ¥) = Lt sX(s) = Lt s 2
s® 0
s® 0
s + 2s + 2
s®¥
c) Given that, X(s) =
1
=
(s - 1)
Lt
s®¥
IJ
K
3. 31
Signals & Systems
2
d) Given that, X(s) =
s + 1
s 2 + 6s + 5
Initial value, x(0) =
Lt
sX(s) =
s®¥
1+
Lt
=
s
e) Given that, X(s) =
1
s2
1+
6
1
1+
+ 2
s
s
s®¥
Final value, x(¥) =
Lt
s®¥
Lt sX(s) =
s® 0
Lt s
s® 0
= ¥ ´
FG
H
s2 1 +
s2 + 1
s 2
=
s + 6s + 5
Lt
s
s®¥
1
¥
6
1
1+
+
¥
¥
FG
H
s2 1 +
= ¥ ´
1
s2
IJ
K
6
1
+ 2
s
s
IJ
K
1+ 0
= ¥
1+ 0 + 0
s2 + 1
0 + 1
= 0 ´
= 0
s + 6s + 5
0 + 0 + 5
2
s + 5
s 2 (s + 9)
Initial value, x(0) =
Lt
sX(s) =
s®¥
Lt
s®¥
s + 5
s 2
=
s (s + 9)
5
1 +
1 +
1
s = 1 ´
9
s®¥ s
¥
1 +
1 +
s
s + 5
Final value, x(¥) = Lt sX(s) = Lt s 2
s® 0
s® 0
s (s + 9)
=
Lt
FG 5 IJ
H sK
s
F 9I
s G1 + J
H sK
s 1 +
Lt
s®¥
3
1
¥ = 0 ´ 1+ 0 = 0
9
1+ 0
¥
s+5
0+5
= Lt
=
= ¥
s ® 0 s 2 + 9s
0+0
Example 3.9
Perform convolution of x1(t) and x2(t) using convolution theorem of Laplace transform.
a) x1(t) = u(t + 5), x2(t) = d(t - 7)
b) x1(t) = u(t – 2), x2(t) = d(t + 6)
c) x1(t) = u(t + 1), x2(t) = r(t – 2) ; where r(t) = t u(t)
Solution
L {u( t)} =
a) Given that, x1(t) = u(t + 5), x2(t) = d(t - 7)
1
and L{d( t)} = 1
s
l
q
if L{x( t)} = X(s) then L x(t ± a ) = e ± as X(s)
x1(t) = u(t + 5)
\ X1(s) = L {x1( t)} = L {u(t + 5)} = e5 s L {u( t )} = e 5 s ´
e5 s
1
=
s
s
.....(1)
x2(t) = d(t - 7)
l
q
l q
\ X 2 (s) = L{x 2 ( t)} = L d( t - 7) = e -7s L d(t ) = e -7s ´ 1 = e -7s
.....(2)
From convolution theorem of Laplace transform,
l
q
L x1( t) * x 2 ( t) = X1(s) X 2 (s)
=
e 5s
e 5s - 7 s
e -2 s
=
´ e -7 s =
s
s
s
RS e UV = L RS 1 UV
Ts W
TsW
-2 s
\ x1( t) * x 2 (t ) = L-1
-1
= u( t) t = t - 2 = u( t - 2)
t = t -2
Using equations (1) and (2)
Chapter 3 - Laplace Transform
3. 32
1
and L{d( t)} = 1
L {u( t)} =
s
b) Given that, x1(t) = u(t – 2), x2(t) = d(t + 6)
x1(t) = u(t – 2)
l
q
if L{x( t)} = X(s) then L x(t ± a ) = e ± as X(s)
l q
l
q
-2 s
l q
\ X1(s) = L x1( t) = L u( t - 2) = e -2 s L u( t) = e -2s
1 e
´ =
s
s
.....(1)
x2(t) = d(t + 6)
l q
l
q
l q
\ X 2 (s) = L x 2 ( t) = L d( t + 6) = e6 s L d( t) = e6 s ´ 1 = e6 s
.....(2)
From convolution theorem of Laplace transform,
l
q
L x1( t) * x 2 ( t) = X1(s) X2 (s)
=
e -2 s
e -2 s + 6 s
e4s
=
´ e6 s =
s
s
s
RS e UV = L RS 1 UV
TsW
TsW
Using equations (1) and (2)
4s
\ x1( t) * x 2 (t ) = L-1
-1
= u(t ) t = t + 4 = u( t + 4)
t= t + 4
c) Given that, x1(t) = u(t + 1), x2(t) = r(t – 2) ; where r(t) = t u(t)
l q
\ X1(s) = L x1( t) = L {u( t + 1)} =
es
s
.....(1)
1
1
and L r( t ) = L t u( t) = 2
s
s
l q l q
if L{x( t)} = X(s) then Llx(t ± a )q = e
L {u( t)} =
x1(t) = u(t + 1)
± as
X(s)
x2(t) = r(t – 2)
l q
l
q
\ X 2 (s) = L x 2 ( t) = L r( t - 2 ) = e -2 s L {r( t)} =
e -2 s
s2
.....(2)
From convolution theorem of Laplace transform,
l
q
L x1( t ) * x 2 ( t) = X1(s) X2 (s)
=
es
e -2 s
e- s
=
´
2
s
s
s3
RS e UV = L RS 1 UV
T s W Ts W
-s
\ x1( t) * x 2 ( t) = L-1
=
2
-1
3
LM t
N2
=
3
t = t -1
O
u( t)P
Q
Using equations (1) and (2)
RS t u(t)UV =
T(n - 1)! W
n - 1
L
t = t -1
1
sn
( t - 1)2
u( t - 1)
2
Example 3.10
Perform convolution of x 1(t) and x2(t) using convolution theorem of Laplace transform and sketch the resultant
waveform, where, x1(t) = u(t) - u (t - 1) and x2(t) = u(t) - u(t - 2).
Solution
x1(t) = u(t) – u(t – 1)
l q l
q l q l
q l q
q l q l
q l q
l q
\ X1(s) = L x1(t ) = L u( t) - u( t - 1) = L u(t ) - L u( t - 1) = L u( t) - e - s L u( t) =
1 e -s
s
s
.....(1)
x2(t) = u(t) – u(t – 2)
l q l
l q
\ X 2 (s) = L x 2 ( t) = L u( t) - u(t - 2) = L u(t ) - L u( t - 2 ) = L u( t) - e -2 s L u(t ) =
1 e -2 s
s
s
.....(2)
3. 33
Signals & Systems
From convolution theorem of Laplace transform,
l
q
L x1( t ) * x 2 ( t ) = X1(s) X 2 (s) =
F1
GH s
-
e-s
s
I F1
JK GH s
-
e -2 s
s
I
JK
Using equations (1) and (2)
e -2 s
e-s
e -3 s
e-s
e -2 s
e -3 s
1
1
+
= 2 +
2
2
2
2
2
2
s
s
s
s
s
s2
s
s
=
RS 1 - e - e + e
s
s
Ts s
RS 1 UV - L RS e UV - L RS e
Ts W
Ts W
Ts
\ x1( t) * x 2 ( t ) = L-1
2
-s
-2 s
-3 s
2
2
2
-s
-1
= L
-1
2
-2 s
-1
2
UV
W
UV + L RS e UV
W Ts W
L {t u( t)} =
l
-3 s
q
L t - a ) u(t - a ) =
-1
2
1
s2
2
e - as
s2
= t u(t) - (t - 1) u(t - 1) - (t - 2) u(t - 2) + (t - 3) u(t - 3)
To sketch the resultant waveform
For t = 0 to 1
When t = 0 to1, u(t) = 1, u(t – 1) = 0,
u(t – 2) = 0, u(t – 3) = 0
\ x1(t) * x2(t) = t ´ 1 – (t – 1) ´ 0 – (t – 2) ´ 0 + (t – 3) ´ 0 = t
x3(t)
For t = 1 to 2
When t = 1 to 2, u(t) =1, u(t – 1) = 1,
u(t – 2) = 0, u(t – 3) = 0
\ x1(t) * x2(t) = t ´ 1 – (t – 1) ´ 1 – (t – 2) ´ 0 + (t – 3) ´ 0 = 1
For t = 2 to 3
When t = 2 to 3, u(t) =1, u(t – 1) = 1,
x3(t) = x1(t) * x2(t)
1
0
1
2
3
t
Fig 1.
u(t – 2) = 1, u(t – 3) = 0
\ x1(t) * x2(t) = t ´ 1 – (t – 1) ´ 1 – (t – 2) ´ 1 + (t – 3) ´ 0 = 3 – t
For t > 3
When t > 3, u(t) = 1, u(t – 1) = 1,
u(t – 2) = 1, u(t – 3) = 1
\ x1(t) * x 2(t) = t ´ 1 – (t – 1) ´ 1 – (t – 2) ´ 1 + (t – 3) ´ 1 = 0
Using the calculated values, the resultant waveform is sketched as shown in fig 1.
3.4 Poles and Zeros of Rational Function of s
Let X(s) be Laplace transform of x(t). When X(s) is expressed as a ratio of two polynomials in s,
then the s-domain signal X(s) is called a rational function of s.
The zeros and poles are two critical complex frequencies at which a rational function of s takes
two extreme values, such as zero and infinity respectively.
Let X(s) is expressed as a ratio of two polynomials in s as shown in equation (3.26).
X(s) =
=
P(s)
Q(s)
b 0 s M + b1 s M - 1 + b 2 s M - 2 + ..... + b M - 1 s + b M
a 0 s N + a 1 s N - 1 + a 2 s N - 2 + ..... + a N - 1 s + a N
.....(3.26)
where, P(s) = Numerator polynomial of X(s)
Q(s) = Denominator polynomial of X(s)
In equation (3.26) let us scale the coefficients of numerator polynomial by b0 and the coefficients
of denominator polynomial by a0 , and the equation (3.26) can be expressed in factorized form as shown
in equation (3.27).
Chapter 3 - Laplace Transform
FG
H
F
a Gs
H
3. 34
b0 sM +
X(S) =
0
= G
N
+
b1 M - 1
b
b
b
+ 2 s M - 2 + ..... + M - 1 s + M
s
b0
b0
b0
b0
a1 N - 1
a
a
a
+ 2 s N - 2 + ..... + N - 1 s + N
s
a0
a0
a0
a0
IJ
K
IJ
K
(s - z1 ) (s - z 2 ) ..... (s - z M )
(s - p1 ) (s - p 2 ) ..... (s - p N )
where, G =
.....(3.27)
b0
= Scaling factor
a0
z1, z2, ....., zM = Roots of numerator polynomial, P(s)
p1, p2, ....., pN = Roots of denominator polynomial, Q(s)
In equation (3.27) if the value of s is equal to any one of the root of numerator polynomial then the
signal X(s) will become zero.
Therefore the roots of numerator polynomial z1, z2, ....., zM are called zeros of X(s). Since s is
complex frequency, the zeros can be defined as values of complex frequencies at which the signal X(s)
becomes zero.
In equation (3.27), if the value of s is equal to any one of the roots of the denominator polynomial
then the signal X(s) will become infinite. Therefore the roots of denominator polynomial p1, p2, ....., pN are
called poles of X(s). Since s is complex frequency, the poles can be defined as values of complex
frequencies at which the signal X(s) become infinite. Since the signal X(s) attains infinte value at poles,
the ROC of X(s) does not include poles.
In a realizable system, the number of zeros will be less than or equal to number of poles. Also for
every zero, we can associate one pole. (When number of finite zeros are less than poles, the missing zeros
are assumed to exist at infinity).
Let zi be the zero associated with the pole pi. If we evaluate |X(s)| for various values of s, then
|X(s)| will be zero for s = zi and infnite for s = pi. Hence the plot of |X(s)| in a three dimensional plane will
look like a pole (or pillar like structure) and so the point s = pi is called a pole.
3.4.1 Representation of Poles and Zeros in s-Plane
We know that, Complex frequency, s = s + jW
where, s = Real part of s, W = Imaginary part of s
jW
­
For example consider the rational function of s shown below.
s-plane
­
Hence the s-plane is a complex plane, with s on real axis and W on
imaginary axis as shown in fig 3.5. In the s-plane, the zeros are marked by
®
-s
small circle “ ” and the poles are marked by letter “X”.
s
­
-jW
X(s) =
(s + 2) (s + 5)
s(s2 + 6s + 13)
Fig 3.5: s-plane.
3. 35
Signals & Systems
The roots of quadratic s2 + 6s + 13 = 0 are,
36 - 4 ´ 13
-6 ± j4
=
= - 3 + j 2, - 3 - j 2
2
2
\ s2 + 6s + 13 = (s + 3 - j2) (s + 3 + j2)
(s + 2) (s + 5)
(s + 2) (s + 5)
=
\ X(s) =
2
s(s + 3 - j2) (s + 3 + j2)
s(s + 6s + 13)
-6 ±
The zeros of the above function are,
.....(3.28)
jW
®
s =
p2
+j2
X
z1 = –2
z1
z2
z2 = –5
p1
X
The poles of the above function are,
s-plane
+j1
-5 -4 -3 -2 -1
0
-j1
X
-j2
p1= 0
®s
p3
p2 = –3 + j2
p3 = –3 – j2
Fig 3.6 : Pole-zero plot of the
function of equation (3.28).
The pole-zero plot of rational function of s of equation (3.28) is shown in fig 3.6.
3.4.2 ROC of Rational Function of s
Case i : Right sided (causal) signal
Let x(t) be a right sided (causal) signal defined as,
x(t) = e - a1t u(t) + e - a 2 t u(t) + e - a 3t u(t) ; where –a1 < –a2 < –a3
Now, the Laplace transform of x(t) is,
X(s) =
1
1
1
N(s)
+
+
=
s + a1
s + a2
s + a3
(s + a 1 ) (s + a 2 ) (s + a 3 )
where, N(s) = (s + a2) (s + a3) + (s + a1) (s + a3) + (s +a1) (s + a2)
jW
The poles of X(s) are,
p1 = –a1 ; p2 = –a2 ; p3 = –a3
Let, s = Real part of s.
Now, the convergence criteria for X(s) are,
s > –a1 ; s > –a2 ; s > –a3
­ s-plane
–a 1 –a 2
X
X
–a 3
scX
ROC
®s
Abscissa of
®
convergence
Fig 3.7 : ROC of a causal signal.
Since -a1 < -a2 < -a3, the X(s) converges, when s > -a3, and does not converge for s < -a3.
\ Abscissa of convergence, sc = -a3.
Therefore ROC of X(s) is all points to the right of abscissa of convergence (or right of the line
passing through -a3) in s-plane as shown in fig.3.7. In terms of poles of X(s) we can say that the ROC is
right of right most pole of X(s), (i.e., right of the pole with largest real part).
Chapter 3 - Laplace Transform
3. 36
Case ii : Left sided (anticausal) signal
Let x(t) be a left sided (anticausal) signal defined as,
x(t) = e - a1t u(- t) + e - a 2t u(- t) + e - a 3t u( - t)
; where -a1 < -a2 < -a3
Now, the Laplace transform of x(t) is,
X(s) = -
1
1
1
N (s)
=
s + a1
s + a2
s + a3
(s + a 1 ) (s + a 2 ) (s + a 3 )
where, N(s) = -(s + a2) (s + a3) - (s + a1) (s + a3) - (s + a1) (s + a2)
The poles of X(s) are,
p1 = -a1 ; p2 = -a2 ; p3 = -a3
Let, s = Real part of s.
Now the convergence criteria for X(s) are,
s < -a1 ; s < -a2 ; s < -a3
Since -a1 < -a2 < -a3, the X(s) converges, when s < -a1,
and does not converge for s > -a1.
\ Abscissa of convergence, sc = -a1.
jW
­
s-plane
ROC
–a 1
X
–a 2
–a 3
X
X
®
sc
s
Therefore ROC of X(s) is all points to the left of abscissa of
of
convergence (or left of the line passing through -a1) in s-plane as
¬Abscissa
convergence
shown in fig.3.8. In terms of poles of X(s) we can say that the ROC
Fig 3.8 : ROC of an anticausal signal.
is left of left most pole of X(s), (i.e., left of the pole with smallest
real part).
Case iii : Two sided signal
Let x(t) be a two sided signal defined as,
x(t) = e - a1t u(t) + e - a 2 t u(t) + e - a 3t u( - t) + e - a 4 t u( - t)
; where -a1 < -a2 < -a3< -a4
Now, the Laplace transform of x(t) is,
1
1
1
1
X(s) =
+
s + a1
s + a2
s + a3
s + a4
N(s)
=
(s + a1 ) (s + a 2 ) (s + a 3 ) (s + a 4 )
where, N(s) = (s + a2) (s + a3) (s + a4) + (s + a1) (s + a3) (s + a4)
- (s + a1) (s + a2) (s + a4) - (s + a 1) (s + a2) (s + a3)
The poles of X(s) are,
p1 = -a1 ; p2 = -a2 ; p3 = -a3 ; p4 = -a4
Let, s = Real part of s.
Now, the convergence criteria for X(s) are,
s > -a1 ; s > -a2 ; s < -a3 ; s < -a4
Since -a1 < -a2 < -a3< -a4, the function X(s) converges, when s lies between -a2 and -a3
(i.e., -a2 < s < -a3 ), and does not converge for s < -a2 and s > -a3.
3. 37
Signals & Systems
\ Abscissa of convergences, sc1 = -a2 and sc2 = -a3.
Therefore ROC of X(s) is all points in the region in between
two abscissa of convergences (or region inbetween the two lines
passing through -a2 and -a3) in s-plane as shown in fig.3.9.
s-plane
–a 1 –a2
jW
ROC
–a 3
­
–a 4
Let X(s) be s-domain representation of a signal with causal
X X ®
X sX
s
c2
sc1
and noncausal part. Let ax be the magnitude of largest pole of
Abscissa of
Abscissa of
causal part of the signal and let ay be the magnitude of smallest pole convergence
®
¬convergence
of anticausal part of the signal and let ax < ay. Now in term of poles
Fig 3.9 : ROC of a two sided signal.
of X(s) we can say that the ROC is the region inbetween two lines
passing through ax and ay, where ax < ay.
3.4.3 Properties of ROC
The various concepts of ROC that has been discussed in section 3.2 and 3.4.2 are summarized as
properties of ROC and given below.
Property-1 : The ROC of X(s) consists of strips parallel to the jW - axis in the s-plane.
Property-2 : If x(t) is of finite duration and is absolutely integrable, then the ROC is the entire s- plane.
Property-3 : If x(t) is right sided, and if the line passing through Re(s) = s0 is in ROC, then all values
of s for which Re(s) > s0 will also be in ROC.
Property-4 : If x(t) is left sided, and if the line passing through Re(s) = s0 is in ROC, then all values of
s for which Re(s) < s0 will also be in ROC.
Property-5 : If x(t) is two sided, and if the line passing through Re(s) = s0 is in ROC, then the ROC
will consists of a strip in the s-plane that includes the line passing through Re(s) = s0 .
Property-6 : If X(s) is rational, (where X(s) is Laplace transform of x(t)), then its ROC is bounded by
poles or extends to infinity.
Property-7 : If X(s) is rational, (where X(s) is Laplace transform of x(t)), then ROC does not include
any poles of X(s).
Property-8 : If X(s) is rational, (where X(s) is Laplace transform of x(t)), and if x(t) is right sided,
then ROC is the region in s-plane to the right of the rightmost pole.
Property-9 : If X(s) is rational, (where X(s) is Laplace transform of x(t)), and if x(t) is left sided, then
ROC is the region in s-plane to the left of the leftmost pole.
Example 3.11
Determine the poles and zeros of the rational function of s given below. Also sketch the pole-zero plot.
X(s) =
d
d
i
i
(s + 1) s2 + 10s + 41
(s + 4) s2 + 4s + 13
Solution
In the given function, both the numerator and denominator polynomials are third order polynomials. Hence the
given function has three zeros and three ploes.
Let z1, z2, z3, be zeros and p1, p2, p3, be poles of X(s).
Chapter 3 - Laplace Transform
3. 38
To Determine Zeros
Consider the numerator polynomial.
(s + 1) (s2 + 10s + 41) = 0
The roots of quadratic s 2 + 10s + 41 = 0 are,
-10 ±
s =
102 - 4 ´ 41
-10 ± -64
-10
j 64
=
=
±
= - 5 ± j4 = - 5 + j4, - 5 - j4
2
2
2
2
\ (s + 1) (s2 + 10s + 41) = (s + 1) (s + 5 – j4) (s + 5 + j4) = 0
\ The roots of numerator polynomial are,
s = –1 ; s = –5 + j4 ; s = –5 – j4
\ Zeros of X(s) are,
z1 = –1 ; z2 = –5 + j4 ; z3 = –5 – j4
To Determine Poles
s-plane
z2
(s + 4) (s2 + 4s + 13) = 0
+j4
P2
2
The roots of quadratic s + 4s + 13 = 0 are,
s =
=
-4 ±
®
jW
Consider the denominator polynomial.
+j3
X
+j2
4 2 - 4 ´ 13
-4 ± -36
=
2
2
P1
X
-4
j 36
±
= - 2 ± j3 = - 2 + j3, - 2 - j3
2
2
–5 –4
z1
–3
–2
–1
P3
\ The roots of denominator polynomial are,
X
®
s
–j3
–j4
z3
\ Poles of X(s) are,
0
–j1
–j2
\ (s + 4) (s2 + 4s + 13) = (s + 4) (s+ 2 – j3) (s + 2 + j3) = 0
s = –4 ; s = –2 + j3 ; s = –2 – j3
+j1
Fig 1 : Pole-zero plot.
p1 = –4 ; p2 = –2 + j3 ; p3 = –2 – j3
Pole-Zero Plot
The pole-zero plot of function X(s) is shown in fig 1. The zeros are denoted by “
” and poles are denoted by “X”.
3.5 Inverse Laplace Transform
The Inverse Laplace transform of X(s) is defined as,
l q
L-1 X(s) = x(t) =
1
2 pj
s = s + jW
z
X(s) e st ds
s = s - jW
Performing inverse Laplace transform by using the above fundamental definition is tedious. But the
inverse Laplace transform by partial fraction expansion method will be much easier. In this method the
s-domain signal is expressed as a sum of first order and second order sections. Then the inverse Laplace
transform is obtained by comparing each section with standard transform pair, (listed in table 3.2).
In the following section the inverse Laplace transform by partial fraction exapansion method is
explained with example.
3. 39
Signals & Systems
3.5.1 Inverse Laplace Transform by Partial Fraction Expansion Method
Let Laplace transform of x(t) be X(s). The s-domain signal X(s) will be a ratio of two polynomials
in s (i.e., rational function of s). The roots of the denominator polynomial are called poles. The roots of
numerator polynomials are called zeros. (For definition of poles and zeros please refer section 3.4). In
signals and systems, three different types of s-domain signals are encountered. They are,
Case i
:
Signals with separate poles.
Case ii
:
Signals with multiple poles.
Case iii :
Signals with complex conjugate poles.
The inverse Laplace transform by partial fraction expansion method of all the three cases are
explained with an example.
Case - i : When s-Domain Signal X(s) has Distinct Poles
Let,
X(s) =
K
s(s + p1 ) (s + p 2 )
.....(3.29)
By partial fraction expansion technique, the equation (3.29) can be expressed as,
K
K2
K3
K
X(s) =
= 1 +
+
s(s + p1 ) (s + p 2 )
s
s + p1
s + p2
.....(3.30)
The residues K1, K2 and K3 are given by,
K1 = X(s) ´ s
s=0
K 2 = X(s) ´ (s + p1 )
s = - p1
K 3 = X(s) ´ (s + p 2 )
We Know that, L{u(t)} =
s = - p2
1
1
, L e - at u(t) =
s
s + a
n
s
By using the above standard Laplace transform pairs the inverse Laplace transform of equation
(3.30) can be obtained as shown below.
RS K + K + K UV
Ts s+p s+pW
R 1 UV + K
R 1U
\ x( t ) = K L S V + K L S
T sW
Ts + p W
l q
L-1 X(s) = L-1
1
2
3
1
-1
1
2
-1
2
3
L-1
1
= K1 u(t) + K2 e- p1t u(t) + K3 e- p2 t u(t)
Example 3.12
Determine the inverse Laplace transform of X(s) =
Solution
Given that, X(s) =
2
s(s + 1) (s + 2)
2
s(s + 1) (s + 2)
RS 1 UV
Ts + p W
2
Chapter 3 - Laplace Transform
3. 40
By partial fraction expansion technique, X(s) can be expressed as,
X(s) =
2
K
K2
K3
+
= 1 +
s(s + 1) (s + 2)
s
s + 1
s + 2
The residue K 1 is obtained by multiplying X(s) by s and letting s = 0.
K1 = X(s) ´ s
=
s=0
2
´s
s(s + 1) (s + 2)
2
(s + 1) (s + 2)
=
s=0
=
2
= 1
1 ´ 2
=
2
= -2
-1( -1 + 2)
s=0
The residue K 2 is obtained by multiplying X(s) by (s + 1) and letting s = –1.
K 2 = X(s) ´ (s +1)
s = -1
=
2
´ (s +1)
s(s + 1) (s + 2)
=
s = -1
2
s(s + 2)
s = -1
The residue K 3 is obtained by multiplying X(s) by (s + 2) and letting s = –2.
K 3 = X(s) ´ (s + 2)
=
s = -2
2
´ (s + 2)
s(s + 1) (s + 2)
=
s = -2
2
s(s + 1)
=
s = -2
2
= 1
-2(-2 + 1)
2
1
2
1
\ X(s) =
=
+
s(s + 1) (s + 2)
s
s + 1
s + 2
Now, x(t) = L-1{X(s)} = L-1
RS 1
Ts
-
2
1
+
s + 1
s + 2
UV
W
= u(t) – 2 e–t u(t) + e–2t u(t)
( x - y)2 = x 2 - 2xy + y 2
= (1 – 2 e–t + e–2t) u(t) = (1 – e–t )2 u(t)
Case - ii : When s-Domain Signal X(s) has Multiple Poles
Let, X(s) =
K
s(s + p1 ) (s + p 2 ) 2
.....(3.31)
By partial fraction expansion technique, the equation (3.31) can be expressed as,
X(s) =
K
K2
K3
K4
K
= 1 +
+
+
s
s + p1
(s + p 2 ) .....(3.32)
(s + p 2 ) 2
s(s + p1 ) (s + p 2 ) 2
The residues K1, K2, K3, and K4 are given by,
K1 = X(s) ´ s
s=0
K 3 = X(s) ´ (s + p 2 ) 2
s = - p2
;
K 2 = X(s) ´ (s + p1 )
;
K4 =
s = - p1
d
X(s) ´ (s + p 2 ) 2
ds
s = - p2
1
1
1
We know that, L{u(t)} = , L e - at u(t) =
, L t e- at u(t) =
s
s+ a
(s + a ) 2
By using the above standard Laplace transform pairs the inverse Laplace transform of
equation (3.31) can be obtained as shown below.
n
s
n
s
m b gr = L RST Ks + s K+ p + (s +Kp ) + s K+ p UVW
R 1 UV + K L RS 1 UV + K
R 1U
\ x(t) = K L S V + K L S
T sW
Ts + p W
T (s + p ) W
L-1 X s
-1
1
2
3
4
2
1
-1
1
-1
-1
2
3
1
= K1 u(t) + K2 e
2
2
- p1 t
u(t) + K3 t e
- p2 t
2
2
- p2 t
u(t) + K4 e
u(t)
4
L-1
RS 1 UV
Ts + p W
2
3. 41
Signals & Systems
In general if the pole p2 in equation (3.31) has multiplicity of q, as shown below,
K
X(s) =
s(s + p1 ) (s + p 2 ) q
then the partial fraction of the above signal can be expressed as,
X(s) =
K (q -1)2
K
K2
K3
K12
K 22
K
= 1+
+
+
+
+ ...... +
q
q -1
q -2
q
s s + p1 ( s + p 2 )
s + p2
(s + p 2 )
(s + p 2 )
s(s + p1 ) (s + p 2 )
The residue K1, K2, K3 are evaluated as shown above. The q – 1 residues K12 , K 22 , .... K (q-1) 2 can
be evaluated using the equation,
1 dr
X(s) ´ (s + 2) q
r ! dq r
Kr2 =
;
r = 1, 2, .... q - 1
Example 3.13
2
s(s + 1) (s + 2)2
Determine the inverse Laplace transform of X(s) =
Solution
Given that, X(s) =
2
s(s + 1) (s + 2)2
By partial fraction expansion technique, X(s) can be expressed as,
K
K
K2
K3
K4
= 1 +
+
+
s
(s + 1)
(s + 2)
s(s + 1) (s + 2)2
(s + 2)2
X(s) =
The residue K 1 is obtained by multiplying X(s) by s and letting s = 0.
K 1 = X(s) ´ s
s=0
2
´s
s(s + 1) (s + 2)2
=
=
s=0
2
(s + 1) (s + 2)2
=
s=0
2
= 0.5
1 ´ 22
The residue K 2 is obtained by multiplying X(s) by (s + 1) and letting s = –1.
K 2 = X(s) ´ (s + 1)
s = -1
=
2
´ (s + 1)
s(s + 1) (s + 2)2
2
= -2
-1(-1 + 2)2
=
s = -1
The residue K 3 is obtained by multiplying X(s) by (s + 2)2 and letting s = –2.
K 3 = X(s) ´ (s + 2)2
s = -2
=
2
´ (s + 2)2
s(s + 1) (s + 2)2
s=
2
= 1
-2( -2 + 1)
=
-2
The residue K 4 is obtained by differentiating the product X(s) ´ (s + 2)2 with respect to s and then letting s = –2.
K4 =
=
d
X(s) ´ (s + 2)2
ds
LM
N
d
2
ds s(s + 1)
OP
Q
=
s = -2
=
s = -2
LM
N
d
2
´ (s + 2)2
ds s(s + 1) (s + 2)2
-2(2s + 1)
s2 (s + 1)2
=
s = -2
OP
Q
s = -2
-2(2(- 2) + 1)
= 1.5
( -2)2 ( -2 + 1)2
2
0.5
2
1
1.5
\ X(s) =
=
+
+
s
s + 1
s + 2
s(s + 1) (s + 2)2
(s + 2)2
0.5
2
1
1.5
Now, x(t) = L-1 X (s) = L-1
+
+
s
s + 1
s + 2
(s + 2)2
l q
RS
T
UV
W
. e -2 t u(t)
= 0.5 u(t) - 2 e - t u(t) + t e -2 t u(t) + 15
d
i
. e -2 t u(t)
= 0.5 - 2 e - t + t e-2 t + 15
FG IJ
H K
d u
=
dx v
u=2
du
dx
v - u dv
dx
v2
v = s(s + 1)
= s2 + s
Chapter 3 - Laplace Transform
3. 42
Case iii : When s-Domain Signal X(s) has Complex Conjugate Poles
Let, X(s) =
K
(s + p1 ) (s2 + bs + c)
.....(3.33)
By partial fraction expansion technique, the equation (3.33) can be expressed as,
X(s) =
K1
K s + K3
K
=
+ 22
s + p1
(s + p1 ) (s2 + bs + c)
s + bs + c
The residue K 1 is given by, K1 = X(s) ´ (s + p1 )
..... (3.34)
s = - p1
The residues K 2 and K 3 are solved by cross multiplying the equation (3.34) and then equating the
coefficients of like power of s.
Finally express X(s) as shown below,
Arranging, s2 + bs ,
K1
K s + K3
+ 2 2
s + p1
s + bs + c
K1
K2 s + K3
=
+
2
s + p1
b
b
+ c s2 + 2 ´ s +
2
2
in the form of ( x + y) 2
X(s) =
FG IJ
H K
=
K1
+
s + p1
=
K1
K2 s + K3
+
2
s + p1
(s + a) 2 + W 0
( x + y) 2 = x 2 + 2 xy + y 2
FG bIJ
H 2K
2
K2 s + K3
FG s + b IJ
H 2K
2
F b IJ
+ Gc H 4K
2
K3
s +
K1
K2
X(s) =
+ K2
s + p1
(s + a) 2 + W 0 2
K1
s + a + K4
+ K2
=
+
s
p1
(s + a) 2 + W 02
Put ,
b
b2
= a and c = W02
4
2
K3
s + a +
- a
K1
K2
=
+ K2
s + p1
(s + a ) 2 + W 0 2
Put ,
K3
- a = K4
K2
W0
K1
s + a
K2 K4
+ K2
+
K K
2
2
W0
s + p1
(s + a) + W 0
(s + a) 2 + W 0 2 Put , 2 4 = K5
W0
W0
K1
s + a
..... (3.35)
\ X(s) =
+ K2
+ K5
s + p1
(s + a) 2 + W 0 2
(s + a ) 2 + W 0 2
=
n
s
We know that, L e - at u(t) =
1
;
s+a
s+a
;
(s + a) 2 + W 02
W0
L e - at sin W 0 t u(t) =
(s + a) 2 + W 0 2
n
s
L e - at cos W 0 t u(t) =
n
s
By using the above standard Laplace transform pairs the inverse Laplace transform of
equation (3.35) can be obtained as shown below.
3. 43
Signals & Systems
l q
R| K + K
S| s + p
T
RS 1 UV + K
Ts + p W
L-1 X(s) = L-1
1
2
1
\ x( t ) = K1 L-1
2
W0
s + a
+ K5
(s + a) 2 + W 02
(s + a ) 2 + W 0 2
L-1
1
RS s + a UV + K
T (s + a) + W W
2
2
= K1 e - p1t u(t) + K2 e- at cos W 0 t u(t) + K5 e
0
- at
5
L-1
U|
V|
W
RS W
UV
T (s + a ) + W W
0
2
2
0
sin W 0 t u(t)
Example 3.14
Determine the inverse Laplace transform of X(s) =
1
(s + 2) (s2 + s + 1)
Solution
Given that, X(s) =
1
(s + 2) (s2 + s + 1)
By partial fraction expansion technique, X(s) can be expressed as,
X(s) =
1
K1
K s + K3
=
+ 22
s + 2
(s + 2) (s 2 + s + 1)
s + s + 1
The residue K 1 is obtained by multiplying X(s) by (s + 2) and letting s = –2.
\ K1 = X(s) ´ (s + 2)
s = -2
=
1
´ (s + 2)
(s + 2) (s2 + s + 1)
=
s = -2
1
1
=
3
(-2)2 - 2 + 1
To solve K 2 and K 3, cross multiply the following equation and substitute the value of K1. Then equate the coefficients
of like power of s.
K s + K3
1
K1
=
+ 22
s + 2
(s + 2) (s 2 + s + 1)
s + s + 1
d
1
ds
3
i
+ s + 1i + K
1 = K1 s 2 + s + 1 + (K 2 s + K 3 ) (s + 2)
1 =
2
2
s2 + 2 K 2 s + K 3 s + 2 K 3
1 =
s2
s
1
+
+
+ K 2 s2 + 2 K 2 s + K 3 s + 2 K 3
3
3
3
1 =
FG 1 + K IJ s
H3 K
2
2
+
FG 1 + 2 K
H3
2
+ K3
IJ s +
K
1
+ 2 K3
3
On equating the coefficients of s2 terms,
1
1
0 =
+ K2
K2 = Þ
3
3
On equating the coefficients of s terms,
1
+ 2 K 2 + K3
3
0 =
Þ
K3 = -
FG IJ
H K
1
1
1
- 2 K2 = - 2´ 3
3
3
=
1
3
1
1
1
-1
1
(s - 1)
s +
1
3 =
3 +
3
3 + 3
\ X(s) =
=
2
2
2
s + s + 1
s+2
s + s + 1
(s + 2) (s + s + 1)
s+2
=
Arranging, s2+s, in
the form of (x+y)2
(x+y)2 = x2+2xy+x2
1
1
1
s -1
1
1
1
s - 1
=
3 s+2
3 (s2 + (2 ´ 0.5s) + 0.5 2 ) + (1 - 0.5 2 )
3 s+2
3 (s + 0.5)2 + 0.75
Chapter 3 - Laplace Transform
3. 44
=
1
1
1
s + 0.5 - 1 - 0.5
3 s+2
3 (s + 0.5)2 +
0.75
=
1
1
1
s + 0.5
1
1.5
0.866
´
+
3 s+2
3 (s + 0.5)2 + 0.866 2
3 0.866 (s + 0.5)2 + 0.866 2
e
=
1
1
1
s + 0.5 - 1.5
3 s+2
3 (s + 0.5)2 + 0.866 2
UV
s + 0.5
0.866
+ 0.577
l q = L RST 31 s +1 2 - 31 (s + 0.5)
(s + 0.5) + 0.866 W
+ 0.866
R
U
U
1
s + 0.5
0.866
R 1 UV - 1 L RS
L S
3
T s + 2 W 3 T (s + 0.5) + (0.866) VW + 0.577 L ST (s + 0.5) + 0.866 VW
\ x(t) = L-1 X(s)
=
j
=
2
-1
2
-1
2
2
-1
2
-1
2
2
2
2
1 -2t
1 -0.5t
e u(t) e
cos 0.866t u(t) + 0.577 e -0.5t sin 0.866t u(t)
3
3
3.5.2 Inverse Laplace Transform Using Convolution Theorem
The convolution theorem of equation (3.16) is useful to evaluate the inverse Laplace transform of
complicated s-domain signals.
Let x(t) be inverse Laplace transform of X(s). Let, the s-domain signal X(s) be expressed as a
product of two s-domain signals X1(s) and X2(s). Let, x1(t) and x2(t) be inverse Laplace transform of
X1(s) and X2(s) respectively.
Now, the inverse Laplace transform of X1(s) and X2(s) are computed separately to get x1(t) and
x2(t), and then the inverse Laplace transform of X(s) is obtained by convolution of x1(t) and x2(t), as
shown below.
X(s) = X1(s) X2(s), L-1{X(s)} = x(t),
Let,
L -1{X1(s)}= x1(t),
L-1{X2(s)} = x2(t).
First determine inverse Laplace transform of X1(s) and X2(s), to get x1(t) and x2(t).
Now, by convolution property,
L{x1(t) * x2(t)} = X1(s) X2(s)
Here, X(s) = X1(s) X2(s).
\X(s) = L{x1(t) * x2(t)}
On taking inverse Laplace transform of the above equation we get,
+¥
x(t) = L-1{X(s)} = x1 (t) * x2 (t) =
z
x1 (l ) x 2 (t - l ) dl
Using equation (3.17)
-¥
Example 3.15
Find the inverse Laplace transform of the s-domain signal, X(s) =
4
d
s2 s 2 + 16
i
Solution
Let, X(s) =
4
d
s 2 s2 + 16
i
=
4
where, X1(s) =
l
q
d
s 2 s2 + 4 2
x1(t) = L-1 X1(s) = L-1
4
s 2 + 42
RS
Ts
2
4
+ 42
i
=
4
1
´ 2
s 2 + 42
s
and
X 2 (s) =
UV = sin4t u(t)
W
1
s2
= X1(s) X 2 (s)
using convolution theorem.
3. 45
Signals & Systems
l q
L lX(s)q
RS 1 UV = t u(t)
Ts W
L lX (s) X (s)q
x 2 (t) = L-1 X 2 (s) = L-1
-1
\ x(t) =
=
2
-1
1
2
By convolution theorem, the inverse Laplace transform of X1(s) X2(s) is given by convolution of x1(t) and x2(t).
Since x1(t) and x2(t) are causal, limits
of integration are changed to 0 to t.
\ x(t) = L–1{X1(s) X2(s)} = x1(t) * x2(t)
+¥
=
z
t
x1 (l ) x 2 (t - l ) dl =
z
t
sin 4l (t - l ) dl =
0
-¥
z
(t - l ) sin 4l dl
0
t
z z z LNM z OQP
0
u= t-l
LM
FG -cos 4l IJ - (-1) FG -cos 4l IJ dl OP
H 4 K
H 4 K Q
N
L
F cos 4l IJ - sin 4l OP
= M-(t - l ) G
H 4 K 16 Q
N
L
F cos 4t IJ - sin 4t + (t - 0) cos 0 + sin 0 OP
= M-(t - t ) G
H 4 K 16
4
16 Q
N
sin 4 t
t
1F
= 0 +
+ 0 =
G t - sin4 4t IJK ; t ³ 0 = 41 FGH t 16
4
4H
z
= (t - l )
uv=u v -
du v
v = sin4 l
t
0
cos 0 = 1, sin 0 = 0
IJ
K
sin 4 t
u(t)
4
Example 3.16
Find the inverse Laplace transform of the following s-domain signals.
a) X(s) =
3s2 + 8s + 23
(s + 3) (s 2 + 2s + 10)
b) X(s) =
8s2 + 11s
(s + 2) (s + 1)3
Solution
a) Given that, X(s) =
3s2 + 8s + 23
(s + 3) (s2 + 2s + 10)
By partial fraction expansion technique, X(s) can be expressed as,
X(s) =
3s2 + 8s + 23
K1
K s + K3
=
+ 2 2
(s + 3) (s 2 + 2s + 10)
s + 3
s + 2s + 10
The residue K 1 is obtained by multiplying X(s) by (s + 3) and letting s = –3.
\ K1 = X(s) ´ (s + 3)
s = -3
3s2 + 8s + 23
´ (s + 3)
(s + 3) (s 2 + 2s + 10)
=
s = -3
27 - 24 + 23 26
3 ´ (-3)2 + 8 ´ ( -3) + 23
=
=
=
=2
9 - 6 + 10
13
( -3)2 + 2 ´ (-3) + 10
To solve K 2 and K 3 cross multiply the following equation and substitute the value of K1. Then equate the coefficients
of like power of s.
K s + K3
3s2 + 8s + 23
K1
=
+ 2 2
s + 2s + 10
(s + 3) (s2 + 2s + 10)
s + 3
d
i
3s2 + 8s + 23 = K1 s2 + 2s + 10 + (K 2 s + K 3 ) (s + 3)
2
2
3s + 8s + 23 = K1 s + 2 K1 s + 10K1 + K 2 s2 + 3 K 2 s + K 3 s + 3 K 3
3s2 + 8s + 23 = (K1 + K 2 )s2 + (2K1 + 3K 2 + K 3 )s + 10K1 + 3 K 3
On equating the coefficients of s2 terms, we get,
3 = K1 + K 2
Þ
K2 = 3 - K1 = 3 - 2 = 1
Chapter 3 - Laplace Transform
3. 46
On equating the coefficients of s terms, we get,
8 = 2 K1 + 3 K 2 + K 3
K 3 = 8 - 2 K1 - 3 K 2 = 8 - 2 ´ 2 - 3 ´ 1 = 1
Þ
2
\ X(s) =
=
2
3s + 8s + 23
s + 1
=
+ 2
(s + 3) (s 2 + 2s + 10)
s+3
s + 2s + 10
2
2
s + 1
s + 1
+ 2
=
+
s+3
s + 2s + 1 + 9
s+3
(s +1)2 + 3 2
s + 1 U
l q RST s +2 3 + (s +1)
V
+3 W
RS 2 UV + L RS s + 1 UV = 2e
T s + 3 W T (s + 1) + 3 W
= L-1 X(s) = L-1
\ x(t)
= L-1
b) Given that, X(s) =
2
2
-1
-3t
2
2
u( t) + e - t cos 3 t u(t )
8s 2 + 11s
(s + 2) (s + 1) 3
By partial fraction expansion technique, X(s) can be expressed as,
X(s) =
8s2 + 11s
K1
K2
K3
K
=
+
+
+ 4
(s + 2) (s + 1)3
s + 2
(s + 1)3 (s +1)2 s +1
The residue K 1 is obtained by multiplying X(s) by (s + 2) and letting s = –2.
\ K1 = X(s) ´ (s + 2)
=
s = -2
8s2 + 11s
´ (s + 2)
(s + 2) (s + 1)3
8 ´ ( -2)2 + 11 ´ (-2)
= -10
( -2 + 1)3
=
s = -2
3
The residue K 2 is obtained by multiplying X(s) by (s + 1) and letting s = –1.
\ K 2 = X(s) ´ (s + 1)3
=
s = -1
8s2 + 11s
´ (s +1)3
(s + 2) (s + 1)3
8 ´ ( -1)2 + 11 ´ (-1)
= -3
( -1 + 2)3
=
s = -1
The residue K 3 is obtained by differentiating the product X(s) ´ (s + 1)3 with respect to s and then letting s = –1.
\ K3 =
d
X(s) ´ (s + 1)3
ds
d
=
ds
LM 8s + 11s OP
N s+2 Q
2
s = -1
=
s = -1
d
ds
LM 8s + 11s
N (s + 2) (s + 1)
2
´ (s + 1)3
3
(16s + 11) (s + 2) - (8s 2 + 11s) ´ 1
=
(s + 2)2
16s + 32s + 11s + 22 - 8s - 11s
(s + 2)2
FG IJ
H K
d u
=
dx v
s = -1
du
dx
v - u dv
dx
v2
s = -1
2
2
=
s = -1
2
2
=
OP
Q
8s + 32s + 22
(s + 2)2
=
s = -1
8 ´ ( -1) - 32 + 22
= -2
(-1 + 2)2
The residue K 4 is obtained by differentiating the product X(s) ´ (s + 1)3 twice with respect to s and then dividing by
2! and letting s = –1.
\ K4 =
=
1 d2
X(s) ´ (s + 1)3
2! ds2
1 d
2 ds
LM 8s + 32s + 22 OP
N (s + 2) Q
=
s = -1
2
=
2
s = -1
LM
N
1 d d
X(s) ´ (s + 1)3
2 ds ds
OP
Q
s= -1
1 (16s + 32) (s + 2)2 - (8s2 + 32s + 22) 2(s + 2)
(s + 2)4
2
2
2
s = -1
1 (16 ´ ( -1) + 32)( -1 + 2) - (8 ´ ( -1) + 32 ´ ( -1) + 22) 2( -1 + 2)
2
( -1 + 2)4
1 -16 + 32 - (8 - 32 + 22) ´ 2
= 10
= ´
1
2
=
3. 47
Signals & Systems
\ X(s) =
8s 2 + 11s
3
2
10
-10
=
+
(s + 2) (s + 1)3 s + 2 (s + 1)3 (s + 1)2 s + 1
l q = L RST s-+102 - (s +31) - (s +21) + s10+ 1UVW
R 1 UV - 3 L RS 1 UV - 2 L RS 1 UV + 10 L RS 1 UV
- 10 L S
Ts + 2 W
Ts + 1 W
T (s + 1) W
T(s + 1) W
\ x(t) = L-1 X(s)
=
-1
3
-1
2
-1
-1
-1
3
2
1 2 -t
t e u(t) - 2 t e -t u(t) + 10 e -t u(t)
2!
. t 2 e - t - 2 t e - t + 10 e -t ) u(t)
- 15
= - 10 e -2t u(t) - 3 ´
= ( -10 e -2t
=
(10 - 2 t - 15
. t 2 )e- t - 10 e -2t u(t)
Example 3.17
Find the inverse Laplace transform of X(s) =
i) – 2 > Re {s} > – 4
4
if the ROC is,
(s + 2) (s + 4)
ii) Re {s} < – 4
iii)
Re {s} > – 2
Solution
Given that , X(s) =
4
(s + 2) (s + 4)
By partial fraction expansion technique, X(s) can be expressed as,
X(s) =
K1
K2
4
=
+
(s + 2) (s + 4) s + 2 s + 4
The residue K 1 is obtained by multiplying X(s) by (s + 2) and letting s = –2.
\ K1 = X(s) ´ (s + 2)
s = -2
=
4
´ (s + 2)
(s + 2) (s + 4)
=
s = -2
4
=2
-2 + 4
The residue K 2 is obtained by multiplying X(s) by (s + 4) and letting s = –4.
\ K 2 = X(s) ´ (s + 4)
\ X(s) =
s = -4
=
4
´ (s + 4)
(s + 2) (s + 4)
=
s = -4
4
= -2
-4 + 2
4
2
2
=
s+4
(s + 2) (s + 4) s + 2
Case i :
Given that ROC lies between lines passing through s = –2 to s = –4. Hence x(t) will be two sided signal.
The term corresponding to the pole, p = –2 will be anticausal signal and the term corresponding to the pole, p = –4
will be causal signal.
\ x(t) = - 2 e-2t u(-t ) - 2 e -4 t u(t )
Case ii :
Given that ROC is left of the line passing through s = –4. Hence x(t) will be anticausal signal.
\ x(t) = - 2 e -2t u( - t ) + 2 e -4 t u(- t ) = 2 e -4 t - e -2t u( - t)
Case iii :
Given that ROC is right of the line passing through s = – 2. Hence x(t) will be causal signal.
\ x(t) = 2 e - t u(t ) - 2 e -4 t u(t ) = 2 e - t - e -4 t u(t)
Chapter 3 - Laplace Transform
3. 48
3.6 Analysis of LTI Continuous Time System Using Laplace Transform
3.6.1 Transfer Function of LTI Continuous Time System
In general, the input-output relation of a LTI (Linear Time Invariant) continuous time system is
represented by the constant coefficient differential equation shown below, (equation (3.36)).
dN
dN - 1
dN - 2
d
y(t) + a 1 N - 1 y(t) + a 2 N - 2 y(t) + ..... + a N - 1
y(t) + a N y(t)
N
dt
dt
dt
dt
dM
dM - 1
dM - 2
d
= b 0 M x( t ) + b1 M - 1 x( t ) + b2 M - 2 x( t ) + ..... + b M - 1
x( t ) + b M x( t ) .....(3.36)
dt
dt
dt
dt
where, N = Order of the system and M £ N.
On taking Laplace transform of equation (3.36) with zero initial conditions we get,
sN Y(s) + a 1 sN - 1 Y(s) + a 2 sN - 2 Y(s) + ..... + a N - 1 s Y(s) + a N Y(s)
= b 0 sM X(s) + b1 sM - 1 X(s) + b2 sM - 2 X(s) + ..... + b M - 1 s X(s) + b M X(s)
d
Y(s) sN + a 1 sN - 1 + a 2 sN - 2 + ..... + a N - 1 s + a N
i
d
= X(s) b0 sM + b1 sM - 1 + b2 sM - 2 + ..... + b M - 1 s + b M
\
i
Y(s)
b sM + b1 sM - 1 + b2 sM - 2 + ..... + b M - 1 s + b M
= 0 N
X(s)
s + a1 sN - 1 + a 2 sN - 2 + ..... + a N - 1 s + a N
.....(3.37)
The transfer function of a continuous time system is defined as the ratio of Laplace transform of
output and Laplace transform of input. Hence the equation (3.37) is the transfer function of an LTI
continuous time system.
The equation (3.37) is a rational function of s (i.e., ratio of two polynomials in s). The numerator
and denominator polynomials of equation (3.36) can be expressed in the factorized form as shown in
equation (3.38).
(s - z1 ) (s - z2 ) (s - z3 ) ..... (s - z M )
Y(s)
= G
.....(3.38)
X(s)
(s - p1 ) (s - p 2 ) (s - p3 ) ..... (s - p N )
where, z1, z2, z3, ..... zM are roots of numerator polynomial
(or zeros of continuous time system)
p1, p2, p3, ..... pN are roots of denominator polynomial
(or poles of continuous time system)
3.6.2 Impulse Response and Transfer Function
Let, x(t) = Input of a LTI continuous time system
y(t) = Output / Response of the LTI continuous time system for the input x(t)
h(t) = Impulse response (i.e., response for impulse input)
Now, the response y(t) of the continuous time system is given by convolution of input and impulse
response. (Refer chapter - 2, section 2.9.1, equation (2.23)).
+¥
y( t ) = x( t ) * h( t ) =
z
-¥
x(l ) h( t - l ) dl
....(3.39)
3. 49
Signals & Systems
On taking Laplace transform of equation(3.39) we get,
If
L{x(t)} = X(s)
and L{h(t)} = H(s)
then by convolution property
L{x(t) * h(t)} = X(s) H(s)
L{y(t)} = L{x(t) * h(t)}
Using convolution property of Laplace transform the above
equation can be written as,
Y(s) = X(s) H(s)
Y(s)
X(s)
(s - z1 ) (s - z 2 ) (s - z 3 ) ..... (s - z M )
Y(s)
\ H(s) =
= G
X(s)
(s - p1 ) (s - p 2 ) (s - p 3 ) ..... (s - p N )
\ H(s) =
.....(3.40)
Using equation (3.38)
From equation (3.40) we can conclude that the transfer function of LTI continuous time system is
also given by Laplace transform of the impulse response.
Alternatively we can say that the inverse Laplace transform of transfer function is the impulse
response of the system.
Y(s)
Using equation (3.40)
\ Impulse response, h(t) = L-1 H(s) = L-1
X(s)
l q
RS
T
UV
W
3.6.3 Response of LTI Continuous Time System Using Laplace Transform
In general, the input-output relation of an LTI (Linear Time Invariant) continuous time system is
represented by the constant coefficient differential equation shown below, (equation (3.41)).
dN
dN - 1
dN - 2
d
y(t)
+
a
y(t)
+
a
y(t) + ..... + a N - 1
y(t) + a N y(t)
1
2
N
N-1
N- 2
dt
dt
dt
dt
dM
dM - 1
dM - 2
d
= b 0 M x( t ) + b1 M - 1 x( t ) + b 2 M - 2 x( t ) + ..... + b M - 1
x( t ) + b M x( t ) .....(3.41)
dt
dt
dt
dt
The solution of the above differential equation (equation (3.41)) is the (total) response y(t) of
LTI system, which consists of two parts.
In signals and systems the two parts of the solution y(t) are called zero-input response yzi(t) and
zero-state response yzs(t).
\ Response, y(t) = yzi(t) + yzs(t)
Zero-Input Response (or Free Response or Natural Response) Using Laplace Transform
The zero-input response yzi(t) is mainly due to initial output (or initial stored energy) in the system.
The zero-input response is obtained from system equation (equation (3.41)) when input x(t) = 0.
On substituting x(t) = 0 and y(t) = yzi(t) in equation (3.41) we get,
dN
dN - 1
dN - 2
d
y zi (t) + a1 N - 1 y zi (t) + a 2 N - 2 y zi (t) + ..... + a N - 1
yzi (t) + a N yzi (t) = 0
N
dt
dt
dt
dt
On taking Laplace transform of the above equation with non-zero initial conditions for output we
can form an equation for Yzi(s).
The zero-input response yzi(t) is given by inverse Laplace transform of Yzi(s).
Chapter 3 - Laplace Transform
3. 50
Zero-State Response (or Forced Response) Using Laplace Transform
The zero-state response yzs(t) is the response of the system due to input signal and with zero initial
output. The zero-state response is obtained from the differential equation governing the system
(equation(3.41)) for specific input signal x(t) for t ³ 0 and with zero initial output.
On substituting y(t) = yzs(t) in equation (3.41) we get,
dN
dN - 1
dN - 2
d
y
(t)
+
a
y
(t)
+
a
y zs (t) + ..... + a N - 1
y zs (t) + a N yzs (t)
zs
1
zs
2
N
N-1
N- 2
dt
dt
dt
dt
dM
dM - 1
dM - 2
d
= b 0 M x( t ) + b1 M - 1 x( t ) + b2 M - 2 x( t ) + ..... + b M - 1
x( t ) + b M x( t )
dt
dt
dt
dt
On taking Laplace transform of the above equation with zero initial conditions for output
(i.e., yzs(t)) and non-zero initial values for input (i.e., x(t)) we can form an equation for Yzs(s).
The zero-state response yzs(t) is given by inverse Laplace transform of Yzs(s).
Total Response
The total response y(t) is the response of the system due to input signal and initial output (or intial
stored energy). The total response is obtained from the differential equation governing the system
(equation(3.41)) for specific input signal x(t) for t ³ 0 and with non-zero initial conditions.
On taking Laplace transform of equation (3.41) with non-zero initial conditions for both input and
output, and then substituting for X(s) we can form an equation for Y(s).
The total response y(t) is given by inverse Laplace transform of Y(s).
Alternatively the total response y(t) is given by sum of zero-input response yzi(t) and zero-state
response yzs(t).
\ Total response, y(t) = yzi(t) + yzs(t)
3.6.4 Convolution and Deconvolution Using Laplace Transform
Convolution
The convolution operation is performed to find the response y(t) of LTI continuous time system
from the input x(t) and impulse response h(t).
If
L{x(t)} = X(s)
\ Response, y(t) = x(t) * h(t)
and L{h(t)} = H(s)
On taking Laplace transform of the above equation we get,
then by convolution property
L{x(t) * h(t)} = X(s) H(s)
L{y(t)} = L{x(t) * h(t)}
\ Y(s) = X(s) H(s)
\ Response, y(t) = L–1{Y(s)}
Procedure : 1.
2.
3.
4.
= L––1{X(s) H(s)}
Take Laplace transform of x(t) to get X(s).
Take Laplace transform of h(t) to get H(s).
Compute the product of X(s)H(s). Let, X(s)H(s) = Y(s).
Take inverse Laplace transform of Y(s) to get y(t).
.....(3.42)
3. 51
Signals & Systems
Deconvolution
The deconvolution operation is performed to extract the input x(t) of an LTI continuous time
system from the response y(t) of the system.
From equation (3.42) get,
Y(s)
X(s) =
H(s)
On taking inverse Laplace transform of the above equation we get,
l q
Input, x(t) = L -1 X(s) = L -1
RS Y(s) UV
T H(s) W
Procedure : 1. Take Laplace transform of y(t) to get Y(s).
2. Take Laplace transform of h(t) to get H(s).
3. Divide Y(s) by H(s) to get X(s) (i.e., X(s) = Y(s) / H(s)).
4. Take inverse Laplace transform of X(s) to get x(t).
3.6.5 Stability in s-Domain
ROC of a Stable LTI System
Let H(s) be Laplace transform of h(t). Now by definition of Laplace transform we get,
+¥
+¥
z
H (s) =
z
h( t ) e - st dt =
h( t ) e - ( s + jW ) t dt
Put, s = s + jW
-¥
-¥
Let us evaluate H(s) for s = 0.
+¥
+¥
\ H(s) =
z
h(t) e - jW t dt =
z
e - jW t h(t) dt = e - jW t
-¥
-¥
For a stable LTI system,
z
\ H(s) = e
- jW t
+¥
-¥
.....(3.43)
h( t ) dt is constant.
Therefore, in equation (3.43), put,
- jW t
e z h(t) dt j - z - jWe e z h(t) dt j dt
z
A + jW e
- jW t
z
h( t) dt = A, where A is constant.
+¥
A dt
z
= A e - jW t + jW e - jW t dt
-¥
z
z z z
uv = u v -
du v
+¥
-¥
.....(3.44)
The evaluation of equation (3.44), is evaluation of Laplace transform for s = jW, (i.e., evaluation
of H(s) along imaginary axis) and so we can say that H(s) exists if the ROC includes the imaginary axis.
Hence for a stable LTI continuous time system the ROC should include the imaginary axis of s-plane.
Location of Poles for Stability of Causal Systems
Let h(t) be impulse response of an LTI causal system. Now if h(t) satisfies the condition,
¥
z
h( t ) dt < ¥
....(3.45)
0
then the LTI continuous time causal system is stable.
Let, h(t) = eat u(t)
¥
Now ,
z
0
¥
z
¥
at
z
h( t ) dt = e u(t) dt = e
0
0
Le O
dt = M P
NaQ
at
at
¥
=
0
1
e a ´¥
ea ´0
e a ´¥
=
a
a
a
a
Chapter 3 - Laplace Transform
3. 52
Let a be negative, and let k = - a, so that k is always positive.
¥
Now ,
z
h( t ) dt =
0
=
1
1
1
e a ´¥
e - k¥
e -¥
=
+ =
+
a
a
k
k
-k
-k
e -¥ = 0
0
1 1
+ = = Constant, and so system is stable.
k k
-k
....(3.46)
Let a be positive.
¥
Now ,
z
h( t ) dt =
0
=
1
ea ´¥
a
a
e¥ = ¥
1 ¥
1
ea¥
- =
- = ¥ , and so system is unstable.
a
a a
a
From the above discussion, the stability condition of equation (3.45)
can be transformed as a condition on location of poles of transfer function
of the LTI continuous time causal system in s-plane.
The transfer function of a continuous time system is given by Laplace
transform of its impulse response.
jW
­
LHP
s-plane
RHP
®
-s
­
Consider the s-plane shown in fig 3.10. The area to the right of
vertical axis is called right half plane (RHP) and the area to the left of vertical
axis is called left half plane (LHP).
.....(3.47)
s
­
-jW
Fig 3.10 : s-plane.
Let, h(t) = eat u(t)
l q
n
s
\ Transfer function, H(s) = L h(t) = L eat u(t) =
1
s - a
Here, the transfer function H(s) has pole at s = a.
If, a < 0, (i.e., if a is negative), then the pole will lie on the left half of s-plane, and
from equation (3.46) we can say that the causal system is stable.
If, a > 0, (i.e., if a is positive), then the pole will lie on the right half of s-plane. and
from equation (3.47) we can say that the causal system is unstable.
Therefore we can say that, for a stable LTI continuous time causal system the poles should lie on
the left half of s-plane.
General Condition for Stability in s-Plane
On combining the condition for location of poles and the ROC we can say that,
1. For a stable LTI continuous time causal system, the poles should lie on the left half of s-plane
and the imaginary axis should be included in the ROC.
2. For a stable LTI continuous time noncausal system, the imaginary axis should be included in
the ROC.
The various types of impulse response of LTI continuous time system and their transfer functions
and the locations of poles are summarized in table 3.4.
3. 53
Signals & Systems
Table-3.4 : Impulse Response and Location of Poles of Transfer Function in s-Plane
Impulse response
h(t)
Transfer function
H(s) = L{h(t)}
Location of poles in
s-plane and ROC
jW
h(t) = A e-at u(t) ; a > 0
h(t)
­
s-plane
H ( s) =
A
t
A
Pole at s = -a.
ROC is s > –a,
where s is real part of s.
ROC
X
s+a
The pole at s = -a, lies on left half of s-plane.
ROC includes imaginary axis. Causal system.
Since pole lies on LHP and the imaginary axis
is included in ROC, the system is stable.
h(t) = A e-at u(-t) ; a > 0
H ( s) = -
h(t)
®s
–a
jW
­
A
ROC
s+a
X –a
A
Pole at s = -a.
ROC is s < –a,
where s is real part of s.
t
s-plane
®s
The pole at s = -a, lies on left half of s-plane.
The ROC does not include imaginary axis.
Noncausal system. Since imaginary axis is not
included in ROC, the system is unstable.
jW
­
h(t) = A eat u(t) ; a > 0
h(t)
H ( s) =
s-a
Pole at s = +a.
ROC is s > +a,
where s is real part of s.
A
t
h(t) = A eat u(-t) ; a > 0
H ( s) = -
h(t)
s-plane
A
A
s-a
+a
X
ROC
®
s
The pole at s = +a, lies on right half of s-plane.
ROC does not include imaginary axis. Causal
system. Since pole lies on RHP and imaginary
axis is not included in ROC, the system is
unstable.
jW
­
s-plane
ROC
Xa
®
s
A
Pole at s = +a.
ROC is s < +a,
where s is real part of s.
t
The pole at s = +a, lies on right half of s-plane.
The ROC includes imaginary axis. Noncausal
system. Since imaginary axis is included in
ROC, the system is stable.
Chapter 3 - Laplace Transform
3. 54
Table-3.4 : Continued....
Impulse response
h(t)
Transfer function
H(s) = L{h(t)}
Location of poles in
s-plane and ROC
jW
h(t) = A e-a½t½ ; a > 0
-
s-a
-a
A
0
t
Poles at s = -a, +a.
ROC is –a < s < +a,
where s is real part of s.
h(t) = A e-at u(t) +A e-bt u(-t)
where a > 0, b > 0, a > b
H ( s) =
h(t)
A
s+ a
-
ROC
s+ b
t
A
H ( s) =
s
­
X
h(t)
0
A
t
0
Pole at s = 0.
ROC is s > 0,
where s is real part of s.
H ( s) =
­
A
s2
X–b
®s
s-plane
ROC
®s
The pole at s = 0 lies on imaginary axis. The
ROC does not include the imaginary axis.
Causal system. Since the imaginary axis is not
included in ROC, the system is unstable.
jW
h ( t ) = A t u(t)
jW
­
s-plane
The poles at s = -a, -b, lie on LHP. The ROC
does not include imaginary axis. Noncausal
system. Since the imaginary axis is not included
in ROC, the system is unstable.
jW
h ( t ) = A u(t)
®s
The pole at s = +a, lies on RHP and pole at
s = -a, lies on LHP. ROC includes imaginary
axis. Noncausal system. Since the imaginary
axis is included in ROC, the system is stable.
A
Poles at s = -a, –b.
ROC is –a < s < –b,
where s is real part of s.
0
s-plane
+a
–aX
A
­
ROC
X
s+a
A
X
H ( s) =
h(t)
A
X
0
s-plane
ROC
®s
h(t)
2A
Double pole at s = 0.
ROC is s > 0,
where s is real part of s.
A
0
1
2
t
The poles at s = 0 lie on imaginary axis. The
ROC does not include the imaginary axis.
Causal system. Since the imaginary axis is not
included in ROC, the system is unstable.
3. 55
Signals & Systems
Table-3.4 : Continued....
Impulse response
h(t)
Transfer function
H(s) = L{h(t)}
h ( t ) = 2A cos bt u(t)
H ( s) =
A
A
+
s + jb
s - jb
Location of poles in
s-plane and ROC
jW
­
+jb
X
h(t)
s-plane
ROC
®s
0
2A
-jb
0
t
-2A
h ( t ) = 2A e
- at
Poles at s = –jb, +jb.
ROC is s > 0,
where s is real part of s.
X
The poles at s = -jb, +jb, lie on imaginary axis.
The ROC does not include the imaginary axis.
Causal system. Since the imaginary axis is not
included in ROC, the system is unstable.
jW
cos bt u(t),
H ( s) =
where a > 0
A
s + a + jb
+
A
­+jb
X
s + a - jb
ROC
–a
h(t)
Poles at s = –a – jb, –a + jb.
ROC is s > –a,
where s is real part of s.
The poles at s = -a-jb, -a+jb, lie on left half
of s-plane. The ROC includes the imaginary
axis. Causal system. Since poles lie on LHP
and the imaginary axis is included in ROC, the
system is stable.
jW
H ( s) =
at
h ( t ) = 2A e cos bt u(t),
A
s - a + jb
+
A
­
+jb
s - a - jb
ROC
®s
+a
-jb
t
h ( t ) = 2A t cos bt u(t)
h(t)
s-plane
X
where a > 0
h(t)
®s
-jb
X
t
s-plane
Poles at s = a – jb, a + jb.
ROC is s > a,
where s is real part of s.
H ( s) =
A
A
+
(s + jb ) 2 (s - jb ) 2
X
The poles at s = a-jb, a+jb, lie on right half of
s-plane. The ROC does not include imaginary
axis. Causal system. Since poles lie on RHP
and the imaginary axis is not included in ROC,
the system is unstable.
jW
­
+jb
X
0
­
®t
Double poles at s = – jb, + jb.
ROC is s > 0,
where s is real part of s.
-jb
s-plane
ROC
®s
X
The poles at s = -jb, +jb, lie on imaginary axis.
The ROC does not include the imaginary axis.
Causal system. Since the imaginary axis is not
included in ROC, the system is unstable.
Chapter 3 - Laplace Transform
3. 56
Example 3.18
Using Laplace transform, determine the natural response of the system described by the equation,
d2 y(t)
dy(t)
dx(t)
+ 6
+ 5 y(t) =
+ 4 x(t) ; y(0) = 1 ;
dt 2
dt
dt
dy(t)
dt
= -2
t=0
Solution
The natural response is the response of the system due to initial values of output alone. Hence for natural
response the input x(t) is considered as zero. Therefore the natural response is also called zero-input response, yzi(t).
On substituting input x(t) = 0 and y(t) = yzi(t) in the given system equation we get,
d2 y zi (t)
dy zi (t)
+ 6
+ 5 y zi (t) = 0
dt 2
dt
On taking Laplace transform of the above equation we get,
s 2 Yzi (s) - s y(0) - y ¢(0) + 6 [sYzi (s) - y(0)] + 5 Yzi (s) = 0
On substituting the given initial conditions of output in the above equation we get,
s2 Yzi(s) – s ´ 1 – (–2) + 6 [s Yzi(s) – 1] + 5 Yzi(s) = 0
s2 Yzi(s) – s + 2 + 6s Yzi(s) – 6 + 5 Yzi(s) = 0
(s2 + 6s + 5) Yzi(s) – s – 4 = 0
The roots of quadratic
s 2 + 6s + 5 = 0 are,
(s2 + 6s + 5) Yzi(s) = s + 4
s=
\ Yzi (s) =
-6 ± 6 2 - 4 ´ 5
2
-6 ± 4
= -1, - 5
=
2
s + 4
s + 4
=
(s + 1) (s + 5)
s2 + 6s + 5
By partial fraction expansion technique, Yzi(s) can be expressed as shown below.
Yzi (s) =
A =
B =
s + 4
A
B
=
+
(s + 1) (s + 5)
s + 1
s + 5
s + 4
´ (s + 1)
(s + 1) (s + 5)
s + 4
´ (s + 5)
(s + 1) (s + 5)
\ Yzi (s) =
=
3
-1 + 4
=
4
-1 + 5
=
1
-5 + 4
=
4
-5 + 1
s = -1
s = -5
3
1
1
1
+
4 s + 1
4 s + 5
On taking inverse Laplace transform of the above equation we get natural response.
\ Natural response
(or zero - input response)
|UV
|W
l q
R 3 1 + 1 1 UV
=L S
T4 s + 1 4 s + 5 W
3
R 1 UV + 1 L RS 1 UV
=
L S
4
T s + 1W 4 T s + 5 W
y zi (t ) = L-1 Yzi (s)
-1
-1
=
-1
3 -t
1 -5 t
e u(t) +
e u(t)
4
4
1
3 e - t + e -5 t u(t)
4
Note : Compare the above result with example 2.8 of chapter - 2.
=
d
i
n
s
L e -at u(t) =
1
s + a
3. 57
Signals & Systems
Example 3.19
Using Laplace transform determine the forced response of the system described by the equation,
5
dy(t)
+ 10 y(t) = 2 x(t) ; for the input, x(t) = 2 u(t)
dt
Solution
The forced response is the response of the system due to input alone. For forced response, the system equation
is solved for the given input with zero initial output. (But initial values of input should be considered).
Input, x(t) = 2 u(t)
\ X(s) = L {x(t)} = L {2 u(t)} =
2
s
.....(1)
On substituting y(t) = yzs(t) in the given system equation we get,
5
dy zs (t)
+ 10 y zs (t) = 2 x(t)
dt
On taking Laplace transform of the above equation with zero initial output conditions we get,
5s Yzs(s) + 10 Yzs(s) = 2 X(s)
5(s + 2) Yzs(s) = 2 X(s)
\ Yzs (s) =
2
X(s)
5(s + 2)
2
2
4
1
´
=
5(s + 2)
s
5 s(s + 2)
By partial fraction expansion technique, Yzs(s) can be expressed as shown below.
=
Yzs (s) =
A =
B =
4
1
4
=
5
5 s(s + 2)
1
´ s
s(s + 2)
=
s=0
1
´ (s + 2)
s(s + 2)
\ Yzs (s) =
4
5
LM 1
N2
LM A
Ns
+
B
s + 2
Using equation (1)
OP
Q
1
1
=
2
0 + 2
1
1
= 2
-2
=
s = -2
1
1
1
2 s + 2
s
OP = 2 LM 1
Q 5 Ns
-
1
s + 2
OP
Q
On taking inverse Laplace transform of the above equation we get forced response.
U|V y
(or zero - state response)|
W
\ Forced response
zs ( t)
m r
2 L R 1U
ML S V - L
5 N Ts W
RS 2 L 1 T 5 MN s
RS 1 UVOP
T s + 2 WQ
= L-1 Yzs (s) = L-1
=
-1
-1
=
2
u(t) - e -2t u(t)
5
=
2
1 - e -2t u(t)
5
d
i
Note : Compare the above result with example 2.9 of chapter - 2.
1
s + 2
OPUV
QW
1
s
l q
L u(t) =
n
s
L e -at u(t) =
1
s + a
Chapter 3 - Laplace Transform
3. 58
Example 3.20
Using Laplace transform determine the complete response of the system described by the equation,
d2 y(t)
dy(t)
dx(t)
+ 5
+ 4 y(t) =
dt 2
dt
dt
; y(0) = 0 ;
dy(t)
dt
= 1, for the input, x(t) = e -2t u( t)
t=0
Solution
Method - I
Input, x(t) = e–2t u(t)
n
s
\ X(s) = L {x(t)} = L e -2t u(t) =
Initial value of input, x(0) = x(t)
t=0
1
s + 2
.....(1)
= e0 u(0) = 1
The given sytem equation is,
d2 y(t)
dy(t)
dx(t)
+ 5
+ 4 y(t) =
dt
dt
dt 2
On taking Laplace transform of the above equation we get,
s 2 Y(s) - s y(0) - y ¢(0) + 5 [s Y(s) - y(0)] + 4 Y(s) = s X(s) - x(0)
On substituting the initial values of output and input in the above equation we get,
s2 Y(s) – s ´ 0 – 1 + 5 [s Y(s) – 0] + 4 Y(s) = s X(s) – 1
s2 Y(s) + 5s Y(s) + 4 Y(s) = s X(s)
ds
2
i
+ 5s + 4 Y(s) = s ´
\ Y(s) =
1
s + 2
Using equation (1)
s
d
(s + 2) s 2 + 5s + 4
i
The roots of quadratic
s 2 + 5s + 4 = 0 are,
s
=
(s + 2) (s + 1) (s + 4)
=
-5 ± 5 2 - 4 ´ 4
2
-5 ± 3
= -1, - 4
=
2
s=
s
(s + 1) (s + 2) (s + 4)
By partial fraction expansion technique, Y(s) can be expressed as shown below.
Y(s) =
A =
s
A
B
C
=
+
+
(s + 1) (s + 2) (s + 4)
s + 1
s + 2
s + 4
s
´ (s + 1)
(s + 1) (s + 2) (s + 4)
s
´ (s + 2)
B =
(s + 1) (s + 2) (s + 4)
C =
s
´ (s + 4)
(s + 1) (s + 2) (s + 4)
\ Y(s) = -
=
-1
-1
1
=
= ( -1 + 2) ( -1 + 4)
1 ´ 3
3
=
-2
-2
=
= 1
-1 ´ 2
( -2 + 1) ( -2 + 4)
=
2
-4
-4
=
= -3 ´ ( -2)
( -4 + 1) ( -4 + 2)
3
s = -1
s = -2
s = -4
1
1
1
2
1
+
3 s + 1
s + 2
3 s + 4
On taking inverse Laplace transform of the above equation we get complete / total response of the system.
3. 59
Signals & Systems
l q
R 1 1 + 1 - 2 1 UV
= L ST 3 s + 1 s + 2 3 s + 4W
1
R 1 UV + L RS 1 UV - 2 L RS 1 UV
= - L S
3
T s + 1W T s + 2 W 3 T s + 4 W
\ Total response, y( t) = L-1 Y(s)
-1
-1
= =
-1
-1
1 -t
2 -4t
e u(t) + e -2t u(t) e u(t)
3
3
FG - 1 e
H 3
-t
+ e -2t -
n
s
L e -at u(t) =
1
s + a
IJ
K
2 -4t
e
u(t)
3
Method - II
Total response, y(t) = yzi(t) + yzs(t)
where, yzi(t) = Zero-input response
y zs(t) = Zero-state response
Zero-input Response
On substituting x(t) = 0 and y(t) = yzi(t) in the system equation we get,
d2 y zi (t)
dy zi (t)
+ 5
+ 4 y zi (t) = 0
dt 2
dt
On taking Laplace transform of the above equation we get,
s 2 Yzi (s) - s y(0) - y ¢(0) + 5 [s Yzi (s) - y(0)] + 4 Yzi (s) = 0
On substituting the initial values of output in the above equation we get,
The roots of quadratic
s 2 + 5s + 4 = 0 are,
s2 Yzi(s) – s ´ 0 – 1 + 5 [s Yzi(s) – 0] + 4 Yzi(s) = 0
\ (s2 + 5s + 4) Yzi(s)= 1
Yzi (s) =
-5 ± 5 2 - 4 ´ 4
2
-5 ± 3
=
= -1, - 4
2
s=
1
1
=
(s + 1) (s + 4)
s 2 + 5s + 4
By partial fraction expansion technique, Yzi(s) can be expressed as shown below.
Yzi (s) =
D =
E =
1
(s + 1) s + 4
b
g
=
E
D
+
s + 4
s + 1
1
´ (s + 1)
(s + 1) (s + 4)
1
´ (s + 4)
(s + 1) (s + 4)
=
1
1
=
3
-1 + 4
=
1
1
= 3
-4 + 1
s = -1
s = -4
1
1
1
1
\ Yzi (s) =
3 s + 1
3 s + 4
On taking inverse Laplace transform of Yzi(s) we get zero-input response, yzi(t)
l
q
\ Zero - input response, y zi ( t) = L-1 Yzi (s) = L-1
RS 1
T3
RS 1 UV T s + 1W
1
1
1
3 s + 4
s + 1
=
1 -1
L
3
=
1 -t
1 -4t
e u(t) e u(t)
3
3
RS
T
1 -1
1
L
3
s + 4
UV
W
UV
W
n
s
L e -at u(t) =
1
s + a
Chapter 3 - Laplace Transform
3. 60
Zero-state Response
Given that, x(t) = e–2t u(t)
\ x(0) = x(t)
t =0
= e0 u(0) = 1
.....(2)
1
s + 2
On substituting y(t) = yzs(t) in the system equation we get,
n
s
X(s) = L {x(t)} = L e -2t u(t) =
.....(3)
d2 y zs (t)
dy zs (t)
dx(t)
+ 4 y zs (t) =
+ 5
dt
dt
dt 2
On taking Laplace transform of the above equation with zero initial output we get,
s2 Yzs(s) + 5s Yzs(s) + 4 Yzs(s) = s X(s) – x(0)
d
i
\ s2 + 5s + 4 Yzs (s) = s ´
1
- 1
s + 2
Using equations (2) and (3)
s - s - 2
(s + 1) (s + 4) Yzs (s) =
s + 2
\ Yzs (s) =
-2
(s + 1) (s + 2) (s + 4)
By partial fraction expansion technique, Yzs(s) can be expressed as shown below.
Yzs (s) =
F =
G =
H =
-2
F
G
H
=
+
+
(s + 1) (s + 2) (s + 4)
s + 1
s + 2
s + 4
-2
´ (s + 1)
(s + 1) (s + 2) (s + 4)
-2
´ (s + 2)
(s + 1) (s + 2) (s + 4)
-2
´ (s + 4)
(s + 1) (s + 2) (s + 4)
\ Yzs (s) = -
=
s = -1
-2
-2
2
=
= ( -1 + 2) (-1 + 4)
1 ´ 3
3
=
-2
-2
=
= 1
-1 ´ 2
( -2 + 1) ( -2 + 4)
=
1
-2
-2
=
= -3 ´ ( -2)
(-4 + 1) ( -4 + 2)
3
s = -2
s = -4
2
1
1
1
1
+
3 s + 1
s + 2
3 s + 4
On taking inverse Laplace transform of Yzs(s) we get zero-state response, yzs(t)
m
r
RS
T
\ Zero - state response, y zs (t ) = L-1 Yzs (s) = L-1 -
2
1
1
1
1
+
3 s + 1
s + 2
3 s + 4
RS 1 UV + L RS 1 UV T s + 1W T s + 2 W
= -
2 -1
L
3
= -
2 -t
1 -4t
e u(t) + e -2t u(t) e u(t)
3
3
-1
RS
T
1 -1
1
L
3
s + 4
UV
W
UV
W
n
s
L e -at u(t) =
Total / Complete Response
Total response, y(t) = yzi(t) + yzs(t)
=
LM 1 e
N3
= -
-t
u(t) -
OP LM- 2 e
Q N 3
1 -4t
e u(t) +
3
-t
u(t) + e -2t u(t) -
1 -t
2 -4t
e u(t) + e -2 t u(t) e u(t) =
3
3
FG - 1 e
H 3
Note : Compare the above result with example 2.10 of chapter - 2.
-t
OP
Q
IJ u(t)
K
1 -4t
e u(t)
3
+ e -2 t -
2 -4t
e
3
1
s + a
3. 61
Signals & Systems
Example 3.21
Determine the impulse response h(t) of the following system. Assume zero initial conditions.
a) y(t) = x(t - t0 )
c)
d2 y( t)
+ y(t) = x(t)
dt 2
b) T0
d 2 y (t )
dy(t)
dx(t)
+ 4
+ 3 y(t) =
+ 2 x(t)
dt
dt
dt 2
d)
d2 y( t)
dy(t)
+ 3
+ 2 y(t) = x(t)
dt
dt 2
Solution
a) Given that, y(t) = x(t – t0)
On taking Laplace transform with zero initial condition we get,
Y(s) = e -st 0 X(s)
Input, x(t) = d(t) ; \ X(s) = L{x(t)} = L{d(t)} = 1
When the input is impulse, the output is denoted by h(t). Let L{h(t)} = H(s).
\ Y(s) = e - st 0 X(s)
Þ
H(s) = e - st 0
Y(s) = H(s) ; X(s) = 1
-1
Impulse response, h(t) = L {H(s)}
L {d(t)} = 1
o t
= L-1 e - st 0
m
b) Given that, T0
r
L d(t - t 0 ) = e - st 0
= d(t - t0 )
d2 y(t)
+ y(t) = x(t)
dt 2
On taking Laplace transform with zero initial condition we get,
T0 s2 Y(s) + Y(s) = X(s)
F
GH
T0 s2 +
I
JK
1
Y(s) = X(s)
T0
Input, x(t) = d(t) ; \ X(s) = L{x(t)} = L{d(t)} = 1
When the input is impulse, the output is denoted by h(t). Let L{h(t)} = H(s).
F
GH
\ T0 s 2 +
1
T0
I
JK
Y(s) = X(s)
F
GH
1
T0
T0 s2 +
Þ
\ H(s) =
I H(s) = 1
JK
F
T Gs
H
U|
|V
1 I|
T JK |
W
0
l q
Impulse response, h( t) = L-1 H(s) = L-1
R|
|S
|| T FGs
T H
1
2
0
= L- 1
R|
||
S|
||
T
+
T0
U|
||
T
V=
F
1 I |
+ G
H T JK ||W
0
2
s
2
0
2
1
T0
+
I
JK
0
1
1
1
Y(s) = H(s) ; X(s) = 1
1
T0
L{sin W 0 t u(t)} =
F
GH
sin
1
T0
t
I
JK
u(t)
W0
s 2 + W02
Chapter 3 - Laplace Transform
3. 62
2
c) Given that,
d y(t)
dy(t)
dx(t)
+ 4
+ 3 y(t) =
+ 2 x(t)
dt
dt
dt 2
On taking Laplace transform with zero initial condition we get,
s2 Y(s) + 4s Y(s) + 3 Y(s) = s X(s) + 2 X(s)
(s2 + 4s + 3) Y(s) = [s + 2] X(s)
Input, x(t) = d(t)
\ X(s) = L{x(t)} = L{d(t)} = 1
When the input is impulse, the output is denoted by h(t). Let L{h(t)} = H(s).
\ (s2 + 4s + 3) Y(s) = (s + 2) X(s) Þ
\ H(s) =
(s2 + 4s + 3) H(s) = s + 2
s + 2
s + 2
=
(s + 1) (s + 3)
s 2 + 4s + 3
The roots of quadratic
s 2 + 4s + 3 = 0 are,
-4 ± 4 2 - 4 ´ 3
2
-4 ± 2
=
= -1, - 3
2
By partial fraction expansion technique, H(s) can be expressed as,
H(s) =
A =
B =
s=
s + 2
A
B
=
+
(s + 1) (s + 3)
s + 1
s + 3
s + 2
´ (s + 1)
(s + 1) (s + 3)
s + 2
´ (s + 3)
(s + 1) (s + 3)
\ H(s) = 0.5
\ Impulse response,
=
1
-1 + 2
= 0.5
=
2
-1 + 3
=
1
-3 + 2
= 0.5
=
2
-3 + 1
s = -1
s = -3
FG
IJ
H
K
R F 1 + 1 IJ UV
h(t) = L {H(s)} = L S0.5 G
T H s + 1 s + 3KW
L R 1 UV + L RS 1 UVOP
= 0.5 M L S
N T s + 1W T s + 3 WQ
= 0.5 e u(t) + e u(t) = 0.5 de + e i u(t)
1
1
1
1
+
= 0.5
+ 0.5
s + 3
s + 1
s + 3
s + 1
-1
-1
-1
-1
-t
d) Given that,
Y(s) = H(s) ; X(s) = 1
-3 t
-t
n
s
L e -at u(t) =
1
s + a
-3 t
d2 y(t)
dy(t)
+ 3
+ 2 y(t) = x(t)
dt
dt2
On taking Laplace transform with zero initial conditions we get,
s2 Y(s) + 3s Y(s) + 2 Y(s) = X(s)
(s2 + 3s + 2) Y(s) = X(s)
Input, x(t) = d(t) ; \ X(s) = L{x(t)} = L{d(t)} = 1
When the input is impulse, the output is denoted by h(t). Let L{h(t)} = H(s).
\ (s2 + 3s + 2) Y(s) = X(s)
\ H(s) =
Þ
(s2 + 3s + 2) H(s) = 1
1
1
=
(s + 1) (s + 2)
s 2 + 3s + 2
By partial fraction expansion technique, H(s) can be expressed as,
H(s) =
1
A
B
=
+
s + 1
(s + 1) (s + 2)
s + 2
Y(s) = H(s) ; X(s) = 1
The roots of quadratic
s 2 + 3s + 2 = 0 are,
-3 ± 3 2 - 4 ´ 2
2
-3 ± 1
=
= -1, - 2
2
s=
3. 63
Signals & Systems
A =
B =
1
´ (s + 1)
(s + 1) (s + 2)
1
´ (s + 2)
(s + 1) (s + 2)
\ H(s) =
=
1
= 1
-1 + 2
=
1
= -1
-2 + 1
s = -1
s = -2
n
s
L e -at u(t) =
1
1
s + 1
s + 2
RS 1 Ts + 1
u(t) = de
\ Impulse response, h(t) = L-1{H(s)} = L-1
= e - t u(t) - e -2 t
-t
UV RS
W T
i u(t)
UV
W
RS
T
1
1
1
- L-1
= L-1
s + 2
s + 1
s + 2
- e -2 t
1
s + a
UV
W
Example 3.22
Perform convolution of x1(t) = e–2t cos 3t u(t) and x2(t) = 4 sin 3t u(t) using Laplace transform.
Solution
Given that, x1(t) = e–2t cos 3t u(t) and x2(t) = 4 sin 3t u(t)
n
s
\ X1(s) = L {x1(t )} = L e -2t cos 3t u(t) =
l
q
s + 2
s + 2
= 2
(s + 2)2 + 32
s + 4s + 13
X2 (s) = L {x 2 ( t)} = L 4 sin 3t u(t) = 4 ´
12
3
= 2
s + 9
s 2 + 32
Now by convolution theorem, x1(t) * x2(t) = L–1{X1(s) X2(s)}
Let, X(s) = X1(s) X2(s)
\ X(s) =
s + 2
12
12(s + 2)
´ 2
=
s 2 + 4s + 13
s + 9
(s2 + 4s + 13) (s 2 + 9)
.....(1)
By partial fraction expansion, X(s) can be expressed as,
X(s) =
12(s + 2)
As + B
Cs + D
= 2
+
(s2 + 4s + 13)(s 2 + 9)
s + 4s + 13
s2 + 9
On cross - multiplying the equation (1) we get,
12(s + 2) = (A s + B) (s2 + 9) + (C s + D) (s2 + 4s +13)
12s + 24 = A s3 + 9A s + B s2 + 9B + C s3 + 4C s2 + 13C s + D s2 + 4D s +13D
12s + 24 = (A + C)s3 + (B + 4C + D)s2 + (9A + 13C + 4D)s + (9B + 13D)
.....(2)
3
On equating the coefficients of s terms of equation (2) we get,
A+C=0
Þ
A = –C
.....(3)
On equating the coefficients of s2 terms of equation (2) we get,
B + 4C + D = 0
.....(4)
On equating the coefficients of s terms of equation (2) we get
9A + 13C + 4D = 12
.....(5)
On equating constants of equation (2) we get,
9B + 13D = 24
.....(6)
On substituting A = –C in equation (5) we get,
9(–C) + 13C + 4D = 12
\ 4C + 4D = 12
Þ C+D=3
Þ C=3–D
.....(7)
Chapter 3 - Laplace Transform
3. 64
On substituting C = 3 – D in equation (4) we get
B + 4 (3 – D) + D = 0
\ B – 3 D = –12
.....(8)
Equation (8) ´ –9
Þ –9B + 27D = 108
Equation (6) ´ 1
Þ
9B + 13D = 24
132
= 3.3
40
\ D =
40D = 132
From equation (7), C = 3 – D = 3 – 3.3 = – 0.3
From equation (3), A = –C = 0.3
From equation (6), B =
\ X(s) =
24 - 13D
24 - 13 ´ 3.3
=
= - 2.1
9
9
As + B
Cs + D
+ 2
s 2 + 4s + 13
s + 9
FG
H
=
0.3s - 2.1
-0.3s + 3.3
+
s 2 + 4s + 13
s2 + 9
Arranging, s2+4s, in
the form of (x+y)2
IJ
K
2.1
-0.3s + 3.3
0.3
=
+
(s2 + (2 ´ 2s) + 2 2 ) + 32
s2 + 9
0.3 s -
=
-0.3s + 3.3
0.3(s - 7)
+
(s + 2)2 + 32
s2 + 9
=
-0.3s + 3.3
0.3(s + 2) - 0.3 ´ 9
+
s2 + 3 2
(s + 2)2 + 32
=
(x+y)2 = x2+2xy+x2
-0.3s + 3.3
0.3(s + 2 - 9)
+
(s + 2)2 + 3 2
s 2 + 32
n
Lne
s + 2
3
- 0.9
= 0.3
(s + 2)2 + 32
(s + 2)2 + 3 2
- 0.3
s (s + as)+ a+ W
W
sinW t u(t)s =
(s + a) + W
L e-at cos W0t u(t) =
On taking inverse Laplace transform of the above equation we get,
2
0
-at
s
3
+ 11
. 2
s 2 + 32
s + 32
2
0
2
0
L{cos W0t u(t)} =
s
s2 + W02
L{sinW0t u(t)} =
W0
s2 + W02
x(t) = 0.3 e–2t cos 3t – 0.9 e–2t sin 3t – 0.3 cos 3t + 1.1 sin 3t ; t ³ 0
2
0
= e–2t (0.3 cos 3t – 0.9 sin 3t) + (1.1 sin 3t – 0.3 cos 3t) u(t)
Note : The result of Example 3.22 can be further simplified as shown below.
x(t) = e–2t (cos 3t ´ 0.3 – sin 3t ´ 0.9) + (sin 3t ´ 1.1 – cos 3t ´ 0.3) ; t ³ 0
Let us construct two right angled triangles as shown below.
11
.
114
.
= 1.14 cos q 2
2
=
cos q2 =
0.
3
0.9
5
=
2
\ 1.1
) q1
0.9
sin q1 =
0.95
\ 0.9 = 0.95 sin q1
1.1 2
+
0.9
+
0.9
2
0.
3
1.1
4
0.3
0.95
\ 0.3 = 0.95 cos q1
cos q1 =
0.3
) q2
1.1
0.3
0.9
-1 0.9
o
tanq =
;
tanq =
; \ q1 = tan
= 716
.
1.1
0.3
0.3
Using the relations obtained from right angled triangle, the x(t) can be written as,
0.3
0.3
114
.
\ 0.3 = 1.14 sin q 2
sin q 2 =
\ q 2 = tan -1
0.3
= 15.3o
1.1
x(t) = e–2t (cos 3t ´ 0.95 cos q1 – sin 3t ´ 0.95 sin q1) + (sin 3t ´ 1.14 cos q2 – cos 3t ´ 1.14 sin q2)
= 0.95 e–2t (cos 3t cos 71.6o – sin 3t sin 71.6o) + 1.14 (sin 3t cos 15.3o – cos 3t sin 15.3o)
= 0.95 e–2t cos(3t + 71.6o) + 1.14 sin(3t – 15.3o) ; t ³ 0
3. 65
Signals & Systems
Example 3.23
Find the transfer function of the systems governed by the following differential equations.
a)
d3 y( t)
d2 y(t )
dy( t)
dx(t )
+4
+3
+ 5y( t) = 2
+ x( t)
3
dt
dt 2
dt
dt
b)
d 3 y (t )
d 2 y( t)
dy( t )
+8
+6
+ 11y( t) = x( t - 2)
3
dt
dt 2
dt
Solution
a) Given that,
d3 y(t)
d2 y(t)
dy(t)
dx(t)
+4
+3
+ 5y(t) = 2
+ x(t)
3
dt
dt
dt
dt 2
On taking Laplace transform of the above equation with zero initial conditions we get,
l q
R d x(t) UV = s X(s)
then, L S
T dt W
s 3 Y(s) + 4s 2 Y(s) + 3sY(s) + 5 Y(s) = 2sX(s) + X(s)
ds
3
2
g
n
+ 4s + 3s + 5 Y(s) = 2s + 1 X(s)
\ Transfer Function,
b) Given that,
b
i
if, L x(t ) = X(s),
n
n
Y(s)
2s + 1
= 3
X(s)
s + 4s2 + 3s + 5
with zero initial conditions
d3 y(t)
d2 y(t)
dy(t)
+8
+6
+ 11y(t) = x(t - 2)
3
dt
dt
dt 2
On taking Laplace transform of the above equation with zero initial conditions we get,
l q
l
s 3 Y(s) + 8s 2 Y(s) + 6sY(s) + 11Y(s) = e -2 s X(s)
ds
3
i
2
+ 8s + 6s + 11 Y(s) = e
\ Transfer Function,
-2 s
if, L x( t) = X(s),
then, L x(t - a) = e - asX(s)
X(s)
q
Y(s)
e -2 s
=
X(s)
s3 + 8s2 + 6s + 11
Example 3.24
Find the transfer function of the systems governed by the following impulse responses.
a)
h(t) = (2 + t) e–3t u(t)
b) h(t) = t2 u(t) - e–4t u(t) + e–7t u(t)
c) h(t) = u(t) + 0.5 e–6t u(t) + 0.2 e–3t cos t u(t)
Solution
a) Impulse response, h(t) = (2 + t) e–3t u(t)
n
l q
= L n(2 + t) e u(t )s = L n2 e
= 2 L n e u( t)s + L lt u(t)q
s
L e - at u(t ) =
The transfer function is given by Laplace transform of Impulse response.
if, L x( t) = X(s),
-3t
-3t
s= s+3
2
1
=
+
s + 3
(s + 3)2
=
l q
l q
n
\ Transfer function, H(s) = L h(t)
-3t
1
1
, L t u( t) = 2
s+a
s
s
n
u( t) + L t e
= 2´
-3t
2(s + 3) + 1
=
(s + 3)2
s
u(t )
1
+
s + 3
1
s2
s
then, L e - at x( t) = X(s + a)
s = s +3
2s + 7
s 2 + 6s + 9
b) Impulse response, h(t) = t2 u(t) - e –4t u(t) + e–7t u(t)
The transfer function is given by Laplace transform of Impulse response.
l q n
= L nt u(t)s - L ne
s
\ Transfer function, H(s) = L h(t) = L t 2 u(t) - e-4t u(t) + e -7tu(t )
2
-4t
s n
s
u( t) + L e-7tu( t)
n
s
L tn u(t ) =
n!
1
; L e - at u( t) =
s+a
sn+1
n
s
Chapter 3 - Laplace Transform
3. 66
=
2
1
1
+
s+4
s+7
s3
=
2(s + 4)(s + 7) - s 3 (s + 7 ) + s 3 (s + 4 )
s3 (s + 4)(s + 7)
=
2(s2 + 7s + 4s + 28) - s4 - 7s3 + s 4 + 4s 3
s3 (s2 + 11s + 28)
-3s3 + 2s 2 + 22s + 56
s5 + 11s 4 + 28s 3
=
c) Impulse response, h(t) = u(t) + 0.5 e–6t u(t) + 0.2 e–3t cos t u(t)
The transfer function is given by Laplace transform of Impulse response.
l q
= L nu(t) + 0.5 e u(t) + 0.2 e cos t u(t)s
= L lu(t)q + 0.5 ´ L n e u(t )s + 0.2 ´ L lcos t u( t)q
\ Transfer function, H(s) = L h(t)
–6t
- at
–3t
–6t
=
1
1
s
+ 0.5 ´
+ 0.2 ´ 2
s
s+6
s + 1 s =s+3
l q 1s
1
L ne u( t)s =
s+a
s
L lcos t u(t )q =
s +1
if, L lx(t )q = X(s),
then, L ne x(t )s = X(s + a)
L u( t ) =
2
s =s +3
- at
=
1
0.5
0.2(s + 3)
+
+
s
s + 6 (s + 3)2 + 1
=
1
0.5
0.2(s + 3)
+
+ 2
s
s+6
s + 6s + 9 + 1
=
1
0.5
0.2(s + 3)
+
+ 2
s
s+6
s + 6s + 10
=
(s + 6)(s 2 + 6s + 10) + 0.5s (s2 + 6s + 10) + 0.2s(s + 3)(s + 6)
s (s + 6) (s2 + 6s + 10)
=
s 3 + 6s 2 + 10s + 6s 2 + 36s + 60 + 0.5s 3 + 3s2 + 5s + 0.2s(s2 + 9s + 18)
s (s 3 + 6s 2 + 10s + 6s2 + 36s + 60)
=
. s2 + 3.6s
s 3 + 6s 2 + 10s + 6s 2 + 36s + 60 + 0.5s 3 + 3s2 + 5s + 0.2s3 + 18
3
2
s(s + 12s + 46s + 60)
=
. s3 + 16.8s2 + 54.6s + 60
17
s 4 + 12s3 + 46s2 + 60s
Example 3.25
The unit step response of continuous time systems are given below. Determine the transfer function of the systems.
a)
s(t) = u(t) + e–2t u(t)
s(t) = t2 u(t) + t e–4t u(t)
b)
c)
s(t) = t u(t) + sin t u(t)
Solution
a) Unit step response, s(t) = u(t) + e –2t u(t)
On taking Laplace transform of unit step response we get,
l q n
Let, L ls(t)q = S(s)
\ S(s) = L nu(t) + e
s
L s(t) = L u(t) + e -2tu(t )
-2t
s l q
n
s
u(t ) = L u(t) + L e -2tu( t)
1
1
s+2+s
2s + 2
=
+
=
=
s s+2
s(s + 2) s(s + 2)
l q
L u( t) =
1
1
; L e - at u( t) =
s
.s + a
n
s
....(1)
3. 67
Signals & Systems
1
; where u( t) is unit step signal.
s
Laplace transform of output
S(s)
Transfer function, H(s) =
=
Laplace transform of input
U(s)
l q
Let, U(s) = L u(t ) =
= S(s) ´
.....(2)
1
2s + 2
2s + 2
=
´s =
U(s)
s(s + 2)
s+2
Using equations (1) and (2)
b) Unit step response, s(t) = t2 u(t) + t e–4t u(t)
On taking Laplace transform of unit step response we get,
l q = L nt u(t) + t e u(t)s
Let, L ls(t)q = S(s)
\ S(s) = L nt u(t) + t e u( t)s = L nt
= L nt u(t)s + L lt u(t )q
L s(t)
2
-4t
2
-4t
2
s= s+4
2
=
2(s + 4) + s
s 3 (s + 4)2
3
2
s
n
n
=
2
1
+ 2
s3
s
s
2
2
1
+
s3 (s + 4)2
=
s= s+4
3
3
l q
n
if, L x( t) = X(s),
.
then, L e - at x( t ) = X(s + a )
s
2
2(s + 8 s + 16) + s
s + 2s + 16s + 32
=
s 3 (s2 + 8s + 16)
s 3 (s2 + 8s + 16)
=
....(1)
1
; where u( t) is unit step signal
s
Laplace transform of output
S(s)
Transfer function, H(s) =
=
Laplace transform of input
U(s)
l q
Let, U(s) = L u(t ) =
= S(s) ´
n!
sn+1
s
L tn u(t ) =
u(t) + L t e -4tu( t)
.....(2)
1
s 3 + 2s2 + 16s + 32
s 3 + 2s2 + 16s + 32
=
´s =
3 2
U(s)
s (s + 8s + 16)
s 4 + 8s 3 + 16s2
Using equations
(1) and (2)
c) Unit step response, s(t) = t u(t) + sint u(t)
On taking Laplace transform of unit step response we get,
l q = L lt u(t) + sin t u(t)q
Let, L ls(t)q = S(s)
\ S(s) = L lt u(t) + sin t u( t)q = L lt u(t)q + L lsin t u( t)q
L s(t)
2
=
2
l q s1
1
L lsin t u(t )q =
s +1
L tu( t) =
.
2
1
1
s + 1+ s
2s + 1
+ 2
= 2 2
= 2 2
s2
s +1
s (s + 1)
s (s + 1)
1
; where u( t) is unit step signal
s
Laplace transform of output
S(s)
Transfer function, H(s) =
=
Laplace transform of input
U(s)
l q
.....(2)
2s 2 + 1
2s2 + 1
1
´s =
=
U(s) s 2 (s2 + 1)
s3 + s
Using equations (1) and (2)
Example 3.26
Find the impulse response of continuous time systems governed by the following transfer functions.
1
1
1
a) H(s) = 2
b) H s =
c) H s = 2
s (s - 2)
s s +1 s - 2
s + s +1
b g b gb
Solution
a) Transfer function, H(s) =
1
s 2 (s - 2)
2
....(1)
Let, U(s) = L u(t ) =
= S(s) ´
2
g
bg
Chapter 3 - Laplace Transform
3. 68
The impulse response is obtained by taking inverse Laplace transform of the transfer function.
l q
\ Im pulse response, h(t) = L -1 H(s) = L - 1
A =
B =
1
´ s2
s 2 (s - 2)
s=0
LM
N
OP
Q
=
s=0
1
´ (s - 2)
C = 2
s (s - 2)
LM
N
d
1
ds s - 2
=
s= 2
RS
T
\ Impulse response, h( t) = L - 1 -
2
L -1
RS A + B + C UV
Ts s s - 2 W
Using partial fraction
expansion technique
2
1
1
=2
0 - 2
=
d
1
´ s2
ds s2 (s - 2)
RS 1 UV =
T s (s - 2) W
OP
Q
=
s=0
-1
(s - 2)2
=
s=0
1
-1
= (0 - 2)2
4
1
1
=
22
4
l q s1
1
L lt u(t )q =
s
1
L ne u( t )s =
s-a
L u( t ) =
1 1 1 1
1 1
+
2 s2 4 s
4 s-2
UV
W
2
1
1
1
1 2t
e - 2t - 1 u( t)
= - tu( t) - u( t) + e2 tu( t) =
2
4
4
4
d
b) Transfer function, H(s) =
at
i
1
s(s + 1)(s - 2)
The impulse response is obtained by taking inverse Laplace transform of the transfer function.
l q
\ Im pulse response, h(t) = L - 1 H(s) = L -1
A =
B =
1
´ s
s(s + 1)(s - 2)
=
s=0
1
´ (s + 1)
s(s + 1)(s - 2)
1
´ (s - 2)
C =
s(s + 1)(s - 2)
RS
T
\ Impulse response, h( t ) = L - 1 = -
c) Transfer function, H(s) =
RS 1 UV = L RS A + B + C UV
T s(s + 1)(s - 2) W T s s + 1 s - 2 W
-1
1
1
=(0 + 1)(0 - 2)
2
Using partial fraction
expansion technique
=
1
1
=
-1 ´ ( -1 - 2)
3
=
1
1
=
2 ´ (2 + 1)
6
s = -1
s=2
1 1
1 1
1 1
+
+
2 s
3 s +1
6 s-2
l q s1
1
L ne u( t)s =
sma
L u(t ) =
UV
W
± at
1
1 -t
1 2t
1 2t
u( t) +
e u( t) +
e u( t ) =
e + 2e - t - 3 u(t )
2
3
6
6
d
i
1
s2 + s + 1
The impulse response is obtained by taking inverse Laplace transform of the transfer function.
l q = L RST s +1s + 1UVW
RS
UV
1
+
´
´
+
+
(s
.
.
)
(
.
)
s
2
0
5
0
5
1
0
5
T
W
R
U|
RS
UV = L |S
1
1
T (s + 0.5) + 0.75 W
|T (s + 0.5) + e 0.75 j V|W
RS
U = L R 1 ´ 0.866 U
1
T (s + 0.5) + 0.866 VW ST 0.866 (s + 0.5) + 0.866 VW
\ Im pulse response, h(t) = L -1 H(s)
= L -1
= L -1
= L -1
-1
2
2
2
2
-1
2
2
2
-1
2
2
2
2
Arranging, s2+ s, in
the form of (x+y)2
(x+y)2 = x2+2xy+ y2
3. 69
Signals & Systems
\ Impulse response, h(t) =
RS
T
1
0.866
´ L -1
(s + 0.5)2 + 0.866 2
0.866
1
e -0.5 t sin(0.866 t) u(t )
=
0.866
UV
W
n
s
L e - at sin W0 t u( t) =
W0
(s + a)2 + W20
Example 3.27
The input and impulse response of continuous time systems are given below. Find the output of the continuous time
systems.
h(t) = e–at u(t)
a) x(t) = d(t),
–3t
c) x(t) = e
b) x(t) = e–2t u(t), h(t) = u(t)
u(t), h(t) = u(t-1)
d) x(t) = cos 4t u(t) + cos 7t u(t), h(t) = d(t-3)
Solution
l q
n
s
L d( t) = 1; L e - at u( t) =
a) Given that, x(t) = d(t) and h(t) = e–at u(t)
l q l q
1
H(s) = L lh( t)q = L ne u( t)s =
s+a
1
s+a
.....(1)
Let, X(s) = L x(t ) = L d(t ) = 1
- at
.....(2)
Response / Output, y(t) = x(t) * h(t)
On taking Laplace transform of the above equation we get,
l q l
q
L y( t) = L x( t) * h( t)
Using convolution theorem
of Laplace transform
= X(s) H(s)
= 1´
1
1
=
s+a
s +a
Using equations (1) and (2)
Response / Output is given by inverse Laplace transform of the above equation.
1
= e - at u( t)
\ Response, y(t) = L -1
s+a
RS
T
UV
W
l q
b) Given that, x(t) = e–2t u(t) and h(t) = u(t)
L u( t ) =
1
Let, X(s) = L x(t ) = L e -2 t u(t ) =
s+2
1
H(s) = L h( t) = L u( t ) =
s
Response / Output, y(t) = x(t) * h(t)
l q
l q
n
s
l q
1
1
; L e - at u( t) =
s
s+a
.....(1)
n
s
.....(2)
On taking Laplace transform of the above equation we get,
l q l
q
L y( t) = L x( t) * h( t)
Using convolution theorem
of Laplace transform
= X(s) H(s)
=
Using equations (1) and (2)
1
1
1
´
=
s+2 s
s(s + 2)
Response / Output is given by inverse Laplace transform of the above equation.
\ Response, y(t) = L -1
A =
B =
RS 1 UV = L RS A + B UV
T s(s + 2) W T s s + 2 W
1
´ s
s (s + 2)
-1
=
s=0
1
´ (s + 2)
s (s + 2)
1
1
=
2
0+2
=
s = -2
1
-2
= -
1
2
Using partial fraction
expansion technique
Chapter 3 - Laplace Transform
3. 70
RS 1 1 - 1 1 UV
T 2 s 2 s+2 W
1
R 1 U 1 L RS 1 UV
L S V2
T s W 2 T s+2 W
\ Response, y(t ) = L - 1
=
-1
-1
1
1 -2 t
1
u(t ) e u( t ) =
1 - e -2 t u(t )
2
2
2
d
=
i
c) Given that, x(t) = e–3t u(t) and h(t) = u(t-1)
1
Let, X(s) = L x(t ) = L e -3 t u( t) =
s+3
l q
n
s
H(s) = L lh( t)q = L lu( t - 1)q = e
-s
.....(1)
l q
´ L u( t) = e - s ´
1 e-s
=
s
s
.....(2)
Response / Output, y(t) = x(t) * h(t)
l q
if L u( t) =
On taking Laplace transform of the above equation we get,
l q l
q
1
e - as
, then L u( t - a) =
s
s
l
q
L y(t ) = L x(t ) * h( t)
Using convolution theorem
of Laplace transform
= X(s) H(s)
1
e -s
e- s
´
=
s+3
s
s(s + 3)
Response / Output is given by inverse Laplace transform of the above equation.
=
RS e UV = L RS 1 UV
T s(s + 3) W T s(s + 3) W
-s
\ Response, y(t) = L -1
-1
1
´ s
A =
s (s + 3)
B =
s=0
1
´ (s + 3)
s (s + 3)
= L -1
t = t -1
RS A + B UV
Ts s + 3W
Using equations (1) and (2)
Using partial fraction
expansion technique
t = t -1
1
1
=
=
3
0+3
=
s = -3
1
-3
= -
l q
l
if, L x( t) = X(s),
then, L x( t - a ) = e - as X(s)
1
3
q
LM 1 L RS 1 UV - 1 L RS 1 UVOP
N 3 T s W 3 T s + 3 WQ
1
1
O 1
L1
L1
= M u( t) = M u( t - 1) e
e
u(t - 1)P = d1 - e
iu(t - 1)
3
3
Q 3
N3
N3
s
L mcos W t u(t )r =
d) Given that, x(t) = cos 4t u(t) + cos 7t u(t) and h(t) = d(t-3)
s +W
s
s
+
Let, X(s) = L lx(t )q = L lcos 4 t u(t ) + cos 7 t u( t)q =
.....(1)
s +4
s +7
H(s) = L lh( t)q = L l d( t - 3)q = e ´ L l d( t)q = e ´ 1 = e
.....(2)
Response / Output, y(t) = x(t) * h(t)
if L ld( t)q = 1, then L ld( t - a)q = e
On taking Laplace transform of the above equation we get,
L ly( t )q = L lx( t) * h( t)q
\ Response, y(t ) = L - 1
RS 1 1 T3 s
UV
W
O
u(t )P
Q
1 1
3 s+3
-1
=
-1
t = t -1
t = t -1
-3 t
-3( t -1)
-3( t -1)
t = t -1
0
2
-3s
2
2
-3 s
2
2
0
2
-3 s
- as
Using convolution theorem
of Laplace transform
= X(s) H(s)
=
LM
Ns
2
OP
Q
LM
N
s
s
s
s
+ 2
´ e -3 s = e -3 s ´ 2
+ 2
+ 42
s + 72
s + 42
s + 72
OP
Q
Using equations (1) and (2)
if, L lx(t )q = X(s),
R
s
s OU
s
s U
L
R
then, L lx(t - a )q = e
\ Response, y(t) = L S e ´ M
+
+
=L S
V
V
P
+
+
+
+
s
4
s
7
s
4
s
7
Q
N
T
W
T
W
L R s UV + L RS s UVOP = cos 4t u(t) + cos 7t u(t)
= ML S
N T s + 4 W T s + 7 WQ
= cos 4( t - 3) u( t - 3) + cos 7( t - 3) u( t - 3) = bcos 4( t - 3) + cos 7(t - 3)g u( t - 3)
Response / Output is given by inverse Laplace transform of the above equation.
-1
-3 s
-1
2
2
2
2
2
2
2
2
t = t -3
-1
-1
2
2
2
2
t = t -3
t= t- 3
- as
X(s)
3. 71
Signals & Systems
Example 3.28
Perform convolution of the following causal signals, using Laplace transform.
a) x1(t) = 2 u(t),
b) x1(t) = e–2t u(t),
x2(t) = u(t)
–5t
c) x1(t) = t u(t),
x2(t) = e
u(t)
x2(t) = e–5t u(t)
d) x1(t) = cos t u(t), x2(t) = t u(t)
l q 1s
1
Llt u( t)q =
s
Solution
L u( t ) =
a) Given that, x1(t) = 2 u(t)
2
x2(t) = u(t)
l q l q
X (s) = Llx (t )q = Llu(t )q =
l q
Let, X1(s) = L x1( t) = L 2 u( t) = 2 L u(t ) = 2 ´
2
2
1
2
=
s
s
.....(1)
1
s
.....(2)
From convolution theorem of Laplace transform,
l
q
L x1(t ) * x 2 (t ) = X1(s) X 2 (s)
=
2
1
2
= 2
´
s
s
s
\ x1( t) * x 2 ( t) = L-1
RS 2 UV =
Ts W
2
Using equations (1) and (2)
2 L-1
RS 1 UV
Ts W
2
= 2 ´ t u( t) = 2 t u( t )
b) Given that, x1(t) = e–2t u(t)
L e - at u( t) =
n
x2(t) = e–5t u(t)
l q n
X (s) = Llx (t)q = Lne
Let, X1(s) = L x1(t) = L e -2t u(t) =
2
-5t
2
s
u(t)s =
s
1
s + 2
1
s + 5
1
s + a
.....(1)
.....(2)
From convolution theorem of Laplace transform,
l
q
L x1( t) * x 2 ( t) = X1(s) X 2 (s)
=
1
1
1
=
´
(s + 2) (s + 5)
s + 2
s + 5
\ x1( t) * x 2 (t ) = L-1
A =
RS
U = L RS A
1
Ts + 2
T (s + 2) (s + 5) VW
-1
1
´ (s + 2)
(s + 2) (s + 5)
1
B =
´ (s + 5)
(s + 2) (s + 5)
\ x1(t) * x 2 (t) = L-1
=
RS 1
T3
1
1
= 3
-5 + 2
-1
-2t
Using partial fraction
expansion technique
RS
T
1 -1
1
L
3
s + 5
UV
W
-5t
1
1
1
L u(t) = ; L t u(t) = 2 ; L e - atu(t) =
s + a
s
s
l q
x2(t) = e–5t u(t)
l q l q s1
1
X (s) = Llx (t)q = Lne u(t)s =
s + 5
Let, X1(s) = L x1(t) = L t u(t) =
-5t
2
=
UV
W
UV = 1 L RS 1 UV W 3 Ts + 2 W
1
u(t) =
de - e i u(t)
3
c) Given that, x1(t) = t u(t)
2
1
1
=
3
-2 + 5
1
1
1
s + 2
3 s + 5
1 -2t
1 -5t
e u(t) e
3
3
B
s + 5
+
=
s = -2
s = -5
Using equations (1) and (2)
2
l q
n
s
.....(1)
.....(2)
3. 72
Chapter 3 - Laplace Transform
From convolution theorem of Laplace transform,
l
q
L x1(t) * x 2 (t) = X1(s) X 2 (s)
1
1
1
= 2
´
s2
s + 5
s (s + 5)
=
RS 1 UV
T s (s + 5) W
RS A + B + C UV
Ts s s + 5 W
\ x1(t) * x 2 (t) = L-1
2
= L-1
2
A =
B =
1
´ s2
s2 (s + 5)
=
s=0
LM
N
d
1
´ s2
ds s2 (s + 5)
OP
Q
1
´ (s + 5)
C = 2
s (s + 5)
\ x1(t) * x 2 (t) = L-1
=
Using equations (1) and (2)
RS
T
Using partial fraction
expansion technique
1
1
=
0 + 5
5
LM
N
d
1
ds s + 5
=
s=0
s = -5
OP
Q
=
s=0
-1
(s + 5)2
=
s=0
1
-1
= (0 + 5)2
25
1
1
=
=
(-5)2
25
1 1
1 1
1
1
+
5 s2
25 s
25 s + 5
UV
W
RS UV - 1 L RS 1 UV + 1 L RS 1 UV
T W 25 T s W 25 T s + 5 W
1 -1 1
L
5
s2
-1
-1
1
1
1 -5t
t u(t) u(t) +
e u(t)
5
25
25
1
=
e -5t + 5t - 1 u(t)
25
=
d
i
d) Given that, x1(t) = cos t u(t)
l q
L t u(t ) =
x2(t) = t u(t)
l q l
q
X (s) = Llx (t)q = Llt u(t)q =
Let, X1(s) = L x1(t) = L cost u(t) =
2
2
s
s2 + 1
m
r
.....(1)
1
s2
From convolution theorem of Laplace transform,
1
s
L x1(t) * x 2 (t) = X1(s) X2 (s) = 2 ´ 2
s
s + 1
1
A
Cs + D
=
=
+
s(s2 + 1)
s
s2 + 1
l
1
s
; L cos W 0 t u( t) = 2
s2
s + W20
.....(2)
q
Using equations (1) and (2)
.....(3)
Using partial fraction
expansion technique
On cross multiplying equation (3) we get,
1 = A (s2 + 1) + (Cs + D)s
\ 1 = As2 + A + Cs2 + Ds
On equating constants of equation (4) we get, A = 1
On equating coefficients of s of equation (4) we get, D = 0
On equating coefficients of s2 of equation (4) we get, A + C = 0 Þ C = –A = –1
l
q
\ L x1(t) * x 2 (t) =
=
A
Cs + D
+
s
s2 + 1
1
s
- 2
s
s + 1
.....(4)
3. 73
Signals & Systems
\ x1(t) * x 2 (t) = L-1
RS 1 Ts s
2
UV
W
RS UV
TW
RS
T
s
1
s
- L-1 2
= L-1
+ 1
s
s + 1
UV
W
l q s1
Lmcos W t u(t )r =
s
L u(t ) =
= u(t) - cos t u(t)
0
Note : Compare the result of Example 3.28 with Example 2.20 of chapter - 2.
2
s
+ W02
Example 3.29
Perform deconvolution operation to extract the signal x1(t).
1 -2 t
e - e -5 t u(t ) ; x 2 (t ) = e -5 t u( t)
3
d
i
a) x1(t) * x2(t) = 2t u(t) ; x2(t) = u(t)
b) x1( t) * x 2 (t ) =
1 -5 t
c) x1( t) * x 2 (t ) =
e + 5 t - 1 u( t) ; x2 ( t) = e -5 t u(t )
25
d) x1(t) * x2(t) = u(t) – cost u(t) ; x2(t) = t u(t)
d
i
Solution
a) Given that, x1(t) * x2(t) = 2t u(t) ; x2(t) = u(t)
Let,
x1(t) * x2(t) = x3(t)
.....(1)
\ x3(t) = 2t u(t)
m r l q
1
X (s) = L lx ( t)q = L lu(t)q =
s
Let, X3 (s) = L x 3 (t ) = L 2t u(t) =
2
2
s2
.....(2)
.....(3)
2
On taking Laplace transform of equation (1) we get,
l
q m r
L x1( t ) * x 2 ( t) = L x 3 ( t)
Using convolution property
of Laplace transform
X 1(s) X 2 (s) = X 3 (s)
X 3 (s)
1
2
2
= X 3 (s) ´
=
´ s=
X 2 (s)
X 2 (s) s 2
s
\ X1(s) =
RS 2 UV = 2 u(t)
Ts W
1
b) Given that, x (t) * x (t) = de - e i u(t) ; x (t) = e
3
l
Using equations (2) and (3)
q
\ x1(t) = L-1 X1(s) = L-1
-2t
1
Let,
-5t
2
2
-5t
u(t)
x1(t) * x2(t) = x3(t)
.....(1)
1
e -2 t
e -5 t
\ x 3 ( t) = e -2 t - e -5 t u(t ) =
u(t ) u( t)
3
3
3
e -2 t
e -5 t
1
1
Let, X 3 (s) = L x 3 (t ) = L
u(t ) u( t) =
3
3
3(s + 2) 3(s + 5)
1
X 2 (s) = L x 2 (t ) = L e -5 t u( t) =
s+5
d
i
m r RST
l q n
UV
W
s
.....(2)
.....(3)
On taking Laplace transform of equation (1) we get,
l
q m r
L x1( t) * x 2 ( t) = L x 3 (t )
X1(s) X 2 (s) = X 3 (s)
\ X1(s) =
Using convolution property
of Laplace transform
1
X 3 (s)
= X 3 (s) ´
X 2 (s)
X 2 (s)
=
LM 1 - 1 OP ´ (s + 5)
N 3(s + 2) 3(s + 5) Q
=
1 s + 5 - (s + 2)
s+5
- =
3(s + 2)
3(s + 2) 3
Using equations (2) and (3)
3. 74
Chapter 3 - Laplace Transform
s+5-s-2
3
1
=
=
3(s + 2)
3(s + 2) s + 2
\ X1(s) =
RS 1 UV = e u(t)
Ts + 2 W
1
c) Given that, x (t) * x (t) =
de + 5t - 1iu(t) ; x (t) = e
25
l
q
-2 t
\ x1( t) = L-1 X1(s) = L-1
-5t
1
Let,
2
2
-5t
u(t)
x1(t) * x2(t) = x3(t)
\ x 3 ( t) =
.....(1)
-5 t
1 -5 t
1
e
t
e + 5 t - 1 u(t ) =
u(t ) + u( t) u( t)
25
25
5
25
d
i
m r RST e25 u(t) + 5t u(t) - 251 u(t)UVW = 25(s1+ 5) + 5s1
1
X (s) = Llx ( t)q = L ne u( t)s =
s+5
-5 t
Let, X 3 (s) = L x 3 ( t) = L
2
-
1
25s
-5 t
2
2
.....(2)
.....(3)
On taking Laplace transform of equation (1) we get,
l
q m r
L x1( t) * x 2 ( t) = L x 3 (t )
Using convolution property
of Laplace transform
X1(s) X 2 (s) = X 3 (s)
\ X1(s) =
X3 (s)
1
= X 3 (s) ´
X2 (s)
X2 (s)
=
LM 1 + 1
N 25(s + 5) 5s
=
s2 + 5(s + 5) - s(s + 5) s2 + 5 s + 25 - s2 - 5 s
25
1
=
=
=
25 s 2
25 s 2
25 s2 s2
l
q
\ x1(t ) = L-1 X1(s) = L-1
2
-
OP
Q
1 s+5 s+5
1
´ (s + 5) =
+
25 5 s2
25 s
25 s
Using equations (2) and (3)
RS 1 UV = t u(t)
Ts W
2
d) Given that, x1(t) * x2(t) = u(t) – cost u(t) ; x2(t) = t u(t)
Let,
x1(t) * x2(t) = x3(t)
.....(1)
\ x3(t) = u(t) – cost u(t)
q
m r l
1
X (s) = L lx (t )q = L lt u(t)q =
s
Let, X3 (s) = L x 3 ( t) = L u(t) - cos t u(t) =
2
2
1
s
s s2 + 1
.....(2)
.....(3)
2
On taking Laplace transform of equation (1) we get,
l
q m r
L x1( t) * x 2 ( t) = L x 3 (t )
Using convolution property
of Laplace transform
X1(s) X 2 (s) = X 3 (s)
\ X1(s) =
LM
N
OP
Q
1
X 3(s)
1
s
= X 3 (s) ´
´ s2
=
X 2 (s)
X 2 (s)
s s2 + 1
=s-
l
q
s(s2 + 1) - s 3 s 3 + s - s3
s3
s
=
=
= 2
s2 + 1
s2 + 1
s +1
s +1
2
\ x1(t ) = L-1 X1(s) = L-1
RS s UV = cos t u(t)
T s + 1W
2
Note : Compare the results of Example 3.29 with Example 3.28.
Using equations (2) and (3)
3. 75
Signals & Systems
Example 3.30
The impulse response of continuous time systems are given below. Determine the unit step response of the
systems using convolution theorem of Laplace transform, and verify the result with Example 2.21 of chapter-2.
a) h(t) = 3t u(t)
b) h(t) = e–5t u(t)
d) h(t) = u(t – 2)
e) h(t) = u(t + 2) + u(t - 2)
c) h(t) = u(t + 2)
Solution
a) Given that, h(t) = 3t u(t)
l q l
1
U(s) = Llu(t)q =
s
q
l q
Let, H(s) = L h(t) = L 3t u(t) = 3 L t u(t) = 3 ´
1
3
= 2
s2
s
.....(1)
; where u(t) is unit step signal
.....(2)
Unit step response, s(t) = h(t) * u(t)
On taking Laplace transform of the above equation we get,
l q
l
q
L s(t) = L h(t) * u(t)
= H(s) U(s)
1
3
3
= 3
´
s
s
s2
Unit step response is given by inverse Laplace transform of the above equation.
=
Using equations (1) and (2)
RS 3 UV = L RS 3 ´ 2 UV
Ts W T 2 s W
3
R 2 UV = 3 t u(t)
L S
2
Ts W 2
\ Unit step response, s(t) = L-1
=
Using convolution theorem
of Laplace transform
-1
3
2+1
-1
n
s
L t n u(t) =
n!
sn + 1
2
2+1
b) Given that, h(t) = e–5t u(t)
l q n
1
U(s) = Llu(t)q =
s
1
s + 5
Let, H(s) = L h(t) = L e -5t u(t) =
s
.....(1)
; where u(t) is unit step signal
.....(2)
Unit step response, s(t) = h(t) * u(t)
On taking Laplace transform of the above equation we get,
l q
l
Using convolution theorem
of Laplace transform
q
L s(t) = L h(t) * u(t) = H(s) U(s)
=
1
1
1
´
=
s + 5
s
s(s + 5)
Using equations (1) and (2)
Unit step response is given by inverse Laplace transform of the above equation.
\ Unit step response, s(t) = L-1
A =
B =
1
´ s
s (s + 5)
RS 1 UV = L RS A
Ts
T s(s + 5) W
-1
=
s=0
1
´ (s + 5)
s (s + 5)
\ Unit step response, s(t) = L-1
=
B
s + 5
+
UV
W
Using partial fraction
expansion technique
1
1
=
0 + 5
5
RS 1
T5
1
-5
= -
1
5
1
1
1
5 s + 5
s
UV
W
=
s = -5
l q
L u(t) =
1
1 -5t
1
u(t) e u(t) =
1 - e -5t u(t)
5
5
5
d
i
1
1
; L e - at u(t) =
s
s + a
n
s
3. 76
Chapter 3 - Laplace Transform
c) Given that, h(t) = u(t + 2)
l q l
1
U(s) = Llu(t)q =
s
q
l q
Let, H(s) = L h(t) = L u(t + 2) = e2s L u(t) = e 2s ´
1
e 2s
=
s
s
.....(1)
; where u(t) is unit step signal
.....(2)
Unit step response, s(t) = h(t) * (t)
On taking Laplace transform of the above equation we get,
l q
l
q
L s(t) = L h(t) * u(t)
Using convolution theorem
of Laplace transform
= H(s) U(s)
Using equations (1) and (2)
1
if L t u(t) = 2 , then
s
eas
L (t + a) u(t + a) = 2
s
= (t + 2) u(t + 2)
e 2s
1
e 2s
´
=
s
s
s2
Unit step response is given by inverse Laplace transform of the above equation.
=
\ Unit step response, s(t) = L-1
RS e UV = L RS 1 UV
T s W Ts W
2s
-1
2
= t u(t)
2
l
q
l
t=t+2
q
t=t +2
To verify the above result with Example 2.21(c) of chapter-2
\ s(t) = t + 2 ; t > –2
When t > –2, u(t + 2) =1,
d) Given that, h(t) = u(t - 2)
Let, H(s) = L h(t) = L u(t - 2) = e -2s L u(t) = e -2s ´
l q l
1
U(s) = Llu(t)q =
s
q
l q
1
e -2s
=
s
s
.....(1)
; where u(t) is unit step signal
.....(2)
Unit step response, s(t) = h(t) * u(t)
On taking Laplace transform of the above equation we get,
l q
l
q
L s(t) = L h(t) * u(t)
= H(s) U(s)
Using convolution theorem
of Laplace transform
e -2s
1
e -2s
´
=
=
s
s
s2
Using equations (1) and (2)
Unit step response is given by inverse Laplace transform of the above equation.
\ Unit step response, s(t) = L-1
RS e UV = L RS 1 UV
T s W Ts W
-2s
-1
2
= t u(t)
2
t=t- 2
t= t -2
1
, then
s2
e - as
L (t - a) u(t - a) = 2
s
l q
if L t u(t) =
= (t - 2) u(t - 2)
l
q
To verify the above result with Example 2.21(d) of chapter-2
When t > 2, u(t - 2) =1,
\ s(t) = t - 2 ; t > 2
e) Given that, h(t) = u(t + 2) + u(t - 2)
Unit step response, s(t) = h(t) * u(t)
= [u(t + 2) + u(t - 2)] * u(t)
= [u(t + 2) * u(t)] + [u(t - 2) * u(t)]
= (t + 2) u(t + 2) + (t - 2) u(t - 2)
To verify the above result with Example 2.21(e) of chapter-2
When t = - 2 to 2,
u(t + 2) = 1, u(t – 2) = 0
\ s(t) = (t + 2) ´ 1 + (t - 2) ´ 0 = t + 2 ; –2 < t < 2
When t > 2, u(t + 2) = 1, u(t – 2) = 1
\ s(t) = (t + 2) ´ 1 + (t - 2) ´ 1 = t + 2 + t - 2 = 2 t ; t < 2
Using the results of (c) and (d)
3. 77
Signals & Systems
Example 3.31
Perform convolution of x1(t) and x2(t) using convolution theorem of Laplace transform and verify the result with
Example 2.23 of chapter-2.
a)
b)
x1(t)
x2(t)
­
1
x1(t)
­
0
1
2
®t
­
0
®t
2
1
0
–1
Fig 3.31.1.
x2(t)
1
1
­
1
®
1
t
3
2
®
0
1
2
3
t
Fig 3.31.2.
Solution
a)
x 1(t)
The mathematical equation of the signal shown in fig 1 is,
1
if L u(t ) = , then
s
e - as
L u( t - a ) =
s
l q
x1(t) = u(t) – u(t – 2)
On taking Laplace transform of above equation we get,
l
X1(s) = L {x1(t)} = L { u(t) - u(t - 2) }
= L {u(t)} - e -2s L {u(t)} =
q
0
1
e -2s
s
s
......(1)
x2(t)
0
+¥
z
2
®t
Fig 2.
1
x 2 (t) e- st dt =
-¥
z
2
1 ´ e -st dt +
0
- st 1
LM e OP
N -s Q
= -
1
–1
By the definition of Laplace transform,
=
­
1
Let, X2(s) = L{x2(t)}
=
t
2
; 0<t<1
= –1 ; 1 < t < 2
X2 (s) =
®
1
Fig 1.
The mathematical equation of the signal shown in fig 2 is,
x2(t) = 1
­
1
0
L e OP
- M
N -s Q
- st
z
1
z
-1 ´ e -st dt =
0
1
2
LM e
N -s
Le
e O
- M
P
-s Q
N -s
-s
=
1
0
-
2
e -st dt -
-2s
z
e -st dt
1
e O
P
-s Q
-s
-
e -s
1
e -2s
e-s
+
+
s
s
s
s
1
2e - s
e -2 s
+
s
s
s
.....(2)
By using convolution property of Laplace transform,
L{x1(t) * x2(t)} = X1(s) X2(s)
l
|RF 1
= L SG
T|H s
R1
= L S
Ts
q
\ x1(t) * x 2 (t) = L-1 X1(s) X1(s)
-1
-1
2
-
e
-2s
s
-
I F1
JK GH s
-
2e- s
e -2 s
+
s
s
I |UV
JK W|
2e - s
e -2 s
e -2 s
2e -3s
e -4 s
+
+
s2
s2
s2
s2
s2
Using equations (1) and (2)
UV
W
3. 78
Chapter 3 - Laplace Transform
1
RS 1 - 2e + 2e - e UV
if L lt u(t)q =
, then
s
s
s
s
s
T
W
e
L l(t - a) u(t - a)q =
R
R
R
1U
2e U
2e U
e U
R
s
= L S V - L S
T s W T s VW + L ST s VW - L ST s VW
R1U
R1U
R1U
R1U
= L S V - 2L S V
+ 2L S V
- L S V
Ts W
Ts W
Ts W
Ts W
\ x1(t) * x 2 (t) = L-1
2
-1
-s
- 3s
2
2
-1
- as
-3 s
-1
2
-1
2
2
-s
2
-4 s
-1
2
-1
-1
2
2
2
t = t -1
t = t -1
t = t -3
+ 2[t u(t)]
2
2
-1
2
= t u(t) - 2[t u(t)]
-4 s
t = t -3
- [t u(t)]
t = t-4
t = t -4
= t u(t) - 2(t - 1) u(t - 1) + 2(t - 3) u(t - 3) - (t - 4) u(t - 4)
To verify the above result with Example 2.23(a) of chapter-2
For t = 0 to 1
When t = 0 to1, u(t) =1, u(t–1) = 0,
u(t–3) = 0, u(t–4) = 0
\ x1(t) * x2(t) = t´1 – 2(t–1)´0 + 2(t–3)´0 - (t–4)´0 = t
For t = 1 to 3
When t = 1 to 2, u(t) =1, u(t–1) = 1,
u(t–3) = 0, u(t–4) = 0
\ x1(t) * x2(t) = t´1 – 2(t–1)´1 + 2(t–3)´0 - (t–4)´0 = t – 2t + 2 = 2 – t
For t = 3 to 4
When t = 3 to 4, u(t) =1, u(t–1) = 1,
u(t–3) = 1, u(t–4) = 0
\ x1(t) * x2(t) = t´1 – 2(t–1)´1 + 2(t–3)´1 - (t–4)´0 = t – 2t + 2 + 2t – 6 = t – 4
For t > 4
When t > 4, u(t) =1, u(t–1) = 1,
u(t–3) = 1, u(t–4) = 1
\ x1(t) * x2(t) = t´1 – 2(t–1)´1 + 2(t–3)´1 - (t–4)´1 = t – 2t + 2 + 2t – 6 - t + 4 = 0
b)
The mathematical equation of the signal shown in fig 1 is,
x1(t)
x1(t) = u(t – 1)
On taking Laplace transform of above equation we get,
e -s
X1(s) = L {u(t – 1)} = e - s L {u(t)} =
.....(1)
s
The mathematical equation of the signal shown in fig 2 is,
0
®
1
3
2
t
Fig 1.
x2(t)
x2(t) = u(t – 2)
On taking Laplace transform of above equation we get,
X1(s) = L {u(t – 2)} = e -2 s L {u(t)} =
­
1
­
1
e -2 s
s
.....(2)
®
0
1
2
3
t
Fig 2.
By using convolution property of Laplace transform,
L{x1(t) * x2(t)} = X1(s) X2(s)
l
RS 1 UV
Ts W
q
\ x1(t) * x 2 (t) = L-1 X1(s) X2 (s) = L-1
= L-1
= t u( t)
2
RS e
Ts
-s
t = t -3
´
e -2s
s
UV = L RS e UV
W
Ts W
= (t - 3) u(t - 3)
t = t-3
To verify the above result with Example 2.23(b) of chapter-2
For t < 3
When t < 3, u(t-3) = 0,
\ x1(t) * x2(t) = (t – 3) ´ 0 = 0
For t ³ 3
When t > 3, u(t-3) = 1
-1
\ x1(t) * x2(t) = (t – 3) ´ 1 = t–3
-3 s
2
Using equations (1) and (2)
l q
l
if, L x(t ) = X(s),
then, L x(t - a ) = e - as X(s)
q
3. 79
Signals & Systems
3.7 Structures for Realization of LTI Continuous Time Systems in s-Domain
In time domain, the input-output relation of a LTI (Linear Time Invariant) continuous time system
is represented by constant coefficient differential equation shown in equation (3.48).
a0
dN
dN - 1
dN - 2
d
dM
y(t)
+
a
y(t)
+
a
y(t)+.....+
a
y(t)
+
a
y(t)
=
b
x( t )
1
N
1
2
N
0
dt
dt N
dt N - 1
dt N - 2
dt M
dM - 1
dM - 2
d
+ b1 M - 1 x( t ) + b2 M - 2 x( t ) +..... bM - 1 x( t ) + b M x( t ) .....(3.48)
dt
dt
dt
In s-domain, the input-output relation of a LTI (Linear Time Invariant) continuous time system is
represented by the transfer function H(s), which is a rational function of s, as shown in equation (3.49).
H(s) =
Y(s)
=
X(s)
b 0 sM + b1 sM - 1 + b2 sM - 2 + ..... + b M - 1 s + b M
a 0 sN + a1 sN - 1 + a 2 sN - 2 + ..... + a N - 1 s + a N
where, N = Order of the system, M £ N and
.....(3.49)
a0 = 1
The above two representations of continuous time system can be viewed as a computational procedure
(or algorithm) to determine the output signal y(t) from the input signal x(t).
The computations in the above equation can be arranged into various equivalent sets of differential
equations, with each set of equations defining a computational procedure or algorithm for implementing
the system.
Table 3.5 : Basic Elements of Block Diagram in Time Domain and s-Domain
Elements of
block diagram
Differentiator
Integrator
(with zero initial
condition)
Constant Multiplier
Time domain
representation
x(t)
x(t)
x(t)
d
dt
s-domain
representation
d
x( t )
dt
z
z x(t) dt
a
a x(t)
+
x2(t)
1
X(s)
X(s)
a
s
s X(s)
s
X( s)
s
a X(s)
X1(s)
x1(t)
Signal Adder
X(s)
x1(t) + x2(t)
+
X2(s)
X1(s) + X2(s)
Chapter 3 - Laplace Transform
3. 80
For each set of equations, we can construct a block diagram consisting of integrators, adders and
multipliers. Such block diagrams are referred to as realization of system or equivalently as structure for
realizing system.The basic elements used to construct block diagrams are listed in table 3.5. (For block
diagram representation of continuous time system refer chapter - 2, section 2.6.2).
Some of the block diagram representations of the system gives a direct relation between time
domain equation and s-domain equation.
The main advantage of rearranging the sets of differential equations is to reduce the computational
complexity and memory requirements.
The different types of structures for realizing continuous time systems are,
1. Direct form-I structure
2. Direct form-II structure
3. Cascade structure
4. Parallel structure
3.7.1 Direct Form-I Structure
Consider the differential equation governing the continuous time system.
dN
dN - 1
dN - 2
d
dM
y(t)
+
a
y(t)
+
a
y(t)
+
.....
+
a
y(t)
+
a
y(t)
=
b
x( t )
N
1
1
2
N
0
dt
dt N
dt N - 1
dt N - 2
dt M
dM - 1
dM - 2
d
+ b1 M - 1 x( t ) + b 2 M - 2 x( t ) + ..... + b M - 1 x( t ) + b M x( t )
dt
dt
dt
\
dN
dN - 1
dN - 2
d
y(t) = -a 1 N - 1 y(t) - a 2 N - 2 y(t) - ..... - a N - 1 y(t) - a N y(t)
N
dt
dt
dt
dt
dM
dM - 1
dM - 2
d
+ b 0 M x( t ) + b1 M - 1 x( t ) + b 2 M - 2 x( t ) + ..... + b M - 1 x( t ) + b M x( t )
dt
dt
dt
dt
On taking Laplace transform of the above equation with zero initial conditions we get,
sN Y(s) = - a 1 sN - 1Y(s) - a 2 s N - 2 Y(s) - ..... - a N - 1s Y(s) - a N Y(s)
+ b 0 sM X(s) + b1s M - 1 X(s) + b 2 s M - 2 X(s) + ..... + b M - 1s X(s) + b M X(s)
On dividing throughout by sN and letting M = N we get,
Y(s)
Y(s)
Y(s)
Y(s)
- a2 2
- ..... - a N - 1 N - 1 - a N N
s
s
s
s
X(s)
X(s)
X(s)
X(s)
+ b 0 X(s) + b1
+ b 2 2 + ..... + b N - 1 N - 1 + b N N
s
s
s
s
The above equation of Y(s) can be directly represented by a block diagram as shown in fig 3.11
and this structure is called direct form-I structure. This structure uses separate integrators for input and
output signals. Hence for realizing this structure more memory is required. The direct form structure
provides a direct relation between time domain and s-domain equations.
Y(s) = - a1
3. 81
Signals & Systems
b0
s
b1
1
X(s)
s
X(s)
s2
........
b2
1
b N–1
1
b N-1
+
1
-a 2
+
+
Y( s)
s2
X(s)
sN -1
+
+
bN
bN
X(s)
sN
s
Y(s)
s
s
-a2
Y(s)
-a N -1 N -1
s
–a N-1
-a N
Y( s)
sN -a
N
Y( s)
s2
s
Y( s)
sN-1
1
s
X( s)
+
Y(s)
s -a
1
1
s
X(s)
sN-1
-a 1
1
s
b2
X(s)
s2
Y(s)
+
........
X( s)
s
b1
+
­
........
1
b0 X(s)
........
X(s)
s
sN
Y(s)
sN
Fig 3.11 : Direct form-I structure of continuous time system.
3.7.2 Direct Form-II Structure
An alternative structure called direct form-II structure can be realized which uses less number of
integrators than the direct form-I structure.
Consider the differential equation governing the continuous time system.
dN
dN - 1
dN - 2
d
dM
y(t)
+
a
y(t)
+
a
y(t)
+
.....
+
a
y(t)
+
a
y(t)
=
b
x( t )
N-1
1
2
N
0
dt
dt N
dt N - 1
dt N - 2
dt M
dM - 1
dM - 2
d
+ b1 M - 1 x( t ) + b 2 M - 2 x( t ) + ..... + b M - 1 x( t ) + b M x( t )
dt
dt
dt
On taking Laplace transform of the above equation with zero initial conditions we get,
sN Y(s) + a1 s N - 1Y(s) + a 2 sN - 2 Y(s) + ..... + a N - 1s Y(s) + a N Y(s) = b 0 sM X(s)
+ b1sM - 1 X(s) + b2 sM - 2 X(s) + ..... + b M - 1 s X(s) + bM X(s)
On dividing throughout by sN and letting M = N we get,
Y(s)
+ a2
s
X(s)
+ b1
+ b2
s
Y(s) + a1
Y(s)
Y(s)
Y(s)
+ ..... + a N - 1 N - 1 + a N N = b 0 X(s)
2
s
s
s
X(s)
X(s)
X(s)
+ ..... + b N - 1 N - 1 + b N N
s2
s
s
Chapter 3 - Laplace Transform
3. 82
LM
N
OP
Q
1
1
1
1
+ a 2 2 + ..... + a N - 1 N - 1 + a N N
s
s
s
s
1
1
1
1
= X(s) b 0 + b1
+ b 2 2 + ..... + b N - 1 N - 1 + b N N
s
s
s
s
1
1
1
1
b 0 + b1
+ b 2 2 + ..... + b N - 1 N - 1 + b N N
Y(s)
s
s
s
s
\
=
1
1
1
1
X(s)
1 + a1
+ a 2 2 + ..... + a N - 1 N - 1 + a N N
s
s
s
s
Y(s) 1 + a 1
LM
N
Let,
OP
Q
Y( s)
W(s)
Y( s)
=
´
X( s)
X( s)
W(s)
where,
and
W(s)
1
=
1
1
1
1
X(s)
+ a 2 2 + ..... + a N - 1 N - 1 + a N N
1 + a1
s
s
s
s
.....(3.50)
Y(s)
1
1
1
1
= b 0 + b1
+ b 2 2 + ..... + b N - 1 N - 1 + b N N .....(3.51)
W(s)
s
s
s
s
On cross multiplying equation (3.50) we get,
W(s) + a1
W(s)
W(s)
W(s)
W(s)
+ a2
+ ..... + a N - 1 N - 1 + a N
= X(s)
2
s
s
s
sN
\ W(s) = - a 1
W(s)
W(s)
W(s)
W(s)
- a2
- ..... - a N - 1 N - 1 - a N N + X(s) .....(3.52)
2
s
s
s
s
W(s)
X(s)
b0
+
1
-a1
+
+
W( s)
s
-a1
1
-a2
W(s)
s
b1
b1
W(s)
s2
b2
W(s)
W( s)
sN-1
sN-1
-aN-1
1
W(s)
-a N N
s
–aN
W( s)
sN
W( s)
s
Y(s)
+
b2
s
W(s)
s2
+
............
...........
+
1
- a N -1
+
s
W(s)
s2
.......
+
W(s)
-a 2 2
s
s
W(s)
s
b0 W(s)
W(s)
W(s)
b N -1 N -1
s
sN-1 b
N- 1
+
s
W(s)
sN
bN
bN
W(s)
sN
Fig 3.12 : Direct form-II structure of continuous time system for N = M.
3. 83
Signals & Systems
On cross multiplying equation (3.51) we get,
W(s)
W(s)
W(s)
W(s)
+ b2
+ ..... + b N - 1 N - 1 + b N N
s
s2
s
s
Y(s) = b 0 W(s) + b1
.....(3.53)
The equations (3.52) and (3.53) represent the continuous time system in s-domain and can be
realized by a direct structure called direct form-II structure as shown in fig 3.12.
Conversion of Direct Form-I Structure to Direct Form-II Structure
The direct form-I structure can be converted to direct form-II structure by considering the direct
form-I structure as cascade of two systems H1 and H2 as shown in fig 3.13. By linearity property the
order of cascading can be interchanged as shown in fig 3.14 and fig 3.15.
In fig 3.15 we can observe that the input to the integrators in H1 and H2 are same and so the output
of integrators in H1 and H2 are same. Therefore, instead of having separate integrators for H1 and H2, a
single set of integrators can be used. Hence the integrators can be merged to combine the cascaded
systems to a single system and the resultant structure will be direct form-II structure as that of fig 3.12.
X(s)
b0
1
1
+
+
s
-a1
1
+
+
–-aN-1
1
s
bN
–aN
H1
s
-a2
...........
.......
.........
b N-1
+
+
.......
s
b2
1
Y(s)
+
s
b1
1
+
s
H2
Fig 3.13 : Direct form-I structure as cascade of two systems.
X(s)
H
Y(s)
Þ
X(s)
H2
H1
Y(s)
Þ
X(s)
H2
H1
Y(s)
Þ
X(s)
Y(s)
H
Fig 3.14 : Conversion of Direct form-I structure to Direct form-II structure.
Chapter 3 - Laplace Transform
3. 84
+
1
1
.........
+
b2
+
s
–a 2
......
+
s
b1
b N–1
–a N–1
1
–a N
s
H2
1
Y(s)
s
–a 1
1
+
1
+
...........
+
s
b0
........
X(s)
+
s
bN
H1
Fig 3.15 : Direct form-I structure after interchanging the order of cascading.
3.7.3 Cascade Structure
The transfer function H(s) of a continuous time system can be expressed as a product of a number
of second order or first order sections, as shown in equation (3.54).
H (s) =
Y(s)
= H 1 (s) ´ H 2 (s) ´ H 3 (s) ..... H m (s)
X(s)
.....(3.54)
m
= P H i (s)
i=1
1
+ c 2i
s
where, H i (s) =
1
+ d 2i
d 0i + d 1i
s
1
c 0i + c1i
s
or
H i (s) =
1
d 0i + d 1i
s
c 0i + c1i
1
s2
1
s2
Second order section
First order section
The individual second order or first order sections can be realized either in direct form-I or direct
form-II structure. The overall system is obtained by cascading the individual sections as shown in
fig 3.16.
X(s)
® H (s)
1
® H (s) ®
2
® H (s) ® Y(s)
m
Fig 3.16 : Cascade structure of continuous time system.
3. 85
Signals & Systems
The difficulty in cascade structure are,
1. Decision of pairing poles and zeros.
2. Deciding the order of cascading the first and second order sections.
3. Scaling multipliers should be provided between individual sections to prevent the system
variables from becoming too large or too small.
3.7.4 Parallel Structure
The transfer function H(s) of a continuous time system can be expressed as a sum of first and
second order sections, using partial fraction expansion technique as shown in equation (3.55).
H (s) =
Y(s)
= C + H 1 (s) + H 2 (s) + H 3 (s) + ..... + H m (s)
X(s)
.....(3.55)
m
= C +
å
H i (s)
i=1
c 0i + c1i
where, H i (s) =
d 0i + d 1i
or
H i (s) =
1
s
Second order section
1
1
+ d 2i 2
s
s
c 0i
d 0i + d 1i
First order section
1
s
The individual first and second order sections can be realized either in direct form-I or direct
form-II structure. The overall system is obtained by connecting individual sections in parallel as shown in
fig 3.17.
X(s)
C
+
H1(s)
+
H2(s)
+
Hm-1(s)
+
Y(s)
Hm(s)
Fig 3.17 : Parallel structure of continuous time system.
Chapter 3 - Laplace Transform
3. 86
Example 3.32
Find the direct form-I and direct form-II structures of the continuous time system represented by the equation,
d2 y(t)
dy(t)
d2 x(t)
dx(t)
+ 0.6
+ 0.7 y(t) =
+ 0.5
+ 0.4 x(t)
2
dt
dt
dt 2
dt
Solution
Direct Form-I Structure
Given that,
d2 y(t)
dy(t)
+ 0.6
+ 0.7 y(t) =
dt
dt 2
d2 x(t)
dx(t)
+ 0.5
+ 0.4 x(t)
dt
dt 2
On taking Laplace transform of the above equation we get,
s 2 Y(s) + 0.6s Y(s) + 0.7 Y(s) = s2 X(s) + 0.5s X(s) + 0.4 X(s)
On dividiing throughout by s2 we get,
Y(s) + 0.6
X(s)
Y(s)
Y(s)
X(s)
+ 0.4 2
+ 0.7 2 = X(s) + 0.5
s
s
s
s
\ Y(s) = - 0.6
.....(1)
Y(s)
X(s)
Y(s)
X(s)
- 0.7
+ X(s) + 0.5
+ 0.4
s
s
s2
s2
.....(2)
The direct form-I structure of the given continuous time system is realized using equation (2) as shown in fig 1.
Direct Form-II Structure
+
X(s)
1
s
0.5
X(s)
s
0.5
X(s)
s
+
-0.6
+
+
1
s
X(s)
s2
1
s
Y(s)
s
–0.6
Y(s)
s
1
s
Y(s)
-0.7 2
s
X(s)
0.4 2
s
0.4
Y(s)
–0.7
Y(s)
s2
Fig 1 : Direct form-I structure.
On rearranging equation (1) we get,
FG
H
Y(s) 1 + 0.6
1
1
+ 0.7 2
s
s
IJ
K
FG
H
= X(s) 1+ 0.5
1
1
+ 0.4 2
s
s
IJ
K
1
1
1 + 0.5
+ 0.4 2
Y(s)
s
s
=
1
1
X(s)
+ 0.7 2
1 + 0.6
s
s
Let,
Y(s)
W(s) Y(s)
=
X(s)
X(s) W(s)
where,
1
W(s)
=
1
1
X(s)
1 + 0.6
+ 0.7 2
s
s
.....(3)
Y(s)
1
1
= 1+ 0.5
+ 0.4 2
W(s)
s
s
.....(4)
Signals & Systems
3. 87
On cross multiplying equation (3) we get,
W(s) + 0.6
W(s)
W(s)
+ 0.7
= X(s)
s
s2
\ W(s) = - 0.6
W(s)
W(s)
- 0.7
+ X(s)
s
s2
.....(5)
On cross multiplying equation (4) we get,
W(s)
W(s)
+ 0.4 2
s
s
Y(s) = W(s) + 0.5
.....(6)
The direct form-II structure of the given system is realized using equations (5) and (6) as shown in fig 2.
X(s)
W(s)
+
+
W (s )
s
W(s)
-0.6
s
–0.6
+
-0.7
1
s
W (s )
s
0.5
0.5
W (s )
s2
W(s )
s2
–0.7
1
s
W (s )
s2
0.4
0.4
W( s )
s
Y(s)
+
W(s)
s2
Fig 2 : Direct form -II structure.
Example 3.33
Find the direct form-I and direct form-II structures of the continuous time system represented by transfer function,
H(s) =
FG
H
8s3 - 4s 2 + 11s - 2
.
1
1
s s2 - s +
4
2
IJ FG
KH
IJ
K
Solution
Direct Form-I Structure
Given that, H(s) =
Let, H(s) =
\
8s3 - 4s 2 + 11s - 2
1
2
- s +
2
FG s - 1 IJ FG s
H 4K H
IJ
K
Y(s)
; where Y(s) = Output in s- domain and X(s) = Input in s - domain.
X(s)
Y(s)
=
X(s)
8s3 - 4s 2 + 11s - 2
1
2
- s +
2
FG s - 1 IJ FG s
H 4K H
IJ
K
8s3 - 4s 2 + 11s - 2
=
5 2
3
1
3
s s s +
4
4
8
=
s3
8s3 - 4s 2 + 11s - 2
1
1 2
1
1
s s - s2 +
s +
2
4
4
8
FG
H
=
IJ
K
1
1
1
+ 11 2 - 2 3
s
s
s
5 1
3 1
1 1
+
s3 1 4 s
4 s2
8 s3
s3 8 - 4
FG
H
IJ
K
Chapter 3 - Laplace Transform
\
3. 88
1
8 - 4
+ 11
Y(s)
s
=
5
3
1
X(s)
1 +
4 s
4
1
1
- 2 3
s2
s
1 1
1
8 s3
s2
.....(1)
On cross multiplying equation (1) we get,
Y(s) -
5 Y(s)
3 Y(s)
1 Y(s)
X(s)
X(s)
X(s)
+
= 8X(s) - 4
+ 11 2 - 2 3
4 s
4 s2
8 s3
s
s
s
\ Y(s) = 8 X(s) - 4
X(s)
X(s)
X(s)
5 Y(s)
3 Y(s)
1 Y(s)
+ 11 2 - 2 3 +
+
s
4 s
4 s3
8 s3
s
s
.....(2)
The direct form-I structure can be obtained from equation (2) as shown in fig 1.
8
X(s)
1
s
8X(s)
-4
X(s)
s
+
X(s)
s
–4
1
s
+
11
X(s)
s2
11
X(s)
s2
-2
X(s)
s3
–2
+
5 Y(s)
4 s
3 Y(s)
4 s2
-
+
1
s
Y(s)
+
+
X(s)
s3
1 Y(s)
8 s3
1
s
Y(s)
s
5/4
1
s
Y(s)
s2
– 3/4
1
s
1/8
Y(s)
s3
Fig 1 : Direct form-I structure.
Direct Form-II Structure
From equation (1) we get,
1
8 - 4
+ 11
Y(s)
s
=
1
5
3
X(s)
1 +
4 s
4
Let,
1
1
- 2 3
s2
s
1
1 1
s2
8 s3
Y(s)
W(s) Y(s)
=
X(s)
X(s) W(s)
where,
W(s)
1
=
5 1
3 1
1 1
X(s)
1 +
4 s
4 s2
8 s3
.....(3)
Y(s)
1
1
1
= 8 - 4
+ 11 2 - 2 3
W(s)
s
s
s
.....(4)
On cross multiplying equation (3) we get,
W(s) -
5 W(s)
3 W(s)
1 W(s)
+
= X(s)
4
4 s2
8 s3
s
\ W(s) = X(s) +
5 W(s)
3 W(s)
1 W(s)
+
4
s
4 s2
8 s3
.....(5)
Signals & Systems
3. 89
On cross multiplying equation (4) we get,
Y(s) = 8 W(s) - 4
W(s)
W(s)
W(s)
+ 11 2 - 2 3
s
s
s
.....(6)
The equations (5) and (6) can be realized by a direct form-II structure as shown in fig 2.
8W(s)
W(s)
+
X(s)
+
+
+
8
5 W(s)
4 s
3 W(s)
4 s2
1 W(s)
8 s3
Y(s)
1
s
W(s)
s
5/4
-4
–4
W(s)
s
+
1
s
W (s )
s2
–3/4
11
11
1
s
1/8
W (s )
s3
-2
–2
W(s )
s2
+
W (s)
s3
Fig 2 : Direct form-II structure.
Example 3.34
Obtain the direct form-I, direct form-II, cascade and parallel structure of the continuous time LTI system governed
by the equation,
d3 y(t)
3 d2 y(t)
3 dy(t)
1
d3 x(t)
d2 x(t)
+
y(t) = 3
+ 2
.
3
2
2
dt
8 dt
32 dt
64
dt
dt
Solution
Direct Form-I Structure
Given that,
d3 y(t)
3 d2 y(t)
3 dy(t)
1
d3 x(t)
d2 x(t)
+
y(t) = 3
+ 2
3
2
2
8
32
dt
64
dt
dt
dt
dt
On taking Laplace transform of the above equation we get,
s 3 Y(s) +
3 2
3
1
s Y(s) s Y(s) Y(s) = 3s3 X(s) + 2s 2 X(s)
8
32
64
.....(1)
On dividing equation (1) throughout by s3 we get,
Y(s) +
X(s)
3 Y(s)
3 Y(s)
1 Y(s)
= 3 X(s) + 2
s
8 s
32 s2
64 s 3
\ Y(s) = -
3 Y(s)
3 Y(s)
1 Y(s)
X(s)
+
+
+ 3 X(s) + 2
8 s
32 s 2
64 s3
s
.....(2)
.....(3)
Chapter 3 - Laplace Transform
3. 90
The direct form-I structure of the given continuous time system is realized using equation (3) as shown in fig 1.
X(s)
3
3 X(s)
1
s
2
2
X(s)
s
+
+
Y(s)
-
X(s)
s
+
+
3 Y(s)
8 s
1
s
–3/8
1
s
3 Y(s)
32 s 2
3/32
1 Y(s)
64 s 3
Y(s)
s
Y(s)
s2
1
s
1/64
Y(s)
s3
Fig 1 : Direct form-I structure.
Direct Form-II Structure
On rearranging equation (2) we get,
FG
H
Y(s) 1 +
3 1
3 1
1 1
8 s
32 s 2
64 s3
IJ
K
FG
H
= X(s) 3 + 2
1
s
IJ
K
1
3 + 2
Y(s)
s
=
3 1
3 1
1 1
X(s)
1 +
8 s
32 s 2
64 s3
Let,
Y(s)
W(s) Y(s)
=
X(s)
X(s) W(s)
where,
1
W(s)
=
3 1
3 1
1 1
X(s)
1+
8 s
32 s 2
64 s3
.....(4)
Y(s)
1
= 3 + 2
W(s)
s
.....(5)
On cross multiplying equation (4) we get,
W(s) +
3 W(s)
3 W(s)
1 W(s)
= X(s)
8
32 s 2
64 s 3
s
\ W(s) = X(s) -
3 W(s)
3 W(s)
1 W(s)
+
+
s
8
32 s 2
64 s 3
.....(6)
On cross multiplying equation (5) we get,
Y(s) = 3 W(s) + 2
W(s)
s
The equations (6) and (7) can be realized by a direct form-II structure as shown in fig 2.
.....(7)
Signals & Systems
3. 91
W(s)
X(s)
3 W(s)
+
+
3
-
+
3 W(s)
8 s
1
s
W(s)
s
–3/8
2
W(s)
s
1
s
3 W(s)
32 s 2
+
2
Y(s)
W (s )
s2
3/32
1
s
1 W(s)
64 s3
W(s)
s3
1/64
Fig 2 : Direct form-II structure.
Cascade Structure
On rearranging equation (1) we get,
FG
H
1
3 2
3
s s 64
8
32
Y(s) s 3 +
\
IJ
K
d
= X(s) 3s 3 + 2s2
i
Y(s)
3s3 + 2s2
=
3
3
1
X(s)
s3 +
s2 s 8
32
64
.....(8)
For cascade realization, the numerator and denominator polynomials of equation(8) should be expressed in the
factorized form, as shown below.
s = –1/8 is one of the root of
denominator polynomial of
2
2
3s s +
equation (8).
Y(s)
3
=
\
1
2
8
X(s)
–1/8 1 3/8 –3/32 –1/64
s s +
s2 +
8
8
64
¯ –1/8 –2/64 +1/64
2
3s 2 s +
1 2/8 –8/64
0
3
=
1
1
1
s s +
s2 +
8
8
4
FG
H
FG
H
FG
H
IJ
K
IJ FG
KH
FG
H
IJ FG
KH
IJ
K
IJ
K
IJ
K
FG
H
IJ
K
1I F
J G s - 41 IJK
2K H
3s 2 s +
=
FG s + 1IJ FG s +
H 8K H
2
3
.....(9)
Since there are three first order factors in the denominator of equation (9), H(s) can be expressed as a product of
three sections as shown in equation (10).
Let,
s
Y(s)
= H(s) =
1
X(s)
s +
8
´
s
1
s +
2
´
3s + 2
= H1(s) ´ H2 (s) ´ H3 (s)
1
s 4
.....(10)
Chapter 3 - Laplace Transform
3. 92
where, H1(s) =
s
1
s +
8
; H2 (s) =
s
and H3 (s) =
1
s +
2
3s + 2
1
s 4
The transfer function H1(s) can be realized in direct form-II structure as shown below.
Let, H1(s) =
1
Y1(s)
s
=
=
1 1
1
X(s)
1+
s +
8 s
8
+
X(s)
Y1(s)
1
s
On cross multiplying the above function we get,
Y1(s) +
1 Y1(s)
= X(s)
s
8
\ Y1(s) = -
-
1 Y1(s)
+ X(s)
8
s
1 Y1(s)
8 s
Y1(s)
s
-1/8
Fig 3 : Direct form-II structure of H1(s).
The direct form-II structure of H1(s) is obtained using the above equation as shown in fig 3.
The transfer function H2(s) can be realized in direct form-II structure as shown below.
Let, H2 (s) =
1
Y2 (s)
s
=
=
1 1
1
Y1(s)
1 +
s +
2 s
2
+
Y1(s)
Y2(s)
1
s
On cross multiplying the above function we get,
Y2 (s) +
1 Y2 (s)
= Y1(s)
s
2
-
1 Y2 (s)
\ Y2 (s) = + Y1(s)
2
s
1 Y2 (s)
2 s
-1/2
Y2 (s)
s
Fig 4 : Direct form-II structure of H2(s).
The direct form-II structure of H2(s) is obtained using the above equation as shown in fig 4.
The transfer function H3(s) can be realized in direct form-II structure as shown below.
1
3s + 2
s
H3 (s) =
=
1
1 1
s 1 4
4 s
Y2(s)
3 + 2
3 W(s)
3
Y(s)
+
1
s
1 W(s)
4 s
1
Y(s)
W(s) Y(s)
s
Let, H3 (s) =
=
=
1 1
Y2 (s)
Y2 (s) W(s)
14 s
3 + 2
where,
W(s)
+
W(s)
1
=
1 1
Y2 (s)
1 4 s
and
1/4
W(s)
s
2
2
W(s)
s
Fig 5 : Direct form-II structure of H3(s).
Y(s)
1
= 3 + 2
W(s)
s
On cross multiplying the above functions we get,
W(s) = Y2 (s) +
1 W(s)
4
s
and Y(s) = 3 W(s) + 2
W(s)
s
The direct form-II structure of H 3(s) is obtained using the above equations as shown in fig 5.
The cascade structure of the given system is obtained by connecting the individual sections shown in fig 3, fig 4 and
fig 5 in cascade as shown in fig 6.
Signals & Systems
3. 93
X(s)
Y1(s)
+
Y2(s)
+
1
s
-
1 Y1(s)
8 s
-1/8
1 Y2 (s)
2 s
1 W(s)
4 s
Y2 (s)
s
-1/2
W(s)
s
1/4
H2(s)
H1(s)
3
Y(s)
+
1
s
1
s
Y1(s)
s
3 W(s)
W(s)
+
2
2
W(s)
s
H3(s)
Fig 6 : Cascade structure of the system.
Parallel Structure
Dividing numerator by denominator
From equation (8) we get,
3
3
2
3s + 2s
Y(s)
=
3 2
3
1
X(s)
3
s +
s s 8
32
64
= 3 +
7 2
9
3
s +
s +
8
32
64
3 2
3
1
s s s3 +
8
32
64
= 3 +
7 2
9
3
s +
s +
8
32
64
1
1
1
s +
s +
s 8
2
4
FG
H
IJ FG
KH
3s3 + 2s2
9
9
3
3s3 + s2 s(-) (-) 8 (+) 32 (+)64
3
7 2 9
s +
s+
64
8
32
3
3
1
s3 + s2 s8
32
64
IJ FG
KH
Using equation (9)
IJ
K
By partial fraction expansion technique the above function can be expressed as shown below.
Y(s)
= H(s) = 3 +
X(s)
A =
FG
H
FG
H
7 2
9
3
s +
s +
8
32
64
1
1
1
s +
s +
s 8
2
4
IJ FG
KH
IJ FG
KH
3
7 2
9
s +
s +
64
8
32
1
1
1
s +
s +
s 8
2
4
IJ FG
KH
IJ FG
KH
IJ
K
´
IJ
K
A
= 3 +
s +
FG s + 1IJ
H 8K
=
7
8
s = - 81
1
8
B
+
s +
B =
FG
H
IJ FG
KH
IJ FG
KH
IJ
K
´
FG s + 1 IJ
H 2K
=
s = - 21
7
8
1
4
IJ = - 13
K 72
FG - 1IJ + 9 FG - 1IJ + 3
H 2 K 32 H 2 K 64
FG - 1 + 1IJ FG - 1 - 1IJ
H 2 8K H 2 4K
7
9
3
8
14 - 9 + 3
+
4
8
32
64
32
64
64
64
=
=
=
=
=
´
9
9
3
3
9
64
9
32
32
8
4
FG IJ FG IJ
H KH K
C
s -
2
FG
H
7 2
9
3
s +
s +
8
32
64
1
1
1
s +
s +
s 8
2
4
+
FG - 1IJ + 9 FG - 1IJ + 3
H 8 K 32 H 8 K 64
FG - 1 + 1IJ FG - 1 - 1IJ
H 8 2K H 8 4K
3
7
9
13
7 - 18 + 24
+
13
64
64
512
512
256
512
=
=
=
=
´ 9
9
3
3
512
9
64
64
8
8
FG IJ FG IJ
H KH K
1
2
2
Chapter 3 - Laplace Transform
C =
FG
H
3. 94
7 2
9
3
s +
s +
8
32
64
1
1
1
s +
s +
s 8
2
4
IJ FG
KH
IJ FG
KH
IJ
K
´
FG s - 1IJ
H 4K
=
s = 41
7
8
FG 1 IJ + 9 FG 1 IJ + 3
H 4 K 32 H 4 K 64
FG 1 + 1 IJ FG 1 + 1 IJ
H 4 8 K H 4 2K
2
22
7
9
3
7 + 9 + 6
+
+
22
32
11
128
128
128
64
128
=
=
=
´
=
=
9
9
3
3
128
9
18
32
32
8
4
4
11
13
A
B
C
9
18
72
+
+
= 3 +
+
+
\ H(s) = 3 +
1
1
1
1
1
1
s +
s +
s s +
s +
s 8
2
4
8
2
4
FG IJ FG IJ
H KH K
Let, H(s) = 3 + H1(s) + H2 (s) + H3 (s)
Y(s)
where, H(s) =
;
X(s)
Now, H(s) =
13
72 ;
H1(s) =
1
s +
8
-
H2 (s) =
4
9
s +
1
2
;
H3 (s) =
11
18
s -
1
4
Y(s)
= 3 + H1(s) + H2 (s) + H3 (s)
X(s)
\ Y(s) = 3X(s) + H1(s) X(s) + H2(s) X(s) + H3 (s) X(s)
Let, Y(s) = 3X(s) + Y1(s) + Y2 (s) + Y3 (s)
.....(11)
13
72 X(s)
1
s +
8
Þ
13
Y1(s)
72
=
1
X(s)
1+
8
Þ
4 1
Y2 (s)
9 s
=
1 1
X(s)
1+
2 s
-
where, Y1(s) = H1(s) X(s) =
Y2 (s) = H2 (s) X(s) =
4
9
s +
Y3 (s) = H2 (s) X(s) =
1
2
11
18
s -
1
4
X(s)
X(s)
Þ
1
s
1
s
11 1
Y3 (s)
s
18
=
1 1
X(s)
1 4 s
In equation (11), the output Y(s) is expressed as sum of four components. The first component is simply the input
multiplied by a constant, and so it is realised as a constant multiplier in the parallel structure shown in fig 10. The other three
components involve first order section of transfer function, and so they are realized by direct form-II structures as shown
below.
The transfer function H1(s) can be realized in direct form-II structure as shown below.
Let,
13
Y1(s)
W1(s) Y1(s)
72
=
=
1
X(s)
X(s) W1(s)
1+
8
where,
1
s
1
s
W1(s)
1
=
1 1
X(s)
1+
8 s
and
13 1
Y1(s)
= 72 s
W1(s)
.....(12)
Signals & Systems
3. 95
On cross multiplying and rearranging the transfer functions of equation(12) we get,
W1(s) = X(s) -
1 W1(s)
8
s
and Y1(s) = -
13 W1(s)
72
s
The direct form-II structure of H1(s) is obtained using the
above equations as shown in fig 7.
X(s)
W1(s)
+
The transfer function H2(s) can be realized in direct form-II
structure as shown below.
Let,
where,
W2 (s)
1
=
1 1
X(s)
1 +
2 s
1
s
-
4 1
Y2 (s)
W2 (s) Y2 (s)
9 s
=
=
1 1
X(s)
X(s) W2 (s)
1+
2 s
Y1(s)
1 W1(s)
8 s
W1(s)
s
-1/8
A
W1(s)
,
s
13
where A = 72
A
Fig 7 : Direct form-II structure of H1(s).
4 1
Y2 (s)
=
9 s
W2 (s)
and
On cross multiplying and rearranging the above functions we get,
W2 (s) = X(s) -
1 W2 (s)
2
s
and Y2 (s) =
4 W2 (s)
9
s
X(s)
The direct form-II structure of H2(s) is obtained using the above
equations as shown in fig 8.
Let,
where,
W3 (s)
1
=
1 1
X(s)
1 4 s
Y2(s)
1
s
The transfer function H3(s) can be realized in direct form-II 1 W (s)
2
structure as shown below.
2 s
11 1
Y3 (s)
W3 (s) Y3 (s)
s
18
=
=
1 1
X(s)
X(s) W3 (s)
1 4 s
W2(s)
+
-1/2
W2 (s)
s
4/9
4 W2 (s)
9 s
Fig 8 : Direct form-II structure of H2(s).
11 1
Y3 (s)
=
18 s
W3 (s)
and
On cross multiplying and rearranging the above functions we get,
W3 (s) = X(s) +
1 W3 (s)
4
s
11 W3 (s)
18
s
and Y3 (s) =
.....(13)
The direct form-II structure of H3(s) is obtained using equation (13) as shown in fig 9.
X(s)
W3(s)
+
Y3(s)
1
s
1 W3 (s)
4 s
W3 (s)
1/4
s
11/18
11 W3 (s)
18 s
Fig 9 : Direct form-II structure of H3(s).
The parallel structure of the given system is obtained by connecting the individual sections shown in fig 7, fig 8 and
fig 9 in parallel as shown in fig 10.
Chapter 3 - Laplace Transform
3. 96
X(s)
X(s)
X(s)
3 X(s)
3
W1(s)
+
Y1(s)
1
s
1 W1(s)
8 s
A
W1(s)
s
-1/8
+
W1(s)
Y(s)
+
,
s
A
where A = -
13
72
H1(s)
X(s)
W2(s)
+
Y2(s)
+
1
s
-
1 W2 (s)
2 s
W2 (s)
s
-1/2
4/9
4 W2 (s)
9 s
H2(s)
X(s)
W3(s)
+
Y3(s)
1
s
1 W3 (s)
4 s
W3 (s)
s
1/4
11/18
11 W3 (s)
18 s
H3(s)
Fig 10 : Parallel structure of the system.
3.8 Summary of Important Concepts
1.
2.
3.
4.
5.
6.
7.
The Laplace transform is used to transform a time domain signal to complex frequency domain.
The signal Kest can be thought of as an universal signal which represents all types of signals.
Complex frequency, s is defined as, s = s + jW.
The Complex frequency plane or s-plane is a two dimensional complex plane with values of s on horizontal
axis and values of W on vertical axis.
In the definition of Laplace transform if the limits of integral is, 0 to +¥, then the Laplace transform is called
one sided Laplace transform or unilateral Laplace transform.
In the definition of Laplace transform if the limits of integral is, -¥ to +¥, then the Laplace transform is called
two sided Laplace transform or bilateral Laplace transform.
A causal signal x(t) is said to be of exponential order if a real, positive constant s (where s is real part of s)
exists such that the function, e-st x( t ) approaches zero as t approaches infinity.
8. For a causal signal, if Lt e -st x( t ) = 0 for s > sc , and if Lt e - st x( t ) = ¥ for s < sc, then sc is called
t®¥
t®¥
abscissa of convergence, (where sc is a point on real axis in s-plane).
+¥
9.
The values of s for which the integral
z x( t ) e
- st
dt converges is called Region Of Convergence (ROC).
-¥
10. For a causal signal, the ROC includes all points on the s-plane to the right of abscissa of convergence.
11. For an anticausal signal, the ROC includes all points on the s-plane to the left of abscissa of convergence.
12. For a two sided signal the ROC includes all points on the s-plane in the region in between two abscissa of
convergences.
13. The convolution theorem of Laplace transform says that, Laplace transform of convolution of two time
domain signals is given by the product of the Laplace transform of the individual signals.
14. When X(s) is expressed as a ratio of two polynomials in s, then the s-domain signal X(s) is called a rational
function of s.
3. 97
Signals & Systems
15. The zeros and poles are two critical complex frequencies at which a rational function of s takes two extreme
values zero and infinity respectively.
16. Since the signal X(s) attains infinte value at poles, the ROC of X(s) does not include poles.
17. In a realizable system, the number of zeros will be less than or equal to number of poles.
18. For a causal signal x(t), in terms of poles of X(s), the ROC is the region to the right of right most pole of X(s),
(i.e., right of the pole with largest real part).
19. For an anticausal signal x(t), in terms of poles of X(s), the ROC is the region to the left of left most pole of X(s),
(i.e., left of the pole with smallest real part).
20. For a signal with causal and anticausal part, in terms of poles of X(s), the ROC is the region in between
largest pole of causal part and smallest pole of anticausal part.
21. The transfer function of a continuous time system is defined as the ratio of Laplace transform of output and
Laplace transform of input.
22. The transfer function of an LTI continuous time system is also given by Laplace transform of the impulse
response.(Alternatively, the inverse Laplace transform of transfer function is the impulse response of the
system).
23. For a stable LTI continuous time system the ROC should include the imaginary axis of s-plane.
24. For a stable LTI continuous time causal system, the poles should lie on the left half of s-plane and the
imaginary axis should be included in the ROC.
25. For a stable LTI continuous time noncausal system, the imaginary axis should be included in the ROC.
3.9 Short Questions and Answers
Q3.1
Find the Laplace transform of x(t) = sin2t.
Solution
2t U 1
l q n s RST1- cos
VW = 2 L l1q - L lcos 2tq
2
1 Ls + 4 - s O
= M
P = s(s 2+ 4)
2 N s(s + 4) Q
X(s) = L x(t ) = L sin2 t = L
2
OP
Q
2
Let x(t) and X(s) be Laplace transform pair. Given that, X(s) =
y(t) = x
LM
N
1 1
s
2 s s2 + 4
2
2
Q3.2
=
FG t IJ - x(2t).
H 5K
s
. Find Laplace transform of
s2 + 1
Solution
RS FG t IJ UV = 5 X (5s) = 5 5s = 25s =
(5s) + 1 25s + 1 s
T H 5KW
L x
2
2
s
2
+
1
25
=
s
s + 0.04
2
l q
If L x(t) = X(s) then by
time scaling property
l q
L x(at) =
s
2
l q = 21 X FGH 2s IJK = 21 F s I = 41 s s = s s+ 4
GH 2 JK + 1 4 + 1
R F tI
U R F t IU
Now, L ly(t)q = L Sx G J - x(2t)V = L SxG J V - L lx(2t)q
H
K
5
T
W T H 5K W
L x(2 t)
2
=
Q3.3
2
2
s
s
s(s 2 + 4) - s(s 2 + 0.04)
3.96 s
- 2
=
= 2
s + 0.04 s + 4
(s 2 + 0.04) ( s 2 + 4)
(s + 0.04 ) ( s 2 + 4 )
2
Find the inverse Laplace transform of X(s) =
LM
N
OP
Q
1 1
s
.
+
2 s s2 + 4
1
X s
a
a
e j
Chapter 3 - Laplace Transform
3. 98
Solution
RS 1 L 1 + s OUV = 1 LM L RS 1 UV + L RS s UVOP
T 2 MN s s + 4 PQW 2 N T s W T s + 4 WQ
l q
x(t ) = L -1 X(s) == L -1
=
Q3.4
-1
-1
2
2
1
1 + cos 2 t
1 + cos 2 t =
= cos2 t
2
2
Let x(t) and X(s) be Laplace transform pair. Given that, x(t) = e–t u(t). Find inverse Laplace transform
of e–3s X(2s).
Solution
l q n s s 1+ 1 Þ X(2s) = 2s1+ 1 = 21
|R 1 1 |UV = 1 e u(t)
Now, L lX(2s)q = L S
|T 2 s + |W 2
1
1
= e
\ L ne X(2ss = L lX(2s)q
= e u(t)
2
2
1
s + 21
X(s) = L x( t) = L e - t u(t) =
-1
-t
-1
2
1
2
-1
-3 s
- 2t
-1
Q3.5
( t -3)
- 2
t=t-3
u(t - 3)
t = t -3
Find inverse Laplace transform of,
i) X(s) =
s
(s 2 + 9)
lq
Re s > 0
ii) X(s) =
s
(s 2 + 9)
lq
Re s < 0
Solution
i) When Re{s} > 0, the x(t) will be causal or right sided signal.
s
s
\ x(t) = L-1 X(s) = L-1 2
= L-1 2
= cos 3t u(t)
s +9
s + 32
RS UV RS
UV
T W T
W
ii) When Re{s} < 0, the x(t) will be anticausal and left sided signal.
R s UV = L RS s UV = cos 3t u(-t)
\ x(t) = L lX(s)q = L S
Ts + 9W Ts + 3 W
l q
-1
-1
-1
2
Q3.6
l q
Given that, L x(t) = X(s) =
2
2
4s + 1
. Determine the initial value, x(0).
s 2 + 6s + 3
Solution
By initial value theorem,
Initial value, x(0) = Lt x(t ) = Lt sX(s) = Lt s ´
s®¥
t ®0
s®¥
FG
H
s 4+
= Lt s ´
s®¥
1
s
4s + 1
s 2 + 6s + 3
IJ
K
4+
Lt
6 3
F 6 3 I = s®¥
1+ +
s G1+ + J
s s
H s sK
l q
Given that, L x(t) = X(s) =
Solution
=
2
4+0
=4
1+ 0 + 0
2
2
Q3.7
1
s
s+4
. Determine the final value, x(¥).
s s + 7s + 10
d
i
2
By final value theorem,
Final value, x(¥) = Lt x( ¥) = Lt sX(s)
t ®¥
= Lt s ´
s®0
Q3.8
Given that, X(s) =
s ®0
s+4
d
s s2 + 7s + 10
i
= Lt
s+4
s®0 s 2 + 7 s + 10
=
0+4
0 + 0 + 10
=
4
10
= 0.4
bm s m + bm - 1 s m - 1 + .....+ b1 s + b0
. Find Lt x(t).
t ®¥
s(s n + an - 1 s n - 1 + .....+ a1 s + a0
Solution
Using final value theorem,
Lt x(t ) = Lt s X(s) = Lt s
t ®¥
s® 0
s®0
bm sm + bm - 1 sm -1 + ...... + b1s + b 0 0 + 0 + ...... + 0 + b0 b 0
=
=
s(sn + an - 1sn - 1 +.....+ a1s + a 0 )
0 + 0 +.....+ 0 + a 0
a0
3. 99
Q3.9
Signals & Systems
2
d
. Find Laplace transform of
x(t) .
s+3
dt
If X(s) =
Solution
Using initial value theorem,
x(0) = Lt s X(s) = Lt s ´
s ®¥
s ®¥
2
= Lt s ´
s + 3 s ®¥
2
2
2
F 3 I = 3 = 1+ 0 = 2
s G1+ J 1+
H sK ¥
Using time differentiation property,
d
2
2s - 2(s + 3)
-6
x(t) = s X(s) - x(0) = s ´
-2=
=
L
dt
s+3
s+3
s+3
RS
T
Q3.10
If X(s) =
UV
W
0.4
. Find Laplace transform of t x(t).
s + 0.2
Solution
Using frequency differentiation property,
l q
L t x(t) = -
Q3.11
OP
Q
LM
N
d
d
0.4
d
0.4
X(s) = =0.4 (s + 0.2)-1 = - 0.4 ´ ( -1) ´ (s + 0.2)-2 =
ds
ds s + 0.2
ds
(s + 0.2)2
The impulse response of a system is e – 4t u(t), and for the input e(t), the response is (1 – e – 4t) u(t). Find
the input x(t).
Solution
Given that, h(t) = e–4t u(t)
l q n
s s +1 4
Given that, y(t) = d1 - e i u(t) = u(t) - e u(t)
1
1
\ Y(s) = L ly(t)q = L nu(t) - e u(t)s = s s+4
\ H(s) = L h( t) = L e -4t u(t) =
-4t
-4t
Using convolution theorem
of Laplace transform
-4t
We know that, y(t) = x(t) * h(t)
On taking Laplace transform of above equation we get,
l q l
q
L y( t) = L x(t ) * h( t)
\ X(s) =
FG
H
IJ b
K
s+4
s+4-s 4
Y(s)
1
1
1
= Y(s) ´
=
´ s+4 =
- 1=
=
s
s
s
H(s)
H(s)
s s+4
l q
Now input, x(t) = L-1 X(s) = L-1
Q3.12
Y(s) = X(s) H(s)
Þ
g
RS 4 UV = 4 u(t)
Ts W
The impulse response of a system is (4e – t – 2e – 3t) u(t). Determine the stability of the system.
Solution
d
i
\ H(s) = L lh(t)q = L n4e
Given that, h(t) = 4e - t - 2e -3t u(t) = 4e - t u(t) - 2e -3 t u(t)
-t
s
u(t) - 2e -3 t u(t)
2
4(s + 3) - 2(s + 1)
2s + 10
4
2(s + 5)
=
=
=
=
(s + 1) (s + 3)
(s + 1) (s + 3) (s + 1) (s + 3)
s +1 s + 3
The poles of H(s) are lying at s = –1 and s = –3, and so they are lying on left half s-plane.
The ROC is the region to the right of right most pole and so imaginary axis is included in ROC.
Since the poles are lying on left half s-plane and the imaginary axis is included in ROC, the given causal system
is stable.
Chapter 3 - Laplace Transform
Q3.13
3. 100
The impulse response of a system is 2e
-5 t
. Determine the stability of the system.
Solution
Given that, h(t) = 2e -5 t = 2e5 t ; t £ 0
= 2e -5 t ; t ³ 0
\ h(t) = 2e5 t u(- t ) + 2e-5 t u(t)
l q n
s
H(s) = L h(t) = L 2e5 t u(- t ) + 2e -5 t u(t)
-2(s + 5) + 2(s - 5)
-20
2
2
=+
=
=
(s - 5) (s + 5)
(s - 5) (s + 5)
s-5 s+5
The poles are lying at s = 5 and s = –5, and so one pole is lying on right half s-plane and another pole lying on
left half s-plane.
The ROC is the region in between the lines passing through s = 5 and s = –5 and so imaginary axis is included
in ROC.
Since imaginary axis is included in ROC, the given noncausal system is stable.
Q3.14
2
Draw the direct form-I structure for the system represented by the transfer function, H(s) = 22s + 3s .
s + 6s + 5
Solution
Let, H(s) =
Y(s)
2s2 + 3s
= 2
X(s) s + 6s + 5
X(s)
+
+
2
Y(s)
On cross m ultiplying the above equation we get,
1
s
1
s
s 2 Y (s) + 6 sY(s) + 5Y (s) = 2s2 X(s) + 3s X(s)
On dividing the above equation by s2 we get,
+
3
–6
1
s
6
5
3
Y (s) + Y (s) + 2 Y(s) = 2X(s) + X(s)
s
s
s
3
6
5
\ Y (s) = 2X (s) + X(s) - Y(s) - 2 Y (s)
s
s
s
–5
Fig Q3.14 : Direct form-I structure.
The above equation can be used to construct the direct form-I structure as shown in fig Q3.14.
Q3.15
Draw the direct form-II structure for the system represented by the transfer function,
H(s) =
0.4s 2 + 0.7s
.
s 2 + 0.5s + 0.9
Solution
0.4s 2 + 0.7s
Y(s)
Let, H(s) =
= 2
=
X(s) s + 0.5s + 0.9
Let,
FG
H
IJ
K
0.7
0.7
0.4 +
s
s
=
0.5 0.9
0.5 0.9
1+
+ 2
s2 1 +
+ 2
s
s
s
s
s2 0.4 +
FG
H
IJ
K
Y(s) W(s) Y(s)
=
´
X(s) X(s) W(s)
1
W(s)
X(s)
..... (1)
=
X(s) 1 + 0.5 + 0.9
s
s2
0.7
Y(s)
= 0.4 +
..... (2)
W(s)
s
On cross multiplying and rearranging equation (1) we get,
where,
0.5
0.9
W(s) = X(s) W(s) - 2 W(s)
s
s
On cross multiplying equation (2) we get,
..... (3)
W(s)
+
0.4
0.4
+
Y(s)
1
s
+
0.7
–0.5
1
s
–0.9
Fig Q3.15 : Direct form -II structure.
0.7
Y(s) = 0.4 W(s) +
W(s)
..... (4)
s
The direct form-II structure is constructed using equations (3) and (4) as shown in fig Q3.15
3. 101
Signals & Systems
3.10 MATLAB Programs
Program
3.1
Write a MATLAB program to find Laplace transform of the following
standard causal signals.
d) cosWt
e) sinhWt
f) e-at sinWt
a) t
b) e-at c) te-at
%****Program
to
find
clear all
syms t real;
syms a real;
syms s complex;
Laplace
%Let
a,
t
%Let
s
be
%(a)
x=t;
disp(‘(a) Laplace
laplace(x)
%(b)
x=exp(-a*t);
disp(‘(b) Laplace
laplace(x)
%(c)
x=t*exp(-a*t);
disp(‘(c) Laplace
laplace(x)
transform
be
any
of
real
complex
some
standard
variable
variable
transform
of
“t”
is’);
transform
of
“exp(-a*t)”
transform
of
“t*exp(-a*t)”
is’);
is’);
%(d)
sg=real(s);
%s - complex frequency ; sg - sigma
o=imag(s);
s=sg+(i*o);
x=cos(o*t);
disp(‘(d) Laplace transform of “cos(o*t)” is’);
laplace(x)
%(e)
x=sinh(o*t);
disp(‘(e) Laplace
laplace(x)
transform
%(f)
x=exp(-a*t)*sin(o*t);
disp(‘(f) Laplace transform
laplace(x)
of
of
“
signals
sinh(o*t)”
(a) Laplace transform of “t” is
ans =
1/s^2
(b) Laplace transform of
“exp(-a*t)” is
ans =
1/(s+a)
(c) Laplace transform of
“t*exp(-a*t)” is
ans =
1/(s+a)^2
o-omega
is’);
“exp(-a*t)*sin(o*t)”
OUTPUT
;
is’);
Chapter 3 - Laplace Transform
3. 102
(d) Laplace transform of “cos(o*t)” is
ans =
s/(s^2-(1/2*s-1/2*conj(s))^2)
(e) Laplace transform of “ sinh(o*t)” is
ans =
-i*(1/2*s-1/2*conj(s))/(s^2+(1/2*s-1/2*conj(s))^2)
(f) Laplace transform of
“exp(-a*t)*sin(o*t)” is
ans =
-i*(1/2*s-1/2*conj(s))/((s+a)^2-(1/2*s-1/2*conj(s))^2)
Program
3.2
Write a MATLAB program to find Laplace transform of the following
standard causal signals.
b) tn-1/(n-1)!
c) tne-at
d) tn-1 e-at/(n-1)!
a) tn
%****Program
to
find
Laplace
transform
clear all
syms t real;
syms a real;
syms n real;
%Let
of
some
standard
t,a ,n
be
a ny
n=input(‘Enter the value of n’); %g et va lue
%(a)
x=t^n;
disp(‘(a) Laplace transform of “t^n” is’);
laplace(x)
of
n
%(b)
x=t^(n-1)/factorial(n-1);
disp(‘(b) Laplace transform
laplace(x)
%(c)
x=(t^n)*exp(-a*t);
disp(‘(c) Laplace transform
laplace(x)
of
“t^(n-1)/(n-1)!”
of
“(t^n)*exp(-a*t)”
signals
r ea l
from
va r ia ble
keyboard
is’);
is’);
%(d)
x=(t^(n-1)*exp(-a*t))/factorial(n-1);
disp(‘(d) Laplace transform of “(t^(n-1)*exp(-a*t))/(n-1)!”
laplace(x)
OUTPUT
Enter the value of n4
(a) Laplace transform
ans =
24/s^5
(b) Laplace transform
ans =
1/s^4
(c) Laplace transform
ans =
24/(s+a)^5
(d) Laplace transform
ans =
1/(s+a)^4
of “t^n” is
of “t^(n-1)/(n-1)!” is
of “(t^n)*exp(-a*t)” is
of “(t^(n-1)*exp(-a*t))/(n-1)!” is
is’);
3. 103
Signals & Systems
Program
3.3
Write a MATLAB program to find Laplace transform of the following
causal signals.
b) 1+0.4e-2tsin3t
c) 3sin2t+3cos2t
a) t2-3t
%****Program
to
clea r a ll
syms t real;
find
%Let
%(a)
x=t^2-(3*t);
disp(‘(a) Laplace
X=laplace(x);
simplify(X)
t
Laplace
be
real
transform
transform
of
given
signals
variable
of
“t^2-(3*t)”
is’);
%(b)
x=1+0.4*exp(-(2*t))*sin(3*t);
disp(‘(b) Laplace transform of
X=laplace(x);
simplify(X)
“1+(0.4*exp(-(2*t)))*sin(3*t)”
%(c)
x=3*sin(2*t)+3*cos(2*t);
disp(‘(c) Laplace transform
X=laplace(x);
simplify(X)
“3*sin(2*t)+3*cos(2*t)”
of
is’);
is’);
OUTPUT
(a) Laplace transform of “t^2-(3*t)” is
ans =
-(-2+3*s)/s^3
(b) Laplace transform of “1+(0.4*exp(-(2*t)))*sin(3*t)” is
ans =
1/5*(5*s^2+26*s+65)/s/(s^2+4*s+13)
(c) Laplace transform of “3*sin(2*t)+3*cos(2*t)” is
ans =
3*(s+2)/(s^2+4)
Program
3.4
Write a MATLAB program to find inverse Laplace transform of the following
s-domain signals.
a) 2/s(s+1)(s+2)
b) 2/s(s+1)(s+2)2
c) 1/(s2+s+1)(s+2)
%****Program
to
determine
inverse
Laplace
transform
clear all;
syms s complex;
X=2/(s*(s+1)*(s+2));
disp(‘Inverse Laplace
x=ilaplace(X);
simplify(x)
transform
X=2/(s*(s+1)*(s+2)^2);
disp(‘Inverse Laplace transform
x=ilaplace(X);
simplify(x)
of
2/(s(s+1)(s+2))
is’);
of
2/(s(s+1)(s+2)^2)
is’);
Chapter 3 - Laplace Transform
3. 104
X=1/((s^2+s+1)*(s+2));
disp(‘Inverse Laplace transform
x=ilaplace(X);
simplify(x)
of
1/((s^2+s+1)*(s+2))
is’);
OUTPUT
Inverse Laplace transform of 2/(s(s+1)(s+2)) is
ans =
1+exp(-2*t)-2*exp(-t)
Inverse Laplace transform of 2/(s(s+1)(s+2)^2) is
ans =
t*exp(-2*t)-2*exp(-t)+3/2*exp(-2*t)+1/2
Inverse Laplace transform of 1/((s^2+s+1)*(s+2)) is
ans =
-1/3*exp(-1/2*t)*cos(1/2*3^(1/2)*t)
+1/3*3^(1/2)*exp(-1/2*t)*sin(1/2*3^(1/2)*t)+1/3*exp(-2*t)
Program
3.5
Write a MATLAB program to perform convolution of signals,x1(t)= t2-3t
and x 2(t)=t,using Laplace transform, and then to perform deconvolution
using the result of convolution to extract
x1(t) and x2(t).
%****Program
for
convolution
&
deconvolution
using
Laplace
transform
clear all;
syms t real;
x1t=(t^2-(2*t));
x2t=t;
X1s=laplace(x1t);
X2s=laplace(x2t);
X3s = X1s*X2s;
con12=ilaplace(X3s);
disp(‘Convolution of x1(t)
simplify(con12)
%product
of
laplace
transform
of
inputs
and x2(t) is’);
% convolution output
decon_X1s=X3s/X1s;
decon_x1t=ilaplace(decon_X1s);
disp(‘The signal x1(t) obtained
simplify(decon_x1t)
decon_X2s=X3s/X2s;
decon_x2t=ilaplace(decon_X2s);
disp(‘The signal x2(t) obtained
simplify(decon_x2t)
by
deconvolution
is’);
by
deconvolution
is’);
OUTPUT
Convolution of x1(t) and x2(t) is
ans =
1/12*t^4-1/3*t^3
The signal x1(t) obtained by deconvolution is
ans =
t
The signal x2(t) obtained by deconvolution is
ans =
t^2-2*t
Program
3.6
Write a MATLAB program to find residues and poles of s-domain signal,
(1.5s3+4.45s 2+4.25s+0.2)/(s3 +3.5s2+3.5s+1)
3. 105
Signals & Systems
%*****Program
to
find
residues
and
poles
of
s-domain
signal
clear all
s=tf(‘s’);
b=[1.5 4.45 4.25
a=[1 3.5 3.5 1];
disp(‘The
f=tf([b],
given
[a])
0.2];
%
%
transfer
Numerator coefficients
Denominator coefficients
function
is,’);
disp(‘The residues, poles and direct terms of given transfer
function are,’);
disp(‘(r - residue ; p - poles ; k - direct terms)’);
[r,p,k]=residue(b,a)
disp(‘The numerator and
r,p,k are,’);
[b,a]=residue(r,p,k)
denominator
coefficients
extracted
from
OUTPUT
The given transfer function is,
Transfer function:
1.5 s^3 + 4.45 s^2 + 4.25 s + 0.2
————————————————s^3 + 3.5 s^2 + 3.5 s + 1
The residues, poles and direct terms of given transfer function are,
(r - residue ; p - poles ; k - direct terms)
r =
-1.6667
2.2000
-1.3333
p =
-2.0000
-1.0000
-0.5000
k =
1.5000
The numerator and denominator coefficients extracted from r,p,k are,
b =
1.5000
4.4500
4.2500
0.2000
a =
1.0000
3.5000
3.5000
1.0000
Program
3.7
Write a MATLAB program to find the impulse response of the LTI system
governed by the transfer function, H(s)=1/(s2+s+1).
%****Program
to
find
impulse
syms s complex;
H=1/(s^2+s+1);
disp(‘Impulse response
h=ilaplace(H);
simplify(h)
h(t)
response
of
continuous
time
is’);
s=tf(‘s’);
H=1/(s^2+s+1);
t=0:0.01:20;
% set a time vector
h=impulse(H,t);
%
impulse
response
of
system
H
LTI
system
Chapter 3 - Laplace Transform
plot(t,h);
xlabel(‘time in
ylabel(‘h(t)’);
3. 106
seconds’);
OUTPUT
Impulse response h(t) is
ans =
2/3*3^(1/2)*exp(-1/2*t)*sin(1/2*3^(1/2)*t)
The sketch of impulse response is shown in fig P3.7.
Fig P3.7 : Impulse response of continuous time LTI system.
Program
Fig P3.8 : Step response of first and
second order system.
3.8
Write a MATLAB program to find the step response of the first and
second order LTI systems governed by the transfer functions,
H(s)=1/(s+2) and H(s)=1/(s2+2.5s+25).
%********Program to find the step response of I and
syms s complex;
H1=1/(s+2);
disp(‘Step response of first order system is’);
h1=ilaplace(H1);
simplify(h1)
H2=1/(s^2+2.5*s+25);
disp(‘Step response of
h2=ilaplace(H2);
simplify(h2)
second
order
system
II
order
systems
is’);
s=tf(‘s’);
H1=1/(s+2);
H2=1/(s^2+2.5*s+25);
t1=0:0.0005:5;
% set a time vector
s1=step(H1,t1);
% step r espo nse o f fir st o r der system
s2=step(H2,t1);
% step r espo nse o f seco nd o r der system
subplot(2,1,1);plot(t1,s1);
xlabel(‘Time in seconds’); ylabel(‘s1(t)’);
subplot(2,1,2);plot(t1,s2);
xlabel(‘Time in seconds’);
ylabel(‘s2(t)’);
3. 107
Signals & Systems
OUTPUT
`
Step response of first order system is
ans =
exp(-2*t)
Step response of second order system is
ans =
4/75*15^(1/2)*exp(-5/4*t)*sin(5/4*15^(1/2)*t)
The sketch of step response of first and second order system are shown in fig P3.8.
3.11 Exercises
I. Fill in the blanks with appropriate words
1.
2.
3.
4.
The Laplace transform will transform a time domain signal to –––––––––– signal.
The imaginary part of complex frequency is called –––––––––– frequency.
The real part of complex frequency is called –––––––––– frequency.
The ROC of Laplace transform of a causal signal includes all points on the s-plane to the –––––––––– of
abscissa of convergence.
5. A signal is said to be –––––––––– if e - st x( t ) approaches zero as t approaches infinity.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
If Laplace transform of x(t) is X(s) then Laplace transform of Kx(t) is –––––––––– .
If Laplace transform of x(t) is X(s) then Laplace transform of x(t – a) is –––––––––– .
When X(s) is a ratio of two polynomials in s, then X(s) is called –––––––––– of s.
The –––––––––– are critical frequencies at which the signal X(s) become infinite.
The –––––––––– are critical frequencies at which the signal X(s) become zero.
The ROC of X(s) consists of stripes parallel to the –––––––––– in s-plane.
The –––––––––– is the ratio of Laplace transform of output and input.
The Laplace transform of –––––––––– gives the transfer function of the system.
For a stable LTI system the ROC should include –––––––––– of s-plane.
For a stable causal system –––––––––– should lie on left half of s-plane.
Answers
1. s-domain
2. radian / real
6.
7.
K X(s)
e–asX(s)
11. jW - axis
12. transfer function
3. neper
4. right
8.
9.
rational function 13. impulse response
poles
14. imaginary axis
5. exponential order
10. zeros
15. poles
II. State whether the following statements are True/False
1. The signal Kest is an universal signal which represents all types of signals.
2. The Laplace transform is used to transform a frequency domain signal to complex frequency domain.
3. The Laplace transform exists only if the signal is of exponential order.
4. The Laplace transform of the sum of two or more signals is equal to sum of transforms of individual
signals.
5. Laplace transform of convolution of two signals is given by the product of the Laplace transform of the
individual signals.
6. For a rational signal X(s), the ROC includes poles of X(s).
7. For a realizable system the number of finite poles and zeros should be equal.
8. For a causal rational signal X(s) the ROC is the region to the right of rightmost pole.
Chapter 3 - Laplace Transform
3. 108
9. For an anticausal rational signal X(s) the ROC is the region to the left of leftmost pole.
10. If x(t) is finite and integrable then the ROC is entire s-plane.
11. While forming transfer functions the initial conditions should not be neglected.
12. Zeros and poles of transfer function are complex frequencies.
13. In a transfer function, if s takes the value of a pole, then the transfer function becomes zero.
14. The inverse Laplace transform of the transfer function gives the impulse response.
15. The direct form-I structure uses less number of integrators than direct form-II structure.
Answers
1. True
2. False
6.
7.
False
False
11. False
12. True
3. True
4. True
8.
9.
True
True
13. False
14. True
10. True
15. False
5. True
III. Choose the right answer for the following questions
1. When W = 0 in the signal x(t) = Ae st , where s = s + jW, the signal x(t) represents,
a) exponential signal
b) step signal
c) ramp signal
d) sinusoidal signal
+¥
z
2. The mathematical operation x(t) e - st dt is called,
-¥
a) Laplace transform
b) two sided Laplace transform
c) one sided Laplace transform
d) generalized frequency domain transform
3. The term W in s = s + jW is called,
a) frequency
b) real frequency
c) complex frequency
d) critical frequency
4. A causal signal x(t) is said to be exponential order for any positive value of s if,
a) Lt e + st x ( t ) = ¥
t ®¥
b) Lt e -st x ( t ) = ¥
t ®¥
5. The ROC of a causal signal x(t) is,
a) entire s-plane
c) Lt e + st x( t ) = 0
t ®¥
d) Lt e - st x( t ) = 0
t ®¥
b) region in between two abscissa of convergence
c) right of abscissa of convergence
d) left of abscissa of convergence
6. The Laplace transform of the causal signal t n u(t) is,
n!
n!
n
a) n+1
b) n
c) n +1
s
s
s
d)
n
sn
7. If x(t) and X(s) are Laplace transform pairs, then Laplace transform of e–at x(t) is,
a) eas X(s)
b) e–as X(s)
c) X(s – a)
d) X(s + a)
8. If x(t) and X(s) are Laplace transform pairs, then Laplace transform of
¥
a)
z
¥
X(s) ds
b)
0
z
¥
X(s) ds
c)
s
1
X(s) ds
s0
z
x(t)
is,
t
d)
¥
1
X(s) ds
ss
z
9. If x(t) is periodic with period T, then Laplace transform of x(t) is defined as,
a)
1
1 - e -sT
T
z
0
x(t) e- st dt
b)
1
1 + e - sT
T
z
0
x(t) e- st dt c)
1
1 - e sT
T
z
0
x(t) e- st dt
d)
1
1 + esT
T
z
0
x(t) e- st dt
3. 109
Signals & Systems
10. If x(t) and X(s) are Laplace transform pair, and X(s) is rational, then which of the following statements are
true.
i) poles and zeros are critical frequencies.
ii) the ROC of X(s) does not include poles.
iii) the number of finite zeros will be less than or equal to number of poles.
a) i and iii only
b) i only
c) ii only
d) all of the above
11. If x(t) and X(s) are Laplace transform pair, and X(s) is rational, then which of the following statements are
true regarding the ROC of X(s).
i) The ROC is bounded by poles or extends to infinity.
ii) If x(t) is right sided, then ROC is right of right most pole.
iii) If x(t) is left sided, then ROC is left of left most pole.
a) i only
b) ii only
c) ii and iii only
12. The inverse Laplace transform of X(s) =
a) x(t) = e - t cos2t
b) x(t) = e - t sin 2t
13. The inverse Laplace transform of X(s) =
a) 4 e -5 t u(t) and 4 e-5 t u( - t)
d) all of the above
2
is,
s 2 + 2s + 5
c) x(t) = e -2 t cos5t
d) x(t) = e -2 t sin 5t
4
for ROC Re{s} > –4 and Re{s} < –4 are respectively.
s+5
b) 4 e5t u(t) and 4 e5 t u( - t)
d) - 4 e -5 t u(t) and - 4 e -5t u(t)
c) 4 e -5t u(t) and - 4 e -5 t u( - t)
14. For a continuous time LTI system which of the following statements are true.
i) The transfer function is ratio of Laplace transform of output and input.
ii) The transfer function is ratio of Laplace transform of input and output.
iii) The transfer function is Laplace transform of impulse response.
a) i only
b) i and iii
c) ii only
d) iii only
15. If x(t), y(t) and h(t) are input, output and impulse response of LTI continuous time system respectively then,
a) h(t) = x(t) * y(t)
b) x(t) = y(t) * h(t)
c) y(t) = x(t) * h(t)
d) y(t) = h(t) * h(t)
16. If X(s), Y(s) and H(s) are Laplace transform of input, output and impulse response of LTI continuous time
system respectively then,
a) x(t) = L-1
RS H(s) UV
T Y(s) W
b) x(t) = L-1
RS Y(s) UV
T H(s) W
c) x(t) = L-1
RS 1 UV
T Y(s) H(s) W
l
17. The convolution of u(t) with u(t) will be equal to,
a) d(t)
b) u(t)
c) t u(t)
d) t 2 u(t)
18. The convolution of e – at u(t) with e – at u(t) will be equal to,
a) t u(t)
b) e -at u(t)
c) e -2at u(t)
d) te -at u(t)
19. Given that H(s) = e
a) d (t - 4 )
-4s
. What is the impulse response of the system?
b) u(t - 4)
c) e-4t u(t)
q
d) x(t) = L-1 Y( s) H(s)
d) e4 t u(t)
20.Which of the following statements are true regarding the stability of continuous time system?
i) For stability of LTI system, the ROC should include imaginary axis.
ii) For stability of causal system the poles should lie on left half s-plane.
iii) For stability of noncausal system there is no restriction on location of poles in s-plane.
a) i and ii only
b) i and iii only
c) ii and iii only
d) all of the above
Chapter 3 - Laplace Transform
3. 110
Answers
1. a
2.
3.
4.
5.
b
b
d
c
6.
a
11. d
16. b
7.
8.
9.
10.
d
b
a
d
12.
13.
14.
15.
17.
18.
19.
20.
b
c
b
c
c
d
a
d
IV. Answer the following questions
1. Define complex frequency.
2. Define Laplace transform of a signal
3. What is the condition to be satisfied for the existence of Laplace transform?
4. What is abscissa of convergence?
5. What is region of convergence(ROC)?
6. Define inverse Laplace transform.
7. Write any two properties of Laplace transform.
8. State and prove initial value theorem.
9. State and prove final value theorem.
10. Define the convolution theorem of Laplace transform.
11. Write any two properties of ROC of Laplace transform.
12. Write a procedure to determine inverse Laplace transform using convolution theorem.
13. Define transfer function of a continuous time system.
14. Define poles and zeros.
15. Write the procedure to perform convolution using Laplace transform.
16. Write the procedure to perform deconvolution using Laplace transform.
17. What is the relation between impulse response and transfer function of the system?
18. What is the condition for stability of a continuous time LTI system?
19. What is the condition for stability of LTI causal system?
20. What are the basic elements of block diagram?
V. Solve the following problems
E3.1
Find the Laplace transform and the ROC of the following signals using the fundamental definition of
Laplace transform.
d
i
a) x(t) = t 2 + 2 t + 4 u(t)
E3.2
b) x(t) = cos2 6t u(t)
d) x(t) = e at +b u(t)
Find the Laplace transform of the following signals.
a) x(t) = (4 e -2t cos5t - 3e -2t sin5t) u(t)
b) x(t) = (t - 2 ) 3 u(t - 2)
c) x(t) = ( 2 - 4t + t 2 ) e - t u(t)
E3.3
c) x(t) = e -5t sin 7 t u(t)
d) x(t) = d (t) + (t e-3t + 2 cos5t) u(t)
Find the Laplace transform of the following signals.
a) x(t)
b) x(t)
c) x(t)
A
2A
¯
Ae–at
d)
x(t)
A
A
A
0
T
3
2T
T
3
Fig E3.3.1.
t
0
T
Fig E3.3.2.
t
t0
1+t 0
Fig E3.3.3.
t
–T
0
T
Fig E3.3.4.
t
3. 111
Signals & Systems
x(t)
E3.4
Find the Laplace transform of the signal shown in fig E3.4.
E3.5
Find the initial value, x(0) in time domain for the following
s-domain signals.
10(s + 1)
2s2 + 2
a) X(s) =
b) X(s) = 3
(s + 2) (s + 6)
8s + 4s2 + 3s + 5
4
1 7 4
c) X(s) = 3
d) X(s) = 3 + 2 +
s
s + 2s2 + 7s + 1
s s
E3.6
0
t
T
T
2
Fig E3.4.
Find the final value, x(¥) in time domain for the following s-domain signals.
a) X(s) =
E3.7
A
10(s + 1)
s(s + 2) (s + 4)
b) X(s) =
s
s +4
c) X(s) =
2
(s + 2) (s + 3)
s2 (s + 5)
d) X(s) =
s(s + 1)
s + 2s + 1
2
Perform the convolution of x1(t) and x2(t) using Laplace transform.
a) x1 (t) = t e-2 t u(t)
; x 2 (t) = 3 sin4t u(t)
b) x1 (t) = (t - 2 ) u(t - 2) ; x2 (t) = e-3t u(t)
c) x1 (t) = 4 t e -5 t u(t) ; x2 (t) = u(t)
d) x1 (t) = t 2 u(t) ; x2 (t) = d (t - 7 )
e) x1 (t) = sin t u(t)
;
x 2 (t) = t e - t u(t)
E3.8
Determine the poles and zeros of the following s-domain signal and sketch the pole-zero plot.
( s + 2) (s2 + 8s + 20)
X(s) =
(s + 3) (s2 + 2s + 2)
E3.9
Find the inverse Laplace transform of the following signals.
s
2s + 1
a) X(s) = 2
b) X(s) = 3
s + 4s + 3
s + 7s2 + 10s
1
(s + 2)
c) X(s) = 3
d) X(s) = 2
2
s + 2s + 2s + 1
s + 4s + 5
E3.10
Find the inverse Laplace transform of the following signals.
s2 + 2s - 2
4s2 + 15s + 62
a) X(s) =
b) X(s) =
s(s + 2) (s - 3)
(s + 1) (s2 + 4s + 20)
c) X(s) =
2s2 + 4s + 34
(s + 2) (s2 + 6s + 25)
d) X(s) =
s3 + 8s2 + 23s + 28
(s + 4s + 13) (s2 + 2s + 5)
2
1
using convolution theorem.
s2 (s2 - a 2 )
E3.11
Find the inverse Laplace transform of X (s) =
E3.12
Find the impulse response of the systems represented by following differential equations.
dy(t)
d 2 y( t )
a)
= x( t - t 0 )
b) 3
+ y( t ) = 0.5 x( t )
dt
dt
d 2 y( t )
dy
dx(t)
c)
+ 0.8 + 015
. y(t) = 0.2
+ x( t )
dt 2
dt
dt
E3.13
Using Laplace transform, determine the natural response of the system represented by following
equations.
d 2 y(t)
dy( t )
dx( t )
dy( t )
a)
.
+ 15
+ 0.36 y(t) = 0.1
+ 0.7 x( t ) ; y(0) = 0.3 ;
= -0.2
dt 2
dt
dt
dt t = 0
b)
d 2 y(t)
dy( t )
+ 10
+ 21 y(t) = 8 x( t )
2
dt
dt
;
y(0) = 2 ;
dy( t )
dt
= -3
t =0
Chapter 3 - Laplace Transform
3. 112
Using Laplace transform, determine the forced response of the system represented by following equations.
E3.14
d 2 y(t)
dy( t )
dx( t )
+9
+ 20 y(t) = 0.2
+ 2 x( t )
; Input x( t ) = 6 u(t)
2
dt
dt
dt
d 3y(t)
dy(t)
d 2 y( t )
b)
+ 18 y(t) = 12 x( t ) ; Input x(t) = e-5t u(t)
+ 10
+ 27
3
dt
dt
dt 2
a)
Using Laplace transform, determine the complete response of the system represented by following
equations.
E3.15
a)
d 2 y(t)
dy( t )
dx( t )
dy( t )
.
+ 14
+ 0.4 y(t) = 0.2
+ 1.25 x( t ) ; y(0) = 1 ;
dt 2
dt
dt
dt
2
b)
E3.16
d y(t)
dy( t )
+ 11
+ 24 y(t) = 3 x( t )
dt 2
dt
;
y(0) = 2
;
dy( t )
dt
= -2 ; for input x(t) = 2e-3t u( t )
t=0
= -0.5 ; for input x(t) = 4 u( t )
t=0
Find the transfer function of the system governed by the following differential equations.
a)
d 3y(t)
d 2 y( t )
dy( t )
dx( t )
+ 0.4
+ 0.3
+ 0.1 y(t) = 0.3
+ x( t - 1)
3
dt
dt 2
dt
dt
d 3 y(t)
d 2y( t )
dy( t )
d 2 x( t )
dx ( t )
+ 14 y(t) = 3
+7
+5
+2
+ 4 x( t )
3
2
dt
dt
dt
dt 2
dt
d 2 y(t)
dy ( t )
c) 4
+2
+ 8 y(t) = x( t - 3)
dt 2
dt
b)
E3.17
Find the transfer function of the system governed by the following impulse responses.
a) h(t) = d (t - 2) + e -4t u(t)
b) h(t) = d (t) + t e-3t u(t) + e -9 t u( t )
c) h(t) = 2 u(t) + 3e-6t u(t) + 4e -7t sin 2t u(t)
E3.18
Determine the transfer function of the system, whose unit step responses are given below.
b) s(t) = [t 2 + 2 e -5t sin 3t] u(t )
a) s(t) = [t + cosh 2t] u(t)
c) s(t) = [3t 2 - 4e -7 t + 4e - t ] u(t)
E3.19
Find the impulse response of the continuous time systems governed by the following transfer functions.
1
4
3
a) H(s) = 2
b) H(s) = 2
c) H(s) = 2
s(s - 16)
s (s + 2 )
s + 18s + 90
E3.20
Find the response of LTI systems whose input and impulse responses are given below.
a) x(t) = 0.5e -2t u(t)
c) x(t) = 3e
E3.21
E3.22
-2 t
cos4t u(t)
;
;
b) x(t) = t e - t u(t)
h(t) = u(t)
h(t) = e
-6 t
;
h(t) = d (t - 2 )
u(t)
Perform deconvolution operation to extract the signal x1(t).
a) x1 (t) * x2 (t) = (1 + t ) u(t)
; x2 (t) = d (t)
b) x1 (t) * x 2 ( t ) = ( t e - t + 3t 2 ) u(t)
;
x 2 (t) = 2e-3t u( t )
c) x1 (t) * x 2 (t) = t 2 u(t)
;
x2 (t) = 2 cosh t u(t)
Find the unit step response of the systems whose impulse responses are given below.
c) h(t) = t e -2 t + e -4t u(t)
a) h(t) = (t + sin t) u(t)
b) h(t) = d (t - 3) + u(t - 3)
d
i
3. 113
E3.23
Signals & Systems
Draw the direct form-I and direct form-II structure of the continuous time systems respresented by the
following equations.
d 2 y(t)
dy( t )
d 2 x( t )
dx( t )
+5
+ 2 y(t) = 3
+4
+ 6 x( t )
2
2
dt
dt
dt
dt
d 3 y(t)
d 2 y( t )
dy(t)
d 2 x(t)
dx(t)
b)
+ 0.9
+ 12
.
+ 2.1 y(t) = 31
.
+ 0.6
+ 0.9 x( t )
3
2
dt
dt
dt
dt 2
dt
a)
E3.24
Draw the direct form-I and direct form-II structure of the continuous time systems governed by the
following transfer functions.
4s2 + 2s - 0.9
2s3 + 3s2 + 7s + 10
a) H(s) = 3
b) H(s) = 3
2
s + 2s - 13s + 0.2
s - 9s2 + 12s + 13
E3.25
Draw the cascade and parallel structure for the LTI system governed by the following transfer functions.
(s + 2) (s - 5)
s2 + 2s
a) H(s) =
b) H(s) =
2
(s + 4) (s + s + 3)
(s + 6) (s2 - 7s + 10)
Answers
4 2
(s + 0.5s + 0.5)
s3
7
c) X(s) =
,
(s + 5) 2 + 49
E3.1 a) X(s) =
ROC : s > 0
,
ROC : s > -5
E3.3 a) X(s) =
c) X(s) =
LM
N
sT
2 sT
A
1 + e 3 - e 3 - e -sT
s
OP
Q
A - st 0
e (1 - e - s )
s
F
GG
H
-
sT
d) X(s) =
eb
s+ a
,
ROC s > -a
b) X(s) =
2
2A 1 - e 2
E3.4 X (s) = 2
sT
Ts
1+ e 2
s2 + 72
, ROC : s > 0
s(s2 + 144)
6 e -2s
s4
4
s + 8s3 + 47s2 + 168s + 250
d) X(s) =
(s + 3)2 (s2 + 25)
4s - 7
s + 2s + 29
2s2
c) X(s) =
(s + 1) 3
E3.2 a) X(s) =
b) X(s) =
b) X(s) =
A
1- e - ( s+ a ) T
(s + a)
d) X(s) =
2A
(cosh sT - 1)
Ts2
I
JJ
K
E3.5 a) x(0) = 10
b) x(0) = 0.25
c) x(0) = 0
d) x(0) = 4
E3.6 a) x(¥) = 1.25
b) x(¥) = 0
c) x(¥) = ¥
d) x(¥) = 0
E3.7 a) x1 ( t ) * x2 ( t ) = (0.6 te -2t + 0.12e -2t - 0.12 cos4t - 0.09 sin4t) u(t)
b) x1 ( t ) * x2 ( t ) =
LM
N
OP
Q
1
1 1
( t - 2 ) - + e -3 ( t - 2 ) u ( t - 2 )
3
3 3
c) x1 (t) * x 2 (t) = 0.16 1 - 5 t e -5t - e -5t u(t)
b g
e) x1 ( t ) * x 2 ( t ) = 0.5 ( t + 1)e - t - cost u(t)
d) x1 ( t ) * x2 ( t ) = t - 7
2
u(t - 7)
E3.8 p 1 = –3, p 2 = –1+j1, p 3 = –1–j
z 1 = –2, z 2 = –4 + j2, z3 = –4 – j2
Chapter 3 - Laplace Transform
E3.9 a) x(t) = [1.5e
-3t
3. 114
-t
- 0.5e ] u(t)
b) x(t) = [0.1 + 0.5e
-2 t
jW
- 0.6 e ] u(t)
b
z2
g
c) x(t) = e - t + e -0.5 t 0.577 sin 0.866t - cos0.866t u(t)
-t
(or) x(t) = e + 1.155e
-0.5t
p1
sin (0.866t - 60 ) u(t)
o
-4
d) x(t) = e -2 t cos t u(t)
E3.10
a) x(t) =
FG 1 - 1 e
H3 5
s-plane
+j2
-5 t
z1
´
-3
-2
p2
´
+j1
-1
p3
0
-j1
´
-2t
+
IJ
K
13 3 t
e u(t)
15
s
-j2
z3
Fig 3.8: Pole-zero plot for exercise
problem E3.8.
b) x(t) = [3e - t + e -2 t cos4t] u(t)
c) x(t) = 2[e -2 t - e -3 t sin 4t] u(t)
d) x(t) = [e -2 t sin 3t + e - t cos2t] u(t)
E3.11 x(t) =
E3.12
1
a2
FG sinhat - tIJ u(t)
H a K
-3 t
e - t - 0.8333 e -0.4 t u(t)
.
E3.15 a) y(t) = 0.25 e + 15833
d
b) y(t) = d0.5 - 0.8 e
a) h(t) = u (t - t 0 )
b) h(t) =
0.5
sin
3
FG 1 tIJ u(t)
H 3K
i
-8 t
+ 2.3 e -3t u(t)
i
c) h(t) = [4.7e-0.3 t - 4.5e-0.5t ] u(t)
E3.13
a) y zi (t) = 0.1778e -0.3t + 01222
.
e -1.2t u(t)
b) y zi (t) =
FG -3 e
H4
-7 t
+
Y(s)
0.3s + e - s
= 3
X(s) s + 0.4s2 + 0.3s + 0.1
Y(s)
3s2 + 2s + 4
b)
= 3
X(s) s + 7s2 + 5s + 14
E3.16 a)
IJ
K
11 -3t
e
u(t)
4
c)
E3.14
a) y zs (t) = 0.6 - 18e
. -4 t + 1.2 e -5t u(t)
b) y zs
d
(t) = d0.3e
i
-t
-e
-3 t
+ 15
. e
-5 t
- 0.8 e -6 t u(t)
i
-2 s
E3.17
E3.19
e (s + 4 ) + 1
(s + 4)
s3 + 16s2 + 70s + 99
b) H(s) = 3
s + 15s2 + 63s + 81
5s3 + 90s2 + 481s + 636
c) H(s) = 3
s(s + 20s2 + 137s + 318)
a) H(s) =
E3.18
1
2 t - 1 + e - 2 t u(t)
2
1
b) h(t) = [cosh 4t - 1] u(t)
4
E3.20
a) h(t) =
b) x1 (t) = ( t + 0.5) e + 3t + 4.5t
F tI
c) x (t) = G t - J u(t)
H 6K
24s3 + 6s2 + 48s + 42
s4 + 8s3 + 7s2
a) y(t) = 0.25 1 - e -2t u(t)
b) y(t) = ( t - 2) e -( t - 2 ) u(t - 2)
c) y(t) = 0.375 e -2 t ( sin 4t + cos4t) - e-6 t u(t)
a) x 1 (t) = (t + 1) u(t)
-t
2s2 - 4
s2 - 4
7s2 + 10s + 34
b) H(s) = 2
s(s +10s + 34)
a) H(s) =
c) H(s) =
c) h(t) = e -9t sin 3t u(t)
E3.21
Y(s)
e -3s
= 2
X(s) 4s + 2s + 8
E3.22
2
u(t)
a) s(t) = [1 + t - cos t ] u(t)
b) s(t) = u(t - 3) + (t - 3) u(t - 3)
3
1
c) s(t) = 0.25 2 - 2 t e -2 t - e -2t - e -4 t u(t)
3. 115
Signals & Systems
E3.23
a)
X(s)
3 X(s)
+
3
X(s)
s
1
s
X(s)
s
4
1
s
6
6
+
X(s)
s2
X(s)
Y(s)
–5
+
4
X(s)
s2
+
–2
Y(s)
s
–5
1
s
Y(s)
s2
–2
1
s
Y(s)
s
1
s
–2 W(s)
s2
1
s
–2
Fig : Direct form-I structure.
E3.23
+
1
s
X(s)
s
3.1
X(s)
s
+
3.1
1
s
X(s)
s2
1
s
0.9
0.9
X(s)
s3
Y(s)
s
–0.9
1
s
Y(s)
s2
–1.2
1
s
–2.1
Y(s)
s3
–2.1
1
s Y(s)
s3
+
–1.2
Y(s)
s2
+
–2.1
+
X(s)
4
s
1
s
X(s)
s
4
X(s)
2
s2
1
s
X(s)
s2
2
1
s
– 0.9
X(s)
s3
-0.9
Y(s)
+
Y(s)
–2
s
+
–2
+
13 Y(s)
s2
+
13
X(s)
s3
–0.2
b)
X(s)
X(s)
s
X(s)
s2
X(s)
s3
W(s)
s2
–1.2
W(s)
1 W(s)
0.6 2
s
s2
s
0.6
+
W(s)
s3
–2.1
W(s)
1 W(s )
0.9 3
s
s3
s
0.9
W(s)
Y(s)
+
1
s
13
Y (s )
s2
+
1
s
+
1
s
3
3
1
s
7
1
s
–0.2
Y(s)
s3
+
+
X(s)
s3
+
X(s)
W(s )
s3
Y(s)
s
9
1
s
Y(s)
s3
–13
1
s
Fig : Direct form-I structure.
+
1
W(s)
s W(s)
–0.9 3
s 3 –0.9 s
Y(s)
s2
Y(s)
s3
2 W(s)
W(s)
+
1
s
Y(s)
–12
s2
+
–12
–13
4
W(s)
s
Fig : Direct form-II structure.
Y(s)
Y(s)
s
4
W(s)
1 W(s)
2 2
s
2
s
s
2
+
–0.2
+
9
X(s)
7 2
s
10
10
X(s)
s
s
W(s)
s2
13
2 X(s)
2
1
s W(s)
W(s)
s
–2
–2
Y(s)
s
Fig : Direct form-I structure.
E3.24
1 W(s)
W(s)
s
3.1
s
s
3.1
+
+
X(s)
1
s
Y(s)
s3
–0.2
W(s)
s2
W( s )
s
–0.9
a)
X(s)
6
6
+
Fig : Direct form-II structure.
Fig : Direct form-I structure.
E3.24
W(s )
s2
W(s)
s
Y(s)
– 0.9
Y(s)
s
–1.2
+
+
0.6
X(s)
s3
+
X(s)
0.6 2
s
4
4
W(s)
+
X(s)
Y(s)
–0.9
W(s)
s
Y(s)
Fig : Direct form-II structure.
b)
X(s)
+
3
W(s)
–5
s
+
–5
Y(s)
s2
3W(s)
W(s)
+
2
9
+
–12
+
–13
W(s)
s
9
1 W(s)
W(s)
s
3
s
s
+
3
W(s)
s2
–12
W(s)
1 W(s )
s
7 2
s2
s
7
+
W(s)
s3
–13
1 W(s)
W(s)
s
10 3
s3
s
10
Y(s)
Fig : Direct form-II structure.
Chapter 3 - Laplace Transform
E3.25
3. 116
a)
X(s)
X(s)
W1(s)
+
+
Y1(s)
W2(s)
+
Y(s)
+
1
s
+
2
–4
–1
+
Y(s)
X(s)
1
s
–3
6/5
+
1
H1(s)
Y1(s)
1
s
1
s
–4
W1(s)
Y2(s)
W2(s)
+
1
s
–5
H2(s)
Fig : Cascade structure.
–1
+
A
+
1 A = –1/5
s B = –17/5
+
–3
B
Fig : Parallel structure.
b)
W1(s)
X(s)
+
W2(s)
Y1(s)
+
+
1
s
W3(s)
Y2(s)
1
s
1
s
–6
-6
5
2
2
H2(s)
H1(s)
Y(s)
+
1
H3(s)
Fig : Cascade structure .
X(s)
Y1(s)
W1(s)
+
+
1
s
A = 3/11
–6
X(s)
A
W2(s)
+
H1(s)
Y2(s)
1
s
B = 35/33
–5
X(s)
B
Y3(s)
W3(s)
+
H2(s)
1
s
C = –1/3
2
C
Fig : Parallel structure.
H3(s)
+
Y(s)
CHAPTER 4
Fourier Series and Fourier Transform of
Continuous Time Signals
4.1 Introduction
The French mathematician Jean Baptiste Joseph Fourier (J.B.J. Fourier) has shown that any periodic
non-sinusoidal signal can be expressed as a linear weighted sum of harmonically related sinusoidal signals.
This leads to a method called Fourier series in which a periodic signal is represented as a function of
frequency.
The Fourier representation of periodic signals has been extended to non-periodic signals by letting
the fundamental period T tend to infinity, and this Fourier method of representing non-periodic signals as
a function of frequency is called Fourier transform. The Fourier represention of signals is also known as
frequency domain representation. In general, the Fourier series representation can be obtained only for
periodic signals, but the Fourier transform technique can be applied to both periodic and non-periodic
signals to obtain the frequency domain representation of the signals.
The Fourier representation of signals can be used to perform frequency domain analysis of signals,
in which we can study the various frequency components present in the signal, magnitude and phase of
various frequency components. The graphical plots of magnitude and phase as a function of frequency
are also drawn. The plot of magnitude versus frequency is called magnitude spectrum and the plot of
phase versus frequency is called phase spectrum. In general, these plots are called frequency spectrum.
4.2 Trigonometric Form of Fourier Series
4.2.1 Definition of Trigonometric Form of Fourier Series
The trigonometric form of Fourier series of a periodic signal, x(t), with period T is defined as,
x(t) =
1
a +
2 0
\ x( t ) =
¥
åa
¥
n
cos nW 0 t +
n=1
åb
n
sin nW 0 t
n=1
1
a + a 1 cos W 0 t + a 2 cos2W 0 t + a 3 cos3W 0 t + .....
2 0
+ b1 sin W 0 t + b 2 sin 2W 0 t + b3 sin 3W 0 t + ...........
2p
= Fundamental frequency in rad/sec
T
F0 = Fundamental frequency in cycles/sec or Hz
where, W0 = 2pF0 =
nW0 = Harmonic frequencies
a0, an, bn = Fourier coefficients of trigonometric form of Fourier series
.....(4.1)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 2
Note : 1. Here a0 2 is the value of constant component of the signal x(t).
2. The Fourier coefficient an and bn are maximum amplitudes of nth harmonic component.
The Fourier coefficients can be evaluated using the following formulae.
a0 =
an =
2
T
2
T
2
T
bn =
+ T /2
z
z
z
(or)
x( t ) dt
2
T
a0 =
- T /2
+T/2
x( t ) cosnW 0 t dt
(or)
2
T
an =
-T/2
+T/2
x( t ) sin nW 0 t dt
(or)
2
T
bn =
-T/2
T
z
z
z
x( t ) dt
.....(4.2)
x( t ) cos nW 0 t dt
.....(4.3)
x( t ) sin nW 0 t dt
.....(4.4)
0
T
0
T
0
In the above formulae, the limits of integration are either –T/2 to +T/2 or 0 to T. In general,
the limit of integration is one period of the signal and so the limits can be from t0 to t0 + T, where t0 is
any time instant.
4.2.2 Conditions for Existence of Fourier Series
The Fourier series exists only if the following Dirichlet’s conditions are satisfied.
1. The signal x(t) is well defined and single valued, except possibly at a finite number of points.
2. The signal x(t) must possess only a finite number of discontinuities in the period T.
3. The signal must have a finite number of positive and negative maxima in the period T.
Note : 1. The value of signal x(t) at t = t0 is x(t0 ) if t = t0 is a point of continuity.
2. The value of signal x(t) at t = t0 is
x(t0+ ) + x(t0- )
if t = t0 is a point of discontinuity.
2
4.2.3 Derivation of Equations for a0, an and bn
Evaluation of a0
The Fourier coefficient a0 is given by,
a0 =
2
T
+ T/ 2
z
x( t ) dt
(or)
a0 =
- T/ 2
2
T
T
z
x( t ) dt
0
Proof :
Consider the trigonometric Fourier series of x(t), (equation (4.1)).
¥
¥
a
x( t) = 0 +
an cosnW 0 t +
bn sinnW 0t
2
n= 1
n= 1
å
å
Let us integrate the above equation between limits 0 to T.
T
\
z
T
x(t ) dt =
0
z
0
=
a0
2
a0
dt +
2
T
z
0
T
an cosnW 0t dt +
åa
T
n=1
n
¥
zå
bn sinnW 0t dt
0 n=1
0 n= 1
¥
dt +
T
¥
zå
z
0
T
¥
cosnW 0t dt +
åb
n=1
n
z
0
sinnW 0t dt
4. 3
Signals & Systems
T
\
z
a0
t
2
x(t ) dt =
0
=
T
a0 +
2
=
2
T
T
z
0
n
0
0
n= 1
0
¥
a0
T - 0 +
2
0
¥
0
0
n
n
0
n= 1
¥
0
+
0
n=1
¥
åa
n
n=1
n
n=1
¥
å
T
¥
0
an
n=1
T
a0 +
2
=
\ a0 =
0
T
a0 +
2
=
å
+
LM sinnW t OP + b LM -cosnW t OP
N nW Q å N nW Q
L sinnW T sin0 O
L -cosnW T
å a MN nW - nW PQ + å b MN nW
LM sinn 2 p T OP
L -cosn 2p T
O
MM 2pT PP + å b MMM 2 pT + 21 p PPP
n
MN n T PQ
MN n T
T PQ
L sinn2 p OP + å b T LM -cosn2p + 1 OP
TM
n2 p Q
N n2p Q
N n2 p
L 1 + 1 OP = T a
T ´ 0 + å b T MN n2 p n2 p Q 2
T
¥
T
¥
an
n= 1
¥
åa
n
n=1
¥
n
n= 1
n
0
cos0
nW 0
OP
Q
2p
T
sin 0 = 0
W0 =
cos 0 = 1
sin n2p = 0 ;
cos n2p = 1
for integer n
n=1
x(t ) dt
0
Evaluation of an
The Fourier coefficient an is given by,
an
2
=
T
+ T/ 2
z
x( t ) cos nW 0 t dt
2
=
T
(or) a n
- T/ 2
T
z
x( t ) cosnW 0 t dt
0
Proof :
Consider the trigonometric form of Fourier series of x(t), (equation (4.1)).
¥
¥
a
x( t) = 0 +
an cosnW 0t +
b n sinnW 0 t
2
n= 1
n= 1
å
=
å
a0
+ a1 cos W 0 t + a2 cos 2W 0t + ..... + ak cos kW 0 t + .....
2
+ b 1 sin W 0 t + b 2 sin2 W 0 t + ..... + bk sinkW 0t + .....
Let us multiply the above equation by cos kW0t
a
\ x( t) coskW 0 t = 0 coskW 0 t + a1 cos W 0t coskW 0 t + a2 cos2W 0 t cos kW 0 t + .....
2
..... + ak cos2 kW 0 t + ..... + b 1 sin W 0t cos k W 0 t + b 2 sin2W 0t cos kW 0 t + .....
..... + bk sinkW 0 t cos k W 0 t + .....
Let us integrate the above equation between limits 0 to T.
T
\
z
T
x(t) coskW 0 t dt =
0
z
0
a0
coskW 0 t dt +
2
T
z
a1 cos W 0 t cosk W 0t dt
0
T
+
z
z
T
a2 cos 2 W 0 t coskW 0t dt + ..... +
0
0
T
+
0
z
z
T
ak cos2 kW 0t dt + ..... +
z
b 1 sin W 0t coskW 0t dt
0
T
b 2 sin2W 0t cosk W 0 t dt + ..... +
0
bk sinkW 0t coskW 0 t dt + .....
.....(4.5)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
T
\
z
T
z
x(t) coskW 0t dt =
0
T
ak cos2 kW 0 t dt = ak
0
z
T
z
0
LM
N
2
T
\ ak =
T
LMt + sin2kW t OP
N 2kW Q
L
sin0 O
ak M
P = 2 MMT +
2kW 0 Q
MN
except
T
a
(1 + cos2kW 0t) dt = k
2
ak
sin2kW 0 T
T +
- 0 2
2kW 0
=
In equation (4.5) all definite
integrals will be zero,
1 + cos2kW 0 t
dt
2
0
a
= k
2
4. 4
ak cos 2 kW 0t dt.
0
0
0
z
0
2p
sin2k
T
T
2p
2k
T
OP
PP =
PQ
2p
T
sin 0 = 0
W0 =
T
ak
2
sin 2k2p = 0
for integer k
T
z
x(t ) cos kW 0 t dt
.....(4.6)
0
th
th
The equation (4.6) gives the k coefficient ak. Hence the n coefficient an is given by,
2
T
an =
T
z
x(t) cosnW 0t dt
0
Evaluation of bn
The Fourier coefficient bn is given by,
bn =
2
T
+ T/ 2
z
x( t ) sin nW 0 t dt (or) b n =
- T/ 2
T
z
2
T
x( t ) sin nW 0 t dt
0
Proof :
Consider the trigonometric form of Fourier series of x(t) (equation (4.1)).
x( t) =
=
a0
+
2
¥
¥
åa
n cos nW 0 t
n= 1
+
åb
n
sin nW 0t
n= 1
a0
+ a1 cos W 0t + a2 cos 2W 0t + ..... + ak cos kW 0 t + .....
2
+ b 1 sin W 0t + b 2 sin 2W 0 t + ..... + bk sin kW 0 t + .....
Let us multiply the above equation by sin kWot
\ x(t ) sin kW 0 t =
a0
sin kW 0 t + a1 cos W 0t sin kW 0 t + a2 cos 2 W 0t sin kW 0t + .....
2
..... + ak cos kW 0t sin kW 0 t + ..... + b 1 sin W 0 t sin kW 0t
+ b2 sin 2W 0 t sin kW 0 t + ..... + bk sin2 kW 0 t + .....
Let us integrate the above equation between limits 0 to T.
T
\
z
T
x(t) sinkW 0t dt =
z
0
0
a0
sinkW 0 t dt +
2
T
z
a1 cos W 0t sinkW 0 t dt
0
T
+
z
T
a2 cos 2W 0 t sinkW 0t dt + ..... +
0
z
z
ak coskW 0 t sink W 0 t dt + .....
0
T
..... +
z
T
b 1 sin W 0 t sinkW 0 t dt +
0
b 2 sin2 W 0 t sinkW 0t dt + .....
0
T
+
z
0
bk sin2 kW0 t dt + .....
.....(4.7)
4. 5
Signals & Systems
T
\
z
T
x(t) sinkW 0 t dt =
z
T
b k sin2 kW 0t dt = bk
0
0
0
T
b
= k
2
=
bk
2
=
bk
2
\ bk =
2
T
z
z
0
sin2kW 0 T
- 0 +
2kW 0
2p
T
T
2p
2k
T
sin2k
OP
PP =
PQ
T
LMt - sin2kW t OP
N 2kW Q
sin0 O
P
2kW 0 Q
b
(1 - cos2kW0 t) dt = k
2
LMT N
LM
MMT MN
In equation (4.7) all definite
integrals will be zero,
1 - cos 2kW 0t
dt
2
T
except
0
0
z
b k sin2 kW 0 t dt.
0
0
T
bk
2
2p
T
sin 0 = 0
W0 =
sin 2k2p = 0
for integer k
T
z
x(t) sinkW 0 t dt
.....(4.8)
0
th
th
The equation (4.8) gives k coefficient b k. Hence the n coefficient bn is given by,
2
T
bn =
T
z
x(t) sinnW 0 t dt
0
4.3 Exponential Form of Fourier Series
4.3.1 Definition of Exponential Form of Fourier Series
The exponential form of Fourier series of a periodic signal x(t) with period T is defined as,
+¥
åc
x( t ) =
n
e jnW o t
.....(4.9)
n = -¥
2p
= Fundamental frequency in rad/sec
T
F0 = Fundamental frequency in cycles/sec or Hz
where, W0 = 2pF0 =
± nW = Harmonic frequencies
0
cn = Fourier coefficients of exponential form of Fourier series.
The Fourier coefficient cn can be evaluated using the following equation.
cn =
1
T
+T/2
z
x(t) e - jnW 0t dt
- T/ 2
(or)
cn =
1
T
T
z
x( t ) e - jnW ot dt
.....(4.10)
0
In equation (4.10), the limits of integration are either –T/2 to +T/2 or 0 to T. In general, the limit
of integration is one period of the signal and so the limits can be from t0 to t0 + T, where t0 is any time
instant.
4.3.2 Negative Frequency
The exponential form of Fourier series representation of a signal x(t) has complex exponential
harmonic components for both positive and negative frequencies. When the positive and negative complex
exponential components of same harmonic are added, it gives rise to real sine or cosine signals.
Alternatively, when the real sine or cosine signal has to be represented in terms of complex exponential
then a signal with negative frequency is required.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 6
Here it should be understood that the signal with negative frequency is not a physically realizable
signal, but it is required for mathematical representation of real signals in terms of complex exponential
signals.
4.3.3 Derivation of Equation for cn
The Fourier coefficient cn is given by,
1
T
cn =
+T/2
z
x(t) e - jnW 0t dt
(or)
cn =
- T/ 2
1
T
T
z
x( t ) e - jnW ot dt
0
Proof :
Consider the exponential form of Fourier series of x(t), (equation (4.9)).
+¥
åc
x( t) =
n
ejnW 0 t = ..... + c- k e - jkW 0 t + . .... + c -2 e - j2W 0 t + c -1 e - jW 0 t
n = -¥
+ c0 + c1 e jW 0t + c2 ej2W 0 t + ..... + ck ejkW 0t + .....
Let us multiply the above equation by e -jkW 0 t .
\ x(t ) e -jkW 0 t = ..... + c-k e -j2kW 0 t + ..... + c-2 e -j2W 0 t e -jkW 0 t + c-1 e -jW 0 t e - jkW 0 t
+ c0 e - jkW 0 t + c1 ejW 0t e -jkW 0 t + c2 ej2W 0t e -jkW 0 t + ..... + ck + .....
Let us integrate the above equation between limits 0 to T.
T
\
z
T
x(t ) e -jkW 0 t dt = ..... +
0
z
z
z
T
c - k e -j2kW 0 t dt + ..... +
0
T
z
c -1 e -jW 0 t e -jkW 0t dt +
0
T
c 0 e - jkW 0 t dt +
0
T
+
c -2 e -j2W 0 t e - jkW 0 t dt
0
T
+
z
0
c 2 ej2W 0 t e -jkW 0 t dt +
..... +
z
0
T
z
c k dt = ck t
c 1 ejW 0 t e - jkW 0 t dt
T
0
=
z
T
0
= ck T - 0 = T ck
ck dt + .....
.....(4.11)
In equation (4.11) all definite
integrals will be zero,
0
1
\ ck =
T
T
z
T
except
z
ck dt
0
x(t ) e -jkW 0 t dt
.....(4.12)
0
The equation (4.12) gives the K th coefficient, ck. Hence the nth coefficient, cn is given by,
cn =
1
T
T
z
x( t) e - jnW 0 t dt
0
4.3.4 Relation Between Fourier Coefficients of Trigonometric and Exponential Form
The relation between Fourier coefficients of trigonometric form and exponential form are given
below.
a0
2
1
c n = (a n - jb n )
2
c0 =
.....(4.13)
for n = 1, 2, 3, 4, ......
.....(4.14)
4. 7
Signals & Systems
1
(a + jb n )
2 n
c- n =
1
2
\ cn =
for –n = -1, -2, -3, -4, ......
.....(4.15)
a 2n + b2n for all values of n, except when n = 0.
.....(4.16)
Proof :
Consider the trigonometric form of Fourier series of x(t), (equation (4.1)).
¥
¥
a
x( t) = 0 +
an cosnW 0t +
b n sinnW 0t
2
n=1
n=1
å
a0
+
2
=
LM e
MN
¥
åa
n
n=1
La
å MN 2
La
+ å M
N2
a
= 0 +
2
jnW 0 t
n
e
n
- j
n=1
n=1
LM e
MN
jnW 0 t
- e -jnW 0 t
2j
OP
PQ
¥
c*n
;
å
n=1
=
LM a
N2
n
+ j
OP
Q
ejnW 0 t +
åc
e jq + e - jq
2
sinq =
e jq - e - jq
2j
1
j
= 2 = -j
j
j
OP
Q
b n -jnW 0 t
e
2
an + jb n
2
*
n
e -jnW 0 t
n=1
-1
¥
åc
n
ejnW 0 t +
n=1
åc
-n
e jnW 0 t
C - n = C*n
n = -¥
-1
¥
åc
-n
ejnW 0 t + c 0 +
åc
n
e jnW 0 t
n=1
n = -¥
+¥
åc
=
cosq =
¥
n
n=1
=
n
OP
Q
¥
åc
= c0 +
¥
b n jnW 0 t
e
+
2
a - jbn
; cn = n
2
\ x(t) = c0 +
OP + b
PQ å
a
b
b
+ n e -jnW 0t - j n e jnW 0 t + j n e -jnW 0 t
2
2
2
jnW 0 t
n=1
a
Let, c 0 = 0
2
+ e -jnW 0 t
2
¥
¥
a0
2
=
å
n
e
e jnW 0 t = 1
for n = 0
jnW 0 t
n = -¥
a0
2
\ c0 =
1
(an - jbn )
2
cn =
c -n =
1
(an + jbn )
2
for
n = 1, 2, 3, ....., ¥
for - n = - 1, - 2, - 3, ....., - ¥
4.3.5 Frequency Spectrum( or Line Spectrum) of Periodic Continuous Time Signals
Let x(t) be a periodic continuous time signal. Now, exponential form of Fourier series of x(t) is,
+¥
x( t ) =
åc
n
e jnW 0t
n = -¥
where, cn is the Fourier coefficient of nth harmonic component.
The Fourier coefficient, cn is a complex quantity and so it can be expressed in the polar form as
shown below.
c n = |c n | Ðc n
where, |cn| = Magnitude of cn ;
Ðc = Phase of c
n
n
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 8
The term, |cn| represents the magnitude of nth harmonic component and the term Ðcn represents
the phase of the nth harmonic component.
The plot of harmonic magnitude / phase of a signal versus "n" (or harmonic frequency nW0) is
called Frequency spectrum (or Line spectrum). The plot of harmonic magnitude versus "n" (or nW0) is
called magnitude (line) spectrum and the plot of harmonic phase versus "n" (or nW0) is called phase
(line) spectrum.
Consider the ramp waveform shown in fig 4.1. The Fourier coefficient cn for this ramp waveform
is given by,
x(t)
A
jA
A
c0 =
,
cn =
2
2np
0
T
2T
3T t
(Please refer example 4.12 for derivation of cn)
j20
j10
Fig 4.1 : Ramp waveform.
Let, A = 20,
=
\ cn =
2np
np
.
.
.
10
= - j1.061 = 1.061 Ð - 90o = 1061
When n = - 3, c -3 = - j
.
Ð-p 2
3p
10
= - j1.592 = 1.592 Ð - 90o = 1592
When n = - 2, c -2 = - j
.
Ð-p 2
2p
10
= - j3.183 = 3.183 Ð - 90o = 3183
When n = - 1, c -1 = - j
.
Ð-p 2
p
20
0, c 0 =
10
=
= 10 Ð 0
When n =
2
10
=
j3.183 = 3.183 Ð + 90o = 3183
When n = 1, c1 = j
.
Ðp 2
p
10
When n =
.
Ðp 2
2, c2 = j
=
j1.592 = 1.592 Ð + 90o = 1592
2p
10
When n = 3, c 3 = j
.
Ðp 2
=
j1.061 = 1.061 Ð + 90o = 1061
3p
.
.
.
Using the above calculated values the magnitude and phase spectrums are sketched as shown in
fig 4.2 and fig 4.3 respectively.
Ðc n
|cn|
+p
2
10
.
.
.
.
.
8
6
–3
–2
–1
0
4
1
2
3
n
–3
.
.
.
.
.
.
.
.
.
2
–2
–1
0
1
2
3
n
Fig 4.2 : Magnitude spectrum of ramp waveform.
-p
2
Fig 4.3 : Phase spectrum of ramp waveform.
4. 9
Signals & Systems
4.4 Fourier Coefficients of Signals With Symmetry
4.4.1 Even Symmetry
A signal, x(t) is called even signal, if the signal satisfies the condition x(-t) = x(t).
The waveform of an even periodic signal exhibits symmetry with respect to t = 0 (i.e., with respect
to vertical axis) and so the symmetry of a waveform with respect to t = 0 or vertical axis is called even
symmetry.
Examples of even signals are,
x(t) = 1 + t2 + t4 + t 6
x(t) = A cos W0t
In order to determine the even symmetry of a waveform, fold the waveform with respect to
vertical axis. After folding, if the waveshape remains same then it is said to have even symmetry.
For even signals the Fourier coefficient a0 is optional, an exists and bn are zero. The Fourier
coefficient a0 is zero if the average value of one period is equal to zero. For an even signal the Fourier
coefficients are given by,
a0 =
an =
4
T
4
T
T /2
z
z
x(t) dt
(or)
0
T/ 2
x(t) cos nW 0 t dt (or)
+ T/ 4
z
z
4
T
a0 =
+ T/ 4
4
T
an =
0
x(t) dt
- T/ 4
x(t) cos nW 0 t dt ;
bn = 0
- T/ 4
Proof :
Consider the equation for a0, (equation (4.2)).
T 2
a0 =
z
z
z
z
0
z
2
2
2
x( t) dt =
x( t) dt +
T -T 2
T -T 2
T
=
z
2
2
x( - t) ( -dt) +
TT2
T
2
T
2
=
T
T 2
x( - t) dt +
0
T 2
2
x( t) dt +
T
0
2
T
Change of integral index,
Let, t = -t ; \dt = –dt
When t = 0, t = –t = 0
When t = –T/2, t = –t= –(–T/2) = T/2
0
x(t ) dt
0
T 2
z
z
x( t) dt
T 2
0
=
Dividing the integral into two parts.
T 2
2
T
x(t ) dt =
0
T 2
z
4
x( t) dt =
T
0
T 2
z
x( - t) dt +
0
2
T
Since t is dummy variable, Let t = t.
T 2
z
x(t ) dt
Since x(t) is even, x(–t) = x(t)
0
T 2
z
x( t) dt
0
Consider the equation for an, (equation (4.3)).
T 2
an =
z
z
z
z
z
0
z
2
2
2
x(t ) cosn W 0 t dt =
x(t ) cosn W 0 t dt +
T -T 2
T -T 2
T
0
=
=
2
2
x( - t) cosn W 0 ( - t ) ( - dt ) +
TT2
T
2
T
2
=
T
=
2
T
T 2
x( - t) cosn W 0 t dt +
0
T 2
2
x( -t ) cosn W 0 t dt +
T
0
T 2
x( t) cosn W 0t dt +
0
2
T
2
T
z
z
x(t ) cosn W 0 t dt
0
x(t ) cosn W 0 t dt
x( t) cosn W 0t dt
0
T 2
T 2
z
T 2
Dividing the integral
into two parts.
Change of integral index,
Let, t = –t ; \dt = –dt
When t = 0, t = –t = 0
When t = –T/2, t = –t= –(–T/2) = T/2
cos(–q) = cosq
0
T 2
z
T 2
z
0
Since t is dummy variable, Let t = t.
x(t ) cosn W 0 t dt
0
x(t ) cosn W 0 t dt =
4
T
T 2
z
0
x(t ) cosn W 0 t dt
Since x(t) is even, x(–t) = x(t)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 10
Consider the equation for bn, (equation (4.4)).
T 2
bn =
0
z
z
z
2
2
2
x( t) sinn W 0t dt =
x(t ) sinn W 0t dt +
T -T 2
T -T 2
T
0
2
2
x( - t ) sinn W 0 ( - t ) ( -dt) +
=
TT2
T
=-
=-
=-
2
T
2
T
2
T
T 2
z
z
z
x( - t ) sinn W 0 t dt +
0
T 2
x( - t) sinn W 0t dt +
0
T 2
x(t ) sinn W 0t dt +
0
2
T
2
T
2
T
T 2
z
0
T 2
z
Change of integral index,
Let, t = -t ; \dt = –dt
When t = 0, t = –t = 0
When t = –T/2, t = –t= –(–T/2) = T/2
x(t ) sinn W 0t dt
0
T 2
z
Dividing the integral
into two parts.
x(t ) sinn W 0 t dt
x( t) sinn W 0t dt
0
sin(–q) = –sinq
T 2
z
x( t) sinn W 0t dt
Since t is dummy variable, Let t = t.
0
T 2
z
Since x(t) is even, x(–t) = x(t)
x( t) sinn W 0 t dt = 0
0
The waveform of some even periodic signals and their Fourier series are given below.
x(t)
The waveform shown in fig 4.4, has even symmetry, half wave
A
symmetry and quarter wave symmetry.Hence for this waveform,
a0 = 0, bn= 0 and an exists only for odd values of n. Therefore the
0
Fourier series consists of odd harmonics of cosine terms. The
-T
–A
2
trigonometric Fourier series representation of the waveform of fig 4.4
T
-T
4
4
is given by equation (4.17). [Please refer example 4.1 for the derivation
Fig
4.4.
of Fourier series]
cos 3W 0 t
cos 5W 0 t
cos 7W 0 t
cos 9W 0 t
4A cos W 0 t
+
- .....
x( t ) =
+
p
1
3
5
7
9
LM
N
A
2A
+
p
2
W0 =
OP
Q
The waveform shown in fig 4.5, has even symmetry and so
bn = 0. If the dc component (a0/2) is substracted from this waveform
then it will have half wave and quarter wave symmetry, and so the
Fourier series has odd harmonics of cosine terms. The trigonometric
Fourier series representation of the waveform of fig 4.5 is given by
equation (4.18). [Please refer example 4.3 for the derivation of Fourier
series]
x( t ) =
t
T
2
LM cos W t
N 1
0
-
.....(4.17)
x(t)
A
-T-T
2 4
0
T
4
t
T
2
Fig 4.5.
W0 =
OP
Q
x(t)
A
T
-T -4
2
0
T
4
t
T
2
Fig 4.6.
2A
4A
+
p
p
LM cos2W t cos4W t cos6W t cos8W t
MN d2 - 1i - d4 - 1i + d6 - 1i - d8 - 1i
0
2
2p
T
cos 3W 0 t
cos 5W 0 t
cos 7W 0 t
cos 9W 0 t
+
+
- ..... .....(4.18)
3
5
7
9
The waveform shown in fig 4.6 has even symmetry and
so bn= 0. The trigonometric Fourier series representation of the
waveform of fig 4.6 is given by equation (4.19).
x( t ) =
2p
T
0
2
0
2
0
2
OP
PQ
+ .....
W0 =
2p
T
.....(4.19)
4. 11
Signals & Systems
The waveform shown in fig 4.7 has even symmetry and
so bn = 0. The trigonometric Fourier series representation of the
waveform of fig 4.7 is given by equation (4.20). [Please refer
example 4.4 for the derivation of Fourier series].
x(t)
A
-T
0
-T
2
T
2
Fig 4.7.
x( t ) =
LM cos2W t cos4W t cos6W t cos8W t
MN d2 - 1i + d4 - 1i + d6 - 1i + d8 - 1i
2A
4A
p
p
0
2
0
2
0
2
The waveform shown in fig. 4.8 has even symmetry
and so bn = 0. If the dc component (a0/2) is subtracted from
this waveform then it will have half wave and quarter wave
symmetry, and so the Fourier series has odd harmonics of
cosine terms. The Fourier series representation of the
waveform of fig 4.8 is given by equation (4.21). [Please
refer example 4.2 for the derivation of Fourier series].
x( t ) =
4A
A
2
p2
LM cos W t
N 1
0
2
+
2p
T
.....(4.20)
+ .....
2
t
W0 =
OP
PQ
0
T
x(t)
A
–2T
-T
0
T
2
T
2T
Fig 4.8.
W0 =
OP
Q
cos3W 0 t
cos5W 0 t
cos7W 0 t
+
+
+ .....
2
2
3
5
72
The waveform shown in fig. 4.9 has even
x(t)
symmetry and so bn = 0. If the dc component
A
(a0/2) is subtracted from this waveform then it will
have half wave and quarter wave symmetry, and
0
-T
T
T
-T
so the Fourier series has odd harmonics of cosine
2
2
terms. The Fourier series representation of the
Fig 4.9.
waveform of fig 4.9 is given by equation (4.22).
[Please refer example 4.11 for the derivation of
Fourier series].
4 A cos W 0 t
A
cos3W 0 t
cos5W 0 t
cos7W 0 t
x( t ) =
+
+
+
+
+ .....
2
12
32
52
72
p2
LM
N
OP
Q
t
2p
T
.....(4.21)
t
W0 =
2p
T
.....(4.22)
4.4.2 Odd Symmetry
A signal, x(t) is called odd signal if it satisfies the condition x(-t) = -x(t).
The waveform of odd periodic signal will exhibit anti-symmetry with respect to t = 0 (i.e., with
respect to vertical axis) and so the anti-symmetry of a waveform with respect to t = 0 or vertical axis is
called odd symmetry.
Examples of odd signals are,
x(t) = t + t3 + t5 + t7
x(t) = A sin W0t
In order to determine the odd symmetry of a waveform, invert either the right side (or the left
side) of the waveform with respect to horizontal axis and then fold the waveform with respect to vertical
axis. After inverting one half and folding, if the waveshape remains same then it is said to have odd
symmetry.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 12
For odd signals a0 and an are zero and bn exists. For odd signal the Fourier coefficients are given by,
a0 = 0 ;
an = 0
T/ 2
z
4
T
bn =
x(t) sin nW 0 t dt
or
+ T/ 4
z
4
T
bn =
0
x(t) sin nW 0 t dt
- T/ 4
Proof :
Consider the equation for a 0, (equation (4.2)).
T 2
a0 =
z
z
z
z
z
0
=
T 2
0
z
2
2
2
x( t) dt =
x(t ) dt +
T -T 2
T -T 2
T
2
2
x( - t) ( - dt ) +
TT2
T
2
=
T
=-
T 2
2
x( - t) dt +
T
0
2
T
T 2
x( t) dt +
0
2
T
T 2
z
Change of integral index,
Let, t = -t ; \dt = –dt
When t = 0, t = –t = 0
When t = –T/2, t = –t= –(–T/2) = T/2
x( t) dt
0
T 2
z
Dividing the integral into two parts.
x(t ) dt
0
2
T
x( t) dt =
0
T 2
z
x( - t ) dt +
0
2
T
T 2
z
x(t ) dt
Since t is dummy variable, Let t = t.
0
T 2
z
Since x(t) is odd, x(–t) =– x(t)
x( t) dt = 0
0
Consider the equation for a n, (equation (4.3)).
T 2
an =
0
z
z
z
z
z
z
2
2
2
x(t ) cosn W 0 t dt =
x(t ) cosn W 0 t dt +
T -T 2
T -T 2
T
0
=
=
=
2
2
x( - t) cosn W 0 ( - t ) ( - dt ) +
TT2
T
2
T
2
T
T 2
x( - t) cosn W 0 t dt +
0
T 2
x( -t ) cosn W 0 t dt +
0
2
T
z
x( t) cosn W 0t dt
z
z
x(t ) cosn W 0 t dt
0
Change of integral index,
Let, t = -t ; \dt = –dt
When t = 0, t = –t = 0
When t = –T/2, t = –t= –(–T/2) = T/2
x(t ) cosn W 0 t dt
cos(–q) = cosq
0
T 2
z
z
Since t is dummy variable, Let t = t.
x(t ) cosn W 0 t dt
0
Since x(t) is odd, x(–t) =– x(t)
T 2
2
x( t) cosn W 0t dt +
T
0
Dividing the integral
into two parts.
0
T 2
T 2
2
T
T 2
2
=T
T 2
x(t ) cosn W 0 t dt = 0
0
Consider the equation for b n, (equation (4.4)).
T 2
bn =
0
z
z
z
2
2
2
x( t) sinn W 0t dt =
x(t ) sinn W 0t dt +
T -T 2
T -T 2
T
0
=
2
2
x( - t ) sinn W 0 ( - t ) ( -dt) +
TT2
T
T 2
z
z
2
=T
2
=T
=
2
T
2
x( - t ) sinn W 0 t dt +
T
0
T 2
2
x( - t) sinn W 0t dt +
T
0
T 2
z
0
x(t ) sin n W 0t dt +
2
T
0
z
x(t ) sinn W 0 t dt
0
T 2
z
x(t ) sinn W 0t dt
0
T 2
z
x( t) sinn W 0t dt
Dividing the integral
into two parts.
Change of integral index,
Let, t = -t ; \dt = –dt
When t = 0, t = –t = 0
When t = –T/2, t = –t= –(–T/2) = T/2
sin(–q) = –sinq
0
T 2
z
Since t is dummy variable, Let t = t.
x( t) sinn W 0t dt
0
T 2
z
T 2
x(t) sinn W 0 t dt =
4
T
T 2
z
0
x( t) sinn W 0t dt
Since x(t) is odd, x(–t) =– x(t)
4. 13
Signals & Systems
The waveform of some odd periodic signals and their Fourier series are given below. Certain
signals will become odd after subtraction of the dc component (a0/2), such a signal waveform is shown
in fig 4.13.
The waveform shown in fig 4.10 has odd symmetry,
half wave symmetry and quarter wave symmetry. Hence for
this waveform, a0= 0, an = 0 and bn exists only for odd values of
n. Therefore the Fourier series consists of odd harmonics of
sine terms. The trigonometric Fourier series representation of
the waveform of fig 4.10 is given by equation (4.23). [Please
refer example 4.5 for derivation of Fourier series].
x(t) =
4A
p
LMsin W t +
N
x(t)
A
0
–T
T
t
–A
Fig 4.10.
W0 =
OP
Q
sin 3W 0 t
sin5W 0 t
sin 7W 0 t
+
+
+ .....
3
5
7
0
2p
T
.....(4.23)
The waveform shown in fig 4.11 has odd symmetry, half wave symmetry and quarter wave
symmetry. Hence for this waveform, a0= 0, an= 0 and bn exists only for odd values of n. Therefore the
Fourier series consists of odd harmonics of sine terms. The trigonometric Fourier series representation
of the waveform of fig 4.11 is given by equation (4.24). [Please refer example 4.6 for derivation of
Fourier series].
x(t)
A
-T
2
T
2
0
–A
W0 =
T
4
t
2p
T
-T
4
Fig 4.11.
x( t ) =
8A
p2
LM sin W t
N 1
0
-
OP
Q
sin 3W 0 t
sin 5W 0 t
sin 7W 0 t
+
+ .....
2
2
3
5
72
.....(4.24)
The waveform shown in fig 4.12 has odd symmetry and so a0 = 0, an = 0, and bn exists for all
values of n. Hence the Fourier series has both even and odd harmonics of sine terms. The trigonometric
Fourier series representation of the waveform of fig 4.12 is given by equation (4.25). [Please refer
example 4.7 for derivation of Fourier series].
x(t)
W0 =
A
0
–T
T
2p
T
t
–A
Fig 4.12.
x( t ) =
2A
p
LM sin W t
N 1
0
-
OP
Q
sin 2W 0 t
sin 3W 0 t
sin 4W 0 t
sin 5W 0 t
+
- ..... .....(4.25)
+
2
3
4
5
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
The waveform shown in fig 4.13 is neither even nor odd. But x(t)
it can be shown that if the dc component (a0/2) is subtracted from
A
this waveform it becomes odd signal. Hence the Fourier coefficients
an = 0 and bn exists for all values of n. Therefore the Fourier series
0
has a dc component and all harmonics (both even and odd harmonics)
of sine terms. The trigonometric Fourier series representation of the
waveform of fig 4.13 is given by equation (4.26). [Please refer example
4.8 for derivation of Fourier series].
x( t ) =
LM sin W t
N 1
A
A
p
2
0
+
4. 14
T
2T
3T
t
Fig 4.13.
W0 =
2p
T
OP
Q
sin 2W 0 t
sin 3W 0 t
sin 4W 0 t
sin 5W 0 t
+
+
+
+ ..... .....(4.26)
2
3
4
5
4.4.3 Half Wave Symmetry (or Alternation Symmetry)
The periodic waveforms in which each period/cycle consists of two equal and opposite half
period/cycle are called alternating waveforms, because this type of waveform will have alternate positive
and negative half cycles. Such waveforms are said to have half wave symmetry or alternation symmetry.
The waveforms with half wave symmetry will satisfy the condition,
FG
H
x t±
T
2
IJ
K
= - x(t)
When a waveform has half wave symmetry, the Fourier series will consist of odd harmonic terms
alone. The waveforms shown in fig 4.4, 4.10 and 4.11 exhibit half wave symmetry. Certain waveform
will exhibit half wave symmetry after subtraction of the dc component (a0/2), such waveforms are
shown in fig 4.5, 4.8 and 4.9.
Some of the waveforms with only half wave symmetry and their Fourier series are given below.
The waveform shown in fig 4.14 has half wave symmetry. Hence the Fourier series consists of
odd harmonic terms alone. The trigonometric Fourier series representation of the waveform of fig 4.14
is given by equation (4.27). [Please refer example 4.10 for the derivation of Fourier series].
x(t)
A
0
–T
T
T
2
-T
2
t
–A
Fig 4.14.
FG cos W t +
H
2A F
G sin W t +
p H
4A
x(t) = - 2
p
+
IJ
K
I
+ .....J
K
0
cos3W 0 t
cos5W 0 t
+
+ .....
2
3
52
0
sin 3W 0 t
sin5W 0 t
+
3
5
W0 =
2p
T
.....(4.27)
4. 15
Signals & Systems
The waveform shown in fig 4.15 has half wave symmemtry. Hence the Fourier series consists of
odd harmonic terms alone. The trigonometric Fourier series representation of the waveform of fig 4.15
is given by equation (4.28).
x(t)
A
–T
-T
0
2
T
2
T
t
–A
Fig 4.15.
x(t) =
FG
H
IJ
K
cos3W 0 t
cos5W 0 t
4A
cos W 0 t +
+
+ .....
32
52
p2
FG
H
W0 =
IJ
K
.....(4.28)
sin 3W 0 t
sin5W 0 t
2A
sin W 0 t +
+
+ .....
3
5
p
-
2p
T
4.4.4 Quarter Wave Symmetry
A waveform with half wave symmetry if in addition has even/odd symmetry then it is said to
have quarter wave symmetry. In a waveform with quarter wave symmetry, each quarter period will
have identical shape, but may have opposite sign. The existence of the type of Fourier coefficients for
waveform with quarter wave symmetry is shown below.
FG
H
IJ
K
T
= - x( t )
2
x(t) has half wave symmerty.
Fourier series has odd harmonic terms.
x t±
FG
H
IJ
K
T
= - x( t )
2
x(t) has even and half wave symmetries.
(i.e., x(t) has quarter wave symmetry).
Fourier series will have odd
harmonics of cosine terms.
x(–t) = x(t) and x t ±
FG
H
IJ
K
T
= - x( t )
2
x(t) has odd and half wave symmetries.
(i.e., x(t) has quarter wave symmetry).
Fourier series will have odd
harmonics of sine terms.
x(–t) = –x(t) and x t ±
The waveforms shown in fig 4.4, 4.10 and 4.11 has quarter wave symmetry. Certain waveforms
will exhibit quarter wave symmetry after subtraction of dc component (a0/2), such waveforms are shown
in fig 4.5, 4.8 and 4.9.
4.5 Properties of Fourier Series
The properties of exponential form of Fourier series coefficients are listed in table 4.1. The proof
of these properties are left as exercise to the readers.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 16
Table 4.1 : Properties of Exponential Form of Fourier Series Coefficients
Note : cn and dn are exponential form of Fourier series coefficients of x(t) and y(t) respectively.
Property
Continuous time
periodic signal
Fourier series
coefficients
Linearity
A x(t) + B y(t)
A cn + B dn
Time shifting
x(t – t0)
c n e - jnW0t 0
Frequency shifting
e - jmW 0 t x(t)
cn – m
Conjugation
x*(t)
c*- n
Time reversal
Time scaling
x(–t)
x(at) ; a > 0
(x(t) is period with period T/a)
c–n
cn
(No change in Fourier coefficient)
+¥
Multiplication
åc
x(t) y(t)
m
dn - m
m = -¥
d
x( t )
dt
Differentiation
z
z
Integration
jnW 0 c n
1
cn
jnW 0
t
x( t ) dt
-¥
(Finite valued and periodic only if a0 =0)
Periodic convolution
x( t) y( t - t ) dt
T cn dn
T
Symmetry of real signals
x(t) is real
cn = c*- n
|cn | = |c- n | ; Ðcn = - Ðc- n
Re{cn} = Re{c- n}
Im{cn} = - Im{c- n}
Real and even
x(t) is real and even
cn are real and even
Real and odd
x(t) is real and odd
cn are imaginary and odd
Parseval's relation
Average power, P of x(t) is defined as,
1
P =
|x(t)|2 dt
T T
z
The average power, P in terms
of Fourier series coefficients is,
+¥
å |c |
2
P =
n
n = -¥
1
T
z
T
+¥
|x(t)|2 dt =
åc
2
n
n = -¥
2
Note : 1. The term cn respresent the power in nth harmonic component of x(t). The total average
power in a periodic signal is equal to the sum of power in all of its harmonics.
2
2. The term cn for n = 0, 1, 2, .... is the distribution of power as a function of frequency and
so it is called power density spectrum or power spectral density of the periodic signal
Signals & Systems
4. 17
4.6 Diminishing of Fourier Coefficients
x1(t)
The basic four waveforms (parabolic, ramp, square and
impulse) related through integration and differentiation are shown
in figure 4.16.
The basic waveform shown in figure 4.16, are related
through integration on going from bottom to top, and they are
related through differentiation on going from top to bottom.
t
It can be stated that the Fourier coefficients of the signals
1
will be proportional to k where k is the number of times, the
n
signal has to be differentiated to produce impulse.
If a waveform has square like structure with jump
continuity (or discontinuity) then it has to be differentiated one
time to produce impulse. In this case the Fourier coefficients
1
an and bn will be proportional to n . (Refer Fourier series of
waveform shown in fig 4.4, 4.5 and 4.10).
If a waveform has ramp like structure (and without jump
continuity) then it has to be differentiated twice to produce
impulse. In this case the Fourier coefficient an and bn will be
1
proportional to 2 .(Refer Fourier series of waveform shown
n
in fig 4.8, 4.9 and 4.11).
t
Integration
Differentiation
x2(t)
x3(t)
jump continuity
t
x4(t)
¥
t
-¥
-¥
Fig 4.16 : Waveform related through
integration and differentiation.
If a waveform has parabolic wiggles (and without jump continuity) then it has to be differentiated
thrice to produce impulse. In this case the Fourier coefficient an and bn will be proportional to 1 .The
n3
above concepts are summarized in table 4.2.
Table : 4.2
Condition
Impulse in
d
x(t)
x(t)
dt
d
d2
x(t)
x(t)
dt
dt 2
d2
d3
x(t)
x(t)
2
dt
dt 3
d k -1
dk
x(t)
x(t)
k -1
dt
dt k
Example
Jump in
Square wave
Triangular wave
Parabolic wave
-
Proportionality
for an and bn
1
n
1
n2
1
n3
1
nk
From the above discussion we can say that the Fourier coefficients will decrease more rapidly with
increasing n for a waveform with parabolic wiggles when compared to ramp or square wave (i.e., the
magnitudes of higher harmonics decreases more rapidly when the waveform has parabolic wiggles).
Similarly when we compare the waveform with ramp like structure and waveform with jumps like
square wave, we can say that the magnitudes of higher harmonics (or Fourier coefficients) decreases
more rapidly with increasing n in waveform with ramp like structure than the waveform with jumps.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 18
4.7 Gibbs Phenomenon
The exponential form of Fourier series of a continuous time periodic signal x(t) is given by,
+¥
x( t ) =
åc
n
e jnW 0t
n =-¥
The above equation is frequency domain representation of the signal x(t) as a sum of infinite series
with each term in the series representing a harmonic frequency component. When the signal x(t) is
reconstructed or synthesised with only N number of terms of the infinite seires, the reconstructed signal
exhibits oscillations (or overshoot or ripples), especially in signals with discontinuities.
x(t)
A
Fig 4.17a.
-
T
2
0
x 1(t)
t
T
2
n=1
A
A
2
Fig 4.17b.
-
T
2
0
x 3(t)
T
2
t
n=3
A
A
2
Fig 4.17c.
-
0
T
2
x 5(t)
T
2
t
n=5
A
A
2
Fig 4.17d.
-
T
2
0
x59(t)
T
2
t
n=59
A
Fig 4.17e.
-
T
2
0
T
2
t
Fig 4.17 : Successively closer approximations to a square pulse signal.
Consider a periodic square pulse signal shown in fig 4.17a. The reconstructed signal using N-terms
of Fourier series are shown in fig 4.17b, c, d and e. (Refer MATLAB program 4.5 in section 4.18). It can
be observed that the reconstructed signal exhibits oscillations and the oscillations are compressed towards
points of discontinuities with increasing value of N. Also it can be observed that, at the points of
discontinuity, the Fourier series converges to average value of the signal on either side of discontinuity.
This phenomenon was named after a famous mathematician, Josiah Gibbs, as Gibbs phenomenon and
the oscillations are called Gibbs oscillations. (Josiah Gibbs is a famous mathematician who first provided
mathematical explanation for this phenomenon).
Also in reconstruction of signal with finite terms of Fourier series it can be observed that the peak
overshoot of the oscillations remains constant irrespective of the value of N, but with increasing N the
peak overshoot shifts towards the point of discontinuity.
Signals & Systems
4. 19
4.8 Solved Problems in Fourier Series
x(t)
Example 4.1
A
Determine the trigonometric form of Fourier series of the
waveform shown in fig 4.1.1.
0
-T
–T
Solution
T
2
T
LM sinn 2p t OP
MM 2pT PP
N nT Q
T/2
2
The waveform shown in fig 4.1.1 has even symmetry,
half wave symmetry and quarter wave symmetry.
\ a 0 = 0, bn = 0 and an =
–A
T /2
z
4
T
t
T
4
-T
4
x(t ) cosnW0 t dt
0
Fig 4.1.1.
The mathematical equation of the square wave is,
T
x(t) = A ; for t = 0 to
4
T
T
to
= –A ; for t =
4
2
Evaluation of an
an =
4
T
T/2
z
x(t) cosnW 0 t dt =
0
4
T
T/4
z
A cosnW 0 t dt +
0
4
T
T/2
zb
g
- A cosnW0 t dt
T/4
L sinn 2p t OP
4A L sinnW t O
4A L sinnW t O
4A M
M
P - T MN nW PQ = T MM 2pT PP - 4TA
T N nW Q
N nT Q
L sinFGn 2p T IJ
L sinFG n 2p T IJ sinFGn 2p T IJ OP
O
sin 0 P
4A M H T 4 K
4A M H T 2 K
M
P - T MM 2p - H 2Tp 4 K PP
2p
2p P
T M
n
n
MN n T
MN n T
PQ
T PQ
T
4A L T
np
O 4A LM T sinnp - T sin np OP
sin
- 0P T MN 2n p
2
2np
2Q
Q T N 2np
T/4
T/4
=
T/2
0
0
0
0
0
T/4
0
=
=
2A
np
2A
np
4A
np
sin
+
sin
=
sin
2
np
2
np
np
2
np
= 0
For even values of n, sin
2
np
= ±1
For odd values of n, sin
2
\ an = 0
; for even values of n
=
an =
4A
np
sin
np
2
; for odd values of n
p
4A
4A
sin
= +
p
1 ´ p
2
4A
4A
3p
=
sin
= 3 ´ p
3p
2
\ a1 =
a3
a5 =
4A
5p
4A
sin
= +
2
5 ´ p
5p
a7 =
7p
4A
4A
sin
= 2
7 ´ p
7p
and so on.
W0 =
2p
T
T/4
sin 0 = 0
sin np = 0
for integer n
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 20
Fourier Series
The trigonometric form of Fourier series of x(t) is,
¥
¥
a
x(t ) = 0 +
an cosnW 0 t +
bn sinnW 0 t
2
n = 1
n = 1
å
å
Here, a0 = 0, bn = 0 and an exists only for odd values of n.
åa
\ x(t) =
n
cosnW 0 t
n = odd
= a1 cos W0 t + a 3 cos 3W0 t + a 5 cos 5W0 t + a 7 cos7W0 t + .....
=
4A
4A
4A
4A
cos W0 t cos3W0 t +
cos5W0 t cos7W0 t + .....
3p
5p
7p
p
=
4A
cos 3W0 t
cos5W0 t
cos7W 0 t
cos W0 t +
+ .....
3
5
7
p
OP
Q
LM
N
x(t)
Example 4.2
A
Find the Fourier series of the waveform shown in fig 4.2.1.
Solution
–2T
–T
The given waveform has even symmetry and so bn = 0
a0 =
4
T
T/2
z
x( t) dt ; an =
0
z
T
2
2T
T
t
Fig 4.2.1.
T/2
4
T
0
x(t ) cosnW 0 t dt ; bn = 0
0
To Find Mathematical Equation for x(t)
Consider the equation of straight line,
Here, y = x(t),
y - y1
x - x1
=
y1 - y 2
x1 - x 2
x = t.
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
\ The equation of straight line can be written as,
Consider points P and Q, as shown in fig 1.
.....(1)
x(t)
Coordinates of point-P = [t1, x(t1)] = [0, 0]
Coordinates of point-Q = [t2, x(t2)] =
LM T , AOP
N2 Q
P
On substituting the coordinates of points P and Q in equation (1) we get,
x(t ) - 0
t - 0
=
T
0 - A
0 2
2A
\ x(t) =
t
T
Þ
-2t
x(t )
=
-A
T
;
for t = 0 to
Þ
x(t) =
T
2
Evaluation of a0
a0 =
=
4
T
T/2
8A
T2
z
x(t ) dt =
0
LM t OP
N2Q
2
T/2
=
0
4
T
T/2
8A
T2
z
0
2A
8A
t dt =
2
T
T
LM T
N8
2
- 0
Q
A
OP = A
Q
T/2
z
0
t dt
2A
t
T
0
T
2
Fig 1.
R
T
t
Signals & Systems
4. 21
Evaluation an
T/2
z
4
T
an =
x(t ) cosnW 0 t dt =
0
T /2
z
4
T
2A
8A
t cosnW 0 t dt =
T
T2
0
T/2
z
t cosn W 0 t dt
0
z
uv = u
LMt sinnW t - 1 ´ F sin nW t I dtOP
G
J
u=t
H nW K PQ
T MN nW
L 2p cos n 2p t OP
F
`
8A M t sin n T t
8A L t sin nW t
- cos nW t I O
T
M
=
- G
H n W JK PPQ = T MMM n 2p + n 4p PPP
T MN nW
T
N T
Q
OP
LM sinn 2p T cos n 2p T
8A T
T 2 +
T 2 - 0 ´ sin 0 - cos 0 P
M
=
2p
2p
T M2
4
p
4p P
n
n
n
MN n T
T
T
T PQ
8A L T
=
M sin np + 4nTp cos np - 4nTp OPPQ = n2Ap cos np - 1
T MN 4np
=
8A
2
T/2
z
0
0
0
0
z z z
v - du v
v = cos W0t
0
T/2
T/2
0
W0 =
0
2
2
2
0
0
2
2
2
0
2p
T
2
0
2
2
2
2
2
2
2
2
2
2
sin 0 = 0
cos 0 = 1
2
2 2
2
2
2 2
sin np = 0
for integer
values of n
For even integer values of n, cos np = + 1
For odd integer values of n, cos np = –1
\ an = 0 ; for even values of n, and
2A
an =
4A
cos np - 1 = - 2 2 ; for odd values of n.
n p
4A
4A
; a3 = - 2 2
; a5 = - 2 2
; and so on.
3 p
5 p
n2 p 2
4A
\ a1 = - 2 2
1p
Fourier Series
The trigonometric form of Fourier series of x(t) is,
x(t) =
a0
+
2
¥
¥
åa
n
cosnW 0 t +
n=1
åb
n
sinnW 0 t
n=1
Here, bn = 0, and an exists only for odd values of n.
\ x(t) =
a0
+
2
åa
n
cosnW 0 t
n = odd
a0
+ a1 cos W 0 t + a 3 cos 3W 0 t + a 5 cos 5W 0 t + .....
2
4A
4A
A
4A
- 2 2 cos W 0 t - 2 2 cos 3W 0 t - 2 2 cos 5W 0 t - ......
=
1p
3 p
5 p
2
=
=
A
4A
2
p2
LMcos W t +
N
0
OP
Q
cos 3W 0 t
cos 5W 0 t
+
+ ......
32
52
Example 4.3
x(t)
Determine the trigonometric form of Fourier series of the
waveform shown in fig 4.3.1.
A
Solution
–T
The waveform of fig 4.3.1 has even symmetry.
\ bn = 0,
a0 =
4
T
T/2
z
0
x(t) dt ; an =
4
T
T /2
z
0
x( t) cosnW 0 t dt
-T -T
2
4
0
T
4
Fig 4.3.1.
T
2
T
t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 22
The mathematical equation of the given periodic rectangular pulse is,
T
x(t) = A ; for t = 0 to
4
T
T
to
4
2
= 0 ; for t =
Evaluation of a0
4
T
T/2
=
4
T
LMA T
N 4
an =
4
T
a0 =
z
4
T
x(t ) dt =
0
- 0
T/4
z
4
At
T
A dt =
0
T/4
0
OP = A
Q
Evaluation of an
4A
=
T
T/2
z
x( t) cosnW 0 t dt =
0
LM sinn 2p t OP
MM 2pT PP
N nT Q
T/4
4A
=
T
0
T/4
z
4
T
A cos nW 0 t dt =
0
LM sinn 2p T
MM 2Tp 4
N nT
sin 0
2p
n
T
4A
T
LM sinnW t OP
N nW Q
T/4
0
0
0
OP
PP
Q
2p
T
sin 0 = 0
W0 =
4A
np
np
T
2A
=
sin
=
sin
´
T
2
2
2np
np
np
= 0
2
np
For odd values of n, sin
= ±1
2
\ an = 0 ; for even values of n, and
For even values of n, sin
np
2A
sin
; for odd values of n.
2
np
2A
p
2A
sin
= +
\ a1 =
2
1 ´ p
p
an =
a3 =
2A
3p
2A
sin
= 2
3 ´ p
3p
a5 =
2A
2A
5p
sin
= +
5 ´ p
5p
2
a7 =
2A
7p
2A
sin
= 7 ´ p
7p
2
Fourier Series
and so on.
The trigonomteric form of Fourier series of x(t) is,
¥
¥
a
x(t ) = 0 +
an cosnW 0 t +
bn sinnW 0 t
2
n = 1
n = 1
Here, bn = 0 and an exists only for odd values of n.
a
\ x(t) = 0 +
a n cosnW 0 t
2
n = odd
å
å
å
a0
+ a1 cos W 0 t + a 3 cos 3W 0 t + a 5 cos 5W 0 t + a 7 cos 7W 0 t + .....
2
A
2A
2A
2A
2A
+
=
cos W 0 t cos 3W 0 t +
cos 5W 0 t cos 7W 0 t + .....
2
3p
5p
7p
p
=
=
A
2A
+
2
p
LMcos W t N
0
OP
Q
cos 3W 0 t
cos 5W 0 t
cos 7W 0 t
+
+ .....
3
5
7
Signals & Systems
4. 23
Example 4.4
x(t)
A
Determine the trigonometric form of Fourier series of
the full wave rectified sine wave shown in fig 4.4.1.
–T
Solution
4
T
T/2
z
x( t) dt ; an =
0
4
T
T
2
T
t
Fig 4.4.1.
The waveform shown in fig 4.4.1 is the output of full
wave rectifier and it has even symmetry.
\ bn = 0, a 0 =
0
-T
2
T /2
z
x( t) cos nW 0 t dt
0
The mathematical equation of full wave rectified output is,
x(t ) = A sin W0t ; for t = 0 to
T
2
and W0 =
2p
T
Evaluation of a0
a0 =
=
T/2
z
4
T
x( t) dt =
0
4
T
LM cos 2p t OP
MM- 2pT PP
N T Q
4A
T
T/2
z
A sin W 0 t dt =
0
T /2
4A
T
=
0
4A
T
LM cos 2p T
MM- 2pT 2
N T
LM- cos W t OP
MN W PQ
O
cos 0 P
+
2p P
P
T Q
T/2
0
0
0
W0 =
2A
2A
4A
- cos p + cos 0 =
=
1 + 1 =
p
p
p
2p
T
cos p = –1 cos 0 = 1
Evaluation of an
an =
=
=
=
T/2
z
4
T
4
T
x( t) cosnW 0 t dt =
0
4A
T
2A
T
T/2
z
z
0
T/2
z
A sin W 0 t cos nW 0 t dt
0
sin(W 0 t + nW 0 t) + sin(W 0 t - nW 0 t )
dt
2
T/2
sin (1 + n) W 0 t dt +
0
2A
T
T/2
z
LM -cos (1 + n) W0t OPT / 2 + 2A LM -cos (1 - n)W0t OPT /2
T N (1 - n) W0 Q
N (1 + n)W0 Q0
0
2p O
L cos(1 + n) 2p t OP
L
2A M -cos(1 - n) T t P
2A M T
M
PP + T MM
PP
2p
T M (1 + n) 2 p
(1 - n)
T
T
N
Q
N
Q
LM -cos(1 + n) 2p T
OP
LM -cos(1 - n) 2p T
cos0
2A
2A
T 2 +
T 2
+
M
M
2p
2p P
2p
T M
T M
P
(1 + n)
(1 + n)
(1 - n)
T
T Q
T
N
N
2A
T
0
=
sin (1 - n) W 0 t dt
0
T/2
=
2 sin A cos B = sin(A + B) + sin(A – B)
= -
A cos (1 + n)p
A
+
(1 + n)p
(1 + n)p
T/2
W0 =
2p
T
0
A cos (1 - n)p
A
+
(1 - n)p
(1 - n)p
+
OP
2p P
P
(1 - n)
T Q cos 0 = 1
cos 0
.....(1)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 24
The equation (1) for an can be evaluated for all values of n except n = 1. For n = 1, an has to be estimated seperately
as shown below.
a1 =
=
T/2
z
z
4
T
0
T/2
4
T
A
0
T/2
z
A sin W 0 t cos W 0 t dt
0
2A
sin 2W 0 t
dt =
2
T
T/2
z
sin 2W 0 t dt
sin 2q = 2 sinq cosq
0
L F 2p T I
LM - cos 2W t OP = 2A MM - cos GH 2 ´ T ´ 2 JK +
2p
T M
N 2W Q
2 ´
MN
T
LM- T cos 2p + T OP = 2A LM- T + T OP = 0
4p Q
T N 4p
4p Q
N 4p
T/2
2A
=
T
=
4
T
x( t) cos W 0 t dt =
cos 0
2p
2 ´
T
0
0
2A
T
0
OP
PP
PQ
W0 =
cos 2p = cos 0 = 1
For values of n > 1, the an are calculated using equation (1) as shown below.
\ an = -
A cos(1 + n)p
A
+
(1 + n)p
(1 + n)p
A cos(1 - n)p
A
+
(1 - n)p
(1 - n)p
When n is even integer, (1 + n) and (1 – n) will be odd,
\ cos( 1 + n)p = –1 ; cos(1 – n)p = –1
When n is odd integer, (1 + n) and (1 – n) will be even,
\ cos(1 + n)p =
1 ; cos(1 – n)p = 1
\ an = 0 ; for odd values of n
A
A
A
A
+
+
+
(1 + n)p
(1 + n)p
(1 - n)p
(1 - n)p
an =
=
\ a2 =
a4 =
a6 =
a8 =
;
for even values of n
2A
2A(1 - n) + 2A (1 + n)
2A
4A
+
=
=
(1 + n)p
(1 - n)p
(1 + n) (1 - n)p
(1 - n2 )p
4A
d
i
1 - 22 p
4A
d1 - 4 ip
2
4A
d1 - 6 ip
2
4A
d1 - 8 ip
2
= -
4A
3p
= -
4A
15p
= -
4A
35p
= -
4A
63p
and so on
Fourier Series
The trigonometric form of Fourier series of x(t) is,
x(t) =
a0
+
2
¥
¥
åa
n
cosnW 0 t +
n=1
åb
n
sinnW 0 t
n=1
Here, bn= 0, and an exists only for even values of n.
a
an cosnW0 t
\ x(t) = 0 +
2
n = even
å
a0
+ a 2 cos 2W0 t + a 4 cos4W 0 t + a 6 cos 6W 0 t + a 8 cos8W 0 t + .....
2
2A
4A
4A
4A
4A
=
cos2W0 t cos4W0 t cos6W0 t cos8W0 t - .....
3p
15p
35p
63p
p
=
=
2A
4A
p
p
LM cos2W t
N 3
0
+
2p
T
OP
Q
cos4W 0 t
cos 6W 0 t
cos 8W 0 t
+
+
+ .....
63
15
35
Signals & Systems
4. 25
Example 4.5
x(t)
Determine the Fourier series of the square wave shown in fig 4.5.1.
Solution
A
–T
The given waveform has odd symmetry, half-wave symmetry
0
- T
2
T
2
T
t
–A
and quarter wave symmetry.
\ a 0 = 0, an = 0, bn =
Fig 4.5.1.
T/2
z
4
T
x( t ) sin nW 0 t dt
0
The mathematical equation of the given waveform is,
T
x(t ) = A ; for t = 0 to
2
T
= -A ; for t =
to T
2
Evaluation of bn
4
T
bn =
T/2
z
0
4A
T
=
4
T
x( t) sin n W 0 t dt =
LM - cos n 2p t OP
MM 2p T PP
N nT Q
z
4A
T
0
cos np = –1,
for n = odd
cos np = +1,
for n = even
\ bn = 0
4A
T
A sin n W 0 t dt =
0
T/2
=
LM - cos nW t OP
N nW Q
O
4A L T
cos 0 P
=
cosnp +
2p P
T MN 2np
P
n
T Q
T/2
LM - cos n 2p T
MM 2pT 2
N nT
+
; for even values of n
OP
Q
LM
N
4A T
T
4A
=
+
=
; for odd values of n
T 2np 2np
np
4A
p
\ b1 =
; b3 =
4A
3p
; b5 =
4A
5p
and so on.
Fourier Series
The trigonometric form of Fourier series of x(t) is,
x(t) =
a0
+
2
¥
¥
åa
n
cosnW 0 t +
n=1
åb
n
sinnW 0 t
n=1
Here, a0 = 0, an = 0 and bn exists only for odd values of n.
\ x(t) =
åb
n
sinnW 0 t
n = odd
= b1 sin W 0 t + b 3 sin3W 0 t + b 5 sin5W 0 t + .....
=
4A
4A
4A
sin W 0 t +
sin3W0 t +
sin5W 0 t + .....
p
3p
5p
=
4A
p
LMsin W0t +
N
OP
Q
sin3W 0t
sin 5W0 t
+
+ .....
3
5
2p
T
cos 0 = 1
T/2
W0 =
0
0
0
T
2np
OP
Q
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
Example 4.6
4. 26
x(t)
Determine the trigonometric form of
Fourier series of the signal shown in fig 4.6.1.
A
-
T
4
T
4
-T
2
Solution
T
2
t
–A
The given signal has odd symmetry, half wave symmetry
and quarter wave symmetry, and so a0 = 0, an = 0,
4
bn =
T
T/2
z
0
4
x( t) sin nW0 t dt (or) bn =
T
Fig 4.6.1.
+T / 4
z
x( t) sin n W0 t dt
-T / 4
Note : Here x(t) is governed by single mathematical equation in the range - T to + T . And so the calculations will
4
4
T
T
be simple, if the integral limit is - to +
4
4
To Find Mathematical Equation for x(t)
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Consider the equation of straight line,
Here, y = x(t),
x = t.
\ The equation of straight line can be written as,
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
.....(1)
Consider points P and Q, as shown in fig 1.
x(t)
OP
LM
Q
N
T
O
L
Coordinates of point-Q = [t , x(t )] = M , AP
N4 Q
T
Coordinates of point-P = [t1, x(t1)] = - , - A
4
2
Q
A
-T
4
2
P
On substituting the coordinates of points P and Q in equation (1) we get,
T
T
t - t +
x( t) - ( - A)
x(t ) + A
4
4
Þ
=
=
T
T
T
-2 A
-A - A
4
4
2
0
T
4
t
–A
Fig 1.
FG IJ
H K
E
2t 1
x( t) 1
= 2A 2
T 2
4A
T
T
\ x( t) =
t ; for t = - to +
T
4
4
Evaluation of bn
bn =
=
4
T
+T/ 4
z
-T /4
16A
T2
+T / 4
4
T
x( t) sinnW0 t dt =
z
-T/ 4
Þ
4A
16A
t sinnW 0 t dt =
T
T2
-
x(t )
2t
=2A
T
Þ
0
z
t sin n W0 t dt
-T / 4
z
0
+T/ 4
-
z
0
1 ´
0
4A
t
T
+T / 4
uv = u
F -cosnW t I dtOP
GH nW JK PQ
u=t
L
2p
2p O
LM- t cosnW t + sinnW t OP = 16A MM-t cos n T t + sin n T t PP
2p
n W QP
T M
4p P
MN nW
n
MN n T
T PQ
LM cos n 2p T sinn 2p T
2p F T I
2p F T I O
cos n
sin n
- J
G
G- J P
H
K
T
T
T
4
T H 4K P
T 4 +
T 4 +
MM2p
2p
PP
4
4p2
4p2
n
n2 2
n2 2
MN 4 n T
T
T
T
Q
LMtF -cos nW t I
MN GH nW JK
x( t) =
z z z
v -
du v
v = sin nW0t
-T/4
+T / 4
=
16A
T2
T/4
0
0
2
2
0
0
2
-T /4
2
2
2
-T /4
=
16A
T2
W0 =
2p
T
4. 27
Signals & Systems
LM
N
16A
T2
np
T2
np
T2
np
T2
np
cos
cos
+
sin
+
+
sin
2
2 2
8np
2
2
8np
2
2
T
4n p
4n2 p 2
bn =
= -
OP
Q
4A
2A
4A
np
np
np
np
2A
cos
cos
+ 2 2 sin
+
+ 2 2 sin
np
np
2
2
2
2
n p
n p
8A
np
sin
n2 p 2
2
=
np
= ± 1
2
np
= 0
For even integer values of n, sin
2
For odd integer values of n, sin
\bn = 0
; for even values of n
8A
np
= 2 2 sin ; for odd values of n
2
n p
8A
\ b1 =
12 p 2
8A
b3 =
32 p 2
8A
b5 =
52 p 2
8A
b7 =
72 p 2
sin
p
8A
= + 2
2
p
sin
3p
8A
= - 2 2
2
3 p
sin
5p
8A
= + 2 2
2
5 p
sin
7p
8A
= - 2 2
2
7 p
and so on.
Fourier Series
The trigonometric form of Fourier series of x(t) is given by,
¥
¥
a
x(t ) = 0 +
a n cos nW 0 t +
bn sin nW0 t
2
n = 1
n =1
å
å
Here, a0 = 0, an = 0 and bn exists only for odd values of n.
åb
\ x( t) =
n
sin nW0 t
n = odd
= b1 sin W0 t + b 3 sin 3W0 t + b 5 sin 5W0 t + b7 sin 7W 0 t + .....
=
=
8A
p2
8A
p
2
sin W0 t -
8A
32 p 2
LMsin W0 t N
sin 3W0 t
2
3
8A
sin 3W0 t +
+
52 p2
sin 5W0 t
2
5
-
sin 5W0 t sin 7W0 t
7
2
8A
72 p2
sin 7W0 t + .....
OP
Q
+ .....
x(t)
Example 4.7
Determine the trigonometric form of Fourier series for
the signal shown in fig 4.7.1.
A
0
T
2
Solution
–A
The given signal has odd symmetry and so a0 = 0, an = 0,
bn =
4
T
T/2
z
0
x(t ) sin nW0 t dt
Fig 4.7.1.
T
t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 28
To Find Mathematical Equation for x(t)
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Consider the equation of straight line,
Here, y = x(t),
x = t.
\ The equation of straight line can be written as,
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
.....(1)
x(t)
Consider points P and Q, as shown in fig 1.
Coordinates of point-P = [t1, x(t1)] = [0, 0]
Q
A
LM
N
OP
Q
T
Coordinates of point-Q = [t2, x(t2)] =
,A
2
On substituting the coordinates of points P and Q in equation (1) we get,
x(t) - 0
t - 0
x(t)
t
=
Þ
=
T
T
0 - A
-A
0 2
2
2At
T
\ x(t) =
; for t = 0 to
T
2
Þ
x(t) =
P
0
–A
2At
T
T
T
2
t
Fig 1.
Evaluation of bn
4
T
bn =
T/2
z
0
LM F
MN GH
8A
- cos nW 0 t
t
n W0
T2
=
I
JK
=
8A
T2
=
8A
T2
2At
8A
sinn W0 t dt =
T
T2
z
0
T/2
z
t sin nW0 t dt
0
z
T/2
F - cos nW t I dtOP
GH nW JK PQ
L n 2p t
sin nW t O
8A M cos T
P = T MM-t 2p
n W PQ
MN n T
-
LM
MN
LM T cos n 2p T
MM- 2 2p T 2
MN n T
LM- T cos np +
N 4np
8A
cos nW 0 t
=
+
-t
n W0
T2
T/2
4
T
x( t) sinnW0 t dt =
z
0
1´
0
u=t
0
0
2
0
2
0
2p
t
T
2
4p
n2 2
T
sin n
T/2
2
+
2p T
sin n
T 2 + 0 ´ cos 0 - sin 0
+
2
2p
4p2
2 4p
n
n2 2
n
2
T
T
T
2
T2
sinnp
4n2 p 2
OP =
Q
-
z z z
uv = u v -
OP
PP
PQ
W0 =
2A
cosnp
np
for n = even
2A
for n = odd
np
2A
2A
2A
2A
2A
\ b1 = +
; b2 = ; b3 = +
; b4 = ; b5 = +
and so on.
p
2p
3p
4p
5p
= +
Fourier Series
The trigonometric form of Fourier series of x(t) is,
x(t ) =
a0
+
2
¥
åa
n = 1
¥
n
cos nW 0 t +
åb
n =1
n
sin nW0 t
2p
T
0
OP
PP
PQ
For odd integer values of n, cos np = –1
2A
np
v = sin nW0t
T/2
For even integer values of n, cos np = +1
\ bn = -
du v
sin 0 = 0
sin np = 0
for integer n
4. 29
Signals & Systems
Here, a0 = 0, an = 0
¥
åb
\ x( t) =
n
sin nW0 t
n=1
= b1 sin W0 t + b 2 sin 2W0 t + b 3 sin 3W0 t + b 4 sin 4W0 t + b 5 sin 5W0 t + .....
=
2A
2A
2A
2A
2A
sin W0 t sin 2W0 t +
sin 3W 0 t sin 4W0 t +
sin 5W0 t - .....
p
2p
3p
4p
5p
=
2A
p
LM sin W t
N 1
0
OP
Q
sin 2W 0 t
sin 3W0 t
sin 4W0 t
sin 5W0 t
+
+
- .....
2
3
4
5
-
x(t)
Example 4.8
Determine the trigonometric form of the Fourier series
of the ramp signal shown in fig 4.8.1.
A
Solution
0
The given signal is neither even nor odd.
\ a0 =
2
T
T
z
0
3T
2T
t
Fig 4.8.1.
T
2
T
x(t ) dt ; an =
T
z
x( t) cos nW 0 t dt ; bn =
0
2
T
T
z
x(t ) sin nW 0 t dt
0
Note : It can be shown that after subtracting a0 /2 from the signal, it becomes odd. Hence an will be equal to zero.
To Find Mathematical Equation for x(t)
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Consider the equation of straight line,
Here, y = x(t),
x = t.
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
\ The equation of straight line can be written as,
.....(1)
x(t)
Consider points P and Q, as shown in fig 1.
Coordinates of point-P = [t1, x(t1)] = [0, 0]
P
0
On substituting the coordinates of points P and Q in equation (1) we get,
x( t) - 0
t - 0
=
0 - A
0 - T
\ x( t) =
At
T
x(t )
t
=
-A
-T
Þ
Q
A
Coordinates of point-Q = [t 2, x(t2)] = [T, A]
Þ
At
x(t) =
T
T
t
Fig 1.
; for t = 0 to T
Evaluation of a0
a0 =
=
2
T
T
z
2
T
x( t ) dt =
0
LM t OP
MN 2 PQ
2
2A
T2
T
=
0
2A
T2
T
T
2A
A
t dt =
T
T2
z
0
LM T
N2
2
- 0
z
t dt
0
OP = A
Q
Evaluation of an
an =
=
2
T
2A
T2
T
z
x( t) cosnW 0 t dt =
0
LMtF sinnW t I
MN GH nW JK
0
0
-
z
2
T
1 ´
T
z
0
2A
At
cos nW 0 t dt =
T
T2
F sinnW t I dt OP
GH nW JK PQ
T
0
0
0
T
z
t cos nW0 t dt
0
z
z z z
uv = u v u=t
du v
v = cos nW0t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
L
O
LMt sinnW t - F - cos nW t I OP = 2A MM t sinn 2Tp t + cos n 2Tp t PP
GH n W JK PQ T M n 2p
4p P
MN nW
n
MN T
PQ
T
L T sinn 2p T cos n 2p T
O
2A M
0 ´ sin 0
cos 0 P
T
T
M 2p + 4p - 2p - 4p PP
T M
n
n
n
MN n T
T
T PQ
T
2A L T
T
T O
sin n2p +
cos n2p - 0 M
P
T N 2np
4n p
4n p Q
2A L
M0 + 4nT p - 4nT p OPQ = 0
T N
0
2
0
0
2
2
2
2
2
sin 0 = 0
cos 0 = 1
2
2
2 2
sin n2p = 0
cos n2p = 1
for integer n
2
2
2
2 2
2
2p
T
0
2
2
2
2
2
2
0
2
2
=
W0 =
0
2
2
=
T
T
2A
an =
T2
=
4. 30
2 2
Evaluation of bn
bn =
=
2
T
2A
T2
2A
=
T2
=
2A
T2
T
z
x(t ) sinnW 0 t dt =
0
T
2
T
z
0
2A
At
sin nW0 t dt =
T
T2
T
z
2A
T2
\ b1 = -
z
T
LMtF - cos nW t I
MN GH nW JK
F - cos nW t I dtOP
GH nW JK PQ
L cos n 2p t
sin nW t O
2A M
P = T MM-t 2pT
n W PQ
MN n T
2p
-
0
0
z
0
u=t
LM-t cos nW t +
MN nW
LM T cos n 2p T sinn T
0
MM- 2pT + 4Tp + 0 ´ 2cos
p
n
n
MN n T
T
T
LM- T cos n2p + T sinn2pOP = - A
4n p
N 2np
Q np
0
2
2
0
2
0
0
2
+
OP
PP
PQ
OP
PP
PQ
T
W0 =
; b2 = -
A
2p
sin 0 = 0
sin n2p = 0
cos n2p = 1
for integer n
2
; b3 = -
A
3p
; b4 = -
A
4p
2p
T
0
2
2
A
p
2p
t
T
2
4p
n2 2
T
sinn
sin 0
4p2
n2 2
T
-
2
2
du v
v = sin nW0t
0
T
0
z z z
uv = u v -
0
1 ´
2
=
t sin nW0 t dt
0
and so on.
Fourier Series
The trigonometric form of Fourier series of x(t) is given by,
¥
¥
a
x(t ) = 0 +
a n cos nW 0 t +
bn sin nW0 t
2
n = 1
n =1
å
an = 0
a
\ x( t) = 0 +
2
å
Here,
¥
åb
n
sin nW 0 t
n = 1
=
a0
+ b1 sin W0 t + b 2 sin 2W0 t + b 3 sin 3W0 t + b 4 sin 4W 0 t + .....
2
=
A
A
A
A
A
sin W 0 t sin 2W 0 t sin 3W0 t sin 4W 0 t - .....
2
p
2p
3p
4p
=
A
A
2
p
LM sin W t
N 1
0
+
POQ
sin 2W0 t
sin 3W 0 t
sin 4W 0 t
+
+
+ .....
2
3
4
Example 4.9
Determine the Fourier series representation of the
half-wave rectifier output shown in fig 4.9.1.
x(t)
A
0
T
2
T
Fig 4.9.1.
2T
t
4. 31
Signals & Systems
Solution
The signal shown in fig 4.9.1 is neither even nor odd.
T
2
T
\ a0 =
z
T
2
T
x(t ) dt ; a n =
0
z
x(t ) cosnW 0 t dt ; bn =
0
T
2
T
z
x( t) sinnW0 t dt
0
The mathematical equation representing half-wave rectified output is,
x(t) = A sin W0 t ; for t = 0 to
= 0
T
2
T
to T
2
; for t =
Evaluation of a0
a0 =
=
T
2
T
z
0
2
T
T/2
z
2A
T
A sin W0 t dt =
0
T/2
LM
OP
MM
PP
N
Q
T
LM- cos p +
N 2p
2p
- cos
t
T
2p
T
2A
=
T
=
x(t ) dt =
=
2A
T
T
2p
OP =
Q
x( t) cosnW0 t dt =
2
T
0
LM
MM
N
2A
T
2p T
-cos
T 2
2p
T
2A
T
LM- T
N 2p
LM -cos W t OP
N W Q
O
cos 0 P
+
2p P
P
T Q
T/2
0
0
0
W0 =
T
2p
´ ( -1) +
OP
Q
2p
T
cos 0 = 1
cos p = –1
2A
T
2A
´
=
T
p
p
Evaluation of an
an =
=
=
T
2
T
z
0
T /2
2A
T
A
T
=
A
T
=
A
T
z
0
T/2
z
A sin W 0 t cos nW0 t dt
2 sin A cos B = sin(A + B) + sin(A – B)
0
sin(W0 t + nW0 t) + sin (W0 t - nW0 t )
A
dt =
T
2
T/2
LM- cos (1 + n) W t - cos (1 - n) W t OP
(1 - n)W
Q
N (1 + n)W
LM -cos (1 + n) 2p T cos(1 - n) 2p T
T 2 T 2
MM
2p
2p
(1 + n)
(1 - n)
T
T
N
LM- T cos (1 + n)p - T cos (1 - n)p +
(1 - n)2p
N (1 + n)2p
0
0
0
0
=
0
A
T
T/2
z
sin(1 + n) W 0 t + sin (1 - n) W0 t dt
0
LM -cos(1 + n) 2p t
T
MM
2p
N (1 + n) T
2p
t
T
2p
(1 - n)
T
cos(1 - n)
-
O
cos 0
cos 0 P
+
+
2p P
2p
P
(1 - n)
(1 + n)
T Q
T
OP
T
T
+
(1 + n)2p
(1 - n)2p Q
W0 =
OP
PP
Q
T/2
0
2p
T
cos 0 = 1
The above expression for an can be evaluated for all values of n except for n = 1. For n = 1, an has to be evaluated
separately as shown below.
a1 =
2
T
T
z
x( t) cos W 0 t dt =
0
2
T
sin 2q = 2 sinq cosq
T
z
A sin W0 t cos W0 t dt
0
T
L cos 4p t OP
sin 2W t
A M2A
A L - cos 2W t O
dt =
M
PQ = T MM 4pT PP
T
2
T N 2W
N T Q
T
A
T
4 pT
A L T
O
LM- + T OP = 0
cos
cos 0P =
+
T
T
4p
T MN 4p
Q
N 4p 4p Q
T
T
=
z
0
0
0
0
W0 =
2p
T
0
0
=
cos 4p = 1
cos 0 = 1
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 32
\ a1 = 0
an = -
A cos(1 + n)p
A cos(1 - n)p
A
A
+
+
(1 + n)2p
(1 - n)2 p
(1 + n)2p
(1 - n)2 p
; for all n except n = 1
When n is even, the terms (n + 1) and (n – 1) are odd,
\ cos (1 + n)p = –1, cos (1 – n)p = –1
When n is odd, the terms (n + 1) and (n – 1) are even,
\ cos (1 + n)p = 1, cos (1 – n)p = 1
\ an = 0 ; for odd values of n
A
A
A
A
+
+
+
(1 + n)2 p
(1 - n)2 p
(1 + n)2p
(1 - n)2 p
an =
; for even values of n
2A
A
A
A(1 - n) + A(1 + n)
+
=
=
(1 + n)p
(1 - n)p
(1 + n) (1 - n)p
(1 - n 2 )p
=
2A
2A
= (1 - 22 )p
3p
\ a2 =
a4 =
2A
2A
= (1 - 4 2 )p
15p
a6 =
2A
2A
= 35p
(1 - 6 2 )p
a8 =
2A
2A
= and so on.
(1 - 82 )p
63p
Evaluation of bn
bn =
=
2
T
z
A
=
T
2
T
x( t) sinnW 0 t dt =
0
T/2
2A
T
A
=
T
=
T
z
0
T/2
z
A sin W 0 t sin nW0 t dt
2 sin A sin B = cos(A – B) – cos(A + B)
0
cos(W0 t - nW0 t) - cos(W0 t + nW0 t)
A
dt =
2
T
LM sin(1 - n)W t
N (1 - n)W
0
0
sin(1 + n)W0 t
(1 + n)W 0
LM sin(1 - n) 2p T
T 2
MM
2p
(1 - n)
T
N
OP
Q
T/2
0
A
=
T
T/2
z
cos(1 - n)W0 t - cos (1 + n)W0 t dt
0
LM sin(1 - n) 2p t
T
MM
2p
(1 - n)
T
N
-
2p T
sin 0
sin 0
T 2 +
2p
2p
2p
(1 + n)
(1 - n)
(1 + n)
T
T
T
sin (1 + n)
-
2p
t
T
2p
(1 + n)
T
sin(1 + n)
OP
PP
Q
OP
PP
Q
T/2
W0 =
2p
T
0
sin 0 = 0
A sin(1 - n)p
A sin(1 + n)p
(1 - n)2p
(1 + n)2p
The above expression for bn can be evaluated for all values of n except for n = 1. For n = 1, bn has to be evaluated
separately as shown below.
b1 =
=
2
T
T
z
x( t) sin W 0 t dt =
0
2A
T
T/ 2
z
0
2
T
T /2
z
A sin W 0 t sin W0 t dt =
0
1 - cos 2W0 t
A
dt =
2
T
2A
T
T /2
z
0
(1 - cos 2W 0 t) dt =
T/ 2
z
sin2q =
sin2 W0 t dt
0
A
T
LMt MN
sin 2W 0 t
2W 0
OP
PQ
T /2
0
1 - cos 2q
2
4. 33
Signals & Systems
A
\ b1 =
T
LM
MMt N
4p
t
T
4p
T
sin
OP
PP
Q
T/2
A
=
T
0
LM
MM T2
MN
sin
-
4p T
T 2 - 0 +
4p
T
sin0
4p
T
OP
PP
PQ
2p
T
sin 0 = 0
W0 =
A
A
A
sin 2 p =
=
2
4p
2
\ b1 =
bn =
sin 2p = 0
A
2
A sin(1 - n)p
A sin(1 + n)p
(1 - n)2 p
(1 + n)2 p
For integer values of n, except when n = 1, sin(1 – n)p = 0.
For integer values of n, sin(1 + n)p = 0.
\ bn = 0 for all values of n except n = 1.
Fourier Series
The trigonometric form of Fourier series of x(t) is given by,
x(t ) =
a0
+
2
¥
¥
åa
n
cos nW 0 t +
n = 1
åb
n
sin nW0 t
n =1
Here, an exists only for even values of n and bn = 0 for all values of n except when n = 1.
\ x( t) =
a0
+
an cos nW0 t + b1 sin nW0 t
2
n = even
å
a0
+ a 2 cos 2W0 t + a 4 cos 4W0 t + a 6 cos6W 0 t + a 8 cos 8W0 t + ..... + b1 sin W0 t
2
A
A
2A
2A
2A
2A
=
cos 2W 0 t cos 4W0 t cos 6W0 t cos 8W0 t - ..... +
sin W0 t
3p
15p
35p
63p
2
p
=
=
A
2A
+
p
p
LM p
N4
sin W0 t -
OP
Q
cos 2W0 t
cos 4W0 t
cos 6W 0 t
cos 8W0 t
- .....
3
15
35
63
Example 4.10
x(t)
A
Find the Fourier series of the signal shown in fig 4.10.1.
T
2
-T
2
T
–A
Solution
The given signal has half-wave symmetry and so a0 will be zero.
The Fourier coefficients an and bn will exist only for odd integer values of n.
\ a 0 = 0 ; an =
2
T
T
z
x(t ) cosnW0 t dt ; bn =
0
2
T
To Find Mathematical Equation for x(t)
Consider the equation of straight line,
Here, y = x(t),
x = t.
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Fig 4.10.1.
T
z
0
x( t) sinnW0 t dt
t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
\ The equation of straight line can be written as,
4. 34
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
.....(1)
Consider points P and Q, as shown in fig 1.
x(t)
Coordinates of point-P = [t1, x(t1)] = [0, 0]
Coordinates of point-Q = [t2, x(t2)] =
Q
A
LM T , AOP
N2 Q
P
On substituting the coordinates of points P and Q in equation (1) we get,
R
0
T
T
2
–A
t - 0
x(t) - 0
=
T
0 - A
0 2
x(t)
t
=
T
-A
2
Þ
2At
x(t) =
T
Þ
S
Fig 1.
Consider points R and S, as shown in fig 1.
Coordinates of point-R = [t3, x(t3)] =
LM T , 0OP
N2 Q
Coordinates of point-S = [t4, x(t4)] = [T, –A]
On substituting the coordinates of points R and S in equation (1) we get,
T
t x(t) - 0
2
=
T
0 - ( - A)
- T
2
T
t x( t )
2
=
T
A
2
Þ
x( t)
2t
= + 1
A
T
Þ
Þ
x(t) = A -
2At
T
Now the mathematical equation of the waveform is given by,
2At
T
x(t) =
; for t = 0 to
2At
T
= A -
; for t =
T
2
T
to T
2
Evaluation of a0
The given signal has half wave symmetry and so a0 = 0.
Proof :
a0 =
=
T
2
T
z
2
T
LM 2AT
MN 8T
x(t) dt =
0
2
T
LM
MN
T 2
z
0
2
- 0 + AT -
2At
dt +
T
2AT
2T
2
-
T
z
T /2
FG A H
IJ
K
2At
dt
T
AT
2AT
+
2
8T
2
OP =
PQ
OP 2 LML 2At O L 2At O OP
PQ = T MNMMN 2T PPQ + MMNAt - 2T PPQ PQ
2 L AT
AT
AT O
+ AT - AT +
= 0
T MN 4
2
4 PQ
2
T 2
0
Evaluation of an
The coefficient an for a signal with half-wave symmetry is,
an =
\ an =
2
T
2
T
T
z
z
x(t) cos nW0 t dt
0
T 2
4A
=
T2
2At
2
cosnW0 t dt +
T
T
0
T 2
z
0
2A
t cosnW0 t dt +
T
z FGH
T
T2
T
z
T2
A -
IJ
K
2At
cosnW0 t dt
T
cosnW0 t dt -
4A
T2
T
z
T 2
t cosnW0 t dt
2
T
T 2
t
4. 35
Signals & Systems
\ an =
LMt ´ F sinnW t I - 1 ´ F sinnW t I dt OP + 2A L sinnW t O
GH nW JK PQ T MMN nW PPQ
MN GH nW JK
F sinnW t I - 1 ´ F sinnW t I dtOP
4A L
t ´ G
M
GH nW JK PQ
T MN
H nW JK
F -cosnW t I OP + 2A LM sinnW t OP
4A L t sinnW t
- G
M
T MN nW
T N nW
H n W JK PQ
Q
F -cosnW t I OP
4A L t sinnW t
- G
M
T NM nW
H n W JK QP
L T 2p T cosn 2p T
OP
4A M 2 sinn T 2
cos
0
´
0
sin0
M 2p + 4pT 2 - 2p - 4p PP
T M
n
n
n
MN n T
T
T PQ
T
2p T O
L 2p
sinn
2A M sinn T T
T 2 P
+
M
PP
2p
T M n 2p
n
T
N T
Q
LM T sinn 2p T cosn 2p T T ´ sin n2p T
4A
M 2pT + 4pT - 2 2p T 2 T M
n
n
MN n T
T
T
4A
T2
0
0
T 2
z
T
0
0
0
T
z
0
2
0
0
0
0
0
0
2
u=t
du v
v = cos W0t
0
2
0
0
z z z
uv = u v -
T
0
2
z
T2
T/2
=
T 2
0
0
T/2
T
0
0
2
2
0
=
2
2
0
T/2
2
2
2
2
2
2
2
2
A
A
A
A
sinnp + 2 2 cosnp - 2 2 +
np
n p
n p
np
2A
A
sinn2p - 2 2 cosn2p
np
n p
A
A
= 0 + 2 2 cosnp - 2 2 + 0 - 0 - 0
n p
n p
2A
2A
2A
= 2 2 cosnp - 2 2 = 2 2 (cosnp n p
n p
n p
When n is even integer, cos np = +1
When n is odd integer, cos np = –1
=
=\ a1 = -
1)
4A
; for odd integer values of n
p 2n2
; a3 = -
4A
32 p 2
; a5 = -
4A
52 p 2
and so on.
Evaluation of bn
The coefficient bn for a signal with half wave symmetry is,
T
2
T
z
\ bn =
2
T
bn =
=
x(t) sinnW0 t dt
0
4A
T2
T2
z
0
T 2
z
0
z FGH
T
2At
2
sinnW0 t dt +
T
T
t sinnW0 t dt +
2A
T
T2
A -
z
IJ
K
2At
sinnW0 t dt
T
T
T2
sin 0 = 0
cos 0 = 1
n2p T
T 2
2
2 4p
n
2
T
cos
A
sinnp
np
A
A
+
sinnp + 2 2 cosnp
np
n p
A
A
- 2 2 + 0 + 2 2 cosnp
n p
n p
sinn2p -
; for even integer values of n
4A
12 p 2
2p
T
2
2
\ an = 0
W0 =
sinnW 0 t dt -
4A
T2
T
z
T 2
t sinnW0 t dt
OP
PP
PQ
For integer n
sin np = 0
sin n2p = 0
cos n2p = 1
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 36
LMt ´ F -cosnW t I - 1 ´ F -cosnW t I dtOP + 2A L -cosnW t O
M
PPQ
GH nW JK PQ
T MN nW
MN GH nW JK
uv = u v - du v
F
W tI
W tI O
4A L
Mt ´ FGH -cosn
JK - 1 ´ GH -cosn
JK dtPPQ
T MN
nW
nW
u=t
v = sin nW t
T /2
T
W0t I O
W0 t O
4A L - t cosnW0 t
M nW - FG -sinn
P + 2AT LM -cosn
=
2 M
2 2 JP
W
n
T N
H n W0 K Q0
0
N 0 PQT /2
T
F
- sinnW 0 t I O
4A L - t cosnW 0 t
M
P
- G
n W0
T 2 MN
H n2W02 JK PQT /2
L T 2p T sinn 2p T
O
2p
W =
4A M - 2 cosn T 2
sin 0 P
0 ´ cos 0
T
T
2
M
PP
=
+
+
2p
2p
sin 0 = 0
p
p
T M
4
4
n
n
n
MN n T
T
T PQ
T
2p
2p T O
L
cos n
2A M cosn T T
T 2 P
+
+
M 2p
PP
2p
T M
n
n
T
T
N
Q
LM -T cosn 2p T sinn 2p T T ´ cosn 2p T sinn 2p T OP
4A
M 2p T + 4pT + 2 2p T 2 - 4Tp 2 PP
T M
n
n
n
T
T
T
NM n T
QP
4A
T2
\ bn =
T 2
z
0
0
T
0
0
0
0
0
0
T2
z
T
z
0
2
0
0
z z z
0
T2
0
2
2
2
2
2
2
2
2
2
2
A
A
cosnp + 2 2 sinnp
np
n p
A
+ 2 2 sinn2p n p
A
A
= cosnp + 0 +
np
np
A
A
A
=
cosnp =
(1
np
np
np
= -
A
A
2A
cosn2p +
cosnp +
cosn2p
np
np
np
A
A
cosnp + 2 2 sinnp
np
n p
A
2A
A
cosnp +
+ 0 cosnp + 0
np
np
np
- cosnp)
When n is even integer, cos np = +1
; for even integer values of n.
2A
=
; for odd integer values of n.
np
2A
2A
2A
; b3 =
; b5 =
and so on.
\ b1 =
3p
5p
p
Fourier Series of x(t)
The Fourier series of x(t) is
¥
¥
a
x(t ) = 0 +
a n cosnW 0 t +
bn sinnW0 t
2
n=1
n=1
å
å
Here a0 = 0 and the Fourier coefficients an and bn exist only for odd values of n.
\ x( t) =
åa
n = odd
n
cosnW 0 t +
åb
n
2
-
When n is odd integer, cos np = –1
\ bn = 0
2
2
2
sinnW0 t
n = odd
= a1 cos W 0 t + a 3 cos 3W 0 t + a 5 cos 5W 0 t + .....
+ b1 sin W0 t + b 3 sin 3W 0 t + b5 sin 5W 0 t + .....
For integer n
sin np = 0
sin n2p = 0
cos n2p = 1
4. 37
Signals & Systems
4A
4A
4A
cos W0 t - 2 2 cos 3W0 t - 2 2 cos 5W0 t - .....
3 p
5 p
p2
\ x(t) = -
2A
2A
2A
sin W 0 t +
sin3W0 t +
sin 5W0 t + .....
3p
5p
p
+
4A
p2
= -
FG cos W t +
H
0
+
2A
p
cos 3W0 t
cos 5W0 t
+
32
52
FG sin W t +
H
0
IJ
K
+ .....
sin 3 W0 t
sin5W0 t
+
3
5
IJ
K
+ .....
x(t)
Example 4.11
Determine the exponential form of the Fourier
series representation of the signal shown in fig 4.11.1. Hence
determine the trigonometric form of Fourier series.
A
-T
2
Solution
To Find Mathematical Equation for x(t)
T
2
t
Fig 4.11.1.
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Consider the equation of straight line,
Here, y = x(t),
0
x = t.
\ The equation of straight line can be written as,
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
.....(1)
Consider points P, Q and R as shown in fig 1.
Coordinates of point-P = [t1, x(t1)] =
LM -T , 0OP
N2 Q
x(t)
Coordinates of point-Q = [t2, x(t2)] = [0, A]
Coordinates of point-R = [t3, x(t3)] =
Q A
LM T , 0OP
N2 Q
P
-T
2
On substituting the coordinates of points P and Q in equation (1) we get,
T
t +
x(t) - 0
2
=
-T
0 - A
- 0
2
Þ
x( t)
-2t
=
- 1 Þ
T
-A
x(t) = A +
2At
T
On substituting the coordinates of points Q and R in equation (1) we get,
x(t) - A
=
A - 0
t - 0
T
0 2
2At
\ x(t) = A +
T
2At
= A T
Þ
x( t)
-2t
- 1 =
A
T
Þ
x(t) = A -
T
to 0
2
T
; for t = 0 to
2
; for t = -
Evaluation of c0
cn =
1
T
+T / 2
z
x(t) e - jnW 0 t dt
-T / 2
When n = 0, c0 =
=
1
T
1
T
+T / 2
z
z
1
T
x(t) e 0 dt =
-T / 2
0
-T /2
FG A +
H
IJ
K
+T / 2
z
x(t) dt
-T / 2
2At
1
dt +
T
T
+T /2
z
0
FG A H
IJ
K
2At
dt
T
2 At
T
R
0
Fig 1.
T
2
t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
0
0
z
z
A
2A
dt +
T -T /2
T2
\ c0 =
A
t
T
=
T /2
z
A
T
t dt +
-T /2
0
LM t OP + A t
T
N2Q
TO
2A L
+
M0 - T8 OPQ +
2 PQ
T N
0
-T/2
+
LM0 +
N
T /2
t dt
0
0
A
T
2
2A
T2
-
-T /2
2
z
LM t OP
N2Q
LM T - 0OP N2 Q
0
2
2A
T2
T/2
2A
T2
dt -
T/2
0
=
A
T
=
A
A
A
A
2A
2A
A
A
+
=
= A =
2
4
2
4
2
4
2
2
2
4. 38
2A
T2
LM T
N8
2
- 0
OP
Q
Evaluation of cn
cn =
=
1
T
1
T
A
=
T
=
+T 2
z
z FGH
z
x(t) e - jnW 0 t dt
-T /2
IJ
K
0
2At
1
e - jnW 0 t dt +
T
T
A +
-T / 2
0
2A
e - jnW 0 t dt +
T2
-T / 2
z
0
t e - jnW 0 t dt
T/2
z
A
T
+
-T / 2
e - jnW 0 t dt
2A
T2
-
0
T/2
z
t e - jnW 0 t dt
0
z
z z z
LM e
O
uv = u v - du v
e
t
1 ´
dt P
- jnW
MN - jnW
PQ
u=t
v = e
OP - 2A LMt e
O
e
A Le
+
1 ´
dt P
M
jn
jn
jn
W
W
W
T MN
T MN
PQ
PQ
OP
OP
OP
e
A Le
2A L e
A Le
M
M
M
t
+
+
jn
jn
jn
W
W
W
T M
T M
T M
PQ
PQ
( - jnW ) PQ
N
N
N
2p
OP
e
2A L t e
W =
M
T
jn
W
T M
( - jnW ) PQ
N
L
L
FG IJ O
FG IJ
FG IJ O
H K P 2A M 0 ´ e
H K
H KP
e
e
e
e
A M e
T
M
P
M
P
+
+
+
2p P
2p
T M - jn 2 p
T M - jn 2 p
4p
2
4p P
- jn
jn
-n
MMN T -n T
P
MN T
T PQ
T
T PQ
LM
OP
L
OP
A Me
e
2A M T e
0 ´ e
e
e
P - T M 2 2p - 4p - 2p + 4p PP
M
+
2p P
T M - jn 2p
- jn
- jn
-n
MMN - jn T -n T
P
MN T
T QP
T
T PQ
A
T
LM e
MN - jnW
OP
PQ
- jnW 0 t
0
0
+
-T 2
- jnW 0 t
2A
T2
- jnW 0 t
0
z
0
0
-T 2
- jnW 0 t
- jn
- jnW 0 t
0
0
- jnW 0 t
- jnW 0 t
2
0
0
- jnW 0 t
0
0
0
2
z
- jnW 0 t
2
2
- jnW0 t
0
0
-T 2
0
- jnW 0 t
T/2
0
=
IJ
K
2At
e - jnW 0 t dt
T
A -
0
- jnW 0 t
=
z FGH
T/ 2
-T 2
T/ 2
0
+T 2
0
T /2
- jnW 0 t
0
2
0
0
2p T
2
T
0
0
- jn
0
2
2
2p T
2
T
- jn
2
2p T
2
T
2
2
2
- jn
2p T
T 2
- jn
0
2p T
T 2
2
2
- jn
2
2p T
T 2
2
2
= -
A
A e jnp
A
A e jnp
A e jnp
A e - jnp
+
- 0 +
2 2
2 2
j2np
j2np
j2np
j2np
2n p
2n p
+
=
A
A e - jnp
A e - jnp
A
+
- 0 +
j2np
j2np
2n2 p 2
2n2 p 2
A
A e jnp
A e - jnp
2
2 2
n p
2n p
2n2 p 2
2
0
0
2
2
2
4. 39
Signals & Systems
We know that,
e ± jnp = cosnp ± jsinnp
= + 1 ± j0 = 1
; for even n
= - 1 ± j0 = - 1 ; for odd n.
\ When n is even,
A
A
A
A
A
cn = 2 2 = 2 2 - 2 2 = 0
n p
2n2 p 2
2n2 p 2
n p
n p
\ When n is odd,
A
A
A
A
A
2A
+
+
= 2 2 + 2 2 = 2 2
n2 p 2
2n2 p2
2n2 p2
n p
n p
n p
cn =
2A
2A
= 2 2
(-1)2 p 2
1p
c1 =
2A
12 p 2
c -3 =
2A
2A
= 2 2
( -3)2 p 2
3 p
c3 =
2A
32 p 2
c -5 =
2A
2A
= 2 2
( -5)2 p 2
5 p
c5 =
2A
52 p2
\ c -1 =
and so on
and so on
Exponential Form of Fourier Series
The exponential form of Fourier series is,
+¥
x(t ) =
-1
cn e jnW 0 t =
å
n = -¥
å
¥
cn e jnW 0 t + c 0 +
n = -¥
cn e jnW 0 t
å
n =1
Here cn exist only for odd values of n.
cn e jnW 0 t + c 0 +
å
\ x( t) =
å
cn e jnW 0 t
n = positive
odd integer
n = negative
odd integer
= ..... + c -5 e - j5W 0 t + c -3 e - j3W 0 t + c -1 e - jW 0 t + c 0 + c1 e jW 0 t + c 3 e j3 W 0 t + c 5 e j5 W 0 t + .....
2A
2A
2A
A
2A
e - j5 W 0 t + 2 2 e - j3W 0 t + 2 2 e - jW 0 t +
+ 2 2 e jW 0 t
52 p 2
3 p
1p
2
1p
2A
2A
+ 2 2 e j3W 0 t + 2 2 e j5 W 0 t + .....
3 p
5 p
x(t) = ..... +
=
2A
p2
FG..... +
H
+
2A
p2
1 - j5 W 0 t
1
1
e
+ 2 e - j3 W 0 t + 2 e - jW 0 t
52
3
1
FG 1
H1
2
IJ
K
+
A
2
IJ
K
1 j3 W 0 t
1
e
+ 2 e j5 W 0 t + .....
32
5
e jW 0 t +
Trigonometric Form of Fourier Series
The trigonometric form of Fourier series can be obtained as shown below.
x(t) =
=
LM 1 ee + e j + 1 ee + e j + 1 ee
3
5
N1
2A L 1
1
1
O
2 cos W t +
2 cos3W t +
2 cos5W t + .....P
3
5
p MN 1
Q
4A L cos W t
cos 3W t
cos5W t
O
+
+
+ .....P
3
5
p MN 1
Q
A
2A
+ 2
2
p
A
+
2
A
=
+
2
2
jW 0 t
- jW 0 t
j3 W 0 t
2
0
2
j5 W 0 t
2
2
0
2
0
2
- j3 W 0 t
2
2
0
2
0
2
j
OP
Q
+ e - j5 W 0 t + .....
0
cosq =
e jq + e - jq
2
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 40
x(t)
Example 4.12
Determine the exponential form of the Fourier series
A
representation of the signal shown in fig 4.12.1. Hence determine
the trigonometric form of Fourier series.
0
3T
2T
T
Solution
t
Fig 4.12.1.
To Find Mathematical Equation for x(t)
y - y1
x - x1
=
y1 - y 2
x1 - x 2
Consider the equation of straight line,
Here, y = x(t),
x = t.
x(t ) - x( t1)
t - t1
=
x( t1) - x(t 2 )
t1 - t 2
\ The equation of straight line can be written as,
.....(1)
Consider points P and Q, as shown in fig 1.
x(t)
Coordinates of point-P = [t1, x(t1)] = [0, 0]
Coordinates of point-Q = [t 2, x(t2)] = [T, A]
On substituting the coordinates of points P and Q in equation (1) we get,
x( t) - 0
t - 0
=
0 - A
0 - T
At
T
\ x( t) =
x(t )
t
=
-A
-T
Þ
Q
A
P
0
At
x(t) =
T
Þ
T
t
Fig 1.
; for t = 0 to T
Evaluation of c0
cn =
T
z
1
T
x(t) e - jnW 0 t dt
0
When n = 0, c0 =
=
T
z
z
1
T
0
T
1
T
0
A
=
T2
2
x(t) dt
0
T
z
At
A
dt =
T
T2
LM T
N2
T
z
1
T
x(t) e 0 dt =
t dt =
0
O
- 0P =
Q
A
T2
LM t OP
N2Q
2
T
0
A
2
Evaluation of cn
cn =
=
1
T
T
z
1
T
x(t) e - jnW 0 t dt =
0
A
T2
A
=
T2
LMt e
MN - jnW
LM
MM t e 2p
- jn
NM T
- jnW 0 t
-
0
- jn
2p
t
T
+
z
e
T
z
1´
- jn
n2
2p
t
T
4p 2
T2
At - jnW 0 t
A
e
dt = 2
T
T
0
e - jnW 0 t
dt
- jnW0
OP
PP
PQ
T
A
=
T2
z
t e - jnW 0 t dt
- jnW 0 t
2
0
0
- jn
2p
T
T
- jn
2
2p
T
T
2
2
0
z z z
uv = u v -
u=t
0
OP = A LM t e
PQ T MN - jnW LM
MM T e 2p + e 4p
- jn
n
T
NM T
T
z
T
e - jnW 0 t
( - jnW0 )2
OP
PQ
du v
v = e - jnW 0 t
T
0
e0
- 0 4p2
n2 2
T
OP
PP
PQ
W0 =
2p
T
A
A
A
= e - jn2p + 2 2 e - jn2 p - 2 2
jn2 p
n 4p
n 4p
= -
A
A
A
A
+ 2 2 - 2 2 = jn2p
jn2 p
n 4p
n 4p
e–jn2p = cos n2p – jsin n2p
= 1 – j0 = 1; for integer n
4. 41
Signals & Systems
\ c -1 =
A
j2p
c1 = -
A
j2p
c -2 =
A
j4 p
c2 = -
A
j4 p
c -3 =
A
j6 p
c3 = -
A
j6 p
and so on.
and so on.
Exponential Form of Fourier Series
The exponential form of Fourier series is,
+¥
å
x(t ) =
-1
cn e jnW 0 t =
n = -¥
å
¥
cn e jnW 0 t + c 0 +
n = -¥
å
cn e jnW 0 t
n=1
= ..... c -3 e - j3 W 0 t + c -2 e - j2 W 0 t + c -1 e - jW 0 t + c0 + c1 e jW 0 t + c 2 e j2W 0 t + c 3 e j3W 0 t + .....
A
A
A
A
A
e - j3 W 0 t +
e - j2W 0 t +
e - jW 0 t +
e jW 0 t
2
j6 p
j4p
j2 p
j2 p
= ..... +
=
A
A
e j2W 0 t e j3W 0 t .....
j4p
j6p
LM
MN
e - j3 W 0 t
e - j2 W 0 t
e - jW 0 t
A
..... +
+
+
j2p
3
2
1
OP +
PQ
A
A
2
j2p
LM e
MN 1
jW 0 t
+
OP
PQ
e j2W 0 t
e j3W 0 t
+
+ .....
2
3
Trigonometric Form of Fourier Series
The trigonometric form of Fourier series can be obtained as shown below.
x(t) =
=
LM1F e - e I + 1 F e - e I + 1 F e
JK 3 GH
2j
MN1GH 2j JK 2 GH
A L sin W t
sin 2W t
sin 3 W t
O
+
+
+ .....P
p MN 1
2
3
Q
jW 0 t
A
A
p
2
A
2
- jW 0 t
0
j2 W 0 t
0
- j2 W 0 t
0
j3 W 0 t
- e - j3 W 0 t
2j
I
JK
OP
PQ
+ .....
sin q =
e jq - e - jq
2j
4.9 Fourier Transform
4.9.1 Development of Fourier Transform From Fourier Series
The exponential form of Fourier series representation of a periodic signal is given by,
+¥
x( t ) =
åc e
n
jnW 0 t
.....(4.29)
n = -¥
T2
where, c n =
z
1
x( t ) e - jnW 0 t dt
T -T 2
.....(4.30)
In the Fourier representation using equation (4.29), the cn for various values of n are the spectral
components of the signal x(t), located at intervals of fundamental frequency W0. Therefore the frequency
spectrum is discrete in nature.
The Fourier representation of a signal using equation (4.29) is applicable for periodic signals. For
Fourier representation of non-periodic signals, let us consider that the fundamental period tends to
infinity. When the fundamental period tends to infinity, the fundamental frequency W0 tends to zero or
becomes very small. Since fundamental frequency W0 is very small, the spectral components will lie very
close to each other and so the frequency spectrum becomes continuous.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 42
In order to obtain the Fourier representation of a non-periodic signal let us consider that the
fundamental frequency W0 is very small.
Let, W0 ® DW
On replacing W0 by DW in equation (4.29) we get,
+¥
x( t ) =
åc e
jnDW t
n
n =-¥
On substituting for cn in the above equation from equation (4.30) (by taking t as dummy variable
for integration) we get,
LM 1
å MT
N
n= -¥
z
OP
PQ
T2
+¥
x(t) =
x(t ) e - jnDWt dt e jnDWt
-T 2
We know that, W 0 = 2pF0 =
2p
;
T
\
.....(4.31)
1 W0
=
T 2p
1 DW
=
T 2p
On substituting for 1 from equation (4.32) in equation (4.31) we get,
T
.....(4.32)
Since W0 ® DW ,
LM DW
å M 2p
N
+¥
x(t) =
n= -¥
z
OP
PQ
T2
-T 2
LM
åM
N
+¥
x( t) e - jnDWt dt e jnDWt =
1
2 p n= -¥
z
T2
OP
PQ
x(t ) e - jnDWt dt e jnDW t DW
-T 2
For non-periodic signals, the fundamental period T tends to infinity. On letting limit T tends to
infinity in the above equation we get,
L
å MM
N
+¥
x( t ) = Lt
T® ¥
When T ® ¥ ;
\ x( t ) =
=
1
2p
1
2p
1
2p n =-¥
å
LM
MN
zz
z
+¥
-¥
®
z
z
OP
PQ
T2
x(t ) e - jnDWt dt e jnDW t DW
-T 2
DW ® W
;
+¥
x( t) e - jnWt dt
-¥
OP e
PQ
jnWt
dW
+¥
X( jW) e jnWt dW
.....(4.33)
-¥
z
+¥
where, X( jW) =
-¥
z
Since t is a dummy variable, Let t = t.
+¥
x(t ) e - jnWt dt =
x( t ) e - jnWt dt
.....(4.34)
-¥
The equation (4.34) is Fourier transform of x(t) and equation (4.33) is inverse Fourier transform
of x(t).
Since the equation (4.34) extracts the frequency components of the signal, transformation using
equation (4.34) is also called analysis of the signal x(t). Since the equation (4.33) combines the frequency
components of the signal, the inverse transformation using equation (4.33) is also called synthesis of the
signal x(t).
4. 43
Signals & Systems
Definition of Fourier Transform
Let,
x(t) = Continuous time signal
X(jW) = Fourier transform of x(t)
The Fourier transform of continuous time signal, x(t) is defined as,
z
+¥
X( jW) =
x(t) e - jWt dt
-¥
Also, X(jW) is denoted as F{x(t)} where "F" is the symbol used to denote the Fourier transform
operation.
z
+¥
l q
\ F x( t ) = X( jW) =
x( t ) e - jWt dt
.....(4.35)
-¥
Note : Sometimes the Fourier transform is expressed as a function of cyclic frequency F, rather than
radian frequency W. The Fourier transform as a function of cyclic frequency F, is defined as,
z
+¥
X(jF) =
x(t) e - j2pFt dt
-¥
Condition for Existence of Fourier Transform
The Fourier transform of x(t) exists if it satisfies the following Dirichlet condition.
1. The x(t) be absolutely integrable.
z
+¥
i. e.,
x( t ) dt < ¥
-¥
2. The x(t) should have a finite number of maxima and minima within any finite interval.
3. The x(t) can have a finite number of discontinuities within any interval.
Definition of Inverse Fourier Transform
The inverse Fourier transform of X(jW) is defined as,
l
q
x( t ) = F -1 X( jW) =
1
2p
z
+¥
X( jW ) e jWt dW
.....(4.36)
-¥
The signals x(t) and X(jW) are called Fourier transform pair and can be expressed as shown
below,
x(t)
F
®
F -1
¬
X(jW)
Note : When Fourier transform is expressed as a function of cyclic frequency F, the inverse Fourier
transform is defined as,
x(t) = F
-1
l X(jF)q =
z
+¥
X(jF) e j2 pFt dF
-¥
4.9.2 Frequency Spectrum Using Fourier Transform
The X(jW) is a complex function of W. Hence it can be expressed as a sum of real part and
imaginary part as shown below.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 44
\ X(jW) = Xr(jW) + jXi (jW)
where, Xr(jW) = Real part of X(jW)
Xi(jW) = Imaginary part of X(jW)
The magnitude of X(jW) is called Magnitude spectrum.
X2r (jW ) + X2i (jW )
\ Magnitude spectrum, X(jW ) =
.....(4.37)
(or)
.....(4.38)
X(jW ) X* (jW )
Magnitude spectrum, X(jW ) =
where, X* (jW) = Conjugate of X(jW )
The phase of X(jW) is called Phase spectrum.
\ Phase spectrum, ÐX(jW ) = tan -1
X i ( jW)
X r ( jW)
.....(4.39)
The magnitude spectrum will always have even symmetry and phase spectrum will have odd
symmetry. The magnitude and phase spectrum together called frequency spectrum.
4.10 Properties of Fourier Transform
1. Linearity
l q
l
q
Let, F x1 ( t ) = X1 ( jW) ; F x 2 ( t ) = X2 ( jW)
The linearity property of Fourier transform says that,
F{a1 x1(t) + a2 x2(t)} = a1 X1(jW) + a2 X2(jW)
Proof :
By definition of Fourier transform,
z
+¥
X 1(jW) =
z
+¥
x 1( t) e - jWt dt
and X 2 (jW ) =
-¥
x2 (t ) e -jWt dt
.....(4.40)
-¥
Consider the linear combination a1 x1(t) + a2 x2(t). On taking Fourier transform of this signal we get,
m
r
z
+¥
F a1 x 1 ( t ) + a2 x 2 (t ) =
-¥
z
z
+¥
x 1( t) e - jWt dt + a2
-¥
x 2 ( t) e
2. Time shifting
The time shifting property of Fourier transform says that,
l q = X( jW) then
F lx(t - t )q = e
X( jW)
- jW 0 t
0
- jWt
a 2 x 2 (t ) e
- jW t
dt
-¥
dt
-¥
= a1 X 1 (jW) + a2 X 2 (jW)
If F x(t )
z
+¥
a1 x 1 (t) e - jWt dt +
-¥
+¥
= a1
z
+¥
a1 x 1(t ) + a2 x 2 ( t) e - jWt dt =
Using equation (4.40)
4. 45
Signals & Systems
Proof :
By definition of Fourier transform,
z
z
z
+¥
m r
F x(t) = X (jW ) =
m
x(t ) e -jWt dt
-¥
+¥
r
\ F x(t - t 0 ) =
x(t - t 0 ) e - jWt dt =
-¥
+¥
=
z
.....(4.41)
Let, t – t0 = t
\ t = t + t0
On differentiating
dt = dt
+¥
x( t) e - jW( t + t 0 ) dt
-¥
z
+¥
x( t ) e
-jWt
´ e
- jWt 0
dt = e - jWt 0
x( t) e - jWt dt
Since t is a dummy
variable for ingetration
we can change t to t.
-¥
-¥
z
+¥
= e -jWt 0
x( t) e - jWt dt = e - jWt 0 X(jW )
Using equation (4.41)
-¥
3. Time scaling
The time scaling property of Fourier transform says that,
l q = X( jW) then
1
F lx(at)q =
Xe j
a
If F x(t )
jW
a
Proof :
By definition of Fourier transform,
z
z
+¥
F { x(t) } = X (jW ) =
x(t ) e -jWt dt
-¥
+¥
\ F { x(at) } =
z
+¥
x(at) e -jWt dt =
-¥
x( t) e
e aj
- jW t
-¥
z
dt
a
Put, at = t
;
e j
If "a" happens to be negative then it can be proved that,
F {x(at)} = -
1
X
a
1
X
a
jW
a
z
-¥
e j
for both positive and negative values of "a"
The time reversal property of Fourier transform says that,
If F {x(t)} = X( jW ) then
F {x( - t)} = X(- jW )
Proof :
From time scaling property we know that,
1
X
a
e j
jW
a
Let, a = –1.
\ F { x( - t) } = X( -jW )
eaj
-j W t
dt
a
dt is
similar to the form of Fourier
transform except that W is
\
jW
a
x( t) e
dt =
-¥
+¥
4. Time reversal
F { x(at) } =
The term
replaced by
e j
Hence in general,
F {x(at)} =
z
;
+¥
+¥
- je W jt
1
1
jW
x( t ) e a dt =
X a
a
a
-¥
The above transform is applicable for positive values of "a".
=
t
a
\ t =
x( t) e
d i.
W
a
e j
-j W t
a
dt = X
e j
jW
a
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 46
5. Conjugation
The conjugation property of Fourier transform says that,
If F x(t) = X( jW ) then
l q
F nx ( t )s = X ( - jW)
*
*
Proof :
By definition of Fourier transform,
m r
R* U
\ F Sx (t)V =
T W
z
+¥
F x(t) = X(jW ) =
x(t ) e -jWt dt
-¥
z
LM
MN z
+¥
x*(t) e -jWt dt
+¥
=
+¥
The term,
-¥
LM
MN
= X * b-jWg
x(t) e jWt dt
-¥
*
= X( -jW )
OP*
PQ
z
+¥
=
x(t) e -j( - W )t dt
-¥
z
x(t ) e - j( - W ) t dt
-¥
OP*
PQ
is similar to the form of Fourier transform
except that W is replaced by – W.
z
+¥
\
b g
x(t ) e -j( - W ) tdt = X - jW
-¥
6. Frequency shifting
The frequency shifting property of Fourier transform says that,
l q
F ne
x( t )s = X(j(W - W ))
If F x(t) = X( jW) then
jW 0 t
0
Proof :
By definition of Fourier transform,
z
+¥
m r
F x(t) = X(jW ) =
x(t ) e -jWt dt
-¥
z
z
+¥
{
\F e
jW 0 t
}
x(t) =
e
x(t) e - j( W - W 0 ) t dt = X j W - W0
gi
e
x(t ) e
dt =
-¥
x(t ) e
- jWt
-¥
+¥
=
z
jW 0 t
- jWt
db
+¥
z
x(t ) e -j( W - W 0 )t dt is similar to
dt the form of-¥Fourier transform except that
W is replaced by W – W0.
The term
+¥
jW 0 t
-¥
+¥
\
z
-¥
db
x(t ) e - j( W - W 0 ) tdt = X j W - W 0
gi
7. Time differentiation
The differentiation property of Fourier transform says that,
l q
Rd U
F S x(t)V = jW X(jW )
T dt W
If F x(t) = X( jW) then
Proof :
Consider the definition of Fourier transform of x(t).
z
+¥
x(t ) e
dt
m r
R d U FG d x(t)IJ e dt
\ F S x(t)V =
T dt W H dt K
F x(t) = X(jW ) =
-jWt
.....(4.42)
-¥
z
+¥
-¥
-jWt
z
+¥
=
-¥
e - jWt
FG d x(t)IJ dt
H dt K
4. 47
Signals & Systems
\F
RS d x(t)UV =
T dt W
+¥
e - jWt x(t)
-¥
z
z
+¥
-
( -jW) e - jWt x(t) dt
z z z
uv = u v -
du v
-¥
z
+¥
= e -¥ x( ¥) - e + ¥ x( -¥) + jW
x(-¥) = 0
e–¥ = 0
x(t) e - jWt dt
-¥
z
+¥
= jW
x(t) e -jWt dt = jW X(jW )
Using equation (4.42)
-¥
8. Time integration
The integration property of Fourier transform says that,
l q
R|
U|
F S x(t) dt V =
|T
W|
If F x(t) = X( jW) and X(0) = 0 then
z
t
1
X(jW)
jW
-¥
Proof :
Consider a continuous time signal x(t). Let X(jW) be Fourier transform of x(t). Since integration and differentiation
are inverse operations, x(t) can be expressed as shown below.
d
dt
LM
MN
z
t
x( t) dt
-¥
OP
PQ
= x(t)
On taking Fourier transform of the above equation we get,
R| d L
OU
S| dt MM x( t) dtPP|V| = F mx(t)r
QW
T N
R
U
|
|
jW F S x( t) dt V = F mx(t )r
|T
|W
R|
U| 1 X(jW)
\ F S x( t ) dt V =
|T
|W jW
z
t
F
-¥
z
z
t
Using time differentiation
property of Fourier transform.
-¥
t
m r
F x(t ) = X(jW )
-¥
9. Frequency differentiation
The frequency differentiation property of Fourier transform says that,
If F{x(t)} = X(jW), then
d
F t x( t ) = j
X ( jW )
dW
l
q
Proof :
By definition of Fourier transform,
z
F
GGH z
+¥
b g
x(t ) e -jWt dt
X jW = F {x(t)} =
-¥
On differentiating the above equation with respect to W we get,
d
d
X jW =
dW
dW
b g
z
+¥
-¥
+¥
=
-¥
x( t) e - jWt dt
x(t )
I
JJK
FG d e IJ dt
H dW K
-jWt
Interchanging the order of
integration and differentiation
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
z
+¥
d
X jW =
dW
b g
\
e
1
j
j
x( t) -jt e -jWt dt =
-¥
zb
+¥
g
t x(t ) e - jWt dt
-j = -j ´
-¥
1
F t x(t)
j
m r
d
\ F mt x( t)r = j
X bjW g
dW
=
4. 48
j
1
=
j
j
Using definition of
Fourier transform.
10. Convolution theorem
The convolution theorem of Fourier transform says that, Fourier transform of convolution of two
signals is given by the product of the Fourier transform of the individual signals.
i.e., if F{x1(t)} = X1(jW) and F{x2(t)} = X2(jW) then,
.....(4.43)
F{x1(t) * x2(t)} = X1(jW) X2(jW)
The equation (4.43) is also known as convolution property of Fourier transform.
With reference to chapter-2, section -2.9 we get,
z
+¥
x1 ( t ) * x 2 ( t ) =
.....(4.44)
x1 ( t) x 2 ( t - t) dt
-¥
where t is a dummy variable used for integration.
Proof :
Let x1(t) and x2(t) be two time domain signals. Now, by definition of Fourier transform,
z
z
+¥
b g
X 1 jW = F {x 1 (t ) } =
x 1( t) e - jWt dt
.....(4.45)
x2 (t ) e -jWt dt
.....(4.46)
-¥
+¥
b g
X 2 jW = F {x 2 ( t } =
-¥
Using definition of Fourier transform we can write,
z
z LMMN z
¥
m
r
F x 1 (t) * x 2 (t) =
x 1(t) * x 2 (t e - jWt dt
-¥
¥
=
-¥
OP
PQ
+¥
x 1 ( t) x2 ( t - t) dt e -jWt dt
-¥
Let, e–jWt = ejWt ´ e–jWt ´ e–jWt = e–jWt ´ e–jW(t – t) = e–jWt ´ e–jWM
where, M = t – t and so, dM = dt
.....(4.47)
.....(4.48)
.....(4.49)
Using equations (4.48) and (4.49), the equation (4.47) can be written as,
m
zz
z
+¥ +¥
r
F x 1 (t) * x 2 (t) =
x 1( t) x 2 (M) e -jWt e -jWM dt dM
-¥ -¥
+¥
=
x 1( t ) e -jWt dt ´
-¥
z
+¥
x2 ( M) e - jWM dM
.....(4.50)
-¥
In equation (4.50), t and M are dummy variables used for integration, and so they can be changed to t.
Therefore equation (4.50) can be written as,
m
r
z
+¥
F x 1 (t) * x2 (t) =
z
+¥
x 1 (t) e - jWt dt ´
-¥
-¥
b g b g
= X 1 jW X 2 jW
x2 (t ) e -jWt dt
Using equations
(4.45) and (4.46)
4. 49
Signals & Systems
11. Frequency convolution
l q
l
q
Let, F x1 ( t ) = X1 ( jW) ; F x 2 ( t ) = X2 ( jW) .
The frequency convolution property of Fourier transform says that,
{
}
F x1 ( t) x 2 ( t ) =
z
l = +¥
1
2p
X1 ( jl ) X 2 ( j(W - l )) dl
l = -¥
Proof :
By definition of Fourier transform,
z
z
+¥
m r
F x(t) = X (jW ) =
m
x(t ) e -jWt dt
-¥
t = +¥
r
\ F x 1(t ) x 2 (t ) =
x 1(t ) x 2 (t ) e -jWt dt
.....(4.51)
t = -¥
By the definition of inverse Fourier transform we get,
m
r
x 1 (t) = F -1 X 1 (jW) =
z
z
W = +¥
1
2p
X 1 (jW) ejWt dW =
W = -¥
1
2p
z
l = +¥
X 1 (jl ) e jlt dl
.....(4.52)
l = -¥
Here W is the vairable
used for ingetration.
Let us change W to l.
On substituting for X1(t) from equation (4.52) in equation (4.51) we get,
m
z
t = +¥
r
F x 1 (t ) x2 (t ) =
t = -¥
=
LM 1
MN 2p
z
z
z
l = -¥
X 1 (jl )
l = -¥
l = +¥
1
=
2p
X 1 (jl ) ejlt dl
LM
MN
L
X (jl ) M
MN
l = +¥
1
2p
l = +¥
1
l = -¥
z
z
OP
PQ
x 2 (t ) e -jWt dt
t = +¥
OP
PQ
O
dt P dl
PQ
Interchanging the
order of integration.
x 2 (t ) e -jWt ejlt dt dl
t = -¥
t = +¥
x 2 (t ) e
- j( W - l ) t
t = -¥
t = +¥
The term,
z
t = +¥
X 1 (jl ) X 2 (j(W - l )) dl
2
-j( W - l )t
dt
is similar to the form of Fourier transform
except that W is replaced by W – l.
l = +¥
1
=
2p
z x ( t) e
t = -¥
\
b
g
x 2 ( t) e - j( W - l ) tdt = X 2 j(W - l )
t = -¥
l = -¥
12. Parseval's relation
The Parseval's relation says that,
l q
If F x(t) = X( jW) then
z
z
+¥
+¥
1
| x( t )| dt =
2p
-¥
2
| X( jW)|2 dW
-¥
Proof :
Let x(t) be a continuous time signal and x*(t) be conjugate of x(t).
Now, |x(t)|2 = x(t) x* (t )
On integrating the above equation with respect to t we get,
z
t = +¥
t = -¥
z
t = +¥
| x(t )|2 dt =
x(t ) x* (t) dt
.....(4.53)
t = -¥
By definition of inverse Fourier transform, we can write,
m r
x( t) = F -1 X (jW ) =
1
2p
z
W = +¥
X (jW ) e jWt dW
W = -¥
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 50
On taking conjugate of the above equation we get,
x * (t ) =
1
2p
z
W = +¥
X * (jW ) e -jWt dW
.....(4.54)
W = -¥
Using equation (4.54) the equation (4.53) can be written as,
z
z
t = +¥
t = +¥
| x(t )|2 dt =
t = -¥
x(t )
t = -¥
1
2p
=
1
=
2p
1
2p
=
z
z
LM 1
MN 2p
z
W = +¥
W = -¥
W = +¥
X * (jW )
W = -¥
OP
PQ
O
dt P dW
PQ
X * (jW ) e - jWt dW dt
LM
MN
z
t = +¥
x(t) e - jWt
t = -¥
Interchanging the
order of integration.
Using definition of
Fourier transform.
W = +¥
X * (jW ) X(jW ) dW
X(jW) X * (jW) = X (jW)
2
W = -¥
z
W = +¥
| X (jW )|2 dW
W = -¥
2
Note : The term X(jW ) represents the distribution of energy as function of W and so it is called
energy density spectrum or energy spectral density of the signal x(t).
13. Duality
l q
l q
If F x1 (t) = X1 ( jW) and F x2 (t) = X2 ( jW )
and if x2 ( t ) º X1 ( jW), i.e., x 2 ( t) and X1 ( jW) are similar functions
then X 2 ( jW) º 2 px1 ( - jW), i.e., X2 ( jW ) are 2 px 2 (- jW) are similar functions
Alternatively duality property is expressed as shown below.
If
x2 ( t ) Û X1 ( jW)
then X2 ( jW) Û 2p x1 (- jW )
Proof :
m r
Let, F x 1 (t ) = X 1 (jW )
and
m r
F x 2 ( t) = X 2 (jW )
Let, x 2 ( t) and X 1(jW ) are similar in form.
\ x 2 ( t) = X 1(jW )
.....(4.55)
jW = t
By definition of inverse Fourier transform,
z
+¥
x 1 (t) =
1
X 1(jW) ejWt dW
2 p -¥
z
+¥
\ x 1( -t ) =
1
X 1 (jW ) e -jWt dW
2 p -¥
ze
Replacing t by –t
+¥
\ x 1( -t )
t = jW
=
1
X 1 (jW )
2 p -¥
jW = t
je
-jWt
z
dW
interchanging jW and t
+¥
\ x 1( -jW ) =
1
x 2 (t ) e - jWt dW
2 p -¥
z
Using equation (4.55)
+¥
\ x 2 (t ) e -jWt dW = 2 p x 1 ( -jW )
-¥
\ X 2 (jW ) = 2 p x 1( - jW )
Note : For even function x 1(–jW) = x1(jW).
\ X2(jW) = 2p x1(jW)
Using definition of
Fourier transform
4. 51
Signals & Systems
14. Area under a time domain signal
z
z
+¥
Area under x(t) =
x(t ) dt
-¥
If x(t) and X(jW ) are Fourier transform pair,
+¥
then,
x(t ) dt = X(0)
-¥
where, X(0) = Lt X(jW )
jW ® 0
Proof :
By definition of Fourier transform,
z
+¥
X (jW ) =
x( t) e - jWt dt
-¥
z
+¥
\ X (0 ) = Lt X(jW ) = Lt
jW ® 0
z
jW ® 0
+¥
=
z
+¥
x(t) e 0 dt =
-¥
z
x(t ) e -jWt dt
-¥
x( t) dt
-¥
+¥
\
x( t) dt = X( 0)
-¥
15. Area under a frequency domain signal
z
+¥
Area under X(jW ) =
X(jW) dW
-¥
If x(t) and X(jW) are Fourier transform pair,
z
+¥
then,
X(jW) dW = 2p x(0)
-¥
where, x(0) = Lt x(t)
t®0
Proof :
By definition of inverse Fourier transform,
z
+¥
x( t) =
1
X (jW ) e jWt dW
2 p -¥
z
+¥
\ x( 0) = Lt x(t ) = Lt
t ®0
t® 0
1
X (jW ) e jWt dW
2 p -¥
z
+¥
=
z
+¥
\
-¥
z
+¥
1
1
X (jW ) e 0 dW =
X(jW ) dW
2 p -¥
2 p -¥
X(jW ) dW = 2 p x(0)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 52
Table 4.3 : Summary of Properties of Fourier Transform
Let, F{x(t)} = X(jW) ; F{x1(t)} = X1(jW) ; F{x2(t)} = X2(jW)
Property
Time domain signal
Frequency domain signal
Linearity
a1 x1(t) + a2 x2(t)
a1 X1(jW) + a2 X2(jW)
Time shifting
x(t – t0)
e - jWt 0 X( jW)
Time scaling
x(at)
1
a
Time reversal
x(–t)
X(–jW)
Conjugation
x*(t)
X*(–jW)
e jW 0 t x( t )
X(j(W – W0))
Frequency shifting
d
dt
Time differentiation
z
d i
X
jW
a
x( t )
jW X(jW)
x(t) dt
X(jW)
= p X(0) d(W)
jW
t
Time integration
-¥
d
j dW
X( jW)
t x(t)
Frequency differentiation
z
+¥
Time convolution
x1 ( t )* x 2 ( t) =
x1 (t ) x2 ( t - t )dt
X1(jW) X2(jW)
-¥
Frequency convolution
(or Multiplication)
1
2p
x1(t) x2(t)
z
l =+¥
X1( jl ) X2 ( j( W - l )) dl
l =-¥
*
X(jW ) = X (jW )
X(jW ) = X( - jW ) ; ÐX(jW ) = -ÐX( - jW )
Symmetry of real signals
x(t) is real
Re{X(jW )} = Re{X( - jW )}
Im{X(jW )} =
- Im{X( - jW )}
Real and even
x(t) is real and even
X(jW) are real and even
Real and odd
x(t) is real and odd
X(jW) are imaginary and odd
Duality
If
x2(t) º X1(jW)
[i.e., x2(t) and X1(jW) are similar functions]
then X2(jW) º 2px1(–jW)
z
[i.e.,X2(jW) and 2px1(–jW) are similar functions]
+¥
Area under a frequency
domain signal
X(jW ) dW = 2p x(0)
-¥
z
+¥
Area under a time
domain signal
x( t ) dt = X( 0)
-¥
Energy in frequency domain is,
Energy in time domain is,
z
+¥
Parseval's relation
E=
-¥
2
x(t) dt
z
+¥
E=
z
+¥
1
2
x( t ) dt =
X(jW) dW
2 p -¥
-¥
2
1
2p
z
+¥
-¥
2
X(jW) dW
4. 53
Signals & Systems
4.11 Fourier Transform of Some Important Signals
Fourier Transform of Unit Impulse Signal
The impulse signal is defined as,
z
+¥
x( t ) = d(t) = ¥ ; t = 0 and
d( t ) dt = 1
-¥
= 0 ; t ¹ 0
By definition of Fourier transform,
z
+¥
X( jW) =
z
+¥
x( t ) e - jWt dt =
-¥
d( t ) e - jWt dt
-¥
= 1 ´ e
- jW t
t=0
= 1 ´ e0 = 1
d(t) exists only for t = 0
\ F{x(t)} = 1
The plot of impulse signal and its magnitude spectrum are shown in fig 4.18 and fig 4.19 respectively.
|X(jW)|
|X(jW)| = 1
x(t)
x(t) = d(t)
1
¥
0
0
t
W
Fig 4.19 : Magnitude spectrum of
impulse signal.
Fig 4.18 : Impulse signal.
Fourier Transform of Single Sided Exponential Signal
The single sided exponential signal is defined as,
x( t ) = A e - at ; for t ³ 0
By definition of Fourier transform,
b g
X jW =
z
z
+¥
z
+¥
x( t ) e - jWt dt =
-¥
A e - at e - jWt dt
0
OP
LA e
=
Ae
dt = M
MN -ba + jWg PQ
L A e - A e OP = A
= M
MN -ba + jWg -ba + jWg PQ a + jW
A
\ F nA e u(t)s =
a + jW
+¥
+¥
- ( a + jW ) t
- ( a + jW ) t
0
0
-¥
- at
0
e–¥ = 0
.....(4.56)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 54
The plot of exponential signal and its magnitude spectrum are shown in fig 4.20 and fig 4.21
respectively.
x(t)
|X(jW)|
X( jW) =
x(t) = A e–at u(t)
A
A
a
0
0
t
A
2
a + W2
W
Fig 4.21 : Magnitude spectrum of single
sided exponential signal.
Fig 4.20: Single sided exponential
signal.
Fourier Transform of Double Sided Exponential Signal
The double sided exponential signal is defined as,
x( t ) = A e - a|t | ; for all t
\ x( t ) = A e + at ; for t = - ¥ to 0
= A e- at ; for t = 0 to + ¥
By definition of Fourier transform,
z
z
+¥
X( jW) =
z
z
-¥
-¥
0
¥
=
A e( a - jW ) t dt +
-¥
0
-¥
0
z
+¥
0
x( t ) e - jWt dt =
A e at e - jWt dt +
A e - at e - jWt dt
0
LM A e OP
MN a - jW PQ
Ae
Ae
-b a + jWg
-ba + jWg
( a - jW ) t
-¥
0
Ae
Ae
+
a - jW
a - jW
A(a + jW) + A(a - jW)
2aA
=
= 2
(a - jW) (a + jW )
a + W2
=
n
\ F A e - a |t |
s=
0
LM A e
OP
MN -ba + jWg PQ
- ( a + jW ) t
A e - ( a + jW ) t dt =
+
-¥
¥
0
A
A
+
a - jW
a + jW
e–¥ = 0
2
2
(a+b)(a-b) = a -b j2 = -1
=
2aA
a2 + W2
.....(4.57)
The plot of double sided exponential signal and its magnitude spectrum are shown in fig 4.22 and
fig 4.23 respectively.
x(t)
A
x(t) = A e–a|t|
|X(jW)|
X( jW ) =
2aA
a2 + W 2
2A
a
0
t
Fig 4.22 : Double sided
exponential signal.
0
W
Fig 4.23 : Magnitude spectrum of double
sided exponential signal.
4. 55
Signals & Systems
Fourier Transform of a Constant
Let, x(t) = A, where A is a constant.
If definition of Fourier transform is directly applied, the constant will not satisfy the condition,
z
+¥
x( t ) dt < ¥
-¥
Hence the constant can be viewed as a double sided exponential with limit "a" tends to 0 as shown
below.
Let x1(t) = Double sided exponential signal.
The double sided exponential signal is defined as,
x1(t) = A e–a|t|
i.e,
x1(t) = A e at ; for t = –¥ to 0
= A e–at ; for t = 0 to +¥
\ x(t) = A Lt x1 (t)
a®0
On taking Fourier transform of the above equation we get,
{
}
F {x( t )} = F Lt x1 ( t )
a®0
F {x( t )} = Lt F {x1 ( t )}
a®0
X( jW) = Lt
a®0
= Lt
a®0
F{x(t)} = X(jW)
X1 ( j W )
2aA
W + a2
F{x1(t)} = X1(jW)
Using equation (4.57)
2
The above equation is 0 for all values of W except at W = 0.
At W = 0, the above equation represents an impulse of magnitude "k".
\ X(jW) = k d(W) ; W = 0
=0
; W ¹0
The magnitude "k" can be evaluated as shown below.
z
+¥
k =
-¥
2aA
dW = 2aA
2
W + a2
z
+¥
-¥
1
dW
W + a2
2
LM 1 tan FG W IJ OP = 2aA L 1 tan
MN a
Na H a K Q
L 1 p - 1 FG - p IJ OP = 2aAFG p IJ
= 2aA M
H aK
Na 2 a H 2KQ
\ F lAq = 2pA d(W)
+¥
= 2aA
-1
-1
(+¥) -
-¥
z
dx
1
x
=
tan -1
2
a
a
x + a
2
1
tan -1 ( -¥)
a
OP
Q
= 2 pA
.....(4.58)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 56
The plot of constant and its magnitude spectrum are shown in fig 4.24 and fig 4.25 respectively.
x(t)
|X(jW)|
x(t) = A
|X(jW)| = 2pA d(W)
A
2pA d(W)
0
t
0
Fig 4.24 : Constant.
W
Fig 4.25 : Magnitude spectrum of constant.
Fourier Transform of Signum Function
The signum function is defined as,
x(t) = sgn(t) = 1 ; t > 0
= –1 ; t < 0
The signum function can be expressed as a sum of two one sided exponential signal and taking
limit "a" tends to 0 as shown below.
e - at u( t ) - eat u( - t )
\ sgn(t) = Lt
a®0
\
e - at u( t ) - eat u( - t )
x(t) = sgn(t) = Lt
a®0
By definition of Fourier transform,
z
+¥
X(jW ) =
x( t ) e
dt =
-¥
Lt
a® 0
z
z
z
z
0
- at
- jWt
a® 0
=
a® 0
+ ( a - jW ) t
-¥
- ( a + jW ) t
¥
( a - jW ) t
0
=
Lt
a® 0
= a Lt
® 0
l
q
\ F sgn( t ) =
- jWt
0
- ( a + jW ) t
0
Lt
at
-¥
0
Lt
e - at u( t ) - eat u(- t) e- jWt dt
LM
O
dt P
e e
dt e e
MN
PQ
LM
O
dt P
e
dt e
MN
PQ
LML e
O
MMM -(a + jW) OPP - LMM bea - jWg OPP PP
Q N
Q Q
NN
LM e
OP
e
e
e
+
MN -(a + jW) -(a + jW) a - jW a - jW PQ
LM 1 - 1 OP = 1 + 1 = 2
N a + jW a - jW Q jW jW jW
+¥
=
Lt
a® 0
-¥
+¥
=
z
+¥
- jWt
-¥
2
jW
0
e0 = 1 ; e–¥ = 0
-¥
0
0
-¥
.....(4.59)
4. 57
Signals & Systems
The plot of signum function and its magnitude spectrum are shown in fig 4.26 and fig 4.27
respectively.
x(t) = +1 ; t > 0
x(t)
= –1 ; t < 0
|X(jW)|
X ( jW ) =
2
W
+1
0
t
–1
0
W
Fig 4.27 : Magnitude spectrum of signum function.
Fig 4.26 : Signum function.
Fourier Transform of Unit Step Signal
The unit step signal is defined as,
u(t) = 1 ; t ³ 0
=0 ; t<0
If can be proved that, sgn(t) = 2u(t) –1
\ x(t) = u( t ) =
Þ
u( t ) =
1
1 + sgn( t )
2
1
1 + sgn( t )
2
On taking Fourier transform of the above equation we get,
RS 1 1 + sgn(t) UV
T2
W
R1U R1
U 1 F {1} + 1 F {sgn(t )}
\ X( jW) = F S V + F S sgn(t )V =
2
2
2
TW T
W 2
1
1 L 2 O
1
=
[ 2 p d (W ) ] +
= p d(W) +
M
P
2
2 N jW Q
jW
l q
F x(t) = F
\ F {u(t)} = p d(W) +
Using equations
(4.58) and (4.59)
1
jW
.....(4.60)
The plot of unit step signal and its magnitude spectrum are shown in fig 4.28 and fig 4.29
respectively.
1
X( jW) = pd ( W) +
x(t) = u(t) = 1 ; t ³ 0
=0 ; t<0
x(t)
1
0
t
Fig 4.28 : Unit step signal.
W
pd(W)
|X(jW)|
0
W
Fig 4.29 : Magnitude spectrum of unit step signal.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 58
Fourier Transform of Complex Exponential Signal
The complex exponential signal is defined as,
x ( t ) = A e jW 0 t
= e jW 0 t A
Frequency shifting property.
If F{x(t)} = X(jW) then,
F e jW 0 t x( t ) = X(j(W - W 0 ))
On taking Fourier transform we get,
l q
n
n
s
F x(t) = F e jW 0t A
= F {A} W = W - W
s
Using frequency shifting property
0
= 2p d(W) W = W - W
Using equation (4.46)
0
= 2p d(W - W 0 )
n
F ne
s = 2pA d(W - W )
As = 2pA d (W + W )
\ F A e jW 0 t
Similarly,
- jW 0 t
.....(4.61)
0
.....(4.62)
0
The signal A e - jW 0 t can be represented by a rotating vector of magnitude, "A", in clockwise
direction in a complex plane with an angular speed of W0t as shown in fig.4.30 . The magnitude
spectrum of Ae - jW 0 t is shown in fig. 4.31.
Im
complex plane
Circle of magnitude A
|X(jW)| = 2p A d(W+W0)
|X(jW)|
2pA d(W+W0)
–W0t
Re
A
0
-W 0
Fig 4.30 : Complex exponential signal.
W
Fig 4.31 : Magnitude spectrum of Ae - jW 0 t .
The signal Ae jW0t can be represented by a rotating vector of magnitude "A", in anticlockwise
direction in a complex plane with an angular speed of W0t as shown in fig 4.32. The magnitude spectrum
of Ae + jW 0 t is shown in fig 4.33.
Im
Complex plane
|X(jW)| = 2pA d(W–W0)
|X(jW)|
2pA d(W–W0 )
A
Circle of
magnitude A
W0 t
Re
0
Fig 4.32 : Complex exponential signal.
W0
W
Fig 4.33: Magnitude spectrum of Ae + jW 0t .
4. 59
Signals & Systems
Fourier Transform of Sinusoidal Signal
The sinusoidal signal is defined as,
A jW 0 t
x(t) = A sin W 0 t =
e
- e - jW 0 t
2j
On taking Fourier transform we get,
d
i
sinq =
e jq - e - jq
2j
RS A de - e iUV = A F ne s - F ne s Using equations
(4.61) and (4.62).
W 2j
T2 j
A
Ap
=
dbW - W g - dbW + W g
2 p dbW - W g - 2 p dbW + W g =
2j
j
Ap
dbW - W g - dbW + W g
\ F lA sin W tq =
.....(4.63)
j
l q
jW 0 t
F x(t) = F
- jW 0 t
jW 0 t
0
- jW 0 t
0
0
0
0
0
0
The plot of sinusoidal signal and its spectrum are shown in fig 4.34 and fig 4.35.
X(jW)
x(t)
X(jW) = Ap d(W–W0) – Ap d(W+W0)
A pd(W-W
A
0
)
-W0
0
t
+W0
W
–A
–Apd(W+W0 )
Fig 4.35 : Spectrum of sinusoidal signal.
Fig 4.34 : Sinusoidal signal.
Fourier Transform of Cosinusoidal Signal
The cosinusoidal signal is defined as,
A
x(t) = A cos W 0 t =
e jW 0 t + e - jW 0 t
2
d
cosq =
i
e jq + e - jq
2
On taking Fourier transform we get,
RS A de + e iUV = A F ne s + F ne s Using equations
(4.61) and (4.62).
T2
W 2
A
2 p dbW - W g + 2p dbW + W g = Ap dbW - W g + dbW + W g
=
2
.....(4.64)
\ F lA cos W tq = Ap d(W - W ) + d(W + W )
l q
jW 0 t
F x(t) = F
- jW 0 t
jW 0 t
0
0
- jW 0 t
0
0
0
0
0
The plot of cosinusoidal signal and its magnitude spectrum are shown in fig 4.36 and fig 4.37.
|X(jW)|
x(t)
|X(jW)| = Ap d(W–W0) + Ap d(W+W0)
Ap d(W–W0 )
Ap d(W+W0 )
t
-W 0
Fig 4.36 : Cosinusoidal signal.
0
W0
W
Fig 4.37 : Magnitude spectrum of cosinusoidal signal.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 60
Table 4.4 : Fourier Transform of Standard Signals and their Magnitude Spectrum
x(t)
X(jW) and Magnitude Spectrum
X(jW) = e - jW t 0
x(t) = d(t-t0)
x(t)
|X(jW)|
1
¥
t0
0
t
Shifted impulse signal
0
x(t) = e jW 0t
Complex plane
W
X( jW) = 2 pd(W - W 0 )
Im
Unit Circle
|X(jW)|
2pd(W -W 0 )
W 0t
Re
0
W
W
0
Complex exponential signal
bg
x t = cosW 0 t
X( jW) = p[d (W - W 0 ) + d (W + W 0 )]
x(t)
|X(jW)|
1
p d(W–W0)
p d(W+W0)
t
–1
0
-W 0
Cosinusoidal signal
bg
X( jW) =
x t = sin W 0 t
W
W0
p
d (W - W 0 ) - d (W + W 0 )
j
X(jW)
x(t)
p d(W–W0)
1
–1
Sinusoidal signal
t
-W 0
0
–p d(W+W0)
W
0
W
Signals & Systems
4. 61
Table 4.4 : Continued.....
x(t)
X(jW) and Magnitude Spectrum
b g
bg
X jW = 2pd W
x(t) = 1
x(t)
|X(jW)|
1
2p d(W)
0
t
0
Constant
x(t) = sgn(t) =
x(t)
t
= 1 ;t > 0
|t|
= -1 ; t < 0
W
X( jW) =
2
jW
|X(jW)|
+1
0
t
–1
W
0
Signum signal
x(t) = u(t) = 1 ; t ³ 0
= 0 ; t < 0
X(jW) = pd(W) +
|X(jW)|
pd(W)
x(t)
1
0
1
jW
t
0
Unit step signal
x(t) = e-at u(t)
X(jW) =
x(t)
W
1
a + jW
|X(jW)|
1
1
a
0
t
Decaying exponential signal
x(t) = t e-at u(t)
0
X(jW) =
x(t)
W
1
(a + jW) 2
|X(jW)|
1
a2
0
t
Product of ramp and decaying exponential signal
0
W
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 62
Table 4.4 : Continued.....
x(t)
X(jW) and Magnitude Spectrum
x(t) = e–a|t|
2a
a 2 + W2
X( jW ) =
x(t)
|X(jW)|
1
2
a
0
-
0
t
1
a
1
a
W
Double exponential signal
X( jW ) =
x(t)
a + jW
(a + jW )2 + W20
+1
|X(jW)|
0
t
–1
0
-W 0
Exponentially decaying cosinusoidal signal
W
W
0
WT
2 = T sinc WT
WT
2
2
F I
H K
sin
X( jW) = T
X(jW)
x(t)
T
1
0
T
- 2
t
T
2
- 4Tp - 2Tp
0
2p
T
W
4p
T
Rectangular pulse
t
T
t
T
x( t ) = 1 +
= 1-
F sin WT I
2
X( jW ) = TG
GH WT JJK
; t = - T to 0
; t = 0 to 0
2
|X(jW)|
x(t)
2
T
1
–T
0
T
Triangular pulse
t
- 4p
T
- 2p
T
0
2p
T
4p
T
W
Signals & Systems
4. 63
Table 4.4 : Continued.....
x(t)
x( t ) =
W0
2p
sinc
X(jW) and Magnitude Spectrum
sin
e tj =
W0
2
1
p
F W0 tI
H2 K
e
j e
W
W
X( jW) = u W + 20 - u W - 20
t
x(t)
j
|X(jW)|
W0
2p
W0 =
t
2p
W0
- W2 p0
1
W
-
2 2
t
W
0
2
2
Sinc pulse
x( t ) = e - a
W
0
2p
T
X( jW) =
x(t)
p
a
e
2
- W
e 2a j
|X(jW)|
1
p
a
t
Gaussian pulse
+¥
x( t ) =
W
0
F
å GH
+¥
å dbt - nTg
X( jW) =
n =-¥
2p
2 pn
d WT n= -¥
T
|X(jW)|
x(t)
2p
T
1
–3T
-2T
–T
IJ
K
0
T
2T
3T
t
Impulse train
-6 p
-4 p
-2 p
T
T
T
0
2p
T
4p
6p
T
T
From table 4.4 the following observations are made.
1. The Fourier transform of a Gaussian pulse will be another Gaussian pulse.
2. The Fourier transform of an impulse train will be another impulse train.
3. The Fourier transform of a rectangular pulse will be a sinc pulse and viceversa.
4. The Fourier transform of a triangular pulse will be a squared sinc pulse.
5. The Fourier transform of a constant will be an impulse and vice-versa.
W
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 64
Table 4.5 : Standard Fourier Tranform Pairs
x(t)
X(jW)
d(t)
1
d(t-t0)
e - jW t 0
A
where, A is constant
u(t)
sgn(t)
t u(t)
tn - 1
u(t)
(n - 1)!
where, n = 1, 2, 3, .....
t n u(t)
where, n = 1, 2, 3, .....
bg
2pAd W
pd(W) +
1
jW
2
jW
1
( jW) 2
1
( jW) n
n!
( jW) n+1
e–at u(t)
1
jW + a
t e–at u(t)
1
( jW + a ) 2
Ae
-a t
Ae jW 0t
2Aa
a2 + W 2
2pA d ( W - W 0 )
sin W 0 t
p
d ( W - W 0 ) - d (W + W 0 )
j
cosW 0 t
p[d (W - W 0 ) + d (W + W 0 )]
Signals & Systems
4. 65
4.12 Fourier Transform of a Periodic Signal
Let, x(t)
= Continuous time periodic signal
l q
X( jW) = F x( t ) = Fourier transform of x(t)
The exponential form of Fourier series representation of x(t) is given by,
+¥
å
x( t ) =
c n e jnW 0t
From equation (4.9)
n = -¥
On taking Fourier transform of the above equation we get,
b g
UV = c F ne s
W å
å c 2p dbW - nW g = 2p å c 2p dbW - nW g
l q
X jW = F x( t ) = F
RS
Tå
+¥
+¥
c n e jnW 0 t
jnW 0 t
n
n = -¥
n = -¥
+¥
=
+¥
n
0
n = -¥
n
Using equation (4.61)
0
n = -¥
= ..... + 2pc -3 d (W + 3W 0 ) + 2pc -2 d (W + 2W 0 ) + 2pc -1 d (W + W 0 )
+ 2pc 0 d (W) + 2 pc1 d (W - W 0 ) + 2pc 2 d (W - 2W 0 )
+ 2 p c 3 d (W - 3W 0 ) +.....
.....(4.65)
The magnitude of each term of equation (4.65) represents an impulse, located at its harmonic
frequency in the magnitude spectrum. Hence we can say that the Fourier transform of a periodic continuous
time signal consists of impulses located at the harmonic frequencies of the signal. The magnitude of
each impulse is 2p times the magnitude of Fourier coefficient, i.e., the magnitude of nth impulse is 2p |cn|.
4.13. Analysis of LTI Continuous Time System Using Fourier Transform
4.13.1 Transfer Function of LTI Continuous Time System in Frequency Domain
The ratio of Fourier transform of output and the Fourier transform of input is called transfer
function of LTI continuous time system in frequency domain.
Let, x(t) = Input to the continuous time system
y(t) = Output of the continuous time system
\ X(jW) = Fourier transform of x(t)
Y(jW) = Fourier transform of y(t)
Now , Transfer function =
Y( jW )
X( jW)
.....(4.66)
The transfer function of LTI continuous time system in frequency domain can be obtained from
the differential equation governing the input-output relation of an LTI continuous time system, (refer
chapter-2, equation (2.13)),
dN
d N -1
d N- 2
d
dM
y(t) + a 1 N - 1 y(t) + a 2 N - 2 y(t) + ....... + a N - 1 y(t) + a N y(t) = b0 M x( t )
N
dt
dt
dt
dt
dt
M -1
M-2
d
d
d
+ b1 M - 1 x( t ) + b 2 M - 2 x( t ) +.......+ b M - 1 x( t ) + b M x( t )
dt
dt
dt
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 66
On taking Fourier transform of the above equation and rearranging the resultant equation as a ratio
of Y(jW) and X(jW), the transfer function of LTI continuous time system in frequency domain is obtained.
Impulse Response and Transfer Function
Consider an LTI continuous time system, H shown in fig 4.38. Let x(t) and y(t) be the input and
output of the system respectively.
system
H
x(t)
system
d(t)
y(t)
H
Impulse
input
Fig 4.38.
h(t)
Impulse
response
Fig 4.39.
For a continuous time system H, if the input is impulse signal d(t) as shown in fig 4.39, then the
output is called impulse response, which is denoted by h(t).
The importance of impulse response is that the response for any input to LTI system is given by
convolution of input and impulse response.
Symbolically, the convolution operation is denoted as,
.....(4.67)
y ( t ) = x( t ) * h(t)
where, " * " is the symbol for convolution.
Mathematically, the convolution operation is defined as,
+¥
y( t ) = x( t ) * h(t) =
z
x( t) h(t - t) dt
-¥
where, t is the dummy variable for integration.
Let, H(jW) = Fourier transform of h(t)
X(jW) = Fourier transform of x(t)
Y(jW) = Fourier transform of y(t)
Now, by convolution property of Fourier transform we get,
l
q
F x( t ) * h(t) = X( jW) H(jW)
.....(4.68)
Using equation (4.67), the equation (4.68) can be written as,
l q
F y( t ) = X( jW) H(jW)
\ Y( jW) = X( jW) H( jW)
\ H( jW) =
Y( jW )
X( jW)
.....(4.69)
From equations (4.66) and (4.69) we can say that the transfer function in frequency domain is
given by Fourier transform of impulse response.
\ H(jW) =
Y( jW)
X( jW)
.....(4.70)
Signals & Systems
4. 67
4.13.2 Response of LTI Continuous Time System Using Fourier Transform
Consider the transfer function of LTI continuous time system, H(jW).
Y( jW)
X( jW)
H(jW) =
Now, response in frequency domain, Y( jW) = H(jW) X( jW)
The response function Y(jW) will be a rational function of jW, and so Y(jW) can be expressed as
a ratio of two factorized polynomial in jW as shown below.
Y( jW) =
( jW + z1 ) ( jW + z 2 ) ( jW + z3 ) .....
( jW + p1 ) ( jW + p2 ) ( jW + p3 ) .....
.....(4.71)
By partial fraction expansion technique the equation (4.71) can be expressed as shown below.
k1
k2
k3
.....(4.72)
Y( jW) =
+
+
+ .....
jW + p1
jW + p 2
jW + p 3
where, k1, k2, k3, ..... are residues.
Now the time domain response y(t) can be obtained by taking inverse Fourier transform of equation
(4.72). The inverse Fourier transform of each term in equation (4.72) can be obtained by comparing the
terms with standard Fourier transform pair listed in table 4.5.
1
From table - 4.5, we get, F e - at u( t ) =
.....(4.73)
a + jW
n
s
Using equation (4.73), the inverse Fourier transform of equation (4.72) can be obtained as shown
below.
y( t ) = k 1 e - p1t u( t ) + k 2 e - p2 t u( t ) + k 3 e- p3t u( t ) + .....
.....(4.74)
Since the transfer function is defined with zero initial conditions, the response obtained by using
equation (4.74) is the time domain steady state (or forced) response of the LTI continuous time system.
Note: Only steady state or forced response alone can be computed via frequency domain
4.13.3 Frequency Response of LTI Continuous Time System
The output y(t) of an LTI continuous time system is given by convolution of h(t) and x(t).
+¥
\ y(t) = x(t) * h(t) = h(t) * x(t) =
z
h(t ) x(t - t ) dt
.....(4.75)
-¥
Consider a special class of input (sinusoidal input),
Ae jW t = A(cos W t + j sin W t )
x(t) = A ejWt
where, A = Amplitude ;
.....(4.76)
W = Angular frequency in rad/sec
\ x(t - t ) = Ae jW (t - t )
On substituting for x(t – t) from equation (4.77) in equation (4.75) we get,
+¥
y( t ) =
z
-¥
h( t) Ae jW (t - t ) dt
.....(4.77)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 68
+¥
\ y( t ) =
z
h( t) Ae jW t e - j W t dt
-¥
+¥
= Ae jW t
z
.....(4.78)
h(t ) e - j W t dt
-¥
By the definition of Fourier transform,
+¥
H(jW) = F {h(t)} =
z
Replace t by t.
+¥
h( t ) e
- jW t
dt =
-¥
z
h( t) e
- jW t
dt
.....(4.79)
-¥
Using equations (4.76) and (4.79), the equation (4.78) can be written as,
y(t) = x(t) H(jW)
.....(4.80)
From equation (4.80) we can say that if a complex sinusoidal signal is given as input signal to an
LTI continuous time system, then the output is also a sinusoidal signal of the same frequency modified
by H(jW). Hence H(jW) is called the frequency response of the continuous time system.
Since the H(jW) is a complex function of W, the multiplication of H(jW) with input produces a
change in the amplitude and phase of the input signal, and the modified input signal is the output signal.
Therefore, an LTI system is characterized in the frequency domain by its frequency response.
The function H(jW) is a complex quantity and so it can be expressed as magnitude function and
phase function.
\H(jW) = |H(jW)| ÐH(jW)
where,
|H(jW)| = Magnitude function
ÐH(jW) = Phase function
The sketch of magnitude function and phase function with respect to W will give the frequency
response graphically.
Let, H(jW) = Hr(jW) + jHi(jW)
where, Hr(jW) = Real part of H(jW)
Hi(jW) = Imaginary part of H(jW)
The magnitude function is defined as,
|H(jW)|2 = H(jW) H*(jW) = [Hr(jW) + jHi(jW)] [Hr(jW) - jHi(jW)]
where, H*(jW) is complex conjugate of H(jW)
\ |H( jW)|2 = H 2r ( jW) + H 2i ( jW )
Þ
The phase function is defined as,
ÐH( jW) = Arg[H( jW)] = tan -1
LM H ( jW) OP
N H ( jW) Q
i
r
|H( jW)| =
H 2r ( jW ) + H 2i ( jW )
Signals & Systems
4. 69
From equation (4.70) we can say that the frequency response H(jW) of an LTI continuous time
system is same as transfer function in frequency domain and so, the frequency response is also given by
the ratio of Fourier transform of output to Fourier transform of input.
\ Frequency response, H( jW ) =
Y( jW)
X( jW)
.....(4.81)
Advantages of frequency response analysis
1.The practical testing of systems can be easily carried with available sinusoidal signal generators
and precise measurement equipments .
2. The transfer function of complicated systems can be determined experimentally by frequency
response tests.
3. The design and parameter adjustment is carried out more easily in frequency domain.
4. In frequency domain, the effects of noise disturbance and parameter variations are relatively
easy to visualize and incorporate corrective measures.
5. The frequency response analysis and designs can be extended to certain nonlinear systems.
4.14 Relation Between Fourier and Laplace Transform
Let x(t) be a continuous time signal, defined for all t.
The definition of Laplace tranform of x(t) is,
+¥
L{x(t)} = X(s) =
z
z
x( t ) e - st dt
-¥
On substituting s = s + jW in the above definition of Laplace tranform we get,
+¥
L{x(t)} = X(s) =
x( t ) e - ( s + jW ) t dt
.....(4.82)
-¥
The definition of Fourier transform of x(t) is,
+¥
l q
F x( t ) = X(jW) =
z
x( t ) e - jWt dt
.....(4.83)
-¥
On comparing equations (4.82) and (4.83) we can say that, the Fourier transform of a continuous
time signal, is obtained by letting s = 0 (i.e., s = jW) in the Laplace transform. Summary of this relation
is presented in table 4.6, for causal signals.
\ X( jW ) = X(s) s = jW
Since s = s + jW, we can say that, the Laplace transform is a generalized transform and Fourier
transform is a particular transform when s = jW. Since s = jW represents the points on an imaginary axis
in the s-plane, we can say that, the Fourier transform is an evaluation of the Laplace transform along the
imaginary axis in the s-plane.
Since Fourier transform is evaluation of Laplace transform along imaginary axis, the ROC of X(s)
should include the imaginary axis. For all causal signals, the imaginary axis is included in ROC. Therefore
for all causal signals the Fourier transform exist.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 70
Table 4.6 : Summary of Laplace and Fourier Tranform for Causal Signals
x(t)
X(s)
for t = 0 to ¥
X(jW)
X( jW ) = X(s) s = jW
d(t)
1
1
u(t)
1
s
1
jW
t u(t)
1
s2
1
( jW) 2
tn - 1
u(t)
(n - 1)!
where, n = 1, 2, 3, .....
1
sn
n!
t n u(t)
where, n = 1, 2, 3, .....
s
n+1
1
( jW) n
n!
( jW) n+1
e–at u(t)
1
s + a
1
jW + a
t e–at u(t)
1
(s + a) 2
1
( jW + a) 2
sin W 0 t u(t)
W0
s + W 20
W0
W
= 2 0 2
2
2
( jW) + W 0
W0 - W
cosW 0 t u(t)
s
s2 + W 20
jW
jW
= 2
( jW) 2 + W 20
W0 - W2
bg
sinh W 0 t u t
bg
cosh W 0 t u t )
2
W0
s - W 20
2
s
s - W 20
2
W0
-W
= 2 02
( jW) 2 - W 20
W + W0
jW
- jW
= 2
2
2
( jW) - W 0
W + W 20
bg
W0
(s + a ) 2 + W 20
W0
( jW + a ) 2 + W 20
bg
s + a
(s + a ) 2 + W 20
jW + a
( jW + a ) 2 + W 20
e - at sin W 0 t u t
e - at cosW 0 t u t
Signals & Systems
4. 71
4.15 Solved Problems in Fourier Transform
Example 4.13
Determine the Fourier transform of following continuous time domain signals.
bg
=0
bg
b) x t = e - at cos W0 t u t
a) x(t) = 1– t2 ; for |t| < 1
; for |t| >1
Solution
a) Given that, x(t) = 1 – t2 ; for |t| < 1
\ x(t) = 1 – t2 ; for t = –1 to + 1
By definition of Fourier transform,
+¥
l q
F x(t ) =
z
+1
x(t ) e - jWt dt =
-¥
z
+1
(1 - t 2 ) e - jWt dt =
-1
L e OP
= M
N - jW Q
L e OP
= M
N - jW Q
L e OP
= M
N - jW Q
L e OP
= MN jW Q
L e OP
= MN jW Q
L e OP
= MN jW Q
z
+1
e - jWt dt -
-1
2
- jWt
-1
- jWt 1
z
-1
- jWt 1
1
- jW t
= -
e
jW
z z z
z
uv = u v -
-1
z
- jWt
2 - jWt
- jWt
2
-1
2 - jW t
1
- jWt
z z z
-1
z
- jWt
-1
- jWt
- jW t
1
2
-1
-1
1
2 - jW t
- jWt
- jW t
2
1
3
-1
-1
jW
- jW
- jW
- jW
2
3
jW
2 e jW
2e jW
2
W
jW 3
2 e - jW
2 e - jW
2 e jW
2 e jW
e - jW
e jW
e - jW
e jW
+
+
+
2
3
2
jW
jW
jW
jW
jW
jW 3
W
W
= -
2
W2
= -
2
2
2 cos W +
2 j sin W
jW 3
W2
= -
4 cos W
4 sin W
+
W2
W3
4
W2
du v
1
= -
=
du v
1
-1
- jWt 1
uv = u v -
- jWt
2 - jW t
- jWt 1
z
-1
z
2 - jWt
-1
- jW
t 2 e - jWt dt
L e - 2t e dtOP
- Mt
- jW
Q
N - jW
L t e + 2 t e dt OP
- MjW
N jW
Q
L t e + 2 F t e - 1 ´ e dtI OP
- M JK PQ
- jW
MN jW jW GH - jW
L t e + 2 e- t e + e dtjOP
- MN jW (jW)
Q
L t e - 2 F - t e + e I OP
- M W GH
- jW JK PQ
MN jW
L t e + 2t e + 2e OP
- M W
jW Q
N jW
L e + 2e + 2e + e +
e
- M +
jW
W
jW
jW
N jW
- jWt 1
- jWt
z
de
jW
FG sin W
H W
i
+ e - jW +
- cos W
IJ
K
2
jW 3
de
jW
- e - jW
i
sin q =
e jq - e - j q
2j
OP
Q
cos q =
e jq + e - j q
2
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
bg
4. 72
bg
(b) Given that, x t = e- at cosW0t u t
Since u(t) = 1, for t ³ 0, we can write,
x(t) = e–at cosW0t
;
for t ³ 0
cos q =
By definition of Fourier transform,
+¥
l q
F x(t ) =
z
¥
z
x( t) e - jWt dt =
-¥
=
1
2
1
=
2
=
¥
e - at cos W0 t e - jWt dt =
0
e - at e jW 0 t e - jWt dt +
0
¥
e - (a - j W 0
+ j W) t
0
1
2
1
dt +
2
+ e - jW 0 t
2
0
dt
e - ( a + jW 0
+jW )t
dt
0
¥
- ( a - jW 0 + j W ) t
- jW t
0
LM e
O 1L e
O
MN -(a - jW + jW) PPQ + 2 MMN -(a + jW + jW) PPQ
OP + 1 LM
1 L
e
e
e
M
2 N -(a - jW + jW)
2 N -(a + jW
-(a - jW + jW) Q
1 L
M0 + a - jW1 + jW OPQ + 21 LMN 0 + a + jW1 + jW OPQ
2 N
OP
1 L
1
1
+
M
2 N (a + jW ) - jW
(a + jW) + jW Q
1
2
Ie
JK
¥
- ( a + j W 0 + j W) t
0
0
¥
0
-¥
0
0
0
=
jW 0 t
e - at e - jW 0 t e - jWt dt
0
=
Fe
GH
¥
z
z
-¥
=
e - at
0
¥
z
z
z
e jq + e - j q
2
0
+ j W)
-
e0
-(a + jW0 + jW)
e–¥ = 0 ; e 0 = 1
0
0
0
(a+b)(a-b) = a2-b2 j2 = -1
LM (a + jW) + jW + (a + jW) - jW OP
(a + jW ) + W
N
Q
=
1
2
=
1
2 (a + jW)
a + jW
=
2 (a + jW)2 + W02
(a + jW )2 + W02
0
0
2
2
0
Example 4.14
x(t)
1
Determine the Fourier transform of the rectangular pulse shown in fig 4.14.1.
Solution
The mathematical equation of the rectangular pulse is,
x(t ) = 1
;
–T
for t = - T to + T
0
T
t
Fig 4.14.1.
By definition of Fourier transform,
+T
+¥
l q
F x(t ) =
z
x(t ) e - jWt dt =
z
1 ´ e - jWt dt =
-T
-¥
d
sin W T
sin WT
= 2T
W
WT
= 2 T sinc WT
- jW t
+T
-T
e - jW T
e jWT
1
e jW T - e - j W T
=
=
- jW
- jW
jW
= 2
LM e OP
MN - jW PQ
i = jW1 2j sin WT
OP
Q
sinq =
e jq - e - j q
2j
sin q
= sinc q
q
Signals & Systems
4. 73
Example 4.15
x(t)
1
Determine the Fourier transform of the triangular pulse shown in fig 4.15.1.
Solution
0
–T
T
t
Fig 4.15.1.
The mathematical equation of triangular pulse is,
t
; for t = - T to 0
T
t
= 1 ; for t = 0 to T
T
(Please refer example 4.11 for the mathematical equation of triangular pulse).
x(t ) = 1 +
By definition of Fourier transform,
+¥
z
z
l q
F x(t ) =
0
z
z
x(t ) e - jWt dt =
-¥
-T
0
=
1
T
e - jWt dt +
-T
LM e OP
N - jW Q
- jWt
=
0
-T
1
e - jWt
= jW
= -
= -
==
1
T
+
1
e - jWt
jW
0
-T
0
0
z FGH
t e- jWt dt +
dt +
z
- jWt
1
jWT
1
jWT
-
LMt e
N
LMt e
N
1
1
e0 - e jWT jW
jWT
- jWt
- jWt
t
e - jWt dt
T
1 -
0
z
-
-
LM0 MN
OP
Q
dt O
QP
e - jW t
dt
- jW
1 ´
z
e - jWt
e - jWt
- jW
0
e
- jW
OP
Q
T
1
T
e - jWt dt -
0
LMt e
N - jW
IJ
K
T
- jW t
T
-T
-
-T
FG1 + t IJ e
H TK
z
t e - jWt dt
+
LM e OP
N - jW Q
z
0
0
- jW t
-T
0
T
0
1
e - jW t
jW
-T
0
1
e - jW t
jW
-T
T
0
+
T
0
z z z
uv = u v -
LMt e N - jW
1 L
+
te
jWT NM
1 L
Mt e jWT N
1
T
- jW t
- jWt
- jW t
z
-
e - jWt
dt
- jW
1 ´
z
e - jW t
- jW
du v
e - jWt dt
OP
Q
OP - 1 e- jWT - e0
e
+ T e jWT +
- jW PQ
jW
1 L - jWT
e - jWT
- 0 +
+
Te
M
- jW
jWT MN
OP
Q
OP
Q
T
0
T
0
T
0
jWT
e0
- jW
OP
PQ
1
1
1
e - jWT
e - jWT
1
e jWT
e jWT
e jWT
e - jWT
- 0 +
- 0 +
+
+
+
2
2
2
W
W
j
j
W
W
jW
jW
j
j
TW
TW2
TW
TW
2
TW
2
1
-
TW
2
(e jWT + e - jWT ) =
2
TW
2
-
1
TW2
2 cos WT
cos q =
e jq + e - j q
2
2
(1 - cos W T)
TW2
Alternatively the above result can be expressed as shown below.
=
l q
F x(t ) =
=
2
TW
2
2
TW 2
(1 - cos WT) =
FG 2 sin2 WT IJ
H
2 K
2
TW
2
FG1 H
FG WT IJ IJ
H 2 KK
cos 2
WT
= T 2 2 sin
= T
2
T W
4
F sin WT I 2
G 2 JJ = T FG sinc WT IJ 2
= TG
GH W2T JK H 2 K
2
FG WT IJ
H2K
FG WT IJ 2
H2K
sin2
sin2 q =
1 - cos 2q
2
sin q
= sinc q
q
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 74
Example 4.16
Determine the inverse Fourier transform of the following functions, using partial fraction expansion technique.
a) X(jW) =
3( jW) + 14
( jW )2 + 7( jW ) + 12
jW + 7
( jW + 3)2
b) X( jW ) =
Solution
a) Given that, X(jW ) =
3(jW ) + 14
(jW ) 2 + 7(jW ) + 12
=
3( jW) + 14
( jW + 3) ( jW + 4 )
By partial fraction expansion technique we can write,
X( j W ) =
k1 =
k2 =
3( jW) + 14
k1
k2
=
+
( j W + 3 ) ( j W + 4)
jW + 3
jW + 4
3( jW) + 14
´ ( jW + 3 )
( jW + 3) ( jW + 4)
3( jW) + 14
´ ( jW + 4)
( jW + 3) ( jW + 4)
=
3(-3) + 14
= 5
-3 + 4
=
3(-4) + 14
= -2
-4 + 3
jW = - 3
jW = - 4
5
2
\ X(jW) =
jW + 3
jW + 4
.....(1)
1
jW + a
.....(2)
n
s
We know that, F e - at u( t) =
Using equation (2), the inverse Fourier transform of equation (1) is,
x(t ) = 5 e -3t u(t ) - 2 e -4t u( t)
b) Given that, X(jW ) =
jW + 7
(jW + 3) 2
By partial fraction expansion technique X(jW) can be written as,
\ X(jW) =
k1 =
k2 =
k1
k2
+
jW + 3
( jW + 3)2
jW + 7
( jW + 3)2
= -3 + 7 = 4
jW = - 3
LM
N
jW + 7
d
´ ( jW + 3)2
d( jW) ( jW + 3)2
\ X(jW) =
OP
Q
=
jW = - 3
d
jW + 7
d( jW)
4
1
+
jW + 3
( jW + 3)2
n
s
We know that, F e - at u( t) =
n
´ ( jW + 3)2
s
F t e - at u(t ) =
1
jW + a
1
( jW + a )2
Using equations (4) and (5), the inverse Fourier transform of equation (3) is,
x(t ) = 4t e -3t u( t) + e -3t u( t) = (4t + 1) e -3t u( t)
= 1
jW = - 3
.....(3)
.....(4)
.....(5)
Signals & Systems
4. 75
Example 4.17
Determine the convolution of x1(t) = e–2t u(t) and x2(t) = e–6t u(t), using Fourier transform.
Solution
Let, X1(jW) = Fourier transform of x1(t)
X2(jW) = Fourier transform of x2(t)
By convolution property of Fourier transform,
l
q
F x1(t ) * x 2 ( t) = X 1( jW ) X 2 ( jW)
Let,
X(jW) = X1( jW) X 2 ( jW )
n
s
n
s
= F e -2t u(t) ´ F e -6t u(t)
1
1
´
jW + 2
jW + 6
=
By partial fraction expansion technique X(jW) can be expressed as,
X( j W ) =
k1 =
k2 =
k2
1
k1
=
+
jW + 2
jW + 6
(jW + 2) (jW + 6)
1
´ (jW + 2)
( jW + 2) ( jW + 6)
1
´ (jW + 6)
( jW + 2) ( jW + 6)
\ X(jW) =
=
1
1
=
= 0.25
4
-2 + 6
=
1
1
= = - 0.25
4
-6 + 2
jW = - 2
jW = - 6
0.25
0.25
jW + 2
jW + 6
On taking inverse Fourier transform of the above equation we get,
x(t) = 0.25 e–2t u(t) – 0.25 e–6t u(t)
–2t
= 0.25(e
n
s
F e - at u(t) =
–6t
– e ) u(t)
1
jW + a
Example 4.18
The impulse response of an LTI system is h(t) = 2 e–3 t u(t).
Find the response of the system for the input x(t) = 2e–5 t u(t), using Fourier transform.
Solution
Given that, x(t) = 2 e–5 t u(t).
l q
n
s
\ X ( jW) = F x(t ) = F 2 e -5t u(t) =
2
jW + 5
n
Given that, h(t) = 2 e–3 t u(t).
l q
F e
n
\ H( jW) = F h( t) = F 2 e
-3t
2
u(t) =
jW + 3
s
For LTI system, the response, y(t) = = x(t) * h(t)
On taking Fourier transform of equation (3) we get,
F {y(t)} = F {x(t) * h(t)}
.....(1)
- at
1
u(t) =
jW + a
s
.....(2)
.....(3)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 76
Let, F{y(t)} = Y(jW).
\ Y(jW) = F {x(t) * h(t)}
= X(jW) H(jW)
2
2
4
=
=
´
jW + 5
jW + 3
( jW + 5) (jW + 3)
Using convolution property
of Fourier transform.
Using equations (1) and (2)
By partial fraction expansion technique, the above equation can be written as,
4
k1
k2
=
+
( jW + 5) (jW + 3)
jW + 5
jW + 3
Y ( j W) =
4
´ (jW + 5)
( jW + 5) (jW + 3)
k1 =
k2 =
4
´ (jW + 3)
( jW + 5) (jW + 3)
=
4
= -2
-5 + 3
=
4
= 2
-3 + 5
jW = - 5
jW = - 3
2
2
\ Y ( jW) = +
jW + 5
jW + 3
On taking inverse Fourier transform of Y(jW) we get y(t).
l
RS
T
q
y( t) = F -1 Y ( jW) = F -1 -
2
2
+
jW + 5
jW + 3
UV
W
d
i
= - 2 e -5 t u(t) + 2 e -3 t u(t) = 2 e -3 t - e -5 t u(t)
Example 4.19
x(t)
1
Determine the Fourier transform of the
periodic pulse function shown in fig 4.19.1.
Solution
–T
–2T
The mathematical equation for one
period of the periodic pulse function is,
–a 0
- T
2
a
T
2
T
t
2T
Fig 4.19.1.
x(t ) = 1 ; t = - a to + a
T
T
= 0 ; t = - to - a and t = a to
2
2
The Fourier coefficient cn is given by,
1
cn =
T
=
=
1
T
+ T /2
z
x(t ) e
- jnW 0 t
-T /2
LM e
MN - jnW
- jnW0 a
1
dt =
T
jnW0 a
e
- jnW0
0
OP
PQ =
+a
z
e
- jnW 0 t
-a
1 2
T nW0
1
dt =
T
LM e
MN
jnW0 a
LM e OP
MN - jnW PQ
OP
- e
2j
PQ
- jnW 0 t
0
+a
-a
1
=
T
The exponential Fourier series representation of the periodic pulse function is,
+¥
åc
n
e jnW 0 t
n = -¥
On taking Fourier transform of the above equation we get,
l q
F x( t)
|RS å c
|T
+¥
= F
n = -¥
n
e jnW 0 t
|UV
|W
- jnW0t
0
OP
PQ
+a
W0 =
2p
T
-a
- jnW0 a
1 2 T
1
sin(nW0a ) =
sin(anW0 )
T n 2p
np
x(t ) =
LM e
MN - jnW
sinq =
e jq - e - j q
2j
....(1)
Signals & Systems
4. 77
R|S c
|T å
U|V
|W å c F oe t
+¥
\ X(jW) = F
+¥
e jnW 0 t =
n
n = -¥
l q
jnW 0 t
n
F x( t) = X( jW)
n = -¥
o
F e jnW 0 t
+¥
åc
=
n
2 p d(W - nW0 )
n = -¥
+¥
=
å
n = -¥
+¥
=
å
n = -¥
å
2 sin(anW 0 )
d(W - nW0 )
n
2aW 0
sin q
= sinc q
q
F sin anW I d(W - nW )
GH anW JK
0
+¥
=
0
0
n = -¥
å 2aW
0
sin c (anW0 ) d(W - nW0 )
n = -¥
Example 4.20
x(t)
A
Determine the Fourier transform of the periodic
impulse function shown in fig 4.20.1.
–2T
Solution
1
T
+T/2
z
x( t) e - jnW 0 t dt =
-T/2
1
T
T
2T
t
Fig 4.20.1.
W0 =
+T/2
z
0
–T
The mathematical equation for one period of the
periodic impulse function is,
T
T
x(t ) = A d(t) ; for t = to +
2
2
The Fourier coefficient cn is given by,
cn =
0
Substituting for c n
from equation (1).
1
sin(anW 0 ) 2 p d(W - nW0 )
np
+¥
=
t = 2p d(W - nW )
A d(t ) e - jnW 0 t dt =
-T/2
A - jnW 0 t
e
T
t = 0
=
A
T
2p
T
....(1)
The Exponential Fourier series representation of the periodic impulse train is,
+¥
åc
x(t ) =
n
e jnW 0 t
n = -¥
On taking Fourier transform of the above equation we get,
|RS å c
|T
+¥
l q
F x(t ) = F
n
n = -¥
e jnW 0 t
|UV
|W
+¥
\ X(jW) =
å c F oe t
n
jnW 0 t
l q
F x( t) = X( jW)
n = -¥
+¥
=
å
o
F e jnW 0 t
c n 2p d(W - nW0 )
t = 2p d(W - nW )
0
n = -¥
+¥
=
On substituting for
cn from equation (1)
+¥
å
A
2 p d(W - nW 0 ) =
T
n =
å AW
n = -¥
0
d(W - nW 0 )
-¥
The magnitude spectrum of X(jW) is shown in fig 1, which is also a periodic impulse function of W.
|X(jW)|
AW0
–3W 0
–2W 0
-W0
0
Fig 1.
W0
2W 0
3W 0
W
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 78
Example 4.21
Find the Fourier transform and sketch the magnitude and phase spectrum for the signal, x(t) = e–at u(t).
Solution
Given that, x(t) = e–at u(t).
The Fourier transform of x(t) is, (refer section 4.11 for Fourier transform of e–at u(t)).
l q n
s
X( jW) = F x( t) = F e - at u(t) =
1
a + jW
=
1
a - jW
a - jW
=
´
a + jW a - jW (a + jW) (a - jW )
=
a - jW
a
W
=
-j 2
a 2 + W2 a 2 + W2
a + W2
The X(jW) is calculated for a = 0.5 and a = 1.0 and tabluted in table 1and table 2 respectively. Using the values listed
in table 1and table 2, the magnitude and phase spectrum are sketched as shown in fig 1 and fig 2 respectively.
Note : The function X(jW) is calculated using complex mode of calculator, the magnitude and phase are calculated
using rectangular to polar conversion technique.
Table 1 : Frequency Spectrum for a = 0.5
X(jW)
W
|X(jW)|
ÐX(jW)
in rad
–8
0.0080 + j 0.12 = 0.12 Ð1.506
= 0.12 Ð0.48p
0.12
0.48p
–6
0.014 + j 0.167 = 0.166 Ð1.4866
= 0.166 Ð0.473p
0.166
0.473p
–4
0.03 + j 0.246
= 0.248 Ð1.45
= 0.248 Ð0.46p
0.248
0.46p
–3
0.054 + j 0.324 = 0.328 Ð1.40
= 0.328 Ð0.45p
0.328
0.45p
–2
0.118 + j 0.47
= 0.485 Ð1.325
= 0.485 Ð0.422p
0.485
0.422 p
–1
0.4 + j 0.8
= 0.89 Ð1.11
= 0.89 Ð0.353p
0.89
0.353p
0
2+ j 0
= 2 Ð0
= 2 Ð0
2.0
0
1
0.4 – j 0.8
= 0.89 Ж1.11
=0.89 Ж0.353p
0.89
–0.353p
2
0.118 – j 0.47
= 0.485 Ж1.325
= 0.485 Ж0.422p
0.485
–0.422p
3
0.054 – j 0.324 = 0.328 Ж1.40
= 0.282 Ж0.45 p
0.282
–0.45p
4
0.03 – j 0.246
= 0.248 Ж1.45
= 0.248 Ж0.46 p
0.248
–0.46p
6
0.014 – j 0.167 = 0.166 Ж1.4866 = 0.078 Ж 0.473p
0.166
–0.473p
8
0.0080 – j 0.12 = 0.12 Ж1.506
0.12
–0.48 p
= 0.12 Ж0.48 p
Signals & Systems
4. 79
Table 2 : Frequency Spectrum for a = 1
X(jW)
W
|X(jW)|
ÐX(jW)
in rad
–8
0.015 + j 0.123 = 0.124 Ð1.45
= 0.124 Ð0.461p
0.124
0.461p
–6
0.03 + j 0.162
= 0.165 Ð1.39
= 0.165 Ð0.442 p
0.165
0.442 p
–4
0.059+ j 0.24
= 0.25 Ð1.33
= 0.25 Ð0.423p
0.25
0.423p
–3
0.1 + j 0.3
= 0.316 Ð1.25
= 0.316 Ð0.398p
0.316
0.398p
–2
0.2 + j 0.4
= 0.45 Ð1.11
= 0.45 Ð0.353p
0.45
0.353p
–1
0.5 + j 0.5
= 0.707 Ð0.785 = 0.707 Ð0.25p
0.707
0.25p
0
1+ j 0
= 1 Ð0
1.0
0
1
0.5 – j 0.5
= 0.707 Ж0.785 = 0.707 Ж0.25p
0.707
–0.25 p
2
0.2 – j 0.4
= 0.45 Ð –1.11 = 0.45 Ж0.353p
0.45
–0.353 p
3
0.1 – j 0.3
= 0.316 Ж1.25 = 0.316 Ж0.398p
0.316
–0.398p
4
0.059 – j 0.24
= 0.25 Ж1.33 = 0.25 Ж0.423p
0.25
–0.423p
6
0.03 – j 0.162
= 0.165 Ж1.39 = 0.165 Ж0.442p
0.165
–0.442p
8
0.015 – j 0.123 = 0.124 Ж1.45 = 0.124 Ж0.461p
0.124
–0.461p
= 1 Ð0
X( jW)
2.0
1.8
1.6
1.4
¬ a = 0.5
1.2
1.0
¬ a =1.0
0.8
0.6
0.4
0.2
–W
–9
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
Fig 1 : Magnitude spectrum of X(jW).
5
6
7
8
9
W
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 80
ÐX(jW)
0.5 p
a = 0.5
0.4 p
a = 1.0
0.3 p
0.2 p
0.1 p
–W
–9
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
5
4
6
7
8
9
W
–0.1 p
–0.2 p
–0.3 p
–0.4 p
–0.5 p
a = 1.0
a = 0.5
–ÐX(jW)
Fig 2: Phase spectrum of X(jW).
4.16 Summary of Important Concepts
1. The Fourier series is frequency domain representation of periodic signals.
2. The Fourier series exists only if Dirichlet’s conditions are satisfied.
3. The signals with negative frequency are required for mathematical representation of real signals in terms of
complex exponential signals.
4. In exponential form of Fourier series, cn represents the magnitude of n th harmonic component.
5. In exponential form of Fourier series, Ðcn represents the phase of the nth harmonic component.
6. The plot of harmonic magnitude/phase versus harmonic number “n” (or harmonic frequency) is
called frequency spectrum.
7. The frequency spectrum obtained from Fourier series is also called line spectrum.
8. The plot of magnitude versus n (or n W 0 ) is called magnitude (line) spectrum.
4. 81
Signals & Systems
9. The plot of phase versus n (or n W 0 ) is called phase (line) spectrum.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
For signals with even symmetry, the Fourier coefficients bn are zero.
For signals with odd symmetry, the Fourier coefficients a0 and an are zero.
For signals with half wave symmetry, the Fourier series will consists of odd harmonic terms alone.
A signal with half wave symmetry, if in addition has even/odd symmetry then it is said to have quarter wave
symmetry.
For signals with quarter wave symmetry, the Fourier series will consists of either odd harmonics of
sine terms or odd harmonics of cosine terms.
The Fourier transform has been developed from Fourier series by considering the fundamental
period T as infinity.
The Fourier transform is used to obtain the frequency domain representation of non-periodic as well as
periodic signals.
The Fourier transform of a signal exists only if the signal is absolutely integral.
The Fourier transform of a signal is also called analysis of the signal.
The inverse Fourier transform of a signal is also called synthesis of the signal.
The frequency spectrum of non-periodic signals will be continuous, whereas frequency spectrum of periodic
signals will be discrete.
The magnitude spectrum will have even symmetry and phase spectrum will have odd symmetry.
The Fourier transform of a periodic continuous time signal will have impulses located at the harmonic
frequencies of the signal.
The ratio of Fourier transform of output and input signal of a system is called transfer function in
frequency domain.
The Fourier transform of impulse response gives the frequency domain transfer function.
The Fourier transform is evaluation of Laplace transform along imaginary axis in s-plane.
4.17 Short Questions and Answers
Q4.1
What is the value of x(t) at t = t0 in the waveform
shown in fig Q4.1.
Solution:
x(t)
4
In the waveform shown in fig Q4.1, t = t 0 is a point of
discontinuity. If t = t0 is a point of discontinuity, then the
value of x(t) at t = t 0 is given by,
x(t 0 ) =
Here,
Q4.2
x(t 0+ )
x( t 0+ )
t0
0
Fig Q4.1.
t
+ x( t 0- )
2
x( t 0- ) = 0
= 4,
\ x(t 0 ) =
4+0
=2
2
Find the constant component of the periodic pulse signal
shown in fig Q4.2.
x(t)
2
Solution:
The constant component of any periodic time domain signal is
a0/2, where,
2
a0 =
T
z bg
T
x t dt
0
bg
;
\ a0 =
z
Here, x t = 2
2
10
for t = 0 to 1 ms, and T = 10 ms
1
2 dt =
0
\ Constant component =
2
2t
10
a0
2
=
1
0
0.4
2
=
2
10
= 0.2
2 - 0 =
4
10
= 0.4
0
1 ms
10 ms
Fig Q4.2.
t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
Q4.3
4. 82
Determine the magnitude of the fundamental frequency component of the periodic pulse signal
shown in fig Q4.3.
x(t)
Solution:
10
The Fourier coefficient of nth harmonic component is given by,
z bg
z bg
T
2p
1
- jnW 0 t
0
xt e
dt ; where W 0 =
T0
T
The Fourier coefficient of fundamental component is obtained when n = 1.
cn =
T
\ c1 =
1
xt e
T0
\ c1 =
z
=
1
10
20 0
5
- jp
F
GH
t
20
Fig Q4.3.
- j2 pt
T dt
Here, x(t) = 10 ; for t = 0 to 2,
2
2
- j2 p
t
e 20 dt
and T = 20
=
1
2
I
JK
p
-j
e 5
5
- jp
-1 =
z
2
-jpt
e 10 dt
0
FG cos p
H 5
LM
e
1 M -jp
=
M
2 M
MM 10
N
I
- 1J
K
LM OP
1 Me
P
=
2 M -jp P
MN 10 PQ
- jpt
10
- j sin
b
p
5
- jp 2
10
2
OP
PP
10 P
PP
Q
0
e
-jp
0
g
= j 1.5915 0.8090 - j 0.5878 - 1 = 0.9355 - j 0.304 = 0.9836 Ð - 0.314 rad
Magnitude of fundamental component, |c1| = 0.9836
Q 4.4
What is Magnitude and Phase spectrum?
Let x(t) be a time domain signal and X(jW) be the Fourier transform of x(t).
The X(jW) is a complex function of W and so it can be expressed as,
b g
b g b g
X jW = X jW ÐX jW
where, |X(jW)| = Magnitude function or Magnitude spectrum.
ÐX(jW) = Phase function or phase spectrum.
Q 4.5
The Fourier transform of the signal shown in fig Q4.5.1 is,
b g
1
X jW =
W
2
de
jW
- jW e
jW
i
x1(t)
x(t)
-1 .
1
x2(t)
1
1
Using the properties of Fourier transform find the
Fourier transform of the signal shown in fig Q4.5.2
–1
and Q4.5.3.
1 t
0
Fig Q4.5.2.
0 t
Fig Q4.5.1.
Solution:
–1
0 t
Fig Q4.5.3.
The signal shown in fig Q4.5.2 is the folded version of the signal shown in fig Q4.5.1.
i.e., x1( t) = x( -t)
m b gr = XbjWg = W1 de
Given that , F x t
2
jW
i
- jWe jW - 1
Using time reversal property of Fourier transform we can write,
m b gr = X bjWg = Xb- jWg = W1 de
F x1 t
1
2
- jW
i
+ je- jW - 1
The signal shown in fig Q4.5.3 is the shifted version of the signal shown in fig Q4.5.2.
i.e., x2(t) = x1(t – t0) ; and t0 = –1
Using time shifting property of Fourier transform we can write,
m b gr = X b jWg = e
F x2 t
2
jWt 0
b g
X1 jW = e - jW
1
e - jW - jW
e - jW + je - jW - 1 =
e + je - jW - 1
2
W
W2
d
i
d
i
4. 83
Q 4.6
Signals & Systems
If Fourier transform of e – t u(t) is
1
1
then find the Fourier transform of
using duality
1 + jW
1+t
property.
Solution:
Given, x1(t) = e–t u(t) and X1( jW) =
x 2 ( t) =
1
1 + jW
1
1+ t
Here, x2(t) and X1(jW) are similar functions.
\ By duality property,
e
X 2 ( jW) = 2p x1( t)
Q 4.7
t =- jW
j = 2p FH e
u(t)
t = - jW
IK = 2p e
jW
u( - jW)
Find the Fourier constant a0 for the continuous time signal defined as,
T
x(t) = Kt
; 0£t£
2
T
= K(T - t) ;
£ t £T
2
Solution:
Given that,
x(t) = Kt
; 0£t£
2
a0 =
T
=
z
T
0
2
x(t ) dt =
T
T
2
z
Kt dt +
0
T
2
2
T
z
T
K(T - t) dt
T
2
LM OP + 2 LMKTt - Kt OP = 2 LM KT
2 Q
TN 8
N Q TN
2 L KT O 2 L KT O
M P + M P = 4 KT = KT2
TN 8 Q TN 8 Q T 8
2 Kt 2
T 2
2
T
OP
Q
2
-0 +
T
2
0
2
=
T
2
T
£t£T
2
= K(T - t) ;
Q 4.8
-t
2
LM
N
2
KT 2 KT 2 KT 2
KT 2 +
2
2
8
T
OP
Q
2
A periodic signal x(t) is defined as x(t) = (1 – t) 2 ; 0 £ t £ T . Find the Fourier coefficient bn.
Solution
Given that,
x(t) = (1 – t)2 ; 0 £ t £ T.
Now, x(–t) = (1–(–t)2) = 1– t2.
Here x(t) = x(–t) and so the given signal is even signal.
For even signals, bn = 0.
Q 4.9
A continuous time signal varies exponentially in the interval 0 to T. Find the Fourier constant
signal.
Solution
Given that,
x(t) = et ; 0 £ t £ T.
z
T
Now, a 0 =
\
z
T
2
2 t
2
x( t) dt =
e dt = e t
T0
T0
T
a0 eT - 1
=
2
T
T
0
=
2 T
2 T
e - e0 =
e -1
T
T
d
i
a0
of the
2
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
Q 4.10
Find the Fourier transform of the signal e
Solution
x(t) = e
-3 t
-3 t
4. 84
u(t) .
= e -3t for t > 0
= e3t for t < 0
The Fourier transform of x(t) is,
z
z
x( t) e - jWt dt =
-¥
( 3 - jW )t
0
-¥
0
-¥
OP = LM e - e OP + LM- e + e OP
LM e
W
+
(
)
3
j
Q N 3 - jW 3 - jW Q N 3 + jW 3 + jW Q
N
- (3 + jW) t
+
z
¥
e( 3 - jW)t dt + e -( 3 + jW) t dt
¥
0
-¥
0
-¥
e–¥ = 0 ; e0 = 1
0
1
3 + jW + 3 - jW
1
6
+
=
= 2
3 - jW 3 + jW
3 2 + W2
3 + W2
=
Q 4.11
0
z
0
¥
e 3t e - jWt dt + e -3t e - jWt dt =
-¥
LM e OP
N 3 - jW Q
=
z
0
¥
X(jW) =
(a+b)(a-b) = a2-b2 j2 = -1
Determine the Fourier transform of x(t) using time shifting property, x(t) = e
Solution
l q {
= F {e
} + F {e
-3 t - t 0
-3 t - t 0
+e
3 t +t0
3 t +t0
}
{
F e
3´2
3´2
6
´ e - jWt 0 + 2
´ e jWt 0 = 2
e - jWt 0 + e jWt 0
32 + W 2
3 + W2
3 + W2
12 cos W t 0
=
3 2 + W2
e
z
-a t - t 0
j
.
-a t
}= a
cos q =
2
2
2
2
e - jWt 0
+ W2
e jq + e - jq
2
x(t)
2
+¥
For the signal shown in fig Q4.12. Find a) X(j0) b)
3 t + t0
By time shifting property,
}
=
+e
{ } = a 2+aW
F e
X(jW) = F x(t) = F e
Q 4.12
-3 t - t0
X(jW ) dW .
-¥
Solution
1
z
+¥
a) X(j0) =
x(t) dt = Area of the signal
-¥
–1
0
1
2
Fig Q4.12.
1
= Area of rectangle – Area of triangle = 4 ´ 2 - ´ 2 ´ 1 = 8 - 1 = 7
2
b) By definition of inverse Fourier transform,
z
z
3
t
+¥
x(t ) =
1
X( jW) e jWt dW
2 p -¥
On letting t = 0 in the above equation we get,
x(0) =
1
2p
e0 = 1
+¥
X( jW) e 0 dW
from fig Q4.12 , x(0) = 2
-¥
z
+¥
\ X ( jW) dW = 2 p ´ x(0) = 2 p ´ 2 = 4 p
-¥
Q 4.13
Find energy in frequency domain for the signal shown in
fig Q4.13.1.
x(t)
3
Solution
2
The square of the given signal is shown in fig Q4.13.2.
1
0
1
3
2
Fig Q4.13.1.
4 t
4. 85
Signals & Systems
The energy E in frequency domain is given by,
x2(t)
9
z
+¥
E=
1
2
X( jW) dW
2p -¥
z
8
7
+¥
=
2
x( t) dt
Using Parseval’s relation
6
-¥
5
= Area of x 2 (t) = Area of rectangle - Area of triangle
1
= 9 ´ 4 - ´ 2 ´ 5 = 36 - 5 = 31 joules
2
Q4.14
4
3
Determine which of the following real signals shown in fig Q4.14
have Fourier transforms that satisfy the following conditions.
2
a) Re{X(jW )} = 0
1
b) Im{X(jW )} = 0
z
0
+¥
c)
3
2
4 t
Fig Q4.13.2.
X(jW ) dW = 0
x3(t)
-¥
x2(t)
x1(t)
1
x4(t)
t
t
t
t
Fig a.
Fig b.
Fig c.
Fig d.
Fig Q4.14.
Solution
a) Given that Re{X(jW)} = 0
In order to satisfy the given condition, the time domain signal should be real and odd.
\ The signals shown in fig c and d will satisfy the condition, Re{X(jW)} = 0 .
b) Given that Im{jW} = 0
In order to satisfy the given condition, the time domain signal should be real and even.
\ The signals shown in fig a and b will satisfy the condition, Im{X(jW} = 0
z
+¥
c) Given that X(jW) dW = 0
-¥
z
z
+¥
We know that,
X(jW)dW = 2p x(0)
-¥
+¥
\ If x(0) = 0, then
X(jW)dW = 0
-¥
z
+¥
Here the signals shown in fig b, c and d has x(0) = 0, and so they will satisfy the condition
X(jW)dW = 0
-¥
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
Q 4.15
4. 86
For the signal shown in fig Q4.15.1., find energy in frequency domain. x(t)
2
Solution
The square of the given signal is shown in fig Q4.15.2.
1
The energy E in frequency domain is given by,
z
0
3 t
1
2
Fig Q4.15.1.
0
1
3 t
2
Fig Q4.15.2.
+¥
E=
1
2
X( jW) dW
2p -¥
z
x(t)
4
+¥
=
2
Using Parseval’s relation
x( t ) dt
-¥
3
= Area of x 2 (t ) = Area of triangle in fig Q4.15.2.
=
Q 4.16
2
1
´ 2 ´ 4 = 4 joules
2
1
The impulse response of a system in frequency domain is,
H(jW ) = + j
; W <0
= -j ; W >0
Find the response for the input, x(t) = cos t.
Solution
Given that, x(t) = cos t
l q l q
\ X( jW) = F x(t ) = F cos t = p d(W + 1) + d(W - 1)
\ X( jW) = p d(W + 1) ; W < 0
= p d(W - 1) ; W > 0
Let y(t) be response and h(t) be impulse response in time domain.
We know that, y(t) = x(t) * h(t)
On taking Fourier transform of above equation we get,
l
q
Y ( jW) = F x( t) * h(t)
= X( jW) H( jW)
Using convolution property
of Fourier transform
\ Y( jW) = j ´ p d(W + 1) ; W < 0
= - j ´ p d(W - 1) ; W > 0
\ Y( jW) = j p d(W + 1) - jp d(W - 1) =
l
p
d(W - 1) - d(W + 1)
j
q
Response in time domain, y(t) = F -1 Y ( jW) = F -1
Q 4.17
RS p d(W - 1) - d(W + 1) UV = sin t
Tj
W
Find the output of LTI system with impulse response h(t) =
a f
a f
a f
a f
sin 5 t + 1
sin 5 t - 1
and input x(t) =
p t +1
p t -1
Solution
l q RST sinp (5t(-t -1)1) UVW = F RST sinp t5t UVW e-jW
Using time shifting property
H( jW) = F h( t) = F
= 1 ´ e - jW ; W < 5 = e - jW ; W < 5
F
Similarly, X( jW) = e + jW ; W < 5
0
0
0
We know that,
Response, y(t) = x(t) * h(t)
RS sin W t UV = F RS W
T pt W T p
RW
= FS
Tp
.....(1)
sin W0 t
W0t
UV
W
sin c W0 t
= 1 ; W < W0
UV
W
4. 87
Signals & Systems
On taking Fourier transform of equation (1) we get,
l
q
Y(jW) = F x(t ) * h(t) = X ( jW) H( jW)
=e
jW
e
- jW
l
; W <5
=1 ; W <5
5
5 sin 5t sin 5t
sin c 5t =
=
p
p 5t
pt
q
\ y( t) = F -1 Y(jW) =
Q 4.18
Using convolution property
Find the inverse Fourier transform of, X(jW ) =
Solution
Let,
X ( jW) =
1
b4 + jWg
2
where, X1( jW) =
1
b4 + jW g
2
. Using convolution property.
1
1
´
= X 1( jW) ´ X 2 ( jW)
4 + jW 4 + jW
=
1
1
and X 2 ( jW) =
4 + jW
4 + jW
l
q
\ x1(t ) = F -1 X 1( jW) = F -1
RS 1 UV = e
T 4 + jW W
-4t
u(t)
x 2 (t ) = e -4t u(t)
By time convolution property,
z
+¥
x(t) = x1( t) * x 2 (t ) =
x1( t) x1( t - t) dt
-¥
z
z
t
z
t
t
= e -4 t e -4( t - t) dt = e -4 t e -4 te 4 t dt
0
= e -4 t e -4 t + 4 t dt
0
z
0
t
= e -4 t dt = e -4 t T
t
0
= e -4 t t - 0 = t e -4 t ; t ³ 0
Using frequency
shifting property
= t e -4 t u(t)
0
Q 4.19
If x(t) and X(jW) are Fourier transform pair. Determine the Fourier transform of, x(t)sin W0 t.
Solution
r |RS|
T
m
F x(t ) sin W0 t = F x( t)
=
Q 4.20
|UV
|W
e jW0 t - e - jW0 t
1
1
= F x( t) e jW0 t - F x( t) e - jW0 t
2j
2j
2j
o
t
o
t
1
1
1
X( W - W 0 ) - X( W + W 0 ) =
X( W - W 0 ) - X( W + W 0 )
2j
2j
2j
Determine the exponential Fourier series representation of the following signals.
a) x(t) = cos 4t + sin6t
b) x(t) = sin 2 t
Solution
a) x(t) = cos 4t + sin6t
=
e j4 t + e - j4 t e j6 t - e - j6 t
1
1
1
1
+
= - e - j6 t + e - j4 t + e j4 t + e j6 t
2
2j
2j
2
2
2j
b) x(t) = sin2 t
F
GH
=
1 - cos2t 1 1
1 1 e j2t + e - j2t
= - cos2t = 2
2 2
2 2
2
=
1 1 j2t 1 - j2t
1
1 1
- e - e = - e - j2t + - e j2t
2 4
4
4
2 4
I
JK
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 88
4.18 MATLAB Programs
Program
4.1
Write a MATLAB program to find Fourier transform of the following
signals.
d) At e-btu(t)
e) A cosW0t
a) A
b) u(t)
c) Ae-tu(t)
%
Program
%Let
syms
to
find
fourier
transform
of
given
t, A, b, o be any real variables
t real; syms A real;
syms b real;
%(a)
x = A;
disp(‘(a) Fourier
X=fourier(x)
%(b)
x = heaviside(t);
disp(‘(b) Fourier
X=fourier(x)
transform
of
syms
o
%(d)
x=A*t*exp(-b*t)*heaviside(t);
disp(‘(d) Fourier transform of
X=fourier(x)
transform
of
“A
exp(-t)
“At
“A
domain
signals
real;
is’);
%heaviside(t) is unit step
transform of
“u(t)” is’);
%(c)
x = A*exp(-t)*heaviside(t);
disp(‘(c) Fourier transform of
X=fourier(x)
%(e)
x=A*cos(o*t);
disp(‘(e) Fourier
X=fourier(x)
“A”
time
u(t)”
exp(-b*t)
cos(o*t)”
OUTPUT
(a) Fourier transform of “A” is
X =
2*i*pi*dirac(1,w)
(b) Fourier transform of
“u(t)” is
X =
pi*dirac(w)-i/w
(c) Fourier transform of
“A exp(-t) u(t)” is
X =
A/(1+i*w)
(d) Fourier transform of
“At exp(-b*t) u(t)” is
X =
A/(b+i*w)^2
(e) Fourier transform of “A cos(o*t)” is
X =
A*pi*(dirac(w-o)+dirac(w+o))
signal
is’);
u(t)”
is’);
is’);
4. 89
Signals & Systems
Program
4.2
Write a MATLAB program to find inverse Fourier transform of the following
frequency domain signals.
a) X1 ( jW) = Ap[d (W - W 0 ) + d (W + W 0 )]
c) X 3 ( jW) =
%
Program
%Let
syms
to
A
1 + jW
find
inverse
Fourier
Ap
d (W - W 0 ) - d (W + W 0 )
j
3( jW) + 14
d) X 4 (jW) =
( jW) 2 + 7( jW) + 12
b) X 2 ( jW) =
transform
of
t, A, b, o, w be any real variables
t real; syms A real; syms o real; syms
%(a)
X1=A*pi*(dirac(w-o)+dirac(w+o));
disp(‘(a) Inverse Fourier transform
x1=ifourier(X1,t)
%(b)
X2=A*pi*(dirac(w-o)-dirac(w+o))/i;
disp(‘(b) Inverse Fourier transform
x2=ifourier(X2,t)
%(c)
X3=A/(1+i*w);
disp(‘(c) Inverse Fourier
x3=ifourier(X3,t)
transform
%(d)
X4=(3*w+14)/(w^2+7*w+12);
disp(‘(d) Inverse Fourier transform
x4=ifourier(X4,t)
w
frequency
domain
signals
real;
of
X1
is’);
of
X2
is’);
of
X3
is’);
of
X4
is’);
OUTPUT
(a) Inverse Fourier transform of X1 is
x1 =
A*cos(o*t)
(b) Inverse Fourier transform of X2 is
x2 =
A*sin(o*t)
(c) Inverse Fourier transform of X3 is
x3 =
A*exp(-t)*heaviside(t)
(d) Inverse Fourier transform of X4 is
x4 =
1/2*i*(2*heaviside(t)-1)*(-2*exp(-4*i*t)+5*exp(-3*i*t))
Program
4.3
Write a MATLAB program to determine response of LTI system whose
impulse response is h(t)= 2e -3tu(t) for the input is x(t)=2e-5tu(t),using
Fourier transform.
%
Program
to
find
the
response
syms t real;
h=2*exp(-3*t).*heaviside(t);
x=2*exp(-5*t).*heaviside(t);
of
LTI
system
using
Fourier
transform
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
disp(‘Fourier
H=fourier(h)
transform
of
the
impulse
disp(‘Fourier
X=fourier(x)
transform
of
the
input
Y=H*X;
disp(‘The response of
y_res=ifourier(Y,t)
the
given
4. 90
response
is’);
is’);
system
is’);
OUTPUT
Fourier transform of the impulse response is
H =
2/(3+i*w)
Fourier transform of the input is
X =
2/(5+i*w)
The response of the given system is
y_res =
2*heaviside(t)*(exp(-3*t)-exp(-5*t))
Program
4.4
Write a MATLAB program to perform convolution of signals,x1(t)= e-2tu(t)
and x2(t)=e-6tu(t),using Fourier transform.
%
Program
to
perform
convolution
using
Fourier
syms t real;
x1=exp(-2*t).*heaviside(t); %heaviside(t)
x2=exp(-6*t).*heaviside(t);
disp(‘Fourier transform
X1=fourier(x1)
disp(‘Fourier transform
X2=fourier(x2)
Y=X1*X2;
of
x1(t)
is’);
of
x2(t)
is’);
disp(‘Let x3 be convolution
x3=ifourier(Y,t)
of
x1(t)
is
and
transform
unit
step
signal
x2(t).’);
OUTPUT
Fourier transform of x1(t) is
X1 =
1/(2+i*w)
Fourier transform of x2(t) is
X2 =
1/(6+i*w)
Let x3 be convolution of x1(t) and x2(t).
x3 =
1/4*heaviside(t)*(exp(-2*t)-exp(-6*t))
Program
4.5
Write a MATLAB program to reconstruct the following periodic signal
represented by its Fourier series, by considering only 3,5 and 59 terms.
x(t) =
1
+
2
¥
åb
n=1
n
sin nW 0 t ; b n =
2
; W 0 = 2 pF; F = 1.
pn
4. 91
Signals & Systems
%
%
Program
pa r tia l
syms
t
to reconstruct periodic
sum o f fo ur ier ser ies
square
pulse
signal
using
real;
N=input(‘Enter number of signals to reconstruct’);
n_har=input(‘Enter no. of harmonics in each signal as
array’);
t=-1:.002:1;
omega_o=2*pi;
for k=1:N
n=[];
n=[1:2:n_har(k)];
b_n=2 . /(pi*n);
L_n=leng th(n);
x=0.5+b_n*sin(omega_o*n’*t);
subplot(N,1,k),plot(t,x),xlabel(‘t’),ylabel(‘recons
axis( [-1 1 -0.5 1.5]);
text(.55, 1.0,[‘no. of har. =’,num2str(n_har(k))]);
end
signal’);
OUTPUT
Enter number of signals to reconstruct 3
Enter no. of harmonics in each signal as array [3 5 59]
Fig P4.5 : Reconstructed signals using only 3, 5 and 59 terms of Fourier series.
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 92
4.19 Exercises
I. Fill in the blanks with appropriate words
2.
3.
In Fourier series representation of a signal, if W0 is the fundamental frequency then nW0 are called _________
frequencies.
Fourier series is useful for frequency domain analysis of __________ signals.
The plot of magnitude of Fourier coefficient cn with respect to "n" is called__________.
4.
5.
6.
For a signal x(t), the condition to be satisfied for half wave symmetry is__________.
For a signal x(t), the condition to be satisfied for quarter wave symmetry is__________ and __________.
If x(t) and X(jW) are Fourier transform pair then magnitude of X(jW) is called__________.
7.
8.
If x(t) and X(jW) are Fourier transform pair then phase of X(jW) is called__________.
The ratio of Fourier transform of output and input of a system is called__________.
1.
9. The Fourier transform of impulse response will be equal to__________.
10. If x(t) is real and odd, then X(jW) will be__________.
11. Fourier transform of periodic continuous time signal consists of __________ located at __________
frequencies of the signal.
12. Fourier transform is evalution of Laplace transform along the __________ axis in s-plane.
Answers
1. harmonic
4. x(t ± T/2) = –x(t),
3. magnitude line spectrum
6. magnitude spectrum
7. phase spectrum
2. periodic
5. x(t ± T/2) = –x(t);
x(–t) = ± x(t)
8. transfer function
10. imaginary and odd
11. impulses, harmonic
12. imaginary
9. transfer function
II. State whether the following statements are True/False
1.
The magnitude line spectrum is symmetric and phase line spectrum is anti-symmetric.
2.
3.
For waveforms with even symmetry, the Fourier coefficient a0 is always zero.
For waveforms with odd symmetry, the Fourier coefficients a0 and an are always zero.
4.
5.
6.
An alternating waveform will always have even harmonics only.
An even signal with half wave symmetry will always have even harmonics of cosine terms.
An odd signal with half wave symmetry will always have odd harmonics of sine terms.
7.
8.
The shifting of vertical axis of a waveform, will not affect the even/odd symmetry.
In frequency spectrum using Fourier transform, the magnitude spectrum will have even symmetry and
phase spectrum will have odd symmetry.
9. The Fourier transform of a periodic signal will be summation of impulses.
10. The total average power in a periodic signal is equal to the sum of power in all of its harmonics.
11. Fourier transform is useful for frequency domain analysis of both periodic and non-periodic signals.
12. Laplace transform is a generalized transform and Fourier transform is a particular transform.
Answers
1. True
7. False
2. False
8. True
3. True
9. True
4. False
10. True
5. False
11. True
6. True
12. True
4. 93
Signals & Systems
III. Choose the right answer for the following questions
1. The Fourier coefficient ‘an’ can be evaluated as,
+¥
a) a n =
2
x ( t ) cosnW 0 t dt
T -¥
c) a n =
2 2
x ( t ) sin nW0 t dt
T -T
T
z
z
b) a n =
2
x ( t ) cosnW0 t dt
T0
d) a n =
2
x( t ) cosnW 0 t dt
T0
z
z
T
¥
2
2. The exponential Fourier coefficient of x(t) = sin2t is,
1
1
, C0 = 0, C-1 = p
p
a) C1 =
1
1
1
b) C -1 = - , C0 = , C1 = 4
2
4
1
1
c) C -1 = - , C0 = 1, C1 =
2
2
d) C-1 = -2, C0 = 1, C1 = -2
3. Fourier transform of the signum function, x(t)=sgn(t) is,
b) 2 / jW
a) 2 / W
c) 2 / (jW) 2
d) - 2 / jW
4. The Unit step signal u(t) can also be expressed as,
a) 1 + sgn(t)
b) 1 + d(t)
c)
1 1
+ sgn(t)
2 2
d) 2sgn(t)
5. The necessary and sufficient condition for x(t) to be real is,
a) X* ( jW) = X(jW)
b) X*( jW) = X( - jW)
c) X ( jW) = X*( - jW)
d) none of the above
6. Fourier series of any periodic signal x(t) can only be obtained if,
a) finite number of discontinuities within finite time interval, T.
b) finite number of positive and negative maxima in the period, T.
c) well defined at infinite number of points.
d) both (a) and (b).
7. The inverse Fourier Transform of X(jW) is defined as,
-¥
¥
a)
1
X ( jW) e jWt dW
2p 0
z
b)
1
X ( jW) e - jWt dt
2p +¥
z
¥
c)
1
X ( jW) e jWt dW
2p -¥
z
¥
1
X ( jW) e - jWt dW
2p 0
z
d)
l
q is given by,
8. The frequency convolution property of Fourier transform says that, F x1 (t) x 2 (t)
l = +¥
a) X1(jW) X2(jW)
c)
1
2p
b)
1
X1 ( jl ) X 2 j( W - l ) dt
2p l = -¥
d)
1
2p
¥
z
X1 ( jW) X 2 ( jW) dW
-¥
z
b
g
¥
z
X1 ( jW)
2
2
X 2 ( jW) dW
-¥
9. Fourier transform of a sinusoidal signal 2sinW0t is,
2p
d ( W - W0 ) - d ( W + W 0 )
j
2p
c)
d ( W + W0 ) - d ( W - W 0 )
j
a)
p
d ( W - W0 ) - d ( W + W0 )
2j
p
d)
d ( W - W 0 ) - d ( W + W0 )
j
b)
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 94
th
10. In the Fourier transform of a periodic signal, the magnitude of n impulse is,
C
C
a) n
d) n
b) 2p C n
c) C n
2p
2
11. Fourier transform of the causal signal, x(t) = cosW0 t u(t) is,
jW
-W
W
a) 2 0 2
b) 2
c) 2 0 2
W 0 - W2
W + W0
W0 - W
d)
jW
W20 + W2
12. If the signal x(t) has odd and half wave symmetry, then, the Fourier series will have only,
a) odd harmonics of sine terms.
b) constant term and odd harmonics of cosine terms.
c) even harmonics of sine terms
d) odd harmonics of cosine terms.
FG
H
13. Differentiation of signum function is, i.e.,
a) 2d( t )
IJ
K
d
sgn(t)
dt
c)
b) d (t)
1
d( t )
2
d) 2 u(t)
14. Fourier transform of cosW0 t is,
a) X(W – W0) + X(W + W0)
1
1
c)
X ( W - W 0 ) + X ( W + W0 )
2
2
b) p[d(W –W0) + d(W + W0)]
p
d)
X ( W - W 0 ) + X ( W + W0 )
2
15. The Fourier transform of x(t) exists only if,
z
+¥
+¥
¥
a)
x ( t ) dt < ¥
b)
z
x( t ) dt > 0
c)
z
d)
-¥
-¥
0
+¥
x ( t ) e jWt dt < ¥
z
x ( t ) dt < ¥
-¥
16. By convolution property of Fourier transform, F{x(t)* h(t)} is,
+¥
1
a)
X ( jW) H( j W) dW
b) X(jW) H(jW)
2 p -¥
c) X(jW) * H(jW)
d) None of the above
z
R|1 t < T
S|0 T < t < T . The constant component of x(t) is
2
T 2T
1
17. A periodic signal x(t) of period T0 is given by, x(t) =
0
1
a)
T0
T1
b)
T1
T0
c)
1
d)
T0
T1
2T0
18. For two periodic waveforms, square and triangular, the magnitude of the nth Fourier series coefficients,
for n > 0, are respectively proportional to,
a) n -3 and n -2
b) n -2 and n -3
c) n -1 and n -2
d) n -4 and n -2
19. When the waveform has parabolic structure/wiggles, the magnitude of higher harmonics,
a) increases rapidly
b) decreases more rapidly
c) remain same
d) become zero
20. The initial value of a continuous time signal in frequency domain is,
¥
¥
z
a) X(0) = x ( t ) dt
0
b) X(0) =
1
x ( jW) dt
2p -¥
z
¥
c) X(0) =
1
x ( t ) dW
2p -¥
z
¥
z
d) X(0) = x ( t ) dt
-¥
4. 95
Signals & Systems
21. Which of the following cannot be the Fourier series expansion of a periodic signal?
a) x(t) = 2 cost + 3 cos3t
b) x(t) = 2 cospt + 7 cost
c) x(t) = cost + 0.5
d) x(t) = 2 cos1.5pt + sin 3.5pt
22. Let Cn be a Fourier coefficient in exponential form. It is given that, Cn = 2 + j7 then C–n is,
a) –2 – j7
b) 2 – j7
c) –2 + j7
d) 2 + j7
d
x(t) will be,
dx
X ( W)
d)
jW
23. The Fourier transform of a function x(t) is X(jW). Then, the Fourier transform of
a)
dX(W)
dF
b) j2 p W X( W)
24. If Fourier transform of the signal e
-t
c) jW X(jW)
is
2
2
then, the Fourier transform of the signal
, using
1 + W2
1 + t2
duality property is,
a) 2 p e jW
b) 2 p e - jW
25. Fourier transform of gaussian pulse will be,
a) another gaussian pulse
c) sinc pulse
c) 2p ( - jW)
d) 2p ( jW)
b) squared sinc pulse
d) impulse train
26. The signal x(t) = u(6t) has a Fourier transform of,
a)
1
+ 6p d( W )
jW
b)
6
+ 6p d( W)
jW
c)
1
+ p d( W)
jW
d)
1
+ p d( 6W)
j6W
27. Consider the signal, x(t) = t(T-t); 0 £ t £ T. The Fourier coefficient a0 for this signal is,
a)
T3
4
b)
T2
3
c)
3
T2
28. The Fourier transform of the signal x(t) = e7t u(–t) is,
1
7
7
a)
b)
c)
7 + jW
1 + jW
1 - jW
29. For a signal x(t) = e
a)
2
1 + W2
-t
d)
T3
12
d)
1
7 - jW
d)
1
2 + W2
, the Fourier transform is,
b)
2
1 - W2
c)
1
2 - W2
a
a
and Fourier transform of x2(t) is
, then F{x1(t)* x2(t)} is,
W -a
W +a
2
W+a
a
a2
b)
c) 2
d)
W-a
W - a2
W2 + a 2
30. If Fourier transform of x1(t) is
a)
W-a
W+a
31. If Fourier transform of x(t) is X(W) then Fourier transform of e - jW0 t x(t) is,
a)
X( W)
W0
b) X( W - W0 )
c) X( W + W 0 )
d) W 0 X(W )
32. If a signal x(t) is differentiated ‘m’ times to produce an impulse then its Fourier coefficients will be
proportional to,
1
1
a) nm
b) m-1
c) nm – 1
d) m
n
n
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
2a
d2
x(t) is,
2 , then Fourier transform of
dt 2
a +W
33. If Fourier transform of x(t) is
a)
-2 a W 2
a 2 + W2
b)
2
4a 2
(a + W2 ) 2
FG
H4
2
IJ
K
4
e - jWt 0
+ W2
b)
c)
2
34. The Fourier transform of the signal x(t) = e
a)
4. 96
-4
e - jWt 0
4 + W2
2
-4 t - t0
2 aW
( a + W2 ) 2
d)
a 2 + W2
2a
8
e - jWt 0
4 + W2
d)
-8
e - jWt 0
4 + W2
2
is,
c)
2
2
+¥
35. A signal x(t) has a magnitude of 5 at t =0.Then value of
z
X(jW )dW
will be,
-¥
a) 10
b) 10p
c) 5p
d)
10
p
Answers
1. b
6. d
11. b
16. b
21. b
26. c
31. b
2. b
7. c
12. a
17. c
22. b
27. b
32. d
3. b
8. b
13. a
18. c
23. c
28. d
33. a
4. c
9. a
14. b
19. b
24. b
29. a
34. c
5. a
10. b
15. d
20. d
25. a
30. c
35. b
IV. Answer the following questions
1. Write the conditions for existence of Fourier series.
2. Write the trigonometric form of Fourier series representation of a periodic signal and explain.
3. Write the exponential form of Fourier series representation of a periodic signal and explain.
4. What is the relation between Fourier coefficients of trigonometric and exponential form?
5. Write a short note on negative frequency.
6. Define frequency spectrum or line spectrum.
7. Write short note on Fourier coefficients of signals with even symmetry and odd symmetry.
8. What is the effect of half wave symmetry on Fourier coefficients of a signal?
9. What is the effect of quarter wave symmetry on Fourier coefficients of a signal?
10. Write any two properties of Fourier series.
11. Write the Parseval’s relation for continuous time periodic signal.
12. Explain how Fourier transform is obtained from Fourier series.
13. Define Fourier transform and inverse Fourier transform of a signal.
14. Write any two properties of Fourier transform.
15. State and prove time differentiation property of Fourier transform.
16. State and prove frequency differentiation property of Fourier transform.
17. Write the convolution theorem of Fourier transform.
18. What is the relation between Fourier transform and Laplace transform?
19. Show that Fourier transform of a periodic signal will be impulses.
20. Define the frequency domain transfer function.
4. 97
Signals & Systems
V. Solve the following problems
E 4.1
Determine the trigonometric Fourier series representation of the signal shown in fig E4.1.
x(t)
x(t)
A
A
-T
2
0
T
2
t
T
T
2
0
–T
Fig E4.1.
T
t
–A
Fig E4.2.
E 4.2
Determine the trigonometric Fourier series representation of the signal shown in fig E4.2.
E 4.3
Determine the exponential Fourier series representation of the signal shown in fig E4.3 and hence obtain
the trigonometric Fourier series.
x(t)
x(t)
A
A
0
2T
T
3T t
0
Fig E4.3.
T
2
t Fig E4.4.
T
E 4.4
Determine the exponential Fourier series representation of the signal shown in fig E4.4 and hence
obtain the trigonometric Fourier series.
E 4.5
Determine the exponential Fourier series representation of the signal shown in fig E4.5.
x(t)
1
–3
–2
e–t
0
–1
1
2
3 t
Fig E4.5.
E 4.6
Find the Fourier transform of the time domain signals given below.
; t = 0 to + ¥
i) x(t) = t sinW0t
ii) x(t) = t cos W0t ;
t = 0 to + ¥
E 4.7
Determine the Fourier transform of the rectangular pulse shown in fig E4.7.
E 4.8
Determine the Fourier transform of the triangular pulse shown in fig E4.8.
x(t)
x(t)
x(t)
A
A
A
T
–T
0
T
Fig E4.7.
E 4.9
t
0
T
2
T t
Fig E4.8.
Determine the Fourier transform of the signal shown in fig E4.9.
E 4.10 Determine Fourier transform of the periodic signal shown in fig E4.10.
–A
Fig E4.9.
t
Chapter 4 - Fourier Series and Fourier Transform of Continuous Time Signals
4. 98
x(t)
A
-T
2
0
–a
a
t
T
2
Fig E4.10.
Answers
A
2A
- 2
4
p
E4.1 x(t) =
FG cos W t +
H 1
0
0
2
0
2
- j2 W 0 t
0
- j5W 0 t
0
- j3 W 0 t
0
d
E4.5 x(t) = 1 - e -1
E4.6 i) X(jW) =
- jW 0 t
j2 W 0 t
j3 W 0 t
0
- jW 0 t
0
jW 0 t
j3 W 0 t
j5 W 0 t
0
i LMN......+ 1e- j6p + 1e- j4p + 1e- j2 p + 1 + 1e+ j2 p + 1e+ j4 p + 1e+ j6p +....OPQ
- j3 W 0 t
- j2 W 0 t
-2WW 0
( W2 + W 20 ) 2
E4.7 X(j W) =
A
(1 - e - jWT )
jW
E4.8 X(jW) =
2A
( 2 e- jWT /2 - e - jWT - 1)
TW 2
E4.9 X(jW) = jWAT2 sinc
+¥
E4.10 X(j W) =
jW0 t
A
å 2p 2n2
n=-¥
FG WT IJ
H 2p K
1 - cos2 np
- jW 0 t
ii) X(jW) =
jW 0 t
j2 W 0 t
W02 - W 2
( W 02 + W2 ) 2
IJ
K
sin 2W 0 t
sin 3W 0 t
+
.....
2
3
0
2
- j3W 0 t
E4.4
FG sin W t H 1
FG cos W t + cos 3W t + cos 5W t +.........IJ
H 1
K
3
5
A F
x(t) =
G .......- e 3 - e 2 - e 1 + jp + e 1 + e 2 + e 3 +.......IJK
j2 p H
A A F sin W t
sin 2W t
sin 3W t
I
+
x(t) =
+
+
+....J
G
K
2
3
2 p H 1
I
A F
e
e
e
jp e
e
e
x(t) =
.....+ +
+
+
+.....J
G
jp H
5
3
1
2
1
3
5
K
A 2 A F sin W t
sin 3W t
sin 5W t
I
x(t) =
+
+
+
+....J
G
K
2
p H 1
3
5
8A
p2
E4.2 x(t) =
E4.3
IJ
K
cos 3W 0 t
cos 5W 0 t
A
+
+..... +
32
52
p
0
2
j3 W 0 t
CHAPTER 5
State Space Analysis of
Continuous Time Systems
5.1 Introduction
The state variable approach is a powerful tool / technique for the analysis and design of systems.
The analysis and design of any system such as linear system, non-linear system, time invariant system,
time varying system, multiple input and multiple output system can be carried out using state space
method. The state space analysis can be carried with or without initial conditions.
The state space analysis is a modern approach and is also easy to analyse using digital computers.
The conventional (or old) methods of analysis employ the transfer function of the system. The drawbacks
in the transfer function model and analysis are,
1.
2.
3.
4.
Transfer function is defined under zero initial conditions.
Transfer function is applicable to linear time invariant systems.
Transfer function analysis is restricted to single input and single output systems.
Transfer function does not provide information regarding the internal state of the system.
5.2 State Model of a Continuous Time System
State and State Variables
The state of a continuous time system refers to the condition of the system at any time instant,
which is described using a minimum set of variables called state variables. The knowledge of these
variables at t = 0 together with knowledge of inputs for t > 0, completely describes the behaviour of the
system for t ³ 0.
The state variables are a set of variables that completely describe the state or condition of a system
at any time instant.
State Equations
Consider a continuous time system with M-inputs, N-state variables, and P-outputs.
Let, q1(t), q2(t), q3(t), ....., qN(t) be N-state variables of the continuous time system,
x1(t), x2(t), x3(t), ....., xM(t) be M-inputs of the continuous time system,
y1(t), y2(t), y3(t), ....., yP(t) be P-outputs of the continuous time system.
Now, the continuous time system can be represented by the block diagram shown in fig 5.1.
R| x (t)
|| x (t)
M number of inputs S x (t )
|| MM
|T x (t)
1
®
®
®
SYSTEM
® y1 ( t)
® y2 ( t )
® y3 ( t )
U|
||
V|
||
W
N number of
State variables
3
P number of outputs
M
M q1 ( t ), q 2 (t ), q 3 ( t ), ..... M
M
M
M
....., q N ( t )
y
®
®
M
P ( t)
Fig 5.1 : State space representation of a continuous time system.
2
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 2
The state equations can be formed by taking first derivative of state variables as function of state
variables and inputs. Therefore, the N-state variables can be expressed as N-number of first order differential
equations as shown below.
q& 1 (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N ( t ), x1 (t ), x 2 (t ), x3 (t ), ....., x M (t )
q& 2 (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N ( t ), x1 (t ), x2 (t ), x3 (t ), ....., x M (t )
q& 3 (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N (t ), x1 (t ), x2 (t ), x3 (t ), ....., x M ( t )
M
M
q& N (t) = F q 1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N ( t ), x1 (t ), x2 (t ), x 3 (t ), ....., x M (t )
Mathematically, the above functional form of first derivative of the N-state variables can be expressed
as shown below.
q& 1 (t) = a 11 q1 ( t ) + a12 q 2 ( t ) + a 13 q 3 ( t ) + ......
+ b11 x1 ( t ) + b12 x2 ( t ) +
&q 2 (t) = a 21 q 1 ( t ) + a 22 q 2 ( t ) + a 23 q 3 ( t ) + ......
+ b21 x1 ( t ) + b22 x2 ( t ) +
q& 3 (t) = a 31 q 1 ( t ) + a 32 q 2 ( t ) + a 33 q 3 ( t ) + ......
+ b31 x1 ( t ) + b32 x2 ( t ) +
+ a1N q N ( t )
b13 x3 ( t ) + ...... + b1M xM ( t )
+ a 2 N q N (t )
b23 x3 ( t ) + ..... + b 2 M x M ( t )
+ a 3N q N ( t )
b33 x3 ( t ) + ..... + b3M xM ( t )
M
M
q& N (t) = a N1 q1 ( t ) + a N 2 q 2 ( t ) + a N 3 q 3 ( t ) + ...... + a NN q N ( t )
+ b N1 x1 ( t ) + bN 2 x2 ( t ) + bN 3 x3 ( t ) + ...... + bNM xM ( t )
The above equations of first derivative of the N-state variables are called state equations and can
be expressed in the matrix form as shown in the matrix equation (5.1).
LMq& ( t) OP LM a a a ...OP
MMq&& (t)PP = MMa a a ...PP
MNq M(t)PQ MNa M a M a M...PQ
B
B
1
11
12
13
2
21
22
23
3
31
32
33
& (t)
Q
A
where, A
B
LMq (t) OP LM b b b ...OP
MMq (t)PP + MMb b b ...PP
MNq M(t)PQ MNb M b M b M...PQ
B
B
1
11
12
13
2
21
22
23
3
31
32
33
Q( t )
B
LM x (t) OP
MMx (t)PP
MNx M(t)PQ
B
1
2
3
....(5.1)
X( t )
= System Matrix
= Input Matrix
Q(t) = State Vector
X(t) = Input Vector
Therefore, the Matrix form of the state equation can be written as shown in equation (5.2).
& (t) = A Q(t) + B X(t)
....(5.2)
Q
Signals & Systems
5. 3
Output Equations
The output equations can be formed by taking the outputs as function of state variables and inputs.
Therefore, the P-outputs can be expressed as shown below.
y1 (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N ( t ), x1 (t ), x2 (t ), x 3 (t ), ....., x M (t )
y 2 (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N (t ), x1 (t ), x 2 (t ), x 3 ( t ), ....., x M (t )
y 3 (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N ( t ), x1 (t ), x2 (t ), x3 (t ), ....., x M (t )
M
M
y P (t) = F q1 ( t ), q 2 ( t ), q 3 ( t ), ....., q N ( t ), x1 (t ), x2 ( t ), x3 (t ), ....., x M (t )
Mathematically, the above functional form of the outputs can be expressed as shown below.
y1 (t) = c11 q 1 ( t ) + c12 q 2 ( t ) + c13 q 3 ( t ) + ..... + c1N q N ( t )
+ d11 x1 ( t ) + d12 x2 ( t ) + d13 x3 ( t ) + ..... + d1M xM ( t )
y 2 (t) = c21 q1 ( t ) + c 22 q 2 ( t ) + c23 q 3 ( t ) + ..... + c2 N q N ( t )
+ d 21 x1 ( t ) + d 22 x2 ( t ) + d 23 x3 ( t ) + ..... + d 2 M x M ( t )
y 3 (t) = c 31 q1 ( t ) + c 32 q 2 ( t ) + c 33 q 3 ( t ) + ..... + c 3N q N ( t )
+ d 31 x1 ( t ) + d 32 x2 ( t ) + d 33 x3 ( t ) + ..... + d3M xM ( t )
M
M
y P (t) = c P1 q 1 ( t ) + c P 2 q 2 ( t ) + c P 3 q 3 ( t ) + ..... + c PN q N ( t )
+ d P1 x1 ( t ) + d P 2 x2 ( t ) + d P 3 x3 ( t ) + ..... + d PM x M ( t )
The above equations are called output equations and can be expressed in the matrix form as
shown in equation (5.3).
LM y (t) OP LMc
MMy (t)PP = MMc
MNy M( t)PQ MNc
B
1
2
3
OP
PP
PQ
c12 c13 ......
c
21 22 c 23 ......
31 c 32 c 33 ......
M M M
11
B
Y( t )
1
2
3
B
D
= Output Matrix ;
Q(t) = State Vector
OP
PP
PQ
d 12 d 13 ......
d
21 22 d 23 ......
31 d 32 d 33 ......
M M M
11
Q( t )
C
where, C
LMq (t) OP LMd
MMq (t)PP + MMd
MNq M(t)PQ MNd
B
;
D
LM x (t) OP
MMx (t)PP
MN x M(t)PQ
B
1
2
3
....(5.3)
X( t )
= Transmission Matrix
X(t) = Input Vector
Therefore the Matrix form of output equation can be written as shown in equation (5.4).
Y ( t ) = C Q( t ) + D X ( t )
....(5.4)
State Model
The state model of a continuous time system is given by state equations and output equations.
State equations :
Output equations :
& (t) = A Q(t) + B X(t)
Q
Y ( t ) = C Q ( t ) + D X( t )
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 4
5.3 State Model of a Continuous Time System from Direct Form-II Structure
Consider the equation of a 3rd order continuous time LTI system,
d3
d2
d
d3
d2
y(t) + a1 2 y(t) + a 2
y(t) + a 3 y(t) = b0 3 x( t ) + b1 2 x( t )
3
dt
dt
dt
dt
dt
d
+ b2
x( t ) + b3 x( t )
dt
The direct form-II structure of the system described by the above equation is shown in fig 5.2.
b q& (t)
X(s)
x(t)
+
1
+
–a1 q3(t)
–a 1
0
b0
q& 3 (t)
3
+
y(t)
Y(s)
s
q3(t)
b1
b1 q3(t)
+
q& 2 (t)
11 s
s
+
–a2 q2(t)
–a 2
q2(t)
b2
b2 q2(t)
+
q& 1 (t)
11 s
s
q1(t)
–a3 q1(t)
–a 3
Q1(s)
b3 q1(t)
b3
Fig 5.2 : Direct form - II structure of a continuous time system.
In a state model of a system, the choice of the number of state variables is equal to the number of
integrators. The direct form-II structure shown in fig 5.2 has one input x(t), one output y(t) and three
integrators, and so let us choose three state variables. Let, q1(t), q2(t) and q3(t) be the three state variables.
Assign state variables at the output of every integrator. Hence the first derivative of a state variable will be
available at the input of the integrator.
State Equations
The state equations are formed by equating the sum of incoming signals of the integrator to first
derivative of the state variable, as shown below.
U|
V|
W
q& 1 (t) = q 2 ( t )
State equations
q& 2 (t) = q 3 ( t )
q& 3 (t) = - a 3 q 1 ( t ) - a 2 q 2 ( t ) - a1 q 3 ( t ) + x( t )
E
U|
b gV
|W
q& 1 (t) = 0 ´ q1 ( t ) + 1 ´ q 2 ( t ) + 0 ´ q 3 ( t ) + 0 ´ x( t )
q& 2 (t) = 0 ´ q1 ( t ) + 0 ´ q 2 ( t ) + 1 ´ q 3 ( t ) + 0 ´ x t State equations
q& 3 (t) = -a 3 ´ q 1 ( t ) - a 2 ´ q 2 ( t ) - a1 ´ q 3 ( t ) + 1 ´ x( t )
Signals & Systems
5. 5
On arranging the state equations in the matrix form we get,
LMq& (t) OP LM 0
MMq& (t)PP = MM 0
Nq& (t)Q N-a
OP LMq (t) OP LM0OP
0
1 P Mq ( t ) P + M0P
MN1PQ
- a -a PQ MNq ( t ) PQ
BA
B
B
1
1
2
3
0
1
x( t )
2
3
2
1
3
Output Equations
The output equation y(t) is formed by equating the incoming signals of output node / summing
point to y(t) as shown below.
y( t ) = b 0 q& 3 ( t ) + b 3 q 1 ( t ) + b 2 q 2 ( t ) + b1 q 3 ( t )
On substituting for q& 3 ( t ) from state equation we get,
y( t ) = b 0 -a 3 q 1 ( t ) - a 2 q 2 ( t ) - a 1 q 3 ( t ) + x( t ) + b 3 q1 ( t ) + b2 q 2 ( t ) + b1 q 3 ( t )
= (b3 - b 0 a 3 ) q1 ( t ) + ( b 2 - b 0 a 2 ) q 2 ( t ) + (b1 - b0 a1 ) q 3 ( t ) + b0 x( t )
On arranging the output equation in the matrix form we get,
LMq (t) OP
MMq (t)PP +
Nq ( t ) Q
1
y ( t ) = b 3 - b 0 a 3 b 2 - b 0 a 2 b1 - b 0 a 1
2
3
B
C
b0
x( t )
B
D
State Model
State model of a continuous time system is given by state equations and output equations.
& (t) = A Q (t) + B X(t)
Q
State equations :
LMq& (t) OP LM 0
MMq& (t)PP = MM 0
Nq& (t)Q N-a
1
1
2
0
3
3
E
OP LMq (t) OP LM0OP
1 P Mq (t)P + M0P
MN1PQ
-a PQ MNq (t) PQ
0
1
2
-a 2
1
x(t)
3
Output equations : Y (t) = C Q (t) + D X (t)
E
LMq (t) OP
MMq (t)PP +
Nq (t)Q
1
y(t) = b 3 - b 0 a 3
b2 - b0 a 2
b1 - b 0 a 1
2
3
b0
x(t)
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 6
5.4 Transfer Function of a Continuous Time System from State Model
Consider the state equation,
& (t) = A Q (t) + B X(t)
Q
On taking Laplace transform of the above equation with zero initial condition we get,
sQ(s) = A Q(s) + B X(s)
sQ(s) – A Q(s) = B X(s)
(sI – A) Q(s) = B X(s), where I is the unit matrix.
On premultiplying the above equation by (sI – A)–1, we get,
Q(s) = (sI – A)–1 B X(s)
.....(5.5)
Consider the output equation,
Y(t) = C Q(t) + D X(t)
On taking Laplace transform of the above equation we get,
Y(s) = C Q(s) + D X(s)
.....(5.6)
On substituting for Q(s) from equation (5.5) in (5.6) we get,
Y(s) = C (sI – A)–1 B X(s) + D X(s)
\ Y(s) = [C (sI – A)–1 B + D] X(s)
For single input and single output system,
Y(s)
= C (sI - A ) -1 B + D
X( s)
.....(5.7)
The equation (5.7) is the transfer function of the continuous time system.
5.5 Solution of State Equations and Response of Continuous Time System
Solution of State Equations in Time Domain
The state equation of Nth order continuous time system is given by,
& (t) = A Q(t) + B X(t)
Q
& (t) - A Q (t) = B X(t)
\Q
On premultiplying the above equation by e-At we get,
& (t) - A Q(t) = e - At B X(t)
e - At Q
& (t) + e - At ( - A) Q(t) = e - At B X(t)
e - At Q
Consider the differential of e -At Q (t).
d - At
& (t) + e - At ( - A ) Q(t)
e Q(t) = e - At Q
dt
d
i
.....(5.8)
d(uv) = u dv + du v
.....(5.9)
Signals & Systems
5. 7
Using equation (5.9), the equation (5.8) can be written as shown below.
d - At
e Q(t) = e - At B X(t)
dt
\ d e - At Q(t) = e - At B X(t) dt
d
i
d
i
On integrating the above equation between limits 0 to t we get,
t
e - At Q(t) =
z
e - At B X(t ) dt + Q(0)
Q(0) is initial condition vector
t = Dummy variable substituted for t
0
On premultiplying the above equation by eAt we get,
t
e At e - At Q(t) = e At Q(0) + e At
z
e - At B X(t ) dt
0
t
\ Q(t) = e At Q(0) + e At
z
e - At B X( t ) dt
e At e - At = e At - At = e 0 = I = Unit matrix
0
t
= e At Q(0) +
z
z
z
e At e - At B X( t) dt
0
e At is independent of
integral variable t.
t
= e At Q(0) +
e A( t - t ) B X( t) dt
0
t
At
(0)
Q(t) = e1
Q4
42
3 +
Solution due to
initial condition
e A( t - t ) B X( t) dt
0 44424443
1
.....(5.10)
Solution due to input
The equation (5.10) is the time domain solution of state equations of the continuous time system.
Here, the matrix eAt is called state transition matrix of continuous time system.
Solution of State Equations using Laplace Transform
Consider the state equation of the continuous time system.
& (t) = A Q (t) + B X(t)
Q
On taking Laplace transform of the above equation we get,
sQ(s) – Q(0) = A Q(s) + B X(s)
Q(0) is initial condition vector
sQ(s) – A Q(s) = Q(0) + B X(s)
(sI – A) Q(s) = Q(0) + B X(s)
I is the unit matrix
On premultiplying the above equation by (sI – A)–1, we get,
Q(s) = (sI – A)–1 Q(0) + (sI – A)–1 B X(s)
......(5.11)
The equation (5.11) is the solution of state equations of the continuous time system in s-domain.
The solution of state equations in time domain is given by the inverse Laplace transform of Q(s).
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 8
On taking inverse Laplace transform of equation (5.11) we get,
l q
\ Q(t) = L n(sI - A) Q(0) + (sI - A ) B X(s)s
Q(t) = L n(sI - A ) s Q(0) + L n( sI - A ) B Xb sgs
1444
424444
3
1444424444
3
Q(t) = L-1 Q(s)
-1
-1
-1
-1
-1
-1
-1
......(5.12)
Solution due to
input
Solution due to
initial condition
The equation (5.12) is the time domain solution of state equations of the continuous time system.
On comparing equations (5.10) and (5.12) we get,
n
State Transition Matrix, e At = L-1 (sI - A) -1
s
......(5.13)
The equation (5.13) is used to compute state transition matrix via Laplace transform.
Response of Continuous Time System
The response,Y(t) of continuous time system can be computed by substituting the solution of state
equations Q(t), from equation (5.10) or (5.12), in the output equation shown below.
Response of continuous time system, Y(t) = C Q(t) + D X(t)
......(5.14)
The response, when there is no input is called zero-input response, Yzi(t) and this can be obtained
from equations (5.12) and (5.14) by taking X(t) and X(s) as zero. The zero-input response is due to initial
conditions.
Zero-input response , Yzi(t) = C Q(t)
-1
where, Q(t) = L
......(5.15)
lQ(s)q = L n(sI - A) s Q(0)
-1
-1
The response, when there is no initial conditions is called zero-state response, Yzs(t) and this can be
obtained from equations (5.12) and (5.14) by taking Q(0) as zero. The zero-state response is due to
input.
Zero-state response , Yzs(t) = C Q(t) + D X(t)
-1
where, Q(t) = L
lQ(s)q = L n(sI - A)
-1
......(5.16)
-1
s
B X(s)
5.6 Solved Problems in State Space Analysis
Example 5.1
Determine the state model of the continuous time system governed by the equation,
d3
d2
d
d3
d2
d
x( t) + 2x( t)
y(t) + 3 2 y(t) - 2
y(t) + 0.5 y(t) = 4 3 x(t ) + 2.5 2 x(t ) + 1.5
3
dt
dt
dt
dt
dt
dt
Solution
The direct form-II structure of the continuous time system described by the given equation is shown in fig 1.
The choice of number of state variables is equal to number of integrators. The direct form-II structure has one input
x(t), one output y(t) and three integrators and so let us choose three state variables.
Let, q1(t), q2(t) and q3(t) be the three state variables. Assign state variables at the output of every integrator. Hence
the first derivative of state variable will be available at the input of integrator.
Signals & Systems
5. 9
4 q& (t)
X(s)
+
x(t)
4
3
+
q& 3 (t)
y(t)
Y(s)
1
s
+
–3 q3(t)
q3(t)
–3
2.5 q3(t)
2.5
q& (t)
+
2
1
s
+
2 q2(t)
q2(t)
2
1.5
1.5 q 2(t)
+
q& (t)
1
1
s
q1(t)
–0.5 q1(t)
–0.5
Q1(s)
2
2 q1(t)
Fig 1 : Direct form - II structure.
State Equations
The state equations are formed by equating the sum of incoming signals of integrator to the first derivative of state
variable, as shown below.
q& 1( t) = q2 ( t)
q& 2 ( t ) = q3 ( t)
q& 3 ( t ) = - 0.5 q1( t) + 2 q2 ( t) - 3 q3 ( t) + x( t)
On arranging the state equations in the matrix form we get,
LM q& (t)OP
MMq& (t)PP
Nq& (t)Q
1
=
2
3
LM 0
MM 0
N-0.5
1
0
2
OP LM q (t)OP LM0OP
P Mq (t)P + MM0PP
-3PQ MNq ( t)PQ
N1Q
0
1
1
2
x( t)
3
Output Equations
Output equation y(t) is formed by equating incoming signals of output node / summing point to y(t) as shown below.
y(t) = 4 q& 3 (t ) + 2 q1(t) + 1.5 q2 (t) + 2.5 q3 (t)
On substituting for q& 3 (t ) from state equation we get,
y( t) = 4 -0.5 q1( t) + 2 q2 ( t) - 3 q3 ( t) + x(t ) + 2 q1( t) + 1.5 q2 (t ) + 2.5 q3 ( t)
= - 2 q1( t) + 8 q2 (t ) - 12 q3 (t ) + 4 x( t) + 2 q1(t ) + 15
. q2 ( t) + 2.5 q3 ( t)
\ y( t) = 9.5 q2 ( t) - 9.5 q3 (t ) + 4 x(t )
On arranging the output equation in the matrix form we get,
LM q (t) OP
MMq (t)PP +
Nq (t)Q
1
y(t ) = 0
9.5
- 9.5
2
4 x( t)
3
State Model
State model of a continuous time system is given by state equations and output equations.
State equations
& (t) = A Q(t) + B X(t)
: Q
LM q& (t) OP
MMq& (t)PP
Nq& (t)Q
1
2
3
=
E
LM 0
MM 0
N-0.5
1
0
2
OP LM q (t) OP LM0OP
P Mq (t)P + MM0PP
-3PQ MNq (t )PQ
N1Q
0
1
1
2
3
x(t )
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 10
Output equations : Y(t) = C Q(t) + D X(t)
E
LM q (t) OP
MMq (t)PP +
Nq (t)Q
1
y(t ) = 0
9.5
- 9.5
2
4
x(t )
3
Example 5.2
The transfer function of a continuous time system is given by,
H(s) =
15
. s3 + 2 s2 + 3 s + 2
s3 + 3 s2 + 2 s + 4
Determine the state model of the continuous time system.
Solution
15
. s 3 + 2 s2 + 3 s + 2
s3 + 3 s2 + 2 s + 4
On dividing the numerator and denominator by s3 we get,
Given that,
H(s) =
1
1
1
+ 3 2 + 2 3
s
s
s
1
1
1
1+ 3
+ 2 2 + 4 3
s
s
s
15
. + 2
H(s) =
Y(s)
Let, H(s) =
=
X(s)
1
1
1
+ 3 2 + 2 3
s
s
s
1
1
1
1+ 3
+ 2 2 + 4 3
s
s
s
15
. + 2
On cross multiplication we get,
Y (s) + 3
Y(s)
Y(s)
Y(s)
X(s)
X(s)
X(s)
. X(s) + 2
+ 2
+ 4
= 15
+ 3
+ 2
s
s
s2
s3
s2
s3
Y(s)
Y (s)
Y(s)
X(s)
X(s)
X(s)
. X(s) + 2
- 2
- 4
+ 15
+ 3
+ 2
s
s2
s3
s
s2
s3
The direct form-II structure of the continuous time system described by the above equation is shown in fig 1.
\ Y (s) = - 3
The choice of number of state variables is equal to the number of integrators. The direct form-II structure has one
input x(t), one output y(t) and three integrators, and so let us choose three state variables. Let, q1(t), q2(t) and q3(t) be the
three state variables. Assign state variables at the output of every integrator. Hence the first derivative of state variable will
be available at the input of integrator.
X(s)
x(t)
+
1.5
1.5 q& 3 (t)
+
q& 3 (t)
1
s
+
–3 q3(t)
q3(t)
2
–3
2 q3(t)
+
q& (t)
2
1
s
+
–2 q2(t)
q2(t)
3
–2
3 q2(t)
q& (t)
1
1
s
–4 q1(t)
q1(t)
–4
Q1(s)
2
2 q1(t)
Fig 1 : Direct form - II structure.
+
y(t)
Y(s)
Signals & Systems
5. 11
State Equations
The state equations are formed by equating the sum of incoming signals of integrator to first derivative of state
variable, as shown below.
q& 1( t) = q2 (t )
q& 2 ( t) = q3 (t )
q& 3 ( t) = - 4 q1( t) - 2 q2 ( t) - 3 q3 ( t) + x(t )
On arranging the state equations in the matrix form we get,
LM q& (t) OP
MMq& (t)PP =
Nq& (t)Q
1
2
3
LM 0
MM 0
N- 4
OP LM q (t) OP LM0OP
PP MMq (t)PP + MM0PP
-3Q Nq (t )Q
N1Q
1
0
-2
0
1
1
2
x( t)
3
Output Equations
Output equation y(t) is formed by equating incoming signals of output node / summing point to y(t) as shown below.
y(t) = 15
. q& 3 ( t) + 2 q1(t) + 3 q2 (t) + 2 q3 (t)
On substituting for q& 3 (t ) from state equation we get,
y( t) = 15
. -4 q1(t ) - 2 q2 (t ) - 3 q3 (t ) + x( t) + 2 q1( t) + 3 q2 (t ) + 2 q3 ( t)
= - 6 q1( t) - 3 q2 ( t) - 4.5 q3 ( t) + 15
. x(t ) + 2 q1( t) + 3 q2 (t ) + 2 q3 ( t )
= - 4 q1(t ) - 2.5 q3 ( t) + 15
. x(t )
On arranging the output equation in the matrix form we get,
LM q (t)OP
MMq (t)PP +
Nq (t)Q
1
y( t) = -4
0
- 2.5
2
15
. x(t )
3
State Model
State model of a continuous time system is given by state equations and output equations.
& (t) = A Q(t) + B X(t)
State equations : Q
E
LM q& (t) OP L 0
MMq& (t)PP = MM 0
Nq& (t)Q MN-4
Output equations :
1
1
2
0
3
-2
OP LM q (t) OP LM0OP
PP MMq (t)PP + MM0PP
-3Q Nq (t )Q
N1Q
0
1
1
2
x(t )
3
Y(t) = C Q(t) + D X(t)
E
LM q (t) OP
MMq (t)PP +
Nq (t)Q
1
y( t) = -4
0
- 2.5
2
3
1.5 x( t)
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 12
Example 5.3
The state space representation of a continuous time system is given by,
A =
LM2 -1OP
N3 1 Q
; B =
LM1OP
N2Q
; C = 1 3
; D = 3
Derive the transfer function of the continuous time system.
Let, P be a square matrix.
Transpose of Cofactor Matrix of P
Now, P -1 =
Deter min ant of P
If, P is a square matrix of size 2 ´ 2, then its cofactor
matrix is obtained by interchanging the elements of
main diagonal and changing the sign of other two
elements as shown in the following example.
Solution
Transfer function of a continuous time system is given by,
Y (s)
= C (sI - A)-1 B + D
X(s)
L1 0OP - LM2 -1OP
sI - A = s M
MN0 1PQ MN3 1PQ
s
L 0OP - LM2 -1OP = LMs - 2
= M
MN0 sPQ MN3 1 PQ MN -3
(sI - A)-1 =
LMs - 1
MN 3
1
s-2
1
-3
s -1
P =
\ P -1 =
OP
s - 1PQ
-1 O
P =
s - 2PQ
LM
MN
s - 1 -1
1
= 2
s - 3s + 5
3
s-2
1
OP
PQ
2
= 1 3
s 2 - 3s + 5
-1
s 2 - 3s + 5
s-2
s 2 - 3s + 5
LM s - 1
+5
MM s - 3s
3
MN
2
=
s 2 - 3s + 5
OP 1
PP LMN2OPQ +
PQ
OP LM1OP +
Q MN2PQ
OP
Q
LM
N
p 22
1
´
p11 p12
-p 21
p 21 p 22
-1
s 2 - 3s + 5
s - 2
s 2 - 3s + 5
OP
PQ
OP
PP
PQ
3
LM s - 1 + 9
N s - 3s + 5
=
s - 1 + 9 + 2( -1 + 3(s - 2)) + 3(s 2 - 3s + 5)
s 2 - 3s + 5
=
s - 1 + 9 - 2 + 6s - 12 + 3s 2 - 9s + 15
s 2 - 3s + 5
-1 + 3(s - 2)
s 2 - 3s + 5
21
p12
p 22
LM
MN
=
2
11
s - 1 -1
1
(s - 2) (s - 1) - ( -3) ´ 1 3
s-2
Y (s)
\
= C (sI - A )-1 B + D
X(s)
LM s - 1
MM s - 33s + 5
MN
LMp
Np
3 =
=
LM s - 1 + 9
N s - 3s + 5
2
+
2(-1 + 3(s - 2))
s2 - 3s + 5
OP +
Q
3
3s 2 - 2s + 9
s2 - 3s + 5
Example 5.4
Find the state transition matrix for the continuous time system parameter matrix, A =
Solution
n
0O
Ls
= M
-2PQ
N0
s
0O
L-3
- M
s PQ
N0
The state transition matrix, e At = L-1 (s I - A )-1
sI - A = s
LM1 0OP
N0 1Q
-
LM-3
N0
0
-2
OP = LMs + 3
Q N0
0
s+2
OP
Q
LM-3
N0
0
-2
OP
Q
-p12
p11
OP
Q
Signals & Systems
5. 13
(sI - A)-1 =
LMs + 2
N0
1
s+3
0
0
s+2
0
s+3
LM
N
=
0
s+ 2
1
0
s+3
(s + 3) (s + 2)
=
LM 1
MM s + 3
MN 0
OP
Q
Let, P be a square matrix.
OP
Q
If, P is a square matrix of size 2 ´ 2, then its
cofactor matrix is obtained by interchanging the
elements of main diagonal and changing the sign
of other two elements as shown in the following
example.
p11 p12
P =
p 21 p 22
OP
1 P
P
s + 2 PQ
0
LM
N
R|L 1 0 OU|
n
s
S|MMM s + 3 1 PPPV|
|TMN 0 s + 2 PQ|W
LML RS 1 UV 0 OP
MM T s + 3 W R 1 UPP = LMe 0u(t) e
MN 0 L ST s + 2 VWPQ N
\ P -1 =
e At = L-1 (s I - A)-1 = L-1
-1
=
Transpose of Cofactor Matrix of P
Deter min ant of P
Now, P -1 =
-3t
-1
0
-2t
u(t)
OP
Q
LM
N
p 22
1
´
p11 p12
-p 21
p 21 p 22
-p12
p11
OP
Q
OP
Q
Example 5.5
LM q& (t) OP =
MNq& (t)PQ
LM1
MN1
OP LM q (t) OP
PQ MNq (t)PQ
L q (0) OP = LM1OP
Compute the solution of state equation by assuming the initial state vector as, M
MNq (0)PQ MN0PQ
The state equation of an LTI continuous time system is given by,
1
2
0
1
1
2
1
2
Solution
The solution of state equations are given by,
n
s
n
s
Q(t) = L-1 (sI - A)-1 Q(0) + L-1 (sI - A )-1 B X(s)
Here X(s) = 0, (because there is no input).
n
s
L1 0OP - LM1 0OP
sI - A = s M
MN0 1PQ MN1 1PQ
Ls 0OP - LM1 0OP = LMs - 1 0 OP
= M
MN0 sPQ MN1 1PQ MN -1 s - 1PQ
LMs - 1 0 OP
1
(sI - A) =
s -1 0
N 1 s - 1Q
\ Q(t) = L-1 (sI - A )-1 Q(0)
-1
-1
1
(s - 1)2
=
LM 1
MM s -1 1
MN (s - 1)
2
Now, P -1 =
Transpose of Cofactor Matrix of P
Deter min ant of P
If, P is a square matrix of size 2 ´ 2, then its
cofactor matrix is obtained by interchanging the
elements of main diagonal and changing the sign
of other two elements as shown in the following
example.
p11 p12
P =
p 21 p 22
LM
N
\ P -1 =
s -1
=
Let, P be a square matrix.
LMs - 1
N1
0
OP
Q
s -1
OP
P
1 P
s - 1PQ
0
OP
Q
LM
N
p 22
1
´
p11 p12
-p 21
p 21 p 22
-p12
p11
OP
Q
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 14
Now, Q(s) = (sI – A )–1 Q(0)
LM 1
MM s 1- 1
MN (s - 1)
=
LM 1 OP
OP
1O
L
s - 1
P
1 P M0P = M
1
M
P
N
Q
s - 1PQ
MN (s - 1) PPQ
LM L RS 1 UV OP
L
O
lQ(s)q = MM R T s -1 1WUPP = MMe u(t) PP
MNL ST (s - 1) VWPQ Nt e u(t)Q
0
2
2
-1
t
-1
Q( t) = L
t
-1
2
Example 5.6
The state model of a continuous time system is given by,
A =
LM2 0OP
N3 1Q
; B =
LM1OP
N2Q
; C = 1 3
; D = 3
Find the response of the system for unit step input. Assume zero initial condition.
Solution
The response, Y(t) = C Q(t) + D X(t)
n
s
n
s
-1
-1
-1
-1
where, Q(t) = L (sI - A) Q(0) + L (sI - A ) B X(s)
Here Q(0) = 0, (because initial conditions are zero).
n
L1 0OP
sI - A = s M
MN0 1PQ
s
\ Q(t) = L-1 (sI - A)-1 B X(s)
(sI - A )-1 =
1
s-2
-3
LM2 0OP = LMs 0OP - LM2 0OP = LMs - 2 0 OP
MN3 1PQ MN0 sPQ MN3 1PQ MN -3 s - 1PQ
Let, P be a square matrix.
LMs - 1 0 OP
0
Transpose of Cofactor Matrix of P
MN 3 s - 2PQ
Now, P =
-1
Deter min ant of P
s -1
LM
MN
=
0
s -1
1
(s - 2) (s - 1)
3
s-2
=
LM 1
MM s 3- 2
MN (s - 1) (s - 2)
OP
P
1 P
s - 1PQ
0
OP
PQ
If, P is a square matrix of size 2 ´ 2, then its
cofactor matrix is obtained by interchanging the
elements of main diagonal and changing the sign
of other two elements as shown in the following
example.
p11 p12
P =
p 21 p 22
Given that, x(t) = u(t).
1
\ X(s) = L x(t ) = L u( t ) =
s
l q
l q
LM
N
\ P -1 =
Now, Q(s) = (sI – A )–1 B X(s)
=
=
LM 1
LM s -1 2 + 0
0 O L1O
P
s-2
1
O
L
MM 3
P M P = MM 3
1 P M P MN s PQ
MN (s - 1) (s - 2) s - 1PQ MN2PQ
MN (s - 1) (s - 2) +
LM 1 OP
LM s -1 2 OP
1O
L
MM 3 + 2 (s - 2) PP MN s PQ = MM s(s2s -- 2)1 PP
MN s(s - 1) (s - 2) PQ
MN(s - 1) (s - 2) PQ
OP
P L 1O
2 P MN s PQ
s - 1PQ
OP
Q
1
p11 p12
p 21 p 22
´
LM p
N -p
22
21
-p12
p11
OP
Q
Signals & Systems
5. 15
l q
Q(t) = L-1 Q(s) = L-1
Let,
LM 1 OP
MM s2s(s -- 12) PP =
MN s(s - 1) (s - 2) PQ
-1
-1
1
k
k2
= 1 +
s(s - 2)
s
s - 2
1
´ s
s(s - 2)
where, k1 =
k2 =
\
Let,
LM L RS 1 UV OP
MM T s (s - 2) W PP
MML RS s(s -2s1) -(s1- 2) UVPP
N T
WQ
=
s=0
1
´ (s - 2)
s(s - 2)
1
1
= 0-2
2
=
s=2
1
2
1
1 1
1
1
= +
s(s - 2)
2 s
2 s-2
k
k2
k3
2s - 1
= 1 +
+
s(s - 1) (s - 2)
s
s - 1
s-2
2s - 1
´ s
s(s - 1) (s - 2)
where, k 1 =
k2 =
k3 =
0 - 1
1
= 2
(0 - 1) (0 - 2)
=
s=0
2s - 1
´ (s - 1)
s(s - 1) (s - 2)
2s - 1
´ (s - 2)
s(s - 1) (s - 2)
=
2 ´ 1 - 1
= -1
1 ´ (1 - 2)
=
2 ´ 2 - 1
3
=
2 ´ (2 - 1)
2
s =1
s=2
2s - 1
1 1
1
3
1
\
= +
s(s - 1) (s - 2)
2 s
s - 1
2 s - 2
LM L RS 1 UV OP
T s(s - 2) W
Q(t) = M
MM R 2s - 1 U PPP =
MNL ST s(s - 1) (s - 2) VW PQ
LM - 1 u(t) + 1 e u(t)
2
2
= M
MN- 1 u(t) - e u(t) + 3 e
2
2
-1
-1
2t
t
2t
LM L RS- 1 1 + 1 1 UV OP
MM R 1 T1 2 s 1 2 s 3- 2 W 1 UPP
MNL ST- 2 s - s - 1 + 2 s - 2 VWPQ
OP LM 1 de - 1i u(t) OP
PP = MM 1 2
P
3e - e - 1i u(t)P
u(t)
d
Q N2
Q
-1
-1
2t
2t
t
Now, the response of the continuous time system is given by,
Response,
Y(t) = C Q(t) + D X(t)
LM 1 de - 1i u(t) OP
2
MM 1 3e
P+
- e - 1i u(t)P
d
N2
Q
2t
= 1 3
2t
d
3 u(t)
t
i
d
i
= 0.5 e2t - 1 u(t) + 1.5 3e2t - e t - 1 u(t) + 3u(t)
d
\ Response, y(t ) = d5e
i
d
i
= 0.5e 2t - 0.5 + 4.5e2t - 15e
. t - 15
. + 3 u(t) = 5 e2t - 15
. e t + 1 u(t )
2t
i
- 15e
. t + 1 u(t)
Chapter 5 - State Space Analysis of Continuous Time Systems
5.7
5. 16
Summary of Important Concepts
1. The state of a continuous time system refers to the condition of continuous time system at any time instant.
2. The state variables of a continuous time system are a set of variables that completely describe the state of
continuous time system at any time instant.
3. The state equations of a continuous time system are a set of N-number of first order differential equations.
4. The state equations of a continuous time system are formed by taking first derivative of state variables as
function of state variables and inputs.
5. The output equations of a continuous time system are a set of P-number of algebraic equations formed by
taking outputs as function of state variables and inputs.
6. The state model of a continuous time system is given by state equations and output equations.
7. The transfer function of a continuous time system can be obtained from its state model using the equation
C(sI – A)–1 B + D.
8. The state transition matrix of a continuous time system is, eAt = L–1{(sI – A)–1}.
9. The solution of state equations of a continuous time system due to initial condition (and with no input) is,
Q(t) = eAt Q(0).
10. The solution of state equations of a continuous time system due to input (and with zero initial conditions)
is, Q(t) = L –1{(sI – A)–1 B X(s)}.
11. The solution of state equations of a continuous time system due to input and initial conditions is,
Q(t) = eAt Q(0) + L–1{(sI – A)–1 B X(s)}.
12. The response of a continuous time system using its state model can be obtained from the equation,
Y(t) = C Q(t) + D X(t)
5.8 Short Questions and Answers
Q5.1
What are the advantages of state space analysis?
1.
Q5.2
2.
3.
The state space analysis is applicable to any type of systems. They can be used for modelling and
analysis of linear and non-linear systems, time invariant and time variant systems and multiple
input and multiple output systems.
The state space analysis can be performed with initial conditions.
The variables used to represent the system can be any variables in the system.
4.
Using this analysis the internal states of the system at any time instant can be predicted.
What is a state vector?
The state vector is an ( N ´ 1) column matrix (or vector) whose elements are state variables of the
system, (where N is the order of the system). It is denoted by Q(t).
Q5.3
What is state space?
The set of all possible values which the state vector Q(t) can have (or assume) at time t forms the state
space of the system.
Q5.4
What is input and output space?
The set of all possible values which the input vector X(t) can have (or assume) at time t forms the input
space of the system.
The set of all possible values which the output vector Y(t) can have (or assume) at time t forms the
output space.
5. 17
Q5.5
Signals & Systems
Determine system matrix of the system governed by the differential equation,
dy
d 2y
+2
+ 4y = 0.
dt
dt 2
Solution
d2 y
dy
+2
+ 4y = 0. The given system is a second order system and so let us choose two state
dt 2
dt
variables q1 and q2. Let us equate the state variables to the system variable y and its derivative dy/dt.
\ q1 = y
Þ q& 1 = y& = q2
Given that,
q2 = y&
Þ q& 2 = &&
y
On substituting the state variables and their derivatives in the given system equation we get,
q& 2 + 2q2 + 4q1 = 0
q& 2 = - 4q1 - 2q2
Þ
Now the state equations are,
q& 1 = q2
q& 2 = - 4q1 - 2q2
Now, the state equations in the matrix form is,
q1
0
1
q& 1
=
Þ
-4 -2 q 2
q& 2
LM OP
MN PQ
LM
MN
OP LM
PQ MN
OP
PQ
Therefore the system matrix is, A =
Q5.6
LM 0
MN-4
& = AQ
Q
OP
-2PQ
1
Determine state model of the system shown in fiq Q5.6.1.
Solution
X(s)
Let us assign first derivative of state variables at the input of the
integrator as shown in fig Q5.6.2, so that the state variables exists at
the output of integrators.
The state equations are formed by equating the incoming signals to
input of the integrator to the first derivative of state variable as shown
below.
+
1
+
1
s
Y(s)
s
-a
-b
Fig Q5.6.1.
q& 1(t) = q2 (t)
q& 2 (t) = - b q1(t) - a q2 (t) + x(t)
Here the output equation is, y(t) = q1(t).
X(s)
Now, the state model in matrix form is,
x(t)
LM q& (t) OP
MNq& (t)PQ
1
2
y(t)
=
LM 0
MN-b
= 1
OP LM q (t)OP + LM0OP x(t)
-a PQ MNq (t)PQ MN 1PQ
Lq (t) OP
0 M
MNq (t)PQ
1
1
+
+
q& 2 (t)
– a q2(t)
1
q& 1 (t)
s
q 2(t)
1
LM s
Given that, f(s) = L oe t = M s + 5
MN s -+5 5
-a
– b q 1(t)
-b
OP
PP
Q
1
s 2 + 5 . Determine the system matrix.
s
2
s +5
2
At
2
Solution
We know that , f(s) = [ sI – A ] -–1
On taking inverse on both sides of above equation we get,
[f(s)] – 1 = sI – A
f(s)
-1
Y(s)
s q (t)
1
2
2
Q5.7
1
LM
MM
MN
s
1 s2 + 5
=
5
f(s)
s2 + 5
A = sI – [f(s)] – 1
Þ
-1
s2 + 5
s
s2 + 5
OP
PP
PQ
Fig Q5.6.2.
y(t)
Chapter 5 - State Space Analysis of Continuous Time Systems
Here, f(s) =
\ f(s)
-1
=
-5
s
s
1
s2 + 5
1
´ 2
- 2
´ 2
=
= 2
2
s +5 s +5 s +5 s +5
s +5
s2 + 5
2
ds
d
2
+5
L
i MMM
MN
s
s2 + 5
5
s2 + 5
\ System matrix, A = sI - f(s)
Q5.8
5. 18
-1
s2 + 5
s
s2 + 5
-1
=s
i
OP Ls -1O
PP = MMN5 s PPQ
PQ
LM1
MN0
0
OP - LMs
PQ MN5
OP
PQ
-1
1
s
=
LMs
MN0
0
s
OP - LMs
PQ MN5
OP
PQ
-1
s
=
LM 0
MN-5
OP
PQ
1
0
Determine the transfer function of the system described by the following state model.
&
q(t)
= - 2 q(t) + 0.5 x(t)
y(t) = 0.7 q(t)
Solution
& = -2 q(t) + 0.5x(t)
Given that, q(t)
On taking Laplace transform of above equation we get,
s Q(s) = - 2 Q(s) + 0.5 X(s)
Þ
Q(0) = 0
s Q(s) + 2 Q(s) = 0.5 X(s)
Þ
(s + 2) Q(s) = 0.5X(s)
0.5
\ Q(s) =
X(s)
s+2
.....(1)
Given that, y( t) = 0.7 q( t)
On taking Laplace transform of above equation we get,
Y(s) = 0.7 Q(s)
0.5
X(s)
s+2
From equations (1) and (2) we can write,
Y(s)
0.5
0.35
\ Transfer function,
= 0.7 ´
=
X(s)
s+2
s+2
= 0.7 ´
Q5.9
Using equation (1)
The output equation of a system is, y(t) = [ 1 2 ] Q(t).
If the state transition matrix, e A t =
LM e
MN-e
-t
e- 2t
-2t
0
OP
PQ
and the initial condition vector, Q(0) =
LM2OP
N 1Q ,
Find the zero-input response of the system.
Solution
When the input is zero, the state equations, Q(t) = eAt Q(0).
\ Zero - input response, y zi (t) = 1 2 Q(t) = 1 2 eA t Q(0)
LM e
MN-e
-t
1 2
=
d
-2t
OP L2O = e
MP
0 PQ MN 1PQ
e -2t
-t
- 2e -2t
e -2t
LM2OP
MN1PQ
i
i u(t)
= 2 e - t - 2e -2t + e -2t = 2e - t - 3e-2t ; for t > 0
d
-t
= 2e - 3e
Q5.10
-2 t
The output equation of a system is, y(t) = [ 1 2 ] Q(t) + [ 2 ] X(t).
LMe
MN e
-2t
If the state vector, Q(t) =
-t
OP
PQ
.Find the zero-state response of the system for unit step input.
5. 19
Signals & Systems
Solution
LMe OP +
MN e PQ
-2 t
Zero - state response, y zs ( t) = 1 2 Q( t ) + 2 X( t) =
1 2
-t
2 u(t )
= e -2 t + 2e -t + 2 ´ 1 = e -2 t + 2e - t + 2 ; for t > 0
d
i
= e -2 t + 2e - t + 2 u( t)
5.9
MATLAB Programs
Program
5.1
Write a MATLAB program to find the state model of the system governed
by the transfer function, H(s)= (1.5s3+2s2+3s+2)/(s3+3s2+2s+4).
%Program
to
determine
the
state
model
from
the
clear all
s=tf(‘s’);
H=(1.5*s^3+2*s^2+3*s+2)/(s^3+3*s^2+2*s+4);
disp(‘The given transfer function is’);
tf(H)
%display the given
transfer
transfer
function
function
disp(‘Enter numerator coefficients of given transfer function’);
b=input(‘’);
disp(‘Enter denominator coefficients of given transfer function’);
a=input(‘’);
[A,B,C,D]=tf2ss(b,a)
%compute
matrices
A,B,C,D
OUTPUT
The given transfer function is
Transfer function:
1.5 s^3 + 2 s^2 + 3 s + 2
------------------------s^3 + 3 s^2 + 2 s + 4
Enter numerator coefficients of given transfer function
[1.5 2 3 2]
Enter denominator coefficients of given transfer function
[1 3 2 4]
A =
-3
1
0
-2
0
1
-4
0
0
B =
1
0
0
C =
-2.5000
D =
1.5000
0
-4.0000
of
state
model
Chapter 5 - State Space Analysis of Continuous Time Systems
Program
5. 20
5.2
Write a MATLAB program to determine the transfer function of a system
by getting the state model from the user through keyboard.
%Program
to
determine
the
transfer
disp(‘Enter elem ents o f the
disp(‘A=’);
A=input(‘’);
disp(‘Enter elem ents o f the
disp(‘B=’);
B=input(‘’);
disp(‘Enter elem ents o f the
disp(‘C=’);
C=input(‘’);
disp(‘Enter elements of the
disp(‘D=’);
D=input(‘’);
[b,a]=ss2tf(A,B,C,D)
disp(‘The transfer function
tf(b,a)
system
input
function
of
Nth
m a tr ix
of
size
m a tr ix
o utput
the
B
m a tr ix
transmission
of
A
state
of
C
of
matrix
model
order
size
NxN’);
Nx1 ’);
size
D
system.
of
1 xN’);
size
1x1’);
is’);
OUTPUT
Enter elements of the
A =
[2 -1;
Enter elements of the
B =
[1; 2]
Enter elements of the
C=
[1 3]
Enter elements of the
D =
[3]
system matrix A of size NxN
3 1]
input matrix B of size Nx1
output matrix C of size 1xN
transmission matrix D of size 1x1
B =
3.0000
-2.0000
9.0000
A =
1.0000
-3.0000
5.0000
The transfer function of the state model is
Transfer Function:
3 S^2 - 2 S + 9
---------------S^2 - 3 S + 5
5.10 Exercises
I. Fill in the blanks with appropriate words
1.
2.
3.
4.
A set of variables that completely describe a system at any time instant are called –––––––––.
The ––––––––– of a continuous time system consists of the state equations and output equations.
The pictorial representation of the state model of the system is called ––––––––– .
The number of state variables in the state diagram of a continuous time system is equal to the number
of –––––––––.
5. In state space modelling the number of state variables will decide ––––––––– of the system.
5. 21
Signals & Systems
6. In the matrix form of state equation, A represents –––––––––.
7. In the matrix form of state equation, B represents –––––––––.
8. In the matrix form of output equation, C represents –––––––––.
9. In the matrix form of output equation, D represents –––––––––.
10. The transfer function model does not provide information regarding ––––––––– of the system.
Answers
1. state variables 2. state model
3. state diagram
4. integrators
5. order
6. system matrix 7. input matrix
8.output matrix
9. transmission matrix
10. internal state
II. State whether the following statements are True/False
1. The state variable analysis can be applied for any type of system (linear/nonlinear and time varying/invarying
system).
2. The transfer function analysis is applicable to nonlinear systems.
3. The transfer function analysis can be carried with initial conditions.
4. The state space analysis can be carried with initial conditions and on multiple input and output systems.
5. In transfer function analysis only one input and output is considered at any one time.
6. In state space analysis multiple inputs and outputs can be allowed at any one time.
7. The state model of a system is nonunique but the transfer function is unique.
8. The state and output equations of a system are functions of state variables and inputs.
9. The state diagram of a continuous time system cannot be used for simulation of the system in analog
computers.
10. The state diagram of a continuous time system provides a direct relation between time domain and s-domain.
Answers
1. True
2. False
3. False
4. True
5. True
6. True
7. True
8. True
9. False
10. True
III. Choose the right answer for the following questions
& = A Q(t) + B X(t) the input matrix is,
1. In the state equation of a continuous time system, Q(t)
a) B
b) X(t)
c) A
d) Q(t)
2. A continuous time system is described by four numbers of first order differential equations. The number of
integrators present in the direct form-II structure is,
a) 2
b) 4
c) 1
d) 8
3. If there is no input, the time domain solution of state equation of a continuous time system is equal to,
a) Q(t) = eAt
b) Q(t) = e At X(0)
4. Inverse Laplace transform of (sI – A)–1 is called
a) state equation in matrix form
c) transfer function
c) Q(t) = e At Q(0)
d) Q(t) = e- At Q(0)
b) state transition matrix
d) response of continuous time system
5. The transfer functions analysis has a drawback that it cannot be applied to
a) single input and single output system
b) linear time varying system
c) linear time invariant system
d) none of the above
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 22
6. The transfer function of a continuous time system is described by,
Y(s) s 4 + 4s + 3
=
. The
X(s) s 5 + 4s 3 + 1
number of state variables in the state model of the system is,
a) 1
b) 5
c) 4
d) 2
7. The transfer function of the continuous time system having the following state variable description is,
1 0
1
A=
; C= 1 0
; B=
-2 1
0
LM
N
a)
OP
Q
LM OP
NQ
2
(s - 1) 2
b)
1
s+1
c)
1
s-1
d)
1
(s + 1)2
LMt e u(t) 0 OP. The value of (sI - A) is,
N e u(t) 2 t e u(t)Q
O
LM 1
LM 2
O
LM 2
O
LM 1
O
0 P
0 P
0 P
0 P
(s - 4)
(s + 4)
(s + 4)
(s - 4)
a) M
P b) M 1
P
c) M
P d) MM 1
P
2 P
2 P
2 P
2 P
MM 1
1
M
M
MN (s + 1) (s - 1) PQ
MN (s - 1) (s + 1) PQ
MN (s - 1) (s + 1) PQ
N (s + 1) (s - 1) PQ
L1 0 OP . The state transition matrix is,
9. The system matrix of a continuous time system is given by, A = M
N1 1Q
LM 1
O
LM 1 0 OP
LM 1 0 OP
LM 1 1 OP
0 P
(s + 1)
s-1
s
1
s
1
s
1
d) M
a) M
c) M
P
b) M
1 P
MM 1
MM (s --11) s -1 1 PPP
MN -1 s -1 1PPQ
MN(s + 1) s1 PPQ
Q
N
N (s - 1) (s +1) QP
Le 0 OP , and initial condition vector is L2O .
10. The state transition matrix of a system is M
MN3PQ
MN 0 e PQ
-4t
-1
8. The state transition matrix of a system is given by, e At =
2
-4t
t
2
2
2
2
2
2
2
2
2
2
2
-2 t
-t
The state of the system at the end of 1-second without any external input is,
a)
LM0.271OP
N1.100Q
b)
LM 0.135 OP
N0.368Q
c)
LM0.271OP
N0.736Q
d)
LM0.135OP
. Q
N1100
Answers
1. a
6. b
2. b
7. c
3. c
8.d
4. b
9. c
5. b
10. a
IV. Answer the following questions
1. Compare state space analysis and transfer function model analysis of continuous time system.
2. What are the drawbacks in transfer function model analysis of continuous time system?
3. Define state and state variables of continuous time system.
4. What is state model of a continuous time system?
5. Write the state model of Nth order continuous time system.
6. Write the state equations of Nth order continuous time system.
7. Write the output equations of Nth order continuous time system.
5. 23
Signals & Systems
8. How will you determine the transfer function of a continuous time system from state model ?
9. Write the expression to compute state transition matrix of a continuous time system via Laplace transform.
10. Write the expression for time domain solution of state equations of a continuous time system in terms of state
transition matrix and using Laplace transform.
V. Solve the following problems
E 5.1.
E 5.2.
Determine the state model of the continuous time systems governed by the following differential
equations.
a)
d 3y ( t )
d 2 y( t )
dy( t )
d 3x ( t )
d 2 x(t )
dx( t )
+ 3.5
+2
- 4 y(t) = 2.5
+ 12
.
-5
+ x( t )
3
2
3
dt
dt
dt
dt
dt 2
dt
b)
d 2 y( t )
dy( t )
dx( t )
+6
+ 4 y(t) = 4
+ x( t )
dt 2
dt
dt
Compute the state transition matrix eAt for the following system matrices.
a) A =
E 5.3.
LM 0 1 OP
N -5 -6 Q
1
2
s + 3s + 2
LM2 -1OP ;
N1 0 Q
L 0 1OP ;
b) A = M
N-1 1Q
LM-2 1 OP
N 1 -2 Q
d) A =
LM-2 0 OP
N 0 -1Q
b) H(s) =
s+3
s2 + 4s + 3
c) H(s) = 2
s + 3s + 2
s + 9 s + 20
2
d) H(s) =
s2 + 6s + 8
( s + 3) (s2 + 2s + 2)
LM1OP ;
N0Q
L0O
B= M P ;
N2Q
B=
C= 3 1 ; D= 2
C= 3 0 ; D= 0
Compute the solution of the following state equations representing LTI continuous time systems.
a)
b)
E 5.6.
c) A =
Determine the transfer function of the continuous time systems having the following state variable
description.
a) A =
E 5.5.
LM-1 0 OP
N 0 -3Q
Obtain the state model of the continuous time systems whose transfer functions are given below.
a) H(s) =
E 5.4.
b) A =
LMq& (t) OP = LM3 1 OP LMq (t) OP ;
Nq& (t)Q N0 -4Q Nq (t)Q
LMq& (t) OP = LM-1 0 OP LMq ( t) OP ;
Nq& ( t)Q N 2 -5Q Nq ( t)Q
1
1
2
2
1
1
2
2
LMq (0) OP = LM 0 OP
Nq (0)Q N-2Q
LMq (0) OP = LM-0.5OP
Nq (0)Q N 1 Q
1
2
1
2
Find the response of LTI continuous time systems having the state model and inputs as given below.
Assume zero initial conditions.
LM1
N0
L1
b) A = M
N4
a) A =
OP ; B = LM3OP ;
Q
N5Q
0O
L -1O
; B= M P ;
3PQ
N -2 Q
-1
1
C = 2 -1 ; D = 1 ; Input, x(t) = e-2t u(t)
C= 4 3 ;
D = 2 ; Input, x1 (t) = 2 u(t) and x2 (t) = e t u(t)
Chapter 5 - State Space Analysis of Continuous Time Systems
5. 24
Answers
E5.1 a) State equation
0
q& 1 ( t )
LM OP LM
MMq&& (t)PP = MM0
Nq ( t)Q N4
OP LMq (t) OP LM0OP
1 P Mq ( t ) P + M0P
-3.5PQ MNq ( t ) PQ MN1PQ
1
0
2
0
3
-2
1
2
x(t)
3
b) State equation
q& 1 ( t )
0 1
=
-4 -6
q& 2 ( t )
Output equation
LM
N
OP LM
Q N
OP LMq (t) OP + LM0OP
Q Nq ( t)Q N1Q
Output equation
LMq ( t) OP
y(t) = 11 -10 -7.55 Mq ( t )P + 2.5 x(t)
MNq ( t)PQ
LM 1 d5e - e i 1 de - e i OP
4
4
a) e = M
u( t )
5
1
MN 4 d-e + e i 4 d5e - e iPPQ
Le cosh t e sinh t OP u(t)
c) e = M
N e sinh t e cosh t Q
1
x(t)
2
LMq ( t) OP
Nq ( t)Q
y(t) = 1 4
1
1
2
2
3
-t
E5.2
-5 t
-t
-5t
At
-t
-5 t
-5 t
-2t
-2t
-2t
-2t
-t
At
E5.3 a) State equation
q& 1 ( t )
0 1 q1 ( t )
0
=
+
&q 2 ( t )
-2 -3 q 2 ( t )
1
Output equation
q1 ( t )
y(t) = 1 0
q 2 ( t)
LM
N
OP LM
Q N
OP LM
QN
LM OP
N Q
c) State equation
q& 1 ( t )
0
q& 2 ( t ) = 0
LM OP
MM & PP
Nq ( t ) Q
3
LM
MM
N-20
OP LM OP
Q NQ
x(t)
OP LMq ( t) OP LM0OP
1P Mq ( t )P + M0P
0PQ MNq ( t ) PQ MN1PQ
1
0
0
-9
3
Output equation
-t
-2 t
d) e At
OP
Q
0O
P u(t)
e Q
0
u(t)
e -3 t
-t
b) State equation
q& 1 ( t )
0 1
=
q& 2 ( t )
-2 -3
LM
N
OP LM
Q N
Output equation
y(t) = 3
OP LMq ( t) OP + LM0OP
Q Nq ( t)Q N1Q
L q (t) OP
1 M
Nq ( t ) Q
1
2
x(t)
LM OP LM 0
MMq&& (t)PP = MM 0
N q ( t ) Q N -6
OP LMq ( t) OP LM0OP
P Mq (t)P + M0P
-5PQ MNq ( t ) PQ MN1PQ
1
2
0
3
-8
LMq (t) OP
MM PP
Nq ( t ) Q
1
2
3
LMq (t) OP
1 Mq ( t ) P
MNq (t) PQ
2
3
Y(s) 2 s2 - s + 3
=
(s - 1) 2
X(s)
Y(s)
6
b)
=
X(s) s2 - s + 1
LM d
iOP u(t)
MN
PQ
OP u(t)
1 L -2e
b) Q(t) = M
4 N-e + 5e Q
2 -4 t
e - e 3t
E5.5 a) Q(t) = 7
-2e -4 t
E5.4 a)
-t
-t
LM 1 FG -5t e + 14 e - 14 e
3
3
a) Q(t) = M 3 H
5
MM
de - e i
3
N
L 10 13 e - 34 e
y(t) = M- t e +
N 3 9 9
t
t
-2 t
-2 t
t
t
1
1
1
y(t) = 8 6
3
E5.6
0
Output equation
y (t) = 3 4 1 q 2 ( t )
t
-2t
IJ OP
K P u( t)
PP
Q
OP u(t)
Q
x(t)
2
1
d) State equation
q& 1 ( t )
1
2
LMe
N0
Le
=M
N0
b) e At =
-5 t
LM 2d1 - e i OP
b) Q(t) = M F 1 2
I u( t)
MN4GH - 3 - 3 e + e JK PPQ
3t
3t
t
y(t) = 8 + 12e t - 16 e3t u( t )
x(t)
CHAPTER 6
Discrete Time Signals and Systems
6.1 Discrete and Digital Signals
The discrete signal is a function of a discrete independent variable. The independent variable is
divided into uniform intervals and each interval is represented by an integer. The letter "n" is used to
denote the independent variable. The discrete or digital signal is denoted by x(n). The discrete signal is
defined for every integer value of the independent variable "n". The magnitude (or value) of discrete signal
can take any discrete value in the specified range. Here both the value of the signal and the independent
variable are discrete.
When the independent variable is time t, the discrete signal is called discrete time signal. In
discrete time signal, the time is quantized uniformly using the relation t = nT, where T is the sampling time
period. (The sampling time period is inverse of sampling frequency). The discrete time signal is denoted
by x(n) or x(nT).
The digital signal is same as discrete signal except that the magnitude of the signal is quantized. The
magnitude of the signal can take one of the values in a set of quantized values. Here quantization is
necessary to represent the signal in binary codes.
The discrete or digital signals have a sequence of numbers (or values) defined for integer values of
the independent variable. Hence the discrete or digital signals are also known as discrete sequence. In this
book the term sequence and signal are used synonymously. Also in this book the discrete signal is referred
as discrete time signal.
6.1.1 Generation of Discrete Signals
A discrete signal can be generated in the following three methods.
The methods 1 and 2 are independent of any time frame but method 3 depends critically on time.
1. Generate a set of numbers and arrange them as a sequence.
Example :
The numbers 0, 1, 2, ...., (N – 1) form the ramp like sequence and can be expressed as,
x(n) = n ; 0 £ n £ (N – 1)
2. Evaluation of a numerical recursion relation will generate a discrete signal.
Example :
x(n) =
x(n - 1)
with initial condition x(0) = 1, gives the sequence,
3
0
When n = 0 ; x(0 ) = 1 ( initial condition)
When n = 1 ; x(1) =
x(1 - 1)
=
3
x(0 )
1
=
3
3
FG 1 IJ
H 3K
F1 I
= G J
H3 K
=
1
x(n) =
FG 1 IJ
H 3K
n
; 0 £ n < ¥
Chapter 6 - Discrete Time Signals and Systems
6. 2
When n = 2 ; x(2) =
x(2 - 1)
x(1)
1
=
=
3
3
9
When n = 3 ; x(3) =
x(3 - 1)
x(2)
1
=
=
3
3
27
\ x(n) =
FG 1 IJ
H 3K
FG 1 IJ
H 3K
F 1I
= G J
H 3K
2
=
3
and so on
n
; 0 £ n < ¥
3. A third method is by uniformly sampling a continuous time signal and using the amplitudes of
the samples to form a sequence.
Let, x(t) = Continuous time signal
Now, Discrete signal, x(nT) = x(t) t = nT ; - ¥ < n < ¥
where, T is the sampling interval
The generation of discrete signal by sampling an analog ramp signal is shown in fig 6.1.
x(t)
x(nT)
......
0
0
T
.
1 2 3 4 5
N-1
n
t
Fig 6.1a. Analog signal.
Fig 6.1b. Discrete time signal.
Fig 6.1 : Generation of a discrete time signal.
6.1.2 Representation of Discrete Time Signals
The discrete time signal can be represented by the following methods.
1. Functional representation
In functional representation the signal is represented as mathematical equation, as shown in the
following example.
x(n) = – 1
=
2
=
1.5
= – 0.9
=
1.4
=
1.6
=
0
;
;
;
;
;
;
;
n = – 2
n = – 1
n =
0
n =
1
n =
2
n =
3
other n
x(n)
2. Graphical representation
2
1.6
1.5
In graphical representation the signal is represented
in a two dimensional plane. The independent variable is
represented in the horizontal axis and the value of the signal
is represented in the vertical axis as shown in fig 6.2.
–2
–1
–1
1.4
0
1
–0.9
2
3
n
Fig 6.2 : Graphical representation of a
discrete time signal.
6. 3
Signals & Systems
3. Tabular representation
In tabular representation, two rows of a table are used to represent a discrete time signal. In the
first row the independent variable "n" is tabulated and in the second row the value of the signal for each
value of "n" are tabulated as shown in the following example.
n
........... –2 -1
x(n)
........... –1
2
0
1
2
3
..............
1.5 –0.9 1.4 1.6 ..............
4. Sequence representation
In sequence representation, the discrete time signal is represented as one dimensional array as
shown in the following examples.
An infinite duration discrete time signal with the time origin, n = 0, indicated by the symbol ­ is represented as,
x(n) = { ..... – 1, 2, 1.5, – 0.9, 1.4, 1.6, ..... }
­
An infinite duration discrete time signal that satisfies the condition x(n) = 0 for n < 0 is represented as,
x(n) = { 1.5, – 0.9, 1.4, 1.6, ... }
­
or
x(n) = {1.5, – 0.9, 1.4, 1.6, ... }
A finite duration discrete time signal with the time origin, n = 0, indicated by the symbol ­ is represented as,
x(n) = { – 1, 2, 1.5, – 0.9, 1.4, 1.6 }
­
A finite duration discrete time signal that satisfies the condition x(n) = 0 for n < 0 is represented as,
x(n) = { 1.5, – 0.9, 1.4, 1.6 }
­
or
x(n) = { 1.5, – 0.9, 1.4, 1.6}
6.2 Standard Discrete Time Signals
1. Digital impulse signal or Unit sample sequence
Impulse signal, d( n) = 1 ; n = 0
d(n)
u(n)
1
1
=0 ; n ¹ 0
0
n
1
2
4
3
Unit step signal, u( n) = 1 ; n ³ 0
= 0; n < 0
5
n
Fig 6.4 : Unit step signal.
Fig 6.3 : Digital
impulse signal.
2. Unit step signal
.....
0
5
ur(n)
4
3
3. Ramp signal
2
.....
1
Ramp signal, u r ( n) = n ; n ³ 0
= 0 ;n < 0
0
1
2
3
4
5
Fig 6.5 : Ramp signal.
n
Chapter 6 - Discrete Time Signals and Systems
6. 4
4. Exponential signal
n
Exponential signal, g( n) = a ; n ³ 0
= 0 ;n < 0
g(n)
g(n)
a>1
0<a<1
.....
.....
0
1
2
3
4
0
6
5
1
2
3
4
n
Fig 6.6b : Increasing exponential signal.
n
Fig 6.6a : Decreasing exponential signal.
Fig 6.6 : Exponential signal.
5. Discrete time sinusoidal signal
The discrete time sinusoidal signal may be expressed as,
x n = A cos w 0 n + q ; for n in the range -¥ < n < +¥
bg
b
g
xb ng = A sin bw n + qg ; for n in the range -¥ < n < +¥
0
where, w0 = Frequency in radians/sample ;
q = Phase in radians
w0
= Frequency in cycles/sample
f0 =
2p
x(n)
–9 –8 –7 –6 –5 –4
4
–3 –2 –1
0 1
2
5
6
7
8 9
n
3
Fig 6.7a : Discrete time sinusoidal signal represented
by equation x(n) = A cos(w0n).
x(n)
–6 –5 –4 –3 –2 –1
7
0 1
2
x(n)
1
–5 –4 –3
2
3
4
5
6
7
n
-2 -1 0
4
5 6
8
9 10 11 12
n
Fig 6.7b : Discrete time sinusoidal signal represented by
equation x(n) = Asin(w0n).
14243
p
3
Fig 6.7c : Discrete time sinusoidal signal represented by equation
x(n)= A cos p n + p ; w 0 = p ; q = p .
6
3
6
3
FH
3
IK
Fig 6.7 : Discrete time sinusoidal signals.
6. 5
Signals & Systems
Properties of Discrete Time Sinusoid
1. A discrete time sinusoid is periodic only if its frequency f is a rational number, (i.e., ratio of two
integers).
2. Discrete time sinusoids whose frequencies are separated by integer multiples of 2p are identical.
\ x(n) = A cos[( w0 + 2pk ) n + q ],
for k = 0,1,2...........are identical in the interval
- p £ w0 £ p and so they are indistinguishable.
Proof :
cos[( w0 + 2pk) n + q] = cos(w0n + 2pnk + q) = cos[(w0n + q) + 2pnk]
= cos(w0n + q) cos 2pnk - sin (w0n + q) sin 2pnk
Since n and k are integers, cos 2pnk =1 & sin 2pnk = 0
\ cos[( w0 + 2pk) n + q] = cos(w0n + q),
for k = 1, 2, 3, .....
Conclusion
1.The sequences of any two sinusoids with frequencies in the range, -p £ w o £ p , or,
-1/2 £ f0 £ 1/2, are distinct.
by 2p
[-p £ w £ p ¾divide
¾¾¾
¾® -1/2 £ f £ 1/2]
2. Any discrete time sinusoid with frequency w0 > |p| (or f0 > |1/2|) will be identical to another
discrete time sinusoid with frequency w0 < |p| (or f0 < |1/2|).
6. Discrete time complex exponential signal
The discrete time complex exponential signal is defined as,
x(n) = a n e j( w 0 n + q) = an [cos(w0n + q) + j sin(w0n + q)]
= an cos(w0n + q) + j an sin(w0n + q) = xr(n) + j xi(n)
where, xr(n) = Real part of x(n) = an cos(w0n + q)
xi(n) = Imaginary part of x(n) = an sin(w0n + q)
The real part of x(n) will give an exponentially increasing cosinusoid sequence for a > 1 and exponentially
decreasing cosinusoid sequence for 0 < a < 1.
xr(n)
xr(n)
0<a<1
a>1
...... n
n
.....
Fig 6.8a : The discrete time sequence represented by the
equation, xr(n) = an cos w0n for 0 < a < 1.
Fig 6.8b : The discrete time sequence represented by
the equation xr(n) = an cos w0n for a > 1.
Fig 6.8 : Real part of complex exponential signal.
Chapter 6 - Discrete Time Signals and Systems
6. 6
The imaginary part of x(n) will give rise to an exponentially increasing sinusoid sequence for a > 1 and
exponentially decreasing sinusoid sequence for 0 < a < 1.
xi(n)
xi(n)
0<a<1
a>1
.....
.....
n
n
Fig 6.9b : The discrete time sequence
represented by the equation
xi(n) = an sin w0n for a > 1.
Fig 6.9a : The discrete time sequence
represented by the equation,
xi(n) = an sin w0n for 0 < a < 1.
Fig 6.9 : Imaginary part of complex exponential signal.
6.3 Sampling of Continuous Time (Analog) Signals
The sampling is the process of conversion of a continuous time signal into a discrete time signal.
The sampling is performed by taking samples of continuous time signal at definite intervals of time.
Usually, the time interval between two successive samples will be same and such type of sampling is
called periodic or uniform sampling.
The time interval between successive samples is called sampling time (or sampling period or
sampling interval), and it is denoted by “T”. The unit of sampling period is second (s). (The lower units are
milli-second (ms) and micro-second (ms)).
The inverse of sampling period is called sampling frequency (or sampling rate), and it is denoted
by Fs. The unit of sampling frequency is Hertz (Hz). (The higher units are kHz and MHz).
Let, xa(t) = Analog / Continuous time signal.
x(n) = Discrete time signal obtained by sampling xa(t).
Mathematically the relation between x(n) and xa(t) can be expressed as,
x(n) = xa (t) t = nT = xa (nT) = xa
F nI
GH F JK ;
for n in the range -¥ < n < ¥
s
where, T = Sampling period or interval in seconds
1
= Sampling rate or Sampling frequency in Hertz
Fs =
T
bg
b
Example : Let, x a t = A cos W 0t + q
g
b
= A cos 2 pF0 t + q
g
where, W0 = Frequency of analog signal in rad/s
F0 =
W0
= Frequency of analog signal in Hz
2p
Let xa (t) be sampled at intervals of T seconds to get x(n), where T =
1
Fs
6. 7
Signals & Systems
\ x(n) = x a (t)
t = nT
= A cos(W 0t + q)
t = nT
F
I
= A cos(W nT + q) = A cos G 2 pF n + qJ = A cos d2pf n + qi = A cos dw n + qi
o
o
HF
K
0
0
s
where, f 0 =
F0
= Frequency of discrete sinusoid in cycles/sample
Fs
w0 = 2pf 0= Frequency of discrete sinusoid in radians/sample
6.3.1 Sampling and Aliasing
In section 6.2 it is observed that any two sinusoid signals with frequencies in the range
-1/2 £ f £ +1/2 are distinct and a discrete sinusoid with frequency, f > 1/2 will be identical to another
discrete sinusoid with frequency, f < 1/2 . Therefore we can conclude that range of discrete frequency
is -1/2 to +1/2 . But the range of analog frequency is -¥ to +¥ . While sampling analog signals, the infinite
frequency range continuous time signals are mapped or converted to finite range discrete time signals.
The relation between analog and digital frequency is,
f =
F
Fs
.....(6.1)
The range of discrete frequency is,
-
1
1
£ f £
2
2
.....(6.2)
On substituting for f from equation (6.1) in equation (6.2) we get,
-
1
F
1
£
£
2
Fs
2
.....(6.3)
On multiplying equation (6.3) by Fs we get,
Fs
F
.....(6.4)
£ F £ s
2
2
From equation (6.4) we can say that when an analog signal is sampled at a frequency Fs, the
highest analog frequency that can be uniquely represented by a discrete time signal will be Fs/2.
The continuous time signal with frequency above Fs/2 will be represented as a signal within the range
+ Fs/2 to - Fs/2 . Hence the signal with frequency above Fs/2 will have an identical signal with frequency
below Fs/2 in the discrete form.
-
Hence infinite number of high frequency continuous time signals will be represented by a single
discrete time signal. Such signals are called alias. While sampling at Fs, the frequency above Fs/2 will
have alias with frequency below Fs/2. Hence the point of reflection is Fs/2, and the frequency Fs/2 is called
folding frequency.
The discrete time sinusoids, A sin((2pf + 2pk)n), will be alias for integer values of k. It is also
observed that, a sinusoidal signal with frequency F1 will be an alias of sinusoidal signal with frequency F2
if it is sampled at a frequency Fs = F1 - F2. In general if the sampling frequency is any multiple of F1 - F2,
(i.e., Fs = k(F1 - F2) where k = 1, 2, 3, ........) the signal with frequency F2 will be an alias of the signal
with frequency F1. The phenomenon of high frequency component getting the identity of low frequency
component during sampling is called aliasing.
Chapter 6 - Discrete Time Signals and Systems
6. 8
Let, Fmax be maximum analog frequency that can be uniquely represented as discrete time signal
when sampled at a frequency Fs.
Now, Fmax =
Fs
2
.....(6.5)
.....(6.6)
\ Fs = 2Fmax
The equation (6.6) gives a choice for selecting sampling frequency. From equation (6.6) we can
say that for unique representation of analog signal with maximum frequency Fmax, the sampling frequency
should be greater than 2Fmax.
i.e., to avoid aliasing Fs ³ 2Fmax
..... (6.7)
When sampling frequency Fs is equal to 2Fmax, the sampling rate is called Nyquist rate.
It is observed that a nonshifted sinusoidal signal when sampled at Nyquist rate, will produce zero
sample sequence (i.e., discrete sequence with all zeros), (because the sinusoidal signal is sampled at its
zero crossings). (Refer example 6.3). Hence to avoid zero sampling of sinewave, the sampling frequency
Fs should be greater than 2Fmax, where Fmax is the maximum frequency in the analog signal.
A discrete signal obtained by sampling can be reconstructed to analog signal, only when it is
sampled without aliasing. The above concepts of sampling analog signals are summarized as the sampling
theorem, given below.
Sampling Theorem : A band limited continuous time signal with highest frequency (bandwidth) Fm hertz
can be uniquely recovered from its samples provided that the sampling rate Fs is
greater than or equal to 2Fm samples per second.
Note : The effects of aliasing in frequency spectrum are discussed in Chapter-8.
Example 6.1
Consider the analog signals, x1(t) = 2 cos 2p(10t) and x2(t) = 2 cos 2p(50t).
Find a sampling frequency so that 50Hz signal is an alias of the 10Hz signal?
Solution
Let, the sampling frequency, Fs = 50 - 10 = 40Hz.
\ x1(n) = x1(t)
t = nT =
= 2 cos
x 2 (n) = x 2 (t)
= 2 cos 2p(10t)
n
t=
Fs
= 2 cos2 p
FG 10 ´ n IJ
H 40 K
p
n
2
t = nT =
= 2 cos
n
Fs
n
Fs
= 2 cos 2p(50t)
t=
FG
H
p
5p
n = 2 cos 2pn +
n
2
2
IJ
K
n
Fs
= 2 cos 2p
= 2 cos
FG 50 ´ n IJ
H 40 K
For integer values of n
p
n
2
cos(2pn + q) = cos q
From the above analysis we observe that x1(n) and x2(n) are identical , and so x2(t) is an alias of x1 (t) when sampled
at a frequency of 40Hz.
6. 9
Signals & Systems
Example 6.2
Let an analog signal, xa(t) = 10 cos 100 pt. If the sampling frequency is 75Hz, find the discrete time signal x(n). Also
find an alias frequency corresponding to Fs = 75Hz.
Solution
x(n) = x a (t)
t = nT =
= 10 cos
= 10 cos100pt
n
Fs
t =
n
Fs
= 10 cos100p ´
FG
H
n
Fs
IJ
K
100p ´ n
4p
2p
2p
1
n = 10 cos 2p n = 10 cos
n = 10 cos 2p n
= 10 cos
75
3
3
3
3
We know that the discrete time sinusoids whose frequencies are separated by integer multiple of 2p are identical.
8p
4
2p
2p
n = 10 cos 2p n
n = 10 cos
\10cos
+ 2 p n = 10 cos
3
3
3
3
4
2p
4
Now, 10 cos2 p n is an alias of 10cos
n. Here the frequency of the signal, 10 cos 2p n is,
3
3
3
4
f =
cycles / sample
3
F
4
We know that, f =
Þ F = f Fs =
´ 75 = 100Hz
Fs
3
FG
H
IJ
K
\ When, Fs = 75Hz, F = 100 Hz is an alias frequency.
Example 6.3
Consider the analog signal, xa(t) = 5 cos50pt + 2 sin 200pt – 2 cos100pt .
Determine the minimum sampling frequency and the sampled version of analog signal at this frequency. Sketch the
waveform and show the sampling points. Comment on the result.
Solution
The given analog signal can be written as shown below.
xa(t) = 5 cos50pt + 2 sin 200pt – 2 cos100pt = 5 cos 2p F1t + 2 sin 2p F2t – 2cos 2p F3t
Where, 2p F1 = 50p
Þ F1 = 25Hz
2p F2 = 200p
Þ F2 = 100Hz
2p F3 = 100p
Þ F3 = 50Hz
The maximum analog frequency in the signal is 100Hz. The sampling frequency should be twice that of this maximum
analog frequency.
i.e., Fs ³ 2 Fmax Þ
Fs ³ 2 ´ 100
Let, sampling frequency, Fs = 200 Hz
\ x a (nT) = x a (t ) t = nT = x a (t)
= 5 cos
t=
n
Fs
50pn
200pn
100 pn
pn
pn
+ 2 sin
- 2 cos
+ 2 sin pn - 2 cos
= 5 cos
200
200
200
4
2
For integer values of n, sinpn = 0.
pn
pn
- 2 cos
2
4
The components of analog waveform and the sampling points are shown in fig1.
Comment : In the sampled version of analog signal xa (nT), the component 2 sin 200pt will give always zero samples when
sampled at 200Hz for any value of n. This is the drawback in sampling at Nyquist rate (i.e., sampling at Fs = 2Fmax).
\ x a (nT) = 5 cos
Chapter 6 - Discrete Time Signals and Systems
6. 10
1
= 0.04 sec
F1
5 cos 50pt ; F1 = 25Hz ; T1 =
2 sin 200 pt ; F2 = 100Hz ; T2 =
1
= 0.01 sec
F2
2 cos100pt; F3 = 50Hz ; T3 =
1
= 0.02 sec
F3
Fs = 200Hz ; T =
1
= 0.005 sec
Fs
0.005
0.01
0.02
0.04
Fig 1. Sampling points of the components of the signal xa(t).
6.4 Classification of Discrete Time Signals
The discrete time signals are classified depending on their characteristics. Some ways of
classifying discrete time signals are,
1. Deterministic and nondeterministic signals
2. Periodic and aperiodic signals
3. Symmetric and antisymmetric signals.
4. Energy signals and power signals
5. Causal and noncausal signals
6.4.1 Deterministic and Nondeterministic Signals
The signals that can be completely specified by mathematical equations are called deterministic
signals.The step, ramp, exponential and sinusoidal signals are examples of deterministic signals.
The signals whose characteristics are random in nature are called nondeterministic signals.
The noise signals from various sources are best examples of nondeterministic signals.
6.4.2 Periodic and Aperiodic Signals
A signal x(n) is periodic with periodicity of N samples (where N is an integer) if and only if
x(n + N) = x(n)
; for all n
The smallest value of N for which the above equation is true is called fundamental period. If there
is no value of N that satisfies the above equation, then it is called aperiodic or nonperiodic signal. When
N is the fundamental period, the periodic signals will also satisfy the condition x(n + kN) = x(n), where k
is an integer. Periodic signals are power signals. The sinusoidal and complex exponential signals are
periodic signals when their fundamental frequency, f0 is a rational number.
x(n)
N=5
2
2
1
1
1
1
1
2
2
1
1
1
0
–1
–1
n
–1
–1
–1
–1
x(n) = {.......1, 1, 2, –1, –1, 1, 1, 2, –1, –1, 1, 1, 2, –1, –1, ........}
­
Fig 6.10 : Periodic discrete time signal.
–1
–1
6. 11
Signals & Systems
Example 6.4
Determine whether following signals are periodic or not. If periodic find the fundamental period.
a) x(n) = sin
FG 6p n + 1IJ
H7
K
b) x(n) = cos
FG n
H8
- p
IJ
K
p
c) x(n) = cos n2
8
d) x(n) = e j7 pn
Solution
a) Given that, x(n) = sin
FG 6p n + 1IJ
H7
K
Let N and M be two integers.
6p
6pn
6p
N
(n + N) + 1 = sin
Now, x(n + N) = sin
+ 1 +
7
7
7
FG
H
IJ
K
Since sin(q + 2pM) = sin q, for periodicity
FG
H
6p
IJ
K
N should be integral multiple of 2p.
7
Let,
6p
N = M ´ 2p, where M and N are integers.
7
\ N = M ´ 2p ´
7
7M
=
6p
3
Here N is an integer if, M = 3, 6, 9, 12, .......
Let,
M = 3;
\N = 7
When N = 7 ; x(n + N) = sin
FG 6pn
H7
+ 1 +
IJ
K
6p
´ 7
7
= sin
FG 6pn
H7
+ 1 + 6p
IJ
K
= sin
Hence x(n) is periodic with fundamental period of 7 samples.
b) Given that, x(n) = cos
FG n
H8
- p
IJ
K
Let N and M be two integers.
b
g
Now, x n + N = cos
FG n + N
H 8
- p
IJ
K
= cos
Since cos(q + 2pM) = cos q, for periodicity
Let,
FG n
H8
+
N
- p
8
IJ
K
= cos
FG n
H8
- p +
N
8
IJ
K
N
8 should be equal to integral multiple of 2p.
N
= M ´ 2p ; where M and N are integers.
8
\ N = 16 pM
Here N cannot be an integer for any integer value of M and so x(n) will not be periodic.
c) Given that, x(n) = cos
FG p n IJ
H8 K
2
The general form of discrete time cosinusoid is x(n) = cos(2pfon).
p 2
n
8
p 2
1
n
n ´
=
\ fo =
8
2 pn
16
Let, 2pfon =
Since n is an integer, f0 is a rational number and so cos
FG p n IJ is periodic.
H8 K
2
FG 6pn
H7
IJ
K
+ 1 = x(n)
Chapter 6 - Discrete Time Signals and Systems
6. 12
Let N be the fundamental period and M be an integer.
p 2
8
= 16M
Now for periodicity,
N = M ´ 2 p Þ N2 = M ´ 2 p ´
p
8
Þ
N = 4 M
Here N is integer, when M = 12, 22, 32, .....
Let, M = 1 ; \ N = 4
Hence x(n) is periodic with fundamental period of 4 samples.
d) Given that, x(n) = ej7pn
Let N and M be two integers.
Now, x(n + N) = ej7p(n + N)
= ej7pn ej7pN
Since e j2pM = 1, for periodicity 7p N should be integral multiple of 2p.
Let, 7pN = M ´ 2p,
\ N = M ´ 2p ´
1
2M
=
7p
7
Here, N is integer, when M = 7, 14, 21, ......
When M = 7 ; N = 2
\ x(n) is periodic with fundamental period of 2 samples.
6.4.3 Symmetric (Even) and Antisymmetric (Odd) Signals
The signals may exhibit symmetry or antisymmetry with respect to n = 0. When a signal exhibits
symmetry with respect to n = 0 then it is called an even signal. Therefore the even signal satisfies the
condition,
x(-n) = x(n)
When a signal exhibits antisymmetry with respect to n = 0, then it is called an odd signal. Therefore
the odd signal satisfies the condition,
x(-n) = -x(n)
x(n)
x(n)
3
3
2
2
–4
2
2
1
2
1
–3
–2
–1
1
1
0
1
2
3
4
–4
n
1
–3 –2 –1
0
1
–1 –1
x(n) = {1, 2, 3, 2, 1, 2, 3, 2, 1}
­
Fig 6.11a : Symmetric (or even) signal.
1
2
3
4
n
–1
–2
x(n) = {2, 1, –1, –1, 0, 1, 1, –1, –2}
­
Fig 6.11b : Antisymmetric (or odd) signal.
Fig 6.11 : Symmetric and antisymmetric discrete time signal.
A discrete signal x(n) which is neither even nor odd can be expressed as a sum of even and
odd signal.
Let,
x(n) = xe (n) + xo (n)
bg
where, x e n = Even part of x(n) ;
bg
xo n = Odd part of x(n)
Note : If x(n) is even then its odd part will be zero. If x(n) is odd then its even part will be zero.
6. 13
Signals & Systems
Now, it can be proved that,
1
xe (n) =
x(n) + x(- n)
2
1
xo (n) =
x(n) - x(- n)
2
Proof :
Let, x(n) = xe(n) + xo(n)
......(6.8)
On replacing n by –n in equation (6.8) we get,
x(–n) = xe(–n) + xo(–n)
......(6.9)
Since xe (n) is even, xe(–n) = xe(n)
Since xo (n) is odd, xo(–n) = – xo(n)
Hence the equation (6.9) can be written as,
x(–n) = xe(n) – xo(n)
......(6.10)
On adding equation (6.8) and (6.10) we get,
x(n) + x(–n) = 2 xe(n)
1
x(n) + x( -n)
2
\ x e (n) =
On subtracting equation (6.10) from equation (6.8) we get,
x(n) – x(–n) = 2 xo(n)
1
x(n) - x( -n)
2
\ x o (n) =
Example 6.5
Determine the even and odd parts of the signals.
a) x(n) = a n
b) x(n) = 2 e
p
j n
3
c) x(n) = {4, - 4, 2, - 2}
Solution
a) Given that, x(n) = an
\ x(-n) = a -n
1
1 n
Even part, x e (n) =
[ x(n) + x( -n)] =
[a + a - n ]
2
2
1
1 n
Odd part, x o ( -n) =
[ x(n) - x(-n)] =
[ a - a -n ]
2
2
b) Given that, x(n) = 2 e
p
j n
3
j
p
n
3
-j
p
n
3
x(n) = 2 e
\ x(-n) = 2 e
Even part, xe (n) =
=
= 2 cos
p
p
n + j2 sin n
3
3
p
p
= 2 cos n - j2 sin n
3
3
1
[ x(n) + x(-n)]
2
LM
N
OP
Q
LM
N
OP
Q
1
1
p
p
p
p
p
p
2 cos n + j2sin n + 2 cos n - j2sin n =
4 cos n = 2 cos n
2
2
3
3
3
3
3
3
Chapter 6 - Discrete Time Signals and Systems
6. 14
1
Odd part, xo (n) =
[ x(n) - x(-n)]
2
=
OP
Q
LM
N
OP
Q
LM
N
1
1
p
p
p
p
p
p
j4 sin n = j2 sin n
2 cos n + j2sin n - 2 cos n + j2sin n =
2
2
3
3
3
3
3
3
c) Given that, x(n) = {4, –4, 2, –2}
­
\ x(0) = 4
Given that, x(n) = {4, –4, 2, –2},
­
; x(1) = –4 ; x(2) = 2
; x(3) = –2
\ x(–3) = –2 ; x(–2) = 2 ; x(–1) = –4 ; x(0) = 4
x(–n) = {–2, 2, –4, 4},
­
1
[x(n) - x(-n)]
2
Even part, xe (n) =
Odd part, xo (n) =
At n = – 3 ;
1
[x(n) + x(-n)]
2
x(n) + x(–n) = 0 + (–2) = – 2
At n = – 3
;
x(n)
– x(–n)
=
0 – (–2) =
At n = – 2 ;
x(n) + x(–n) = 0 + 2
2
At n = – 2
;
x(n)
– x(–n)
=
0–2
=–4
At n = – 1
;
x(n)
– x(–n)
=
0 – (–4) =
4
4–4
0
At n = – 1 ;
=
x(n) + x(–n) = 0 + (–4)
At n =
0 ;
x(n) + x(–n) = 4 + 4
=
8
At n = 0
;
x(n)
– x(–n)
=
At n =
1 ;
x(n) + x(–n) = – 4 + 0
=–4
At n = 1
;
x(n)
– x(–n)
=–4–0
At n =
2 ;
x(n) + x(–n) = 2 + 0
=
2
At n = 2
;
x(n)
– x(–n)
=
At n =
3 ;
x(n) + x(–n) = – 2 + 0
=–2
At n = 3
;
x(n)
– x(–n)
= –2 – 0
2–0
\ x(n) + x(–n) = {–2, 2, –4, 8, –4, 2, –2}
­
1
\ xe (n) =
[x(n) + x(-n)]
2
\ x(n) – x(–n) = {2, –2, 4, 0, –4, 2, –2}
­
1
[x(n) - x(-n)]
\ xo (n) =
2
= {-1, 1, - 2, 4, - 2, 1, - 1}
­
= {1, - 1, 2, 0, - 2, 1, - 1}
­
= –2
=
=–4
=
The energy E of a discrete time signal x(n) is defined as,
¥
å
x(n)
2
.....(6.11)
n = -¥
The energy of a signal may be finite or infinite, and can be applied to complex valued and
real valued signals.
If energy E of a signal is finite and non-zero, then the signal is called an energy signal. The
exponential signals are examples of energy signals.
The average power of a discrete time signal x(n) is defined as,
Power, P = lim
N ®¥
1
2N + 1
N
å
x(n)
2
.....(6.12)
n = -N
If power P of a signal is finite and non-zero, then the signal is called a power signal. The periodic
signals are examples of power signals.
For energy signals, the energy will be finite and average power will be zero. For power signals the
average power is finite and energy will be infinite.
i.e., For energy signal, 0 < E < ¥ and P = 0
For power signal, 0 < P < ¥ and E = ¥
2
=–2
6.4.4 Energy and Power Signals
Energy, E =
2
6. 15
Signals & Systems
Example 6.6
Determine whether the following signals are energy or power signals.
FG 1 IJ n u(n)
H 4K
a) x(n) =
b) x(n) = cos
FG p nIJ
H3 K
c) x(n) = u(n)
Solution
F 1 I u(n)
GH 4 JK
F 1I
Here, x(n) = G J u(n) for all n.
H 4K
F 1I
\ x(n) = G J = 0.25 ;
H 4K
n
a) Given that, x(n) =
n
n
n
+¥
å
Energy, E =
¥
2
x(n) =
n = -¥
å
2
Lt
N® ¥
Infinite geometric series
sum formula.
¥
n
1
C =
1 - C
n= 0
¥
=
å
å
(0.25 2 )n
n=0
1
= 1.067 joules
1 - 0.0625
(0.0625)n =
n=0
Power, P =
(0.25)n
n=0
¥
å
=
n ³ 0
+N
1
2N + 1
å
2
x(n) =
n = -N
1
= Lt
N ®¥ 2N + 1
N
1
2N + 1
Lt
N®¥
1
(0.252 )n = Lt
N ®¥ 2N + 1
å
n = 0
Using infinite geometric series sum formula
N
å
(0.25)n
n = 0
N
å
(0.0625)n
n = 0
1
(0.0625)N+1 - 1
2N + 1
0.0625 - 1
1
0.0625 - 1
=
´
=0
0.0625 - 1
¥
=
2
Lt
Using finite geometric series sum formula
N ®¥
Finite geometric series
sum formula.
¥
n
CN + 1 - 1
C =
C - 1
n= 0
å
Here E is finite and P is zero and so x(n) is an energy signal.
b) Given that, x(n) = cos
F p nI
GH 3 JK
+¥
x(n)
2
+¥
cos 2
=
n = -¥
n = -¥
n = -¥
+¥
=
2p
FG p nIJ = å 1 + cos 3 n
å
å
H3 K
2
1F
2p I I
1F
2p I
F
å GH1 + cos 3 nJK JK = 2 GH å 1 + å cos 3 nJK
2 GH
+¥
Energy, E =
+¥
+¥
n
n = -¥
n = -¥
n = -¥
=
1
( ¥ + 0) = ¥
2
Note : Sum of infinite 1's is infinity. Sum of samples of one period of sinusoidal signal is zero.
Power, P =
=
=
Lt
N ®¥
Lt
N ®¥
Lt
N ®¥
1
2N + 1
1
2N + 1
1
2N + 1
N
å
x(n)
n = -N
2
=
N
1
Lt
N® ¥
2N + 1
n = -N
FG1 + cos 2p nIJ
H
3 K
å
2
1 L
2p O
n
1 + å cos
nP
M
å
3 PQ
2 MN
N
n = -N
N
N
n = -N
n = -N
å
cos2
pn
3
Chapter 6 - Discrete Time Signals and Systems
=
=
6. 16
LM
MN
1
1
11+44
1 +.......+
1 +31 + 0
24431 + 1 + 11+.......+
44244
2
2N + 1
N terms
N terms
Lt
N ®¥
1
1
2N + 1 =
2N + 1 2
Lt
N ®¥
Lt
N ®¥
OP
PQ
1
1
=
watts
2
2
Since P is finite and E is inifinite, x(n) is a power signal.
2p
2p
n for two periods are
n is periodic with periodicity of 3 samples. Samples of cos
3
3
given below. It can be observed that sum of samples of a period is zero.
2p
2p
2p
When n = 0 ; cos
n = 1, When n = 1 ; cos
n = - 0.5,
When n = 2 ; cos
n = - 0.5
3
3
3
2p
2p
2p
When n = 3 ; cos
n = 1, When n = 4 ; cos
n = - 0.5,
When n = 5 ; cos
n = - 0.5
3
3
3
Note: The term cos
c) Given that, x(n) = u(n)
+¥
E =
å
x(n)
2
+¥
=
n = -¥
P =
=
Lt
N ®¥
Lt
N® ¥
å
bu(n)g
2
+¥
=
å u(n) = 1 + 1 + 1 ......... ¥ = ¥
n=0
n=0
1
2N + 1
1
N
å
x(n) 2 =
n = -N
1
Lt
N® ¥
2N + 1
FG
H
F
N G2 +
H
N 1 +
(N + 1) = Lt
2N + 1
N ®¥
IJ
NK
1I
J
NK
N
å
u(n) = Lt
N ®¥
n=0
1
F
I
11+
1 +42
1 4444
+.........+1
G
444
3JJ
G
2N + 1 H
K
N + 1 terms
1
1
¥ = 1 + 0 = 1 watts
1
2 + 0
2
2+
¥
1 +
=
Since P is finite and E is infinite, x(n) is a power signal.
6.4.5 Causal, Noncausal and Anticausal signals
A signal is said to be causal, if it is defined for n ³ 0. Therefore if x(n) is causal, then x(n) = 0
for n < 0.
A signal is said to be noncausal, if it is defined for either n £ 0, or for both n £ 0
and n > 0. Therefore if x(n) is noncausal, then x(n) ¹ 0 for n < 0. A noncausal signal can be converted to
causal signal by multiplying the noncausal signal by a unit step signal, u(n).
When a noncausal signal is defined only for n £ 0, it is called an anticausal signal.
x(n) = {1, –1, 2, –2, 3, –3}
­
x(n) = { 2, 2, 3, 3,.............}
­
x(n) = {1, –1, 2, –2, 3, –3}
­
x(n) = {............,2, 2, 3, 3}
­
x(n) = {2, 3, 4, 5, 4, 3, 2}
­
123 123
Examples of Causal and Noncausal Signals
x(n) = {......, 2, 3, 4, 5, 4, 3, 2,......}
­
Causal signals
Noncausal signals
6. 17
Signals & Systems
6.5 Mathematical Operations on Discrete Time Signals
6.5.1 Scaling of Discrete Time Signals
Amplitude Scaling or Scalar Multiplication
Amplitude scaling of a signal by a constant A is accomplished by multiplying the value of every
signal sample by the constant A.
Example :
Let y(n) be amplitude scaled signal of x(n), then y(n) = A x(n)
When n = 0
; y(0) = A x(0) = 0.1 ´ 20
= 2.0
; n =1
When n = 1
; y(1) = A x(1) = 0.1 ´ 36
= 6.6
40
; n=2
When n = 2
; y(2) = A x(2) = 0.1 ´ 40
= 4.0
= –15
; n=3
When n = 3
; y(3) = A x(3) = 0.1 ´ (– 15) = –1.5
Let, x(n) =
20
; n=0
=
36
=
and A = 0.1,
Time Scaling (or Down Sampling and Up Sampling)
There are two ways of time scaling a discrete time signal. They are down sampling and up sampling.
In a signal x(n), if n is replaced by Dn, where D is an integer, then it is called down sampling.
In a signal x(n), if n is replaced by
n
, where I is an integer, then it is called up sampling.
I
Example :
If x(n) = an ; n ³ 0 ; 0 < a < 1
then x1(n) = x(2n) will be a down sampled version of x(n) and x2(n) = x n will be an up sampled version on x(n).
2
When n = 0 ; x1(0) = x(0) = a0
When n = 0 ; x 2 (0) = x 0 = x(0) = a 0
2
2
When n = 1 ; x1(1) = x(2) = a
When n = 1 ; x 2 (1) = x 1 = 0
2
When n = 2 ; x1(2) = x(4) = a4
ej
ej
ej
When n = 2 ; x (2) = x e 2 j = x(1) = a
2
When n = 3 ; x (3) = x e 3 j = 0
2
When n = 4 ; x (4 ) = x e 4 j = x(2) = a and so on.
2
1
x 1(3) = x(6) = a 6 and so on.
When n = 3 ;
2
2
2
2
x(n)
x1(n)
a0
x1(n) = x(2n)
a0
x2(n)
x 2 (n) = x n
2
ej
a0
a1
a1
a2
a2
a3
a4
0
1
2
3
4
a
5
5
a6
6
....
.... n
a2
a4
0
1
2
a3
a 6 ....
3
.... n
0
Fig 6.12a : A discrete time signal x(n). Fig 6.12b : Down sampled signal of x(n).
1
2
3
4
5
6
.....
..... n
Fig 6.12c : Up sampled signal of x(n).
Fig 6.12 : A discrete time signal and its time scaled version.
Chapter 6 - Discrete Time Signals and Systems
6. 18
6.5.2 Folding (Reflection or transpose) of Discrete Time Signals
The folding of a signal x(n) is performed by changing the sign of the time base n in x(n). The
folding operation produces a signal x(–n) which is a mirror image of x(n) with respect to time
origin n = 0.
Example :
Let x(n) = n ; –3 £ n £ 3. Now the folded signal, x1(n) = x(–n) = –n ; –3 £ n £ 3
x(n)
x1(n)
3
3
1
1
–1
–2
–3
0
1
x1(n) = x(–n)
2
2
2
3
–2
–3
n
–1
2
1
0
3
n
–1
–1
–2
–2
–3
–3
Fig 6.13a : A discrete time signal x(n).
Fig 6.13b : Folded signal of x(n).
Fig 6.13 : A discrete time signal and its folded version.
6.5.3 Time shifting of Discrete Time Signals
A signal x(n) may be shifted in time by replacing the independent variable n by n – m, where m is
an integer. If m is a positive integer, the time shift results in a delay by m units of time. If m is a negative
integer, the time shift results in an advance of the signal by |m| units in time. The delay results in shifting
each sample of x(n) to the right. The advance results in shifting each sample of x(n) to the left.
Example :
Let, x(n) = 1
=2
=3
=0
; n=2
; n=3
; n=4
; for other n
Let, x1(n) = x(n – 2), where x1(n) is delayed signal of x(n)
When n = 4 ; x1(4) = x(4 – 2) = x(2) = 1
When n = 5 ; x1(5) = x(5 – 2) = x(3) = 2
When n = 6 ; x1(6) = x(6 – 2) = x(4) = 3
The sample x(2) is available at n = 2 in the original
sequence x(n), but the same sample is available at n = 4
in x1(n). Similarly every sample of x(n) is delayed by two
sampling times.
x(n)
The sample x(2) is available at n = 2 in the original sequence
x(n), but the same sample is available at n = 0 in x2(n). Similarly
every sample of x(n) is advanced by two sampling times. Hence
the signal x2(n) is an advanced version of x(n).
x1(n) = x(n – 2)
x1(n)
3
Let, x2(n) = x(n + 2), where x2(n) is an advanced signal of x(n)
When n = 0 ; x2(n) = x(0 + 2) = x(2) = 1
When n = 1 ; x 2(n) = x(1 + 2) = x(3) = 2
When n = 2 ; x2(n) = x(2 + 2) = x(4) = 3
2
3
2
0
1
2
3
4
5
6
I sampling time ®
IInd sampling time ®
IIIrd sampling time ®
IVth sampling time ®
Vth sampling time ®
VIth sampling time ®
1
Time origin ®
1
st
x2(n) = x(n + 2)
x2(n)
3
2
n
0
1
2
3
4
1
5
6
n
Fig 6.14b : Delayed signal of x(n).
0
1
2
3
4
5
6
n
Fig 6.14c : Advanced signal of x(n).
Fig 6.14a : A discrete time signal x(n).
Fig 6.14 : A discrete time signal and its shifted version.
6. 19
Signals & Systems
Delayed Unit Impulse Signal
x3(n)
The unit impulse signal is defined as,
x3(n) = d(n – m)
1
d(n) = 1 ; for n = 0
m
0
= 0 ; for n ¹ 0
n
Fig 6.15 : Delayed unit impulse.
The unit impulse signal delayed by m units of time is denoted as d(n – m).
Now, d(n – m) = 1 ; n = m
x4(n)
x4(n) = u(n – m)
=0;n¹m
m+4
m+2
u(n) = 1 ; for n ³ 0
m+3
0
m
The unit step signal is defined as,
m+1
1
Delayed Unit Step Signal
n
Fig 6.16 : Delayed unit step signal.
= 0 ; for n < 0
The unit step signal delayed by m units of time is denoted as u(n – m).
Now, u(n – m) = 1 ; n ³ m
=0;n<m
6.5.4 Addition of Discrete Time Signals
The addition of two signals is performed on a sample-by-sample basis.
The sum of two signals x1(n) and x2(n) is a signal y(n), whose value at any instant is equal to the
sum of the samples of these two signals at that instant.
i.e., y(n) = x1(n) + x2(n) ; -¥ < n < ¥.
Example :
Let, x1(n) = {1, 2, –1} and x2(n) = {–2, 1, 3}
When n = 0 ; y(0) = x 1(0) + x2(0) = 1 + (–2) = –1
When n = 1 ; y(1) = x1(1) + x2(1) = 2 + 1 = 3
When n = 2 ; y(2) = x1(2) + x2(2) = –1 + 3 = 2
\ y(n) = x1(n) + x2(n) = {–1, 3, 2}
6.5.5 Multiplication of Discrete Time Signals
The multiplication of two signals is performed on a sample-by-sample basis.The product of two
signals x1(n) and x2(n) is a signal y(n), whose value at any instant is equal to the product of the samples
of these two signals at that instant. The product is also called modulation.
Example :
Let, x1(n) = { 1, 2, –1 } and x2(n) = { –2, 1, 3 }
When n = 0 ; y(0) = x1(0) ´ x2(0) = 1 ´ (–2) = –2
When n = 1 ; y(1) = x1(1) ´ x2(1) = 2 ´ 1
= 2
When n = 2 ; y(2) = x1(2) ´ x2(2) = –1 ´ 3 = –3
\ y(n) = x1(n) ´ x2(n) = {–2, 2, –3}
Chapter 6 - Discrete Time Signals and Systems
6. 20
6.6 Discrete Time System
A discrete time system is a device or algorithm that operates on a discrete time signal, called the
input or excitation, according to some well defined rule, to produce another discrete time signal called
the output or the response of the system. We can say that the input signal x(n) is transformed by the
system into a signal y(n), and the transformation can be expressed mathematically as shown in equation
(6.13).The diagrammatic representation of discrete time system is shown in fig 6.17.
Response, y(n) = H{x(n)}
.....(6.13)
where, H denotes the transformation (also called an operator).
Input signal
or
Excitation
x(n)
Discrete time
system
®H
y(n)
®
Output signal
or
Response
Fig 6.17 : Representation of discrete time system.
A discrete time system is linear if it obeys the principle of superposition and it is time invariant if
its input-output relationship do not change with time. When a discrete time system satisfies the properties
of linearity and time invariance then it is called an LTI system (Linear Time Invariant system).
Impulse Response
When the input to a discrete time system is a unit impulse d(n) then the output is called an impulse
response of the system and is denoted by h(n).
\ Impulse Response, h(n) = H{d(n)}
d(n)
®H
.....(6.14)
h(n)
®
Fig 6.18 : Discrete time system with impulse input.
6.6.1 Mathematical Equation Governing Discrete Time System
The mathematical equation governing the discrete time system can be developed as shown below.
The response of a discrete time system at any time instant depends on the present input, past
inputs and past outputs.
Let us consider the response at n = 0. Let us assume a relaxed system and so at n = 0, there is no
past input or output. Therefore the response at n = 0, is a function of present input alone.
i.e., y(0) = F[x(0)]
Let us consider the response at n =1. Now the present input is x(1), the past input is x(0) and past
output is y(0). Therefore the response at n = 1, is a function of x(1), x(0), y(0).
i.e., y(1) = F[y(0), x(1), x(0)]
Let us consider the response at n = 2. Now the present input is x(2), the past inputs are x(1) and
x(0), and past outputs are y(1) and y(0). Therefore the response at n = 2, is a function of x(2), x(1),
x(0), y(1), y(0).
i.e., y(2) = F[y(1), y(0), x(2), x(1), x(0)]
6. 21
Signals & Systems
Similarly, at n = 3, y(3) = F[y(2),y(1), y(0), x(3), x(2), x(1), x(0)]
at n = 4, y(4) = F[y(3),y(2), y(1), y(0), x(4), x(3), x(2), x(1), and so on.
In general, at any time instant n,
y(n) = F[y(n – 1), y(n – 2), y(n – 3), .....y(1), y(0), x(n), x(n – 1),
x(n – 2), x(n – 3) ..... x(1), x(0)]
.....(6.15)
For an LTI system, the response y(n) can be expressed as a weighted summation of dependent
terms. Therefore the equation (6.15) can be written as,
y(n) = – a1 y(n – 1) – a2 y(n – 2) – a3 y(n – 3) – ...........
+ b0 x(n) + b1 x( n – 1) + b2 x(n – 2) + b3 x(n – 3) +........
..... (6.16)
where, a1, a2, a3, .... and b0, b1, b2, b3, ..... are constants.
Note : Negative constants are inserted for output signals, because output signals are feedback
from output to input. Positive constants are inserted for input signals, because input
signals are feed forward from input to output.
Practically, the response y(n) at any time instant n, may depend on N number of past outputs,
present input and M number of past inputs where M £ N. Hence the equation (6.16) can be written as,
y(n) = – a1 y(n – 1) – a2 y(n – 2) – a3 y(n – 3) – ........ – a N y(n – N)
+ b0 x(n) + b1 x(n – 1) + b2 x(n – 2) + b3 x(n – 3) + ....... +bM x(n – M)
M
N
\
bg
yn = -
å a yb n - mg + å b xb n - mg
m
m
m=1
.....(6.17)
m=0
The equation (6.17) is a constant coefficient difference equation, governing the input-output
relation of an LTI discrete time system.
In equation (6.17) the value of "N" gives the order of the system.
If N = 1, the discrete time system is called 1st order system
If N = 2, the discrete time system is called 2nd order system
If N = 3, the discrete time system is called 3 rd order system , and so on.
The general difference equation governing 1st order discrete time LTI system is,
y(n) = – a1 y(n – 1) + b0 x(n) + b1 x(n – 1)
The general difference equation governing 2nd order discrete time LTI system is,
y(n) = – a2 y(n – 2) – a1 y(n – 1) + b0 x(n) + b1 x(n – 1) + b2 x(n – 2)
Chapter 6 - Discrete Time Signals and Systems
6. 22
6.6.2 Block Diagram and Signal Flow Graph Representation of Discrete Time System
The discrete time system can be represented diagrammatically by block diagram or signal flow
graph. These diagrammatic representations are useful for physical implementation of discrete time system
in hardware or software.
The basic elements employed in block diagram or signal flow graph are Adder, Constant multiplier,
Unit delay element and Unit advance element.
Adder
: An adder is used to represent addition of two discrete time sequences.
Constant Multiplier
: A constant multiplier is used to represent multiplication of a scaling factor
(constant) to a discrete time sequence.
Unit Delay Element
: A unit delay element is used to represent the delay of samples of a discrete
time sequence by one sampling time.
Unit Advance Element : A unit advance element is used to represent the advance of samples of a
discrete time sequence by one sampling time.
The symbolic representation of the basic elements of block diagram and signal flow graph are listed
in table 6.1.
Table 6.1 : Basic Elements of Block Diagram and Signal Flow Graph
Element
Block diagram
representation
x1(n)
Signal flow
graph representation
x1(n) + x2(n)
+
x1(n)
1
x1(n) + x2(n)
Adder
x2(n)
x2(n)
ax (n)
x(n)
x(n)
a
Constant multiplier
x(n)
Unit delay element
x (n -1)
z-1
x (n + 1)
x(n)
Unit advance element
z
x(n)
x(n)
1
a
z-1
z
ax (n)
x (n -1)
x (n + 1)
6. 23
Signals & Systems
Example 6.7
Construct the block diagram and signal flow graph of the discrete time systems whose input-output relations are
described by the following difference equations.
a) y(n) = 0.5 x(n) + 0.5 x(n – 1)
b) y(n) = 0.5 y(n – 1) + x(n) – 2 x(n – 2)
c) y(n) = 0.25 y(n – 1) + 0.5 x(n) + 0.75 x(n – 1)
Solution
a) Given that, y(n) = 0.5 x(n) + 0.5 x(n – 1)
The individual terms of the given equation are 0.5 x(n) and 0.5 x(n – 1). They are represented by basic elements
as shown below.
Block diagram representation
x(n)
0.5 x(n)
0.5
z –1
0.5 x(n)
Signal flow graph representation
0.5
x(n)
z –1
0.5 x(n)
0.5 x(n – 1)
0.5 x(n)
0.5 x(n – 1)
The input to the system is x(n) and the output of the system is y(n). The above elements are connected as shown
below to get the output y(n).
0.5 x(n)
x(n)
0.5
z –1
0.5 x(n – 1)
+
y(n)
x(n)
0.5
y(n)
0.5 x(n) z –1
1
0.5 x(n)
Fig 1 : Block diagram of the system
y(n) = 0.5 x(n) + 0.5 x(n – 1).
Fig 2 : Signal flow graph of the system
y(n) = 0.5 x(n) + 0.5 x(n – 1).
b) Given that, y(n) = 0.5 y(n – 1) + x(n) – 2 x(n – 2)
The individual terms of the given equation are 0.5 y(n – 1) and – 2 x(n – 2). They are represented by basic
elements as shown below.
Block diagram representation
Signal flow graph representation
y(n)
x(n)
y(n)
x(n)
z –1
z –1
x(n – 1)
y(n –1)
–1
zz–1
x(n – 2)
0.5 y(n – 1)
– 2
z –1
x(n – 1)
0.5
– 2 x(n – 2)
z –1
0.5 y(n – 1)
0.5
y(n – 1)
z –1
x(n – 2)
–2
– 2 x(n – 2)
The input to the system is x(n) and the output of the system is y(n). The above elements are connected as shown
below to get the output y(n).
Chapter 6 - Discrete Time Signals and Systems
x(n)
6. 24
y(n)
+
+
+
x(n)
1
1
y(n)
z –1
1
z –1
z –1
z –1
1
1
1
1
z –1
0.5
z –1
0.5
–2
–2
Fig 3 : Block diagram of the system
described by the equation
y(n) = 0.5 y(n – 1) + x(n) – 2 x(n – 2).
Fig 4 : Signal flow graph of the system
described by the equation
y(n) = 0.5 y(n – 1) + x(n) – 2 x(n – 2).
c) Given that, y(n) = 0.25 y(n – 1) + 0.5 x(n) + 0.75 x(n – 1)
The individual terms of the given equation are 0.25 y(n – 1), 0.5 x(n) and 0.75 x(n – 1). They are represented by
basic elements as shown below.
Block diagram representation
x(n)
0.5
Signal flow graph representation
0.5 x(n)
0.5
x(n)
0.5 x(n)
x(n)
x(n)
z –1
0.75
x(n – 1)
0.75 x(n – 1)
z –1
0.75 x(n – 1)
5
0.7
x(n – 1)
y(n)
0.25 y(n – 1)
y(n)
z –1
0.25 y(n – 1)
z –1
y(n – 1)
0.25
y(n – 1)
The input to the system is x(n) and the output of the system is y(n). The above elements are connected as shown
below to get the output y(n).
x(n)
0.5
+
y(n)
+
x(n) 1
z –1
0.5
z –1
0.75
z
0.75
–1
1
y(n)
1
z –1
0.25
0.25
Fig 6 : Signal flow graph of the system described
Fig 5 : Block diagram of the system described
by the equation
by the equation
y(n) = 0.25 y(n – 1) + 0.5 x(n) + 0.75 x(n – 1).
y(n) = 0.25 y(n – 1) + 0.5 x(n) + 0.75 x(n – 1).
6. 25
Signals & Systems
6.7 Response of LTI Discrete Time System in Time Domain
The general equation governing an LTI discrete time system is,
M
N
å
bg
yn = -
g å b xbn - mg
b
am y n - m +
m=1
m
m=0
N
\
y(n) +
M
å
b
g
am y n - m
=
m=1
N
(or )
å b xbn - mg
m
m= 0
M
å
g å b xbn - mg with a
b
am y n - m =
m= 0
m
o
=1
.....(6.18)
m= 0
The solution of the difference equation (6.18) is the response y(n) of LTI system, which consists
of two parts. In mathematics, the two parts of the solution y(n) are homogeneous solution yh(n) and
particular solution yp(n).
\ Response, y(n) = yh(n) + yp(n)
.....(6.19)
The homogeneous solution is the response of the system when there is no input.The particular
solution yp(n) is the solution of difference equation for specific input signal x(n) for n ³ 0.
In signals and systems, the two parts of the solution y(n) are called zero-input response yzi(n) and
zero-state response yzs(n).
\ Response, y(n) = yzi(n) + yzs(n)
.....(6.20)
The zero-input response is mainly due to initial conditions (or initial stored energy) in the system.
Hence zero-input response is also called free response or natural response. The zero-input response is
given by homogeneous solution with constants evaluated using initial conditions.
The zero-state response is the response of the system due to input signal and with zero initial
condition. Hence the zero-state response is called forced response.The zero-state response or forced
response is given by the sum of homogeneous solution and particular solution with zero initial conditions.
6.7.1 Zero-Input Response or Homogeneous Solution
The zero-input response is obtained from homogeneous solution yh(n) with constants evaluated
using initial condition.
\ Zero - input response, yzi ( n) = yh ( n) with constants evaluated using initial conditions
The homogeneous solution is obtained when x(n) = 0. Therefore the homogeneous solution is the
solution of the equation,
N
å
b
g
am y n - m = 0
m=0
Let us assume that the solution of equation (6.21) is in the form of an exponential.
i.e., y(n) = ln
.....(6.21)
Chapter 6 - Discrete Time Signals and Systems
6. 26
On substituting y(n) = ln in equation (6.21) we get,
N
åa
m
ln - m = 0
m= 0
On expanding the above equation (by taking a0 = 1), we get,
ln + a1 ln – 1 + a2 ln – 2 + ... + aN – 1 ln – (N – 1) + aN ln – N = 0
ln – N (lN + a1 lN – 1 + a2 lN – 2 + ... + aN – 1 l + aN) = 0
Now, the characteristic polynomial of the system is given by,
lN + a1 lN – 1 + a2 lN – 2 + ... + aN – 1 l + aN = 0
The characteristic polynomial has N roots, which are denoted as l1, l2,...lN.
The roots of the characteristic polynomial may be distinct real roots, repeated real roots or complex.
The assumed solutions for various types of roots are given below.
Distinct Real Roots
Let the roots l1, l2, l3, ... lN be distinct real roots. Now the homogeneous solution will be in the
form,
y h ( n) = C1 ln1 + C 2 ln2 + C3 ln3 + ...... + C N lnN
where, C1 , C2 , C3 ,...... C N are constants that can be evaluated using initial conditions.
Repeated Real Roots
Let one of the real roots l1 repeats p times and the remaining (N – p) roots are distinct real roots.
Now, the homogeneous solution is in the form,
y h ( n) = (C1 + C2 n + C3 n2 + ..... + C p n p - 1 ) ln1 + C p + 1 lnp + 1 + .... + C N lnN
where, C1 , C2 , C3 ,...... C N are constants that can be evaluated using initial conditions.
Complex Roots
Let the characteristic polynomial has a pair of complex roots l and l* and the remaining (N – 2)
roots be distinct real roots. Now, the homogeneous solution will be in the form,
yh(n) = rn [C1 cos nq + C2 sin nq] + C3 l3n + C4 l4n + ... + CN lNn
b
a
C1, C2, C3 ... CN are constants that can be evaluated using initial conditions.
where, l = a + jb, l* = a - jb, r =
a 2 + b 2 , q = tan-1
6.7.2 Particular Solution
The particular solution, yp(n) is the solution of the difference equation for specific input signal
x(n) for n ³ 0. Since the input signal may have different form, the particular solution depends on the form
or type of the input signal x(n).
If x(n) is constant, then yp(n) is also a constant.
Example :
Let, x(n) = u(n) ;
now, yp(n) = K u(n)
6. 27
Signals & Systems
If x(n) is exponential, then yp(n) is also an exponential.
Example :
Let, x(n) = an u(n) ;
now, yp(n) = K an u(n)
If x(n) is sinusoid, then yp(n) is also a sinusoid.
Example :
Let, x(n) = A cos w0n ; now, yp(n) = K1 cos w0n + K2 sin w0n
The general form of particular solution for various types of inputs are listed in table 6.2.
Table 6.2 : Particular Solution
Input signal, x(n)
Particular solution, y p(n)
A
K
AB
K Bn
A nB
K0 nB + K1 n(B–1) + ..... + KB
An nB
An (K0 nB + K1 n(B–1) + .....+ KB)
A cos w0n
A sin w0n
123
n
K1 cos w0n + K2 sin w0n
6.7.3 Zero-State Response
The zero-state response or forced response is obtained from the sum of homogeneous solution and
particular solution and evaluating the constants with zero initial conditions.
\ Zero - state response, y zs ( n) = yh ( n) + y p ( n)
with constants C1 , C2 ,.... C N evaluated with zero initial conditions
6.7.4 Total Response
The total response of discrete time system can be obtained by the following two methods.
Method-1
The total response is given by sum of homogeneous solution and particular solution.
\ Total response, y(n) = yh(n) + yp(n)
Procedure to Determine Total Response by Method-1
1. Determine the homogeneous solution yh(n) with constants C1, C2, .....CN.
2. Determine the particular solution yp(n) and evaluate the constants K for any value of
n ³ 1 so that no term of y(n) vanishes.
3. Now the total response is given by the sum of yh(n) and yp(n).
\ Total response, y(n) = yh (n) + yp(n)
4. The total response will have N number of constants C1, C2, .....CN. Evaluate the given equation
and the total response for n = 0, 1, 2, ....N – 1 and form two sets of N number of equations and
solve the constants C1, C2, .....CN.
Chapter 6 - Discrete Time Signals and Systems
6. 28
Method-2
The total response is given by sum of zero-input response and zero-state response.
\ Total response, y(n) = yzi(n) + yzs(n)
Procedure to Determine Total Response by Method-2
1. Determine the homogeneous solution yh(n) with constants C1, C2, ....CN.
2. Determine the zero-input response, which is obtained from the homogeneous solution yh(n)
and evaluating the constants C1, C2, ....CN using the initial conditions.
3. Determine the particular solution yp(n) and evaluate the constants K for any value of n ³ 1 so
that no term of y(n) vanishes.
4. Determine the zero-state response, yzs(n) which is given by sum of homogeneous solution
and particular soulution and evaluating the constants C1, C2, ....CN with zero initial conditions.
5. Now, the total response is given by sum of zero input response and zero state response.
\ Total response, y(n) = yzi(n) + yzs(n)
Example 6.8
Determine the response of first order discrete time system governed by the difference equation,
y(n) = – 0.5 y(n – 1) + x(n)
When the input is unit step, and with initial condition a) y(–1) = 0
b) y(–1) = 1/3.
Solution
Given that, y(n) = – 0.5 y(n – 1) + x(n)
\ y(n) + 0.5 y(n – 1) = x(n)
.....(1)
Homogeneous Solution
The homogeneous equation is the solution of equation (1) when x(n) = 0.
\ y(n) + 0.5 y(n – 1) = 0
.....(2)
Put, y(n) = ln in equation (2).
\ ln + 0.5 l(n – 1) = 0
l(n – 1) (l + 0.5) = 0
Þ
l = – 0.5
The homogeneous solution yh(n) is given by,
yh(n) = C ln = C (– 0.5)n ;
for n ³ 0
.....(3)
Particular Solution
Given that the input is unit step and so the particular solution will be in the form,
y(n) = K u(n)
.....(4)
On substituting for y(n) from equation (4) in equation (1) we get,
K u(n) + 0.5 K u(n – 1) = u(n)
.....(5)
In order to determine the value of K, let us evaluate equation (5) for n = 1, ( Q we have to evaluate equation (5) for
any n ³ 1, such that none of the term vanishes).
6. 29
Signals & Systems
From equation (5) when n = 1, we get,
K + 0.5 K = 1
1.5 K = 1
\
K =
1
10
2
=
=
15
.
15
3
The particular solution yp(n) is given by,
2
u(n) ; for all n
3
2
=
; for n ³ 0
3
y p (n) = K u(n) =
Total Response
The total response y(n) of the system is given by sum of homogeneous and particular solution.
\ Response, y(n) = yh(n) + yp(n)
2
= C( -0.5)n +
3
for n ³ 0
;
.....(6)
At n = 0, from equation (1), we get, y(0) + 0.5 y(–1) = 1
\ y(0) = 1 – 0.5 y(–1)
At n = 0, from equation (6), we get,
On equating (7) and (8) we get,
y(0) = C +
C +
2
3
=
1
- 0.5 y( -1)
3
On substituting for C from equation (9) in equation (6) we get,
FG 1
H3
IJ
K
- 0.5 y(-1) (-0.5)n +
2
3
a) When y(–1) = 0
y( -1) = 0
\ y(n) =
1
2
(-0.5)n +
3
3
;
b) When y(–1) = 1/3
1
3
1
1
2
\ y(n) =
- 0.5 ´
( -0.5)n +
3
3
3
y(-1) =
FG
H
IJ
K
0.5
2
(-0.5)n +
3
3
1
2
n
=
(-0.5) +
; for n ³ 0
6
3
=
.....(8)
2
= 1 - 0.5 y( -1)
3
\ C = 1 - 0.5 y(- 1) -
y(n) =
....(7)
for
n ³ 0
2
3
.....(9)
Chapter 6 - Discrete Time Signals and Systems
6. 30
Example 6.9
Determine the response y(n), n ³ 0 of the system described by the second order difference equation,
y(n) – 2 y(n – 1) – 3 y(n – 2) = x(n) + 4 x(n – 1),
when the input signal is, x(n) = 2n u(n) and with initial conditions y( – 2) = 0, y( –1) = 5.
Solution
Given that, y(n) – 2 y(n – 1) – 3 y(n – 2) = x(n) + 4 x(n – 1)
.....(1)
Homogeneous Solution
The homogeneous equation is the solution of equation (1) when x(n) = 0.
\ y(n) – 2 y(n – 1) – 3 y(n – 2) = 0
.....(2)
n
Put y(n) = l in equation (2).
\ ln – 2 ln – 1 – 3 ln – 2 = 0
ln – 2 (l2 – 2l – 3) = 0
The characteristic equation is,
l2 – 2 l – 3 = 0
Þ
(l – 3) (l + 1) = 0
\ The roots are, l = 3, –1
The homogeneous solution, yh(n) is given by,
yh (n) = C1 ln1 + C 2 ln2
= C1(3)n + C 2 ( -1)n ;
.....(3)
for n ³ 0
Particular Solution
Given that the input is an exponential signal, 2n u(n) and so the particular solution will be in the form,
y(n) = K 2n u(n)
.....(4)
On substituting for y(n) from equation (4) in equation (1) we get,
K 2n u(n) –2 K 2(n – 1) u(n – 1) – 3 K 2(n – 2) u(n – 2) = 2n u(n) + 4 ´ 2(n – 1) u(n – 1)
.....(5)
In order to determine the value of K, let us evaluate equation (5) for n = 2, ( Q we have to evaluate equation (5) for
any n ³ 1, such that none of the term vanishes).
From equation (5) when n = 2, we get,
K 22 – 2K ´ 21 – 3K ´ 20 = 22 + 4 ´ 21
4K – 4K – 3K = 12
– 3K = 12
12
= -4
3
The particular solution yp(n) is given by,
\ K = -
yp (n) = K 2n u(n) = (–4) 2n u(n)
Total Response
The total response y(n) of the system is given by sum of homogeneous and particular solution.
\ Response, y(n) = yh (n) + yp (n)
= C1 3n + C2 (– 1)n + (– 4) 2n ; for n ³ 0
.....(6)
6. 31
Signals & Systems
When n = 0,
From equation (1) we get,
y(0) – 2 y(–1) – 3 y(–2) = x(0) + 4 x(–1)
.....(7)
Given that, y(–1) = 5, y(–2) = 0
x(n) = 2n u(n), \ x(0) = 20 = 1
x(–1) = 0
On substituting the above conditions in equation (7) we get,
y(0) – 2 ´ 5 – 3 ´ 0 = 1 + 0
\ y(0) = 11
.....(8)
When n = 1,
From equation (1) we get,
y(1) – 2 y(0) – 3 y(–1) = x(1) + 4 x(0)
.....(9)
We know that, y(0) = 11, y(–1) = 5, y(–2) = 0
Given that,
x(n) = 2n u(n),
\ x(0) = 20 = 1
x(1) = 21 = 2
On substituting the above conditions in equation (9) we get,
y(1) – 2 ´ 11 – 3 ´ 5 = 2 + 4 ´ 1
\ y(1) = 6 + 37 = 43
.....(10)
When n = 0,
From equation (6) we get,
y(0) = C1 30 + C2 (–1)0 + (–4) 20 = C1 + C2 – 4
.....(11)
From equations (8) and (11) we can write,
C1 + C2 – 4 = 11
\ C1 + C2 = 15
.....(12)
When n = 1,
From equation (6) we get,
y(1) = C1 ´ 3 + C2 (–1) + (–4) 2 = 3 C1 – C2 – 8
.....(13)
From equations (10) and (13) we can write,
3 C1 – C2 – 8 = 43
\ 3 C1 – C2 = 51
.....(14)
On adding equations (12) and (14) we get,
4C1 = 66
\ C1 =
66
33
=
4
2
33
30 - 33
3
=
= 2
2
2
33
3
y(n) =
(3)n ( -1)n + (- 4) 2n ; for n ³ 0
2
2
From equation (12), C 2 = 15 - C1 = 15 -
\
=
LM 33 3
N2
n
-
OP
Q
3
(-1)n - 4 (2)n u(n) ; for all n.
2
Chapter 6 - Discrete Time Signals and Systems
6. 32
6.8 Classification of Discrete Time Systems
The discrete time systems are classified based on their characteristics. Some of the classifications
of discrete time systems are,
1. Static and dynamic systems
2. Time invariant and time variant systems
3. Linear and nonlinear systems
4. Causal and noncausal systems
5. Stable and unstable systems
6. FIR and IIR systems
7. Recursive and nonrecursive systems
6.8.1 Static and Dynamic Systems
A discrete time system is called static or memoryless if its output at any instant n depends at most
on the input sample at the same time but not on the past or future samples of the input. In any other case,
the system is said to be dynamic or to have memory.
y(n) = a x(n)
123 123
Example :
Static systems
y(n) = n x(n) + 6 x3(n)
å x(n - m)
m=0
¥
y(n) =
å x(n - m)
m=0
123
N
y(n) =
123
y(n) = x(n) + 3 x(n – 1)
Finite memory is required
Dynamic systems
Infinite memory is required
6.8.2 Time Invariant and Time Variant Systems
A system is said to be time invariant if its input-output characteristics do not change with time.
Definition : A relaxed system H is time invariant or shift invariant if and only if
H
H
x(n) ¾¾
® y(n) implies that, x(n - m) ¾ ¾
® y(n - m)
for every input signal x(n) and every time shift m.
i.e., in time invariant systems, if y(n) = H{x(n)} then y(n – m) = H{x(n – m)}.
Alternative Definition for Time Invariance
A system H is time invariant if the response to a shifted (or delayed) version of the input is
identical to a shifted (or delayed) version of the response based on the unshifted (or undelayed) input.
i.e., In a time invariant system, H{x(n - m)} = z-m H{x(n)}; for all values of m .....(6.22)
The operator z-m represents a signal delay of m samples.
The diagrammatic explanation of the above definition of time invariance is shown in fig 6.19.
Signals & Systems
6. 33
Procedure to test for time invariance
1. Delay the input signal by m units of time and determine the response of the system for this
delayed input signal. Let this response be y1 (n).
2. Delay the response of the system for undelayed input by m units of time. Let this delayed
response be y2(n).
3. Check whether y1 (n) = y2(n). If they are equal then the system is time invariant.
Otherwise the system is time variant.
x(n)
Input signal
x(n - m)
Delayed input
z-m
H
Delay
x(n)
Input signal
H
System
System
y(n)
Response for
undelayed input
y1(n)
Response for
delayed input
y2(n)
Delayed
response
z-m
Delay
If, y1(n) = y2(n), then the system is time invariant
Fig 6.19 : Diagrammatic explanation of time invariance.
Example 6.10
Test the following systems for time invariance.
a) y(n) = x(n) – x(n – 1)
b) y(n) = n x(n)
c) y(n) = x(–n)
d) y(n) = x(n) – b x(n – 1)
Solution
a) Given that, y(n) = x(n) – x(n – 1)
Test 1 : Response for delayed input
x(n)
Input signal
x(n - m)
z -m
Delayed input
Delay
H
System
y1(n) = x(n – m) – x(n – m – 1)
Response for
delayed input
Test 2 : Delayed response
x(n)
Input signal
H
y(n) = x(n) – x(n – 1)
Response for
System undelayed input
z -m
Delay
y2(n) = x(n – m) – x(n – m – 1)
Delayed response
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
b) Given that, y(n) = n x(n)
Test 1 : Response for delayed input
x(n)
Input signal
z -m
Delay
x(n - m)
Delayed input
H
System
y1(n) = (n – m) x(n – m)
Response for
delayed input
Test 2 : Delayed response
x(n)
Input signal
H
System
y(n) = n x(n)
Response for
undelayed input
z -m
Delay
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
y2(n) = n x(n – m)
Delayed response
Chapter 6 - Discrete Time Signals and Systems
6. 34
c) Given that, y(n) = x(–n)
Test 1 : Response for delayed input
x(n)
Input signal
x(n - m)
z -m
Delayed input
Delay
y1(n) = x(–(n – m)) = x(–n + m)
H
Response for
delayed input
System
Test 2 : Delayed response
x(n)
Input signal
y(n) = x(- n)
Response for
System undelayed input
H
y2(n) = x(–n – m)
z -m
Delayed response
Delay
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
d) Given that, y(n) = x(n) – b x(n – 1)
Test 1 : Response for delayed input
x(n)
Input signal
x(n - m)
z -m
Delayed input
Delay
y1(n) = x(n – m) – b x(n – m – 1)
H
Response for
delayed input
System
Test 2 : Delayed response
x(n)
Input signal
H
y(n) = x(n) – b x(n – 1)
System
Response for
undelayed input
y2(n) = x(n – m) – b x(n – m – 1)
z -m
Delayed response
Delay
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
Example 6.11
Test the following systems for time invariance.
M
a) y(n) = x(n) + C
b) y(n) = n x2(n)
c) y(n) = ax(n)
d) y(n) =
åb
k=0
Solution
N
k
x(n - k) -
åa
k
y(n - k)
k=1
a) Given that, y(n) = x(n) + C
Test 1 : Response for delayed input
x(n)
Input signal
z -m
Delay
x(n - m)
Delayed input
H
System
y1(n) = x(n – m) + C
Response for
delayed input
Test 2 : Delayed response
x(n)
Input signal
H
y(n) = x(n) + C
Response for
System undelayed input
z -m
Delay
y2(n) = x(n – m) + C
Delayed response
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
b) Given that, y(n) = n x2(n)
Test 1 : Response for delayed input
x(n)
Input signal
z -m
Delay
x(n - m)
Delayed input
H
System
Test 2 : Delayed response
x(n)
Input signal
H
y(n) = n x2(n)
Response for
System undelayed input
z -m
Delay
Conclusion : Here, y1(t) ¹ y2(t), therefore the system is time variant.
y1(n) = (n – m) x2(n – m)
Response for
delayed input
y2(n) =n x2(n – m)
Delayed response
Signals & Systems
6. 35
c) Given that, y(n) = ax(n)
Test 1 : Response for delayed input
x(n)
x(n - m)
z -m
Input signal
Delayed input
Delay
y1(n) = ax(n – m)
H
Response for
delayed input
System
Test 2 : Delayed response
x(n)
y(n) = ax(n)
Response for
System undelayed input
H
Input signal
y2(n) = ax(n – m)
z -m
Delayed response
Delay
Conclusion : Here, y1(t) = y2(t), therefore the system is time invariant.
M
N
åb
d) Given that, y(n) =
k
åa
x(n - k) -
k=0
k
y(n - k)
k=1
Test 1 : Response for delayed input
x(n)
Input signal
x(n - m)
z -m
Delayed input
Delay
y1(n)
H
System
Response for
delayed input
M
Response for delayed input, y1(n) = H{x(n – m)} =
Input signal
k
x(n - m - k) -
y(n)
Response for
System undelayed input
H
Response for undelayed input = H{x(n)} = y(n) =
Delayed response, y2(n)
–m
=z
LM b
MN å
M
N
åb
k
x(n - k) -
k=0
åb
k=0
åa
k
y(n - k)
k=1
åa
k
OP
PQ
y(n - k)
k=1
N
M
=
x(n - k) -
N
k
y(n - m - k)
Delayed response
Delay
M
= z-m
k
y2(n)
z -m
k=0
H{x(n)}
åa
k=1
k=0
Test 2 : Delayed response
x(n)
N
åb
k
x(n - m - k) -
åa
k
y(n - m - k)
k=1
Conclusion : Here, y1(n) = y2(n), therefore the system is time invariant.
6.8.3 Linear and Nonlinear Systems
A linear system is one that satisfies the superposition principle. The principle of superposition
requires that the response of the system to a weighted sum of the signals is equal to the corresponding
weighted sum of the responses of the system to each of the individual input signals.
Definition : A relaxed system H is linear if
.....(6.23)
H{a1 x1(n) + a2 x2(n)} = a1 H{x1(n)} + a2 H{x2(n)}
for any arbitrary input sequences x1(n) and x2(n) and for any arbitrary constants a1 and a2.
If a relaxed system does not satisfy the superposition principle as given by the above definition, the
system is nonlinear.The diagrammatic explanation of linearity is shown in fig 6.20.
Procedure to test for linearity
1.
2.
3.
4.
Let x1(n) and x2(n) be two inputs to system H, and y1(n) and y2(n) be corresponding responses.
Consider a signal, x3(n) = a1 x1(n) + a2 x2(n) which is a weighed sum of x1(n) and x2(n).
Let y3(n) be the response for x3(n).
Check whether y3(n) = a1 y1(n) + a2 y2(n). If they are equal then the system is linear, otherwise it is nonlinear.
Chapter 6 - Discrete Time Signals and Systems
x1(n)
a1
6. 36
a1x1(n)
+
x2(n)
a2
x1(n)
a1
H
H{a1x1(n) + a2x2(n)}
a1H{x1(n)}
+
x2(n)
H
a2x2(n)
H{x1(n)}
H
a1x1(n) + a2x2(n)
H{x2(n)}
a2
a1H{x1(n)} + a2H{x2(n)}
a2H{x2(n)}
The system, H is linear if and only if, H{a1 x1(n) + a2 x2(n)} = a1 H{x1(n)} + a2 H{x2(n)}
Fig 6.20 : Diagrammatic explanation of linearity.
Example 6.12
Test the following systems for linearity.
a) y(n) = n x(n),
b) y(n) = x(n2),
c) y(n) = x2(n),
d) y(n) = A x(n) + B,
e) y(n) = e x(n).
Solution
a) Given that, y(n) = n x(n)
Let H be the system represented by the equation, y(n) = n x(n) and the system H operates on x(n)
to produce, y(n) = H{x(n)} = n x(n).
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = n x1(n)
y2(n) = H{x2(n)} = n x2(n)
\ a1 y1(n) + a2 y2(n) = a1 n x1(n) + a2 n x2(n)
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = n[a1 x1(n) + a2 x2(n)] = a1 n x1(n) + a2 n x2(n)
.....(2)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) = a1 y1(n) + a2 y2(n). Hence the system is linear.
b) Given that, y(n) = x(n2)
Let H be the system represented by the equation, y(n) = x(n2) and the system H operates on x(n) to
produce, y(n) = H{x(n)} = x(n2).
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = x1(n2)
y2(n) = H{x2(n)} = x2(n2)
\ a1 y1(n) + a2 y2(n) = a1 x1(n2) + a2 x2(n2)
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
6. 37
Signals & Systems
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = a1 x1(n2) + a2 x2(n2)
.....(2)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) = a1 y1(n) + a2 y2(n). Hence the system is linear.
c) Given that, y(n) = x2(n)
Let H be the system represented by the equation, y(n) = x2(n) and the system H operates on x(n) to
produce, y(n) = H{x(n)} = x2(n).
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = x12(n)
y2(n) = H{x2(n)} = x22(n)
\ a1 y1(n) + a2 y2(n) = a1 x12(n) + a2 x22(n)
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = [a1 x1(n) + a2 x2(n)]2
= a12 x12 (n) + a22 x22(n) + 2 a1 a2 x1(n)x2(n)
.....(2)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y 3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
d) Given that, y(n) = A x(n) + B
Let H be the system represented by the equation, y(n) = A x(n) + B and the system H operates on x(n)
to produce, y(n) = H{x(n)} = A x(n) + B.
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = A x1(n) + B
y2(n) = H{x2(n)} = A x2(n) + B
\ a1 y1(n) + a2 y2(n) = a1[A x1(n) + B] + a2[A x2(n) + B]
= A a1 x1(n) + B a1 + A a2 x2(n) + B a2
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = A [a1 x1(n) + a2 x2(n)] + B
= A a1 x1(n) + A a2 x2(n) + B
.....(2)
The condition to be satisfied for linearity is, y 3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y 3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
e) Given that, y(n) = ex(n)
Let H be the system represented by the equation, y(n) = ex(n) and the system H operates on x(n) to produce,
y(n) = H{x(n)} = ex(n).
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{ x1(n)} = e x1(n)
y 2 (n) = H{ x 2 (n)} = e x 2 (n)
Chapter 6 - Discrete Time Signals and Systems
\ a1 y1(n) + a 2 y 2 (n) = a1 e
x 1(n)
+ a2 e
6. 38
x 2 (n)
..... (1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)}
= e[a1 x1(n) + a 2 x 2 (n)] = e a1 x1(n) ea 2 x 2 (n)
.....(2)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
Example 6.13
Test the following systems for linearity.
a) y(n) = x(n) + C,
b) y(n) = ax(n), c) y(n) = n x2(n).
Solution
a) Given that, y(n) = x(n) + C
Let H be the system represented by the equation, y(n) = x(n) + C and the system H operates on x(n) to
produce, y(n) = H{x(n)} = x(n) + C.
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = x1(n) + C
y2(n) = H{x2(n)} = x2(n) + C
\ a1 y1(n) + a2 y2(n) = a1 x1(n) + a1 C + a2 x2(n) + a2 C
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = a1 x1(n) + a2 x2(n) + C
.....(2)
The condition to be satisfied for linearity is, y 3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
b) Given that, y(n) = ax(n)
Let H be the system represented by the equation, y(n) = a x(n) and the system H operates on x(n)
to produce, y(n) = H{x(n)} = ax(n).
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{ x1(n)} = a x1(n)
y 2 (n) = H{ x 2 (n)} = a x 2 (n)
\ a1 y1(n) + a 2 y 2 (n) = a1 a x1(n) + a 2 a x 2 (n)
.....(1)
Consider a linear combination of inputs, a1x1(n) + a2x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y 3 (n) = H{ a1 x1(n) + a 2 x 2 (n)} = a[a 1 x1(n) + a 2 x 2 (n)]
= a a1 x1(n) a a 2 x 2 (n)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
.....(2)
6. 39
Signals & Systems
2
c) Given that, y(n) = n x (n)
Let H be the system represented by the equation, y(n) = n x 2 (n) and the system H operates
on x(n) to produce, y(n) = H{x(n)} = n x2(n) .
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = n x12(n)
y2(n) = H{x2(n)} = n x22(n)
\ a1 y1(n) + a2 y2(n) = a1 n x12(n) + a2 n x22(n)
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = n[a1 x1(n) + a2 x2(n)]2
= n a12 x12(n) + n a22 x22(n) + 2 n a1 a2 x1(n) x2(n)
.....(2)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
Example 6.14
Test the following systems for linearity.
g
a y(n) = 2 x(n) +
1
x(n - 1)
N
M
g
g
b y(n) = x(n) - b x(n - 1) c y(n) =
åb
m
x(n - m) -
åc
m
y(n - m)
m = 1
m = 0
Solution
a) Given that, y(n) = 2 x(n) +
1
x(n - 1)
1
Let H be the system represented by the equation, y(n) = 2 x(n) +
and the system H operates
x(n - 1)
1
on x(n) to produce, y(n) = H {x(n)} = 2 x(n) +
x(n - 1)
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H {x1(n)} = 2 x1(n) +
1
x1(n - 1)
y 2 (n) = H {x 2 (n)} = 2 x 2 (n) +
1
x 2 (n - 1)
F
GH
\ a1 y1(n) + a 2 y 2 (n) = a1 2 x1(n) +
1
x1(n - 1)
I
JK
F
GH
+ a 2 2 x 2 (n) +
1
x 2 (n - 1)
I
JK
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2 (n)}
= 2[a1 x1(n) + a 2 x 2 (n)] +
1
a1 x1(n - 1) + a 2 x 2 (n - 1)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (2) we can say that, y3(n) ¹ a1 y1(n) + a2 y2(n). Hence the system is nonlinear.
.....(2)
Chapter 6 - Discrete Time Signals and Systems
6. 40
b) Given that, y(n) = x(n) - b x(n - 1)
Let H be the system represented by the equation, y(n) = x(n) – b x(n –1) and the system H operates
on x(n) to produce, y(n) = H{x(n)} = x(n) – b x(n –1) .
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
\ y1(n) = H{x1(n)} = x1(n) – b x1(n – 1)
y2(n) = H{x2(n)} = x2(n) – b x2(n – 1)
\ a1 y1(n) + a2 y2(n) = a1 x1(n) – a1 b x1(n – 1) + a2 x2(n) – a2 b x2(n – 1)
.....(1)
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)} = a1 x1(n) + a2 x2(n) – b[a1 x1(n – 1) + a2 x2(n – 1)]
= a1 x1(n) – a1 b x1(n – 1) + a2 x2(n) – a2 b x2(n – 1)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
.....(2)
From equations (1) and (2) we can say that, y3(n) = a1 y1(n) + a2 y2(n). Hence the system is linear.
M
c) Given that, y(n) =
N
åb
åc
x(n - m) -
m
m= 0
m
y(n - m)
m= 1
Let H be the system represented by the equation,
N
M
y(n) =
åb
m
åc
x(n - m) -
m
y(n - m)
m=1
m=0
The response of the system
H for the input x(n)
U| = H{ x(n)} = y(n) = b
VW|
å
N
M
m
x(n - m) -
åc
m
y(n - m)
m=1
m=0
Consider two signals x1(n) and x2(n).
Let y1(n) and y2(n) be the response of the system H for inputs x1(n) and x2(n) respectively.
M
\ y1(n) = H{x1(n)} =
N
åb
m
åc
x1(n - m) -
m=0
M
y2(n) = H{x2(n)} =
m
y1(n - m)
m=1
N
åb
m
x 2 (n - m) -
m=0
åc
m
y 2 (n - m)
m =1
F b x (n - m) - c y (n - m)I
å
GH å
JK
F
I
+ a G å b x (n - m) - å c y (n - m)J
H
K
M
N
\ a1 y1(n) + a 2 y 2 (n) = a1
m
1
m=1
M
2
1
m
m=0
N
m
2
m
m=0
2
.....(1)
m =1
Consider a linear combination of inputs, a1 x1(n) + a2 x2(n). Let the response of the system for this linear combination
of inputs be y3(n).
\ y3(n) = H{a1 x1(n) + a2 x2(n)}
M
=
N
åb
m
(a1 x1(n - m) + a 2 x 2 (n - m)) -
m =0
m
y 3 (n - m)
m =1
M
= a1
åc
åb
m=0
M
m
x1(n - m) + a 2
åb
N
m
x 2 (n - m) -
m =0
åc
m
y 3 (n - m)
m =1
By time invariant property,
If y3(n) = H{a1 x1(n) + a2 x2(n)} then y3(n – m) = H{a1 x1(n – m) + a2 x2(n – m)}
If y2(n) = H{x2(n)} then y2(n – m) = H{x2(n – m)}
If y1(n) = H{x1(n)} then y1(n – m) = H{x1(n – m)}
.....(2)
6. 41
Signals & Systems
\ y3(n – m) = H{a1 x1(n – m) + a2 x2(n – m)} = a1 H{x1(n – m)} + a2 H{x2(n – m)}
.....(3)
= a1 y1(n – m) + a2 y2(n – m)]
Using equation (3), the equation (2) can be written as,
M
M
åb
y 3 (n) = a1
m
m = 0
åb
åc
F b
GH å
bm x 2 (n - m) - a1
m = 0
M
åc
x1(n - m) -
m = 0
N
åc
m
y1(n - m) - a 2
m = 1
I
JK
N
m
[a1 y1(n - m) + a 2 y 2 (n - m)]
N
å
x1(n - m) + a 2
m
m = 1
M
m
m = 0
= a1
bm x 2 (n - m) -
m = 0
M
= a1
N
å
x1(n - m) + a 2
m
y1(n - m) + a 2
m = 1
F
GH å
åc
m
y 2 (n - m)
m = 1
M
N
bm x 2 (n - m) -
m = 0
åc
m
I
JK
y 2 (n - m)
m = 1
.....(4)
The condition to be satisfied for linearity is, y3(n) = a1 y1(n) + a2 y2(n).
From equations (1) and (4) we can say that the condition for linearity is satisfied. Therefore the system is linear.
6.8.4 Causal and Noncausal Systems
Definition : A system is said to be causal if the output of the system at any time n depends only on
the present input, past inputs and past outputs but does not depend on the future inputs and outputs.
If the system output at any time n depends on future inputs or outputs then the system is called
noncausal system.
The causality refers to a system that is realizable in real time. It can be shown that an LTI system
is causal if and only if the impulse response is zero for n < 0, (i.e., h(n) = 0 for n < 0).
Let, x(n) = Present input and y(n) = Present output
\x(n-1), x(n-2), ......, are past inputs
y(n-1), y(n-2), ......, are past outputs
In mathematical terms the output of a causal system satisfies the equation of the form
y(n) = F [x(n), x(n-1), x(n-2), .... , y(-1), y(n-2) ....]
where, F[.] is some arbitrary function.
Example 6.15
Test the casuality of the following systems.
n
a) y(n) = x(n) – x(n – 1)
b) y(n) =
å x(m)
c) y(n) = a x(n)
d) y(n) = n x(n)
m = -¥
Solution
a) Given that, y(n) = x(n) – x(n – 1)
When n = 0, y(0) = x(0) – x(–1)
Þ
The response at n = 0, i.e., y(0) depends on the
present input x(0) and past input x(–1)
When n = 1, y(1) = x(1) – x(0)
Þ
The response at n = 1, i.e., y(1) depends on the
present input x(1) and past input x(0).
From the above analysis we can say that for any value of n, the system output depends on present and past inputs.
Hence the system is causal.
Chapter 6 - Discrete Time Signals and Systems
6. 42
n
b) Given that, y(n) =
å x(m)
m = -¥
0
å x(m)
When n = 0, y(0) =
m = -¥
Þ
= ... x(–2) + x(–1) + x(0)
The response at n = 0, i.e., y(0) depends on the present
input x(0) and past inputs x(–1), x(–2),.....
1
å x(m)
When n = 1, y(1) =
m = -¥
= ... x(–2) + x(–1) + x(0) + x(1)
Þ
The response at n = 1, i.e., y(1) depends on the present
input x(1) and past inputs x(0), x(–1), x(–2),..
From the above analysis we can say that for any value of n, the system output depends on present and past inputs.
Hence the system is causal.
c) Given that, y(n) = a x(n)
When n = 0, y(0) = a x(0)
Þ
The response at n = 0, i.e., y(0) depends on the present input x(0).
When n = 1, y(1) = a x(1)
Þ
The response at n = 1, i.e., y(1) depends on the present input x(1).
From the above analysis we can say that the response for any value of n depends on the present input. Hence the
system is causal.
d) Given that, y(n) = n x(n)
When n = 0, y(0) = 0 ´ x(0)
When n = 1, y(1) = 1 ´ x(1)
When n = 2, y(2) = 2 ´ x(2)
Þ
Þ
Þ
The response at n = 0, i.e., y(0) depends on the present input x(0).
The response at n = 1, i.e., y(1) depends on the present input x(1).
The response at n = 2, i.e., y(2) depends on the present input x(2).
From the above analysis we can say that the response for any value of n depends on the present input. Hence the
system is causal.
Example 6.16
Test the causality of the following systems.
a) y(n) = x(n) + 3 x(n + 4)
b) y(n) = x(n2)
c) y(n) = x(2n)
d) y(n) = x(–n)
Solution
a) Given that, y(n) = x(n) + 3 x(n + 4)
W hen n = 0, y(0) = x(0) + 3 x(4)
When n = 1, y(1) = x(1) + 3 x(5)
Þ
The response at n = 0, i.e., y(0) depends on the
present input x(0) and future input x(4).
Þ
The response at n = 1, i.e., y(1) depends on the
present input x(1) and future input x(5).
From the above analysis we can say that the response for any value of n depends on present and future inputs.
Hence the system is noncausal.
b) Givent that, y(n) = x(n2)
When n = –1 ; y(–1) = x(1)
Þ
The response at n = –1, depends on the future input x(1).
When n = 0 ; y(0) = x(0)
Þ
The response at n = 0, depends on the present input x(0).
When n = 1 ; y(1) = x(1)
Þ
The response at n = 1, depends on the present input x(1).
When n = 2 ; y(2) = x(4)
Þ
The response at n = 2, depends on the future input x(4).
From the above analysis we can say that the response for any value of n (except n = 0 and n = 1) depends on future
inputs. Hence the system is noncausal.
c) Given that, y(n) = x(2n)
When n = –1 ; y(–1) = x(–2)
Þ
The response at n = –1, depends on the past input x(–2).
When n = 0 ; y(0) = x(0)
Þ
The response at n = 0, depends on the present input x(0).
When n = 1 ; y(1) = x(2)
Þ
The response at n = 1, depends on the future input x(2).
From the above analysis we can say that the response of the system for n > 0, depends on future inputs. Hence the
system is noncausal.
6. 43
Signals & Systems
d) Given that, y(n) = x(–n)
When n = –2 ; y(–2) = x(2)
Þ
The response at n = –2, depends on the future input x(2).
When n = –1 ; y(–1) = x(1)
Þ
The response at n = –1, depends on the future input x(1).
When n = 0
; y(0) = x(0)
Þ
The response at n = 0, depends on the present input x(0).
When n = 1
; y(1) = x(–1)
Þ
The response at n = 1, depends on the past input x(–1).
From the above analysis we can say that the response of the system for n < 0 depends on future inputs. Hence the
system is noncausal.
6.8.5 Stable and Unstable Systems
Definition : An arbitrary relaxed system is said to be BIBO stable (Bounded Input-Bounded Output
stable) if and only if every bounded input produces a bounded output.
Let x(n) be the input of discrete time system and y(n) be the response or output for x(n).The term
bounded input refers to finite value of the input signal x(n) for any value of n. Hence if input x(n) is
bounded then there exits a constant Mx such that |x(n)| £ Mx and Mx < ¥, for all n.
Examples of bounded input signal are step signal, decaying exponential signal and impulse signal.
Examples of unbounded input signal are ramp signal and increasing exponential signal.
The term bounded output refers to finite and predictable output for any value of n. Hence if output
y(n) is bounded then there exists a constant My such that |y(n)| £ My and My < ¥, for all n.
In general, the test for stability of the system is performed by applying specific input. On applying
a bounded input to a system if the output is bounded then the system is said to be BIBO stable.For LTI
(Linear Time Invariant) systems the condition for BIBO stability can be transformed to a condition on
impulse response as shown below.
Condition for Stability of LTI System
By convolution sum formula, the response of LTI system is given by,
¥
å h(m) x(n - m)
y( n) =
.....(6.24)
m = -¥
Taking absolute values on both sides of equation (6.24) we get,
¥
¥
å
|y(n)| =
å
h(m) x(n - m) =
m = -¥
|h(m)| |x(n - m)|
.....(6.25)
m = -¥
If the input is bounded then |x(n - m)| = Mx. Hence equation (6.25) can be written as,
¥
å
|y(n)| =
¥
|h(m)| M x = M x
m = -¥
å |h(m)|
.....(6.26)
m = -¥
From equation (6.26) we can say that, the output is bounded if the impulse response of the system
satisfies the condition,
¥
å | h(m)| < ¥
m = -¥
Since m is a dummy variable used in convolution operation we can change m by n in the above
equation.
+¥
\
å
|h(n)| < ¥
.....(6.27)
n = -¥
Hence from equation (6.27) we can say that, an LTI system is stable if the impulse response is
absolutely summable.
Chapter 6 - Discrete Time Signals and Systems
6. 44
Example 6.17
Test the stability of the following systems.
a) y(n) = cos[x(n)]
b) y(n) = x(–n – 2)
c) y(n) = n x(n)
Solution
a) Given that, y(n) = cos [x(n)]
The given system is nonlinear system, and so the test for stability should be performed for specific inputs.
The value of cos q lies between –1 to +1 for any value of q. Therefore the output y(n) is bounded for any value of
input x(n). Hence the given system is stable.
b) Given that, y(n) = x(–n – 2)
The given system is time variant system, and so the test for stability should be performed for specific inputs.
The operations performed by the system on the input signal are folding and shifting. A bounded input signal will remain
bounded even after folding and shifting. Therefore in the given system, the output will be bounded as long as input is bounded.
Hence the given system is BIBO stable.
c) Given that, y(n) = n x(n)
The given system is time variant system, and so the test for stability should be performed for specific inputs.
If x(n) tends to infinity or constant, as "n" tends to infinity, then y(n) = n x(n) will be infinite as "n" tends to
infinity. So the ststem is unstable.
Case i :
Case ii : If x(n) tends to zero as "n" tends to infinity, then y(n) = n x(n) will be zero as "n" tends to infinity. So the
system is stable.
Example 6.18
Determine the range of values of "a" and "b" for the stability of LTI system with impulse response,
h(n) = bn
= a
n
;
n<0
;
n³0
Solution
¥
The condition to be satisfied for the stability of the system is,
å
| h(n)| < ¥.
n = -¥
Given that, h(n) = an ; n ³ 0
= bn ; n < 0
¥
\
å
-1
|h(n)| =
n = -¥
¥
å
| b|n +
n = -¥
å
¥
| b|-n +
n =1
å
å
¥
| b|-n - 1 +
n = 0
¥
=
å
n =0
| a|n
n = 0
¥
=
| a|n
n = 0
¥
=
å
å
|b|0 = 1
| a|n
n = 0
n
e| b| j
-1
¥
-1+
å
| a|n
n = 0
The summation of infinite terms in the above equation converges if, 0 < |a| < 1 and 0 < |b|–1 < 1. Hence by using infinite
geometric series formula,
Infinite qeometric
+¥
1
1
series sum formula
+
|h(n)| =
1
1 - | a|
1- | b|-1
¥
n = -¥
1
cn =
= constant
1
c
n = 0
Therefore, the system is stable if |a| < 1 and |b| > 1.
if 0 < c < 1
å
å
6. 45
Signals & Systems
6.8.6 FIR and IIR Systems
In FIR system (Finite duration Impulse Response system), the impulse response consists of finite
number of samples. The convolution formula for FIR system is given by,
N-1
å
y( n) =
h( m) x( n - m)
.....(6.28)
m= 0
where, h(n) = 0 ; for n < 0 and n ³ N
From equation (6.28) it can be concluded that the impulse response selects only N samples of the
input signal.In effect, the system acts as a window that views only the most recent N input signal samples
in forming the ouput. It neglects or simply forgets all prior input samples. Thus a FIR system requires
memory of length N. In general, a FIR system is described by the difference equation,
N-1
y(n) =
å
b m x(n - m)
m= 0
where, bm = h(m) ; for m = 0 to N -1
In IIR system (Infinite duration Impulse Response system), the impulse response has infinite
number of samples. The convolution formula for IIR systems is given by,
¥
y( n) =
å h(m) x(n - m)
m= 0
Since this weighted sum involves the present and all the past input sample, we can say that the IIR
system requires infinite memory. In general, an IIR system is described by the difference equation,
M
N
y(n) = -
å
m=1
a m y(n - m) +
åb
m
x(n - m)
m=0
6.8.7 Recursive and Nonrecursive Systems
A system whose output y(n) at time n depends on any number of past output values as well as
present and past inputs is called a recursive system. The past outputs are y(n–1), y(n–2), y(n– 3), etc.,.
Hence for recursive system, the output y(n) is given by,
y(n) = F [y(n-1), y(n-2),...y(n- N), x(n), x(n -1),...x(n- M)]
A system whose output does not depend on past output but depends only on the present and past
input is called a nonrecursive system.
Hence for nonrecursive system, the output y(n) is given by,
y(n) = F [x(n), x(n – 1) ,....., x(n – M)]
In a recursive system, in order to compute y(n0), we need to compute all the previous values y(0),
y(1) ,......., y(n0– 1) before calculating y(n0). Hence the output samples of a recursive system has to be
computed in order [i.e., y(0), y(1), y(2), ....]. The IIR systems are recursive systems.
In nonrecursive system, y(n0) can be computed immediately without having y(n 0- 1),
y(n0- 2)..... Hence the output samples of nonrecursive system can be computed in any order [i.e. y(50),
y(5), y(2), y(100),....]. The FIR systems are nonrecursive systems.
Chapter 6 - Discrete Time Signals and Systems
6. 46
6.9 Discrete or Linear Convolution
The Discrete or Linear convolution of two discrete time sequences x1(n) and x2(n) is defined as,
+¥
x 3 ( n) =
+¥
å
x1 ( m) x2 (n - m)
or
x 3 ( n) =
m = -¥
å
x2 (m) x1 (n - m)
.....(6.29)
m = -¥
where, x3(n) is the sequence obtained by convolving x1(n) and x2(n)
m is a dummy variable
If the sequence x1(n) has N1 samples and sequence x2(n) has N2 samples then the output sequence
x3(n) will be a finite duration sequence consisting of "N1+N2–1" samples. The convolution results in a
nonperiodic sequence. Hence this convolution is also called aperiodic convolution.
The convolution relation of equation (6.29) can be symbolically expressed as
x3(n) = x1(n) * x2(n) = x2(n) * x1(n)
where, the symbol * indicates convolution operation.
..... (6.30)
Procedure For Evaluating Linear Convolution
Let, x1(n) = Discrete time sequence with N1 samples
x2(n) = Discrete time sequence with N2 samples
Now, the convolution of x1(n) and x2(n) will produce a sequence x3(n) consisting of N1+N2–1
samples. Each sample of x3(n) can be computed using the equation (6.29). The value of x3(n) at n = q
is obtained by replacing n by q, in equation (6.29).
+¥
\ x 3 (q) =
å x (m) x (q - m)
1
2
.....(6.31)
m = -¥
The evaluation of equation (6.31) to determine the value of x3(n) at n = q, involves the following
five steps.
1. Change of index : Change the index n in the sequences x1(n) and x 2(n), to get the
sequences x1(m) and x2(m).
2. Folding
3. Shifting
: Fold x2(m) about m = 0, to obtain x2(-m).
: Shift x2(-m) by q to the right if q is positive, shift x2(-m) by q to the left
if q is negative to obtain x2(q - m).
4. Multiplication
: Multiply x1(m) by x2(q - m) to get a product sequence. Let the product
sequence be vq(m). Now, vq(m) = x1(m) ´ x2(q - m).
5. Summation
: Sum all the values of the product sequence vq(m) to obtain the value of
x3(n) at n = q. [i.e., x3(q)].
The above procedure will give the value x3(n) at a single time instant say n = q. In general, we are
interested in evaluating the values of the seqence x3(n) over all the time instants in the range -¥ < n < ¥.
Hence the steps 3, 4 and 5 given above must be repeated, for all possible time shifts in the range
-¥ < n < ¥.
6. 47
Signals & Systems
In the convolution of finite duration sequences it is possible to predict the start and end of the
resultant sequence. If x1(n) starts at n = n1 and x2(n) starts at n = n2 then, the initial value of n for x3(n)
is "n = n1 + n2". The value of x1(n) for n < n1 and the value of x2(n) for n < n2 are then assumed to be
zero.The final value of n for x3(n) is "n = (n1 + n2) + (N1 + N2 – 2)".
6.9.1 Representation of Discrete Time Signal as Summation of Impulses
A discrete time signal can be expressed as summation of impulses and this concept will be useful to
prove that the response of discrete time LTI system can be determined using discrete convolution.
Let, x(n) = Discrete time signal
d(n) = Unit impulse signal
d(n - m) = Delayed impulse signal
We know that, d(n) = 1 ; at n = 0
= 0 ; when n ¹ 0
and, d(n – m) = 1 ; at n = m
= 0 ; when n ¹ m
If we multiply the signal x(n) with the delayed impulse d(n - m) then the product is non-zero only
at n = m and zero for all other values of n. Also at n = m, the value of product signal is mth sample x(m)
of the signal x(n).
\ x(n) d(n - m) = x(m)
Each multiplication of the signal x(n) by an unit impulse at some delay m, in essence picks out the
single value x(m) of the signal x(n) at n = m, where the unit impulse is non-zero. Consequently if we
repeat this multiplication for all possible delays in the range -¥ < m < ¥ and add all the product sequences,
the result will be a sequence that is equal to the sequence x(n).
For example, x(n) d(n - (-2)) = x(-2)
x(n) d(n - (-1)) = x(-1)
x(n) d(n)
= x(0)
x(n) d(n - 1)
= x(1)
x(n) d(n - 2)
= x(2)
From the above products we can say that each sample of x(n) can be expressed as a product of the
sample and delayed impulse, as shown below.
\ x(-2) = x(-2) d(n-(-2))
x(-1) = x(-1) d(-(-1))
x(0) = x(0) d(n)
x(1) = x(1) d(n - 1)
x(2) = x(2) d(n - 2)
Chapter 6 - Discrete Time Signals and Systems
6. 48
\ x(n) = ..... + x(-2) + x(-1) + x(0) + x(1) + x(2) + ...........
= ..... + x(-2) d(n - (-2)) + x(-1) d(n - (-1)) + x(0) d(n) + x (1) d(n - 1)
+ x(2) d(n - 2) + ...........
+¥
å
=
x(m) d (n - m)
.....(6.32)
m = -¥
In equation (6.32) each product x(m) d(n – m) is an impulse and the summation of impulses gives
the sequence x(n).
6.9.2 Response of LTI Discrete Time System Using Discrete Convolution
In an LTI system, the response y(n) of the system for an arbitrary input x(n) is given by convolution
of input x(n) with impulse response h(n) of the system. It is expressed as,
+¥
y( n) = x( n) * h( n) =
å x( m) h(n - m)
.....(6.33)
m = -¥
where, the symbol * represents convolution operation.
Proof :
Let y(n) be the response of system H for an input x(n)
\ y(n) = H{x(n)}
.....(6.34)
From equation (6.32) we know that the signal x(n) can be expressed as a summation of impulses,
+¥
å
i.e., x(n) =
x(m) d(n - m)
.....(6.35)
m = -¥
where, d(n – m) is the delayed unit impulse signal.
From equation (6.34) and (6.35) we get,
|RS å
|T
+¥
y(n) = H
x(m) d(n - m)
m = -¥
|UV
|W
.....(6.36)
The system H is a function of n and not a function of m. Hence by linearity property the equation (6.36) can be
written as,
+¥
y(n) =
å
x(m) H { d(n - m)}
m = -¥
.....(6.37)
Let the response of the LTI system to the unit impulse input d(n) be denoted by h(n),
\ h(n) = H{d(n)}
Then by time invariance property the response of the system to the delayed unit impulse input d(n – m) is given by,
h(n – m) = H{d(n – m)}
.....(6.38)
Using equation (6.38), the equation (6.37) can be expressed as,
+¥
y(n) =
å
x(m) h(n - m)
m = -¥
The above equation represents the convolution of input x(n) with the impulse response h(n) to yield the output
y(n). Hence it is proved that the response y(n) of LTI discrete time system for an arbitrary input x(n) is given by
convolution of input x(n) with impulse response h(n) of the system.
6. 49
Signals & Systems
6.9.3 Properties of Linear Convolution
The Discrete convolution will satisfy the following properties.
Commutative property : x1(n) * x2(n) = x2(n) * x1(n)
Associative property
: [x1(n) * x2(n)] * x3(n) = x1(n) * [x2(n) * x3(n)]
Distributive property
: x1(n) * [x2(n) + x3(n)] = [x1(n) * x2(n)] + [x1(n) * x3(n)]
Proof of Commutative Property :
Consider convolution of x1(n) and x2(n).
By commutative property we can write,
x1(n) * x2(n) = x2 (n) * x1(n)
(LHS)
(RHS)
LHS = x1(n) * x2(n)
+¥
å
=
x 1( m) x 2(n - m)
.....(6.39)
m = -¥
where, m is a dummy variable used for convolution operation.
Let, n – m = p
when m = –¥, p = n – m = n + ¥ = +¥
\m=n–p
when m = +¥, p = n – m = n – ¥ = –¥
On replacing m by (n – p) and (n – m) by p in equation (6.39) we get,
+¥
å
LHS =
x 1(n - p) x 2 (p)
p = -¥
+¥
å
=
x 2 ( p) x1(n - p)
p = -¥
= x2 (n) * x1(n)
p is a dummy variable used for convolution operation
= RHS
Proof of Associative Property :
Consider the discrete time signals x1(n), x2(n) and x3(n). By associative property we can write,
[x1 (n) * x2(n)] * x3(n) = x1(n) * [x2 (n) * x3(n)]
LHS
Let,
RHS
y1(n) = x1(n) * x2(n)
.....(6.40)
Let us replace n by p
\ y1(p) = x1(p) * x2(p)
+¥
=
å
x 1(m) x 2(p - m)
.....(6.41)
m = -¥
Let,
y2(n) = x 2(n) * x3(n)
.....(6.42)
+¥
\ y 2 (n) =
å
q = -¥
x 1( q) x 2(n - q)
Chapter 6 - Discrete Time Signals and Systems
6. 50
+¥
å
\ y 2 (n - m) =
x 1( q) x 2(n - q - m)
.....(6.43)
q = -¥
where p, m and q are dummy variables used for convolution operation.
LHS = [x1(n) * x2(n)] * x3(n)
= y1(n) * x3(n)
Using equation (6.40)
+¥
å
=
y 1( p) x 3(n - p)
p = -¥
=
+¥
+¥
å
å
p = -¥
m = -¥
x 1( m) x 2 (p -m) x 3 (n - p)
+¥
+¥
å
=
Using equation (6.41)
å
x 1 (m)
m = -¥
x 2 (p - m) x 3 (n - p)
.....(6.44)
p = -¥
Let, p – m = q
when p = –¥, q = p – m = –¥ – m = –¥
when p = +¥, q = p – m = +¥ – m = +¥
\p=q+m
On replacing (p – m) by q, and p by (q + m) in the equation (6.44) we get,
+¥
LHS =
+¥
å
x 1(m)
m = -¥
å
x 2 ( q) x 3 (n - q - m)
q = -¥
+¥
=
å
x 1( m) y 2 (n - m)
Using equation (6.43)
m = -¥
= x1 (n) * y2(n)
= x 1(n) * [x2(n) * x3(n)]
Using equation (6.42)
= RHS
Proof of Distributive Property :
Consider the discrete time signals x1(n), x2(n) and x3(n). By distributive property we can write,
x1 (n) * [x2(n) + x3(n)] = [x1(n) * x2 (n)] + [x1(n) * x3 (n)]
LHS
RHS
LHS = x1(n) * [x2(n) + x3(n)]
= x 1(n) * x4(n)
x 4(n) = x2(n) + x3(n)
+¥
=
å
x1( m) x 4 (n - m)
m is a dummy variable used for convolution operation
m = -¥
+¥
=
å
x1( m) [x2 (n - m) + x3 (n - m)]
m = -¥
+¥
=
å
m = -¥
+¥
x1( m) x 2(n - m) +
å
m = -¥
= [x1 (n) * x2(n)] + [x 1(n) * x3 (n)]
= RHS
x1( m) x 3(n - m)
Signals & Systems
6. 51
6.9.4 Interconnections of Discrete Time Systems
Smaller discrete time systems may be interconnected to form larger systems. Two possible basic
ways of interconnection are cascade connection and parallel connection. The cascade and parallel
connections of two discrete time systems with impulse responses h1(n) and h2(n) are shown in fig 6.21.
h1(n)
x(n)
h1(n)
y(n)
y1(n)
y(n)
x(n)
+
h2(n)
h2(n)
Fig 6.21a : Cascade connection.
Fig 6.21b : Parallel connection.
Fig 6.21 : Interconnection of discrete time systems.
Cascade Connected Discrete Time System
Two cascade connected discrete time systems with impulse response h1(n) and h2(n) can be replaced
by a single equivalent discrete time system whose impulse response is given by convolution of individual
impulse responses.
x(n)
h1(n)
y1(n)
h2(n)
y(n)
Þ
x(n)
y(n)
h1(n)*h2(n)
Fig 6.22 : Cascade connected discrete time systems and their equivalent.
Proof:
With reference to fig 6.22 we can write,
y1(n) = x(n) * h1(n)
.....(6.45)
y(n) = y1(n) * h2(n)
.....(6.46)
Using equation (6.45), the equation (6.46) can be written as,
y(n) = x(n) * h1(n) * h2(n)
= x(n) * [h1(n) * h2(n)]
= x(n) * h(n)
.....(6.47)
where, h(n) = h1(n) * h2(n)
From equation (6.47) we can say that the overall impulse response of two cascaded discrete
time systems is given by convolution of individual impulse responses.
Parallel Connected Discrete Time Systems
Two parallel connected discrete time systems with impulse responses h1(n) and h2(n) can be replaced
by a single equivalent discrete time system whose impulse response is given by sum of individual impulse
responses.
h1(n)
y1(n)
y(n)
x(n)
+
h2(n)
Þ
x(n)
h1(n)+h2(n)
y2(n)
Fig 6.23 : Parallel connected discrete time systems and their equivalent.
y(n)
Chapter 6 - Discrete Time Signals and Systems
6. 52
Proof:
With reference to fig 6.23 we can write,
y1(n) = x(n) * h1(n)
.....(6.48)
y2(n) = x(n) * h2(n)
.....(6.49)
y(n) = y1(n) + y 2(n)
.....(6.50)
On substituting for y1(n) and y2(n) from equations (6.48) and (6.49) in equation (6.50) we get,
y(n) = [ x(n) * h1(n)] + [ x(n) * h2(n)]
.....(6.51)
By using distributive property of convolution the equation (6.51) can be written as shown below,
y(n) = x(n) * [h1(n) + h2(n)]
= x(n) * h(n)
.....(6.52)
where, h(n) = h1(n) + h2(n)
From equation (6.52) we can say that the overall impulse response of two parallel connected discrete time
system is given by sum of individual impulse responses.
Example 6.19
Determine the impulse reponse for the cascade of two LTI systems having impulse responses,
n
h1(n) =
n
FG 1 IJ u(n) and h (n) = FG 1IJ u(n).
H 2K
H 4K
2
Solution
Let h(n) be the impulse response of the cascade system. Now h(n) is given by convolution of h1(n) and h2(n).
\ h(n) = h1(n) * h2(n)
=
+¥
å h1(m) h2 (n - m)
m = -¥
where, m is a dummy variable used for convolution operation
The product h1(m) h2(n – m) will be non-zero in the range 0 £ m £ n. Therefore the summation index in the above
equation is changed to m = 0 to n.
n
\ h(n) =
n
å
h1(m) h2 (n - m) =
m = 0
FG 1 IJ
H 4K
F 1I
= G J
H 4K
F 1I
= G J
H 4K
F 1I
= G J
H 4K
F 1I
= G J
H 4K
n
=
m = 0
n
FG 4 IJ
H 2K
å
m= 0
n
å
m
m= 0
n
n -m
n
=
å
m = 0
m
n
FG 1IJ FG 1 IJ FG 1 IJ
H 2K H 4K H 4K
-m
n
=
n
FG 1IJ å FG 1 IJ
H 4K
H 2K
m
4m
m= 0
m
n
å2
m
FG 1IJ FG 1 IJ
H 2K H 4K
F 2 - 1I
GH 2 - 1 JK
Finite geometric series
sum formula
N
CN + 1 - 1
Cn =
C -1
n= 0
å
n+1
n
(2n + 1 - 1) ; for n ³ 0
n
(2n + 1 - 1) u(n) ; for all n
Using finite geometric series sum formula
Signals & Systems
6. 53
Example 6.20
Determine the overall impulse response of the interconnected discrete time system shown below,
where, h1(n) =
FG 1IJ
H 3K
n`
u(n), h2 (n) =
FG 1IJ
H 2K
n
u(n) and h3 (n) =
FG 1IJ
H 5K
n
u(n).
h1(n)
x(n)
+
+
h2(n)
y(n)
h3(n)
Solution
The given system can be redrawn as shown below.
h1(n)
y(n)
+
x(n)
h1(n)
h3(n)
+
h2(n)
The above system can be reduced to single equivalent system as shown below.
y1(n)
h1(n)
x(n)
y(n)
x(n)
y(n)
Þ
h1(n) + h2(n)
h1(n) + [(h1(n) + h2(n)) * h3(n)]
Þ
+
h3(n)
x(n)
h(n)
y(n)
Here, h(n) = h1(n) + [(h1(n) + h2(n)) * h3(n)]
= h1(n) + [h1(n) * h3(n)] + [h2(n) * h3(n)]
Using distributive property
Let us evaluate the convolution of h1(n) and h3(n).
¥
å h (m) h (n - m)
h1 (n) * h3 (n) =
1
3
m = -¥
The product of h1(m) h3(n – m) will be non-zero in the range 0 £ m £ n. Therefore the summation index in the above
equation can be changed to m = 0 to n.
n
\ h1 (n) * h3 (n) =
å h (m) h (n - m)
1
3
m= 0
m
n
FG 1 IJ FG 1 IJ
H 3 K H 5K
F 1I
F 1I
= G J å G J
H 5K
H 3K
=
å
n-m
m= 0
n
n
m= 0
m
n
n
FG 1 IJ FG 1 IJ FG 1 IJ
H 3K H 5 K H 5 K
F 1I
F 5I
= G J å G J
H 5K
H 3K
=
å
m= 0
n
m
5m
n
m= 0
m
-m
Chapter 6 - Discrete Time Signals and Systems
6. 54
n +1
FG 5 IJ - 1
1I H 3 K
F
= G J
H 5K 5 - 1
n
Using finite geometric series sum formula
3
Finite geometric series
n
FG 5 IJ 5 - 1
1I H 3 K 3
F
F 1I L 3 F 5 I 5 3 O
= G J
H 5 K 5 - 3 = GH 5 JK MMN 2 GH 3 JK 3 - 2 PPQ
3
5 F 1I F 5 I
=
G J G J - 23 FGH 51IJK = 52 FGH 31IJK - 32 FGH 51 JIK
2 H 5K H 3K
5 F 1I
=
G J u(n) - 32 FGH 51IJK u(n) ; for all n
2 H 3K
n
n
n
n
n
n
sum formula
n
n
N
åC
m
=
m= 0
CN+1 - 1
C -1
n
; for n ³ 0
n
Let us evaluate the convolution of h2(n) and h3(n).
+¥
å h (m) h (n - m)
h2 (n) * h3 (n) =
2
3
m = -¥
The product of h2(m) and h3(n–m) will be non-zero in the range 0 £ m £ n. Therefore the summation index in the above
equation can be change to m = 0 to n.
n
\ h2 (n) * h3 (n) =
å
h2 (m) h3 (n - m)
m = 0
å
m = 0
F 1I
= G J
H 5K
n -m
m
n
=
n
FG 1IJ FG 1 IJ
H 2K H 5K
F 1I
å GH 2 JK 5
FG 5 IJ - 1
H 2K
n
FG 1IJ FG 1 IJ FG 1 IJ
H 2K H 5K H 5K
F 1I
F 5I
= G J å G J
H 5K
H 2K
å
-m
Finite geometric series
m = 0
m
n
m
n
=
n
m
m = 0
sum formula
m
n
N
åC
m = 0
m= 0
m
=
CN+1 - 1
C -1
n + 1
=
GHF 51IJK
n
Using finite geometric series sum formula
5
-1
2
n
FG 5 IJ 5 - 1
1I H 2 K 2
F
F 1I L 2 F 5 I 5 2 O
= G J
H 5 K 5 - 2 = GH 5 JK MMN 3 GH 2 JK 2 - 3 PPQ
2
5 F 1I F 5 I
=
G J G J - 23 FGH 51 IJK = 53 FGH 21IJK - 23 FGH 51IJK
3 H 5K H 2K
5 F 1I
=
G J u(n) - 23 FGH 51IJK u(n) for all n
3 H 2K
n
n
n
n
n
n
n
n
n
for n ³ 0
n
Now, the overall impulse response h(n) is given by,
h(n) = h1(n) + h1(n) * h3 (n) + h2 (n) * h3 (n)
n
n
n
n
FG 1IJ u(n) + 5 FG 1 IJ u(n) - 3 FG 1IJ u(n) + 5 FG 1IJ u(n) - 2 FG 1 IJ
H 3K
2 H 3K
2 H 5K
3 H 2K
3 H 5K
F 5 I F 1I
F 3 + 2 IJ FG 1 IJ u(n) + 5 FG 1IJ u(n)
= G1 + J G J u(n) - G
H 2K H 3K
H 2 3K H 5K
3 H 2K
L 7 F 1I 13 F 1I + 5 F 1 I OP u(n)
= M G J MN 2 H 3 K 6 GH 5 JK 3 GH 2 JK PQ
=
n
n
n
n
n
n
n
u(n)
Signals & Systems
6. 55
Example 6.21
Find the overall impulse response of the interconnected system shown below. Given that h1(n) = an u(n),
h2(n) = d(n – 1), h3(n) = d(n – 2).
h2(n)
h1(n)
y(n)
x(n)
+
h1(n)
h3(n)
Solution
The given system can be reduced to single equivalent system as shown below.
h1(n) * h2(n)
x(n)
+
y(n)
h3(n) * h1(n)
Þ
x(n)
[h1(n) * h2(n)] + [h3(n) * h1(n)]
y(n)
Þ
h(n)
x(n)
y(n)
Here, h(n) = [h1(n) * h2(n)] + [h3(n) * h1(n)]
Let us evaluate the convolution of h1(n) and h2(n).
¥
h1(n) * h2 (n) =
å h (m) h (n - m)
1
2
m = -¥
¥
=
å h (m) h (n - m)
2
Using commutative property
1
m = -¥
¥
=
å d(m - 1) a
¥
(n - m)
m = -¥
å d(m - 1) a
=
n
a -m
m = -¥
¥
å d(m - 1) a
= an
-m
m = -¥
The product of d(m – 1) and a–m in the above equation will be non-zero only when m = 1.
\ h1(n) * h2(n) = an a–1 = an – 1 ; for n ³ 1
= an – 1 u(n – 1) ; for all n.
Let us evaluate the convolution of h3(n) and h1(n).
¥
h3 (n) * h1(n) =
å h (m) h (n - m)
3
m = -¥
1
Chapter 6 - Discrete Time Signals and Systems
6. 56
¥
¥
å d(m - 2) a
h3 (n) * h1(n) =
(n - m)
å d(m - 2) a
=
n
a -m
m = -¥
m = -¥
¥
= an
å d(m - 2) a
-m
m = -¥
The product of d(m – 2) and a–m in the above equation will be non-zero only when m = 2.
\ h1(n) * h2(n) = an a–2 = an – 2
=a
n–2
; for n ³ 2
u(n – 2) ; for all n
Now, the overall impulse response h(n) is given by,
h(n) = [h1(n) * h2(n)] + [h3(n) * h1(n)]
= a(n – 1) u(n – 1) + a(n – 2) u(n – 2)
6.9.5 Methods of Performing Linear Convolution
Method -1: Graphical Method
Let x1(n) and x2(n) be the input sequences and x3(n) be the output sequence.
1. Change the index "n" of input sequences to "m" to get x1(m) and x2(m).
2. Sketch the graphical representation of the input sequences x1(m) and x2(m).
3. Let us fold x 2(m) to get x 2(–m). Sketch the graphical representation of the folded
sequence x2(–m).
4. Shift the folded sequence x2(–m) to the left graphically so that the product of x1(m) and
shifted x2(–m) gives only one non-zero sample. Now multiply x1(m) and shifted x2(–m) to get
a product sequence, and then sum-up the samples of product sequence, which is the first
sample of output sequence.
5. To get the next sample of output sequence, shift x2(–m) of previous step to one position right
and multiply the shifted sequence with x1(m) to get a product sequence. Now the sum of the
samples of product sequence gives the second sample of output sequence.
6. To get subsequent samples of output sequence, the step-5 is repeated until we get a
non-zero product sequence.
Method -2: Tabular Method
The tabular method is same as that of graphical method, except that the tabular representation of
the sequences are employed instead of graphical representation. In tabular method, every input sequence,
folded and shifted sequence is represented by a row in a table.
Method -3: Matrix Method
Let x1(n) and x2(n) be the input sequences and x3(n) be the output sequence. In matrix method one
of the sequences is represented as a row and the other as a column as shown below.
Multiply each column element with row elements and fill up the matrix array.
Now the sum of the diagonal elements gives the samples of output sequence x3(n). (The sum of the
diagonal elements are shown below for reference).
Signals & Systems
6. 57
x 2(0)
x 2(1)
x 2(2)
x 2(3)
x 1(0)x 2(0)
x 1(0)x 2(1)
x 1(0)x 2(2)
x 1(0)x 2(3)
x 1(1)
x 1(1)x 2(0)
x 1(1)x2(1)
x 1(1)x 2(2)
x 1(1)x 2(3)
x 1(2)
x 1(2)x 2(0)
x 1(2)x 2(1)
x 1(2)x 2(2)
x 1(2)x 2(3)
x 1(3)
x 1(3)x 2(0)
x 1(3)x 2(1)
x 1(3)x 2(2)
x 1(3)x 2(3)
......
x 1(0)
x3(0) = ..... + x1(0) x2(0) + .....
x3(1) = ..... + x 1(1) x2(0) + x1(0 ) x2(1) + .....
x3(2) = ..... + x 1(2) x2(0) + x1(1) x2(1) + x1(0) x2(2) + .....
......
x3(3) = ..... + x1(3) x 2(0) + x1(2) x 2(1) + x1(1) x2(2) + x1(0) x2(3) + .....
\ x3(n) = {..... x3(0), x3(1), x3(2), x3(3), .....}
Example 6.22
Determine the response of the LTI system whose input x(n) and impulse response h(n) are given by,
x(n) = {1, 2, 3, 1} and h(n) = {1, 2, 1, –1}
­
­
Solution
The response y(n) of the system is given by convolution of x(n) and h(n).
+¥
y(n) = x(n) * h(n) =
å x(m) h(n - m)
m = -¥
In this example the convolution operation is performed by three methods.
The Input sequence starts at n = 0 and the impulse response sequence starts at n = –1. Therefore the output
sequence starts at n = 0 + (–1) = –1.
The input and impulse response consists of 4 samples, so the output consists of 4 + 4 – 1 = 7 samples.
Chapter 6 - Discrete Time Signals and Systems
6. 58
Method 1 : Graphical Method
The graphical representation of x(n) and h(n) after replacing n by m are shown below. The sequence h(m) is folded
with respect to m = 0 to obtain h(–m).
h(m)
x(m)
h(–m)
3
1
1
1
0
2
2
2
2
1
3
1
–1
m
1
2
1
0
1
0
–1
–1
Fig 1 : Input sequence.
–1
–2
m
1
Fig 2 : Impulse response.
m
Fig 3 : Folded impulse response.
The samples of y(n) are computed using the convolution formula,
y(n) =
+¥
+¥
m = -¥
m = -¥
å x(m) h(n - m) = å x(m) h (m) ; where h (m) = h(n - m)
n
n
The computation of each sample using the above equation are graphically shown in fig 4 to fig 10. The graphical
representation of output sequence is shown in fig 11.
+¥
å x(m) h(-1- m)
When n = - 1 ; y(-1) =
m = -¥
+¥
å x(m) h
=
åv
=
m = -¥
-1(m)
m = -¥
v–1(m)
x(m)
h–1(m)
+¥
-1(m)
3
X
2
1
1
–3 –2
–1
Þ
2
1
0
–1
1
0
m
1
1
2
3
0
m
Fig 4 : Computation of y(–1).
+¥
When n = 0 ; y(0) =
+¥
=
0
m = -¥
m
å
v 0 (m)
m = -¥
v0(m)
x(m)
h0(m)
3
2
+¥
å x(m) h(0 - m) = å x(m) h (m)
m = -¥
1
The sum of product sequence v–1(m)
gives y(–1). \ y(–1) = 1
3
X
2
1
–2
–1
1
1
0
Þ
2
1
m
2
2
1
0
1
2
3
–1
Fig 5 : Computation of y(0).
m
0
1
2
3
m
The sum of product sequence v0(m)
gives y(0). \ y(0) = 2 + 2 = 4
6. 59
Signals & Systems
+¥
+¥
+¥
å x(m) h(1 - m) = å x(m) h (m)
When n = 1 ; y(1) =
m = -¥
å v (m)
=
1
1
m = -¥
h1(m)
m = -¥
v1(m)
x(m)
4
3
X
2
–1
0
1
Þ
2
1
1
3
1
2
–1
0
m
1
1
1
2
3
0
m
Fig 6 : Computation of y(1).
+¥
+¥
m = -¥
h2(m)
3
m
+¥
å x(m) h(2 - m) = å
When n = 2 ; y(2) =
2
1
The sum of product sequence v1(m)
gives y(1). \ y(1) = 1 + 4 + 3 = 8
åv
x(m) h2 (m) =
m = -¥
2 (m)
m = -¥
x(m)
v2(m)
6
3
X
2
0
2
1
1
1
2
1
3
0
m
Þ
1
1
2
3
2
1
m
–1
0
1
3
2
m
–1
The sum of product sequence v2(m)
gives y(2). \ y(2) = –1 + 2 + 6 + 1 = 8
Fig 7 : Computation of y(2).
+¥
+¥
å x(m) h(3 - m) = å x(m) h (m)
When n = 3 ; y(3) =
3
m = -¥
+¥
m = -¥
h3(m)
å
=
v 3 (m)
m = -¥
v3(m)
x(m)
3
X
2
1
1
0
3
2
1
1
2
3
2
4
1
0
m
Þ
1
2
3
1
0
m
2
3
m
–1
-2
The sum of product sequence v3(m)
gives y(3). \ y(3) = – 2 + 3 + 2 = 3
Fig 8 : Computation of y(3).
+¥
When n = 4 ; y(4) =
+¥
å x(m) h(4 - m) = å x(m) h (m)
4
m = -¥
m = -¥
h4(m)
+¥
å v (m)
=
4
m = -¥
v4(m)
x(m)
3
X
2
1
1
2
Þ
1
1
1
0
0
2
1
3
4
5
m
0
1
2
3
2
1
3
m
m
-1
Fig 9 : Computation of y(4).
-3
The sum of product sequence v4(m)
gives y(4). \ y(4) = –3 + 1 = –2
Chapter 6 - Discrete Time Signals and Systems
6. 60
+¥
+¥
å x(m) h(5 - m)
When n = 5 ; y(5) =
=
m = -¥
+¥
å x(m ) h (m ) = å v (m)
5
5
m = -¥
h5(m)
m = -¥
x(m)
v5(m)
3
X
2
1
1
Þ
2
1
1
0
3
2
1
-1
0
1
2
3
-1
The output sequence, y(n) =
4
5
6
0
m
1
2
3
m
Fig 10 : Computation of y(5).
RS1, 4, 8, 8, 3,
T A
UV
W
- 2, - 1
m
The sum of product sequence
v5(m) gives y(5). \ y(5) = –1
y(n)
8
8
4
3
1
–2
–1
0
1
2
4
3
5
6
7
–1
n
–2
Fig 11 : Graphical representation of y(n).
Method - 2 : Tabular Method
The given sequences and the shifted sequences can be represented in the tabular array as shown below.
Note : The unfilled boxes in the table are considered as zeros.
m
–3
–2
–1
0
1
1
2
–1
1
2
1
1
2
1
–1
1
2
1
–1
1
2
1
–1
1
2
1
–1
1
2
x(m)
h(m)
h(–m)
h(–1 – m) = h–1(m)
h(0 – m) = h0(m)
h(1 – m) = h1(m)
h(2 – m) = h2(m)
h(3 – m) = h3(m)
h(4 – m) = h4(m)
h(5 – m) = h5(m)
–1
1
2
3
2
3
1
1
–1
–1
4
5
6
1
1
2
1
–1
1
2
1
6. 61
Signals & Systems
Each sample of y(n) is computed using the convolution formula,
+¥
+¥
m = -¥
m = -¥
å x(m) h(n - m) = å x(m) h (m),
y(n) =
n
where hn (m) = h(n - m)
To determine a sample of y(n) at n = q, multiply the sequence x(m) and hq(m) to get a product sequence (i.e., multiply
the corresponding elements of the row x(m) and hq(m)). The sum of all the samples of the product sequence gives y(q).
3
When n = -1 ; y( -1) =
å x(m) h
Q The product is valid only for m = -3 to + 3
-1(m)
m = -3
= x(–3) h–1(–3) + x(–2)h–1(–2) + x(–1)h–1(–1) + x(0) h–1(0) + x(1) h–1(1)
+ x(2) h–1(2) + x(3) h–1(3)
= 0+0+0+1+0+0+0=1
The samples of y(n) for other values of n are calculated as shown for n = –1.
3
When n = 0 ; y(0) =
å x(m) h (m) =
0
0 + 0 + 2 + 2 + 0 + 0 = 4
m = -2
3
When n = 1 ; y(1) =
å x(m) h (m)
1
= 0 + 1 + 4 + 3 + 0 = 8
m = -1
3
When n = 2 ; y(2) =
å x(m) h (m)
2
= -1 + 2 + 6 + 1 = 8
m = 0
4
When n = 3 ; y(3) =
å x(m) h (m)
3
= 0 - 2 + 3 + 2 + 0 = 3
m = 0
5
When n = 4 ; y(4) =
å x(m) h (m)
4
= 0 + 0 - 3 + 1 + 0 + 0 = -2
m = 0
6
When n = 5 ; y(5) =
å x(m) h (m)
5
= 0 + 0 + 0 - 1 + 0 + 0 + 0 = -1
m = 0
The output sequence, y(n) =
l 1, 4, 8, 8, 3,
A
- 2, - 1
q
Method - 3 : Matrix Method
The input sequence x(n) is arranged as a column and the impulse response is arranged as a row as shown below.
The elements of the two dimensional array are obtained by multiplying the corresponding row element with the column
element. The sum of the diagonal elements gives the samples of y(n).
x(n)
¯
h(n) ®
1
2
1
1
1´1
1´2
1 ´ 1 1 ´ (–1)
2
2´1
2´ 2
2 ´ 1 2 ´ (–1)
3
3´1
3´ 2
3 ´ 1 3 ´ (–1)
1
1´1
1 ´2
1 ´ 1 1 ´ (–1)
h(n)® 1
x(n)
1
1
¯
–1
Þ
2
2
1
1
1
–1
2
2
4
2
–2
3
3
6
3
–3
1
1
2
1
–1
y(–1) = 1
y(0) = 2 + 2 = 4
y(3) = 2 + 3 + (–2) = 3
y(1) = 3 + 4 + 1 = 8
y(4) = 1 + (–3) = –2
y(2) = 1 + 6 + 2 + (–1) = 8
y(5) = –1
\ y(n) = {1, 4, 8, 8, 3, –2, –1}
­
Chapter 6 - Discrete Time Signals and Systems
6. 62
Example 6.23
Determine the output y(n) of a relaxed LTI system with impulse response,
h(n) = an u(n) ; where |a| < 1 and
When input is a unit step sequence, i.e., x(n) = u(n).
Solution
The graphical representation of x(n) and h(n) after replacing n by m are shown below. Also the sequence x(m) is
folded to get x(–m)
x(–m)
x(m)
h(m)
1
1
1
a
a2
a3
0
1
3
2
0
m
Fig 1 : Impulse response.
1
2
3
–3
m
Fig 2 : Input sequence.
–2
–1
0
m
Fig 3 : Folded input sequence.
Here both h(m) and x(m) are infinite duration sequences starting at n = 0. Hence the output sequence y(n) will also
be an infinite duration sequence starting at n = 0
By convolution formula,
¥
¥
m = -¥
m =0
å h(m) x(n - m) = å h(m) x (m) ;
y(n) =
n
where xn (m) = x(n - m)
The computation of some samples of y(n) using the above equation are graphically shown below.
¥
¥
¥
å h(m) x(m) = å h(m) x (m) = å v (m)
When n = 0 ; y(0) =
0
m= 0
0
m= 0
m= 0
h(m)
x0(m)
v0(m)
1
1
1
a
Þ
a2
a3
0
1
2
3
m
–3
-2
0
-1
1
m
0
¥
When n = 1 ; y(1) =
¥
å h(m) x(1 - m) = å
m = 0
h(m)
2
1
m
y(0) = 1
Fig 4 : Computation of y(0).
¥
h(m) x1(m) =
m = 0
å
v 1(m)
m = 0
v1(m)
x1(m)
1
1
1
a
a
a
Þ
X
2
a3
0
1
2
3
m
–2
–1
0
Fig 5 : Computation of y(1).
1
m
–1
0
1
2
y(1) = 1 + a
m
Signals & Systems
6. 63
¥
When n = 2 ; y(2) =
å
¥
å h(m) x (m) = å v
h(m) x(2 - m) =
m = 0
¥
2
m = 0
h(m)
v2(m)
x2(m)
1
2 (m)
m = 0
1
1
a
a
Þ
X
a2
a2
a3
0
1
2
3
–1
m
0
1
2
0
m
1
2
m
3
y(2) = 1 + a + a2
Fig 6 : Computation of y(2).
Solving similarly for other values of n, we can write y(n) for any value of n as shown below.
n
y(n) = 1 + a + a2 +......+ an =
åa
p
; for n ³ 0
1+ a + a2
1+ a
y(n)
1+ a + a2 + a 3
p=0
1
0
1
2
n
3
Fig 7 : Graphical representation of y(n).
6.10 Circular Convolution
6.10.1 Circular Representation and Circular Shift of Discrete Time Signal
Consider a finite duration sequence x(n) and its periodic extension xp(n). The periodic extension of
x(n) can be expressed as xp(n) = x(n + N), where N is the periodicity. Let N = 4. The sequence x(n) and
its periodic extension are shown in fig 6.24.
Let, x(n) =
=
=
=
1;
2;
3;
4;
n
n
n
n
=
=
=
=
0
1
2
3
xp(n)
x(n)
4
2
–4
2
2
1
1
1
0
2
1
3
n
Fig 6.24a : Finite duration sequence x(n).
3
3
3
2
1
4
4
4
3
–3
–2
–1
0
1
2
3
4
5
6
7
n
Fig 6.24b : Periodic extension of x(n).
Fig 6.24 : A finite duration sequence and its periodic extension.
Let us delay the periodic sequence xp(n) by two units of time as shown in fig 6.25(a). (For delay
the sequence is shifted right). Let us denote one period of this delayed sequence by x1(n). One period of
the delayed sequence is shown in figure 6.25(b).
Chapter 6 - Discrete Time Signals and Systems
xp(n – 2)
6. 64
x1(n)
4
2
2
–2
–1
0
1
1
2
1
2
2
1
1
3
3
3
3
x1(n) = xp((n – 2))4
4
4
4
3
6
5
4
7
8
0
n
1
2
3
n
Fig 6.25b : One period of xp(n – 2).
Fig 6.25a : xp(n) delayed by two units of time.
Fig 6.25 : Delayed version of xp(n).
The sequence x1(n) can be represented by xp(n – 2, (mod 4)), or xp((n – 2))4, where mod 4 indicates
that the sequence repeats after 4 samples. The relation between the original sequence x(n) and one period
of the delayed sequence x1(n) are shown below.
x1(n) = xp(n – 2, (mod 4)) = xp((n – 2))4
\ When n = 0; x1(0) = xp((0 – 2))4 = x p((– 2))4= x(2) = 3
When n = 1; x 1(1) = xp((1 – 2))4 = xp((– 1))4= x(3) = 4
When n = 2; x1(2) = xp((2 – 2))4 = xp((0))4 = x(0) = 1
When n = 3; x1(3) = xp((3 – 2))4 = x p((1))4 = x(1) = 2
The sequences x(n) and x1(n) can be represented as points on a circle as shown in fig 6.26. From
fig 6.26 we can say that, x1(n) is simply x(n) shifted circularly by two units in time, where the counter
clockwise (anticlockwise) direction has been arbitrarily selected for right shift or delay.
x(1) = 2
x1(1) = 4
x(n)
x(0) = 1
­
x(2) = 3
Rotate
anticlockwise
x1(2) = 1
x1(n)
= xp((n – 2))4
x1(0) = 3
x1(3) = 2
Fig 6.26b : Circular representation of x1(n).
x(3) = 4
Fig 6.26a : Circular representation of x(n).
Fig 6.26 : Circular representation of a signal and its delayed version.
Let us advance the periodic sequence xp(n) by three units of time as shown in fig 6.27(a). Let us
denote one period of this advanced sequence by x2(n). One period of the advanced sequence is shown in
fig 6.27(b).
x2(n)
xp(n + 3)
4
4
3
3
2
–3
1
-2
-1
0
3
3
2
2
1
x2(n) = xp((n + 3))4
4
4
2
1
1
1
2
3
4
5
6
7
8
Fig 6.27a : xp(n) advanced by three units of time.
n
0
1
2
3
n
Fig 6.27b : One period of xp(n + 3).
Fig 6.27 : Advanced version of xp(n).
6. 65
Signals & Systems
The sequence x2(n) can be represented by xp(n + 3, (mod 4)) or xp((n + 3))4, where mod 4 indicates
that the sequence repeats after 4 samples. The relation between the original sequence x(n) and one period
of the advanced sequence x2(n) are shown below.
x2(n) = xp(n + 3, (mod 4)) = xp((n + 3))4
\ When n = 0; x2(0) = xp((0 + 3))4 = xp((3))4 = x(3) = 4
When n = 1; x2(1) = xp((1 + 3))4 = xp((4))4 = x(0) = 1
When n = 2; x2(2) = xp((2 + 3))4 = xp((5))4 = x(1) = 2
When n = 3; x2(3) = xp((3 + 3))4 = xp((6))4 = x(2) = 3
The sequences x(n) and x2(n) can be represented as points on a circle as shown in fig 6.28. From
fig 6.28 we can say that x2(n) is simply x(n) shifted circularly by three units in time where clockwise
direction has been selected for left shift or advance.
x(1) = 2
x(n)
x(0) = 1
Rotate
clockwise
x(3) = 4
Fig 6.28a : Circular representation of x(n).
­
x(2) = 3
x2(1) = 1
x2(2) = 2
x2(n)
x2(0) = 4
= xp((n + 3))4
x2(3) = 3
Fig 6.28b : Circular representationof x2(n).
Fig 6.28 : Circular representationof a signal and its advanced version.
Thus we conclude that a circular shift of an N-point sequence is equivalent to a linear shift of its
periodic extension and viceversa. If a nonperiodic N-point sequence is represented on the circumference
of a circle then it becomes a periodic sequence of periodicity N. When the sequence is shifted circularly,
the samples repeat after N shifts. This is similar to modulo-N operation. Hence, in general, the circular
shift may be represented by the index mod-N. Let x(n) be an N-point sequence represented on a circle and
x¢(n) be its shifted sequence by m units of time.
Now, x¢(n) = x(n – m, mod N) º x((n – m))N
..... (6.53)
When m is positive, the equation (6.53) represents delayed sequence and when m is negative, the
equation (6.53) represents advanced sequence.
6.10.2 Circular Symmetrics of Discrete Time Signal
The circular representation of a sequence and the resulting periodicity gives rise to new definitions
for even symmetry, odd symmetry and the time reversal of the sequence.
An N-point sequence is called even if it is symmetric about the point zero on the circle. This implies
that,
x(N - n) = x(n) ; for 0 £ n £ N - 1
...... (6.54)
Chapter 6 - Discrete Time Signals and Systems
6. 66
An N-point sequence is called odd if it is antisymmetric about the point zero on the circle.
This implies that,
x(N - n) = - x(n) ; for 0 £ n £ N - 1
...... (6.55)
The time reversal of a N-point sequence is obtained by reversing its sample about the point zero on
the circle. Thus the sequence x(–n, (mod N) ) is simply written as,
x (-n, (mod N)) = x(N - n) ; for 0 £ n £ N - 1
...... (6.56)
This time reversal is equivalent to plotting x(n) in a clockwise direction on a circle, as shown in
fig 6.29.
x(6)
x(2)
x(5)
x(1)
x(3)
x(n)
x(4)
x(0)
x(5)
x(4)
x(7)
x(-n)
x(3)
x(7)
x(6)
x(0)
x(1)
x(2)
Fig 6.29 : Circular representation of an 8 point sequence and its folded sequence.
6.10.3 Definition of Circular Convolution
The Circular convolution of two periodic discrete time sequences x1(n) and x2(n) with periodicity
of N samples is defined as,
N-1
x 3 ( n) =
å
m=0
N-1
x1 ( m) x 2 ((n - m))
or
N
x3 ( n) =
å
x2 ( m) x1 ((n - m)) N
.....(6.57)
m= 0
where, x3(n) is the sequence obtained by circular convolution,
x1((n – m))N represents circular shift of x1(n)
x2((n – m))N represents circular shift of x2(n)
m is a dummy variable.
The output sequence x3(n) obtained by circular convolution is also a periodic sequence with periodicity
of N samples. Hence this convolution is also called periodic convolution.
The convolution relation of equation (6.57) can be symbolically expressed as
x3(n) = x1(n) * x2(n) = x2(n) * x1(n)
..... (6.58)
where, the symbol * indicates circular convolution operation.
The circular convolution is defined for periodic sequences. But circular convolution can be performed
with non-periodic sequences by periodically extending them.The circular convolution of two sequences
requires that, at least one of the sequences should be periodic. Hence it is sufficient if one of the sequences
is periodically extended in order to perform circular convolution.
6. 67
Signals & Systems
The circular convolution of finite duration sequences can be performed only if both the sequences
consist of the same number of samples. If the sequences have different number of samples, then convert
the smaller size sequence to the length of larger size sequence by appending zeros.
Circular convolution basically involves the same four steps as that for linear convolution, namely,
folding one sequence, shifting the folded sequence, multiplying the two sequences and finally summing
the values of the product sequence. Like linear convolution, any one of the sequence is folded and rotated
in circular convolution.
The difference between the two is that, in circular convolution the folding and shifting (rotating)
operations are performed in a circular fashion by computing the index of one of the sequences by
modulo-N operation. In linear convolution there is no modulo-N operation.
6.10.4 Procedure for Evaluating Circular Convolution
Let, x1(n) and x2(n) be periodic discrete time sequences with periodicity of N-samples. If x1(n)
and x2(n) are non-periodic then convert the sequences to N-sample sequences and periodically extend the
sequence x2(n) with periodicity of N-samples.
Now the circular convolution of x1(n) and x2(n) will produce a periodic sequence x3(n) with
periodicity of N-samples. The samples of one period of x3(n) can be computed using the equation (6.57).
The value of x3(n) at n = q is obtained by replacing n by q, in equation (6.57).
N-1
\ x 3 (q ) =
å x (m) x ((q - m))
1
2
N
.....(6.59)
m= 0
The evaluation of equation (6.59) to determine the value of x3(n) at n = q involves the following
five steps.
1. Change of index : Change the index n in the sequences x1(n) and x2(n), in order to get the
sequences x1(m) and x2(m). Represent the samples of one period of the
sequences on circles.
2. Folding
: Fold x2(m) about m = 0, to obtain x2(-m).
3. Rotation
: Rotate x2(-m) by q times in anti-clockwise if q is positive, rotate x2(-m)
by q times in clockwise if q is negative to obtain x2((q – m))N.
4. Multiplication
: Multiply x1(m) by x2((q – m))N to get a product sequence. Let the product
sequence be vq(m). Now, vq(m) = x1(m) ´ x2((q – m))N.
5. Summation
: Sum up the samples of one period of the product sequence vq(m) to
obtain the value of x3(n) at n = q. [i.e., x3(q)].
The above procedure will give the value of x3(n) at a single time instant say n = q. In general we are
interested in evaluating the values of the sequence x3(n) in the range 0 < n < N - 1. Hence the steps
3 , 4 and 5 given above must be repeated, for all possible time shifts in the range 0 < n < N - 1.
Chapter 6 - Discrete Time Signals and Systems
6. 68
6.10.5 Linear Convolution via Circular Convolution
When two numbers of N-point sequences are circularly convolved, it produces another N-point
sequence. For circular convolution, one of the sequence should be periodically extended. Also the resultant
sequence is periodic with period N.
The linear convolution of two sequences of length N1 and N2 produces an output sequence of
length N1 + N2 -1. To perform linear convolution via circular convolution both the sequences should be
converted to N1 + N2 -1 point sequences by padding with zeros. Then perform circular convolution of
N1 + N2 -1 point sequences. The resultant sequence will be same as that of linear convolution of N1 and
N2 point sequences.
6.10.6 Methods of Computing Circular Convolution
Method 1 : Graphical Method
In graphical method the given sequences are converted to same size and represented on circles. In
case of periodic sequences, the samples of one period are represented on circles. One of the sequence is
folded and shifted circularly. Let x1(n) and x2(n) be the given sequences. Let x3(n) be the sequence
obtained by circular convolution of x1(n) and x2(n). The following procedure can be used to get a sample
of x3(n) at n = q.
1. Change the index n in the sequences x1(n) and x2(n) to get x1(m) and x2(m) and then represent
the sequences on circles.
2. Fold one of the sequence. Let us fold x2(m) to get x2(–m).
3. Rotate (or shift) the sequence x2(–m), q times to get the sequence x2((q – m))N. If q is positive
then rotate (or shift) the sequence in anticlockwise direction and if q is negative then rotate (or
shift) the sequence in clockwise direction.
4. The sample of x3(q) at n = q is given by,
N-1
å
x 3 (q) =
m=0
N-1
x1 (m) x2 ((q - m)) N =
å
x1 (m) x2,q (m)
m=0
where, x2,q (m) = x2 ((q - m)) N
Determine the product sequence x1 ( m) x 2,q ( m) for one period.
5. The sum of all the samples of the product sequence gives the sample x3(q) [i.e., x3(n) at n = q].
The above procedure is repeated for all possible values of n to get the sequence x3(n).
Method 2 : Tabular Method
Let x1(n) and x2(n) be the given N-point sequences. Let x3(n) be the N-point sequence obtained by
circular convolution of x1(n) and x2(n). The following procedure can be used to obtain one sample of
x3(n) at n = q.
6. 69
Signals & Systems
1. Change the index n in the sequences x1(n) and x2(n) to get x1(m) and x2(m) and then represent
the sequences as two rows of tabular array.
2. Fold one of the sequence. Let us fold x2(m) to get x2(–m).
3. Periodically extend x2(–m). Here the periodicity is N, where N is the length of the given sequences.
4. Shift the sequence x2(–m), q times to get the sequence x2((q – m))N. If q is positive then shift
the sequence to the right and if q is negative then shift the sequence to the left.
5. The sample of x3(q) at n = q is given by
N-1
x 3 (q) =
å x (m) x ((q - m))
1
2
N-1
N
=
å x (m) x
1
2,q
(m)
m=0
m=0
where x2,q (m) = x 2 ((q - m)) N
Determine the product sequence x1 ( m) x 2,q ( m) for one period.
6. The sum of the samples of the product sequence gives the sample x3(q) [i.e., x3(n) at n = q].
The above procedure is repeated for all possible values of n to get the sequence x3(n).
Method 3: Matrix Method
Let x1(n) and x2(n) be the given N-point sequences. The circular convolution of x1(n) and x2(n)
yields another N-point sequence x3(n).
In this method an (N ´ N) matrix is formed using one of the sequence as shown below. Another
sequence is arranged as a column vector (column matrix) of order (N ´ 1). The product of the two
matrices gives the resultant sequence x3(n).
LMx (0) x (N - 1)
MMxx ((12)) xx ((10))
MM M
M
x
(
N
2
)
x
(
N
MMx (N - 1) x (N -- 32))
N
2
2
2
2
2
2
2
2
2
2
x 2 ( N - 2)
x2 ( N - 1)
x2 (0)
M
x 2 ( N - 4)
x 2 ( N - 3)
..... x 2 (2)
..... x2 (3)
..... x2 (4)
M
..... x2 (0)
..... x2 (1)
LM
MM
MM
MM
MN
OP
PP
PP
PP
Q
Example 6.24
Perform circular convolution of the two sequences,
x1(n) = {2, 1, 2, 1} and x2(n)= {1, 2, 3, 4}
­
­
Solution
Method 1:Graphical Method of Computing Circular Convolution
Let x3(n) be the sequence obtained by circular convolution of x1(n) and x2(n).
The circular convolution of x1(n) and x2(n) is given by,
OP
PP
PP
PP
PQ
x1 (0)
x 2 (1)
x1 (1)
x2 (2)
x1 (2)
x2 (3)
=
M
´
M
M
x2 ( N - 1)
x1 ( N - 2)
x2 (0)
x1 ( N - 1)
LM x (0) OP
MM xx ((12)) PP
MM M PP
MM x (NM - 2)PP
MN x (N - 1) PQ
3
3
3
3
3
Chapter 6 - Discrete Time Signals and Systems
6. 70
N -1
N- 1
å x (m) x ((n - m))
x 3 (n) =
1
2
N
=
m = 0
å x (m) x
1
2,n (m )
m = 0
where x 2,n (m) = x 2 ((n - m))N and m is the dummy variable used for convolution.
The index n in the given sequences are changed to m and each sequence is represented as points on a circle as
shown below. The folded sequence x2(–m) is also represented on the circle.
x1(1)=1
x1(2)=2
x2(1)=2
x1(0)=2
x1(m)
x2(2)=3
x2(3)=4
x2(0)=1
x2(m)
x1(3)=1
x2(3)=4
Fig 1.
Fig 2.
x2(2)=3
x2(–m)
x2(0)=1
x2(1)=2
Fig 3.
The given sequences are 4-point sequences . \ N = 4.
Each sample of x3(n) is given by sum of the samples of product sequence defined by the equation,
3
å x (m)
x 3 (n) =
1
3
å v (m)
x 2,n (m) =
; where v n (m) = x1(m) x 2,n (m)
n
m = 0
m = 0
Using the above equation, graphical method of computing each sample of x3(n) are shown in fig 4 to fig 7.
3
When n = 0 ; x 3 (0) =
2
1
m = 0
1
x1(m)
2
X
3
å x (m) x (-m)
1
å x (m) x
1
4
1
Þ
2 ´ 3=6
3
å x (m) x (1 - m)
1
2
m = 0
1
2
1
3
=
å x (m) x
1
2
X
4
3
2,1(m)
m = 0
1
x1(m)
x 2,1(m)
2 ´ 1=2
v0(m)
1 ´2=2
The sum of samples of v0(m) gives x3(0)
\ x3(0) = 2 + 4 + 6 + 2 = 14
Fig 4 : Computation of x3(0).
=
0
m = 0
1´ 4 = 4
2
When n = 1 ; x 3 (1)
3
å v (m)
=
2,0 (m)
m = 0
x2,0(m)
3
=
2
=
å v (m)
1
m = 0
1´ 1 = 1
2
Þ
3
Fig 5 : Computation of x3(1).
2´ 4=8
v1(m)
2 ´2=4
1 ´3=3
The sum of samples of v1(m) gives x3(1)
\ x3(1) = 4 + 1 + 8 + 3 = 16
6. 71
Signals & Systems
3
å x (m) x (2 - m)
When n = 2 ; x 3 (2) =
1
2
3
m = 0
2
1´ 2 = 2
x2, 2(m)
1
X
1
3
Þ
2 ´ 1=2
1 ´4=4
The sum of samples of v2(m) gives x3(2)
Fig 6 : Computation of x3(2).
3
å x (m) x (3 - m)
2
m = 0
1
x1(m)
2
1
2
3
x2, 3(m)
1
3
4
Þ
åv
x1(m) x 2,3 (m) =
m = 0
3
2
X
\ x3(2) = 6 + 2 + 2 + 4 = 14
å
=
2 ´3=6
v 2(m)
4
When n = 3 ; x 3 (3) =
2 (m )
m = 0
2
x1(m)
åv
x1(m) x 2,2 (m) =
m = 0
1
2
3
å
=
3 (m )
m = 0
1´ 3 = 3
2 ´ 2=4
1
v3(m)
2´ 4=8
1 ´1=1
The sum of samples of v3(m) gives x3(3)
Fig 7 : Computation of x3(3).
\ x3(3) = 8 + 3 + 4 + 1 = 16
\ x3(n) = {14, 16, 14, 16}
­
Method 2 : Circular Convolution Using Tabular Array
The index n in the given sequences are changed to m and then, the given sequences can be represented in the
tabular array as shown below. Here the shifted sequences x2,n are periodically extended with a periodicity of N = 4. Let x3(n)
be the sequence obtained by convolution of x1(n) and x2(n). Each sample of x3(n) is given by the equation,
N - 1
x 3 (n) =
å x (m) x ((n - m))
1
2
N
m = 0
N - 1
å x (m) x
=
1
2,m (m ),
where x 2,m (m) = x 2 ((n - m))N
m = 0
Note : The bold faced numbers are samples obtained by periodic extension.
m
0
1
2
3
x1(m)
2
1
2
1
x2(m)
1
2
3
4
x2(–m) = x2,0(m)
x2(1– m) = x2,1(m)
x2(2 – m) = x2,2(m)
x3(3 – m) = x2,3(m)
–3
4
–2
–1
3
2
1
4
3
2
4
3
2
1
4
3
4
3
2
1
4
4
3
2
1
To determine a sample of x3(n) at n = q, multiply the sequence, x1(m) and x 2,q (m), to get a product sequence
x1(m) x 2,q (m). (i.e., multiply the corresponding elements of the row x1(m) and x 2,q (m) ). The sum of all the samples of the
product sequence gives x3(q).
Chapter 6 - Discrete Time Signals and Systems
6. 72
3
å x (m) x
When n = 0 ; x 3 (0) =
1
2,0 (m)
m=0
= x1(0) x 2.0 (0) + x1(1) x 2.0 (1) + x1(2) x 2.0 (2) + x1(3 ) x 2,0 (3)
= 2 + 4 + 6 + 2 = 14
The samples of x3(n) for other values of n are calculated as shown for n = 0.
3
When n = 1; x 3 (1) =
å x (m)
1
x 2,1(m) = 4 + 1 + 8 + 3 = 16
m=0
3
When n = 2; x 3 (2) =
å x (m)
1
x 2,2 (m) = 6 + 2 + 2 + 4 = 14
m=0
3
When n = 3; x 3 (3) =
å x (m)
1
x 2,3 (m) = 8 + 3 + 4 + 1 = 16
m=0
\ x 3 (n) =
RS14, 16, 14, 16UV
TA
W
Method 3 : Circular Convolution Using Matrices
The sequence x1(n) can be arranged as a column vector of order N ´ 1 and using the samples of x2(n) the N ´ N
matrix is formed as shown below. The product of the two matrices gives the sequence x3(n).
LMx (0)
MMx (1)
MMx (2)
Nx (3)
LM1 4
MM2 1
MN34 23
2
x 2 (3)
x 2 (2)
2
x 2 (0)
x 2 (3 )
2
x 2 (1)
x 2 (0)
2
x 2 (2)
x 2 (1)
3
4
1
2
2
3
4
1
OP
PP
PQ
LM2OP
MM1 PP
MN12PQ
OP LMx (0)OP
P Mx (1) P
x (3)PP MMx (2)PP
x (0)PQ MNx (3)PQ
LM14OP
16
= M P
MM14PP
N16Q
x 2 (1)
1
x 2 (2)
1
2
1
2
1
LMx (0)OP
MMx (1) PP
MMx (2)PP
Nx (3)Q
3
=
3
3
3
\ x3(n) = {14, 16, 14, 16}
­
Example 6.25
Perform the circular convolution of the two sequences x1(n) and x2(n), where,
x1(n) = 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6
­
l
q
x 2 (n) = 0.1, 0.3, 0.5, 0.7, 0.9, 1.1, 1.3, 1.5
­
l
q
Solution
Let x3(n) be the result of the circular convolution of x1(n) and x2(n). The given sequences consists of eight samples.
The x3(n) will also have 8 samples.
The sequences are represented in the tabular array as shown below after replacing n by m. The sequence x2(m) is
folded and shifted.
The shifted sequences x2,m(m) are periodically extended with a periodicity of N = 8.
6. 73
Signals & Systems
Note : The bold faced numbers are samples obtained by periodic extension.
m
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
x1(m)
0.2 0.4
0.6
0.8 1.0 1.2
1.4 1.6
x2(m)
0.1 0.3
0.5
0.7 0.9 1.1
1.3 1.5
x2(–m) = x2,0(m)
1.5
1.3 1.1
0.9
0.7
0.5
0.3
0.1 1.5
1.3
1.1 0.9 0.7
0.5 0.3
x2(1 – m) = x2,1(m)
1.5 1.3
1.1
0.9
0.7
0.5
0.3 0.1
1.5
1.3 1.1 0.9
0.7 0.5
x2(2 – m) = x2,2(m)
1.5
1.3
1.1
0.9
0.7
0.5 0.3
0.1
1.5 1.3 1.1
0.9 0.7
1.5
1.3
1.1
0.9
0.7 0.5
0.3
0.1 1.5 1.3
1.1 0.9
1.5
1.3
1.1
0.9 0.7
0.5
0.3 0.1 1.5
1.3 1.1
1.5
1.3
1.1 0.9
0.7
0.5 0.3 0.1
1.5 1.3
1.5
1.3 1.1
0.9
0.7 0.5 0.3
0.1 1.5
1.5 1.3
1.1
0.9 0.7 0.5
0.3 0.1
x2(3 – m) = x2,3(m)
x2(4 – m) = x2,4(m)
x2(5 – m) = x2,5(m)
x2(6 – m) = x2,6(m)
x2(7 – m) = x2,7(m)
Each sample of x3(n) is given by the equation,
7
x 3 (n) =
å x (m) x ((n - m))
1
2
8
7
å x (m) x
=
m=0
1
2,n (m)
; where x 2,n (m) = x 2 ((n - m))8
m=0
The samples of x3(0) are calculated as shown below.
7
7
When n = 0 ; x 3 (0) =
å x (m)
1
å x (m)
x 2 ( -m) =
x 2,0 (m)
1
m=0
m =0
= x1(0) x 2,0 (0) + x1(1) x 2,0 (1) + x1(2) x 2,0 (2) + x1(3) x 2,0 (3)
+ x1(4) x 2,0 (4) + x1(5) x 2,0 (5) + x1(6) x 2,0 (6) + x1(7) x 2,0 (7)
= 0.02 + 0.6 + 0.78 + 0.88 + 0.9 + 0.84 + 0.7 + 0.48 = 5.20
The samples of x3(n) for other values of n are calculated as shown for n = 0.
7
7
When n = 1; x 3 (1) =
å x (m)
1
x 2 (1 - m) =
å x (m)
1
x 2 (2 - m) =
7
å x (m)
1
å x (m) x (4 - m)
1
2
å x (m) x
1
2,3 (m)
= 6.64
å x (m) x
2,4 (m)
= 6.48
1
m=0
7
å x (m) x (5 - m)
1
2
7
=
m=0
å
x1(m) x 2,5 (m) = 6.00
m=0
7
7
å x (m) x (6 - m)
1
2
=
å x (m)
1
x 2,6 (m) = 5.20
m =0
m =0
7
å x (m) x (7 - m)
1
m =0
= 6.48
7
=
m =0
When n = 7; x 3 (7) =
2,2 (m)
m =0
7
When n = 6; x 3 (6) =
1
7
x 2 (3 - m) =
m =0
When n = 5; x 3 (5) =
å x (m) x
m =0
m =0
When n = 4; x 3 (4) =
= 6.00
2,1(m)
7
7
When n = 3; x 3 (3) =
1
m =0
m =0
When n = 2; x 3 (2) =
å x (m) x
2
7
=
å x (m) x
1
= 4.08
2,7 (m)
m =0
\ x 3 (n) = 5.20, 6.00, 6.48, 6.64, 6.48, 6.00, 5.20, 4.08
l
A
q
Chapter 6 - Discrete Time Signals and Systems
6. 74
Example 6.26
Find the linear and circular convolution of the sequences, x(n) = 1, 0.5
l
q
l
q
and h(n) = 0.5, 1 .
­
­
Solution
Linear Convolution by Tabular Array
¥
å x(m) h(n - m)
Let, y(n) = x(n) * h(n) =
; where m is a dummy variable for convolution.
m = -¥
Since both x(n) and h(n) starts at n = 0, the output sequence y(n) will also start at n = 0.
The length of y(n) is 2 + 2 – 1 = 3.
Let us change the index n to m in x(n) and h(n). The sequences x(m) and h(m) are represented in the tabular array
as shown below.
Note : The unfilled boxes in the table are considered as zeros.
m
0
1
x(m)
–1
1
0.5
h(m)
0.5
1
h(–m) = h0(m)
1
2
0.5
h(1 – m) = h1(m)
1
0.5
h(2 – m) = h2(m)
1
0.5
Each sample of y(n) is given by the relation,
¥
å x(m) h(n - m)
y(n) =
m = -¥
¥
=
å x(m) h (m)
n
; where hn (m) = h(n - m)
m = -¥
¥
1
å x(m) h(-m)
When n = 0 ; y(0) =
å x(m) h (m)
=
0
m = -¥
m = -1
= x( -1) h0 ( -1) + x(0) h0 (0) + x(1) h0 (1)
¥
å x(m) h(1 - m)
When n = 1 ; y(1) =
å x(m) h (m)
=
1
m = -¥
¥
2
=
m = -¥
\ y(n) =
= 1 + 0.25 = 125
.
m = 0
å x(m) h(2 - m)
When n = 2 ; y(2) =
= 0 + 0.5 + 0 = 0.5
1
å x(m) h (m)
2
= 0 + 0.5 + 0 = 0.5
m = 0
l0.5, 1.25, 0.5q
­
Circular Convolution by Tabular Array
N- 1
å x(m) h((n - m))
Let, y(n) = x(n) * h(n) =
N
; where m is a dummy variable for convolution.
m =0
The index n in the sequences are changed to m and the sequences are represented in the tabular array as shown
below. The shifted sequence hn(m) is periodically extended with periodicity N = 2.
Note : The bold faced number is the sample obtained by periodic extension.
m
–1
x(m)
h(m)
h(–m) = h0(m)
h(1 – m) = h1(m)
1
0
1
1
0.5
0.5
1
0.5
1
1
0.5
6. 75
Signals & Systems
Each sample of y(n) is given by the equation,
N - 1
N - 1
å x(m) h((n - m))
y(n) =
=
N
m= 0
å x(m) h (m); where h (m)
n
n
= h((n - m))N
m= 0
N - 1
å
When n = 0 ; y(0) =
1
å x(m) h (m)
x(m) h( -m) =
0
m= 0
m = 0
= x(0) h0 (0) + x(1) h0 (1) = 0.5 + 0.5 = 10
.
N - 1
When n = 1 ; y(1) =
å
1
å
x(m) h(1 - m) =
m = 0
x(m) h1(m)
m = 0
= x(0) h1(0) + x(1) h1(1) = 1 + 0.25 = 125
.
\ y(n) = 10
. , 1.25
l
q
A
Example 6.27
The input x(n) and impulse response h(n) of a LTI system are given by,
x(n) = {–1, 1, 2, –2) ; h(n) = {0.5, 1, –1, 2, 0.75}
­
­
Determine the response of the system a) Using linear convolution and b) using circular convolution.
Solution
a) Response of LTI system using linear convolution
Let y(n) be the response of LTI system. By convolution sum formula,
+¥
å x(m) h(n - m)
y(n) = x(n) * h(n) =
;
where m is a dummy variable used for convolution.
m = -¥
The sequence x(n) starts at n = 0 and h(n) starts at n = –1. Hence y(n) will start at n = 0 + (–1) = –1. The length of x(n)
is 4 and the length of h(n) is 5. Hence the length of y(n) is (4 + 5 – 1) = 8. Also y(n) ends at n = 0 + (–1) + (4 + 5 –2) = 6.
Let us change the index n to m in x(n) and h(n). The sequences x(m) and h(m) are represented on the tabular array
as shown below. Let us fold h(m) to get h(–m) and shift h(–m) to perform convolution operation.
Note : The unfilled boxes in the table are considered as zeros.
m
–4
–3
–2
–1
x(m)
h(m)
0.5
h(–m)
h(–1 – m) = h–1(m)
h(–m) = h0(m)
h(1 – m) = h1(m)
h(2 – m) = h2(m)
h(3 – m) = h3(m)
h(4 – m) = h4(m)
h(5 – m) = h5(m)
h(6 – m) = h6(m)
0.75
0
1
2
3
–1
1
2
–2
1
–1
2
0.75
0.5
4
5
6
0.75
2
–1
1
2
–1
1
0.5
2
–1
1
0.75
2
–1
1
0.5
0.75
2
–1
1
2
–1
1
0.75
2
–1
1
0.5
0.75
2
–1
1
0.5
0.75
2
–1
1
0.75
0.75
7
0.5
0.5
0.5
0.5
Chapter 6 - Discrete Time Signals and Systems
6. 76
Each sample of y(n) is given by summation of the product sequence, x(m) h(n – m). To determine a sample of y(n)
at n = q, multiply the sequence x(m) and hq(m) to get a product sequence (i.e., multiply the corresponding elements of the row
x(m) and hq(m)). The sum of all the samples of the product sequence gives y(q).
+¥
i. e., y(n) =
å x(m) h(n - m)
m = -¥
+¥
å x(m) h(m)
=
m = -¥
3
å x(m) h
When n = - 1 ; y(- 1) =
-1(m)
m = -4
=
x(–4) h–1(–4) + x(–3) h–1(–3) + x(–2) h–1(–2) + x(–1) h–1(–1) + x(0) h–1(0)
+ x(1) h–1(1) + x(2) h–1(2) + x(3) h–1(3)
= 0 + 0 + 0 + 0 + (–0.5) + 0 + 0 + 0 = –0.5
The samples of y(n) for other values of n are calculated as shown for n = –1.
3
When n = 0 ; y(0) =
å x(m) h (m)
0
= 0 + 0 + 0 + ( -1) + 0.5 + 0 + 0 = - 0.5
m = -3
3
When n = 1 ; y(1) =
å x(m) h (m)
1
= 0 + 0 + 1 + 1 + 1 + 0 = 3
m = -2
3
When n = 2 ; y(2) =
å x(m) h (m)
2
= 0 + ( -2) + ( - 1) + 2 + ( -1)
= -2
m = -1
4
When n = 3 ; y(3) =
å x(m) h (m)
3
= - 0.75 + 2 + ( -2) + ( -2) + 0 = - 2.75
m= 0
5
When n = 4 ; y(4) =
å x(m) h (m)
4
= 0 + 0.75 + 4 + 2 + 0 + 0 = 6.75
m= 0
6
When n = 5 ; y(5) =
å x(m) h (m)
5
= 0 + 0 + 1.5 + (- 4) + 0 + 0 + 0
= - 2. 5
m= 0
7
When n = 6 ; y(6) =
å x(m) h (m)
6
= 0 + 0 + 0 + ( -1.5) + 0 + 0 + 0 + 0
= - 1.5
m= 0
The response of LTI system y(n) is,
y(n) = {–0.5, –0.5, 3, –2, –2.75, 6.75, –2.5, –1.5}
­
b) Response of LTI System Using Circular Convolution
The response of LTI system is given by linear convolution of x(m) and h(m). Let y(n) be the response sequence of
LTI system. To get the result of linear convolution from circular convolution, both the sequences should be converted to the
size of y(n) and perform circular convolution of the converted sequences. Also the converted sequences should start and
end at the same value of n as that of y(n).
The length of x(n) is 4 and the length of h(n) is 5. Hence the length of y(n) is (4 + 5 – 1) = 8. Therefore both the
sequences should be converted to 8-point sequences.
The x(n) starts at n = 0 and h(n) starts at n = –1. Hence y(n) will start at n = 0 + (–1) = –1. The y(n) will end at
n = [0 +(–1)] + (4 + 5 – 2) = 6. Therefore the converted sequences should start at n = –1 and end at n = 6.
\ x(n) = {0, –1, 1, 2, –2, 0, 0, 0} and h(n) = {0.5, 1, –1, 2, 0.75, 0, 0, 0}
­
­
The converted sequences x(n) and h(n) are represented on the tabular array after replacing the index n by m as
shown below. The sequence h(m) is folded and shifted.
The shifted sequences hn(m) are periodically extended with a periodicity of N = 8.
Signals & Systems
6. 77
Note : The bold faced numbers are samples obtained by periodic extension of the sequences.
m
–1
0
1
2
3
4
5
6
x(m)
–7 –6
0
–1
1
2
–2
0
0
0
h(m)
0.5
1
–1
2
0.75
0
0
0
–1
1
0.5
h(–m)
–5
0
–4
0
0
–3
–2
0.75
2
7
h(–1 – m) = h–1(m) 0
0
0
0.75
2
–1
1
0.5
0
0
0
0.75
2
–1
1
h(–m) = h0(m)
0
0
0
0.75
2
–1
1
0.5
0
0
0
0.75
2
–1
0
0
0
0.75
2
–1
1
0.5
0
0
0
0.75
2
0
0
0.75
2
–1
1
0.5
0
0
0
0.75
0
0
0
0.75
2
–1
1
0.5
0
0
0
0
0
0
0.75
2
–1
1
0.5
0
0
0
0
0
0.75
2
–1
1
0.5
0
0.5
0
0
0
0.75
2
–1
1
0.5
h(1 – m) = h1(m)
h(2 – m) = h2(m)
0
h(3 – m) = h3(m)
h(4 – m) = h4(m)
h(5 – m) = h5(m)
h(6 – m) = h6(m)
0
0
0.75
2
–1
1
Let y(n) be the sequence obtained by circular convolution of x(n) and h(n). Each sample of y(n) is given by,
6
y(n) =
å x(m) h((n - m))
6
8
=
m = -1
å x(m) h (m)
n
; where hn (m) = h((n - m))8
m = -1
To determine a sample of y(n) at n = q, multiply the sequence x(m) and hq(m) to get a product sequence x(m) hq(m),
(i.e., multiply the corresponding elements of the row x(m) and hq(m)). The sum of all the samples of the product sequence
gives y(q).
6
x(m) h-1(n) = x( -1) h-1( -1) + x(0) h-1(0) + x(1) h-1(1) + x(2) h-1(2)
When n = - 1 ; y( -1) =
å
m = -1
+ x(3) h-1(3) + x(4) h-1(4) + x(5) h-1(5) + x(6) h-1(6)
= 0 + ( -0.5) + 0 + 0 + 0 + 0 + 0 + 0 = - 0.5
The samples of y(n) for other values of n are calculated as shown for n = –1.
6
When n = 0 ; y(0) =
å x(m) h m
0
= 0 + (-1) + 0.5 + 0 + 0 + 0 + 0 + 0 = - 0.5
m = -1
6
When n = 1 ; y(1) =
å x(m) h m
1
= 0 + 1 + 1 + 1 + 0 + 0 + 0 + 0 = 3
m = -1
6
When n = 2 ; y(2) =
å x(m) h m
2
= 0 + (-2) + ( -1) + 2 + (- 1) + 0 + 0 + 0 = - 2
m = -1
6
When n = 3 ; y(3) =
å x(m) h m
3
= 0 + (-0.75) + 2 + ( -2) + ( -2) + 0 + 0 + 0 = - 2.75
m = -1
6
When n = 4 ; y(4) =
å x(m) h m
4
= 0 + 0 + 0.75 + 4 + 2 + 0 + 0 + 0 = 6.75
m = -1
6
When n = 5 ; y(5) =
å x(m) h m
5
= 0 + 0 + 0 + 1.5 + (-4) + 0 + 0 + 0 = - 2.5
m = -1
6
When n = 6 ; y(6) =
å x(m) h m
6
. + 0 + 0 + 0 = - 1.5
= 0 + 0 + 0 + 0 + ( -15)
m = -1
The response of LTI system y(n) is,
y(n) = {–0.5, –0.5, 3, –2, –2.75, 6.75, –2.5, –1.5}
­
Note : 1. Since circular convolution is periodic, the convolution is performed for any one period.
2. It can be observed that the results of both the methods are same.
Chapter 6 - Discrete Time Signals and Systems
6. 78
6.11 Sectioned Convolution
The response of an LTI system for any arbitrary input is given by linear convolution of the input
and the impulse response of the system. If one of the sequences (either the input sequence or impulse
response sequence) is very much larger than the other, then it is very difficult to compute the linear
convolution for the following reasons.
1. The entire sequence should be available before convolution can be carried out. This makes long
delay in getting the output.
2. Large amounts of memory is required to store the sequences.
The above problems can be overcome in the sectioned convolutions. In this technique the larger
sequence is sectioned (or splitted) into the size of smaller sequence. Then the linear convolution of each
section of longer sequence and the smaller sequence is performed. The output sequences obtained from
the convolutions of all the sections are combined to get the overall output sequence. There are two
methods of sectioned convolutions. They are overlap add method and overlap save method.
6.11.1 Overlap Add Method
In overlap add method the longer sequence is divided into smaller sequences. Then linear convolution of each section of longer sequence and smaller sequence is performed. The overall output sequence
is obtained by combining the output of the sectioned convolution.
Let,
N1 = Length of longer sequence
N2 = Length of smaller sequence
Let the longer sequence be divided into sections of size N3 samples.
Note : Normally the longer sequence is divided into sections of size same as that of smaller sequence.
N3 + N 2 – 1
N3
N2 – 1
N3 + N 2 – 1
N2 – 1
N3– (N2–1)
N2 – 1
N3 + N 2 – 1
Overlapped region
Note : Samples in the shaded region are added
N2 – 1 N3– (N2–1) N2 – 1
Overlapped region
Fig 6.30 : Overlapping of output sequence of sectioned convolution by overlap add method.
The linear convolution of each section with smaller sequence will produce an output sequence of
size N3 + N2 –1 samples. In this method the last N2 –1 samples of each output sequence overlaps with the
first N2 –1 samples of next section. (i.e., there will be a region of N2 –1 samples over which the output
Signals & Systems
6. 79
sequence of qth convolution overlaps the output sequence of (q +1)th convolution). While combining the
output sequences of the various sectioned convolutions, the corresponding samples of overlapped
regions are added and the samples of non-overlapped regions are retained as such.
6.11.2 Overlap Save Method
In overlap save method the results of linear convolution of the various sections are obtained using
circular convolution. In this method the longer sequence is divided into smaller sequences. Each section
of the longer sequence and the smaller sequence are converted to the size of the output sequence of
sectioned convolution. The circular convolution of each section of the longer sequence and the smaller
sequence is performed. The overall output sequence is obtained by combining the outputs of the
sectioned convolution.
Let,
N1 = Length of longer sequence
N2 = Length of smaller sequence
Let the longer sequence be divided into sections of size N3 samples.
Note : Normally the longer sequence is divided into sections of size same as that of smaller sequence.
In overlap save method the results of linear convolution are obtained by circular convolution.
Hence each section of longer sequence and the smaller sequence are converted to the size of output
sequence of size N3 + N2 – 1 samples.The smaller sequence is converted to size of N3 + N2 –1 samples, by
appending with zeros.The convertion of each section of longer sequence to the size N3 + N2 –1 samples
can be performed in two different methods.
Method-1
In this method the first N2 –1 samples of a section is appended as last N2 –1 samples of the previous
section (i.e., the overlapping samples are placed at the beginning of the section). The circular convolution
of each section will produce an output sequence of size N3 + N2 –1 samples. In this output the first N2 –1
samples are discarded and the remaining samples of the output of sectioned convolutions are saved as the
overall output sequence.
N3 + N2 – 1
N3
N2 – 1
N2 – 1
N2 – 1
N3 + N 2 – 1
N3–(N2–1)
N2 – 1
N2 – 1
N2 – 1
N3 + N2 – 1
N3–(N2–1)
Fig 6.31 : Appending of sections of input sequence in
method-1 of overlap save method.
N2 – 1
Appended
with zero
Chapter 6 - Discrete Time Signals and Systems
6. 80
N3 + N 2 – 1
N2 – 1
N3–(N2–1)
N2–1
N3 + N 2 – 1
N2 – 1
N3–(N2–1)
N2–1
N3 + N 2 – 1
N2 – 1
Overlapped region
N3
Note : Samples in the shaded region are discarded
Overlapped region
Fig 6.32 : Overlapping of output sequence of sectioned convolution by
method-1 of overlap save method.
Method-2
In this method the last N2–1 samples of a section is appended as last N2 –1 samples of the next
section (i.e, the overlapping samples are placed at the end of the sections). The circular convolution of
each section will produce an output sequence of size N3 + N2 –1 samples. In this output the last N2 –1
samples are discarded and the remaining samples of the output of sectioned convolutions are saved as the
overall output sequence.
N 3 + N2 – 1
N3–(N2–1) N2 – 1
N2 – 1
N2 – 1
Appended
with zero
N 3 + N2 – 1
N3–(N2–1) N2 – 1
N2 – 1
N2 – 1
N3 + N2 – 1
N3
N2 – 1
Fig 6.33 : Appending of sections of input sequence in
method-2 of overlap save method.
N3 + N 2 – 1
N3
N2 – 1
N3 + N 2 – 1
N2 – 1
N3–(N2–1)
N2 – 1
N3 + N2 – 1
Overlapped
region
Note : Samples in the shaded region are discarded
N2 – 1
N3–(N2–1)
N2 – 1
Overlapped
region
Fig 6.34 : Overlapping of output sequence of sectioned convolution by
method-2 of overlap save method.
Signals & Systems
6. 81
Example 6.28
Perform the linear convolution of the following sequences by a) Overlap add method and b) Overlap save method.
x(n) = {1, –1, 2, –2, 3, –3, 4, –4} ; h(n) = {–1, 1}
Solution
a) Overlap Add Method
In this method the longer sequence is sectioned into sequences of size equal to smaller sequence. Here x(n) is a
longer sequence when compared to h(n). Hence x(n) is sectioned into sequences of size equal to h(n).
Given that, x(n) = {1, –1, 2, –2, 3, –3, 4, –4}
Let x(n) can be sectioned into four sequences, each consisting of two samples of x(n) as shown below.
x1(n) = 1 ; n = 0
x2(n) = 2 ; n = 2
x3(n) = 3
;
n=4
= –1 ; n = 1
= –2 ; n = 3
= –3
;
n=5
x4(n) =
4
; n=6
= –4 ; n = 7
Let y1(n), y2(n), y3(n) and y4(n) be the output of linear convolution of x1(n), x2(n), x3(n) and x4(n) with h(n)
respectively.
Here h(n) starts at n = nh = 0
x1(n) starts at n = n1 = 0,
\ y1(n) will start at n = n1 + nh = 0 + 0 = 0
x2(n) starts at n = n2 = 2,
\ y2(n) will start at n = n2 + nh = 2 + 0 = 2
x3(n) starts at n = n3 = 4,
\ y3(n) will start at n = n3 + nh = 4 + 0 = 4
x4(n) starts at n = n4 = 6,
\ y4(n) will start at n = n4 + nh = 6 + 0 = 6
Here linear convolution of each section is performed between two sequences each consisting of 2 samples.
Hence each convolution output will consists of 2 + 2 – 1 = 3 samples. The convolution of each section is performed by tabular
method as shown below.
Note :
1. Here N1 = 8, N2 = 2, N3 = 2.
\ (N2 – 1) = 2 – 1 = 1
and
(N2 + N3 – 1) = 2 + 2 – 1 = 3
2. The unfilled boxes in the tables are considered as zero.
3. For convenience of convolution operation the index n is replaced by m in x1(n), x2(n), x3(n), x4(n) and h(n).
Convolution of Section - 1
+¥
m
–1
x1(m)
h(m)
h(–m) = ho(m)
h(1 – m) = h1(m)
h(2 – m) = h2(m)
0
1
1
–1
–1
1
2
x1(m) h(n - m)
+¥
å
x1(m) hn (m) ; n = 0, 1, 2
m = -¥
1
where hn (m) = h(n - m)
When n = 0 ; y1(0) =
–1
1
å
m = -¥
=
–1
1
y1(n) = x1(n) * h(n) =
When n = 1 ; y1(1) =
–1
When n = 2 ; y1(2) =
å x (m ) h (m)
å x (m) h (m)
å x (m) h (m)
1
0
= 0 - 1 + 0 = -1
1
1
= 1 + 1
1
2
= 0 - 1 + 0 = -1
= 2
Chapter 6 - Discrete Time Signals and Systems
6. 82
Convolution of Section - 2
m
+¥
–1
0
1
x2(m)
h(m)
–1
h(–m)
1
2
3
2
–2
y 2 (n) = x 2 (n) * h(n) =
4
2
m = -¥
+¥
å
=
1
x 2 (m) hn (m) ; n = 2, 3, 4
m = -¥
where hn (m) = h(n – m)
–1
h(2 – m) = h2(m)
å x (m) h(n - m)
1
When n = 2 ; y 2 (2) =
–1
h(3 – m) = h3(m)
1
When n = 3 ; y 2 (3) =
–1
h(4 – m) = h4(m)
1
–1
When n = 4 ; y 2 ( 4) =
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
2
2
= 0 - 2 + 0 = -2
2
3
= 2 + 2
2
4
= 0 - 2 + 0 = -2
= 4
Convolution of Section - 3
m
–1
0
1
2
3
x3(m)
h(m)
–1
h(–m)
1
4
5
3
–3
6
1
–1
h(4 – m) = h4(m)
1
–1
h(5 – m) = h5(m)
1
–1
h(6 – m) = h6(m)
1
+¥
+¥
å
y 3 (n) = x 3 (n) * h(n) =
–1
x 3 (m) h(n - m) =
m = -¥
å
x3 (m) hn (m) ; n = 4, 5, 6
m = -¥
where hn (m) = h(n – m)
When n = 4 ; y 3 ( 4) =
When n = 5 ; y 3 (5) =
When n = 6 ; y 3 (6) =
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
3
4
= 0 - 3 + 0 = -3
3
5
= 3 + 3
3
6
= 0 - 3 + 0 = -3
= 6
Convolution of Section - 4
m
–1
0
1
2
3
4
5
x4(m)
h(m)
–1
h(–m)
1
6
7
4
–4
8
1
–1
h(6 – m) = h6(m)
1
h(7 – m) = h7(m)
–1
1
h(8 – m) = h8(m)
–1
1
+¥
y 4 (n) = x 4 (n) * h(n) =
å
–1
+¥
x 4 (m) h(n - m) =
m = -¥
å
x 4 (m) hn (m) ; n = 6, 7, 8
m = -¥
where hn (m) = h(n - m)
When n = 6 ; y 4 (6) =
When n = 7 ; y 4 (7) =
When n = 8 ; y 4 (8) =
å
åx
åx
x 4 (m) h6 (m) = 0 - 4 + 0 = -4
4 (m)
h7 (m) = 4 + 4
= 8
4 (m)
h8 (m) = 0 - 4 + 0 = -4
To Combine the Output of Convolution of Each Section
It can be observed that the last sample in an output sequence overlaps with the first sample of next output sequence.
In this method the overall output is obtained by combining the outputs of the convolution of all sections. The overlapped
portions (or samples) are added while combining the output .
Signals & Systems
6. 83
The output of all sections can be represented in a table as shown below. Then the samples corresponding to same
value of n are added to get the overall output.
n
0
1
2
y1(n)
–1
2
–1
y2(n)
–2
3
4
4
–2
y3(n)
–3
5
6
y4(n)
y5(n)
–1
2
–3
4
–5
6
6
7
8
–4
8
–4
–7
8
–4
–3
\ y(n) = x(n) * h(n) = {–1, 2, –3, 4, –5, 6, –7, 8, –4}
b) Overlap Save Method
In this method the longer sequence is sectioned into sequences of size equal to smaller sequence. The number of
samples that will be obtained in the output of linear convolution of each section is determined. Then each section of longer
sequence is converted to the size of output sequence using the samples of original longer sequence.
The smaller sequence is also converted to the size of output sequence by appending with zeros. Then the circular
convolution of each section is performed.
Here x(n) is a longer sequence when compared to h(n). Hence x(n) is sectioned into sequences of size equal to h(n).
Given that, x(n) = {1, –1, 2, –2, 3, –3, 4, –4}
Let x(n) be sectioned into four sequences each consisting of two samples of x(n) as shown below.
x1(n) = 1 ; n = 0
= –1 ; n = 1
x2(n) = 2 ; n = 2
= –2 ; n = 3
x3(n) = 3 ; n = 4
= –3 ; n = 5
x4(n) = 4 ; n = 6
= –4 ; n = 7
Let y1(n), y2(n), y3(n) and y4(n) be the output of linear convolution of x1(n), x2(n), x3(n) and x4(n) with h(n) respectively.
Here linear convolution of each section will result in an output sequence consisting of 2 + 2 – 1 = 3 samples.
The sequence h(n) is converted to 3-sample sequence by appending with zero.
\ h(n) = {–1, 1, 0}
Method - 1
In method - 1 the overlapping samples are placed at the beginning of the sections. Each section of longer sequence
is converted to 3-sample sequence using the samples of original longer sequence as shown below. It can be observed that
the first sample of x2(n) is placed as overlapping sample at the end of x1(n). The first sample of x3(n) is placed as overlapping
sample at the end of x2(n). The first sample of x4(n) is placed as overlapping sample at the end of x3(n). Since there is no
fifth section, the overlapping sample of x4(n) is taken as zero.
x1(n) = 1 ; n = 0
x2(n) = 2 ; n = 2
x3(n) = 3 ; n = 4
x4(n) = 4 ; n = 6
= –1 ; n = 1
= –2 ; n = 3
= –3 ; n = 5
= –4 ; n = 7
= 2; n=2
= 3 ; n=4
= 4 ; n=6
= 0 ;n=8
Now perform circular convolution of each section with h(n). The output sequence obtained from circular convolution
will have three samples. The circular convolution of each section is performed by tabular method as shown below.
Here h(n) starts at n = nh = 0
x1(n) starts at n = n1 = 0,
\ y1(n) will start at n = n1 + nh = 0 + 0 = 0
x2(n) starts at n = n2 = 2,
\ y2(n) will start at n = n2 + nh = 2 + 0 = 2
x3(n) starts at n = n3 = 4,
\ y3(n) will start at n = n3 + nh = 4 + 0 = 4
x4(n) starts at n = n4 = 6,
\ y4(n) will start at n = n4 + nh = 6 + 0 = 6
Note : 1. Here N1 = 8, N2 = 2, N3 = 2.
\ (N2 – 1) = 2 – 1 = 1
and
(N2 + N3 – 1) = 2 + 2 – 1 = 3
2. The bold faced numbers in the tables are obtained by periodic extension.
3. For convenience of convolution operation, the index n in x1(n), x2 (n), x3(n), x4 (n) and h(n)
are replaced by m.
Chapter 6 - Discrete Time Signals and Systems
6. 84
Convolution of Section - 1
m
mf
–2
0
1
2
x1(m)
1
–1
2
h(m)
–1
1
0
1
–1
0
1
0
1
–1
0
When n = 0 ; y1(0) =
0
1
–1
When n = 1 ; y1(1) =
h(–m) = h0(m)
–1
0
h(1 – m) = h1(m)
h(2 – m) = h2(m)
2
1
n
–1
0
h(m)
where hn (m) = h(n - m)
1
–1
0
1
n = 0, 1, 2,
m =0
x2(m)
h(–m)
å x (m) h (m) ;
=
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
When n = 2 ; y1(2) =
–2
x1(m) h((n - m))N
m = mi
Convolution of Section - 2
m
å
y1(n) = x1(n) * h(n) =
2
3
4
2
–2
3
1
0
1
–1
0
1
0
1
–1
0
0
1
–1
1
0
= -1 + 0 + 2 = 1
1
1
=
1 + 1 + 0 = 2
1
2
=
0 - 1 - 2 = -3
–1
h(2 – m) = h2(m)
0
h(3 – m) = h3(m)
h(4 – m) = h4(m)
mf
y 2 (n) = x 2 (n) * h(n) =
å
x 2 (m) h((n - m))N
m = mi
4
=
å x (m) h (m) ;
2
n = 2, 3, 4
n
m = 2
where hn (m) = h((n - m))N
When n = 2 ; y 2 (2) =
When n = 3 ; y 2 (3) =
When n = 4 ; y 2 (4) =
å x (m) h (m) = - 2 + 0 + 3 = 1
å x (m) h (m) = 2 + 2 + 0 = 4
å x (m) h (m) = 0 + -2 - 3 = - 5
2
2
2
3
2
4
Convolution of Section - 3
m
–2
–1
0
1
2
–1
1
0
3
x3(m)
h(m)
h(–m)
0
1
5
6
3
–3
4
–1
h(4 – m) = h4(m)
0
h(5 – m) = h5(m)
1
–1
0
1
0
1
–1
0
0
1
–1
h(6 – m) = h6(m)
mf
y 3 (n) = x 3 (n) * h(n) =
4
å
6
x 3 (m) h((n - m))N =
m = mi
å x (m) h (m) ;
3
n
n = 4, 5, 6
m =4
where hn (m) = h((n - m))N
When n = 4 ; y 3 (4) =
When n = 5 ; y 3 (5) =
When n = 6 ; y 3 (6) =
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
3
4
= -3 + 0 + 4 = 1
3
5
= 3 + 3 + 0 =
3
6
= 0 - 3 - 4 = -7
6
Signals & Systems
6. 85
Convolution of section - 4
m
–2
–1
0
1
2
3
4
5
6
7
8
4
–4
0
1
–1
0
1
0
1
–1
0
0
1
–1
x4(m)
h(m)
–1
h(–m)
0
1
1
0
–1
h(6 – m) = h6(m)
0
h(7 – m) = h7(m)
h(8 – m) = h8(m)
mf
y 4 (n) = x 4 (n) * h(n) =
å
8
å
x 4 (m) h((n - m))N =
m = mi
x 4 (m) hn (m) ; n = 6, 7, 8
m =6
where hn (m) = h((n – m))N
When n = 6 ; y 4 (6) =
When n = 7 ; y 4 (7) =
When n = 8 ; y 4 (8) =
å
åx
åx
x 4 (m) h6 (m) = -4 + 0 + 0 = -4
4 (m)
h7 (m) = 4 + 4 + 0 = 8
4 (m)
h8 (m) = 0 - 4 + 0 = -4
To Combine the Output of the Convolution of Each Section
It can be observed that the last sample in an output sequence overlaps with the first sample of next output sequence.
In overlap save method the overall output is obtained by combining the outputs of the convolution of all sections. While
combining the outputs, the overlapped first sample of every output sequence is discarded and the remaining samples are
simply saved as samples of y(n) as shown in the following table.
n
0
1
2
y1 (n)
1
2
–3
y2 (n)
1
3
4
4
–5
y3 (n)
1
5
6
6
–7
y4 (n)
y(n)
*
2
–3
4
–5
6
7
8
–4
8
–4
–7
8
–4
y(n) = x(n) * h(n) = {*, 2, –3, –4, –5, 6, –7, 8, –4}
Note : Here y(n) is linear convolution of x(n) and h(n). It can be observed that the results of both the methods are
same, except the first N2 – 1 samples.
Method - 2
In method - 2 the overlapping samples are placed at the end of the section.Each section of longer sequence is
converted to 3-sample sequence, using the samples of original longer sequence as shown below. It can be observed that
the last sample of x1(n) is placed as overlapping sample at the end of x2(n). The last sample of x2(n) is placed as overlapping
sample at the end of x3(n). The last sample of x3(n) is placed as overlapping sample at the end of x4(n). Since there is no
previous section for x1(n), the overlapping sample of x1(n) is taken as zero.
x1(n) = 1 ; n = 0
x2(n) = 2 ; n = 2
x3(n) = 3 ;
n=4
x4(n) = 4 ; n = 6
n=5
= –4 ; n = 7
= –1 ; n = 1
= –2 ;
n=3
= –3 ;
= 0; n=2
= –1 ;
n=4
= –2 ;
n=6
= –3 ; n = 8
Now perform circular convolution of each section with h(n). The output sequence obtained from circular convolution
will have three samples. The circular convolution of each section is performed by tabular method as shown below.
Chapter 6 - Discrete Time Signals and Systems
6. 86
Here h(n) starts at n = nh = 0
Note :
x1(n) starts at n = n1 = 0,
\ y1(n) will start at n = n1 + nh = 0 + 0 = 0
x2(n) starts at n = n2 = 2,
\ y2(n) will start at n = n2 + nh = 2 + 0 = 2
x3(n) starts at n = n3 = 4,
\ y3(n) will start at n = n3 + nh = 4 + 0 = 4
x4(n) starts at n = n4 = 6,
\ y4(n) will start at n = n4 + nh = 6 + 0 = 6
\ (N2 – 1) = 2 – 1 = 1
1. Here N1 = 8, N2 = 2, N3 = 2.
and
(N2 + N3 – 1) = 2 + 2 – 1 = 3
2. The bold faced numbers in the tables are obtained by periodic extension.
3. For convenience of convolution the index n is replaced by m in x1(n), x2(n), x3(n), x4(n) and h(n).
mf
Convolution of Section - 1
m
y1(n) = x1(n) * h(n) =
–2
0
1
2
x1(m)
1
–1
0
h(m)
–1
1
0
1
–1
0
1
0
1
–1
0
h(–m) = h0(m)
–1
0
h(1 – m) = h1(m)
h(2 – m) = h2(m)
0
1
–1
m
–2
–1
0
1
x2(m)
h(m)
–1
h(–m)
0
1
2
=
å x (m) h (m) ;
1
0
h(3 – m) = h3(m)
where hn (m) = h((n - m))N
å x (m) h (m) = -1 + 0 + 0 = – 1
= å x (m) h (m) = 1 + 1 + 0 = 2
= å x (m) h (m) = 0 - 1 + 0 = -1
When n = 0 ; y1(0) =
1
0
When n = 1 ; y1(1)
1
1
2
3
4
2
–2
–1
1
0
1
–1
0
1
0
1
–1
0
0
1
–1
h(4 – m) = h4(m)
mf
å
n = 0, 1, 2
n
m = 0
1
2
–1
h(2 – m) = h2(m)
y 2 (n) = x 2 (n) * h(n) =
4
x 2 (m) h((n - m))N =
m = mi
å
x 2 (m) hn (m); n = 2, 3, 4,
m =2
where hn (m) = h((n - m))N
When n = 2 ; y 2 (2) =
When n = 3 ; y 2 (3) =
When n = 4 ; y 2 (4) =
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
2
2
= -2 + 0 - 1 = – 3
2
3
= 2 + 2 + 0 = 4
2
4
= 0 - 2 + 1 = -1
Convolution of Section - 3
m
–2
–1
0
1
2
–1
1
0
3
4
5
6
3
–3
–2
1
–1
0
1
0
1
–1
0
0
1
–1
x3(m)
h(m)
h(–m)
h(4 – m) = h4(m)
h(5 – m) = h5(m)
h(6 – m) = h6(m)
0
1
x1(m) h((n - m))N
m = mi
When n = 2 ; y1(2)
Convolution of Section - 2
å
–1
0
Signals & Systems
6. 87
mf
å
y 3 (n) = x 3 (n) * h(n) =
6
å
x 3 (m) h((n - m))N =
m = mi
x 3 (m) hn (m) ;
n = 4, 5, 6
m = 4
where hn (m) = h((n - m))N
When n = 4 ; y 3 (4) =
When n = 5 ; y 3 (5) =
When n = 6 ; y 3 (6) =
Convolution of Section - 4
m
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
–2
3
4
= -3 + 0 - 2 = –5
3
5
= 3 + 3 + 0 = 6
3
6
= 0 - 3 + 2 = -1
–1
0
1
2
–1
1
0
3
4
5
6
7
8
4
–4
–3
1
–1
0
1
0
1
–1
0
0
1
–1
x4(m)
h(m)
h(–m)
0
1
–1
h(6 – m) = h6(m)
0
h(7 – m) = h7(m)
h(8 – m) = h8(m)
mf
y 4 (n) = x 4 (n) * h(n) =
å
8
å
x 4 (m) h((n - m))N =
m = mi
x 4 (m) hn (m) ; n = 6, 7, 8
m = 6
where hn (m) = h((n – m))N
When n = 6 ; y 4 (6) =
When n = 7 ; y 4 (7) =
When n = 8 ; y 4 (8) =
å
å x (m) h (m)
å x (m) h (m)
x 4 (m) h6 (m) = -4 + 0 - 3 = -7
4
7
= 4 + 4 + 0 = 8
4
8
= 0 - 4 + 3 = -1
To Combine the Output of the Convolution of Each Section
It can be observed that the last sample in an output sequence overlaps with the first sample of next output sequence.
In overlap save method the overall output is obtained by combining the outputs of the convolution of all sections. While
combining the outputs the overlapped last sample of every output sequence is discarded and the remaining samples are
simply saved as samples of y(n) as shown in the following table.
n
0
1
2
y1(n)
–1
2
–1
y2(n)
–3
3
4
4
–1
y3(n)
–5
5
6
y4(n)
y(n)
–1
2
–3
4
–5
6
6
7
8
–1
–7
8
–1
–7
8
*
Note :
Here y(n) is linear convolution of
x(n) and h(n). It can be observed
that the results of both the
methods are same except the
last N2–1 samples.
\ y(n) = x(n) * h(n) = {–1, 2, –3, 4, –5, 6, –7, 8, *}
Example 6.29
Perform the linear convolution of the following sequences by a) Overlap add method and b) Overlap save method.
x(n) = {1, 2, 3, –1, –2, –3, 4, 5, 6} and h(n) = {2, 1, –1}
Solution
a) Overlap Add Method
In this method the longer sequence is sectioned into sequences of size equal to smaller sequence. Here x(n) is a
longer sequence when compared to h(n). Hence x(n) is sectioned into sequences of size equal to h(n).
Given that x(n) = {1, 2, 3, –1, –2, –3, 4, 5, 6}. Let x(n) can be sectioned into three sequences, each consisting of three
samples of x(n) as shown below.
Chapter 6 - Discrete Time Signals and Systems
6. 88
x1(n) = 1 ; n = 0
x2(n) = –1 ; n = 3
x3 (n)= 4 ; n = 6
=2;n=1
= –2 ; n = 4
=5;n=7
=3;n=2
= –3 ; n = 5
=6;n=8
Let y1(n), y2(n) and y3(n) be the output of linear convolution of x1(n), x2(n) and x3(n) with h(n) respectively.
Here h(n) starts at n = nh = 0
x1(n) starts at n = n1 = 0, \ y1(n) will start at n = n1 + nh = 0 + 0 = 0
x2(n) starts at n = n2 = 3, \ y2(n) will start at n = n3 + nh = 3 + 0 = 3
x3(n) starts at n = n3 = 6, \ y3(n) will start at n = n6 + nh = 6 + 0 = 6
Here linear convolution of each section is performed between two sequences each consisting of three samples.
Hence each convolution output will consists of 3 + 3 – 1 = 5 samples. The convolution of each section is performed by tabular
method as shown below.
Note : 1. Here N1 = 9, N2 = 3, N3 = 3,
\ (N2 – 1) = 3 – 1 = 2
and
(N2 + N3 – 1) = 3 + 3 – 1 = 5.
2. The unfilled boxes in the table are considered as zero.
3. For convenience of convolution operation, the index n is replaced by m in x1(n), x2(n), x3(n) and h(n).
Convolution of Section -1
m
0
1
2
x1(m)
1
2
3
h(m)
2
1
–1
h(–m) = h0(m)
–2
–1
–1
h(1 – m) = h1(m)
3
+¥
y1(n) = x1(n) * h(n) =
å x (m) h(n - m)
1
m = -¥
+¥
1
2
–1
1
2
–1
1
2
–1
1
2
–1
1
h(2 – m) = h2(m)
4
=
h(3 – m) = h3(m)
h(4 – m) = h4(m)
å x (m) h (m)
1
for n = 0, 1, 2, 3, 4
where hn (m) = h(n - m)
2
When n = 0 ; y1(0) = å x1(m) ho(m) = 0 + 0 + 2 + 0 + 0 = 2
When n = 1 ; y1(1) = å x1(m) h1(m) = 0 + 1 + 4 + 0
= 5
When n = 2 ; y1(2) = å x1(m) h2(m) = –1 + 2 + 6
= 7
When n = 3 ; y1(3) = å x1(m) h3(m) = 0 – 2 + 3 + 0
= 1
When n = 4 ; y1(4) = å x1(m) h4(m) = 0 + 0 – 3 + 0 + 0 = –3
Convolution of Section - 2
m
–2
–1
0
1
2
x2(m)
h(m)
h(–m) = h0(m)
h(3 – m) = h3(m)
h(4 – m) = h4(m)
h(5 – m) = h5(m)
h(6 – m) = h6(m)
h(7 – m) = h7(m)
2
–1
1
n
m = -¥
1
3
4
5
–1
–2
–3
6
7
–1
2
–1
1
2
–1
1
2
–1
1
–1
2
1
2
–1
1
2
Signals & Systems
6. 89
¥
¥
å
y 2 (n) = x 2 (n) * h(n) =
å
x 2 (m) h(n - m) =
m = -¥
x 2 (m) hn (m); n = 3, 4, 5, 6, 7
m = -¥
where hn (m) = h(n - m)
When n = 3 ; y 2 (3) =
When n = 4 ; y 2 (4) =
When n = 5 ; y 2 (5) =
When n = 6 ; y 2 (6) =
When n = 7 ; y 2 (7) =
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
2
3
= 0 + 0 - 2 + 0 + 0 = –2
2
4
= 0 - 1 - 4 + 0
= -5
2
5
= 1 - 2 - 6
= -7
2
6
= 0 + 2 - 3 + 0
= -1
2
7
= 0 + 0 + 3 + 0 + 0 = 3
Convolution of Section - 3
m
–2
–1
0
1
2
3
4
5
x3(m)
h(m)
2
h(–m) = h0(m)
–1
1
1
6
7
8
4
5
6
10
–1
2
h(6 – m) = h6(m)
–1
h(7 – m) = h7(m)
1
2
–1
1
2
–1
1
h(8 – m) = h8(m)
h(9 – m) = h9(m)
2
–1
h(10 – m) = h10(m)
¥
y 3 (n) = x 3 (n) * h(n)
9
å
=
1
2
–1
1
2
¥
x 3 (m) h(n - m) =
m = –¥
å
x 3 (m) hn (m) ; n = 6, 7, 8, 9, 10
m = –¥
where hn (m) = h(n - m)
When n = 6 ; y 3 (6)
=
When n = 7 ; y 3 (7)
=
When n = 8 ; y 3 (8)
=
When n = 9 ; y 3 (9)
=
When n = 10 ; y 3 (10) =
å
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
x 3 (m) h6 (m) = 0 + 0 + 8 + 0 + 0 =
8
3
7
= 0 + 4 + 10 + 0
=
14
3
8
= – 4 + 5 + 12
= 13
3
9
= 0 – 5 + 6 + 0
= 1
3
10
= 0 + 0 – 6 + 0 + 0 = –6
To Combine the Output of the Convolution of Each Section
It can be observed that the last N2 – 1 sample in an output sequence overlaps with the first N2 – 1 sample of next
output sequence. In this method the overall output is obtained by combining the outputs of the convolution of all sections. The
overlapped portions (or samples) are added while combining the output.
The output of all sections can be represented in a table as shown below. Then the samples corresponding to same
value of n are added to get the overall output.
n
y1(n)
0
2
1
5
2
7
y2(n)
y3(n)
y(n)
2
5
7
3
1
4
–3
5
6
7
8
9
10
–2
–5
–7
–1
8
3
14
13
1
–6
–1
–8
–7
7
17
13
1
–6
\ y(n) = x(n) * h(n) = {2, 5, 7, –1, –8, –7, 7, 17, 13, 1, – 6}
b) Overlap Save Method
In this method the longer sequence is sectioned into sequences of size equal to smaller sequence. The number of
samples that will be obtained in the output of linear convolution of each section is determined. Then each section of longer
sequence is converted to the size of output sequence using the samples of original longer sequences. The smaller
sequence is also converted to the size of output sequence by appending with zeros. Then the circular convolution of each
section is performed.
Chapter 6 - Discrete Time Signals and Systems
6. 90
Here x(n) is a longer sequence when compared to h(n). Hence x(n) is sectioned into sequences of size equal to h(n).
Given that x(n) = {1, 2, 3, –1, –2, –3, 4, 5, 6}.
Let x(n) be sectioned into three sequences each consisting of three samples as shown below.
Let,
N1 = Length of longer sequence
N2 = Length of smaller sequence
N3 = N2 = Length of each section of longer sequence.
x1(n) = 1 ; n = 0
x2(n) = –1 ; n = 3
=2;n=1
= –2 ; n = 4
=5;n=7
=3;n=2
= –3 ; n = 5
=6;n=8
x3 (n) = 4 ; n = 6
Let y1(n), y2(n) and y3(n) be the output of linear convolution of x1(n), x2(n) and x3(n) with h(n) respectively. Here linear
convolution of each section will result in an output sequence consisting of 3 + 3 – 1 = 5 samples.
Hence each section of longer sequence is converted to five sample sequence, using the samples of original longer
sequence as shown below. It can be observed that the first N2 – 1 samples of x2(n) is placed as overlapping sample at the
end of x1(n). The first N2 – 1 samples of x3(n) is placed as overlapping sample at the end of x2(n). Since there is no fourth
section, the overlapping samples of x 3(n) are considered as zeros.
x1(n) = 1 ; n = 0
x2(n) = –1 ; n = 3
x3 (n) = 4 ; n = 6
= 2;n=1
= –2 ; n = 4
=5;n=7
= 3;n=2
= –3 ; n = 5
=6;n=8
= –1 ; n = 3
= 4; n=6
=0;n=9
= –2 ; n = 4
= 5; n=7
= 0 ; n = 10
The sequence h(n) is also converted to five sample sequence by appending with zeros.
\ h(n) = {2, 1, –1, 0, 0}
Now perform circular convolution of each section with h(n). The output sequence obtained from circular convolution
will have five samples. The circular convolution of each section is performed by tabular method as shown below.
Here h(n) starts at n = nh = 0
x1(n) starts at n = n1 = 0, \ y1(n) will start at n = n1 + nh = 0 + 0 = 0
x2(n) starts at n = n2 = 3, \ y2(n) will start at n = n2 + nh = 3 + 0 = 3
x3(n) starts at n = n3 = 6, \ y3(n) will start at n = n3 + nh = 6 + 0 = 6
Note : 1. Here N1 = 9, N2 = 3, N3 = 3
\ (N2 – 1) = 3 – 1 = 2
and
[N2 + N3 – 1] = 3 + 3 – 1 = 5 samples.
2. The bold faced numbers in the table are obtained by periodic extension.
3. For convenience of convolution operation the index n is replaced by m in x1(n), x2(n), x3(n) and h(n).
Convolution of Section - 1
0
1
2
3
4
x1(m)
m
1
2
3
–1
–2
h(m)
2
1
–1
0
0
2
0
0
–1
1
h(–m) = h0(m)
h(1 – m) = h1(m)
h(2 – m) = h2(m)
h(3 – m) = h3(m)
h(4 – m) = h4(m)
–4
0
–3
–2
–1
0
–1
1
0
0
–1
1
2
0
0
–1
0
0
–1
1
2
0
0
0
0
–1
1
2
0
0
0
–1
1
2
Signals & Systems
6. 91
mf
4
å x (m) h((n - m))
y1(n) = x1(n) * h(n) =
1
=
N
m = mi
å
x1(m) hn (m); n = 0, 1, 2, 3, 4
m =0
where h n (m) = h((n - m))N
å x (m)h (m) = 2 + 0 + 0 + 1 – 2 = 1
When n = 1 ; y (0) = å x (m)h (m) = 1 + 4 + 0 + 0 + 2 = 7
When n = 2 ; y (2) = å x (m)h (m) = –1 + 2 + 6 + 0 + 0 = 7
When n = 3 ; y (3) = å x (m)h (m) = 0 – 2 + 3 – 2 + 0 = –1
When n = 4 ; y (4) = å x (m)h (m) = 0 + 0 – 3 – 1 – 4 = – 8
When n = 0 ; y1(0) =
1
o
1
1
1
1
1
2
1
1
3
1
1
4
Convolution of Section - 2
m
–4
–3
–2
–1
0
1
2
x2(m)
h(m)
2
h(–m) = h0(m)
0
0
–1
h(3 – m) = h3(m)
5
6
7
–1
–2
–3
4
5
–1
0
0
0
–1
1
2
0
0
–1
1
0
0
–1
1
2
0
0
–1
0
0
–1
1
2
0
0
0
0
–1
1
2
0
0
0
–1
1
2
2
0
h(5 – m) = h5(m)
h(6 – m) = h6(m)
h(7 – m) = h7(m)
mf
å x (m) h((n - m))
y 2 (n) = x 2 (n) * h(n) =
4
1
1
h(4 – m) = h4(m)
3
2
7
N
=
m = mi
å x (m) h (m); n
2
n
= 3, 4, 5, 6, 7
m =3
where hn (m) = h((n - m))N
When n = 3 ; y 2 (3) =
When n = 4 ; y 2 (4) =
When n = 5 ; y 2 (5) =
When n = 6 ; y 2 (6) =
When n = 7 ; y 2 (7) =
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
å x (m) h (m)
2
3
= -2 + 0 + 0 – 4 + 5 = –1
2
4
= -1 - 4 + 0 + 0 – 5 = - 10
2
5
= 1 - 2 - 6 + 0 + 0 = -7
2
6
= 0 + 2 - 3 + 8 + 0 = 7
2
7
= 0 + 0 + 3 + 4 + 10 = 17
Convolution of Section - 3
m
–4 –3 –2
–1
0
1
2
3
4
5
x3(m)
h(m)
h(–m) = h0(m)
h(6 – m) = h6(m)
h(7 – m) = h7(m)
h(8 – m) = h8(m)
h(9 – m) = h9(m)
h(10 – m) = h10(m)
2
0
0
–1
1
1
–1
6
7
8
9
10
4
5
6
0
0
0
0
0
–1
1
2
0
0
–1
1
0
0
–1
1
2
0
0
–1
0
0
–1
1
2
0
0
0
–1
1
2
0
0
0
–1
1
2
2
0
0
Chapter 6 - Discrete Time Signals and Systems
6. 92
mf
y 3 (n) = x 3 (n) * h(n) =
10
å
å x (m) h (m) ;
x 3 (m) h((n - m))N =
3
m = mi
n
n = 6, 7, 8, 9, 10
m = 6
where hn (m) = h((n - m))N
When n = 6 ; y 3 (6)
=
When n = 7 ; y 3 (7)
=
When n = 8 ; y 3 (8)
=
When n = 9 ; y 3 (9)
=
When n = 10 ; y 3 (10) =
å x (m) h (m) = 8 + 0 + 0 + 0 + 0 = 8
å x (m) h (m) = 4 + 10 + 0 + 0 + 0 = 14
å x (m) h (m) = – 4 + 5 + 12 + 0 + 0 = 13
å x (m) h (m) = 0 – 5 + 6 + 0 + 0 = 1
å x (m) h (m) = 0 + 0 – 6 + 0 + 0 = –6
3
6
3
7
3
8
3
9
3
10
To Combine the Output of Convolution of Each Section
It can be observed that the last N2–1 samples in an output sequence overlaps with the first N2–1 samples of next
output sequence. In overlap save method the overall output is obtained by combining the outputs of the convolution of all
sections. While combining the outputs, the overlapped first N2–1 samples of every output sequence is discarded and the
remaining samples are simply saved as samples of y(n) as shown in the following table.
n
0
1
2
3
4
y1(n)
1
7
7
–1
–8
–1
10
y2(n)
5
6
–7
y3(n)
y(n)
*
*
7
–1
–8
–7
7
8
9
10
7
17
8
14
13
1
–6
7
17
13
1
–6
\ y(n) = x(n) * h(n) = {*, *, 7, –1, –8, –7, 7, 17, 13, 1, –6}
Note : Here y(n) is linear convolution of x(n) and h(n). It can be observed that the results of both the methods are
same except the first N2 – 1 samples.
6.12 Inverse System and Deconvolution
6.12.1 Inverse System
The inverse system is used to recover the input from the response of a system. For a given system,
the inverse system exists, if distinct inputs to a system leads to distinct outputs. The inverse systems
exists for all LTI systems.
The inverse system is denoted by H–1. If x(n) is input and y(n) is the output of a system, then y(n)
is the input and x(n) is the output of its inverse system.
x(n)
®H
y(n)
y(n)
®
®H
–1
w(n) = x(n)
®
Fig 6.35b : Inverse system.
Fig 6.35a : System.
Fig 6.35 : A system and its inverse system.
Let h(n) be the impulse response of a system and h¢(n) be the impulse response of inverse system.
Let us connect the system and its inverse in cascade as shown in fig 6.36.
Identity system
H
x(n)
® h(n)
H–1
® h¢(n)
y(n)
® w(n) = x(n)
Fig 6.36 : Cascade connection of a system and its inverse.
Signals & Systems
6. 93
Now it can be proved that,
h(n) * h¢(n) = d(n)
.....(6.60)
Therefore the cascade of a system and its inverse is identity system.
Proof :
With reference to fig 6.36 we can write,
y(n) = x(n) * h(n)
.....(6.61)
w(n) = y(n) * h¢(n)
.....(6.62)
On substituting for y(n) from equation (6.61) in equation (6.62) we get,
w(n) = x(n) * h(n) * h¢(n)
.....(6.63)
In equation (6.63),
if, h(n) * h¢(n) = d(n), then, x(n) * d(n) = x(n)
In a inverse system, w(n) = x(n), and so,
h(n) * h¢(n) = d(n). Hence proved.
6.12.2 Deconvolution
In an LTI system the response y(n) is given by convolution of input x(n) and impulse response h(n).
i.e., y(n) = x(n) * h(n)
The process of recovering the input from the response of a system is called deconvolution. (or the
process of recovering x(n) from x(n) * h(n) is called deconvolution).
When the response y(n) and impulse response h(n) are available the input x(n) can be computed
using the equation (6.64).
x( n) =
LM
N
1
y(n) h(0)
O
n-1
å x(m) h(n - m)PQ
.....(6.64)
m=0
Proof :
Let x(n) and h(n) be finite duration sequences starting from n = 0. Consider the matrix method of
convolution of x(n) and h(n) shown below.
h(0)
h(1)
h(2)
x(0)h(2)
x(1)h(2)
x(1)h(3)
......
x(2)h(3)
......
x(2)
x(2)h(0)
x(2)h(1)
x(2)h(2)
x(3)
x(3)h(0)
x(3)h(1)
x(3)h(2)
x(3)h(3)
......
......
x(1)h(1)
......
x(1)h(0)
......
......
x(1)
x(0)h(3)
......
x(0)h(1)
......
x(0)h(0)
......
......
x(0)
h(3)
Chapter 6 - Discrete Time Signals and Systems
6. 94
From the above two dimensional array we can write,
y(n) = x(0) h(0)
Þ
x(0) =
y(0)
h(0)
y(1) - x(0) h(1)
h(0)
y(2) - x(0) h(2) - x(1) h(1)
y( 2) = x(2) h(0) + x(1) h(1) + x(0) h(2) Þ x(2) =
h(0)
y(3) - x(0) h(3) - x(1) h(2) - x(2) h(1)
y( 3) = x(3) h(0) + x(2) h(1) + x(1) h(2) + x(0) h(3) Þ x(3) =
h(0)
and so on.
y( 1) = x(1) h(0) + x(0) h(1)
Þ
x(1) =
From the above analysis, in general for any value of n, the x(n) is given by,
x(n) =
\ x(n) =
y(n) - x(0) h(n) - x(1) h(n - 1) - ...... - x(n - 1) h(1)
h(0)
LM
MN
1
y(n) h(0)
O
Q
n- 1
å x(m) h(n - m)PP
m =0
Example 6.30
n
A discrete time system is defined by the equation, y(n) =
å x(m) ;
for n ³ 0. Find the inverse system.
m =0
Solution
n
Given that, y(n) =
å x(m)
m=0
0
When n = 0; y(0) =
å x(m)
= x(0)
m=0
1
When n = 1; y(1) =
å x(m)
= x(0) + x(1) = y(0) + x(1)
m=0
2
When n = 2; y(2) =
å x(m)
= x(0) + x(1) + x(2) = y(1) + x(2)
m=0
3
When n = 3; y(3) =
å x(m)
= x(0) + x(1) + x(2) + x(3) = y(2) + x(3)
m=0
and so on,
From the above analysis we can write,
x(0) = y(0)
;
x(1) = y(1) – y(0)
;
x(2) = y(2) – y(1)
;
x(3) = y(3) – y(2) and so on,
In general for any value of n, the signal x(n) can be written as,
x(n) = y(n) – y(n –1)
Therefore the inverse system is defined by the equation,
x(n) = y(n) – y(n –1)
Example 6.31
When a discrete time system is excited by an input x(n) the response is, y(n) = { 2, 5, 11, 17, 13, 12 }
­
If the impulse response of the system is, h(n) = { 2, 1, 3 }, then what will be the input to the system?
­
Signals & Systems
6. 95
Solution
Let N1 be number of samples in x(n) and N2 be number of samples in h(n), then the number of samples N3 in y(n) is
given by,
N3 = N1 + N2 – 1
\ N1 = N3 – N2 + 1 = 6 – 3 + 1 = 4 samples
Therefore x(n) is 4 sample sequence.
Each sample of x(n) is given by,
x(n) =
LM
MN
O
Q
n - 1
1
y(n) h(0)
å x(m) h(n - m)PP
m=0
When n = 0; x(0) =
y(0)
2
=
= 1
h(0)
2
When n = 1; x(1) =
1
y(1) h(0)
=
When n = 2; x(2 ) =
LM
MN
å x(m)
h(1 - m)
m=0
OP
PQ
1
1
y(1) - x(0) h(1) =
5 - 1 ´ 1 = 2
h(0)
2
LM
MN
1
y(2) h(0)
O
Q
1
å x(m) h(2 - m) PP
m=0
1
=
y(2) - x(0) h(2) - x(1) h(1)
h(0)
=
When n = 3; x(3 ) =
1
11 - 1 ´ 3 - 2 ´ 1
2
LM
MN
1
y(3) h(0)
=3
O
Q
2
å x(m) h(3 - m) PP
m=0
1
=
y(3) - x(0) h(3) - x(1) h(2) - x(2) h(1)
h(0)
1
17 - 1 ´ 0 - 2 ´ 3 - 3 ´ 1 = 4
2
x(n) = {x(0), x(1), x(2), x(3)} = {1, 2, 3, 4}
=
\
A
6.13 Correlation, Crosscorrelation and Autocorrelation
The correlation of two discrete time sequences x(n) and y(n) is defined as,
+¥
rxy (m) =
å
x(n) y(n - m)
.....(6.65)
n = -¥
where rxy(m) is the correlation sequence obtained by correlation of x(n) and y(n) and m is the variable
used for time shift. The correlation of two different sequences is called crosscorrelation and the correlation
of a sequence with itself is called autocorrelation. Hence autocorrelation of a discrete time sequence
is defined as,
+¥
rxx (m) =
å x(n) x(n - m)
n = -¥
.....(6.66)
Chapter 6 - Discrete Time Signals and Systems
6. 96
If the sequence x(n) has N1 samples and sequence y(n) has N2 samples then the crosscorrelation
sequence rxy(m) will be a finite duration sequence consisting of N1 + N2 –1 samples. If the sequence x(n)
has N samples, then the autocorrelation sequence rxx(m) will be a finite duration sequence consisting of
2N –1 samples.
In the equation (6.65) the sequence x(n) is unshifted and the sequence y(n) is shifted by m units of
time for correlation operation. The same results can be obtained if the sequence y(n) is unshifted and the
sequence x(n) is shifted opposite to that of earlier case by m units of time, hence the crosscorrelation
operation can also be expressed as,
+¥
rxy (m) =
å x(n + m) y(n)
....(6.67)
n = -¥
6.13.1 Procedure for Evaluating Correlation
Let, x(n) = Discrete time sequence with N1 samples
y(n) = Discrete time sequence with N2 samples
Now the correlation of x(n) and y(n) will produce a sequence rxy(m) consisting of N1+N2–1
samples. Each sample of rxy(m) can be computed using the equation (6.65). The value of rxy(m) at
m = q is obtained by replacing m by q, in equation (6.65).
+¥
\ rxy (q) =
å x(n) y(n - q)
.....(6.68)
n = -¥
The evaluation of equation (6.68) to determine the value of rxy(m) at m = q involves the following
three steps.
1. Shifting
: Shift y(n) by q times to the right if q is positive, shift y(n) by q times to the
left if q is negative to obtain y(n - q).
2. Multiplication
: Multiply x(n) by y(n - q) to get a product sequence. Let the product
sequence be vq(n). Now, vq(n) = x(n) ´ y(n - q).
3. Summation
: Sum all the values of the product sequence vq(n) to obtain the value of
rxy(m) at m = q. [i.e., rxy(q)].
The above procedure will give the value rxy(m) at a single time instant say m = q. In general we are
interested in evaluating the values of the sequence rxy(m) over all the time instants in the range -¥ < m < ¥.
Hence the steps 1, 2 and 3 given above must be repeated, for all possible time shifts in the
range -¥ < m < ¥.
In the correlation of finite duration sequences it is possible to predict the start and end of the
resultant sequence. If x(n) is N-point sequence and starts at n = n1 and if y(n) is N2-point sequence and
starts at n = n2 then, the initial value of m = mi for rxy(m) is mi = n1 – (n2 + N2 – 1). The value of x(n) for
n < n1 and the value of y(n) for n < n2 are then assumed to be zero.The final value of m = mf for rxy(m) is
mf = mi + (N1+N2– 2).
6. 97
Signals & Systems
The correlation operation involves all the steps in convolution operation except the folding.
Hence it can be proved that the convolution of x(n) and folded sequence y(-n) will generate the
crosscorrelation sequence rxy(m).
i.e., rxy(m) = x(n) * y(-n)
.....(6.69)
The procedure given above can be used for computing autocorrelation of x(n). For computing
autocorrelation using equation (6.68) replace y(n – q) by x(n – q). Similarly when equation (6.69) is used,
replace y(–n) by x(–n).
The autocorrelation of N-point sequence x(n) will give 2N –1 point autocorrelation sequence.
If x(n) starts at n = nx then initial value of m = mi for rxx(m) is mi = – (N –1). The final value of m = mf for
rxx(m) is mf = mi + (2N – 2).
Properties of Correlation
1. The crosscorrelation sequence rxy(m) is simply folded version of ryx(m),
i.e., rxy(m) = ryx(-m)
Similarly for autocorrelation sequence,
rxx(m) = rxx(-m)
Hence autocorrelation is an even function.
2. The crosscorrelation sequence satisfies the condition,
rxy (m) £
rxx (0) ryy (0) =
Ex Ey
where, Ex and Ey are energy of x(n) and y(n) respectively.
On applying the above condition to autocorrelation sequence we get,
rxx (m) £ rxx (0) = E x
From the above equations we infer that the crosscorrelation sequence and autocorrelation
sequences attain their respective maximum values at zero shift/lag.
3. Using the maximum value of crosscorrelation sequence, the normalized crosscorrelation sequence
is defined as,
rxy (m)
rxy (m) £
rxx (0) ryy (0)
Using the maximum value of autocorrelation sequence, the normalized autocorrelation sequence
is defined as,
r xx (m) £
rxx (m)
rxx (0)
Chapter 6 - Discrete Time Signals and Systems
6. 98
Methods of Computing Correlation
Method -1: Graphical Method
Let x(n) and y(n) be the input sequences and rxy(m) be the output sequence.
1. Sketch the graphical representation of the input sequences x(n) and y(n).
2. Shift the sequence y(n) to the left graphically so that the product of x(n) and
shifted y(n) gives only one non-zero sample. Now multiply x(n) and shifted y(n) to get
a product sequence, and then sum-up the samples of product sequence, which is the first
sample of output sequence.
3. To get the next sample of output sequence, shift y(n) of previous step to one position right
and multiply the shifted sequence with x(n) to get a product sequence. Now the sum of the
samples of product sequence gives the second sample of output sequence.
4. To get subsequent samples of output sequence, the step-3 is repeated until we get a
non-zero product sequence.
Method -2: Tabular Method
The tabular method is same as that of graphical method, except that the tabular representation of
the sequences are employed instead of graphical representation. In tabular method, every input sequence
and shifted sequence is represented on a row in a table.
Method -3: Matrix Method
Let x(n) and y(n) be the input sequences and rxy(m) be the output sequence. We know that the
convolution of x(n) and folded sequence y(-n) will generate the crosscorrelation sequence rxy(m). Hence
fold y(n) to get y(-n), and compute convolution of x(n) and y(-n) by matrix method.
In matrix method one of the sequence is represented as a row and the other as a column as shown
below.
y(0)
y(–1)
y(–2)
y(–3)
x(0)y(0)
x(0)y(–1)
x(0)y(–2)
x(0)y(–3)
x(1)
x(1)y(0)
x(1)y(–1)
x(1)y(–2)
x(1)y(–3)
x(2)
x(2)y(0)
x(2)y(–1)
x(2)y(–2)
x(2)y(–3)
x(3)
x(3)y(0)
x(3)y(–1)
x(3)y(–2)
x(3)y(–3)
x(0)
6. 99
Signals & Systems
Multiply each column element with row elements and fill up the matrix array.
Now the sum of the diagonal elements gives the samples of output sequence rxy(m). (The sum of
the diagonal elements are shown below for reference).
:
:
rxy(0) = ..... + x(0) y(0) + .....
rxy(1) = ..... + x(1) y(0) + x(0 ) y(-1) + .....
rxy(2) = ..... + x(2) y(0) + x(1) y(-1) + x(0) y(-2) + .....
rxy(3) = ..... + x(3) y(0) + x(2) y(-1) + x(1) y(-2) + x(0) y(-3) + .....
:
:
Example 6.32
Perform crosscorrelation of the sequences, x(n) = {1, 1, 2, 2} and y(n) = {1, 3, 1}.
Solution
Let rxy(m) be the crosscorrelation sequence obtained by crosscorrelation of x(n) and y(n).
The crosscorrelation sequence rxy(m) is given by,
+¥
rxy =
å x(n) y(n - m)
n = -¥
The x(n) starts at n = 0 and has 4 samples.
\ n1 = 0, N1 = 4
The y(n) starts at n = 0 and has 3 samples.
\ n2 = 0, N2 = 3
Now, rxy(m) will have N1 + N2 – 1 = 4 + 3 – 1 = 6 samples.
The initial value of m = mi = n1 – (n2 + N2 –1)
= 0 – (0 + 3 – 1) = – 2
The final value of m = mf = mi + (N1 + N2 – 2)
= –2 + (4 + 3 – 2) = 3
In this example the correlation operation is performed by three methods.
Method-1 : Graphical Method
The graphical representation of x(n) and y(n) are shown below.
x(n)
y(n)
3
2
2
1
0
1
1
1
1
2
3
n
0
1
2
Fig 2.
Fig 1.
The 6 samples of rxy(m) are computed using the equation,
+¥
rxy (m) =
å x(n) y(n - m)
n = -¥
+¥
=
å
n = -¥
x(n) ym (n) ; where ym (n) = y(n - m)
n
Chapter 6 - Discrete Time Signals and Systems
6. 100
The computation of each sample of rxy(n) using the above equation are graphically shown in fig 3 to fig 8.
The graphical representation of output sequence is shown in fig 9.
+¥
When m = - 2 ; rxy ( -2)
+¥
å x(n) y(n - (-2)) = å x(n) y
=
n = -¥
+¥
-2 (n)
åv
=
n = -¥
x(n)
- 2 (n)
n = -¥
v–2(n)
y–2 (n)
3
2
2
´
1
1
0
2
1
-1
-2
n
3
Þ
1
1
-2 -1 0 1 2 3
n
The sum of product sequence
v–2(n) gives rxy(–2)
n
0
Fig 3 : Computation of rxy(–2).
+¥
When m = - 1 ; rxy ( -1)
å x(n) y(n - (-1))
=
å x(n) y
=
´
1
0
2
1
+¥
-1(n)
åv
=
n = -¥
- 1(n)
n = -¥
y–1(n)
v–1(n)
3
3
2
2
1
\ rxy(–2) = 0 + 0 + 1 + 0 + 0 + 0 = 1
+¥
n = -¥
x(n)
1
3
-1
n
Þ
1
1
1
0
-1
n
+¥
; rxy (0)
=
+¥
å x(n) y(n)
å x(n) y (n)
=
0
n = -¥
x(n)
2
1
3
n
n = -¥
0
n = -¥
v0(n)
3
2
2
´
1
0
2
+¥
3
1
1
å v (n)
=
y 0(n)
2
0
The sum of product sequence
v–1(n) gives rxy(–1)
\ rxy(–1) = 0 + 3 + 1 + 0 + 0 = 4
Fig 4 : Computation of rxy(–1).
When m = 0
1
3
1
0
n
Þ
1
1
2
+¥
=
å x(n) y(n - 1))
1
n = -¥
x(n)
n = -¥
3
n
The sum of product sequence
v0(n) gives rxy(0)
\ rxy(0) =1 + 3 + 2 + 0 = 6
+¥
å x(n) y (n)
=
2
1
0
n
Fig 5 : Computation of rxy(0).
When m = 1 ; rxy (1)
1
+¥
=
å v (n)
1
n = -¥
v1(n)
y 1(n)
6
3
2
2
1
0
´
1
1
2
3
n
0
1
Þ
1
1
2
3
n
Fig 6 : Computation of rxy(1).
2
1
1
3
2
0
n
The sum of product sequence
v1(n) gives rxy(1)
\ rxy(1) = 0 + 1 + 6 + 2 = 9
6. 101
Signals & Systems
+¥
When m = 2 ; rxy (2)
+¥
å x(n) y(n - 2)
=
å x(n) y
=
n = -¥
+¥
2 (n)
å v (n)
=
n = -¥
2
n = -¥
v2(n)
y2(n)
x(n)
6
3
2
2
1
´
1
0
2
1
3
0
n
4
3
2
1
Þ
1
1
0
n
+¥
+¥
å x(n) y(n - 3)
=
å x(n) y
=
n = -¥
x(n)
2
1
3
4
n
The sum of product sequence
v2(n) gives rxy(2)
\ rxy(2) = 0 + 0 + 2 + 6 + 0 = 8
Fig 7 : Computation of rxy(2).
When m = 3 ; rxy (3 )
2
+¥
3 (n)
n = -¥
=
å v (n)
3
n = -¥
v3(n)
y3(n)
3
2
1
0
´
2
1
1
2
3
1
0
n
3
2
Þ
1
1
5
4
0
n
The crosscorrelation sequence, rxy(m) = {1, 4, 6, 9, 8, 2}
­
rxy(m)
9
8
6
4
2
1
-1
0
1
2
3
4
1
2
3
4
5
n
The sum of product sequence v3(n)
gives rxy(3)
\ rxy(3) = 0 + 0 + 0 + 2 + 0 + 0 = 2
Fig 8 : Computation of rxy(3).
-2
2
m
Fig 9 : Graphical representation of rxy(m).
Chapter 6 - Discrete Time Signals and Systems
6. 102
Method - 2: Tabular Method
The given sequences and the shifted sequences can be represented in the tabular array as shown below.
n
–2
–1
x(n)
y(n)
y(n – (–2)) = y–2(n)
1
y(n – (–1)) = y–1(n)
0
1
2
3
1
1
2
2
1
3
1
4
3
1
1
3
1
1
3
1
1
3
1
1
3
1
1
3
y(n) = y0(n)
y(n – 1) = y1(n)
y(n – 2) = y2(n)
y(n – 3) = y3(n)
5
1
Note: The unfilled boxes in the table are considered as zeros.
Each sample of rxy(m) is given by,
rxy (m) =
+¥
+¥
n = -¥
n = -¥
å x(n) y(n - m) = å x(n) y
m (n)
; where ym (n) = y(n - m)
To determine a sample of rxy(m) at m = q, multiply the sequence x(n) and yq(n) to get a product sequence
(i.e., multiply the corresponding elements of the row x(n) and yq(n)). The sum of all the samples of the product sequence
gives rxy(q).
3
When m = -2 ; rxy ( -2) =
å x(n) y
-2 (n)
= 0 + 0 + 1 + 0 + 0 + 0 = 1
-1(n)
= 0 + 3 + 1 + 0 + 0
= 4
= 1 + 3 + 2 + 0
= 6
= 1 + 6 + 2 + 0
= 9
= 0 + 0 + 2 + 6 + 0
= 8
n = -2
3
When m = -1
; rxy ( -1) =
å x(n) y
n = -1
3
When m = 0
; rxy (0)
=
å x(n) y (n)
0
n= 0
3
When m = 1
; rxy (1)
=
å x(n) y (n)
1
n=0
4
When m = 2
; rxy (2)
=
å x(n) y (n)
2
n= 0
5
When m = 3
; rxy (3)
=
å x(n) y (n)
3
= 0 + 0 + 0 + 2 + 0 + 0 = 2
n= 0
\ Crosscorrelation sequence, rxy (m) = {1, 4, 6, 9, 8, 2}
A
Method - 3: Matrix Method
Given that, x(n) = {1, 1, 2, 2}
­
y(n) = {1, 3, 1}
­
\ y(–n) = {1, 3, 1}
­
The sequence x(n) is arranged as a column and the folded sequence y(–n) is arranged as a row as shown below.
The elements of the two dimensional array are obtained by multiplying the corresponding row element with column element.
The sum of the diagonal elements gives the samples of the crosscorrelation sequence, rxy(m).
6. 103
Signals & Systems
y(–n) ®
x(n)
¯
1
3
1
1
1´1
1´3
1´1
1
1´ 1
1´ 3
1´1
2
2´ 1
2´ 3
2´1
2
2´ 1
2´ 3
2´1
y(–n) ®
1
x(n)
¯ 1
1
3
1
3
1
1
1
3
1
2
2
6
2
2
2
6
2
Þ
rxy (–2) = 1
;
rxy (–1) = 1 + 3 = 4
;
rxy(0) = 2 + 3 + 1 = 6
rxy(1) = 2 + 6 + 1 = 9
;
rxy(2) = 6 + 2 = 8
;
rxy(3) = 2
\ rxy(m) = {1, 4, 6, 9, 8, 2}
­
Example 6.33
Determine the autocorrelation sequence for x(n) = {1, 2, 3, 4}.
Solution
Let, rxx(m) be the autocorrelation sequence.
The autocorrelation sequence rxx(m) is given by,
+¥
å x(n) x(n
rxx (m) =
- m)
n = -¥
The x(n) starts at n = 0 and has 4 samples.
\ nx = 0
and N = 4
Now, rxx(m) will have, 2N – 1 = 2 ´ 4 – 1 = 7 samples.
The initial value of m = mi = – (N – 1) = – (4 – 1) = –3
The final value of m = mf = mi + (2N – 2) = –3 + (2 ´ 4 – 2) = 3
The autocorrelation is computed by tabular method. Hence the sequence x(n) and the shifted sequences of x(n) are
tabulated in the following table.
n
–3
–2
–1
0
1
2
3
1
2
3
4
1
2
3
4
1
2
3
x(n)
x(n – (–3)) = x–3(n)
x(n – (–2)) = x–2(n)
x(n – (–1)) = x–1(n)
1
x(n) = x0(n)
4
5
4
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
x(n – 1) = x1(n)
6
x(n – 2) = x2(n)
x(n – 3) = x3(n)
4
Each sample of rxx(m) is given by,
rxx (m) =
+¥
+¥
n = -¥
n = -¥
å x(n) x(n - m) = å x(n) x
m (n)
; where xm (n) = x(n - m)
To determine a sample of rxx(m) at m = q, multiply the sequence x(n) and xq(n) to get a product sequence
(i.e., multiply the corresponding elements of the row x(n) and xq(n)). The sum of all the samples of the product sequence
gives rxx(q).
Chapter 6 - Discrete Time Signals and Systems
6. 104
3
When m = -3
rxx ( -3) =
;
å x(n) x
-3 (n)
= 0 + 0 + 0 + 4 + 0 + 0 + 0 = 4
å x(n) x
-2 (n)
= 0 + 0 + 3 + 8 + 0 + 0
= 11
= 0 + 2 + 6 + 12 + 0
= 20
= 1 + 4 + 9 + 16
= 30
= 0 + 2 + 6 + 12 + 0
= 20
= 0 + 0 + 3 + 8 + 0 + 0
= 11
n = -3
3
When m = -2
rxx ( -2) =
;
n = -2
3
When m = -1
rxx ( -1)
;
å x(n) x
=
-1(n)
n = -1
3
When m = 0
;
rxx (0)
=
å x(n) x (n)
0
n=0
4
When m = 1
;
rxx (1)
=
å x(n) x (n)
1
n= 0
5
When m = 2
;
rxx (2)
=
å x(n) x (n)
2
n= 0
6
When m = 3
;
rxx (3)
=
å x(n) x (n)
3
= 0 + 0 + 0 + 4 + 0 + 0 + 0 = 4
n= 0
\ Autocorrelation sequence, rxx (m) = {4, 11, 20, 30, 20, 11, 4}
A
6.14 Circular Correlation
The circular correlation of two periodic discrete time sequences x(n) and y(n) with periodicity of
N samples is defined as,
N-1
rxy ( m) =
å x( n) y ((n - m))
n=0
*
N
.....(6.70)
where, rxy ( m) is the sequence obtained by circular correlation
y*((n – m))N represents circular shift of y*(n)
m is a variable used for time shift
The circular correlation of two different sequences is called circular crosscorrelation and the
circular correlation of a sequence with itself is called circular autocorrelation. Hence circular autocorrelation
of a discrete time sequence is defined as,
N-1
rxx ( m) =
å x(n) x ((n - m))
n=0
*
N
.....(6.71)
The output sequence obtained by circular correlation is also periodic sequence with periodicity of
N samples. Hence this correlation is also called periodic correlation.The circular correlation is defined for
periodic sequences. But circular correlation can be performed with non-periodic sequences by periodically
extending them.The circular correlation of two sequences requires that, at least one of the sequences
should be periodic. Hence it is sufficient if one of the sequences is periodically extended in order to
perform circular correlation.
The circular correlation of finite duration sequences can be performed only if both the sequences
consists of same number of samples. If the sequences have different number of samples, then convert
the smaller size sequence to the size of larger size sequence by appending zeros.
In the equation (6.70), the sequence x(n) is unshifted and the sequence y*(n) is circularly shifted
by m units of time for correlation operation. The same results can be obtained if the sequence y*(n) is
unshifted and the sequence x(n) is circularly shifted opposite to that of earlier case by m units of time,
hence the circular correlation operation can also be expressed as,
6. 105
Signals & Systems
N-1
rxy ( m) =
å x(( n + m))
n=0
N
y* (n)
.....(6.72)
Circular correlation basically involves the same three steps as that for correlation namely shifting
one of the sequence, multiplying the two sequences and finally summing the values of product sequence.
The difference between the two is that in circular correlation the shifting (rotating) operations are
performed in a circular fashion by computing the index of one of the sequences by
modulo-N operation. In correlation there is no modulo-N operation.
6.14.1 Procedure for Evaluating Circular Correlation
Let, x(n) and y(n) be periodic discrete time sequences with periodicity of N-samples. If x(n) and
y(n) are non-periodic then convert the sequences to N-sample sequence and periodically extend the
sequence y(n) with periodicity of N-samples.
Now the circular correlation of x(n) and y(n) will produce a periodic sequence rxy ( m) with periodicity
of N-samples. The samples of one period of rxy ( m) can be computed using the equation (6.70).
The value of rxy ( m) at m = q is obtained by replacing m by q, in equation (6.70), as shown in
equation (6.73).
N-1
rxy (q ) =
å x( n) y ((n - q))
n=0
*
N
.....(6.73)
The evaluation of equation (6.73) to determine the value of rxy ( m) at m = q involves the
following four steps.
1. Conjugation
: Take conjugate of y(n) to get y*(n). If y(n) is a real sequence then y*(n)
will be same as y(n). Represent the samples of one period of the sequences
on circles.
2. Rotation
: Rotate y*(n) by q times in anticlockwise if q is positive, rotate y*(n) by
q times in clockwise if q is negative to obtain y*((n – q))N.
3. Multiplication
: Multiply x(n) by y*((n – q))N to get a product sequence. Let the product
sequence be vq(m). Now, vq(m) = x(n) ´ y*((n – q))N.
4. Summation
: Sum up the samples of one period of the product sequence vq(m) to
obtain the value of rxy ( m) at m = q. [i.e., rxy (q ) ].
The above procedure will give the value of rxy ( m) at a single time instant say m = q. In general we
are interested in evaluating the values of the sequence rxy ( m) in the range 0 < m < N - 1. Hence the
steps 2 , 3 and 4 given above must be repeated, for all possible time shifts in the range 0 < m < N - 1.
6.14.2 Methods of Computing Circular Correlation
Method 1 : Graphical Method
In graphical method the given sequences are converted to same size and represented on circles. In
case of periodic sequences, the samples of one period are represented on circles. Let x(n) and y(n) be the
given real sequences. Let rxy ( m) be the sequence obtained by circular correlation of x(n) and y(n). The
following procedure can be used to get a sample of rxy ( m) at m = q.
Chapter 6 - Discrete Time Signals and Systems
6. 106
1. Represent the sequences x(n) and y(n) on circles.
2. Rotate (or shift) the sequence y(n), q times to get the sequence y((n – q))N. If q is positive then
rotate (or shift) the sequence in anticlockwise direction and if q is negative then rotate (or shift)
the sequence in clockwise direction.
3. The sample of rxy (q ) at m = q is given by,
N-1
rxy (q) =
N-1
å
x(n) y((n - q)) N =
å
x(n) yq (n)
n=0
n=0
where, yq (n) = y((n - q)) N
Determine the product sequence x(n)yq(n) for one period.
4. The sum of all the samples of the product sequence gives the sample rxy (q )
[i.e., rxy ( m) at m = q].
The above procedure is repeated for all possible values of m to get the sequence rxy ( m) .
Method 2 : Using Tabular Array
Let x(n) and y(n) be the given real sequences. Let rxy ( m) be the sequence obtained by circular
correlation of x(n) and y(n). The following procedure can be used to get a sample of rxy ( m) at m = q.
1. Represent the sequences x(n) and y(n) as two rows of tabular array.
2. Periodically extend y(n). Here the periodicity is N, where N is the length of the given sequences.
3. Shift the sequence y(n), q times to get the sequence y((n – q))N. If q is positive then shift the
sequence to the right and if q is negative then shift the sequence to the left.
4. The sample of rxy (q ) at m = q is given by,
N-1
rxy (q) =
å x(n) y((n - q))
n=0
N-1
N
=
å x(n) y
q ( n)
n=0
where, y q (n) = y((n - q)) N
Determine the product sequence x(n)yq(n) for one period.
5. The sum of all the samples of the product sequence gives the sample rxy (q ) [i.e., rxy ( m)
at m = q].
The above procedure is repeated for all possible values of m to get the sequence rxy ( m) .
Method 3: Using Matrices
Let x(n) and y(n) be the given N-point sequences. The circular correlation of x(n) and y(n) yields
another N-point sequence rxy ( m) .
In this method an N´ N matrix is formed using the sequence y(n) as shown below. The sequence
x(n) is arranged as a column vector (column matrix) of order N ´1. The product of the two matrices
gives the resultant sequence rxy ( m) .
6. 107
Signals & Systems
LMy(0)
MMyy((NN)- 1)
MM M
MMy(2)
Ny(1)
y(1)
y(0)
y( N)
M
y(3)
y(2)
y(2)
y(1)
y(0)
M
y(4)
y(3)
LM
MM
MM
MM
MM
N
OP
PP
PP
PP
PP
Q
x(0)
y( N - 1) y( N )
x(1)
y( N - 2) y( N - 1)
x(2)
y( N - 3) y( N - 2)
M
´
=
M
M
M
y(0)
y(1)
x( N - 2)
y( N)
y(0)
x( N - 1)
OP
PP
PP
PP
Q
.....
.....
.....
.....
.....
LM r (0) OP
MM r (1) PP
MM r M(2) PP
MM M PP
MM r (N - 2)PP
N r (N - 1) Q
xy
xy
xy
xy
xy
Example 6.34
Perform circular correlation of the two sequences,
x(n) = {1, 1, 2, 1} and y(n)= {2, 3, 1, 1}
­
­
Solution
Method 1:Graphical Method of Computing Circular Correlation
The given sequences are represented as points on circles as shown in fig 1 and 2.
x(1)=1
x(2)= 2
y(1)=3
x(0)=1
x(n)
y(2)=1
x(3)=1
y(n)
y(0)=2
y(3)=1
Fig 2.
Fig 1.
Let rxy (m) be the sequence obtained by correlation of x(n) and y(n). The given sequences are 4 sample sequences
and so N = 4. Each sample of rxy (m) is given by the equation,
N - 1
rxy (m) =
N- 1
å x(n) y((n - m))
=
N
n = 0
å x(n) y
m (n),
where ym (n) = y((n - m))N
n = 0
Using the above equation, graphical method of computing each sample of rxy (m) are shown in fig 3 to fig 6.
3
When m = 0 ; rxy (0) =
å x(n) y(n)
3
=
n = 0
x(n)
1
0
n = 0
3
=
å v (n)
0
n = 0
3
1
2
å x(n) y (n)
1
X
1
y0(n)
1´ 3 = 3
2
Þ
1
Fig 3 : Computation of rxy (0) .
2 ´ 1=2
v0(n)
1 ´ 2=2
1 ´1=1
The sum of samples of v0(n) gives rxy (0)
\ rxy (0) = 2 + 3 + 2 + 1 = 8
Chapter 6 - Discrete Time Signals and Systems
6. 108
3
3
å x(n) y(n - 1)
When m = 1 ; rxy (1) =
n = 0
1
1
3
å v (n)
=
1
n = 0
n = 0
2
x(n)
2
å x(n) y (n)
=
1
y1(n)
3
X
1´ 2 = 2
1
1
Þ
2´ 3=6
v1(n)
1
1 ´1=1
The sum of samples of v1(n) gives rxy (1)
Fig 4 : Computation of rxy (1) .
3
å x(n) y(n - 2)
When m = 2 ; rxy (2) =
1
\ rxy (1) = 1 + 2 + 6 + 1 = 10
3
=
n = 0
2
å x(n) y (n)
2
3
=
n = 0
åv
1
X
1´ 1 = 1
y 2(n)
2
1
2 (n)
n = 0
1
x(n)
1 ´1 = 1
1
Þ
2 ´ 2=4
v2(m)
3
1 ´1=1
1 ´3=3
The sum of samples of v2(m) gives rxy (2)
Fig 5 : Computation of rxy (2) .
\ rxy (2) = 1 + 1 + 4 + 3 = 9
3
When m = 3 ; rxy (3) =
å x(n) y(n - 3)
3
=
n = 0
1
2
å x(n) y (n)
3
3
=
n = 0
1
x(n)
1
X
1
3 (n)
n = 0
1´ 1 = 1
y3(n)
1
åv
3
Þ
2
Fig 6 : Computation of rxy (3) .
2 ´ 1=2
v3(m)
1´ 3=3
1 ´2=2
The sum of samples of v3(m) gives rxy (3)
\ rxy (3) = 3 + 1 + 2 + 2 = 8
\ rxy (m) = {8, 10, 9, 8}
­
Method 2 : Circular Correlation Using Tabular Array
The given sequences are represented in the tabular array as shown below. Here the shifted sequences ym(n) are
periodically extended with a periodicity of N = 4. Let rxy (m) be the sequence obtained by correlation of x(n) and y(n). Each
sample of rxy (m) is given by the equation,
N - 1
rxy (m) =
å x(n) y((n - m))
N
n = 0
N- 1
=
å x(n) y
m (n),
where ym (n) = y((n - m))N
n = 0
Note : The bold faced numbers are samples obtained by periodic extension.
6. 109
Signals & Systems
n
0
1
2
3
4
5
6
x(n)
1
1
2
1
y(n)
2
3
1
1
y0(n – 0) = y0(n)
2
3
1
1
2
3
1
y1(n – 1) = y1(n)
1
2
3
1
1
2
3
y2(n – 2) = y2(n)
1
1
2
3
1
1
2
y3(n – 3) = y3(n)
3
1
1
2
3
1
1
To determine a sample of rxy (m) at m = q, multiply the sequence, x(n) and y q (n), to get a product sequence
x(n) x q (n). (i.e., multiply the corresponding elements of the row x(n) and y q (n) ). The sum of all the samples of the product
sequence gives rxy (m) .
3
W he n m = 0 ; rxy ( 0 ) =
å
x (n ) y 0 (n )
n = 0
= x ( 0 ) y 0 ( 0 ) + x (1) y 0 (1) + x (2) y 0 (2) + x ( 3 ) y 0 ( 3 )
= 2 + 3 + 2 + 1 = 8
The samples of rxy (m) for other values of m are calculated as shown for m = 0.
3
When m = 1; rxy (1) =
å x(n)
y1(n) = 1 + 2 + 6 + 1 = 10
n=0
3
When m = 2; rxy (2) =
å x(n)
y 2 (n) = 1 + 1 + 4 + 3 = 9
n=0
3
When m = 3; rxy (3 ) =
å x(n)
y 3 (n) = 3 + 1 + 2 + 2 = 8
n=0
\ rxy (m) =
RS8, 10, 9, 8UV
TA
W
Method 3 : Circular Correlation Using Matrices
The sequence rxy (m) can be arranged as a column vector of order N ´ 1 and using the samples of y(n) the N ´ N
matrix is formed as shown below. The product of the two matrices gives the sequence rxy (m) .
LMy(0)
MMy(3)
MMy(2)
Ny(1)
LM2 3
MM1 2
MM1 1
N3 1
y(1)
y(2)
y(0)
y(1)
y(3)
y(0)
y(2)
y(3)
1
3
2
1
OP
P
3P
P
2 PQ
1
1
\ rxy (m) = {8, 10, 9, 8}
­
OP
y(2)P
y(1) P
P
y(0)PQ
LM1 OP
MM1 PP =
MM2PP
N1 Q
y ( 3)
LMx(0)OP
MMx(1) PP
MMx(2)PP
Nx(3)Q
LM8 OP
MM10PP
MM9 PP
N8 Q
LMr
MMr
MMr
MNr
OP
(1) P
P
(2)P
P
(3)PQ
xy (0 )
=
xy
xy
xy
Chapter 6 - Discrete Time Signals and Systems
6. 110
6.15. Summary of Important Concepts
1. The discrete signal is a function of a discrete independent variable.
2. In a discrete time signal, the value of discrete time signal and the independent variable time are discrete.
3. The digital signal is same as discrete signal except that the magnitude of the signal is quantized.
4. A discrete time sinusoid is periodic only if its frequency is a rational number.
5. Discrete time sinusoids whose frequencies are separated by an integer multiple of 2p are identical.
6. The sampling is the process of conversion of continuous time signal into discrete time signal.
7. The time interval between successive samples is called sampling time or sampling period.
8. The inverse of sampling period is called sampling frequency.
9. The phenomenon of high frequency component getting the identity of low frequency component during
sampling is called aliasing.
10. For analog signal with maximum frequency Fmax, the sampling frequency should be greater than 2Fmax.
11. When sampling frequency Fs is equal to 2Fmax, the sampling rate is called Nyquist rate.
12. The signals that can be completely specified by mathematical equations are called deterministic signals.
13. The signals whose characteristics are random in nature are called nondeterministic signals.
14. A signal x(n) is periodic with periodicity of N samples if x(n + N) = x(n).
15. When a signal exhibits symmetry with respect to n = 0 then it is called an even signal.
16. When a signal exhibits antisymmetry with respect to n = 0, then it is called an odd signal.
17. When the energy E of a signal is finite and non-zero, the signal is called energy signal.
18. When the power P of a signal is finite and non-zero, the signal is called power signal.
19.
20.
21.
22.
For energy signals, the energy will be finite and average power will be zero.
For power signals the average power is finite and energy will be infinite.
A signal is said to be causal, if it is defined for n ³ 0.
A signal is said to be noncausal, if it is defined for both n £ 0 and n > 0.
23. A discrete time system is a device or algorithm that operates on a discrete time signal.
24. When a discrete time system satisfies the properties of linearity and time invariance, it is called an LTI system.
25. When the input to a discrete time system is unit impulse d(n), the output is called impulse response, h(n).
26. In a static or memoryless system, the output at any instant n depends on input at the same time.
27. A system is said to be time invariant if its input-output characteristics do not change with time.
28. A linear system is one that satisfies the superposition principle.
29. A system is said to be causal if the output does not depends on future inputs/outputs.
30. When a system output at any time n depends on future inputs/outputs, it is called noncausal system.
31. System is said to be BIBO stable if and only if every bounded input produces a bounded output.
32. When a system output at any time n depends on past outputs, it is called a recursive system.
33. A system whose output does not depends on past outputs is called a nonrecursive system.
34. The convolution of N1 sample and N2 sample sequence produce a sequence consisting of N1+N2–1 samples.
35. In an LTI system, response for an arbitrary input is given by convolution of input with impulse response h(n).
36. The output sequence of circular convolution is also periodic sequence with periodicity of N samples.
37. The inverse system is used to recover the input from the response of a system.
38. The process of recovering the input from the response of a system is called deconvolution.
39. The correlation of two different sequences is called crosscorrelation.
40. The correlation of a sequence with itself is called autocorrelation.
6. 111
Signals & Systems
6.16. Short Questions and Answers
Q6.1
Perform addition of the discrete time signals, x1(n) = {2, 2, 1, 2} and x2(n) = {–2, –1, 3, 2}.
Solution
In addition operation the samples corresponding to same value of n are added.
When n = 0,
When n = 1,
x1(0) + x2(0) = 2 + (–2) = 0
x 1(1) + x2(1) = 2 + (–1) = 1
When n = 2,
When n = 3,
x1(2) + x2(2) = 1 + 3 = 4
x1(3) + x2(3) = 2 + 2 = 4
\ x1(n) + x2(n) = {0, 1, 4, 4}
Q6.2
Perform multiplication of discrete time signals, x1(n) = {2, 2, 1, 2} and x2(n) = {–2, –1, 3, 2}.
Solution
In multiplication operation the samples corresponding to same value of n are multiplied.
When n = 0,
When n = 1,
\ x1(n)
Q6.3
x1(0)
x1(1)
´ x2(0) = 2 ´ (–2) = –4
´ x2(1) = 2 ´ (–1) = –2
When n = 2,
When n = 3,
x1(2)
x1(3)
´ x2(2) = 1 ´ 3 = 3
´ x2(3) = 2 ´ 2 = 4
´ x2(n) = {–4, –2, 3, 4}
Express the discrete time signal x(n) as a summation of impulses.
If we multiply a signal x(n) by a delayed unit impulse d(n – m), then the product is x(m), where x(m) is
the signal sample at n = m (because d(n – m) is 1 only at n = m and zero for other values of n). Therefore,
if we repeat this multiplication over all possible delays in the range –¥ < m < ¥ and sum all the product
sequences, then the result will be a sequence that is equal to the sequence x(n).
\ x(n) = ... x(–2) d(n + 2) + x(–1) d(n + 1) + x(0) d(n) + x(1) d(n – 1) + x(2) d(n – 2) + ...
+¥
å x(m) d(n - m)
=
m = -¥
Q6.4
What are the basic elements used to construct the block diagram of discrete time system?
The basic elements used to construct the block diagram of discrete time system are Adder, Constant
multiplier and Unit delay element.
x1(n)
+
x1 (n) + x2(n)
x 1(n)
a
x(n)
ax1(n)
x 2(n)
Fig a: Adder.
Q6.5
The x(n) can be represented on the circle as shown
in fig Q6.5a. The x(n – 2, mod4) is circularly shifted
sequence of x(n) by two units of time as shown in
fig Q6.5b.(Here, mod 4 stands for periodicity of 4).
\ x(n – 2, mod4) = {3, 4, 1, 2}
Q6.6
Fig c: Unit delay element.
Fig b: Constant multiplier.
2
Let, x(n) = {1, 2, 3, 4}, be one period of a periodic
sequence. What is x(n – 2, mod4)?
3
x(n – 1)
z –1
x(n)
4
1
1 x(n–2, mod4) 3
4
2
Fig Q6.5a.
Fig Q6.5b.
Why linear convolution is important in signals and systems?
The response or output of an LTI discrete time system for any input x(n) is given by linear
convolution of the input x(n) and the impulse response h(n) of the system. (This means that if
the impulse response of a system is known, then the response of the system for any input can
be determined by convolution operation).
Chapter 6 - Discrete Time Signals and Systems
Q6.7
6. 112
In y(n) = x(n) * h(n), how will you determine the start and end point of y(n)? What will be
the length of y(n)?
Let, length of x(n) be N1 and starts at n = nx. Let, length of h(n) be N2 and starts at n = nh.
Now, y(n) will start at n = nx + nh
y(n) will end at n = (nx + nh) + (N1 + N2 – 2)
The length of y(n) is N1 + N2 – 1.
Q6.8
What is zero padding? Why it is needed?
Appending zeros to a sequence in order to increase the size or length of the sequence is called
zero padding.
In circular convolution, when the two input sequences are of different size, then they are
converted to equal size by zero padding.
Q 6.9
List the differences between linear convolution and circular convolution.
Linear convolution
1.
2.
Q 6.10
Circular convolution
The length of the input sequence
can be different
Zero padding is not required.
1.
The length of the input sequences
should be same.
If the length of the input sequences are
different, then zero padding is required.
Atleast one of the input sequence
should be periodic or should be
periodically extended.
The output sequence is periodic.
The periodicity is same as that of
input sequence.
The length of the input and output
sequences are same.
2.
3.
The input sequences need not be
periodic.
3.
4.
The output sequence is non-periodic.
4.
5.
The length of output sequence will
be greater than the length of input
sequences.
5.
Perform the circular convolution of the two sequences x1(n) = {1, 2, 3} and x2(n) = {4, 5, 6}.
Solution
Let x3(n) be the sequence obtained from circular convolution of x1(n) and x2(n). The sequence x1(n) can be
arranged as a column vector of order 3 ´1 and using the samples of x2(n) a 3 ´3 matrix is formed as shown
below. The product of two matrices gives the sequence x3(n).
LMx (0)
MMx (1)
Nx (2)
2
2
2
x 2 (2) x 2 (1)
x 2 (0) x 2 (2)
x 2 (1) x 2 (0)
OP
PP
Q
LMx (0)OP
MMx (1)PP =
Nx (2)Q
1
1
1
LMx (0)OP
MMx (1)PP
Nx (2)Q
3
3
3
Þ
LM4
MM5
N6
OP
PP
Q
6 5
4 6
5 4
LM 1OP
MM2PP =
N3 Q
LM 31OP
MM 31PP
N28Q
\ x3(n) = x1(n) * x2(n) = {31, 31, 28}.
Q6.11
Perform the linear convolution of the two sequences x1(n) = {1, 2} and x2(n) = {3, 4} via circular
convolution.
Solution
Let x 3(n) be the sequence obtained from linear convolution of x 1 (n) and x 2 (n). The length of
x3(n) will be 2 + 2 – 1 = 3. Let us convert x1(n) and x2(n) into three sample sequences by padding with zeros
as shown below.
x1(n) = {1, 2, 0} and x2(n) = {3, 4, 0}
Now the circular convolution of x1(n) and x2(n) will give x3(n). The sequence x1(n) is arranged as a column
vector and using the sequence x2(n), a 3 ´3 matrix is formed as shown below. The product of the two matrices
gives the sequence x3(n).
6. 113
Signals & Systems
LMx (0)
MMx (1)
Nx (2)
2
2
2
x 2 (2) x 2 (1)
x 2 (0) x 2 (2)
x 2 (1) x 2 (0)
OP
PP
Q
LMx (0)OP
MMx (1)PP =
Nx (2)Q
1
1
1
LMx (0)OP
MMx (1)PP
Nx (2)Q
LM3
MM4
N0
3
3
Þ
3
0 4
3 0
4 3
OP LM1OP
PP MM2PP =
Q N0Q
LM 3 OP
MM10PP
N 8Q
\ x3(n) = x1(n) * x2(n) = {3, 10, 8}
Q6.12
Compare the overlap add and overlap save method of sectioned convolutions.
Overlap add method
Overlap save method
1. Linear convolution of each section of
longer sequence with smaller sequence
is performed.
1.
2. Zero padding is not required.
2.
3. Overlapping of samples of input
sections are not required.
3.
4. The overlapped samples in the output
of sectioned convolutions are added
to get the overall output.
4.
Circular convolution of each section of
longer sequence with smaller sequence
is performed. (after converting them to
the size of output sequence).
Zero padding is required to convert
the input sequences to the size of
output sequence.
The N2–1 samples of an input section of
longer sequence is overlapped with next
input section.
Depending on method of overlapping
input samples, either last N2–1 samples
or first N2–1 samples of output sequence
of each sectioned convolution are
discarded.
Q6.13
In what way zero padding is implemented in overlap save method?
In overlap save method, the zero padding is employed to convert the smaller input sequence to
the size of the output sequence of each sectioned convolution. The zero padding is also employed
to convert either the last section or the first section of the longer input sequence to the size of the
output sequence of each sectioned convolution. (This depends on the method of overlapping input
samples).
Q 6.14
List the similarities and differences in convolution and correlation of two sequences.
Similarities
1. Both convolution and correlation operation involves shifting, multiplication and summation of
product sequence.
2. Both convolution and correlation operation produce same size of output sequence.
Differences
1. Correlation operation does not involve change of index and folding of one of the input sequence.
2. The convolution operation is commutative, [i.e., x(n)*y(n)=y(n)* x(n)], whereas in correlation
operation in order to satisfy commutative property, while performing correlation of y(n) and x(n),
the shifting has to performed in opposite direction to that of performing correlation of x(n) and y(n).
Q6.15
Let rxy(m) be the correlation sequence obtained by correlation of x(n) and y(n) , how will you
determine the start and end point of rxy(m)? What will be the length of rxy(m) ?
Let, length of x(n) be N1 and starts at n = n1. Let length of y(n) be N2 and starts at n = n2.
Now, rxy(m) will start at m i = n1 – (n2 + N2 – 1)
rxy(m) will end at m f = mi + (N1 + N2 – 2)
The length of rxy(m) is N1 + N2 – 1.
Chapter 6 - Discrete Time Signals and Systems
Q 6.16
6. 114
What are the differences between crosscorrelation and autocorrelation.
1. Crosscorrelation operation is correlation of two different sequences, whereas autocorrelation is
correlation of a sequence with itself.
2. Autocorrelation operation is an even function, whereas crosscorrelation is not an even function.
Q 6.17
Perform the correlation of the two sequences, x(n) = {1, 2, 3} and y(n) = {2, 4, 1}.
Solution
\ y(–n) = {1, 4, 2 }
­
Given that, x(n) = {1, 2, 3 } and y(n) = {2, 4, 1}.
­
­
The sequence x(n) is arranged as a column and the folded sequence y(–n) is arranged as a row as shown
below. The elements of the two dimensional array are obtained by multiplying the corresponding row element with
column element. The sum of the diagonal elements gives the samples of the crosscorrelation sequence, rxy(m).
y(–n) ®
y(–n) ®
x(n)
¯
x(n)
¯
1
4
2
1
1´1
1´4
1´2
2
2´ 1
2´4
2´2
3
3´1
3´4
3´2
rxy (–2) = 1;
rxy (–1) = 2 + 4 = 6;
Þ
rxy(0) = 3 + 8 + 2 = 13;
1
4
2
1
1
4
2
2
2
8
4
3
3
12
6
rxy(1) = 12 +4 = 16;
rxy(2) = 6 ;
\ rxy(m) = {1, 6, 13, 16, 6}
­
Q 6.18
Perform the circular correlation of the two sequences, x(n) = {1, 2, 3} and y(n) = {2, 4, 1}.
Solution
Let rxy(m) be the sequence obtained from circular correlation of x(n) and y(n). The sequence x(n) can be
arranged as a column vector of order 3 ´1 and using the samples of y(n) a 3´3 matrix is formed as shown
below. The product of two matrices gives the sequence rxy(m).
LMy(0)
MMy(2)
Ny(1)
OP
P
y(0)PQ
y(1) y(2)
y(0) y(1)
y(2)
LMx(0)OP
MMx(1)PP =
Nx(2)Q
LMr
MMr
MNr
OP
PP
(2)PQ
xy (0)
Þ
xy (1)
xy
LM2
MM1
N4
OP
P
2 PQ
4 1
2 4
1
LM1OP
MM2PP =
N3Q
LM13OP
MM17PP
N12Q
\ rxy(m) = {13, 17, 12}
Q6.19
Perform circular autocorrelation of the sequence, x(n) = {1, 2, 3, 4}.
Solution
Let rxx(m) be the sequence obtained from circular autocorrelation of x(n). The sequence x(n) can be arranged
as a column vector of order 4´1 and again by using the samples of x(n) a 4 ´4 matrix is formed as shown
below. The product of two matrices gives the sequence rxx(m).
LMx(0)
MMx(3)
MMx(2)
Nx(1)
OP
x(2)P
P
x(1)P
x(0)PQ
x(1) x(2) x(3)
x(0) x(1)
x(3) x(0)
x(2) x(3)
LMx(0)OP
MMx(1)PP =
MMx(2)PP
Nx(3)Q
LMr
MMr
MMr
Nr
OP
(1)P
(2)PP
(3)PQ
xx (0)
xx
xx
xx
Þ
LM1
MM4
MM3
N2
OP
3P
P
2P
1PQ
2 3 4
1 2
4
1
3 4
LM1OP
MM2PP =
MM3PP
N4Q
LM 30OP
MM 24PP
MM 22PP
N 24Q
\ rxx(m) = {30, 24, 22, 24 }
Q 6.20
What is the difference between circular crosscorrelation and circular autocorrelation.
Circular crosscorrelation operation is circular correlation of two different sequences, whereas
circular autocorrelation is circular correlation of a sequence with itself.
6. 115
Signals & Systems
6.17 MATLAB Programs
Program
6.1
Write a MATLAB program to generate the standard discrete time signals
unit impulse, unit step and unit ramp signals.
%*******************
n=-20
:
1
:
program
20;
to
plot
%specify
some
the
standard
r a ng e
of
signals
n
%******************* unit impulse signal
x1=1;
x2=0;
x=x1.*(n==0)+x2.*(n~=0);
%generate unit impulse signal
subplot(3,1,1);stem(n,x);
%plot the generated unit impulse
xlabel(‘n’);ylabel(‘x(n)’);title(‘unit impulse signal’);
%******************* unit step signal
x1=1;
x2=0;
x=x1.*(n>=0)+x2.*(n<0);
%generate unit step signal
subplot(3,1,2);stem(n,x);
%plot the generated unit step
xlabel(‘n’);ylabel(‘x(n)’);title(‘unit step signal’);
%******************* unit ramp signal
x1=n;
x2=0;
x=x1.*(n>=0)+x2.*(n<0);
%generate unit ramp signal
subplot(3,1,3);stem(n,x);
%plot the generated unit ramp
xlabel(‘n’);ylabel(‘x(n)’);title(‘unit ramp signal’);
signal
signal
signal
OUTPUT
The output waveforms of program 6.1 are shown in fig P6.1.
Program
6.2
Write a MATLAB program to generate the standard discrete time signals
exponential and sinusoidal signals.
%*******************
n=-20
:
1
:
20;
program
to
plot
%specify
some
the
standard
r a ng e
of
signals
n
%******************* exponential signal
A=0.95;
x=A.^n;
%generate exponential signal
subplot(2,1,1);stem(n,x);
%plot the generated exponential
xlabel(‘n’);ylabel(‘x(n)’);title(‘exponential signal’);
%******************* sinusoidal signal
N=20;
%declare periodicity
f=1/20;
%compute frequency
x=sin(2*pi*f*n);
%generate sinusoidal signal
subplot(2,1,2);stem(n,x);
%plot the generated sinusoidal
xlabel(‘n’);ylabel(‘x(n)’);title(‘sinusoidal signal’);
OUTPUT
The output waveforms of program 6.2 are shown in fig P6.2.
signal
signal
Chapter 6 - Discrete Time Signals and Systems
6.116
Fig P6.1 : Output waveforms of program 6.1.
Program
Fig P6.2 : Output waveforms of program 6.2.
6.3
Write a MATLAB program to find the even and odd part of the signal
x(n)=0.8n .
%To
find
the
n= -5 :1 :5;
A=0.8;
x1=A.^n;
x2=A.^(-n);
even
and
%specify
parts
the
%generate
%generate
if(x2==x1)
disp(‘“The given
else if (x2==(-x1))
disp(‘“The given
else
disp(‘“The given
end
end
xe=(x1+x2)/2;
xo=(x1-x2)/2;
odd
the
the
of
the
r a ng e
of
x(n)=
0.8^n
n
given signal
folded signal
signal
is
even
signal
is
odd
signal
is
neither
%compute
%compute
signal,
signal”’);
signal”’);
even
nor
odd
signal”’);
even part
odd part
subplot(2,2,1);stem(n,x1);
xlabel(‘n’);ylabel(‘x1(n)’);title(‘signal
x(n)’);
subplot(2,2,2);stem(n,x2);
xlabel(‘n’);ylabel(‘x2(n)’);title(‘signal
x(-n)’);
subplot(2,2,3);stem(n,xe);
xlabel(‘n’);ylabel(‘xe(n)’);title(‘even
part
of
x(n)’);
subplot(2,2,4);stem(n,xo);
xlabel(‘n’);ylabel(‘xo(n)’);title(‘odd
part
of
x(n)’);
OUTPUT
“The given signal is neither even nor odd signal”
The output waveforms of program 6.3 are shown in fig P6.3.
6. 117
Signals & Systems
Fig P6.3 : Output waveforms of program 6.3.
Program
Fig P6.4 : Output waveforms of program 6.4.
6.4
Write a MATLAB program to perform Amplitude scaling and Time shift on
the signal x(n) = 1+n; for n = 0 to 2.
Program to declare the given signal as function y(n)
%
decla r e
the
g iven
functio n x = y(n)
x=(1.0 + n).*(n>=0
sig na l
&
as
functio n
y(n)
n<=2);
Note: The above program should be stored as a separate file in the
current working directory
Program to perform amplitude scaling and time shift on y(n)
%To Perform Amplitude scaling and Time shift on signal x(n)=1+n;
n= 0 to 2
%include y.m file in current work directory which declare given
signal as function y(n)
n=-5:1:5;
%specify
y0
y1
y2
y3
y4
%a ssig n the g iven sig na l a s y0
%compute the amplified version of x(n)
%compute the attenuated version of x(n)
%compute the delayed version of x(n)
%compute the advanced version of x(n)
=y(n);
=1.5*y(n);
=0.5*y(n);
=y(n-2);
=y(n+2);
range
of
n
%plot the given signal and amplitude scaled signal
subplot(2,3,1);stem(n,y0);
xlabel(‘n’);ylabel(‘x(n)’);title(‘Signal x(n)’);
subplot(2,3,2);stem(n,y1);
xlabel(‘n’);ylabel(‘x1(n)’);title(‘Amplified signal 1.5x(n)’);
subplot(2,3,3);stem(n,y2);
xlabel(‘n’);ylabel(‘x2(n)’);title(‘Attenuated signal 0.5x(n)’);
%plot the given signal and
subplot(2,3,4);stem(n,y0);
time
shifted
signal
for
Chapter 6 - Discrete Time Signals and Systems
6.118
xlabel(‘n’);ylabel(‘x(n)’);title(‘Signal x(n)’);
subplot(2,3,5);stem(n,y3);
xlabel(‘n’);ylabel(‘x3(n)’);title(‘Delayed signal x(n-2)’);
subplot(2,3,6);stem(n,y4);
xlabel(‘n’);ylabel(‘x4(n)’);title(‘Advanced signal x(n+2)’);
OUTPUT
The input and output waveforms of program 6.4 are shown in fig P6.4.
Program
6.5
Write a MATLAB program to perform convolution of the following two
discrete time signals.
x1(n)=1; 1<n<10
x2(n)=1;
2<n<10
%******************Program to
%******************x1 (N)=1 ; n=
perform convolution of two signals
1 to 1 0 a nd x2(n)=1 ; n= 2 to 1 0
n = 0 : 1 : 15;
%specify
x1=1.*(n>=1 & n<=10);
x2=1.*(n>=2 & n<=10);
N1=length(x1);
N2=length(x2);
x3=conv(x1,x2);
n1=0 : 1 : N1+N2-2;
%generate
%generate
range
of
signal
signal
n
x1(n)
x2(n)
%perform convolution of signals x1(n)
%specify r a ng e o f n fo r x3(n)
subplot(3,1,1);stem(n,x1);
xlabel(‘n’);ylabel(‘x1(n)’);
title(‘signal x1(n)’);
subplot(3,1,2);stem(n,x2);
xlabel(‘n’);ylabel(‘x2(n)’);
title(‘signal x2(n)’);
subplot(3,1,3);stem(n1,x3);
xlabel(‘n’);ylabel(‘x3(n)’);
title(‘signal, x3(n) = x1(n)*x2(n)’);
OUTPUT
The input and output waveforms of program 2.5 are shown in fig P2.5.
Fig P6.5 : Output waveforms of program 6.5.
and
x2(n)
6. 119
Signals & Systems
6.18 Exercises
I Fill in the blanks with appropriate words
1.
2.
3.
A signal x(n) may be shifted in time by m units by replacing the independent variable n by _______.
The _______ of a signal x(n) is performed by changing the sign of the time base n.
If the average power of a signal is finite then it is called _______.
4.
5.
6.
The smallest value of N for which x(n + N) = x(n) is true is called _______.
In a discrete time signal x(n), if x(n) = x(–n) then it is called _______ signal.
In a discrete time signal x(n), if x(–n) = –x(n) then it is called _______ signal.
7.
8.
The output of the system with zero input is called _______.
A discrete time system is _______ if it obeys the principle of superposition.
9. A discrete time system is _______ if its input-output relationship do not change with time.
10. The response of an LTI system is given by _______ of input and impulse response.
11. If the output of a system depends only on present input then it is called _______.
12.
13.
14.
15.
16.
A system is said to be _______ if the output does not depends on future inputs and outputs.
An LTI system is causal if and only if its impulse response is _______ for negative values of n.
When a system output at any time n depends on past output values, it is called _______ system.
An N-point sequence is called _______ if it is symmetric about point zero on the circle.
An N-point sequence is called _______ if it is antisymmetric about point zero on the circle.
17. The _______ is called aperiodic convolution.
18. The _______ is called periodic convolution.
19. Appending zeros to a sequence in order to increase its length is called _______.
20. The two methods of sectioned convolutions are _______ and _______method.
21. In _______ method of sectioned convolution, overlapped samples of output sequences are _______.
22. In _______ method, the overlapped samples in one of the output sequences are discarded.
23. The correlation of two different discrete time sequences is called _______ .
24. The cascade of a system and its inverse is _______.
25. The process of recovering the input from the response of a system is called ______ .
Answers
1. n – m
2. folding
8. linear
9. time invariant
15. even
16. odd
22. overlap save
23. cross correlation
3. power signal
4. fundamental period
5. symmetric
10. convolution
11. memoryless or static
12. causal
17. linear convolution
18. circular convolution
19. zero padding
24. identity system
25. deconvolution
6. antisymmetric
13. zero
20. overlap add, overlap save
7. natural response
14. recursive
21. overlap add, added
II. State whether the following statements are True/False
1.
2.
3.
The discrete signals are continuous function of an independent variable.
In digital signal the magnitudes of the signal are unquantized.
A discrete time signal x(n) is defined for noninteger values of n.
4.
5.
An impulse signal has a nonzero sample only for one value of n.
When we multiply a discrete time signal by unit step signal, the signal is converted to one sided signal.
Chapter 6 - Discrete Time Signals and Systems
6. 120
6.
Shifting a signal to left is called delay and shifting to right is called advance.
7.
8.
9.
Any discrete time signal can be expressed as a summation of impulses.
Periodic signals are power signals.
When the energy of a signal is infinite, it is called energy signal.
10. The output of a system for impulse input is called impulse response.
11. A system can be realized in real time only if it is noncausal and stable.
12. Dynamic systems does not require memory but static systems require memory.
13. A system is time invariant if the response to a shifted version of the input is identical to a shifted version of
the response based on the unshifted input.
14. An LTI system is unstable if the impulse response is absolutely summable.
15. A system whose output depends only on the present and past input is called a recursive system.
16. The circular shift of an N-point sequence is equivalent to a linear shift of its periodic extension.
17. For an N-point sequence represented on a circle, the time reversal is obtained by reversing its sample about
the point zero on the circle.
18. When a nonperiodic N-point sequence is represented on a circle then it becomes periodic with
periodicity N.
19. In linear convolution the length of the input sequences should be same.
20. In circular convolution the length of the input sequences need not be same.
21. In circular correlation the length of the input and output sequences are same.
¥
22. The correlation operation, rxy ( m) =
¥
å x( n) y( n - m)
is not same as rxy ( m) =
n = -¥
å x( n + m) y( n) .
n = -¥
23. The cross correlation sequence rxy(m) is folded version of ryx(m).
24. The inverse systems exist for all LTI systems.
25. The final value of m in autocorrelation sequence of N-point sequence is, mf = mi + (2N – 1).
Answers
1.
False
6.
False
11.
False
16.
True
21.
True
2. False
7.
True
12.
False
17.
True
22.
False
3. False
8.
True
13.
True
18.
True
23.
True
4. True
9.
False
14.
False
19.
False
24.
True
5. True
10.
True
15.
False
20.
False
25.
False
III. Choose the right answer for the following questions
1. x(n) =
x(n - 1)
with initial condition x(0) = –1, gives the sequence,
4
a) x(n) =
2.
FG 1 IJ
H 4K
b) x(n) = -
FG 1 IJ
H 4K
n
c) x(n) =
FG 1 IJ
H 4K
-n
d) x(n) =
FG -1IJ
H4K
-n
The process of conversion of continuous time signal into discrete time signal is known as,
a) aliasing
3.
n
b) sampling
c) convolution
d) none of the above
If Fs is sampling frequency then the relation between analog frequency F and digital frequency f is,
a) f =
F
2Fs
b) f =
Fs
F
c) f =
F
Fs
d) f =
2F
Fs
6. 121
Signals & Systems
4.
If Fs is sampling frequency then the highest analog frequency that can be uniquely represented in its
sampled version of discrete time signal is,
1
F
d)
a) s
b) 2Fs
c) Fs
Fs
2
5.
The sampling frequency of the following analog signal, x(t) = 4 sin150pt + 2 cos50pt should be,
a) greater than 75 Hz b) greater than 150 Hz
c) less than 150 Hz
d) greater than 50 Hz
6.
Which of the following signal is the example for deterministic signal?
a) step
b) ramp
c) exponential
d) all of the above
7.
For Energy signals, the energy will be finite and the average power will be,
a) infinite
b) finite
c) zero
d) cannot be defined
8.
In a signal x(n), if 'n' is replaced by
n
, then it is called,
3
b) folded version
c) downsampling
a) upsampling
9.
d) shifted version
The unit step signal u(n) delayed by 3 units of time is denoted as,
a) u(n + 3) = 1; n ³ 3
= 0; n < 3
b) u(3 - n ) = 1; n ³ 3
= 0; n < 3
c) u(n - 3) = 1; n ³ 3
= 0; n < 3
d) u(3n) = 1; n > 3
= 0; n < 3
10. The zero input response (or) natural response is mainly due to,
a) Initial stored energy in the system
b) Initial conditions in the system
c) Specific input signal
d) both a and b
11. If x(n) = an u(n) is the input signal, then the particular solution yp(n) will be,
a) Kn an u(n)
b) K an u(n)
n
n
c) K1 a u(n) + K2 a u(n)
d) K a–n u(n)
12. The discrete time system, y(n) = x(n–3) – 4x(n–10) is a,
a) dynamic system
b) memoryless system
c) time varying system
d) none of the above
13. An LTI discrete time system is causal if and only if,
a) h(n) ¹ 0 for n < 0
b) h(n) = 0 for n < 0
c) h(n) ¹ ¥ for n < 0
d) h(n) ¹ 0 for n > 0
14. Which of the following system is causal?
a) h(n) = n
FG 1 IJ
H 2K
n
u( n + 1) b) y(n) = x2(n) – x(n+1)
c) y(n) = x(–n) +x(2n–1) d) h(n) = n
FG 1 IJ
H 2K
n
u( n)
15. An LTI system is stable, if the impulse response is,
¥
a)
å h( n) = 0
n = -¥
¥
b)
å h( n) < ¥
n = -¥
¥
c)
å h( n) ¹ 0
d) either a or b
n= -¥
16. The system y(n) = sin[x(n)] is,
a) stable
b) BIBO stable
c) unstable
d) none of the above
17. Two parallel connected discrete time systems with impulse responses h1(n) and h2(n) can be replaced by
a single equivalent discrete time system with impulse response,
a) h1(n) * h2(n)
b) h1(n) + h2(n)
c) h1(n) – h2(n)
d) h1(n) * [h1(n) + h2(n)]
18. Sectioned convolution is performed if one of the sequence is very much larger than the other in order to
overcome,
a) long delay in getting output
b) larger memory space requirment
c) both a and b
d) none of the above
Chapter 6 - Discrete Time Signals and Systems
6. 122
19. In overlap save method, the convolution of various sections are performed by,
a) zero padding
b) linear convolution
c) circular convolution d) both b and c
20. If x(n) is N1-point sequence starts at n = n1,if y(n) is N2-point sequence starts at n = n2, if rxy(m) is the
correlation sequence then the value of m corresponding, to last sample of rxy(m) is,
a) mf = mi + (N1 + N2 – 2) b) mf = mi + (2N – 2)
c) mf = mi + (N1 + N2 – 1) d) mf = mi + (2N + 1)
21. For a system, y(n) = nx(n), the inverse system will be,
bg
a) y 1
n
ej
b) 1 y n
n
d) n –1y(n)
c) ny(n)
22. For a system y(n) = x(n–3) the impulse response of the system and the inverse system will be –––––– and
–––––– respectively.
b) h( n) = d (3n), h¢ ( n) = d n
3
d) h(n) = d(n + 3), h¢(n) = d(n – 6)
ej
a) h(n) = d(n + 3), h¢(n) = d(n – 3)
c) h(n) = d(n – 3), h¢(n) = d(n + 3)
23. The circular correlation rxy (q) of the sequence x1(n) and x2(n) of length 'N' can be defined by the
equation,
¥
a)
N -1
å x ( n) x ( n - q )
1
b)
2
n= -¥
1
*
2
(n - q)
n=0
¥
N -1
c)
å x ( n) x
å x ( n ) x b( n - q ) g
*
2
1
n=0
d)
N
å x ( n) x b( n - q )g
*
2
1
n = -¥
N
24. The evaluation of correlation involves,
a) shifting, rotating and summation
c) change of index, folding and summation
b) shifting, multiplication and summation
d) change of index, folding, shifting & multiplication
25. The circular correlation of N-point sequences is evaluated in the range,
a) – N < m < N
b) – N < m < 0
c) 0 < m < N
d) 0 < m < N– 1
Answers
1.
2.
3.
4.
5.
b
b
c
a
b
6. d
7. c
8. a
9. c
10. a
11.
12.
13.
14.
15.
b
a
b
d
d
16.
17.
18.
19.
20.
a
b
c
c
a
21.
22.
23.
24.
25.
b
c
c
b
d
IV. Answer the following questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
Define discrete and digital signal.
Explain briefly, the various methods of representing discrete time signal with examples.
Define sampling and aliasing.
What is Nyquist rate?
State sampling theorem.
Define the impulse and unit step signal.
Express the discrete time signal x(n) as a summation of impulses.
How will you classify the discrete time signals?
What are energy and power signals?
25.25
6. 123
Signals & Systems
10. When a discrete time signal is called periodic?
11. What is discrete time system?
12. What is impulse response? Explain its significance.
13. Write the difference equation governing the Nth order LTI system.
14. Write the expression for discrete convolution.
15. List the various methods of classifying discrete time systems.
16. Define time invariant system.
17. What is linear and nonlinear systems?
18. What is the importance of causality?
19. What is BIBO stability? What is the condition to be satisfied for stability?
20. What are FIR and IIR systems?
21. Write the convolution sum formula for FIR and IIR systems.
22. What is recursive and non recursive system? Give example.
23. Write the properties of linear convolution.
24. Prove the distributive property of linear convolution.
25. What are the two ways of interconnecting LTI systems ?
26. Define circular convolution.
27. What is the importance of linear and circular convolution in signals & systems ?
28. How will you perform linear convolution via circular convolution?
29. What is sectioned convolution? Why it is performed?
30. What are the two methods of sectioned convolution?
31. What is inverse system? What is its importance?
32. Define deconvolution.
33. Define cross correlation and autocorrelation?
34. What are the properties of correlation?
35. What is circular correlation?
V Solve the following Questions
E6.1
Determine whether the following signals are periodic or not. If periodic find the fundamental period.
3p
5n
4pn
a) x(n) = sin
n+4
b) x(n) = cos
c) x(n) = cos
+p
8
3
12
FG
IJ
H
K
p I
F
d) x(n) = sin G n J
H 18 K
2
E6.2
E6.3
FG
H
IJ
K
FG
H
IJ
K
e) x(n) = e j5n
Determine the even and odd part of the signals.
p
-j n
2
a) x(n) = n
b) x(n) = 7 e 5
a
l
A
q
c) x(n) = 3, -6, 2, -4
a) Consider the analog signal x(t) = 5 sin50pt. If the sampling frequency is 60Hz, find the sampled
version of discrete time signal x(n). Also find an alias frequency corresponding to Fs = 60Hz.
Consider the analog signals, x1 ( t ) = 3cos 2p( 40t ) and x2 ( t ) = 3cos 2p(5t ). Find a sampling frequency
so that 40Hz signal is an alias of 5Hz signal.
c) Consider the analog signal, x( t ) = 2 sin 40p t - 3sin100pt + cos 50pt. Determine the minimum
sampling frequency and the sampled version of analog signal at this frequency. Sketch the
waveform and show the sampling points. Comment on the result.
b)
Chapter 6 - Discrete Time Signals and Systems
E6.4
6. 124
Determine whether the following signals are energy or power signals.
a) x(n) =
FG 3IJ
H 8K
FG
H
n
IJ
K
3p
n
4
b) x(n) = sin
u( n)
c) x(n) = u(2n)
d) x(n) = 2 u(3 – n)
E6.5
Construct the block diagram and signal flow graph of the discrete time systems whose input-output
relations are described by the following difference equations.
a) y(n) = 2y(n – 1) + 2.5 x(n – 2) + 0.5 x(n – 3)
b) y(n) = 3.2 x(n –2) + 0.7 x(n) + 5y(n – 1) + 0.35y(n – 2)
E6.6
Determine the response of the discrete time systems governed by the following difference equations.
a) y(n) = 0.2y(n – 1) + x(n – 1) + 0.5x(n) ; x(n) = 2–n u(n) ; y(–1) = –1
b) y(n) + 2.3y(n – 1) + 0.6y(n – 2) = x(n) + 0.56x(n –1) ; x(n) = u(n) ; y(–2) = 1; y(–1) = –3
E.6.7 Test the following systems for time invariance.
a) y(n) = x(n) + x(n + 2)
b) y(n) = nax(n)
E6.8
d) y(n) = (n –1) x2(n) + C
Test the following systems for linearity.
a) y(n) = x(n) + x2(n – 1)
d) y(n) = x(n) +
E6.9
c) y(n) = x2(n –1) + C
1
x(n)
N
e) y(n) =
M
åb
m
x( n + m) + å c m y( n + m)
m= -1
m=0
Test the causality of the following systems.
a) y(n) = x(n) – x(–n – 1) + x(n –1)
b)
n
c) y(n) =
c) y(n) = a x(n) + b x( n)
b) y(n) = bx(n) + nex(n)
y(n) = a x(2n) + x(n2)
n
2
å x( m) + å x( 2m)
m= -1
d)
h(n) = (0.1)n u(n + 2)
m= -¥
e) y(n) =
å x( n - k )
k = -4
E6.10 Test the stability of the following discrete time systems.
b) y(n) = nx(n –1)
c) h(n) = (0.2)n u(n + 3)
a) y(n) = x2(n) + x(n + 1)
d) h(n) = (4)n u(4 – n)
e) y(n) = x(n – 5)
E6.11 Determine the range of values of 'a' and 'b' for the stability of an LTI system with impulse response,
h( n) =
RS(-2a)
T b
n
; n³0
; n<0
-n
E6.12 a) Determine the impulse response for the cascade of two LTI systems having impulse responses,
FG 1IJ
H 5K
n
u ( n ) and
h 2 ( n) = d ( n - 2 )
® h2(n)
b) Determine the overall impulse response of
the interconnected discrete time system
shown in fig E6.12.
x(n)
+
®
® h1(n)
FG 1 IJ u(n) ; h ( n) = FG 1IJ u( n) ;
H 7K
H 3K
n
Take, h1 ( n) =
® ®
h1 ( n ) =
n
FG 1 IJ u( n)
H 9K
n
h3 ( n) =
2
®+
y(n)
®
h3(n)
Fig E6.12.
E6.13 Determine the response of an LTI system whose impulse response h(n) and input x(n) are given by,
l
q
a) h ( n ) = 1, 2 , 1, -2 , -1
b) h( n) =
R|S1 ;
|T0 ;
A
0£ n£ 3
n³4
l
q
,
x( n) = 1, 2, 3, -1, -3
,
x( n) = a n u ( n );
A
a <1
6. 125
Signals & Systems
E6.14 Perform circular convolution.
a) x1 ( n) = 1, 2, -1, 1 ;
b)
l
x ( n ) = l0,
1
q
l
l
q
x 2 ( n ) = 2, 4 , 2, 1
q
q
x 2 ( n ) = -2, 2 , 0.4, 0.6, 0.8
0.5, 1, 15
., 2 ;
E6.15 The input x(n) and impulse response h(n) of an LTI system are given by,
l
q
l
x( n) = -1, 1, -2, -1, 1, 2 ;
A
q
h( n) = -0.5, 0.5, -1, 0.25, -1, -2
Find the response of the system using
A
a) Linear convolution,
b) Circular convolution.
E6.16 Perform linear convolution of the following sequences by,
a) Overlap add method
b) Overlap save method
l
q
x( n) = -1, 1, 2, -1, 1, 2, -1, 1, -1
l
q
h( n) = 2, 3, -2
;
E6.17 Perform crosscorrelation.
l
q
x( n) = -1, 1 3, -4, ;
A
l
q
h( n) = 2, -1, -2,
A
l
q
E6.18 Determine autocorrelation sequence for x ( n ) = 1, 4, -3, 2, 1 .
A
E6.19 Find the inverse system for the discrete time system,
n
y( n) =
åa
m
x (m - 1) ; for n ³ 0
m= 0
l
A
q
E6.20 The response of a discrete time system is, y( n) = 4, 9, 6, 7.5, 3, 30, -7 when excited by an
l
A
q
input x(n). If the impulse response of the system is h( n) = 2, 4, -1 then what will be the input to the
system?
E6.21 Preform circular correlation of the sequences,
l
q
x( n) = -1, 1, 2, -5
l
q
and y(n) = 4, 3, -1, 1 .
Answers
E6.1 a) periodic; N = 16
E6.2 a) x e(n) = an + a–n
x0(n) = a–n – an
b) nonperiodic
c) periodic; N = 6
p
b) xe ( n) = 7 cos n
5
7
p
x 0 ( n) = sin n
j
5
d) periodic; N = 6
e) nonperiodic.
l
q
A
x ( n) = l2, - 1, 3, 0, - 3, 1, - 2q
A
c) xe ( n) = -2, 1, - 3, 3, - 3, 1, - 2
0
5pn
2 sin 40pt
; Alias frequency = 85Hz
b) Fs = 35Hz
6
2pn
p
c) Fs,min = 100Hz ; x( nT) = 2 sin
+ cos n (sin pn = 0, for integer n)
5
2
3 sin 100pt
The component 3sin100pt will give always zero samples when
sampled at 100Hz for any value of n (Refer fig E6.3c).
E6.3 a) x( n) = 5sin
E6.4 a) E = 1.6 ; P = 0
b) E = ¥ ; P = 0.5
; Energy signal.
; Power signal.
c) E = ¥ ; P = 0.25 ; Power signal.
d) E = ¥ ; P = 2
; Power signal.
cos 50pt
0.01
0.04
0.02
0.05
Fig E6.3c : Sampling points.
Chapter 6 - Discrete Time Signals and Systems
®2.5
®
2.5
®
®
0.35
®
®
z
®
®
0.7
1
®
®
5
–1
z –1
0.35
® y(n)
z –1
z –1
®
®
® 3.2
z –1
y(n)
z –1
1
®
z –1
5
1
®
® ®
+
x(n)
®
® ®
z –1
y(n)
®
®+
®
®+
2
®0.5
E6.5 b)
®0.7
1
®
®
®
®
Fig E6.5a.2 : Signal flow graph. z –1
x(n)
1
®
®
z –1
3.2
z –1
z –1
®
2
®
1
® 0.5
®
z –1
®
®
®
Fig E6.5a.1 : Block diagram.
x(n)
y(n)
®+ ® +
®
z –1
®
®
z –1
®
®
®
x(n)
®
E6.5 a)
6. 126
Fig E6.5b.2 : Signal flow graph.
z –1
Fig E6.5b.1 : Block diagram.
E6.6
a) y(n) =
LM 1 FG 1 IJ
MN 2 H 2 K
OP
PQ
n
- (0.2) n u(n)
E6.7 a) c) Time invariant.
E6.8 a) e) Linear
E6.9 a) Causal
E6.10 a) c)
b) d) Time variant
b) c) d) Nonlinear
b) c) d) e) Noncausal
d) e) Stable system
E6.11 For stability, 0 < a <
FG 1IJ
H 5K
a) y ( n ) = l1, 4,
b) y(n) = 0.4 - 0.0176 ( -0.3) n + 6.9176 ( -2) n u(n)
Unstable system
b)
1
and 0 < b < 1.
2
(n - 2)
E6.12 a) h( n) =
u(n - 2)
E6.13
8, 5, -7, -15, -4,
l
b) h( n) =
A
LM 5 F 1I
MN 2 GH 3JK
7, 3q
n
FG 1 IJ + FG 9 IJ FG 1 IJ OP u(n)
H 9 K H 2 K H 7 K PQ
n
-4
n
n
b) y( n) =
q
l
; for n ³ 0
q
a) x 3 ( n) = 6, 9 , 9 , 3
E6.15
y( n) = 0.5 -1 2.5 -1.75 2.25 1 -0.25 3.25 15
. - 4 -4
l
k
k = n -3
E6.14
A
åa
b ) x 3 ( n ) = 5.6, 15
. , 1.4, 0.8, -0.3
A
q
A
l
l
q
q
E6.16 a) Overlap add method : y( n) = -2, -1, 9, 2, -5, 9, 2, -5, 3, -5, 2
b) Overlap save method : y( n) = *, *, 9, 2, -5, 9, 2, -5, 3, -5, 2
l
A
( m) = l1, 6, 2,
q
E6.17
rxy ( m) = 2, -1, -9, 7, 10, -8,
E6.18
rxx
l
A
q
-12, 31, -12, 2, 6, 1
E6.20 x( n) = 2, 0.5, 3, -2,
A
7q
E6.19 x( n) =
1
[ y( n + 1) - y( n)]
a n+1
l
q
E6.21 rxy ( m) = -8, 14, -5, -22
CHAPTER 7
Z-Transform
7.1 Introduction
Transform techniques are an important tool in the analysis of signals and systems. The Laplace
transforms are popularly used for analysis of continuous time signals and systems. Similarly Z-transform
plays an important role in analysis and representation of discrete time signals and systems. The Z-transform
provides a method for the analysis of discrete time signals and systems in the frequency domain which is
generally more efficient than its time domain analysis.
The Z-transform of x(n) will convert the time domain signal x(n) to z-domain signal X(z), where
the signal becomes a function of complex variable z.
The complex variable z is defined as,
z = u + jv = r ejw
where, u = Real part of z ; v = Imaginary part of z
jv
z-plane
r1
z1
–u
u
r = u 2 + v 2 = Magnitude of z
–jv
v
w = tan -1 = Phase or Argument of z
Fig 7.1 : z-plane.
u
The u and v take values from –¥ to +¥. A two dimensional complex plane with values of u on
horizontal axis and values of v on vertical axis as shown in fig 7.1 is called z-plane. A circle with radius r1
in z-plane represents all values of z1 having same magnitude r1 with variable phase (i.e., w = 0 to 2p).
History of Z-Transform
A transform of a sampled signal or sequence was defined in 1947 by W. Hurewicz as,
¥
z[ f (kT)] =
å f (kT) z
-k
k =0
which was later denoted in 1952 as Z-transform by a sampled-data control group at Columbia
University led by professor John R. Raggazini and including L.A. Zadeh, E.I. Jury, R.E. Kalman, J.E.
Bertram, B. Friedland and G.F. Franklin, (Source : www.ling.upenn.edu).
Definition of Z-Transform
Let,
x(n) = Discrete time signal
X(z) = Z-transform of x(n)
The Z-transform of a discrete time signal, x(n) is defined as,
¥
X( z) =
å x(n) z
-n
;
where, z is a complex variable.
n = -¥
The Z-transform of x(n) is symbolically denoted as,
Z{x(n)}
;
where, Z is the operator that represents Z-transform.
+¥
l q å x(n) z
\ X( z) = Z x(n) =
n = -¥
-n
.....(7.1)
Chapter 7 - Z - Transform
7. 2
Since the time index n is defined for both positive and negative values, the discrete time signal x(n)
in equation (7.1) is considered to be two sided and the transform is called two sided Z-transform. If the
signal x(n) is one sided signal, (i.e., x(n) is defined only for positive value of n) then the Z-transform is
called one sided Z-transform.
The one sided Z-transform of x(n) is defined as,
+¥
X( z) = Z{x( n)} =
å x( n ) z
.....(7.2)
-n
n= 0
The computation of X(z) involves summation of infinite terms which are functions of z. Hence it
is possible that the infinite series may not converge to finite value for certain values of z. Therefore for
every X(z) there will be a set of values of z for which X(z) can be computed. Such a set of values will lie
in a particular region of z-plane and this region is called Region Of Convergence (ROC) of X(z).
Inverse Z-Transform
Let, X(z) be Z-transform of x(n). Now the signal x(n) can be uniquely determined from X(z) and
its region of convergence (ROC).
The inverse Z-transform of X(z) is defined as,
1
.....(7.3)
X(z) z n -1 dz
x( n) =
2 pj c
z
The inverse Z-transform of X(z) is symbolically denoted as,
;
where, Z–1 is the operator that represents the inverse Z-transform
Z–1 {X(z)}
1
X(z) z n -1 dz
\ x(n) = Z -1{X( z)} =
2 pj c
We also refer x(n) and X(z) as a Z-transform pair and this relation is expressed as,
z
Z
x(n) ¬
Z-1
® X(z)
Proof :
Consider the definition of Z-transform of x(n),
+¥
X (z )
=
+¥
å x(n) z
-n
å x(k) z
=
n = -¥
+¥
X (z ) z n - 1 =
-k
Let n ® k
k = -¥
å x(k) z
-k
Multiply both sides by z n - 1
zn- 1
k = -¥
Let us integrate the above equation on both sides over a closed contour "C" within the ROC of X(z) which
encloses the origin.
\
z
z
X(z) z n - 1 dz =
c
+¥
å x(k) z
c k = -¥
+¥
å
=
x(k)
z
n - 1- k
+¥
å
= 2 pj
By Cauchy integral theorem, k =
z
z n - 1 - k dz = 1
Interchanging the order of
summation and integration
z n - 1-k dz
c
k = -¥
1
2 pj
dz
x(k)
-¥
1
2 pj
z
Multiply and divide by 2pj
z n - 1- k dz
.....(7.4)
c
; k = n
c
= 0
; k ¹ n
On applying Cauchy integral theorem the equation (7.4) reduces to,
z
X(z) z n - 1 dz = 2 pj x(n)
+¥
å x(k)
c
k = -¥
\ x(n) =
1
2 pj
z
c
X(z) z n- 1 dz
= x(n)
n=k
7. 3
Signals & Systems
Geometric Series
The Z-transform of a discrete time signal involves convergence of geometric series. Hence the
following two geometric series sum formula will be useful in evaluating Z-transform.
1. Infinite geometric series sum formula.
If C is a complex constant and 0 < |C|< 1, then,
¥
åC
n
n= 0
=
1
1- C
.....(7.5)
2. Finite geometric series sum formula.
If C is a complex constant and,
N -1
When C ¹ 1,
åC
n
=
n= 0
1 - CN
CN - 1
=
1- C
C-1
.....(7.6)
N- 1
When C = 1,
åC
n
= N
.....(7.7)
n= 0
Note : The infinite geometric series sum formula requires that the magnitude of C be strictly
less than unity, but the finite geometric series sum formula is valid for any value of C.
7.2 Region of Convergence
Since the Z-transform is an infinite power series, it exists only for those values of z for which the
series converges. The region of convergence, (ROC) of X(z) is the set of all values of z, for which X(z)
attains a finite value. The ROC for the following six types of signals are discussed here.
Case i :
Finite duration, right sided (causal) signal
Case ii : Finite duration, left sided (anticausal) signal
Case iii : Finite duration, two sided (noncausal) signal
Case iv : Infinite duration, right sided (causal) signal
Case v : Infinite duration, left sided (anticausal) signal
Case vi : Inifinite duration, two sided (noncausal) signal
Case i : Finite duration, right sided (causal) signal
Let, x(n) be a finite duration signal with N-samples, defined in the range 0 £ n £ (N – 1).
\ x(n) = {x(0), x(1), x(2),.....x(N–1)}
Now, the Z - transform of x(n) is,
N- 1
X( z) =
å x(n) z
-n
n=0
= x(0) + x(1)z -1 + x(2)z-2 + ........x(N - 1) z- (N -1)
x(1)
x(2)
x(N - 1)
= x(0) +
+
+ ..........+
2
z
z
zN - 1
Chapter 7 - Z - Transform
7. 4
In the above summation, when z = 0, all the terms except the first term become infinite. Hence the
X(z) exists for all values of z, except z = 0. Therefore, the ROC for finite duration right sided (or causal
signal) is entire z-plane except z = 0.
Case ii : Finite duration, left sided (anticausal) signal
Let, x(n) be a finite duration signal with N-samples, defined in the range –(N–1) £ n £ 0.
\ x(n) = {x(–(N–1)),.....,x(–2), x(–1), x(0)}
Now, the Z-transform of x(n) is,
0
X( z) =
å x(n) z
-n
n = - (N -1)
= x(- (N - 1)) z (N -1) + ....... + x( -2)z 2 + x(-1) z + x(0)
In the above summation, when z = ¥ , all the terms except the last term become infinite.
Hence the X(z) exists for all values of z, except, z = ¥. Therefore, the ROC of X(z) is entire z-plane,
except z = ¥.
Case iii : Finite duration, two sided (noncausal) signal
Let, x(n) be a finite duration signal with N-samples, defined in the range –M £ n £ + M,
N -1
where, M =
2
l
q
\ x(n) = x( - M),....... , x( -2), x(-1), x(0), x(1), x(2), ........x( M )
Now, the Z-transform of x(n) is,
+M
X( z) =
å x(n) z
-n
n = -M
= x( - M) z M + ....... + x( -2) z2 + x(-1) z + x(0) + x(1)z-1 + x(2)z-2 +
....... + x( M) z - M
x(1)
x(2)
= x( - M) z M + ........ + x( -2) z 2 + x(-1) z + x(0) +
+
+
z
z2
x( M)
...... +
zM
In the above summation, when z = 0, the terms with negative power of z attain infinity and when
z = ¥, the terms with positive power of z attain infinity. Hence X(z) converges for all values of z, except
z = 0 and z = ¥. Therefore, the ROC is entire z-plane, except z = 0 and z = ¥.
Case - iv : Infinite duration, right sided (causal) signal
Let, x(n) = r1n ; n ³ 0
Now, the Z-transform of x(n) is,
+¥
X( z) =
å x(n) z
n = -¥
¥
-n
=
¥
år
n
1
n= 0
z- n =
å dr
1
n= 0
z -1
i
n
7. 5
Signals & Systems
¥
If, 0 < |r1 z -1 | < 1, then
å (r z
1
-1 n
)
n=0
=
1
1 - r1 z -1
Using infinite geometric
series sum formula
¥
1
1
C
n=0
if , 0 < |C | < 1
åC
1
1
1
z
\ X(z) =
=
=
=
r1
z - r1
1 - r1 z -1
z
r1
1z
z
n
=
Here the condition to be satisfied for the convergence of X(z) is,
0 < |r1 z–1| < 1
jv
\ |r1 z–1| < 1
| r1 |
< 1
| z|
Þ
z-plane
r1
|z| > |r1|
u
ROC of
The term |r1| represent a circle of radius r1 in z-plane as shown in
x(n) = r1n ; n ³ 0
fig 7.2. From the above analysis we can say that, X(z) converges for all
points external to the circle of radius r1 in z-plane. Therefore, the ROC of Fig 7.2 : ROC of infinite duration
right sided signal.
X(z) is exterior of the circle of radius r1 in z-plane as shown in fig 7.2.
Case v : Infinite duration, left sided (anticausal) signal
Let, x(n) = r2n ; n £ 0
Now, the Z-transform of x(n) is,
+¥
X( z)
+¥
0
å
=
x(n) z- n =
n = -¥
å
n = -¥
r2n z- n =
å
+¥
r2- n zn =
n= 0
1
If , 0 < |r z| < 1, then
(r z) =
r2-1 z
1
n=0
r2
1
1
1
\ X(z) =
=
=
=
-1
z
r
z
r
1 - r2 z
2
2 - z
1r2
r2
r2
= z - r2
å
-1
2
n
Here the condition to be satisfied for the convergence of X(z) is,
0 < |r2–1 z–1| < 1
\ |r2–1 z| < 1
| z|
< 1
| r2 |
Þ
-1
2
z) n
n= 0
¥
-1
2
å (r
Using infinite geometric
series sum formula
¥
1
1
C
n=0
if , 0 < |C | < 1
åC
n
jv
=
z-plane
r2
|z| < |r2 |
The term |r2| represent a circle of radius r2 in z-plane as
shown in fig 7.3. From the above analysis we can say that
X(z) converges for all points internal to the circle of radius r2
in z-plane. Therefore, the ROC of X(z) is interior of the circle
of radius r2 as shown in fig 7.3.
ROC of
x(n) = r2n ; n £ 0
u
Fig 7.3 : ROC of infinite duration left
sided signal.
Chapter 7 - Z - Transform
7. 6
Case vi: Infinite duration, two sided (anticausal) signal
Let, x(n) = r1n u(n) + r2n u(– n)
Now, the Z-transform of x(n) is,
+¥
X( z) =
å
n = -¥
å
å
å
Infinite geometric
series sum formula
¥
1
Cn =
1
C
n=0
if , 0 < |C | < 1
r1n z- n
n=0
å
+¥
r2- n z n +
n=0
+¥
=
å
r2n z - n +
n = -¥
+¥
=
+¥
0
x(n) z - n =
å
r1n z - n
n= 0
+¥
-1
2
(r
n
z) +
n=0
å
(r1 z-1 ) n
n= 0
1
1
=
+
-1
1 - r2 z
1 - r1 z -1
Using infinite geometric series sum formula
if, 0 < |r2-1 z| < 1, and, 0 < |r1 z-1 | < 1
¥
The term
¥
å
(r2- 1 z) n converges if,
å
The term
n= 0
n= 0
0 < r2- 1 z < 1
\
(r1 z -1 ) converges if,
0 < r1- 1 z < 1
r2- 1 z < 1
z
r2
< 1 Þ
r1
< 1
z
|z | < |r2 |
The term |r2| represents a circle of radius r2
and |r1| represents a circle of radius r1 in z-plane. If
|r2| > |r1| then there will be a region between two
circles as shown in fig 7.4. Now the X(z) will
converge for all points in the region between two
circles (because the first term of X(z) converges
for |z| < |r2| and the second term of X(z) converges
for |z| > |r1|). Hence the ROC is the region between
two circles of radius r1 and r2 as shown in fig 7.4.
jv
Þ
|z | > |r1 |
z-plane
r2
r1
ROC of
x(n) = r1n u(n) + r2n u(–n)
if |r2|> |r1|
u
Table 7.1 Summary of ROC of Discrete Time Signals
Sequence
r1 z -1 < 1
\
Fig 7.4 : ROC of infinite duration two
sided signal..
ROC
Finite, right sided (causal)
Entire z-plane except z = 0
Finite, left sided (anticausal)
Entire z-plane except z = ¥
Finite, two sided (noncausal)
Entire z-plane except z = 0 and z = ¥
Infinite, right sided (causal)
Exterior of circle of radius r1, where |z| > r1
Infinite, left sided (anticausal)
Interior of circle of radius r2, where |z| < r2
Infinite, two sided (noncausal) The area between two circles of radius r2 and r1
where, r2 > r1, and r1< |z| < r2, (i.e., |z| >r1, and, |z| < r2)
7. 7
Signals & Systems
Table-7.2 : Characteristic Families of Signals and Corresponding ROC
Signal
ROC in z-plane
Finite Duration Signals
x (n)
Entire z-plane
except z = 0
jv
Right sided
(or causal)
z-plane
u
0
n
jv
x (n)
Left sided
(or anticausal)
Entire z-plane
except z = ¥
z-plane
u
0
n
jv
x (n)
Two-sided
(or noncausal)
Entire z-plane
except z = 0
and z = ¥
z-plane
u
0
n
Infinite Duration Signals
x (n)
jv
Right sided
(or causal)
z-plane
r1
u
0
|z| > r1
n
jv
z-plane
Left sided x (n)
(or anticausal)
u
r2
0
n
jv
Two-sided
(or noncausal)
|z| < r2
z-plane
r1 < |z| < r2
[|z| > r1
and |z| < r2]
x (n)
r1
r2
0
n
u
Chapter 7 - Z - Transform
7. 8
Example 7.1
Determine the Z-transform and their ROC of the following discrete time signals.
a) x(n) = {3, 2, 5, 7}
­
b) x(n) = {6, 4, 5, 3}
­
c) x(n) = {2, 4, 5, 7, 3}
­
Solution
a) Given that, x(n) = {3, 2, 5, 7}
­
i.e., x(0) = 3 ; x(1) = 2 ; x(2) = 5 ; x(3) = 7 ; and x(n) = 0 for n < 0 and for n > 3.
By the definition of Z-transform,
¥
Z {x(n)} = X (z) =
å x(n) z
-n
n = -¥
The given sequence is a finite duration sequence defined in the range n = 0 to 3, hence the limits of summation is
changed to n = 0 to n = 3.
3
\ X( z ) =
å x(n) z
-n
n = 0
= x(0) z0 + x(1) z -1 + x(2) z -2 + x(3) z -3
= 3 + 2z -1 + 5z -2 + 7z -3
= 3 +
2
5
7
+ 2 + 3
z
z
z
In X(z), when z = 0, except the first terms all other terms will become infinite. Hence X(z) will be finite for all values of
z, except z = 0. Therefore, the ROC is entire z-plane except z = 0.
b) Given that, x(n) = {6, 4, 5, 3}
­
i.e, x(–3) = 6 ; x(–2) = 4 ; x(–1) = 5 ; x(0) = 3 ; and x(n) = 0 for n < –3 and for n > 0.
By the definition of Z-transform,
¥
Z {x(n)} = X (z) =
å x(n) z
-n
n = -¥
The given sequence is a finite duration sequence defined in the range n = –3 to 0, hence the limits of summation is
changed to n = –3 to 0.
0
\ X (z) =
å x(n) z
-n
n = -3
= x( -3) z 3 + x( -2) z2 + x(-1) z + x(0)
= 6z 3 + 4 z 2 + 5z + 3
In X(z), when z = ¥, except the last term all other terms become infinite. Hence X(z) will be finite for all values of z,
except z = ¥. Therefore, the ROC is entire z-plane except z = ¥.
c) Given that, x(n) = {2, 4, 5, 7, 3}
A
i.e, x(–2) = 2 ; x(–1) = 4
; x(0) = 5 ; x(1) = 7 ; x(2) = 3 and x(n) = 0 for n < –2 and for n > 2.
By the definition of Z-transform,
¥
Z{x(n)} = X(z ) =
å x(n) z
n = -¥
-n
7. 9
Signals & Systems
The given sequence is a finite duration sequence defined in the range n = –2 to +2, hence the limits of summation is
changed to n = –2 to n = 2.
2
\ X (z) =
å x(n) z
-n
n = -2
= x( -2) z2 + x( -1) z1 + x(0 ) z 0 + x(1) z -1 + x(2) z -2
= 2z2 + 4z + 5 + 7z -1 + 3z -2
7
3
+ 2
z
z
In X(z), when z = 0, the terms with negative power of z will become infinite and when z = ¥, the terms with positive
power of z will become infinite. Hence X(z) will be finite for all values of z except when z = 0 and z = ¥.Therefore, the ROC
is entire z-plane except z = 0 and z = ¥.
= 2z2 + 4z + 5 +
Example 7.2
Determine the Z-transform and their ROC of the following discrete time signals.
b) x(n) = 0.5n u(n)
a) x(n) = u(n)
c) x(n) = 0.2n u(–n –1)
d) x(n) = 0.5n u(n) + 0.8n u(–n – 1)
Solution
a) Given that, x(n) = u(n)
The u(n) is a discrete unit step signal, which is defined as,
u(n) = 1 ; for n ³ 0
Infinite geometric series sum formula
¥
= 0 ; for n < 0
å
By the definition of Z-transform,
n = 0
¥
å x(n) z
Z{x(n)} = X(z) =
-n
¥
å u(n) z
=
n = -¥
¥
åz
=
-n
1
1- C
;
if, 0 < |C| < 1
-n
n = 0
¥
å
=
(z -1)n =
n = 0
n = 0
Cn =
1
1 - z -1
Using infinite geometric series sum formula
1
z
=
=
1- 1 / z
z -1
Here the condition for convergence is, 0 < |z–1| < 1.
\ |z -1| < 1
1
< 1
|z|
Þ
Þ
|z| > 1
The term |z| = 1 represents a circle of unit radius in z-plane. Therefore, the ROC is exterior of unit circle in z-plane.
b) Given that, x(n) = 0.5n u(n)
The u(n) is a discrete unit step signal, which is defined as,
; for
n³0
; for
n<0
\ x(n) = 0.5 ; for
n³0
u(n) = 1
=0
n
= 0
; for n < 0
By the definition of Z-transform,
¥
Z{x(n)} = X (z) =
å x(n) z
n = -¥
¥
=
å d0.5z
n = 0
-n
¥
=
å 0.5
n
z -n
n = 0
-1 n
i
=
1
1 - 0.5z -1
Using infinite geometric series sum formula
Chapter 7 - Z - Transform
\ X(z) =
7. 10
1
=
1
1 - 0.5
z
z
z - 0.5
Here the condition for convergence is, 0 < |0.5 z–1| < 1.
\ |0.5 z -1| < 1
0.5
< 1
|z|
Þ
Þ
|z| > 0.5
The term |z| = 0.5 represents a circle of radius 0.5 in z-plane. Therefore, the ROC is exterior of circle with radius 0.5
in z-plane.
c) Given that, x(n) = 0.8n u(–n –1)
The u(–n –1) is a discrete unit step signal, which is defined as,
u(–n – 1) = 0
=1
\
x(n) = 0
; for n ³ 0
; for n £ –1
; for n ³ 0
= 0.8 n ; for n £ –1
By the definition of Z-transform,
¥
Z{x(n)} = X(z) =
å x(n) z
n = -¥
¥
=
å 0.8
-n
-n
zn =
n = 1
=
=
-1
=
å 0.8
n = -¥
¥
å (0.8
n
-1
n = 1
z -n
¥
z)n =
å (0.8
-1
z)n - 1
n = 0
1
-1
1 - (0.8 -1 z)
1
z
1 0.8
-1 =
(0.8–1 z)0 = 1
Using infinite geometric series
sum formula
z
0.8
0.8 - 0.8 + z
z
=
= -1 =
0.8 - z
0.8 - z
0.8 - z
z - 0.8
Here the condition for convergence is, 0 < |0.8–1 z| < 1.
|z|
< 1 Þ |z| < 0.8
\ |0.8 -1 z| < 1 Þ
0.8
The term |z| = 0.8, represents a circle of radius 0.8 in z-plane. Therefore, the ROC is interior of the circle of radius
0.8 in z-plane.
d) Given that, x(n) = 0.5n u(n) + 0.8n u(–n – 1)
X(z) = Z {x(n)} = Z{0.5n u(n) + 0.8n u(–n –1)}
= Z{0.5n u(n)} + Z{0.8n u( -n - 1)}
z
z
=
z - 0.5
z - 0.8
=
z(z - 0.8) - z(z - 0.5)
z 2 - 0.8z - z2 + 0.5z
= 2
=
(z - 0.5) (z - 0.8)
z - 0.8z - 0.5z + 0.4
Using linearity property
Using the results of (b) and (c)
-0.3z
z 2 - 1.3z + 0.4
Here the condition for convergence of 0.5n u(n) is,
0 < |0.5 z–1| < 1 Þ
|z| > 0.5
and the condition for convergence of 0.8n u(–n –1) is,
0 < |0.8–1 z| < 1 Þ
|z| < 0.8
The term |z| = 0.8, represents a circle of radius 0.8 in z-plane and the term |z| = 0.5 represents a circle of radius 0.5
in z-plane. Hence the common region of convergence for both the terms of x(n) is the region in between the circles of radius
|z| = 0.8 and |z| = 0.5 in the z-plane.
7. 11
Signals & Systems
7.3 Properties of Z-Transform
1. Linearity property
The linearity property of Z-transform states that the Z-transform of linear weighted combination of
discrete time signals is equal to similar linear weighted combination of Z-transform of individual discrete
time signals.
Let, Z{x1(n)} = X1(z) and Z{x2(n)} = X2(z) then by linearity property,
Z{a1x1(n) + a2x2(n)} = a1X1(z) + a2X2(z)
;
where, a1 and a2 are constants.
Proof :
By definition of Z-transform,
+¥
å x (n) z
X 1( z ) = Z {x 1(n)} =
-n
.....(7.8)
1
n = -¥
+¥
å x (n) z
X 2 ( z ) = Z {x 2 (n)} =
-n
.....(7.9)
2
n = -¥
+¥
+¥
a1x 1(n) + a2 x 2 (n) z -n =
å
Z {a 1x1 (n) + a2 x 2 (n)} =
n = -¥
n = -¥
+¥
+¥
å
=
a1x 1(n) z -n + a2 x 2 (n)] z -n
å
a1x 1 (n) z -n +
n = -¥
+¥
å a x (n)z
2 2
n = -¥
-n
= a1
å
+¥
n = -¥
= a1 X 1 (z ) + a2 X 2 ( z )
å x (n)z
x 1 (n) z -n + a2
-n
2
n = -¥
Using equations (7.8) and (7.9)
2. Shifting property
Case i: Two sided Z-transform
The shifting property of Z-transform states that, Z-transform of a shifted signal shifted by m-units
of time is obtained by multiplying zm to Z-transform of unshifted signal.
Let, Z{x(n)} = X(z)
Now, by shifting property,
Z{x(n–m)} = z–m X(z)
Z{x(n+m)} = zm X(z)
Proof :
By definition of Z-transform,
+¥
å x(n) z
X (z ) = Z {x(n)} =
-n
.....(7.10)
n = -¥
+¥
å x(n - m) z
Z {x(n - m)} =
-n
Let, n – m = p, \ n = p + m
when n ® -¥, p ® -¥
when n ® +¥, p ® +¥
n = -¥
+¥
=
å x(p) z
- (m + p)
p = -¥
+¥
=
å x(p) z
p = -¥
= z -m
-m
z -p
+¥
å x(p) z
p = -¥
= z - m X(z)
+¥
-p
= z -m
å x(n) z
-n
Let, p ® n
n = -¥
Using equation (7.10)
Chapter 7 - Z - Transform
7. 12
+¥
å x(n + m) z
Z{x(n + m)} =
-m
n = -¥
+¥
å x(p) z
=
Let, n + m = p, \ n = p – m
when n ® -¥, p ® -¥
when n ® +¥, p ® +¥
- (p - m)
p = -¥
+¥
å x(p) z
=
-p
zm
p = -¥
+¥
m
+¥
å x(p) z
= z
-p
å x(n) z
= zm
p = -¥
-n
Let, p ® n
n = -¥
= z m X(z)
Using equation (7.10)
Case ii: One sided Z-transform
Let x(n) be a discrete time signal defined in the range 0 < n < ¥.
Let, Z{x(n)} = X(z)
Now by shifting property,
l
m
q
å
Z x(n - m) = z - m X(z) +
l
x(-i) z- ( m- i)
i=1
m-1
q
å
Z x(n + m) = z m X(z) -
x(i) z( m- i)
i=0
Proof :
By definition of one sided Z-transform,
+¥
X (z ) = Z {x(n)} =
å x(n) z
-n
.....(7.11)
n= 0
+¥
å
Z {x(n - m)} =
x(n - m) z - n
n= 0
+¥
=
å
Multiply by zm and z –m
x(n - m) z -n z m z -m
n= 0
+¥
= z -m
å
x(n - m) z - (n - m)
0
n =+¥
= z -m
Let, n – m = p,
when n ® 0, p ® –m
when n ® +¥, p ® +¥
x(p) z - p
å
p = -m
-1
+¥
= z -m
å
x(p) z -p + z - m
p= 0
å
m
x(p) z -p + z - m
p= 0
å
å
m
x(n) z -n + z - m
n= 0
å
i= 1
m
m
= z X(z) +
x( -p) zp
p= 1
+¥
= z -m
x(p) z - p
p = -m
+¥
= z -m
å
å
x( -i) z
-( m - i)
x( -i) zi
Let p = n, in first summation
Let p = i, in second summation
Using equation (7.11)
.....(7.12)
i= 1
Note : In equation (7.12) if x(–i) for i = 1 to m are zero then the shifting property for delayed
signal will be same as that for two sided Z-transform.
7. 13
Signals & Systems
+¥
å
Z {x(n + m)} =
x(n + m) z - n
n= 0
+¥
å
=
Multiply by zm and z –m
x(n + m) z -n z m z - m
n= 0
+¥
= zm
å
x(n + m) z -(n + m)
Let, n + m = p,
when n ® 0, p ® m
when n ® + ¥, p ® + ¥
n= 0
+¥
= z
m
å
x(p) z
-p
p=m
m -1
+¥
= zm
å
x(p) z - p - z m
p=0
x(p) z - p
m -1
+¥
= zm
å
p=0
x(n) z -n - z m
å
n=0
å
Let p = n, in first summation
Let p = i, in second summation
x(i) z -i
i=0
m -1
= z m X(z) -
å
Using equation (7.11)
x(i) z m - i
.....(7.13)
i=0
Note : In equation (7.13) if x(i) for i = 0 to m –1 are zero then the shifting property for advanced
signal will be same as that for two sided Z-transform.
3. Multiplication by n (or Differentiation in z-domain)
If Z{x(n)} = X(z)
l
q=
then Z nx(n)
In general,
n
d
X(z)
dz
-z
s FGH -z dzd IJK X(z)
d F
d F
d
d F F
= -z
- z G ..... G -z
-z
G
G
H
H
dz H dz H
dz
dz
m
Z n m x(n) =
144444424444443
IJ IJ .....
KK
X(z)
IJ I
K JK
m - times
Proof :
By definition of Z-transform,
+¥
å x(n) z
X (z ) = Z {x(n)} =
+¥
Z {n x(n)} =
-n
.....(7.14)
n = -¥
n x(n) z - n
å
n = -¥
+¥
=
å
Multiply by z and z –1
n x(n) z - n z z -1
n = -¥
+¥
= -z
å
x(n) -n z - n - 1
n = -¥
+¥
= -z
å
x(n)
n = -¥
= -z
= -z
d
dz
LM d z OP
N dz Q
-n
d -n
z = - n z -n - 1
dz
+¥
å x(n) z
n = -¥
d
X(z)
dz
-n
Interchanging summation
and differentiation
Using equation (7.14)
Chapter 7 - Z - Transform
7. 14
4. Multiplication by an exponential sequence, an (or Scaling in z-domain)
If
Z{x(n)} = X(z)
n
s
then Z a n x(n) = X(a -1z)
Proof :
By definition of Z-transform,
+¥
å x(n) z
Z{x(n)} =
-n
.....(7.15)
n = -¥
+¥
Z {an x(n)} =
an x(n) z - n
å
n = -¥
+¥
å
=
x(n) (a -1z)- n
.....(7.16)
n = -¥
The equation (7.16) is similar
to the form of equation (7.15)
= X(a -1z)
5. Time reversal
If
Z{x(n)} = X(z)
then Z{x(–n)} = X(z–1)
Proof :
By definition of Z-transform,
+¥
å x(n) z
Z{x(n)} =
-n
.....(7.17)
Let, p = –n
when n ® -¥,
when n ® +¥,
n = -¥
+¥
å
Z {x( -n)} =
x( -n) z - n
n = -¥
p ® +¥
p ® -¥
+¥
x(p) z p
å
=
p = -¥
+¥
x(p) (z -1 )- p
å
=
.....(7.18)
p = -¥
The equation (7.18) is similar to
the form of equation (7.17)
= X(z -1 )
6. Conjugation
If
Z{x(n)} = X(z)
then Z{x*(n)} = X*(z*)
Proof :
By definition of Z-transform,
+¥
X(z) = Z {x(n)} =
å
-n
.....(7.19)
n = -¥
+¥
Z {x * (n)} =
å x(n) z
x* (n) z -n
n = -¥
LM
MN å
+¥
=
x(n) (z * )- n
n = -¥
= X(z * )
= X * (z * )
*
OP
PQ
*
.....(7.20)
The equation (7.20) is similar
to the form of equation (7.19)
7. 15
Signals & Systems
7. Convolution theorem
If
Z{x1(n)} = X1(z)
and Z{x2(n)} = X2(z)
then Z{x1(n) * x2(n)} = X1(z) X2(z)
+¥
å
where, x1 (n) * x2 (n) =
x1 (m) x2 (n - m)
.....(7.21)
m = -¥
Proof :
By definition of Z-transform,
+¥
å x (n) z
X 1(z) = Z {x 1 (n)} =
-n
.....(7.22)
1
n = -¥
+¥
å x (n) z
X 2 (z) = Z {x 2 (n)} =
-n
.....(7.23)
2
n = -¥
+¥
å
Z {x 1(n) * x2 (n)} =
x 1(n) * x 2 (n) z -n
n = -¥
LM
MN å
+¥
n = -¥
+¥
x 1(m) x 2 (n - m) z -n
Using equation (7.21)
x 1(m) x 2 (n - m) z -n z -m zm
Multiply by zm and z –m
m = -¥
+¥
å
å
n = -¥
m = -¥
=
OP
PQ
+¥
å
=
+¥
+¥
å
=
x 1(m) z -m
m = -¥
Let, n – m = p
when n ® -¥, p ® -¥
when n ® +¥, p ® +¥
+¥
å
x 1(m) z
-m
m = -¥
LM
MN å
å
x 2 (p) z
-p
p = -¥
OP LM
PQ MN å
+¥
+¥
=
x 2 (n - m) z - (n - m)
n = -¥
+¥
=
å
x 1(n) z -n
n = -¥
x 2 (n) z -n
n = -¥
= X 1 (z ) X 2 ( z )
OP
PQ
Let m = n, in first summation
Let p = n, in second summation
Using equations (7.22) and (7.23)
8. Correlation property
If
Z{x(n)} = X(z) and
Z{y(n)} = Y(z)
–1
then Z{rxy(m)} = X(z) Y(z )
+¥
where, r (m) =
xy
å
x(n) y(n - m)
.....(7.24)
n = -¥
Proof :
By definition of Z-transform,
+¥
X(z) = Z {x(n)} =
å x(n) z
-n
.....(7.25)
-n
.....(7.26)
n = -¥
+¥
Y(z) = Z {y(n)} =
å y(n) z
n = -¥
Chapter 7 - Z - Transform
7. 16
+¥
å
Z { rxy (m)} =
rxy (m) z -m
m = -¥
m = -¥
LM
MN å
+¥
+¥
+¥
=
=
OP
PQ
+¥
å
x(n) y(n - m) z - m
n = -¥
å å
Using equation (7.24)
x(n) y(n - m) z -m z -n z n
Multiply by z n and z –n
m = -¥ n = -¥
+¥
=
=
å
+¥
x(n) z -n
n = -¥
m = -¥
+¥
+¥
å
LM
MN å
å
x(n) z -n
n = -¥
Let, n – m = p
\m=n–p
when m ® -¥, p ® +¥,
when m ® +¥, p ® -¥.
y(p ) z p
p = -¥
+¥
=
y(n - m) z (n - m)
å
OP LM
PQ MN å
+¥
x(n) z -n
n = -¥
y(p ) (z -1 ) -p
p = -¥
OP
PQ
Using equations (7.25)
and (7.26)
-1
= X(z) Y(z )
9. Initial value theorem
Let x(n) be an one sided signal defined in the range 0 £ n £ ¥.
Now, if Z{x(n)} = X(z),
then the initial value of x(n) (i.e., x(0)) is given by,
x(0) =
Lt X(z)
z®¥
Proof :
By definition of one sided Z - transfrom,
+¥
X (z ) =
å x(n) z
-n
n=0
On expanding the above summation we get,
X ( z ) = x(0) + x(1) z -1 + x(2) z -2 + x(3) z -3 + ......
\ X (z ) = x(0) +
x(1)
x(2)
x(3)
+ 2 + 3 + ......
z
z
z
On taking limit z ® ¥ in the above equation we get,
Lt
z®¥
X( z ) =
Lt
z®¥
LMx(0) +
N
= x(0) + 0 + 0 + 0 + ......
\ x(0) =
Lt
z®¥
X (z )
OP
Q
x(1)
x(2)
x(3)
+ 2 + 3 + ......
z
z
z
7. 17
Signals & Systems
10. Final value theorem
Let x(n) be an one sided signal defined in the range 0 £ n £ ¥.
Now, if Z {x(n)} = X(z),
then the final value of x(n) (i.e., x(¥)) is given by,
Lt (1 - z -1 ) X(z) or
x( ¥ ) =
z® 1
x( ¥ ) =
Lt
z® 1
FG z - 1IJ X(z)
H zK
Proof :
By definition of one sided Z - transfrom,
+¥
m r å x(n) z
Z x(n) =
-n
.....(7.27)
n=0
+¥
m
r å
\ Z x(n - 1) - x(n) =
(RHS)
n=0
x(n - 1) - x(n) z -n
(LHS)
m
r
= Z mx(n - 1)r - Z mx(n)r
RHS = Z x(n - 1) - x(n)
= z
-1
X( z) + x( -1) - X(z)
Using linearity property
Using shifting property and equation 7.27
= x( -1) - (1 - z -1 ) X(z)
=
Lt
z®1
x( -1) - (1 - z -1 ) X(z)
Taking limit z ® 1
= x( -1) - z Lt
(1 - z -1 ) X(z)
®1
.....(7.28)
+¥
LHS =
å
x(n - 1) - x(n) z -n
n=0
+¥
=
Lt
z®1
x(n - 1) - x(n) z - n
å
Taking limit z ® 1
n=0
+¥
=
å
x(n - 1) - x(n)
n=0
p
=
=
=
å
Lt
p®¥
x(n - 1) - x(n)
n=0
On applying limit z ® 1, the term z–n
becomes unity
Changing the summation index from
0 to p and then taking limit p ® ¥
Lt
LM x( -1) - x(0) + x(0) - x(1) + x(1) - x(2) + ..... OP
MN ..... + x(p - 2) - x(p - 1) + x(p - 1) - x(p) PQ
Lt
x( -1) - x(p)
p®¥
p®¥
= x( -1) - x(¥)
.....(7.29)
On equating equation (7.29) with (7.28) we get,
x( - 1) - x( ¥) = x( -1) - Lt (1 - z -1 ) X(z)
z®1
\ x( ¥) =
Lt (1 - z -1 ) X(z)
z®1
Chapter 7 - Z - Transform
7. 18
11. Complex convolution theorem (or Multiplication in time domain)
Let, Z {x1(n)} = X1(z)
and Z {x2(n)} = X2(z).
Now, the complex convolution theorem states that,
l
z
1
2 pj
q
Z x1 ( n) x 2 ( n) =
X1 ( v) X 2 z v -1 dv
v
ej
C
where, v is a dummy variable used for contour integration
Proof :
Let, Z {x1(n)} = X1(z)
and Z {x2(n)} = X2(z).
Now, by definition of inverse Z-transform,
x 1 (n) =
1
2 pj
z
1
2pj
X 1( z) zn - 1 dz =
C
z
X 1 ( v) vn - 1 dv
let, z = v
.....(7.30)
C
Now, by definition of Z-transform,
+¥
X 2 (z ) =
å x (n) z
-n
.....(7.31)
2
n = -¥
Using the definition of Z-transform, the Z {x1(n) x2(n)} can be written as,
m
+¥
r
Z x 1(n) x 2 (n) =
å x (n) x (n) z
1
-n
2
n = -¥
LM 1
MN 2pj
+¥
=
å
n = -¥
=
`
=
=
1
2 pj
1
2pj
1
2 pj
z
z
z
C
OP
PQ
X 1( v ) v n - 1 dv x 2 (n) z - n
C
Using equation (7.30)
+¥
X 1 ( v)
å
x2 (n) z -n v n v -1 dv
n = -¥
C
C
z
LM
F zI O
MN å x (n) GH v JK PPQ v
X ( v ) X F z I v dv
H vK
-n
+¥
X 1( v)
2
-1
dv
n = -¥
-1
1
2
12. Parseval’s relation
If Z {x1(n)} = X1(z)
and Z {x2(n)} = X2(z).
Then the Parseval’s relation states that,
+¥
å
n = -¥
x1 (n) x*2 (n) =
1
2 pj
z
C
X1 (z) X*2 1* z -1 dz
z
e j
Using equation (7.31)
7. 19
Signals & Systems
Proof :
Let, Z {x1(n)} = X1(z)
and Z {x2(n)} = X2(z).
Now, by definition of inverse Z-transform,
x 1 (n) =
1
2pj
z
z
1
2 pj
X 1(z ) zn - 1 dz =
C
X 1( v) vn - 1 dv
let, z = v
.....(7.32)
C
Now, by definition of Z-transform,
+¥
m r
å x (n) z
Z x 2 (n) =
-n
.....(7.33)
2
n = -¥
o
t
Using the definition of Z - transform, the Z x 1(n) x*2 (n) can be written as,
+¥
o
t
å x (n) x (n) z
Z x 1 (n) x*2 (n) =
*
2
1
-n
.....(7.34)
n = -¥
On substituting for x 1(n) from equation (7.32) in equation (7.34) we can write,
+¥
+¥
å x (n) x (n) z
1
*
2
-n
=
n = -¥
å
n = -¥
=
=
=
1
2 pj
1
2 pj
1
2 pj
1
=
2 pj
LM 1
MN 2pj
z
OP
PQ
X 1( v) vn - 1 dv x*2 (n) z - n
C
LM x (n) z v OP v dv
MN å
PQ
L
F zI O
X ( v) M å x (n) G J P v dv
H v K PQ
MN
L
Fz I O
X ( v) M å x (n) G J P v dv
MN
H v K PQ
F I
X ( v) X G z J v dv
Hv K
z
z
z
z
+¥
-n
*
2
X 1 ( v)
n
-1
n = -¥
C
-n
+¥
-1
*
2
1
n = -¥
C
+¥
-n
*
*
-1
1
2
*
n = -¥
C
C
*
*
2
1
-1
*
using equation (7.33)
Let us take limit z ® 1 in the above equation,
+¥
\
lt
Z®1
å
x 1 (n) x*2 (n) z - n =
n = -¥
+¥
å
x1 (n) x2* (n) =
n = -¥
+¥
\
å
n = -¥
x 1(n) x*2 (n) =
lt
Z®1
1
2 pj
1
2 pj
z
z
C
C
1
2pj
z
C
FG IJ
H K
*
X 1( v ) X *2 z * v -1 dv
v
FG 1 IJ v
Hv K
F I
X (z ) X G 1 J z
Hv K
-1
X 1( v) X 2*
*
*
2
*
1
-1
dv
dz
let v = z
Chapter 7 - Z - Transform
7. 20
Table-7.3 : Summary of Properties of Z-Transform
Note : X(z) = Z{x(n)} ; X1(z) = Z{x1(n)} ; X2(z) = Z{x2(n)} ; Y(z) = Z {y(n)}
Property
Discrete time signal
Linearity
Z-transform
a1 x1(n) + a2 x2(n)
a1 X1(z) + a2 X2(z)
x(n – m)
z - m X(z) +
m
å x(-i) z
- ( m- i )
i=1
m-1
Shifting x(n) for n ³ 0
å x(i) z
z m X(z) -
x(n + m)
m- i
i= 0
(m ³ 0)
x(n - m)
z-m X(z)
x(n + m)
z
nm x(n)
FG -z d IJ
H dz K
Scaling in z-domain
(or multiplication by an)
an x(n)
X(a-1 z)
Time reversal
x(-n)
X(z-1)
Conjugation
x *(n)
X *(z * )
Convolution
x1(n) * x2(n) =
x(n) for all n
Multiplication by nm
(or differentiation in
z-domain)
m
X(z)
m
X(z)
+¥
å x (m) x (n - m)
1
2
X1(z) X2(z)
m = -¥
+¥
Corrrelation
å x(n) y(n - m)
rxy ( m) =
X(z) Y(z–1 )
m = -¥
Initial value
x(0) = Lt X(z)
z®¥
x( ¥ ) =
Final value
Complex convolution
theorem
Lt (1 - z -1 ) X(z)
z® 1
( z - 1)
X(z)
z® 1
z
if X(z) is analytic for |z| > 1
= Lt
1
2pj
x1(n) x2(n)
+¥
Parseval’s relation
1
å x (n) x (n) = 2pj
1
n = -¥
*
2
z
C
F I
H K
z
C
X1 ( z) X*2 1* z -1 dz
z
X1 ( v) X 2 z v -1 dv
v
ej
7. 21
Signals & Systems
Example 7.3
Find the one sided Z-transform of the following discrete time signals.
a) x(n) = n a(n – 1)
b) x(n) = n2
Solution
a) Given that, x(n) = n a(n – 1)
Let, x1(n) = an
By definition of one sided Z-transform,
¥
å x (n) z
X 1(z) =
-n
1
n = 0
¥
¥
åa
=
n
-1 n
i
n = 0
n = 0
=
å da z
z -n =
1
z
=
1 - a z -1
z - a
Using infinite geometric
series sum formula
Let, x1(n – 1) = an – 1
By shifting property,
Z{x1(n - 1)} = z -1 X1(z) = z -1
z
1
=
z - a
z - a
Given that, x(n) = n a n - 1
If Z{x(n)} = X(z)
Z{x(n)} = Z{n a n - 1} = Z{n x1 (n - 1)} = - z
= -z
d
X1(z)
dz
then Z{n x(n)} = - z
d
X(z)
dz
-1
d
1
z
= -z ´
=
dz z - a
(z - a)2
(z - a)2
b) Given that, x(n) = n2
Let us multiply the given discrete time signal by a discrete unit step signal,
\ x(n) = n2 u(n)
Note : Multiplying a one sided sequence by u(n) will not alter its value.
By the property of Z-transform, we get,
Z {nmu(n)}
=
FG -z d IJ
H dz K
m
U(z)
where, U(z) = Z {u(n)} =
z
z - 1
LM d FG z IJ OP = - z LM z - 1 - z OP = z
N dz H z - 1K Q
N (z - 1) Q (z - 1)
FG -z d IJ U(z) = - z d LM-z d U(z)OP
H dz K
dz
dz N
Q
F (z - 1) - z ´ 2(z - 1) I
I
d F
z
= -z G
= -z
JK
dz GH (z - 1) JK
(z - 1)
H
F (z - 1) (z - 1 - 2z) I = - z F -(z + 1) I = z(z + 1)
= -z G
JK
GH (z - 1) JK (z - 1)
H (z - 1)
F d IJ U(z) = z(z + 1)
\
Z {x(n)} = Z {n u(n)} = G -z
H dz K
(z - 1)
\ -z
d
U(z) = - z
dz
2
2
2
2
2
4
4
3
2
2
3
3
d
u v du - u dv
=
v
v2
Chapter 7 - Z - Transform
7. 22
Example 7.4
Find the one sided Z-transform of the discrete time signals generated by mathematically sampling the following
continuous time signals.
c) cos Wot
b) sinWot
a) t2
Solution
a) Given that, x(t) = t2
The discrete time signals is generated by replacing t by nT, where T is the sampling time period.
\ x(n) = (nT)2 = n2 T2 = n2 g(n)
where, g(n) = T2
By the definition of one sided Z-transform we get,
¥
n
G(z) = Z {g(n)} = Z {T 2 } = å T 2 z - n = T 2 å (z -1)n = T 2
n = 0
n = 0
By the property of Z-transform we get,
X(z) = Z{x(n)} = Z{n2 g(n)} =
FG -z d IJ
H dz K
2
d
dz
G(z) = - z
=
-1
T2 z
z - 1
FG -z d G(z)IJ
H dz K
F -z d T z I = - z d F -z ´ (z - 1) T - T z I
JK
GH dz z - 1JK
dz GH
(z - 1)
d F zT I
(z - 1) T - zT ´ 2(z - 1)
-z
= -z ´
dz GH (z - 1) JK
(z - 1)
2
d
dz
= -z
2
2
2
2
2
=
FG 1 IJ
H1 - z K
2
2
2
4
-zT 2 - T 2
(z - 1) (zT 2 - T 2 - 2zT 2 )
z T 2 (z + 1)
=
= -z ´
4
3
(z - 1)
(z - 1)
(z - 1)3
= -z ´
b) Given that, x(t) = sinWot
The discrete time signals is generated by replacing t by nT, where T is the sampling time period.
\ x(n) = sin (W0nT) = sin wn ; where w = W0T
By the definition of one sided Z-transform,
¥
Z {x(n)} = X(z) =
=
¥
å x(n) z
å
=
sin wn ´ z
n = 0
n = 0
¥
1
e jwn - e - jwn -n
z =
2j
2j
å
n = 0
=
-n
¥
1
2j
å de
n = 0
jw
z -1
n
i
-
sin q =
-n
¥
åe
jwn
z -n -
n = 0
1 ¥
e - jw z -1
2j n = 0
åd
1
2j
¥
åe
- jwn
z -n
n = 0
n
i
=
1
1
1
1
2j 1 - e jw z -1
2j 1 - e - jw z -1
=
z
1
z
1
2j z - e - jw
2 j z - e jw
=
z (z - e - jw ) - z (z - e jw )
z 2 - z e - j w - z 2 + z e jw
=
2 j (z - e jw ) (z - e - jw )
2 j (z 2 - z e - jw - z e jw + e jw e - jw )
=
=
z
2
z (e jw - e - jw ) / 2 j
- z (e jw + e - jw ) + 1
z sin wT
z2 - 2z cos wT + 1
;
e jq - e - jq
2j
where w = W 0 T
Using infinite geometric
series sum formula
sin q =
e jq - e - jq
2j
cos q =
e jq + e - jq
2
7. 23
Signals & Systems
c) Given that, x(t) = cos Wot
The discrete time signal is generated by replacing t by nT, where T is the sampling time period.
\ x(n) = cos(W0nT) = cos wn ; where w = W0T
By the definition of one sided Z-transform,
¥
Z {x(n)} = X(z) =
å
n=0
¥
=
e
å
jwn
n = 0
=
1
2
e jq + e - jq
2
¥
å
x(n) z - n =
+ e
2
cos wn ´ z -n
n =0
- jwn
¥
å (e
jw
1
2
z -1)n +
n = 0
¥
1
2
z -n =
cos q =
åe
jwn
1
2
z-n +
n = 0
¥
åe
- jwn -n
z
n = 0
¥
å (e
- jw
z -1)n
n = 0
=
1
1
1
1
+
2 1 - e - jw z -1
2 1 - e jw z -1
=
z
1
z
1
+
2 z - e - jw
2 z - e jw
=
z (z - e - jw ) + z (z - e jw )
z2 - z e - j w + z 2 - z e j w
=
- jw
jw
2
2 (z - e ) (z - e )
2 (z - z e - jw - z e jw + e jw e - jw )
=
2z2 - z (e jw + e - jw )
z 2 - z(e jw + e - jw ) / 2
= 2
- jw
2
jw
2 [z - z (e + e ) + 1]
z - z (e jw + e - jw ) + 1
=
z (z - cos w )
; where w = W 0T
z2 - 2 z cos w + 1
Using infinite geometric
series sum formula
e jq + e - jq
2
cos q =
Example 7.5
Find the one sided Z-transform of the discrete time signals generated by mathematically sampling the following
continuous time signals.
a) e– a t cos Wot
b) e– a t sinWot
Solution
a) Given that, x(t) = e– a t cos Wot
The discrete time signal x(n) is generated by replacing t by nT, where T is the sampling time period.
\ x(n) = e– a n T cos W0nT = e–anT cos wn ; where w = W0T
By the definition of one sided Z-transform we get,
¥
X(z) = Z {x(n)} =
¥
åe
- anT
cos wn z -n =
n=0
=
=
1
2
å
e - anT
n=0
¥
å de
-a T
jw -1 n
e z
i
n= 0
+
1
2
Fe
GH
jwn
+ e - jwn
2
¥
å de
- aT
n=0
1
1
1
1
+
2 1 - e - aT e jw z -1
2 1 - e -aT e - jw z -1
1
1
1
1
+
=
2 1 - e jw / z eaT
2 1 - e - jw / z eaT
=
in
e - jw z -1
1
2
LM z e
Nz e - e
aT
aT
jw
+
z eaT
z eaT - e - jw
OP
Q
Iz
JK
-n
cos q =
e jq + e - jq
2
Using infinite
geometric series
sum formula
¥
åC
n=0
n
=
1
1- C
Chapter 7 - Z - Transform
7. 24
1 L z e dz e - e i + z e dz e - e i O
PP
M
=
2 M
z
e
e
z
e
e
d
i
d
i
Q
N
L
OP
ze
+ ze - e
ze - e
M
=
MMN dz e i - z e e - z e e + e e PPQ
2
OP
2z e - de + e i
ze L
M
=
2 Mz e
- z e de + e i + 1 P
N
Q
LM z e dz e - coswi OP
=
MN z e - 2z e cosw + 1PQ ; where w = W T
aT
- jw
aT
jw
aT
aT
- jw
aT
2
2aT
aT
- jw
aT
jw
aT
aT - jw
aT
jw
aT
aT
aT
aT 2
2
aT
aT jw
jw
- jw
jw
- jw
jw
- jw
aT
2aT
cosq =
0
aT
e jq + e - jq
2
b) Given that, x(t) = e– a t sinWot
The discrete time signal x(n) is generated by replacing t by nT, where T is the sampling time period.
\ x(n) = e– a nT sinW0 n T = e–anT sin wn ; where w = W0T
By the definition of one sided Z-transform we get,
¥
X(z) = Z {x(n)} =
¥
åe
- anT
sin wn z
-n
=
e
n=0
¥
1
2j
å
=
n=0
- anT
å de
- aT
e jw z -1
n
i
Fe
GH
jwn
I
JK
- e - jwn
z -n
2j
FH
IK
1 ¥
- w -1
å e aT e j z
2j n = 0
-
n=0
=
1
1
1
1
2 j 1 - e - aT e jw z -1
2 j 1 - e - aT e - jw z -1
=
1
1
1
1
2 j 1 - e jw / z eaT
2 j 1 - e - jw / z eaT
=
1
1
z eaT
z eaT
jw
aT
aT
2j z e - e
2 j z e - e - jw
=
aT
=
aT
aT 2
2
2aT
aT
jw
aT
aT
=
- jw
aT
å
jw
aT
- jw
aT
- jw
aT
aT - jw
jw
aT
aT
jw
jw
jw
- jw
- jw
jw
- jw
z eaT sin w
z2 e2aT - 2zeaT cos w + 1
;
e jq - e - jq
2j
Infinite geometric series sum
¥
1
Cn =
formula
1
C
n=0
LM z e dz e - e i - z e dz e - e i OP
MN
PQ
dz e - e i dz e - e i
L dz e i [z e - e - z e + e ] OP
1 M
P
2 j M dz e i - z e e
- ze e + e e P
MN
Q
LM z e e - e / 2j OP
MN z e - z e de + e i + 1PQ
1
2j
aT
=
n
sin q =
where w = W0 T
Example 7.6
Find the initial value, x(0) and final value, x(¥) of the following z-domain signals.
1
1
2z -1
a) X(z) =
b)
c) X(z) =
-2
-1
-2
1 - z
1 + 2z - 3z
1 - 1.8z -1 + 0.8z -2
7. 25
Signals & Systems
Solution
a) Given that, X(z) =
1
1 - z -2
By initial value theorem of Z-transform we get,
x(0) =
Lt
z®¥
X(z) =
Lt
z®¥
1
=
1 - z-2
Lt
z®¥
1
=
1
1 - 2
z
1
1
1 ¥
=
1
= 1
1 - 0
By final value theorem of Z-transform we get,
x(¥) =
=
Lt (1 - z -1) X(z) =
z® 1
Lt (1 - z -1)
z® 1
Lt (1 - z -1)
z®1
1
=
(1 - z -1) (1 + z -1)
1
1 - z -2
Lt
z®1
1
1
1
=
=
2
(1 + z -1)
1 + 1-1
1
b) Given that, X(z) =
-1
-2
1 + 2z - 3z
By initial value theorem of Z-transform we get,
x(0) =
=
Lt
z®¥
X(z) =
Lt
z®¥
1
=
1 + 2z -1 - 3z -2
Lt
z®¥
1
2
3
1 +
- 2
z
z
1
1
=
= 1
2
3
1
0
+ 0
+
1 +
¥
¥
By final value theorem of Z-transform we get,
x(¥) =
=
Lt (1 - z -1) X(z) =
z® 1
Lt
z® 1
Lt (1 - z -1)
z® 1
z -1(z - 1)
=
z (z2 + 2z - 3)
-2
Lt
z®1
1
1 + 2z -1 - 3z -2
z(z - 1)
=
(z - 1) (z + 3)
Lt
z®1
z
1
1
=
=
z + 3
1 + 3
4
2 z -1
c) Given that, X(z) =
-2
1 - 1.8z-1 + 0.8z
By initial value theorem of Z-transform we get,
x(0) =
=
Lt
z®¥
X(z) =
Lt
z®¥
2z-1
=
1 - 1.8z -1 + 0.8z -2
Lt
z®¥
2
z
1.8
0.8
1 + 2
z
z
2
0
¥
=
= 0
1.8
0.8
1
0 + 0
1 +
¥
¥
By final value theorem of Z-transform we get,
x(¥) =
=
=
Lt (1 - z-1) X(z) =
z®1
Lt
z®1
Lt
z®1
Lt (1 - z -1)
z® 1
2z-1
1 - 1.8z -1 + 0.8z -2
2(z - 1)
z -1(z - 1) 2z-1
= Lt
z ® 1 (z - 1) (z - 0.8)
z -2 (z 2 - 1.8z + 0.8)
2
2
= 10
=
1 - 0.8
z - 0.8
Chapter 7 - Z - Transform
7. 26
Table - 7.4 : Some Common Z-transform Pairs
X(z)
x(t)
x(n)
With positive
power of z
d(n)
1
Entire z-plane
u(n) or 1
z
z -1
1
1 - z -1
|z| > 1
an u(n)
z
z-a
1
1 - az -1
|z| > |a|
az
|z| > |a|
n a u(n)
( z - a) 2
az -1
(1 - az -1 ) 2
n2 an u(n)
az (z + a)
( z - a) 3
az -1 (1 + az -1 )
(1 - az -1 ) 3
|z| > |a|
- an u(–n–1)
z
z-a
1
1 - az -1
|z| < |a|
–nan u(–n–1)
az
( z - a) 2
az -1
(1 - az -1 ) 2
|z| < |a|
Tz
Tz -1
(1 - z -1 ) 2
|z| > 1
T 2 z -1 (1 + z -1 )
(1 - z -1 ) 3
|z| > 1
z
z - e -aT
1
1 - e - aT z -1
|z| > |e–aT|
z T e - aT
z -1 T e - aT
(1 - e - aT z -1 ) 2
|z| > |e–aT|
t u(t)
nT u(nT)
t2 u(t)
(nT)2 u(nT)
e- at u(t)
e- anT u(nT)
te
- anT
u(t)
nTe
bg
sin W 0 t u t
bg
cosW 0 t u t
ROC
1
n
- at
With negative
power of z
u(nT)
( z - 1) 2
T2 z (z +1)
( z - 1)
(z - e
3
- aT 2
)
sin W 0 nT u(nT)
= sin wn u(nT)
where, w = W0T
z sin w
z - 2z cos w + 1
z -1 sin w
1 - 2z -1 cos w + z -2
cosW 0 nT u(nT)
= cos wn u(nT)
where, w = W0T
z (z - cos w )
z - 2z cos w + 1
1 - z -1 cos w
1 - 2z -1 cos w + z -2
2
2
|z| > 1
|z| > 1
Note : 1. The signals multiplied by u(n) are causal signals (defined for n ³ 0).
2. The signals multiplied by u(–n – 1) are anticausal signals (defined for n £ 0).
7. 27
Signals & Systems
7.4 Poles and Zeros of Rational Function of z
Let, X(z) be Z-transform of x(n). When X(z) is expressed as a ratio of two polynomials in z or z–1,
then X(z) is called a rational function of z.
Let X(z) be expressed as a ratio of two polynomials in z, as shown below.
X( z) =
N(z)
b + b1z -1 + b2 z -2 + b3 z -3 + ..... + b M z- M
= 0
D(z)
a 0 + a1z -1 + a 2 z -2 + a 3 z -3 + ..... + a N z - N
.....(7.35)
where, N(z) = Numerator polynomial of X(z)
D(z) = Denominator polynomial of X(z)
In equation (7.35) let us scale the coefficients of numerator polynomial by b0 and that of denominator
polynomial by a0, and then convert the polynomials to positive power of z as shown below.
b
b
b
b
b 0 1 + 1 z -1 + 2 z -2 + 3 z -3 + ..... + M z - M
b0
b0
b0
b0
X( z) =
a
a
a
a
a 0 1 + 1 z -1 + 2 z -2 + 3 z -3 + ..... + N z - N
a0
a0
a0
a0
FG
H
FG
H
IJ
K
IJ
K
FG
H
FG z
H
z- M z M +
= G
b 1 M -1
b
b
b
z
+ 2 z M - 2 + 3 z M - 3 + ..... M
b0
b0
b0
b0
a 1 N -1
a
a
a
z
+ 2 z N - 2 + 3 z N - 3 + ..... N
a0
a0
a0
a0
(z - z1 ) (z - z2 ) (z - z 3 ) ..... (z - zN )
= G
(z - p1 ) (z - p2 ) (z - p 3 ) ..... (z - p N )
z- N
N
+
IJ
K
IJ
K
Let, M = N
.....(7.36)
where, z1, z2, z3, .....zN are roots of numerator polynomial
p1,p2, p3, .....pN are roots of denominator polynomial
G is a scaling factor.
In equation (7.36) if the value of z is equal to one of the roots of the numerator polynomial, then the
function X(z) will become zero.
Therefore the roots of numerator polynomial z1, z2, z3, .....zN are called zeros of X(z). Hence the
zeros are defined as values z at which the function X(z) become zero.
In equation (7.36) if the value of z is equal to one of the roots of the denominator polynomial then
the funcion X(z) will become infinite. Therefore the roots of denominator polynomial p1, p2, p3, .....pN are
called poles of X(z). Hence the poles are defined as values of z at which the function X(z) become infinite.
Since the function X(z) attains infinte values at poles, the ROC of X(z) does not include poles.
In a realizable system, the number of zeros will be less than or equal to number of poles. Also for
every zero, we can associate one pole (the missing zeros are assumed to exist at infinity).
Let zi be the zero associated with the pole pi. If we evaluate |X(z)| for various values of z, then
|X(z)| will be zero for z = zi and infnite for z = pi. Hence the plot of |X(z)| in a three dimensional plane will
look like a pole (or pillar like structure) and so the point z = pi is called a pole.
Chapter 7 - Z - Transform
7. 28
7.4.1 Representation of Poles and Zeros in z-Plane
The complex variable, z is defined as,
z = u + jv
where, u = Real part of z
v = Imaginary part of z
Hence the z-plane is a complex plane, with u on real axis and v on imaginary axis (Refer fig 7.1 in
section 7.1). In the z-plane, the zeros are marked by small circle " " and the poles are marked by
letter "X".
For example consider a rational function of z shown below.
1.25 - 1.25 z -1 + 0.2 z-2
2 + 2 z -1 + z -2
0.2 -2
1.25 1 - z -1 +
z
1.25
=
1 -2
2 1 + z -1 +
z
2
X( z) =
FG
H
FG
H
=
IJ
K
IJ
K
=
0.625 (1 - z -1 + 0.16 z -2 )
(1 + z -1 + 0.5 z -2 )
0.625 z -2 (z2 - z + 0.16)
0.625 (z - 0.8) (z - 0.2)
=
(z + 0.5 + j0.5) (z + 0.5 - j0.5)
z -2 ( z 2 + z + 0.5)
.....(7.37)
The roots of quadratic, z2 – z + 0.16 = 0 are,
z =
1 ±
\
1 - 4 ´ 0.16
1 ± 0.6
=
= 0.8, 0.2
2
2
z 2 - z + 0.16 = (z - 0.8) (z - 0.2)
The roots of quadratic, z2 + z + 0.5 = 0 are,
z =
-1 ±
\
1 - 4 ´ 0.5
-1 ± j
=
= - 0.5 ± j 0.5
2
2
z2 + z + 0.5 = (z + 0.5 + j0.5) (z + 0.5 - j0.5)
jv
The zeros of X(z) are roots of numerator
polynomial, which has two roots.
p2
x
j0.5
z-plane
Therefore, the zeros of X(z) are,
z1 = 0.8, z2 = 0.2
The poles of X(z) are roots of denominator
polynomial, which has two roots.
Therefore, the poles of X(z) are,
p1 = – 0.5 – j0.5, p2 = – 0.5 + j0.5
z2
z1
–0.5
0.2
0.8
p1 x
–j0.5
u
Fig 7.5 : Pole-zero plot of X(z) of equation (7.37).
The pole-zero plot of X(z) is shown in fig 7.5.
7. 29
Signals & Systems
7.4.2 ROC of Rational Function of z
Case i: Right sided (causal) signal
Let x(n) be a right sided (causal) signal defined as,
x( n) = r1n u(n) + r2n u(n) + r3n u(n)
;
where r1 < r2 < r3
Now, the Z-transform of x(n) is,
z
z
z
X( z) =
+
+
z - r1
z - r2
z - r3
N ( z)
=
(z - r1 ) (z - r2 ) (z - r3 )
z
z-a
with ROC |z| > |a|
Z{a n u(n)} =
where, N(z) = z(z – r2) (z – r3) + z(z – r1) (z – r3) + z(z – r1) (z – r2)
jv
The poles of X(z) are,
z-plane
ROC of x(n)
p1 = r1, p2 = r2, p3 = r3
r3
The convergence criteria for X(z) are,
r1
r1
|z| > |r1| ; |z| > |r2| ; |z| > |r3|
Since r1 < r2 < r3, the ROC is exterior of the circle
of radius r3 in z-plane as shown in fig.7.6. In terms of
poles of X(z) we can say that the ROC is exterior of a
circle, whose radius is equal to the magnitude of outer
most pole (i.e., pole with largest magnitude) of X(z).
r2
x
r1
x
r2
x
r3
u
Fig 7.6 : ROC of x(n) = r1 u(n) + r2 u(n) +
r3 u(n) where r1 < r2 < r3.
Case ii: Left sided (anticausal) signal
Let x(n) be a left sided (anticausal) signal defined as,
x( n) = - r1n u(- n - 1) - r2n u( - n - 1) - r3n u(- n - 1)
Now, the Z-transform of x(n) is,
z
z
z
X( z) =
+
+
z - r1
z - r2
z - r3
N ( z)
=
(z - r1 ) (z - r2 ) (z - r3 )
;
where r1 < r2 < r3
Z{- a n u( - n - 1)} =
z
z-a
with ROC |z| < |a|
where, N(z) = z(z – r2) (z – r3) + z(z – r1) (z – r3) + z(z – r1) (z – r2)
jv
The poles of X(z) are,
z-plane
p1 = r1, p2 = r2, p3 = r3
The convergence criteria for X(z) are,
|z| < |r1| ; |z| < |r2| ; |z| < |r3|
x r x r xr
1
3
2
u
Since r1 < r2 < r3, the ROC is interior of the circle
of radius r1 in z-plane as shown in fig.7.7. In terms of
poles of X(z) we can say that the ROC is interior of a
ROC of x(n)
circle, whose radius is equal to the magnitude of inner
Fig 7.7 : ROC of x(n) = –r1 u(–n – 1) – r2 u(–n –1)
most pole (i.e., pole with smallest magnitude) of X(z).
– r3 u(–n –1), where r1 < r2 < r3.
Chapter 7 - Z - Transform
7. 30
Case iii: Two sided (noncausal) signal
Let x(n) be two sided signal defined as,
x( n) = r1n u(n) + r2n u(n) - r3n u(- n - 1) - r4n u(- n - 1)
;
where r1 < r2 < r3 < r4
Now, the Z-transform of x(n) is,
z
z
z
z
X( z) =
+
+
+
z - r1
z - r2
z - r3
z - r4
N ( z)
=
(z - r1 ) (z - r2 ) (z - r3 ) (z - r4 )
where, N(z) = z(z – r2) (z – r3) (z – r4) + z(z – r1) (z – r3) (z – r4)
+ z(z – r1) (z – r2) (z – r4) + z(z –r1) (z – r2) (z – r3)
jv
The poles of X(z) are,
p1 = r1 ; p2 = r2 ; p3 = r3 ; p4 = r4
z-plane
The convergence criteria for X(z) are,
|z| > |r1| ; |z| > |r2| ; |z| < |r3| ; |z| < |r4|
Since r1 < r2 < r3 < r4, the ROC is the region
xr xr x r xr
inbetween the circles of radius r2 and r3 as shown in
u
fig 7.8. Let rx be the magnitude of largest pole of causal
signal and let ry be the magnitude of smallest pole of
ROC of x(n)
anticausal signal and let rx < ry. Now in term of poles of
X(z) we can say that the ROC is the region in between Fig 7.8 : ROC of x(n) = r u(n) + r u(n)
1
2
– r3 u(–n –1), –r4 u(–n –1).
two circles of radius rx and ry, where rx < ry.
1
2
3
4
7.4.3 Properties of ROC
The various concepts of ROC that has been discussed in sections 7.2 and 7.4.2 are summarized
as properties of ROC and given below.
Property- 1: The ROC of X(z) is a ring or disk in z-plane, with centre at origin.
Property- 2: If x(n) is finite duration right sided (causal) signal, then the ROC is entire z- plane except z = 0.
Property- 3: If x(n) is finite duration left sided (anticausal) signal, then the ROC is entire z-plane
except z = ¥.
Property-4: If x(n) is finite duration two sided (noncausal) signal, then the ROC is entire z-plane
except z = 0 and z = ¥.
Property- 5 : If x(n) is infinite duration right sided (causal) signal, then the ROC is exterior of a circle of radius r1.
Property- 6: If x(n) is infinite duration left sided (anticausal) signal, then the ROC is interior of a
circle of radius r2.
Property- 7: If x(n) is infinite duration two sided (noncausal) signal, then the ROC is the region
in between two circles of radius r1 and r2.
Property- 8: If X(z) is rational, (where X(z) is Z-transform of x(n)), then the ROC does not include
any poles of X(z).
Property- 9: If X(z) is rational, (where X(z) is Z-transform of x(n)), and if x(n) is right sided, then the
ROC is exterior of a circle whose radius corresponds to the pole with largest magnitude.
Property-10: If X(z) is rational, (where X(z) is Z-transform of x(n)), and if x(n) is left sided, then the
ROC is interior of a circle whose radius corresponds to the pole with smallest magnitude.
Property-11: If X(z) is rational, (where X(z) is Z-transform of x(n)), and if x(n) is two sided, then the
ROC is region in between two circles whose radius corresponds to the pole of causal part
with largest magnitude and the pole of anticausal part with smallest magnitude.
7. 31
Signals & Systems
7.5 Inverse Z-Transform
Let X(z) be Z-transform of the discrete time signal x(n).The inverse Z-transform is the process of
recovering the discrete time signal x(n) from its Z-transform X(z). The signal x(n) can be uniquely
determined from X(z) and its ROC.
The inverse Z-transform can be determined by the following three methods.
1. Direct evaluation by contour integration (or residue method)
2. Partial fraction expansion method.
3. Power series expansion method.
7.5.1 Inverse Z-Transform by Contour Integration or Residue Method
Let, X(z) be Z-transform of x(n).
Now by definition of inverse Z-transform,
x( n) =
1
2 pj
z
X(z) z n - 1dz
.....(7.38)
C
Using partial fraction expansion technique the function X(z) zn – 1 can be expressed as shown below.
X( z) z n - 1 =
A1
A2
A3
AN
+
+
+ ..... +
z - p1
z - p2
z - p3
z - pN
.....(7.39)
where, p1, p2, p3, ..... pN are poles of X(z) zn – 1 and A1, A2, A3, ..... AN are residues
The residue A1 is obtained by multiplying the equation (7.39) by (z – p1) and letting z = p1.
Similarly other residues are evaluated.
\ A1 = ( z - p1 ) X(z) z n-1
.....(7.40.1)
z = p1
A 2 = ( z - p 2 ) X(z) z n - 1
A 3 = ( z - p 3 ) X(z) z n -1
.....(7.40.2)
z = p2
.....(7.40.3)
z = p3
M
M
A N = ( z - p N ) X(z) z n-1
.....(7.40.N)
z = pN
Using equation (7.39) the equation (7.38) can be written as,
x( n) =
=
1
2 pj
z LMN
C
A1
A2
A3
AN
+
+
+ ..... +
z - p1
z - p2
z - p3
z - pN
LMA
1 M
2 pj M
MM
N
1
z
C
dz
+ A2
z - p1
z
C
dz
+ A3
z - p2
z
C
OP dz
Q
dz
+ .....
z - p3
+ AN
z
C
dz
z - pN
OP
PP
PP
Q
.....(7.41)
Chapter 7 - Z - Transform
7. 32
1
, then by Cauchy' s integral theorem,
z - p0
2 pj ; if p 0 is a point inside the contour C in z - plane
1
G(z) dz =
dz =
z - p0
0 ; if p 0 is a point outside the contour C in z - plane
C
C
If ,
G(z) =
z
RS
T
z
Using Cauchy's integral theorem, the equation (7.41) can be written as shown below.
1
A 1 2 pj + A 2 2pj + A 3 2pj + ..... + A N 2pj
2 pj
= A1 + A 2 + A 3 + ..... + A N
x( n) =
= Sum of residues of X(z) z
.....(7.42)
n-1
On substituting for residues from equation (7.40.1) to (7.40.N) in equation(7.42) we get,
x( n) = (z - p1 ) X(z) z n -1
+ (z - p 3 ) X(z) z
n -1
N
\ x(n) =
+ (z - p2 ) X(z) zn-1
z = p1
z = p3
å LMN(z - p ) X(z) z
+ ..... + (z - p N ) X(z) zn -1
n -1
i
i=1
z = p2
z = pi
z = pN
OP
Q
.....(7.43)
where, N = Number or poles of X(z) zn – 1 lying inside the contour C.
Using equation (7.43), by considering only the poles lying inside the contour C, the inverse
Z-transform can be evaluated. For a stable system the contour C is the unit circle in z-plane.
7.5.2 Inverse Z-Transform by Partial Fraction Expansion Method
Let X(z) be Z-transform of x(n), and X(z) be a rational function of z. Now the function X(z) can
be expressed as a ratio of two polynomials in z as shown below. (Refer equation 7.35).
X ( z) =
N(z)
D( z)
.....(7.44)
where, N(z) = Numerator polynomial of X(z)
D(z) = Denominator polynomial of X(z)
Let us divide both sides of equation (7.44) by z and express equation (7.44) as shown below.
X( z)
N(z)
=
z
z D( z)
X( z)
Q(z)
\
=
z
D( z)
where , Q(z) =
.....(7.45)
N(z)
z
Note : It is convenient, if we consider
fraction expansion method.
X(z)
rather than X(z) for inverse Z-transform by partial
z
7. 33
Signals & Systems
On factorizing the denominator polynomial of equation (7.45) we get,
X( z)
Q(z)
Q(z)
=
=
z
D( z)
(z - p1 ) (z - p2 ) (z - p3 ) ..... (z - p N )
.....(7.46)
where, p1, p2, p3, .....pN are roots of denominator polynomial (as well as poles of X(z)).
The equation (7.46) can be expressed as a series of sum terms by partial fraction expansion
technique as shown below.
A1
A2
A3
AN
X( z)
=
+
+
+ ..... +
z
z - p1
z - p2
z - p3
z - pN
where, A1, A2, A3, ..... AN are residues.
\ X(z) = A1
z
z - p1
+ A2
z
z - p2
+ A3
z
z - p3
+ AN
+ .....
z
z - pN
.....(7.47)
Now, the inverse Z-transform of equation (7.47) is obtained by comparing each term with standard
Z-transform pair. The two popular Z-transform pairs useful for inverse Z-transform of equation (7.47)
are given below.
If an is a causal (or right sided) signal then,
n
s
Z a n u(n) =
z
; with ROC z > a
z-a
If an is an anticausal (or left sided) signal then,
n
s
Z - a n u(- n - 1) =
z
; with ROC z < a
z-a
Let r1 be the magnitude of the largest pole and let the ROC be |z| > r1 (where r1 is radius of a circle
in z-plane), then each term of equation (7.47) gives rise to a causal sequence, and so the inverse
Z-transform of equation (7.47) will be as shown in equation (7.48).
x(n) = A1 p1n u(n) + A2 p2n u(n) + A3 p3n u(n) + ..... + AN pNn u(n)
.....(7.48)
Let r2 be the magnitude of the smallest pole and let ROC be |z| < r2 (where r2 is radius of a circle in
z-plane), then each term of equation (7.47) give rise to an anticausal sequence, and so the inverse
Z-transform of equation (7.47) will be as shown in equation (7.49).
x(n) = –A1 p1n u(–n –1) – A2 p2n u(–n –1) – A3 p3n u(–n –1) – .....
– AN pNn u(–n –1)
.....(7.49)
Sometimes the specified ROC will be in between two circles of radius rx and ry, where rx < ry.
(i.e., ROC is rx < |z| < ry). Now in this case, the terms with magnitude of pole less than rx will give rise to
causal signal and the terms with magnitude of pole greater than ry will give rise to anticausal signal so that
the inverse Z-transform of X(z) will give a two sided signal. [Refer section 7.4.2, case iii]
Chapter 7 - Z - Transform
7. 34
Evaluation of Residues
The coefficients of the denominator polynomial D(z) are assumed real and so the roots of the
denominator polynomial are real and/or complex conjugate pairs (i.e., complex roots will occur only in
conjugate pairs). Hence on factorizing the denominator polynomial we get the following cases. (The roots
of the denominator polynomial are poles of X(z)).
Case i : When roots (or poles) are real and distinct
Case ii : When roots (or poles) have multiplicity
Case iii : When roots (or poles) are complex conjugate
Case i : When roots (or poles) are real and distinct
X(z)
can be expressed as,
z
In this case
X( z)
Q( z)
Q( z)
=
=
z
D( z)
( z - p1 ) (z - p 2 ) ...... ( z - p N )
A1
A2
AN
=
+
+ ..... +
( z - p1 )
(z - p2 )
(z - pN )
where, A1, A2 ...... AN are residues and p1, p2, ..... pN are poles
X(z)
by (z–p1) and letting z = p1. Similarly
The residue A1 is evaluated by multiplying both sides of
z
other residues are evaluated.
\ A 1 = ( z - p1 )
A 2 = ( z - p2 )
X(z)
z
X(z)
z
z = p1
z = p2
M
A N = (z - pN )
X(z)
z
z = pN
Case ii : When roots (or poles) have multiplicity
Let one of pole has a multiplicity of q. (i.e., repeats q times). In this case
X(z)
is expressed as,
z
X( z)
Q( z)
Q(z)
=
=
z
D(z)
( z - p1 ) (z - p 2 )......( z - p x ) q ......( z - p N )
A1
A2
A x0
=
+
+ ..... +
( z - p1 )
(z - p2 )
(z - px ) q
A x ( q -1)
A x1
AN
+
+ ...... +
+ ..... +
q -1
(z - px )
(z - pN )
(z - px )
where, A x0 , Ax1 , ...... A x(q-1) are residues of repeated root (or pole), z = px
7. 35
Signals & Systems
The residues of distinct real roots are evaluated as explained in case i.
The residue Axr of repeated root is obtained as shown below.
A xr =
1 dr
r ! dz r
LM(z - p )
N
x
q
X(z)
z
OP
Q
; where r = 0, 1, 2,....(q - 1)
z = px
Case iii : When roots (or poles) are complex conjugate
Let
X(z)
X(z)
has one pair of complex conjugate pole. In this case
can be expressed as,
z
z
X( z)
Q( z)
Q(z)
=
=
z
D(z)
( z - p1 ) (z - p 2 )......( z 2 + az + b)......( z - p N )
A1
A2
Ax
A *x
AN
=
+
+ ..... +
+
+ ..... +
z - p1
z - p2
z - (x + jy)
z - (x - jy)
z - pN
The residues of real and non-repeated roots are evaluated as explained in case i.
The residue Ax is evaluated as that of case i and the residue Ax* is the conjugate of Ax.
7.5.3 Inverse Z-Transform by Power Series Expansion Method
Let X(z) be Z-transform of x(n), and X(z) be a rational function of z as shown below.
X( z) =
N(z)
b + b1 z -1 + b2 z -2 + b3 z -3 + ..... + b M z- M
= 0
D(z)
a 0 + a1 z -1 + a 2 z -2 + a 3 z -3 + ..... + a N z- N
On dividing the numerator polynomial N(z) by denominator polynomial D(z) we can express X(z)
as a power series of z. It is possible to express X(z) as positive power of z or as negative power of z or
with both positive and negative power of z as shown below.
Case i :
X( z) =
N(z)
= c0 + c1 z -1 + c2 z -2 + c 3 z -3 + .....
D(z)
.....(7.50.1)
N(z)
= d 0 + d 1 z1 + d 2 z 2 + d 3 z 3 + .....
D(z)
.....(7.50.2)
Case ii :
X( z) =
Case iii :
X( z) =
N(z)
= ..... + e -3 z3 + e -2 z 2 + e -1 z + e 0
D(z)
+ e1 z -1 + e2 z-2 + e3 z-3 + .....
.....(7.50.3)
The case i power series of z is obtained when the ROC is exterior of a circle of radius r in z-plane
(i.e., ROC is |z| > r). The case ii power series of z is obtained when the ROC is interior of a circle of radius
r in z-plane (i.e., ROC is |z| < r). The case iii power series of z is obtained when the ROC is in between
two circles of radius r1 and r2 in z-plane (i.e., ROC is r1 < |z| < r2).
Chapter 7 - Z - Transform
7. 36
By the definition of Z-transform, we get,
¥
å x(n) z
X( z) =
-n
n = -¥
On expanding the summation we get,
X( z) = ........ x( -3) z 3 + x( -2) z 2 + x(-1) z1 + x(0) z 0
+ x(1) z -1 + x(2) z-2 + x(3) z-3 + .........
.....(7.51)
On comparing the coefficients of z of equations (7.50) and (7.51), the samples of x(n) are
determined. [i.e., the coefficient of zi is the ith sample, x(i) of the signal x(n)].
Note : The different methods of evaluation of inverse Z-transform of a funcion X(z) will result in
different type of mathematical expressions. But the inverse Z-transform is unique for a
specified ROC and so on evaluating the expressions for each value of n, we may get a same
signal.
Example 7.7
Determine the inverse Z-transform of the function, X(z) =
prove that the inverse Z-transform is unique.
3 + 2z-1 + z-2
by the following three methods and
1 - 3z -1 + 2z -2
1. Residue Method
2. Partial Fraction Expansion Method
3. Power Series Expansion Method
Solution
Method-1 : Residue Method
3 + 2z-1 + z-2
z-2 (3z2 + 2z + 1)
3z2 + 2z + 1
=
= 2
.
Given that, X(z) =
-1
-2
-2 2
z (z - 3z + 2)
z - 3z + 2
1 - 3z + 2z
Let us divide the numerator polynomial by denominator polynomial and express X(z) as shown below.
X(z) =
3z2 + 2z + 1
11z - 5
= 3 + 2
z2 - 3z + 2
z - 3z + 2
= 3 +
z2 – 3z + 2
11z - 5
(z - 1) (z - 2)
Let, X1(z) = 3 and X 2 (z) =
3
3z2 + 2z + 1
3z2 – 9z + 6
(–)
(+)
(–)
11z – 5
11z - 5
z 2 - 3z + 2
; \ X(z) = X1(z) + X 2 (z)
x(n) = Z -1{X(z)} = Z -1{X1(z)} + Z -1{X 2 (z)}
= Z -1{3} + Z -1{X 2 (z)}
N
= 3 d(n) + å
i=1
LM(z - p ) X (z) z
N
= 3 d(n) + (z - 1)
= 3 d(n) +
n- 1
i
2
z = pi
OP
Q
Using residue theorem
11z - 5
11z - 5
zn - 1
+ (z - 2)
zn - 1
(z - 1) (z - 2)
(z
1) (z - 2)
z=1
z=2
11 - 5
11 ´ 2 - 5 n - 1
(1)n - 1 +
2
2 - 1
1 - 2
\ x(n) = 3 d(n) - 6 u(n - 1) + 17(2)n - 1 u(n - 1) = 3 d(n) + - 6 +17(2)n - 1 u(n - 1)
7. 37
Signals & Systems
When n = 0,
x(0) = 3 – 0 + 0
When n = 1,
x(1)
= 0 – 6 + 17 ´ 20 = 11
=3
When n = 2,
x(2)
= 0 – 6 + 17 ´ 21 = 28
When n = 3,
x(3)
= 0 – 6 + 17 ´ 22 = 62
When n = 4,
x(4) = 0 – 6 + 17 ´ 23 = 130
\ x(n) = {3, 11, 28, 62, 130, .....}
­
Method-2 : Partial Fraction Expansion Method
Given that, X(z) =
\
Let,
3 + 2z-1 + z-2
z-2 (3z2 + 2z + 1)
3z2 + 2z + 1
=
=
.
-1
-2
-2 2
(z - 1) (z - 2)
1 - 3z + 2z
z (z - 3z + 2)
X(z)
3z2 + 2z + 1
=
z
z(z - 1) (z - 2)
X(z)
3z 2 + 2z + 1
A1
A2
A3
=
=
+
+
z(z - 1) (z - 2)
z
z - 1
z - 2
z
Now, A 1 = z
X(z)
3z2 + 2z + 1
1
= z
=
z(z - 1) (z - 2) z = 0
z z=0
(-1) ´ -2)
b g
= 0.5
A 2 = (z - 1)
X(z)
3z2 + 2z +1
3 + 2 + 1
= (z - 1)
=
= -6
z(z - 1) (z - 2) z = 1
z z=1
1 ´ (1 - 2)
A 3 = (z - 2)
3z2 + 2z + 1
3 ´ 22 + 2 ´ 2 + 1
X(z)
=
= 8.5
= (z - 2)
z(z - 1) (z - 2) z = 2
2 ´ (2 - 1)
z z=2
0.5
6
X(z)
=
+
z - 1
z
z
\ X(z) = 0.5 - 6
8.5
z - 2
z
z
+ 8.5
z - 1
z - 2
Z{d(n)} = 1
On taking inverse Z-transform of X(z) we get,
x(n) = 0.5 d(n) – 6 u(n) + 8.5 (2)n u(n) = 0.5 d(n) + [–6 + 8.5(2)n] u(n)
´ 20 = 3
When n = 0,
x(0) = 0.5 – 6 + 8.5
When n = 1,
x(1) = 0 – 6 + 8.5 ´ 21 = 11
When n = 2,
x(2) = 0 – 6 + 8.5
´ 22 = 28
When n = 3,
x(3) = 0 – 6 + 8.5
´ 23 = 62
When n = 4,
x(4) = 0 – 6 + 8.5
´ 24 = 130
\ x(n) = {3, 11, 28, 62, 130, .....}
­
Method-3 : Power Series Expansion Method
Given that, X(z) =
3 + 2z
1 - 3z
-1
-1
+ z
-2
+ 2z
-2
Z{u(n)} =
z
z - 1
Z{ann(n)} =
z
z - a
Chapter 7 - Z - Transform
7. 38
Let us divide the numerator polynomial by denominator polynomial as shown below.
3 + 11z–1 + 28z–2 + 62z–3 + 130z–4 + ......
–1
1 – 3z + 2z
–2
3 + 2z–1 + z–2
3 – 9z–1 + 6z–2
(–) (+)
(–)
11z–1 – 5z–2
11z–1 – 33z–2 + 22z–3
(–)
(+)
(–)
28z–2 – 22z–3
28z–2 – 84z–3 + 56z–4
(–)
(+)
(–)
62z–3 – 56z–4
62z–3 – 186z–4 + 124z–5
(–)
(+)
(–)
130z–4 – 124z–5
:
:
3 + 2z-1 + z-2
= 3 + 11z-1 + 28z-2 + 62z-3 + 130z -4 + .....
1 - 3z-1 + 2z -2
Let x(n) be inverse Z-transform of X(z).
\ X(z) =
.....(1)
Now, by definition of Z-transform,
+¥
X(z) =
å x(n) z
-n
n = -¥
= .....+ x(0) + x(1) z-1 + x(2) z-2 + x(3) z-3 + x(4) z-4 +......
.....(2)
On comparing equations (1) and (2) we get,
x(0) = 3
x(1) = 11
x(2) = 28
x(3) = 62
x(4) = 130 and so on.
\ x(n) = {3, 11, 28, 62, 130, .....}
­
Conclusion : It is observed that the results of all the three methods are same.
Example 7.8
Determine the inverse Z-transform of the following Z-domain functions.
a) X(z) =
3z2 + 2z + 1
b) X(z) =
z2 + 3z + 2
z - 0.4
z2 + z + 2
c ) X(z) =
z - 4
(z - 1) (z - 2)2
Solution
a) Given that, X(z) =
3z 2 + 2z + 1
z 2 + 3z + 2
On dividing the numerator by denominator, the X(z) can be expressed as shown below.
X(z) =
3z2 + 2z + 1
z2 + 3z + 2
= 3 +
-7z - 5
z 2 + 3z + 2
By partial fraction expansion we get, X(z) = 3 +
= 3 +
-7z - 5
(z + 1) (z + 2)
A1
A2
+
z + 1
z + 2
3
z2 + 3z + 2 3z2 + 2z + 1
3z2 + 9z + 6
(–)
(–)
(–)
–7z – 5
7. 39
Signals & Systems
A 1 = (z + 1)
A2
-7z - 5
(z + 1) (z + 2) z =
-7z - 5
= (z + 2)
(z + 1) (z + 2) z =
=
-1
-2
-7z - 5
z + 2
-7z - 5
=
z + 1
=
-7 ´ ( -1) - 5
=2
-1 + 2
=
-7 ´ (-2) - 5
= -9
-2 + 1
z = -1
z = -2
1
z
2
9
1
z
\ X(z) = 3 +
-9
= 3 + 2
z z - (- 2)
z + 1
z + 2
z z - (-1)
Multiply and
dvide by z
z
z
- 9z - 1
z - ( -1)
z - ( -2)
= 3 + 2z -1
On taking inverse Z-transform of X(z) we get,
Z{d(n)} = 1
x(n) = 3 d(n) + 2( -1)n -1 u(n - 1) - 9(-2)n - 1 u(n - 1)
Z{a n u(n)} =
= 3 d(n) + 2(-1)n -1 - 9( -2)n -1 u(n - 1)
z
z - a
z
z - a
When n = 0,
x(0) = 3 + 0 + 0
= 3
If Z {an u(n)} =
When n = 1,
x(1) = 0 + 2 – 9
= –7
then by time shifting property
When n = 2,
x(2) = 0 – 2 – 9 ´ (–2) = 16
When n = 3,
x(3) = 0 + 2 – 9 ´ (–2) = –34
When n = 4,
x(4) = 0 – 2 – 9 ´ (–2)3 = 70
Z{a(n
- 1)
u(n - 1)} = z -1
2
z
z - a
\ x(n) = {3, –7, 16, –34, 70, .....}
­
Alternate Method
X(z) =
\
3z2 + 2z + 1
z 2 + 3z + 2
X(z)
3z2 + 2z + 1
3z2 + 2z + 1
=
=
2
z
z(z + 1) (z + 2)
z(z + 3z + 2)
By partial fraction expansion technique
X(z)
can be expressed as,
z
X(z)
3z2 + 2z + 1
A
A2
A3
=
= 1 +
+
z
z(z + 1) (z + 2)
z
z + 1
z + 3
A1 = z
X(z)
3z2 + 2z + 1
= z
z(z + 1) (z + 2)
z z=0
A 2 = (z + 1)
A 3 = (z + 2)
X(z)
z z=
X(z)
z z=
= (z + 1)
-1
\ X(z) = 0.5 -
z=0
-2
=
z = -1
3z2 + 2z + 1
z(z + 1) (z + 2)
=
z = -2
3(-1)2 + 2( -1) + 1
= -2
-1 ´ ( -1 + 2)
3(- 2)2 + 2(- 2) + 1
= 4.5
-2 ´ (-2 + 1)
4.5
z + 2
2z
4.5 z
+
z + 1
z + 2
= 0.5 - 2
1
= 0.5
1 ´ 2
3z2 + 2z + 1
z(z + 1) (z + 2)
= (z + 2)
0.5
2
X(z)
\
=
+
z + 1
z
z
=
z
z
+ 4.5
z - ( -1)
z - (- 2)
Z{d(n)} = 1
Z{a n u(n)} =
z
z - a
Chapter 7 - Z - Transform
7. 40
On taking inverse Z-transform of X(z), we get,
x(n) = 0.5 d(n) – 2 (–1)n u(n) + 4.5 (–2)n u(n) = 0.5 d(n) + [–2 (–1)n + 4.5 (–2)n] u(n)
When n = 0,
x(0) = 0.5 – 2 + 4.5
= 3
When n = 1,
x(1)
= 0 + 2 + 4.5 ´ (–2) = –7
When n = 2,
x(2)
= 0 – 2 + 4.5 ´ (–2)2 = 16
When n = 3,
x(3)
= 0 + 2 + 4.5 ´ (–2)3 = –34
x(4) = 0 – 2 + 4.5 ´ (–2)4 = 70
When n = 4,
\ x(n) = {3, –7, 16, –34, 70, .....}
­
Note: The closed form expression of x(n) in the two methods look different, but on evaluating x(n) for various
values of n we get same signal x(n).
b) Given that, X(z) =
z - 0.4
The roots of the quadratic
z2 + z + 2
z2 + z + 2 = 0 are,
z - 0.4
z - 0.4
=
X(z) = 2
z + z + 2
(z + 0.5 - j 7 / 2) (z + 0.5 + j 7 / 2)
z =
A
+
z + 0.5 - j 7 / 2
A = (z + 0.5 - j 7 / 2)
=
=
\
A* =
\ X(z) =
A*
z + 0.5 + j 7 / 2
z - 0.4
(z + 0.5 - j 7 / 2) (z + 0.5 + j 7 / 2) z =
z - 0.4
(z + 0.5 + j 7 / 2) z =
-0.9 + j 7 / 2
=
j 7
FG 0.5 +
H
=
- 0.5 + j 7 / 2
-0.9
+
j 7 /2
j 7
IJ
7K
j0.9
*
j 7
FG 0.5 H
=
0.5 + j0.9 / 7
z + 0.5 - j 7 / 2
= (0.5 + j0.9 / 7 )
1 - 4 ´ 2
2
= - 0.5 ± j 7 / 2
By partial fraction expansion we get,
X(z) =
-1 ±
+
- 0.5 + j 7 / 2
-0.5 + j 7 / 2 - 0.4
-0.5 + j 7 / 2 + 0.5 + j 7 / 2
= 0.5 +
j0.9
7
IJ
7K
j0.9
Multiply and
dvide by z
0.5 - j0.9 / 7
z + 0.5 + j 7 / 2
1
1
z
z
+ (0.5 - j0.9 / 7 )
z z + 0.5 - j 7 / 2
z z + 0.5 + j 7 / 2
= (0.5 + j0.9 / 7 )z -1
z
z - ( -0.5 + j 7 / 2)
+ (0.5 - j0.9 / 7 ) z -1
On taking inverse Z-transform of X(z) we get,
x(n) = (0.5 + j0.9 / 7 ) (-0.5 + j 7 / 2)(n- 1) u(n - 1)
+ (0.5 - j0.9 / 7 ) (-0.5 - j 7 / 2)(n - 1) u(n - 1)
n
z
z - ( -0.5 - j 7 / 2)
If Z{a u(n)} =
z
z - a
then by time shifting property,
Z{a (n - 1) u(n - 1)} = z -1
z
z - a
7. 41
Signals & Systems
Alternatively the above result can be expressed as shown below.
Here, 0.5 + j0.9 7 = 0.605 Ð34.2o
= 0.605 Ð0.19p
0.5 – j0.9 7 = 0.605 Ж34.2 = 0.605 Ж0.19p
o
– 0.5 + j 7 2 = 1.414 Ð110.7o = 1.414 Ð0.62p
–0.5 – j 7 2 = 1.414 Ж110.7o = 1.414 Ж0.62p
\ x(n) = [0.605 Ð0.19p] [1.414 Ð0.62p](n – 1) u(n – 1) + [0.605 Ж0.19p] [1.414 Ж0.62p](n – 1) u(n – 1)
= [0.605 Ð0.19p] [1.414(n – 1) Ð0.62p (n – 1)] u(n – 1) + [0.605Ж0.19p] [1.414(n – 1)Ж0.62p (n – 1)] u(n – 1)
= 0.605 (1.414)(n – 1)Ð(0.19p + 0.62pn – 0.62p) u(n – 1) + 0.605 (1.414)(n – 1) Ð(–0.19p – 0.62pn + 0.62p) u(n – 1)
= 0.605 (1.414)(n – 1) [1 Ð(0.62n – 0.43)p + 1 Ж (0.62n – 0.43)p] u(n – 1)
= 0.605 (1.414)(n – 1) [cos ((0.62n – 0.43)p) + j sin((0.62n – 0.43)p) + cos((0.62 n – 0.43)p)
–j sin((0.62n – 0.43)p) ] u(n – 1)
= 0.605 (1.414)(n – 1) 2 cos ((0.62n – 0.43)p) u(n – 1)
= 1.21 (1.414)(n –1) cos ((0.62n – 0.43)p) u(n – 1)
c) Given that, X(z) =
z - 4
(z - 1) (z - 2) 2
By partial fraction expansion we get,
X(z) =
z - 4
A1
A2
A3
+
+
=
z - 1
(z - 2 )
(z - 1) (z - 2)2
( z - 2) 2
A 1 = (z - 1)
z - 4
(z - 1) (z - 2)2
z - 4
A 2 = (z - 2)2
A3 =
(z - 1) (z - 2)2
LM
N
2
OP
Q
=
z =1
z - 4
z - 1z
=
z = 2
3
=
(z - 1)2
z= 2
=
= 2
d
dz
1 - 4
= -3
(1 - 2)2
2 - 4
= -2
2 - 1
LM z - 4 OP
N z - 1Q
z = 2
3
=
= 3
(2 - 1)2
-3
z
z
2z
2
3
1
1
1
+
+ 3
= -3
z - 1
z - 2
z z - 1
z (z - 2)2
z z - 2
(z - 2)2
= - 3z -1
Z {u(n)} =
If
=
z= 2
d
z - 4
(z - 2)2
dz
(z - 1) (z - 2)2
(z - 1) - (z - 4)
=
(z - 1)2
z=
\ X(z) =
z - 4
(z - 2 )2
=
z= 1
Multiply and
dvide by z
z
2z
z
- z -1
+ 3z-1
z - 1
z - 2
(z - 2)2
z
z - 1
;
Z {an u(n)} =
z
z - a
;
Z {na n u(n)} =
az
(z - a )2
Z{x(n)} = X(z) then by time shifting property Z{x(n – 1)} = z–1 X(z)
\ Z {u(n - 1)} = z -1
z
z
; Z {a(n - 1) u(n - 1)} = z -1
z - 1
z - a
and Z{(n - 1) a(n - 1) u(n - 1)} = z -1
az
(z - a)2
On taking inverse Z-transform of X(z) using standard transform and shifting property we get,
x(n) = - 3 u(n - 1) - (n - 1) 2n - 1 u(n - 1) + 3 ´ 2n -1 u(n - 1)
= -3 - (n - 1) 2n -1 + 3(2)n -1 u(n - 1)
Chapter 7 - Z - Transform
7. 42
Example 7.9
Determine the inverse Z-transform of the following function.
a)
c)
X(z) =
X( z ) =
1
1 - 15
. z
-1
1 + z
+ 0.5z
-2
-1
1 - z -1 + 0.5z -2
b)
X(z) =
d)
X(z) =
z2
z
2
- z + 0.5
1
(1 + z -1) (1 - z -1)2
Solution
a) Given that, X(z) =
1
1 - 1.5 z -1 + 0.5z -2
X(z) =
\
1
. z
1 - 15
-1
+ 0.5z
-2
=
z2
z2
1
=
= 2
.
15
0.5
(z - 0.5)
(
)
z
1
. z + 0.5
z - 15
1 + 2
z
z
X(z)
z
=
(z - 1) (z - 0.5)
z
By partial fraction expansion, X(z)/z can be expressed as,
X(z)
A1
A2
+
=
z
z - 1
z - 0.5
A 1 = (z - 1)
A2
X(z)
z
= (z - 1)
z = 1
X( z )
= (z - 0.5)
z z=
0.5
z
(z - 1) (z - 0.5)
=
z= 1
1
= 2
1 - 0.5
z
0.5
= (z - 0.5)
=
= -1
(z - 1) (z - 0.5) z = 0.5
0.5 - 1
X( z )
2
1
\
=
z
z - 1
z - 0.5
2z
z
\ X( z ) =
z - 1
z - 0.5
Z{an u(n)} =
Z{u(n)} =
z
; ROC |z| > |a|
z - a
z
; ROC |z| > 1
z - 1
On taking inverse Z-transform of X(z), we get,
x(n) = 2 u(n) – 0.5n u(n) = [2 – 0.5n] u(n)
b) Given that, X(z) =
z2
z
2
X (z ) =
\
- z + 0.5
z2
z2
=
(z - 0.5 - j0.5) (z - 0.5 + j0.5)
z - z + 0.5
2
X(z)
z
=
z
(z - 0.5 - j0.5) (z - 0.5 + j0.5)
The roots of quadratic
z 2 - z + 0.5 = 0 are,
By partial fraction expansion, we can write,
X(z)
A
A*
=
+
z
z - 0.5 - j0.5
z - 0.5 + j0.5
A = (z - 0.5 - j0.5)
= (z - 0.5 - j0.5)
X(z)
z z=
z =
1 ±
1 - 4 ´ 0.5
2
= 0.5 ± j 0.5
0.5 + j0.5
z
(z - 0.5 - j0.5) (z - 0.5 + j0.5) z
= 0.5 + j0.5
7. 43
Signals & Systems
\ A =
0.5 + j0.5
0.5 + j0.5
=
= 0.5 - j0.5
j10
.
0.5 + j0.5 - 0.5 + j0.5
\ A * = (0.5 - j0.5)* = 0.5 + j0.5
\
X( z )
0.5 - j0.5
0.5 + j0.5
=
+
z
z - 0.5 - j0.5
z - 0.5 + j0.5
(0.5 - j0.5) z
(0.5 + j0.5) z
+
z - (0.5 + j0.5)
z - (0.5 - j0.5)
X (z ) =
z{an u(n)} =
On taking inverse Z-transform of X(z) we get,
z
z - a
;
ROC |z| > |a|
x(n) = (0.5 – j0.5) (0.5 + j0.5)n u(n) + (0.5 + j0.5) (0.5 – j0.5)n u(n)
Alternatively the above result can be expressed as shown below.
Here, 0.5 + j0.5 = 0.707 Ð45o = 0.707 Ð0.25p
0.5 – j0.5 = 0.707 Ж45o = 0.707 Ж 0.25p
\ x(n) = [0.707 Ж 0.25p] [0.707 Ð0.25p]n u(n) + [0.707 Ð0.25p] [0.707 Ж 0.25p]n u(n)
= [0.707 Ж0.25p] [0.707n Ð0.25pn] u(n) + [0.707 Ð0.25p] [0.707n Ж0.25pn] u(n)
= 0.707(n + 1) Ð(0.25p (n – 1)) u(n) + 0.707(n + 1) Ð(–0.25p(n – 1)) u(n)
= 0.707(n + 1) [1 Ð0.25p (n – 1) + 1Ð –0.25p(n – 1)]u(n)
= 0.707(n + 1) [cos (0.25p (n – 1)) + j sin (0.25p(n – 1)) + cos (0.25p(n – 1)) – j sin (0.25p(n – 1))] u(n)
= 0.707(n + 1) 2 cos (0.25p (n – 1)) u(n)
c) Given that, X(z) =
X(z) =
=
1 + z -1
1 - z -1 + 0.5z -2
1 + z -1
1 - z
-1
+ 0.5 z
z (z + 1)
(z
2
- z + 0.5)
-2
=
=
The roots of the quadratic
z2 – z + 0.5 = 0 are,
z -1(z + 1)
-2
z (z
2
- z + 0.5)
1 - 4 ´ 0.5
2
= 0.5 ± j0.5
z (z + 1)
(z - 0.5 - j0.5) (z - 0.5 + j0.5)
1 ±
z =
By partial fraction expansion, we can write,
X(z)
(z + 1)
A
A*
=
=
+
z
(z - 0.5 - j0.5) (z - 0.5 + j0.5)
z - 0.5 - j0.5
z - 0.5 + j0.5
A = (z - 0.5 - j0.5)
= (z - 0.5 - j0.5)
X (z)
z
z = 0. 5 + j0.5
(z + 1)
(z - 0.5 - j0.5) (z - 0.5 + j0.5)
z = 0.5 + j 0.5
. + j0.5
15
0.5 + j0.5 + 1
. + 0.5 = 0.5 - j15
.
=
= - j15
=
j1
0.5 + j0.5 - 0.5 + j0.5
. * = 0.5 + j15
.
A * = (0.5 - j15)
\
X( z )
.
.
0.5 - j15
0.5 + j15
=
+
z
z - 0.5 - j0.5
z - 0.5 + j0.5
.
X (z) = (0.5 - j15)
z
z
.
+ (0.5 + j15)
z - (0.5 + j0.5)
z - (0.5 - j0.5)
Z {a n u(n)} =
z
z - a
Chapter 7 - Z - Transform
7. 44
On taking inverse Z-transform of X(z) we get,
x(n) = (0.5 – j1.5) (0.5 + j0.5)n u(n) + (0.5 + j1.5) ( 0.5 – j0.5)n u(n)
Alternatively the above result can be expressed as shown below.
Here, 0.5 – j1.5 = 1.581 Ж71.6 o = 1.581 Ж0.4p
0.5 + j1.5 = 1.581 Ð71.6 o = 1.581 Ð0.4p
0.5 + j0.5 = 0.707 Ð45o
= 0.707 Ð0.25p
0.5 – j0.5 = 0.707 Ж45
= 0.707 Ж0.25p
o
\ x(n) = [1.581 Ж0.4p] [0.707 Ð0.25p]n u(n) + [1.581 Ð0.4p] [0.707 Ж0.25p]n u(n)
= [1.581 Ж0.4p] [0.707n Ð0.25pn] u(n) + [1.581 Ð0.4p] [0.707n Ж0.25pn] u(n)
= 1.581 (0.707)n [1 Ðp(0.25n – 0.4) + 1 Жp(0.25n – 0.4)] u(n)
= 1.581 (0.707)n [cos (p(0.25n – 0.4)) + j sin (p(0.25n – 0.4)) + cos (p(0.25n – 0.4))
– j sin (p(0.25n – 0.4))] u(n)
= 1.581 (0.707)n 2 cos (p(0.25 n – 0.4)) u(n)
= 3.162 (0.707)n cos (p(0.25n – 0.4)) u(n)
d) Given that, X(z) =
X(z) =
1
(1 + z -1 ) (1 - z -1) 2
1
-1
-1 2
(1 + z ) (1 - z )
=
1
z3
=
-2
2
z (z + 1) z (z - 1)
(z + 1) (z - 1)2
-1
X( z )
z2
\
=
z
(z + 1) (z - 1)2
By partial fraction expansion, we can write,
X( z )
A1
A2
A3
=
+
+
z
z + 1
z - 1
(z - 1)2
X(z)
z
A 1 = (z + 1)
A 2 = (z - 1)
A3 =
2
X(z)
z
LM
N
d L z O
M P
dz MN z + 1PQ
= (z + 1)
z = -1
=
z = -1
z2
= (z - 1)
(z + 1) (z - 1)2
2
z= 1
d
X(z)
(z - 1)2
dz
z
OP
Q
2
=
z2
(z + 1) (z - 1)2
=
z= 1
=
z= 1
LM
MN
z= 1
2
z2
(z - 1)2
z2
=
z + 1
d
z
(z - 1)2
dz
(z + 1) (z - 1)2
(z + 1) 2z - z
(z + 1)2
2
=
z=1
0.25
0.5
0.75
=
+
+
z + 1
z - 1
(z - 1)2
z
z
z
X(z) = 0.25
+ 0.5
+ 0.75
z - 1
z + 1
(z - 1)2
OP
PQ
=
z = -1
=
z= 1
( -1)2
= 0.25
( -1 - 1)2
1
= 0.5
1 + 1
z= 1
3
(1 + 1) ´ 2 - 1
=
= 0.75
4
(1 + 1)2
X( z )
\
z
z
z
z
= 0.25
+ 0.5
+ 0.75
z - ( -1)
z - 1
(z - 1)2
On taking inverse Z-transform of X(z) we get,
x(n) = 0.25(-1)n + 0.5n u(n) + 0.75 u(n)
n
= [0.25( -1) + 0.5n + 0.75] u(n)
Z {a n u(n)} =
z
z - a
Z {n u(n)} =
z
(z - 1)2
Z {u(n)} =
z
z - 1
7. 45
Signals & Systems
Example 7.10
1
1 - 1.5 z -1 + 0.5 z -2
(b) ROC : |z| < 0.5.
Determine the inverse Z-transform of X(z) =
When
(a) ROC : |z| > 1.0 and
Solution
a) Since the ROC is the exterior of a circle, we expect x(n) to be causal signal. Hence we can express X(z) as a power
series expansion in negative powers of z. On dividing the numerator of X(z) by its denominator we get,
X(z) =
1
1 - 1.5 z -1 + 0.5 z -2
= 1 + 1.5 z -1 + 1.75 z -2 + 1.875 z -3 + 1.9375 z -4 + .....
.....(1)
1 + 1.5 z -1 + 1.75 z -2 + 1.875 z -3 + 1.9375 z -4 + .........
1 - 1.5 z
-1
+ 0.5 z
-2
1
1 - 1.5 z -1 + 0.5 z -2
( - ) (+)
( -)
1.5 z -1 - 0.5 z -2
1.5 z -1 - 2.25 z -2 + 0.75 z -3
( -)
(+)
( -)
1.75 z
-2
1.75 z
-2
( -)
- 0.75 z -3
- 2.625 z -3 + 0.875 z -4
( -)
(+)
1.875 z
-3
- 0.875 z -4
1.875 z -3 - 2.8125 z-4 + 0.9375 z-5
( -)
(+ )
( -)
1.9375 z-4 - 0.9375 z-5
M
If X(z) is Z-transform of x(n) then, by the definition of Z-transform we get,
¥
X(z) = Z {x(n)} =
å x(n) z
-n
n = -¥
For a causal signal,
¥
X( z ) =
å x(n) z
-n
n = 0
On expanding the summation we get,
X(z) = x(0) z 0 + x(1) z -1 + x(2) z -2 + x(3) z -3 + x(4) z -4 + ........
.....(2)
On comparing the two power series of X(z) (equations (1) and (2)), we get,
x(0) = 1 ; x(1) = 1.5 ; x(2) = 1.75 ; x(3) = 1.875 ; x(4) = 1.9375 ; .......
l
A
q
x(n) = 1, 15
. , 1.75, 1.875, 1.9375,....
b) Since the ROC is the interior of a circle, we expect x(n) to be anticausal signal. Hence we can express X(z) as a power
series expansion in positive powers of z. Therefore, rewrite the denominator polynomial of X(z) in the reverse order and
then the numerator, is divided by the denominator as shown below.
Chapter 7 - Z - Transform
7. 46
2z2 + 6z 3 + 14z4 + 30z 5 + 62z6 + .....
0.5 z
-2
- 1.5 z
-1
+ 1
1
1 - 3z + 2z2
( - ) (+)
( -)
3z - 2z 2
3z - 9z 2 + 6z3
(- )
( -)
(+)
7z
2
- 6z 3
7z
2
- 21z 3 + 14z4
( -)
( -)
(+)
15z3 - 14z 4
15z3 - 45z 4 + 30z 5
( -)
(- )
(+)
31z
4
- 30z 5
M
\ X(z) =
1
1
=
1 - 1.5 z -1 + 0.5 z -2
0.5 z -2 - 1.5 z -1 + 1
.....(3)
= 2z 2 + 6z 3 + 14z 4 + 30z5 + 62z6 + ......
If X(z) is Z-transform of x(n) then, by the definition of Z-transform we get,
¥
X(z) = Z {x(n)} =
å x(n) z
-n
n = -¥
0
For an anticausal signal, X(z) =
å x(n) z
-n
n = -¥
On expanding the summation we get,
X(z) = ..... x( -6) z6 + x( -5) z 5 + x( -4) z 4 + x( -3) z 3 + x(-2) z2 + x( -1) z + x(0)
.... (4)
On comparing the two power series of X(z) (equations (3) and (4)), we get,
x(0) = 0 ; x(-1) = 0 ; x( -2) = 2 ; x(-3) = 6 ; x( -4) = 14 ; x(-5) = 30 ; x(-6) = 62 ; .....
\ x(n) =
l........, 62, 30, 14, 6, 2, 0, 0 q
A
Example 7.11
Determine the inverse Z-transform of X(z) =
a) if ROC is, |z| > 0.6
1
1 - 0.8 z -1 + 0.12 z -2
b) if ROC is, |z| < 0.2
c) if ROC is, 0.2 < |z| < 0.6
Solution
Given that, X(z) =
\
1
1 - 0.8 z
-1
+ 0.12 z
-2
=
-2
z (z
2
1
z2
=
(z - 0.6) (z - 0.2)
- 0.8z + 0.12)
X(z)
z
=
z
(z - 0.6) (z - 0.2)
By partial fraction expansion technique we get,
X(z)
z
A1
A2
=
=
+
z - 0.6
z - 0.2
z
(z - 0.6) (z - 0.2)
The roots of quadratic
z 2 - 0.8z + 0.12 = 0 are,
0.8 2 - 4 ´ 0.12
2
0.8 ± 0.4
=
= 0.6, 0.2
2
z =
0.8 ±
7. 47
Signals & Systems
A1 = (z - 0.6)
A2
X(z)
z z
= (z - 0.6)
= 0.6
X(z)
= (z - 0.2)
z z=
0.2
z
(z - 0.6) (z - 0.2) z
=
0.6
= 1.5
0.6 - 0.2
=
0.2
= - 0.5
0.2 - 0.6
= 0.6
z
= (z - 0.2)
(z - 0.6) (z - 0.2) z =
0.2
X(z)
1.5
0.5
\
=
z
z - 0.6
z - 0.2
\ X(z) = 1.5
z
z
- 0.5
z - 0.6
z - 0.2
Now, the poles of X(z) are p1 = 0.6, p2 = 0.2
a) ROC is |z| > 0.6
The specified ROC is exterior of the circle whose radius corresponds to the largest pole, hence x(n) will be a causal
(or right sided) signal.
z
Z {a n u(n)} =
; ROC |z| > |a|
\ x(n) = 1.5(0.6)n u(n) – 0.5 (0.2)n u(n)
z - a
b) ROC is |z| < 0.2
The specified ROC is interior of the circle whose radius corresponds to the smallest pole, hence x(n) will be an
anticausal (or left sided) signal.
\ x(n) = 1.5(–(0.6)n u(–n – 1)) – 0.5 (–(0.2)n u(–n – 1))
Z { -an u( -n - 1)} =
= –1.5 (0.6)n u(–n – 1) + 0.5 (0.2)n u(–n – 1)
z
; ROC |z| < |a|
z - a
c) ROC is 0.2 < |z| < 0.6
The specified ROC is the region in between two circles of radius 0.2 and 0.7. Hence the term corresponds to the pole,
p1 = 0.6 will be anticausal signal (because |z| < 0.6) and the term corresponds to the pole, p 2 = 0.2, will be a causal signal
(because |z| > 0.2).
\ x(n) = 1.5(–(0.6)n u(–n –1)) – 0.5 (0.2)n u(n)
= –1.5(0.6)n u(–n –1)) – 0.5 (0.2)n u(n)
7.6 Analysis of LTI Discrete Time System Using Z-Transform
7.6.1 Transfer Function of LTI Discrete Time System
Let x(n) be the input and y(n) be the output of an LTI discrete time system. The mathematical
equation governing the input-output relation of an LTI discrete time system is given by, (refer chapter-6,
equation (6.17)),
N
y( n) = -
åa
M
m
y( n - m) +
m=1
åb
m
.....(7.52)
x(n - m)
m=0
The equation (7.52) is a constant coefficient difference equation and N is the order of the system.
Let us take Z-transform of equation (7.52) with zero initial conditions (i.e., y(n) = 0 for n < 0 and
x(n) = 0 for n < 0).
RS a y(n - m) + b x(n - m)UV
å
W
T å
U
U
R
R
= Z S- å a y( n - m)V + Z S å b x(n - m)V
W T
W
T
N
l q
M
\ Z y(n) = Z -
m
m
m=1
m=0
N
M
m
m
m=1
N
= -
åa
m=1
m=0
M
m
l
q å b Z lx(n - m)q
Z y( n - m) +
m
m=0
.....(7.53)
Chapter 7 - Z - Transform
7. 48
Let y(n) = 0 for n < 0, now if Z{y((n)} = Y(z), then Z{y(n–m)} = z–m Y(z) (Using shifting
property)
Let x(n) = 0 for n < 0, now if Z{x((n)} = X(z), then Z{x(n–m)} = z–m X(z) (Using shifting
property)
Using shifting property of Z-transform, the equation (7.53) is written as shown below.
M
N
Y( z) = -
åa
m=1
m=0
åa
m
åb
m
z - m X(z)
OP = b
Q å
m
z - m X(z)
z - m Y(z) =
m=1
LM
N
Y( z) 1 +
z - m X(z)
m
M
N
Y( z) +
åb
z - m Y(z) +
m
m=0
M
N
åa
m=1
m
z- m
m=0
M
\
Y(z)
=
X(z)
åb
m
z- m
m=0
N
1 +
åa
m
z- m
m=1
On expanding the summations in the above equation we get,
Y( z)
b + b1 z -1 + b 2 z -2 + b 3 z -3 + ..... + b M z - M
= 0
X( z)
1 + a1 z -1 + a 2 z -2 + a 3 z -3 + ..... + a N z - N
.....(7.54)
The transfer function of a discrete time system is defined as the ratio of Z-transform of output
and Z-transform of input. Hence the equation (7.54) is the transfer function of an LTI discrete time
system.
The equation (7.54) is a rational function of z–1 (i.e., ratio of two polynomials in z–1). The numerator
and denominator polynomials of equation (7.54) are converted to positive power of z and then expressed
in the factorized form as shown in equation (7.55). (Refer equation (7.36)).
Let M = N
(z - z1 ) (z - z 2 ) (z - z3 ) ..... (z - z N )
Y( z)
= G
X( z)
(z - p1 ) (z - p2 ) (z - p3 ) ..... (z - p N )
.....(7.55)
where, z1, z2, z3, ..... zN are roots of numerator polynomial (or zeros of discrete time system)
p1, p2, p3, ..... pN are roots of denominator polynomial (or poles of discrete time system)
7.6.2 Impulse Response and Transfer Function
Let, x(n) = Input of an LTI discrete time system
y(n) = Output / Response of the LTI discrete time system for the input x(n)
h(n) = Impulse response (i.e., response for impulse input)
Now, the response y(n) of the discrete time system is given by convolution of input and impulse
response. (Refer chapter-6, equation (6.33)).
+¥
\ y(n) = x(n) * h(n) =
å x(m) h(n - m)
m = -¥
7. 49
Signals & Systems
On taking Z-transform of the above equation we get,
Z {y(n)} = Z {x(n) * h(n)}
Using convolution property of Z-transform, the above equation
can be written as,
Y( z) = X(z) H(z)
If Z{x(n)} = X(z)
and Z{h(n)} = H(z)
then by convolution property,
Z{x(n) * h(n)} = X(z) H(z)
\ H(z) =
Y(z)
X(z)
\ H(z) =
(z - z1 ) (z - z 2 ) (z - z3 ) ..... (z - z N )
Y(z)
Using equation (7.55)
= G
X(z)
(z - p1 ) (z - p 2 ) (z - p3 ) ..... (z - p N )
.....(7.56)
From equation (7.56) we can conclude that the transfer function of an LTI discrete time system is
also given by Z-transform of the impulse response.
Alternatively, we can say that the inverse Z-transform of transfer function is the impulse response
of the system.
Y(z)
Using equation (7.56)
\ Impulse reponse, h(n) = Z -1 H ( z) = Z -1
X(z)
l q
RS
T
UV
W
7.6.3 Response of LTI Discrete Time System Using Z-Transform
In general, the input-output relation of an LTI (Linear Time Invariant) discrete time system is
represented by the constant coefficient difference equation shown below, (equation (7.52)).
N
bg
b
M
g å b xb n - mg
å
å a yb n - mg = å b xbn - mg with a
yn = -
am y n - m +
m=1
N
(or )
m
m=0
M
m
m= 0
m
.....(7.57)
o
= 1
m= 0
The solution of the above difference equation (equation (7.57)) is the (total) response y(n) of LTI
discrete time system, which consists of two parts. In signals and systems the two parts of the solution
y(n) are called zero-input response yzi(n) and zero-state response yzs(n).
\ Response, y(n) = yzi(n) + yzs(n)
.....(7.58)
Zero-input Response (or Free Response or Natural Response) Using Z-Transform
The zero-input response yzi(n) is mainly due to initial output (or initial stored energy) in the system.
The zero-input response is obtained from system equation (equation (7.57)) when input x(n) = 0.
On substituting x(n) = 0 and y(n) = yzi(n) in equation (7.57) we get,
N
åa
m
b
g
y zi n - m = 0 ; with a o = 1
m= 0
On taking Z-transform of the above equation with non-zero initial conditions for output we can
form an equation for Yzi(z). The zero-input response yzi(n) of a discrete time system is given by inverse
Z-transform of Yzi(z).
Zero-State Response (or Forced Response) Using Z-Transform
The zero-state response yzs(n) is the response of the system due to input signal and with zero initial
output. The zero-state response is obtained from the difference equation governing the system
(equation(7.57)) for specific input signal x(n) for n ³ 0 and with zero initial output.
Chapter 7 - Z - Transform
7. 50
On substituting y(n) = yzs(n) in equation (7.57) we get,
N
å
m= 0
b
M
g å b xb n - mg ; with a
a m y zs n - m =
m
o
= 1
m= 0
On taking Z- transform of the above equation with zero initial conditions for output
(i.e., yzs(n)) and non-zero initial values for input (i.e., x(n)) we can form an equation for Yzs(z). The zerostate response yzs(n) of a discrete time system is given by inverse Z-transform of Y zs(z).
Total Response
The total response y(n) is the response of the system due to input signal and initial output (or intial
stored energy). The total response is obtained from the difference equation governing the system
(equation(7.57)) for specific input signal x(n) for n ³ 0 and with non-zero initial conditions.
On taking Z-transform of equation (7.57) with non-zero initial conditions for both input and output,
and then substituting for X(z) we can form an equation for Y(z). The total response y(n) is given by
inverse Z-transform of Y(z). Alternatively, the total response y(n) is given by sum of zero-input response
yzi(n) and zero-state response yzs(n).
\ Total response, y(n) = yzi(n) + yzs(n)
7.6.4 Convolution and Deconvolution Using Z-Transform
Convolution
The convolution operation is performed to find the response y(n) of an LTI discrete time system
from the input x(n) and impulse response h(n).
\ Response, y(n) = x(n) * h(n)
On taking Z-transform of the above equation we get,
Z{y(n)} = Z{x(n) * h(n)}
Using convolution property
\Y(z) = X(z) H(z)
.....(7.59)
\ Response, y(n) = Z––1{Y(z)} = Z––1{X(z) H(z)}
Procedure : 1.
2.
3.
4.
Deconvolution
Take Z-transform of x(n) to get X(z).
Take Z-transform of h(n) to get H(z).
Get the product X(z) H(z).
Take inverse Z-transform of the product X(z) H(z).
The deconvolution operation is performed to extract the input x(n) of an LTI system from the
response y(n) of the system.
From equation (7.59) get,
Y(z)
X( z) =
H(z)
On taking inverse Z-transform of the above equation we get,
Y(z)
Input, x(n) = Z -1 X(z) = Z -1
H(z)
l q
Procedure : 1.
2.
3.
4.
RS
T
UV
W
Take Z-transform of y(n) to get Y(z).
Take Z-transform of h(n) to get H(z).
Divide Y(z) by H(z) to get X(z) (i.e., X(z) = Y(z) / H(z)).
Take inverse Z-transform of X(z) to get x(n).
7. 51
Signals & Systems
7.6.5 Stability in z-Domain
Location of Poles for Stability
Let, h(n) be the impulse response of an LTI discrete time system. Now, if h(n) satisfies the condition,
+¥
å
.....(7.60)
h(n) < ¥
n = -¥
then the LTI discrete time system is stable, (Refer equation 6.27 in chapter - 6).
The stability condition of equation (7.60) can be transformed as a condition on location of poles of
transfer function of the LTI discrete time system in z-plane.
Let, h(n) = an u(n)
+¥
Now ,
å
+¥
å
h(n) =
n = -¥
¥
å
a n u(n) =
n = -¥
an
n= 0
¥
If |a| is such that, 0 < |a| < 1, then
åa
n
n= 0
=
1
= constant, and so the system is stable.
1- a
¥
If |a| > 1, then
åa
n
= ¥ and so the system is unstable.
n= 0
l q
n
s
Now, H(z) = Z h(n) = Z a n u(n) =
z
z - a
Here H(z) has pole at z = a.
If |a| < 1, then the pole will lie inside the unit circle and if |a| > 1, then the pole will lie outside the unit
circle. Therefore we can say that, for a stable discrete time system the pole should lie inside the unit
circle. The various types of impulse response of LTI system and their transfer functions and the locations
of poles are summarized in table 7.5.
ROC of a Stable System
Let, H(z) be Z-transform of h(n). Now, by definition of Z-transform we get,
+¥
å
H ( z) =
h(n) z - n
n = -¥
Let us evaluate H(z) for z = 1.
+¥
\ H ( z) =
å
h(n)
n = -¥
On taking absolute value on both sides we get,
+¥
H ( z) =
å
+¥
Þ
h(n)
H ( z) =
n = -¥
å
h(n)
n = -¥
For a stable LTI discrete time system,
+¥
å
h(n) < ¥
Þ
H ( z) < ¥
n = -¥
Therefore, we can conclude that z = 1 will be a point in the ROC of a stable system. Hence for a
stable discrete time system the ROC should include the unit circle.
General Condition for Stability in z-plane
On combining the condition for location of poles and the ROC we can say that for a stable LTI
discrete time system the poles should lie inside the unit circle and the unit circle should be included in ROC.
Chapter 7 - Z - Transform
7. 52
Table-7.5 : Impulse Response and Location of Poles
Impulse response
h(n)
Transfer function
H(z)
h(n) = an u(n) ; 0 < a < 1
h(n)
Location of poles in
z-plane and ROC
Unit
circle
jv
z
z - a
ROC is | z| > a
z-plane
H ( z) =
xa
pole at z = a
u
ROC
n
+¥
Since 0 < a < 1, the pole z = a, lies inside the
unit circle. The ROC contains the unit circle.
å h(n) < ¥ ; Stable system
n=0
h(n) = (–a)n u(n) ; 0 < |–a| < 1
h(n)
Unit
circle
z
z + a
ROC is | z| > |- a|
H ( z) =
n
jv
z-plane
–a x
u
ROC
pole at z = - a
+¥
å
Since 0 < |–a| < 1, the pole at z = –a, lies inside
the unit circle. The ROC contains the unit
circle.
h(n) < ¥ ; Stable system
n=0
h(n) = an u(n) ; a > 1
h(n)
Unit
circle
z
H ( z) =
z - a
ROC is | z| > a
n
+¥
å
z-plane
xa
pole at z = a
h(n) = ¥ ; Unstable system
jv
u
ROC
Since a > 1, the pole at z = a, lies outside the
unit circle. The ROC does not contain the unit
circle.
jv
Unit
z-plane
circle
n=0
h(n) = (–a)n u(n) ; |–a| > 1
h(n)
z
z + a
ROC is | z| > |- a|
H ( z) =
n
¥
å
n=0
h(n) = ¥ ; Unstable system
pole at z = - a
x
–a
u
ROC
Since |–a| > 1, the pole at z = –a, lies outside
the unit circle. The ROC does not contain the
unit circle.
7. 53
Signals & Systems
Table-7.5 : Continued....
Impulse response
h(n)
Transfer function
H(z)
h(n) = a u(n) ; a > 0 (i.e., a is positive)
az
H ( z) =
z - 1
ROC is | z| > 1
pole at z = 1
h(n)
a
n
Location of poles in
z-plane and ROC
jv
Unit
circle
z-plane
x1
u
ROC
¥
å h(n) = ¥ ; Unstable system
The pole z = 1 lies on the unit circle. The
ROC does not contain the unit circle.
n=0
h(n) = a(–1)n u(n) ; a > 0 (i.e., a is positive)
h(n)
jv
Unit
circle
az
z + 1
ROC is | z| > 1
pole at z = - 1
z-plane
H ( z) =
a
n
x
u
–1
ROC
–a
The pole at z = –1 lies on the unit circle.
The ROC does not contain the unit circle.
+¥
å h(n) = ¥ ; Unstable system
n=0
h(n) = n an u(n) ; 0 < a < 1
h(n)
jv
Unit
circle
az
(z - a) 2
ROC is | z| > a
Two poles at z = a
z-plane
H ( z) =
n
¥
å h(n) < ¥ ; Stable system
n=0
xxa
ROC
Since 0 < a < 1, the two poles at z = a, lie
inside the unit circle. The ROC contains the
unit circle.
h(n) = n (–a)n u(n) ; 0 < |–a| < 1
az
(z + a) 2
ROC is | z| > a
Two poles at z = - a
H ( z) =
n
¥
å h(n) < ¥ ; Stable system
jv
Unit
circle
h(n)
n=0
u
xx
–a
z-plane
u
ROC
Since 0 < |–a| < 1, the two poles at z = –a,
lie inside the unit circle. The ROC contains
the unit circle.
Chapter 7 - Z - Transform
7. 54
Table-7. 5 : Continued....
Impulse response
h(n)
Transfer function
H(z)
h(n) = n an u(n) ; a > 1
h(n)
Location of poles in
z-plane and ROC
Unit
circle
az
(z - a) 2
ROC is | z| > a
Two poles at z = a
jv
z-plane
H ( z) =
¥
n
xxa
u
ROC
Since a > 1, the two poles at z = a, lie outside
the unit circle. The ROC does not contain the
unit circle.
å h(n) = ¥ ; Unstable system
n=0
h(n) = n (–a)n u(n) ; |–a| > 1
h(n)
Unit
circle
az
(z + a) 2
ROC is | z| > |-a|
Two poles at z = - a
H ( z) =
n
jv
z-plane
xx
–a
u
ROC
Since |–a| > 1, the two poles at z = –a, lie
outside the unit circle. The ROC does not
contain the unit circle.
¥
å h(n) = ¥ ; Unstable system
n=0
h(n) = n u(n)
h(n)
z
H ( z) =
(z - 1) 2
ROC is | z| > 1
Two poles at z = 1
¥
n
h(n) = n (–1)n u(n) ; |–a| > 1
h(n)
n
¥
z-plane
xx1
u
ROC
Unit
circle
z
(z + 1) 2
ROC is | z| > 1
H ( z) =
å h(n) = ¥ ; Unstable system
jv
The two poles at z = 1, lie on the unit circle.
The ROC does not contain the unit circle.
å h(n) = ¥ ; Unstable system
n=0
n=0
Unit
circle
Two poles at z = - 1
jv
z-plane
xx
–1
u
ROC
The two poles at z = –1, lie on the unit circle.
The ROC does not contain the unit circle.
7. 55
Signals & Systems
Table-7.5 : Continued....
Transfer function
H(z)
h(n) = rn cos w0n u(n) ; 0 < r < 1
h(n)
Location of poles in
z-plane and ROC
Unit
circle
H ( z)
r
---
r
u
p2
---x
n
p1
z(z - r cosw 0 )
(z - r cosw 0 - jr sinw 0 )
(z - r cosw 0 + jr sinw 0 )
---
=
cos w0n
z-plane
---
rn
jv
x
Impulse response
h(n)
ROC
ROC is |z| > r.
A pair of conjugate poles at
z = p1 = r cos w0 + jr sin w0 Since 0 < r < 1, the conjugate pole pairs
z = p2 = r cos w0 – jr sinw0 lie inside the unit circle. The ROC
contains the unit circle
¥
å h(n) < ¥ ; Stable system
n=0
h(n) = rn cos w0 n u(n) ; r > 1
h(n)
Unit
circle
z-plane
p1
---
---
x
H ( z)
jv
---
---
r
z(z - r cosw 0 )
(z - r cosw 0 - jr sinw 0 )
u
–a
r
(z - r cosw 0 + jr sinw 0 )
p
n ROC is |z| > r.
ROC
A pair of conjugate poles at
z = p1 = r cos w0 + jr sin w0
z = p2 = r cos w0 – jr sinw0 Since r > 1, the conjugate pole pairs
---
---
=
---
---
---
---
---
--- x
2
lie outside the unit circle.
The ROC does not contain the unit
circle.
¥
å h(n) = ¥ ; Unstable system
n=0
h(n) = cos w0 n u(n)
Unit
circle
H ( z)
-----------
å h(n) = ¥ ; Unstable system
n=0
p2
ROC is |z| > 1.
A pair of conjugate poles
The conjugate pole pairs lie on the
on unit circle at,
z = p1 = cos w0 + j sin w0 unit circle.
The ROC does not contain the unit
z = p2 = cos w0 – j sinw0
circle.
¥
---x
ROC
–1
u
---
(z - cosw 0 + j sinw 0 )
n
p1
---
1
z-plane
x
z(z - cosw 0 )
=
(z - cosw 0 - j sinw 0 )
h(n)
jv
Chapter 7 - Z - Transform
7. 56
7.7 Relation Between Laplace Transform and Z-Transform
7.7.1 Impulse Train Sampling of Continuous Time Signal
Consider a periodic impulse train p(t) shown in fig 7.9a, with period T. The pulse train can be
mathematically expressed as shown in equation (7.61).
¥
å d ( t - nT)
p( t ) =
..... (7.61)
n = -¥
When a continuous time signal x(t) is multiplied by the impulse train p(t), the product signal will
have impulses. A continuous time signal x(t) and the product of x(t) and p(t) are shown in fig 7.9b and fig
7.9c respectively. In fig 7.9c, the magnitudes of the impulses are equal to magnitude of x(t), and so the
product signal is impulse sampled version of x(t), with sampling period T. Let us denote the product
signal as xp(t) and it is mathematically expressed as shown in equation (7.62).
¥
xp ( t) =
å x(nT) d( t - nT)
..... (7.62)
n =-¥
where, x(nT) are samples of x(t) at t = nT
p(t)
x(t)
–4T –3T –2T –T 0
T
2T 3T 4T
0
t
Fig 7.9a : Impulse train.
xp(t)
–4T –3T –2T –T 0 T
t
2T 3T 4T
t
Fig 7.9b : Continuous time signal.
Fig 7.9c : Samples of continuous
time signal.
Fig 7.9 : Impulse sampling of continuous time signals.
7.7.2 Transformation From Laplace Transform to Z-Transform
Let x(t) be a continuous time signal, and xp(t) be its impulse sampled version of discrete time
signal. From equation (7.62) we get,
¥
L{d(t)} = 1
xp ( t) =
x(nT) d ( t - nT)
If L{x(t)} = X(s) then
n =-¥
by time shifting property
On taking Laplace transform of the above equation we get,
L{x(t–a)} = e –as X(s)
å
q = X (s) = L RS å x(nT) d(t - nT)UV = å x(nT) Lld(t - nT)q
T
W
\ X (s) = å x(nT) e
= å x(nT) de i
¥
l
L xP ( t)
¥
P
n =-¥
¥
n =-¥
¥
- nsT
sT
-n
p
n =-¥
n =-¥
..... (7.63)
where Xp(s) is Laplace transform of xp(t).
Let us take a transformation, esT = z.
On substituting, esT = z, in equation (7.63) we get,
¥
X p (s) =
å x(nT) z
n =-¥
-n
.
.... (7.64)
7. 57
Signals & Systems
The Z-transform of x(nT), using the definition of Z-transform is given by,
¥
X(z) =
å x(nT) z
-n
.
n =-¥
.... (7.65)
On comparing equations (7.64) and (7.65) we can say that, if a discrete time signal x(nT) is a
sampled version of x(t), then Z-transform of the discrete time signal can be obtained from Laplace
transform of sampled version of x(t), by choosing the transformation, esT = z. This transformation is also
called impulse invariant transformation.
7.7.3 Relation Between s-Plane and z-Plane
Consider a point s1 in s-plane as shown in fig 7.10. Now the transformation,
..... (7.66)
will transform the point s 1 to a corresponding point z1 in z-plane.
Let the coordinates of s1 be s1 and W1 as shown in fig 7.10.
..... (7.67)
\ s1 = s 1 + jW 1
Using equation (7.67) the equation (7.66) can be written as,
z1 = e
(s 1 + jW 1 )T
= e
s 1T
e
jW
LHP
s-plane
RHP
jW 1 ------- s 1
-------
e s1T = z1
s1
s
Fig 7.10 : s-plane.
jW 1T
..... (7.68)
On separating the magnitude and phase of equation (7.68) we get,
..... (7.69)
z1 = e (s 1 + jW1 )T = e s 1T e jW1T
From equation (7.69) the following observations can be made.
1. If s1 < 0 (i.e., s1 is negative), then the point-s1 lies on Left Half (LHP) of s-plane.
In this case, |z1| < 1, hence the corresponding point-z1 will lie inside the unit circle in z-plane.
2. If s1 = 0 (i.e., real part is zero), then the point-s1 lies on imaginary axis of s-pane.
In this case, |z1| = 1, hence the corresponding point-z1 will lie on the unit circle in z-plane.
3. If s1 > 0 (i.e., s1 is positive), then the point-s1 lies on the Right Half (RHP) of s-plane.
In this case |z1| >1, hence the corresponding point-z1 will lie outside the unit circle in z-plane.
The above discussions are applicable for mapping of any point on s-plane to z-plane.
In general all points of s-plane, described by the equation,
2 pk
, for k = 0, ± 1, ± 2 .....
T
map as a single point in the z-plane described by equation,
s = s 1 + jW1 + j
z1 = e
FG s1 + jW1 +
H
IJ
K
j2 pk
T
T
= e s1T e jW1T e j2 pk = e s1T e jW1T
..... (7.70)
e ± j2 pk = 1 ; for integer k
..... (7.71)
The equation (7.70) represents a strip of width 2p/T in the s-plane for values of imaginary part of
s in the range –p/T £ W £ +p/T is mapped into the entire z-plane. Similarly the strip of width 2p/T in the
s-plane for values of imaginary part of s in the range p/T £ W £ 3p/T is also mapped into the entire
z-plane. Likewise the strip of width 2p/T in the s-plane for values of imaginary part of s in the range
-3p/T £ W £ -p/T is also mapped into the entire z-plane.
Chapter 7 - Z - Transform
7. 58
In general any strip of width 2p/T in the s-plane for values of imaginary part of s in the range
(2k – 1)p/T £ W £ (2k + 1) p/T, where k is an integer, is mapped into the entire z-plane. Therefore we
can say that the transformation, esT = z, leads to many-to-one mapping, (and does not provide one-to-one
mapping).
In this mapping, the left half portion of each strip in s-plane maps into the interior of the unit
circle in z-plane, right half portion of each strip in s-plane maps into the exterior of the unit circle in z-plane
and the imaginary axis of each strip in s-plane maps into the unit circle in z-plane as shown in fig 7.11.
jW
jv
3p/T
Unit circle
RHP
LHP
¬
p/T
s
j1
–1
1
u
-p/T
–j1
-3p/T
Fig 7.11a : s-plane.
Fig 7.11b : z-plane.
Fig 7.11 : Mapping of s-plane into z-plane.
Relation Between Frequency of Continuous Time and Discrete Time Signal
Let, W = Frequency of continuous time signal in rad/sec.
w = Frequency of discrete time signal in rad/sec.
Let, z = rejw be a point on z-plane, and s = s + jW, be a corresponding point in s-plane.
Consider the transformation,
z = esT
..... (7.72)
Put, z = r ejw and s = s + jW in equation (7.72)
\ r e jw = e ( s + jW ) T
..... (7.73)
r e jw = e sT e jWT
On equating the imaginary part on either side of equation (7.73) we get,
w
..... (7.74)
T
When the transformation esT = z is employed, the equation (7.74) can be used to compute the
frequency of discrete time signal for a given frequency of continuous time signal and viceversa.The
frequency of discrete time signal w is unique over the range (-p, +p), and so the mapping w = WT
implies that the frequency of continuous time signal in the interval -p/T £ W £ +p/T maps into the
corresponding values of frequency of discrete time signal in the interval -p £ w £ +p.
w = WT
or
W =
The mapping of s-plane to z-plane, using the transformation, esT = z is not one-to-one. Therefore
in general, the interval (2k-1)p/T £ W £ (2k +1)p/T, where k is an integer, maps into the corresponding
values of -p £ w £ +p. Thus the mapping of the frequency of continuous time signal W to the frequency
of discrete time signal w is many-to-one. This reflects the effects of aliasing due to sampling.
7. 59
Signals & Systems
7.8 Structures for Realization of LTI Discrete Time Systems in z-Domain
In time domain, the input-output relation of an LTI (Linear Time Invariant) discrete time system is
represented by constant coefficient difference equation shown in equation (7.75).
N
bg
å
yn = -
b
M
g å b xb n - mg
am y n - m +
m=1
.....(7.75)
m
m=0
In z-domain, the input-output relation of an LTI (Linear Time Invariant) discrete time system is
represented by the transfer function H(z), which is a rational function of z, as shown in equation (7.76).
H ( z) =
b + b1 z -1 + b 2 z -2 + b 3 z -3 + ..... + b M z - M
Y( z)
= 0
X( z)
1 + a 1 z -1 + a 2 z -2 + a 3 z -3 + ..... + a N z - N
.....(7.76)
where, N = Order of the system, M £ N
The above two representations of discrete time system can be viewed as a computational procedure
(or algorithm) to determine the output signal y(n) from the input signal x(n). The computations in the
above equation can be arranged into various equivalent sets of difference equations.
For each set of equations, we can construct a block diagram consisting of Adder, Constant
multiplier, Unit delay element and Unit advance element. Such block diagrams are referred to as realization
of system or equivalently as structure for realizing system.The basic elements used to construct block
diagrams are listed in table 7.6. The block diagram representation of discrete time system in time domain
are discussed in chapter-6, section 6.6.2, and the block diagram representation of discrete time system in
z-domain are discussed in chapter-10.
Table 7.6 : Basic elements of block diagram in time domain and z-domain
Elements of
block diagram
Time domain
representation
x1(n)
Adder
z-domain
representation
x1(n) + x2(n)
+
X1 (z)
x2(n)
Constant multiplier
Unit delay element
a
x(n)
a X (z)
X(z)
a
x (n -1)
z-1
x (n + 1)
x(n)
Unit advance element
X 2(z)
ax (n)
x(n)
X1(z) + X2(z)
+
z
X(z)
z-1X (z)
z-1
z X (z)
X(z)
z
Chapter 7 - Z - Transform
7. 60
Example 7.12
Determine the impulse response h(n) for the system described by the second order difference equation,
y(n) – 4y(n – 1) + 4y(n – 2) = x(n – 1).
Solution
The difference equation governing the system is,
y(n) – 4y(n – 1) + 4y (n – 2) = x(n – 1)
Let us take Z-transform of the difference equation governing the system with zero initial conditions.
If Z {x(n)} = X(z)
then by shifting property
Z {x(n – m} = z–m X(z)
\ Z{y(n) – 4y(n – 1) + 4y (n – 2)} = Z{x(n – 1)}
Z{y(n)} – 4 Z{y(n – 1)} + 4 Z{y(n – 2)} = Z{x(n – 1)}
Y(z) – 4z–1 Y(z) + 4z–2 Y(z) = z–1 X(z)
–1
–2
If Z {y(n)} = Y(z)
then by shifting property
Z {y(n – m} = z–m Y(z)
–1
(1 – 4z + 4z ) Y(z) = z X(z)
Y(z)
z -1
=
-1
X(z) Y(z)1 - 4z + 4 z -2
We know that,
= H(z)
X( z )
\
\ H(z) =
z -1
1 - 4z
-1
+ 4z
-2
=
z -1
-2
z (z
2
- 4z + 4)
=
z
(z - 2)2
=
1 2z
2 ( z - 2) 2
The impulse response h(n) is given by inverse Z-transform of H(z).
Impulse response, h(n) = Z -1 {H(z)} = Z -1
RS 1 2z UV
T 2 ( z - 2) W
2
Z {na n u(n)} =
az
(z - a )2
= (1/2) n2n u(n) = n2(n – 1) u(n)
Example 7.13
Find the transfer function and unit sample response of the second order difference equation with zero initial condition,
y(n) = x(n) – 0.25y(n – 2).
Solution
The difference equation governing the system is,
y(n) = x(n) – 0.25 y(n – 2)
Let us take Z-transform of the difference equation governing the system with zero initial condition.
Z{y(n)} = Z{x(n – 0.25 y(n – 2)}
Z{y(n)} = Z{x(n)} – 0.25 Z{y(n – 2)}
-2
Y(z) = X(z) – 0.25 z Y(z)
Y(z) + 0.25z–2 Y(z) = X(z)
(1 + 0.25z–2) Y(z) = X(z)
\ Transfer function,
Y(z)
1
=
X(z)
1 + 0.25z -2
Z {x(n)} = X(z)
Z {y(n)} = Y(z)
Z {y(n – 2)} = z–2 Y(z)
(Using shifting property)
7. 61
Signals & Systems
We know that,
Y(z)
= H(z)
X(z)
\ H(z) =
(a + b) (a - b) = a 2 - b 2
j2 = -1
1
1
z2
= -2 2
=
-2
(z + j 0.5) (z - j 0.5)
1 + 0.25 z
z (z + 0.25)
By partial fraction expansion we can write,
H(z)
z
A
A*
=
=
+
z
(z + j0.5) (z - j0.5)
z + j 0.5
z - j0.5
H(z)
z
A = ( z + j0.5)
=
z
z – j0.5 z =
= (z + j0.5)
z = - j0.5
=
- j0.5
; where A* is conjugate of A.
z
(z + j0.5) (z - j0.5) z =
- j0.5
- j0.5
- j0.5
1
=
=
= 0.5
- j0.5 - j0.5
2(- j0.5)
2
\ A* = 0.5
0.5
0.5
A*
A
H(z)
+
+
=
=
z - j0.5
z + j0.5
z - j0.5
z + j0.5
z
\ H(z) =
0.5z
0.5z
0.5z
0.5z
=
+
+
z + j0.5
z - j0.5
z - (- j0.5)
z - j0.5
The impulse response is obtained by taking inverse Z-transform of H(z).
RS 0.5z + 0.5z UV
T z - (- j0.5) z - j0.5 W
RS z UV + Z RS z UVOP
T z - (- j0.5) W
T z - j0.5 WPQ
\ Impulse response, h(n) = Z -1 H(z) = Z -1
l
L
= 0.5 MZ
MN
q
-1
-1
Z{an u(n)} =
= 0.5 (- j0.5)n u(n) + ( j0.5)n u(n)
Alternatively the impulse response can be expressed as shown below.
Here, - j0.5 = 0.5Ð - 90° = 0.5Ð - p / 2 = 0.5 Ð - 0.5p
+ j0.5 = 0.5Ð 90°
\
= 0.5Ðp / 2
= 0.5 Ð0.5p
n
h(n) = 0.5 (0.5Ð - 0.5p) + (0.5Ð0.5p )n u(n)
= 0.5 0.5n Ð - 0.5np + 0.5n Ð 0.5np u(n)
= 0.5 (0.5)n cos 0.5np - jsin 0.5np + cos 0.5np + jsin 0.5np u(n)
= 0.5 (0.5)n 2 cos 0.5np u(n)
= 0.5n cos (0.5np) u(n)
Example 7.14
Determine the impulse response sequence of the discrete time LTI system defined by,
y(n) – 2y(n – 1) + y(n – 2) = x(n) + 3x(n – 3).
Solution
The difference equation governing the LTI system is,
y(n) – 2y(n – 1) + y(n – 2) = x(n) + 3x(n – 3)
z
z - a
Chapter 7 - Z - Transform
7. 62
Let us assume that the initial conditions are zero.
On taking Z-transform of the difference equation governing
the system we get,
Z{x(n)} = X(z) ,
\
Z{ax(n – m} = az–mX(z)
Z{y(n)} = Y(z) ,
\
Z{ay(n – m} = az–mY(z)
Z{y(n) – 2y(n – 1) + y(n – 2)} = Z{x(n) + 3x(n – 3)}
Z{y(n)} – 2 Z{y(n – 1)} + Z{y(n – 2)} = Z{x(n)} + 3 Z{x(n – 3)}
l q
Zln u(n)q =
Y(z) – 2z–1 Y(z) + z–2 Y(z) = X(z) + 3z–3 X(z)
Z u(n) =
(1 – 2z–1 + z–2) Y(z) = (1 + 3z–3) X(z)
\
Y (z)
1 + 3z -3
=
X( z )
1 - 2z -1 + z-2
We know that,
l
=
z
(z - 1)2
Z (n + 1) u(n + 1) = z
Y(z)
= H(z)
X(z)
\ H(z) =
q
z
z - 1
z
(z - 1)2
Z (n - 2) u(n - 2) = z -2
l
q
z
(z - 1)2
1 + 3z -3
1 + 3z-3
z 2 + 3z -1
= -2 2
=
-1
-2
1 - 2z + z
z (z - 2z + 1)
(z - 1)2
z2
2
(z - 1)
+
3z -1
(z - 1)2
z
= z
(z - 1)2
+ 3z -2
z
(z - 1)2
The impulse response is obtained by taking inverse Z-transform of H(z).
R| z + 3z
S| bz - 1g
T
R|z z U| + 3Z
S| bz - 1g V|
T
W
\ Impulse r esponse, h(n) = Z-1 H(z) = Z-1 z
l q
= Z-1
-2
2
U|
bz - 1g V|W
R|z z
S| bz - 1g
T
2
-1
2
z
-2
2
U|
V|
W
= (n + 1) u(n + 1) + 3(n – 2) u(n – 2)
Example 7.15
Find the impulse response of the system described by the difference equation,
y(n) – 3y(n – 1) – 4y (n – 2) = x(n) + 2x(n – 1).
Solution
The difference equation governing the LTI system is,
Z{y(n)} = Y(z) ; \
y(n) – 3y(n – 1) – 4y(n – 2) = x(n) + 2x(n – 1)
Z{y(n – m)} = z–m Y(z)
Z{x(n)} = X(z) ; \ Z{x(n – m)} = z–m X(z)
On taking Z-transform we get,
Y(z) – 3z–1Y(z) – 4 z–2Y(z) = X(z) + 2z–1 X(z)
The roots of the quadratic,
(1 – 3z – 1 – 4z – 2) Y(z) = (1 + 2z–1) X(z)
\
z 2 - 3z - 4 = 0 are,
Y (z)
1 + 2z -1
=
X( z )
1 - 3z-1 - 4z -2
We know that
z =
Y (z)
= H(z)
X( z )
\ H(z) =
1 + 2z -1
1 - 3z
-1
- 4z
-2
=
z -2 (z 2 + 2z)
-2
z (z
2
- 3z - 4 )
=
z 2 + 2z
(z - 4) (z + 1)
3 ± 32 + 4 ´ 4
= 4 or – 1
2
7. 63
Signals & Systems
By partial fraction expansion technique,
H(z)
z + 2
A
B
=
=
+
z
(z - 4 )(z + 1)
z - 4
z + 1
A = (z - 4)
B = (z + 1)
H(z)
z z=
H(z)
z z=
= (z - 4)
4
= (z + 1)
-1
z + 2
(z - 4) (z + 1) z
=
= 4
z + 2
(z - 4) (z + 1) z =
=
-1
z + 2
z + 1z=
z + 2
z - 4z=
=
4
=
-1
4 + 2
6
.
=
= 12
4 + 1
5
1
-1 + 2
=
= - 0.2
-1 - 4
-5
H(z)
A
B
12
.
0.2
\
=
+
=
z
z - 4
z + 1
z - 4
z + 1
\ H(z) = 12
.
FG
H
z
z
z
- 0.2
= 12
.
z - 4
z + 1
z - 4
IJ
K
- 0.2
FG z IJ
H z - (-1) K
Z
RS z UV = a
Tz - a W
n
The impulse response is obtained by taking inverse Z-transform of H(z).
\ Impulse response, h(n) = 1.2(4)n u(n) – 0.2(–1)n u(n)
Example 7.16
Determine the steady state response for the system with impulse function h(n) = (j0.5)n u(n) for an input
x(n) = cos (pn) u(n).
Solution
Let y(n) be the steady state response of the system, which is given by convolution of x(n) and h(n).
\ Steady state response, y(n) = x(n) * h(n)
On taking Z-transform of the above equation we get,
Z{y(n)} = Z{x(n) * h(n)}
\ Y(z) = X(z) H(z)
Using convolution property
\ y(n) = Z–1{X(z) H(z)}
Given that, h(n) = (j0.5)n u(n)
n s = z -z a
Zlcos(w n) u(n)q =
z
Z a n u(n)
z
z - j 0.5
\ H(z) = Z{h(n)} =
2
z(z - cos w )
- 2z cos w + 1
Given that, x(n) = cos(pn) u(n)
\ X(z) = Z{x(n)} =
z(z - cos p)
z(z + 1)
z(z + 1)
z
= 2
=
=
z 2 - 2z cos p + 1
z + 2z + 1
(z + 1)2
z + 1
\ Y(z) = X(z) H(z) =
z
z
z2
´
=
z + 1
(z + 1) (z - j0.5)
z - j0.5
By partial fraction expansion technique we can write,
Y (z)
z
A
B
=
=
+
z
(z + 1) (z - j0.5)
z + 1
z - j0.5
A = (z + 1)
Y(z)
z z=
= (z + 1)
-1
z
(z + 1) (z - j0.5) z =
=
-1
z
z - j0.5 z =
-1 + j0.5
–1
1 - j0.5
1 - j0.5
=
= 2
=
= 0.8 - j0.4
´
-1 + j0.5
–1 – j0.5
1 + 0.52
125
.
=
-1
-1
-1 - j0.5
Chapter 7 - Z - Transform
7. 64
B = (z - j0.5)
=
\
Y( z )
z
z
j0.5
= (z - j0.5)
=
=
z z = j0.5
(z + 1) (z - j0.5) z = j0.5
z + 1 z = j0.5
j0.5 + 1
j0.5 - (j0.5)2
j0.5
1 - j0.5
0.25 + j0.5
´
=
= 0.2 + j0.4
=
1 + j0.5
1 - j0.5
125
.
12 + 0.52
Y (z)
A
B
0.8 - j0.4
0.2 + j0.4
+
+
=
=
z
z + 1
z - j0.5
z + 1
z - j0.5
\ Y (z) = (0.8 - j0.4)
= (0.8 - j0.4)
z
z
+ (0.2 + j0.4 )
z + 1
z - j0.5
Z{an u(n)} =
z
z
+ (0.2 + j0.4)
z - j0.5
z - ( -1)
z
z - a
The steady state response is obtained by taking inverse Z-transform of Y(z).
\ Steady state response, y(n) = (0.8 – j0.4) (–1)n u(n) + (0.2 + j0.4) (j0.5)n u(n)
Alternatively the steady state response can be expressed as shown below.
Here, 0.8 - j0.4 = 0.894 Ð - 26.6o = 0.894 Ð - 015
. p
0.2 + j0.4 = 0.447 Ð63.4 o
o
- 1 = 1 Ð180
= 0.447 Ð0.35p
= 1 Ðp
j0.5 = 0.5 Ð90o = 0.5 Ð0.5p
\ y(n) = 0.894 Ð - 015
. p [1 Ðp ]n u(n) + 0.447 Ð0.35p
[0.5Ð0.5p ]n u(n)
= 0.894 Ð - 0.15p 1n Ð np u(n) + 0.447Ð 0.35p
0.5n Ð 0.5np u(n)
= 0.894 Ð(n - 0.15)p u(n) + 0.447 (0.5)n Ð(0.5n + 0.35)p u(n)
Example 7.17
Obtain and sketch the impulse response of shift invariant system described by,
y(n) = 0.4 x(n) + x(n – 1) + 0.6 x(n – 2) + x(n – 3) + 0.4 x(n – 4).
Solution
The difference equation governing the system is,
y(n) = 0.4 x(n) + x(n – 1) + 0.6 x(n – 2) + x(n – 3) + 0.4 x(n – 4)
If Z{x(n)} = X(z) then by shifting property
On taking Z-transform we get,
Z {x(n - k )} = z -k X(z)
Y(z) = 0.4X(z) + z–1X(z) + 0.6z–2X(z) + z–3X(z) + 0.4z–4X(z)
Y(z) = [0.4 + z–1 + 0.6z–2 + z–3 + 0.4z–4] X(z)
\
Y (z )
= 0.4 + z -1 + 0.6z -2 + z-3 + 0.4z -4
X (z)
We know that,
Y(z)
= H(z) ;
X(z)
\ H(z) = 0.4 + z -1 + 0.6z -2 + z -3 + 0.4z -4
.....(1)
7. 65
Signals & Systems
By the definition of one sided Z-transform we get,
h(n)
+¥
å h(n)z
H(z) =
-n
1.0
1.0
n = 0
= h(0) z0 + h(1) z–1 + h(2) z–2 + h(3) z–3 + h(4) z–4 +.... .....(2)
On comparing equations (1) & (2) we get,
h(0) = 0.4
h(3) = 1
h(1) = 1
h(4) = 0.4
h(2) = 0.6
h(n) = 0
1.0
0.8
0.6
0.6
0.4
0.4
0.2
; for n < 0 and n > 4
0
\ Impulse response, h(n) = {0.4, 1.0, 0.6, 1.0, 0.4}
­
1
2
3
4
5
n
6
Fig 1 : Graphical representation of
impulse response h(n).
Example 7.18
Determine the response of discrete time LTI system governed by the difference equation y(n) = – 0.5 y(n – 1) + x(n),
when the input is unit step and initial condition, a) y(–1) = 0 and b) y(–1) = 1/3.
Solution
Given that, x(n) = u(n)
m r m r
\ X(z) = Z x(n) = Z u(n) =
;
Given that, y(n) = – 0.5 y(n – 1) + x(n)
\ y(n) + 0.5 y(n – 1) = x(n)
z
z -1
If Z{y(n)}=Y(z)
On taking Z-transform of above equation we get,
then Z{y(n – 1)}
Y(z) + 0.5 z -1 Y (z) + y( -1) = X(z)
Y(z) 1+ 0.5 z -1 + 0.5 y(-1) =
.....(1)
z
z -1
= z –1 Y(z) – y(–1)
Using equation (1)
FG 0.5 IJ = z - 0.5 y(-1)
H z K z -1
F z + 0.5 IJ = z - 0.5 y(-1)
Y( z ) G
H z K z -1
Y( z ) 1 +
\ Y( z ) =
Let, P(z) =
Let,
1 z y( -1)
z2
(z - 1) (z + 0.5) 2 z + 0.5
z2
(z - 1) (z + 0.5)
P(z)
z
=
z
(z - 1) (z + 0.5)
Þ
z
A
B
=
+
(z - 1) (z + 0.5) z - 1 z + 0.5
A=
B=
z
´ (z - 1)
(z - 1) (z + 0.5)
z =1
z
´ (z + 0.5)
(z - 1) (z + 0.5)
=
1
1
2
=
=
.
1 + 0.5 15
3
z= -0.5
=
-0.5
-0.5
5
1
=
=
=
-0.5 - 1 -15
.
15 3
1
1
2 z
1
P(z) 2 1
z
=
+
Þ
+
P(z) =
3 z - 1 3 z + 0.5
3 z - 1 3 z + 0.5
z
2 z
1
1 z y( -1)
z
\ Y( z ) =
+
3 z - 1 3 z + 0.5 2 z + 0.5
\
.....(2)
a) When y(–1) = 0
From equation (2), when y(–1) = 0, we get,
Y(z) =
z
2 z
1
+
3 z - 1 3 z + 0.5
\Response, y(n) = Z-1 Y(z) = Z-1
l q
RS 2
T3
z
z
1
+
z - 1 3 z + 0.5
UV =
W
2
1
1
u(n) + (-0.5)n u(n) = 2 + (-0.5)n u(n)
3
3
3
Chapter 7 - Z - Transform
7. 66
b) When y(–1) = 1/3
From equation (2), when y(–1) = 1/3, we get,
2 z
1
z
1
+
- ´
3 z - 1 3 z + 0.5 2
2 z
1 1
z
=
+ ´
=
3 z - 1 2 3 z + 0.5
Y(z) =
l q
\Response, y(n) = Z-1 Y(z) = Z-1
RS 2
T3
1
z
3 z + 0.5
2 z
1
z
+
3 z - 1 6 z + 0.5
z
z
1
+
z - 1 6 z + 0.5
UV
W
2
1
1
u(n) + (-0.5)n u(n) = 4 + (-0.5)n u(n)
3
6
6
Note : Compare the results of example 7.18 with example 6.8 of chapter-6.
=
Example 7.19
Determine the response of LTI discrete time system governed by the difference equation,
y(n) – 2 y(n – 1) – 3 y(n – 2) = x(n) + 4 x(n – 1) for the input x(n) = 2n u(n) and with initial condition y(–2)= 0, y(–1) = 5.
Solution
l q n
s
\ X(z) = Z x(n) = Z 2n u(n) =
;
Given that, x(n) = 2n u(n)
z
z-2
.....(1)
Given that, y(n) – 2 y(n – 1) – 3 y(n – 2) = x(n) + 4 x(n – 1)
On taking Z-transform of above equation we get,
Y (z) - 2 z -1 Y (z) + y(-1) - 3 z -2 Y (z ) + z -1 y (-1) + y (-2) = X(z) + 4 z -1 X(z ) + x( -1)
l q
l
l
.....(2)
q
q
If Z y(n) = Y (z), then Z y(n - 1) = z -1 Y (z) + y( -1)
and Z y(n - 2 = z -2 Y(z ) + z -1 y (-1) + y( -2)
Given that, y(–2)= 0,
y(–1) = 5
x(n) = 2n u(n) = 2n for n ³ 0
=0
for
Þ
n<0
x(–1) = 0
On substituting the above initial conditions in equation (2) we get,
Y (z) - 2 z -1 Y (z) - 2 ´ 5 - 3 z -2 Y (z) - 3 z -1 ´ 5 + 0 = X(z) + 4 z -1 X(z ) + 0
Y (z) -
2
3
15
4
Y(z) - 10 - 2 Y(z) = X ( z ) + X ( z)
z
z
z
z
Y (z)
Fz
GH
Y(z)
(z - 3) (z + 1) z + 4 15 + 10 z
=
+
z2
z-2
z
Y(z)
(z - 3) (z + 1) z 2 + 4 z + 15 z - 30 + 10 z 2 - 20 z
=
z (z - 2)
z2
Y(z)
(z - 3) (z + 1) 11z2 - z - 30
=
z2
z (z - 2)
Y(z) =
2
I FG
JK H
Þ
IJ FG
K H
- 2z - 3
15 + 10 z
z
=
z
z-2
z2
IJ FG z + 4 IJ
KH z K
Þ
Y(z)
FG
H
Y ( z) 1 -
IJ FG
K H
IJ
K
FG
H
2 3
15
4
+ 10 = X (z) 1 +
z z2
z
z
IJ
K
Using equation (1)
(z - 3) (z + 1) z (z + 4) + (15 + 10 z) (z - 2)
=
z2
z (z - 2)
z2
z (11z 2 - z - 30)
11z 2 - z - 30
´
=
z (z - 2)
(z - 3) (z + 1) (z - 2) (z - 3) (z +1)
7. 67
Signals & Systems
Let,
11z 2 - z - 30
Y( z )
A
B
C
=
=
+
+
(z - 2) (z - 3 ) (z + 1) z - 2 z - 3 z + 1
z
A=
11z2 - z - 30
´ ( z - 2)
(z - 2) (z - 3 ) (z + 1)
11z 2 - z - 30
´ (z - 3 )
B=
(z - 2) (z - 3 ) (z + 1)
11z2 - z - 30
´ (z + 1)
C=
(z - 2) (z - 3) (z + 1)
\
=
z =2
11z2 - z - 30
(z - 3) (z + 1)
11z2 - z - 30
=
(z - 2) (z + 1)
z =3
11z 2 - z - 30
=
z = -1
(z - 2) (z - 3)
Y (z)
33
1
3 1
-4
=
+
z
z-2
2 z - 3 2 z +1
RS
T
l q
\ Response, y(n) = Z -1 Y(z) = Z -1 -4
Þ
=
11´ 22 - 2 - 30 12
=
= -4
-3
(2 - 3) (2 + 1)
=
11´ 32 - 3 - 30 66 33
=
=
4
2
(3 - 2) (3 + 1)
=
11´ ( -1)2 + 1- 30 -18 -3
=
=
12
2
( -1 - 2) (-1 - 3)
z=2
z=3
z = -1
z
33
z
3 z
+
z-2
2 z - 3 2 z +1
Y(z) = -4
UV
W
z
33
z
3 z
+
z - 2 2 z - 3 2 z +1
= -4 (2)n u(n) +
33 n
3
(3) u(n) - ( -1)n u(n) =
2
2
LM-4 (2)
N
n
+
OP
Q
33 n 3
(3) - ( -1)n u(n)
2
2
Note : Compare the result of example 7.19 with example 6.9 of chapter-6.
Example 7.20
Find the response of the time invariant system with impulse response h(n) = {1, 2, 1, –1} to an input signal
x(n) = {1, 2, 3, 1}.
Solution
Let y(n) = Response or Output of an LTI system.
The response of an LTI system is given by the convolution of input signal and impulse response.
\ y(n) = x(n) * h(n)
On taking Z-transform we get,
Z{y(n)} = Z{ x(n) * h(n)}
By convolution property
\ Y(z) = X(z) H(z)
Z{x(n) * h(n)} = X(z) H(z)
Given that, x(n) = {1, 2, 3, 1}
By definition of one sided Z-transform,
¥
X(z) =
å x(n) z
-n
3
=
å x(n) z
-n
= x(0) z0 + x(1) z -1 + x(2) z -2 + x(3) z -3
n = 0
n = 0
= 1 + 2z -1 + 3z-2 + z -3
Given that, h(n) = { 1, 2, 1, –1}
By definition of one sided Z-transform,
¥
H(z) =
å
n = 0
h(n) z -n =
3
å h(n) z
-n
= h(0) z0 + h(1) z -1 + h(2) z -2 + h(3) z -3
n = 0
= 1 + 2z -1 + z -2 - z -3
\ Y(z) = X(z) H(z)
= [1 + 2z–1 + 3z–2 + z–3] [1 + 2z–1 + z–2 – z–3]
= 1 + 2z–1 + z–2 – z–3
+ 2z–1 + 4z–2 + 2z–3 – 2z–4
+ 3z–2 + 6z–3 + 3z–4 – 3z–5
+ z–3 + 2z–4 + z–5 – z–6
= 1 + 4z–1 + 8z–2 + 8z–3 + 3z–4 – 2z–5 – z–6
.....(1)
Chapter 7 - Z - Transform
7. 68
By definition of one sided Z-transform we get,
¥
å y(n) z
Y (z) =
-n
n = 0
= y(0) z0 + y(1) z–1 + y(2) z–2 + y(3) z–3 + y(4) z–4 + y(5) z–5+...
On comparing equations (1) and (2) we get,
y(0) = 1
y(2) = 8
y(4) = 3
y(1) = 4
y(3) = 8
.....(2)
y(6) = –1
y(5) = –2
\ The response of the system, y(n) = {1, 4, 8, 8, 3, –2, –1}
­
Example 7.21
l
q
Using Z-transform, perform deconvolution of the response y(n) = 1, 4, 8, 8, 3, -2, -1 and impulse
l
q
response h(n) = 1, 2, 1, -1 to extract the input x(n).
Solution
l
q
\ Y(z) = Zly(n)q = å y(n) z = å y(n) z
Given that, y(n) = 1, 4, 8, 8, 3, -2, -1
+¥
6
-n
n= -¥
-n
n =0
= y(0) + y(1) z -1 + y(2) z -2 + y(3) z -3 + y(4) z -4 + y(5) z -5 + y(6) z -6
= 1+ 4 z -1 + 8 z -2 + 8 z -3 + 3 z -4 - 2 z -5 - z -6
l
q
\ H(z) = Zlh(n)q = å h(n) z = å h(n) z
Given that, h(n) = 1, 2, 1, -1
+¥
3
-n
n= -¥
We know that, H(z) =
\ X( z ) =
-n
= h(0) + h(1) z -1 + h(2) z-2 + h(3) z -3 = 1+ 2z -1 + z -2 - z -3
n=0
Y(z)
X(z)
Y(z) 1+ 4 z -1 + 8 z -2 + 8 z -3 + 3 z -4 - 2 z -5 - z -6
=
H(z)
1+ 2z -1 + z -2 - z -3
= 1+ 2z -1 + 3 z -2 + z -3
-1
1+ 2z + 3 z
1 + 2 z -1 + z -2 - z -3
-2
.....(1)
+z
-3
1 + 4 z -1 + 8 z -2 + 8z -3 + 3 z -4 - 2 z -5 - z -6
1 + 2 z -1(–)
+ z -2 (+)
- z -3
(–) (–)
2 z -1 + 7 z -2 + 9 z -3 + 3 z -4
+ 4 z -2 (–)
- 2 z -4
+ 2z -3 (+)
2 z -1 (–)
(–)
3 z -2 + 7 z -3 + 5 z -4 - 2z -5
+ 3 z -4 (+)
- 3 z -5
+ 6 z -3 (–)
3 z -2 (–)
(–)
z -3 + 2 z -4 + z -5 - z -6
(–)
+ 2 z -4 (–)
+ z -5 (+)
- z -6
z -3 (–)
0
By the definition of Z-transform,
+¥
l q å x(n) z
X(z) = Z x(n) =
-n
n = -¥
On expanding the above summation we get,
X(z) =....... + x(0) + x(1) z -1 + x(2) z -2 + x(3) z -3 +.......
On comparing equations (1) and (2) we get,
x(0) = 1 ; x(1) = 2
;
x(2) = 3 ;
x(3) = 1
l
q
\ Input, x(n) = 1, 2, 3, 1
.....(2)
7. 69
Signals & Systems
Example 7.22
An LTI system is described by the equation y(n) = x(n) + 0.8 x(n – 1) + 0.8 x(n – 2) – 0.49 y(n – 2). Determine the
transfer function of the system. Sketch the poles and zeros on the z-plane.
Solution
Z{y(n)} = Y(z); \ Z{y(n – m)} = z–mY(z)
Given that, y(n) = x(n) + 0.8 x(n – 1) + 0.8 x(n – 2) – 0.49 y(n – 2)
Z{x(n)} = X(z); \ Z{x(n – m)} = z–mX(z)
On taking Z-transform we get,
Y(z) = X(z) + 0.8z–1 X(z) + 0.8z–2 X(z) – 0.49z–2 Y(z)
Y(z) + 0.49z–2 Y(z) = X(z) + 0.8z–1 X(z) + 0.8z–2 X(z)
(1 + 0.49z–2) Y(z) = (1 + 0.8z–1 + 0.8z–2) X(z)
\
Y(z)
1 + 0.8z -1 + 0.8z -2
=
X(z)
1 + 0.49z -2
.....(1)
The equation(1) is the transfer function of the LTI system.
jv
®
H(z) =
Y(z)
z -2 (z2 + 0.8z + 0.8)
1 + 0.8z-1 + 0.8z -2
=
=
-2
X(z)
z -2 (z2 + 0.49)
1 + 0.49z
z-plane
j1
z2 + 0.8z + 0.8
z2 + 0.49
The poles are the roots of the denominator polynomial,
z1
=
j0.8
j0.7 Xp1
unit circle
j0.6
j0.4
j0.2
z2 + 0.49 = 0
–1 –0.8 –0.6–0.4–0.2
\ z2 = –0.49
0.2 0.4 0.6 0.8
–j0.2
1
®u
–j0.4
z = ± -0.49 = ± j0.7
–j0.6
z2
\ The poles are p1 = j0.7, p2 = –j0.7
–j1
The zeros are the roots of the numerator polynomial,
Fig 1 : Pole-zero plot of LTI system.
2
z + 0.8z + 0.8 = 0
z =
–j0.7 X p2
–j0.8
-0.8 ± 0.82 - 4 ´ 0.8
-0.8 ± -2.56
-0.8 ± j0.16
=
=
= - 0.4 ± j0.8
2
2
2
\ The zeros are z1 = –0.4 + j0.8 and z2 = –0.4 – j0.8
\ H(z) =
z 2 + 0.8z + 0.8
z2 + 0.49
=
(z + 0.4 - j0.8) (z + 0.4 + j0.8)
(z - j0.7) (z + j0.7)
The fig1 Shows the location of poles and zeros on the z-plane. The poles are marked as “X ” and
Zeros as " ".
Example 7.23
Determine the step response of an LTI system whose impulse response h(n) is given by h(n) = a–n u(–n); 0 < a < 1.
Solution
Q u( -n) = 1 ; n £ 0
On taking Z-transform of impulse response h(n) we get,
¥
l q
H(z) = Z h(n) =
å h(n) z
n = -¥
-n
0
=
åa
=0 ; n<0
z
n = -¥
1a
1
= 1 - az
z - 1a
The step input, u(n) = 1 ; n ³ 0
=
= 0 ;n > 0
¥
-n
-n
=
åaz
n n
n=0
;
¥
=
å (az)
n
n=0
ROC |z| < |1/ a|
.....(1)
Note : Since a < 1, 1/a > 1, and so
ROC includes unit circle
Chapter 7 - Z - Transform
7. 70
On taking Z-transform of unit step signal we get,
¥
å u(n) z
l q
U(z) = Z u(n) =
¥
-n
åz
=
n = -¥
¥
-n
=
n = 0
å dz
-1 n
i
=
n=0
1
z
=
; ROC |z| > |
1 - z -1
z - 1
.....(2)
Let y(n) be step response. Now the step response is given by convolution of step input u(n) and impulse
response h(n)
\ y(n) = u(n) * h(n)
On taking Z-transform we get,
Z{y(n)} = Z{u(n) * h(n)}
By convolution property
\ Y(z) = U(z) H(z)
Z{u(n) * h(n)} = U(z) H(z)
On substituting for U(z) and H(z) from equations (1) and (2) respectively we get,
Y (z) = U(z) H(z) =
FG z IJ FG -1 a IJ
H z - 1K H z - 1 a K
By partial fraction expansion we can write,
-1 / a
Y (z)
A
B
+
=
=
z
(z - 1) (z - 1 / a )
z - 1
z - 1/ a
A = (z - 1)
Y (z)
z
B = (z - 1 / a )
=
z=1
Y(z)
z z
-1 / a
z - 1/ a
=
= 1/ a
=
z =1
-1 / a
-1 / a
-1
1
=
=
=
a - 1
a - 1
1 - 1/ a
1 - a
a
-1 / a
-1 / a
-1 / a
-1
=
=
=
1- a
z - 1 z = 1/ a
1/ a - 1
1- a
a
Y(z)
1
1
1
1
=
z
(1 - a ) (z - 1)
(1 - a) (z - 1 / a)
1
z
1
z
\ Y(z) =
(1 - a) (z - 1)
(1 - a) (z - 1 / a)
\
z
z - 1
z
Z{-bn u(-n - 1)} =
z - b
Z{-u( -n - 1)} =
Note : Since impulse response is anticausal, the step response is also anticausal.
On taking inverse Z-transform of Y(z) we get step response.
\ Step response, y(n) = -
1
1
u( -n - 1) +
1 - a
1- a
FG 1IJ
H aK
n
LMF 1I - 1OP F 1 I u(-n - 1)
MNGH a JK PQ GH 1 - a JK
n
u( -n - 1) =
Example 7.24
Test the stability of the first order system governed by the equation, y(n) = x(n) + b y(n – 1), where |b| < 1.
Z{y(n)} = Y(z) ;
Solution
Z{y(n – 1)} = z––1Y(z) ;
Z{x(n)} = X(z)
Given that, y(n) = x(n) + b y(n – 1)
On taking Z-transform we get,
Y(z) = X(z) + b z–1 Y(z)
\
Þ
Y(z) – b z–1 Y(z) = X(z)
Y (z )
1
=
X( z )
1 - b z -1
We know that, Y(z)/X(z) is equal to H(z).
\ H(z) =
1
1
z
= -1
=
z - b
1 - b z -1
z (z - b)
Þ
(1 – b z–1) Y(z) = X(z)
7. 71
Signals & Systems
On taking inverse Z-transform of H(z) we get the impulse response h(n).
\ Impulse response,h(n) = bn u(n)
¥
The condition to be satisfied for the stability of the system is,
¥
¥
å
å
h(n) =
n = -¥
n = -¥
¥
b
n
=
n= 0
å
å |h(n)|< ¥
n
|b|
n= 0
Since |b| < 1, using the infinite geometric series sum formula we can write,
¥
Infinite geometric series sum formula
1
1-|b|
å |b| =
n
n= 0
¥
å
¥
1
h(n) =
= constant
|b|
1
-¥
å
\
n=
Cn =
n = 0
1
1- C
;
if, 0 < |C| < 1
The term 1/(1–|b|) is less than infinity and so the system is stable.
Example 7.25
Using Z-transform, find the autocorrelation of the causal sequence x(n) = an u(n), –1 < a < 1.
Solution
Given that, x(n) = an u(n)
l q n
s
\ X(z) = Z x (n) = Z a nu(n) =
\ X(z -1) = X(z)
z=z 1
=
1
z
=
1 - a z -1 z - a
1
1
1
=a z -1a
1- a z
Let, rxx (m) be autocorrelation sequence
By correlation property of Z - transform,
l
q
Z rxx (m) = X(z) X(z -1) =
Let,
1
1
z
z
1
´
=a ( z - a ) (z - 1 a )
z - a -a z - 1 a
z
B
A
=
+
( z - a ) (z - 1 a ) z - a z - 1 a
a
a2
a
= 2
= 2
a -1a a -1 a -1
z =a
a
z
1a
1a
1
-1
z
B=
´ ( z - 1 a) z =1 a =
=
=
=
=
( z - a ) (z - 1 a )
z - a z =1 a 1 a - a 1- a 2 1- a 2 a 2 - 1
a
1
1
1
1
1
1
1
a2
a
=
\ Z rxx (m) =- 2
+
a a2 - 1 z - a a2 - 1 z - 1 a
a - 1 z - a a(a 2 - 1) z - 1 a
A=
z
´ (z - a)
( z - a ) (z - 1 a )
l
\ rxx
z =a
=
z
z -1 a
=
F
GH
I
JK
R a 1 + 1
1 U
a
R 1 UV + 1 Z RSz z UV
(m) = Z S=Z Sz
V
1
a
z
z
a
1
(
1
)
1
a
a
a
a
T z - a W a(a - 1) T z - 1 a W
T
W
1
a
FG 1 IJ u(n - 1)
=a
u(n - 1) +
z z=z =1
a -1
a(a - 1) H a K
If Z {x(n)} = X(z)
O
O
1 L 1 F 1I
1 LF 1 I
then by shifting property
=
- a (a) P u(n - 1) =
- a P u(n - 1)
M
M
G
G
J
J
a - 1 MN a H a K
a - 1 NMH a K
Z {x(n – m} = z X(z)
PQ
QP
q
-1
-1
2
2
-1
2
-1
-1
2
n -1
(n -1)
2
–1
2
n -1
n
n -1
2
2
0
n
–m
Chapter 7 - Z - Transform
7. 72
7.9 Summary of Important Concepts
1. The Z-transform provides a method for analysis of discrete time signals and systems in frequency domain.
2. The region of convergence (ROC) of X(z) is the set of all values of z, for which X(z) attains a finite value.
3. The zeros are defined as values of z at which the function X(z) becomes zero.
4. The poles are defined as values of z at which the function X(z) becomes infinite.
5. In a realizable system, the number of zeros will be less than or equal to number of poles.
6. The ROC of X(z) is a ring or disk in z-plane, with centre at origin.
7. If x(n) is finite duration right sided (causal) signal, then the ROC is entire z-plane except z = 0.
8. If x(n) is finite duration left sided (anticausal) signal, then the ROC is entire z-plane except z = ¥.
9. If x(n) is finite duration two sided (noncausal) signal, then the ROC is entire z-plane except z = 0 and z = ¥.
10. If x(n) is infinite duration right sided (causal) signal, then the ROC is exterior of a circle of radius r1.
11. If x(n) is infinite duration left sided (anticausal) signal, then the ROC is interior of a circle of radius r2.
12. If x(n) is infinite duration two sided (noncausal) signal, then the ROC is the region in between two circles of
radius r1 and r2.
13. If X(z) is rational, (where X(z) is Z-transform of x(n)), then the ROC does not include any poles of X(z).
14. If X(z) is rational, (where X(z) is Z-transform of x(n)), and if x(n) is right sided, then ROC is exterior of a circle
whose radius corresponds to pole with largest magnitude.
15. If X(z) is rational, (where X(z) is Z-transform of x(n)), and if x(n) is left sided, then ROC is interior of a circle
whose radius corresponds to pole with smallest magnitude.
16. If X(z) is rational, (where X(z) is Z-transform of x(n)), and if x(n) is two sided, then ROC is region in between
two circles whose radii corresponds to pole of causal part with largest magnitude and pole of anticausal part
with smallest magnitude.
17. The inverse Z-transform is the process of recovering the discrete time signal x(n) from its Z-transform X(z).
18. The transfer function of an LTI discrete time system is defined as the ratio of Z-transform of output and
Z-transform of input.
19. The transfer function of an LTI discrete time system is also given by Z-transform of the impulse response.
20. The inverse Z-transform of transfer function is the impulse response of the system.
21. The zero-input response yzi(n) is mainly due to initial output (or initial stored energy) in the system.
22. The zero-state response yzs(n) is the response of the system due to input signal and with zero initial output.
23. The total response y(n) is the response of the system due to input signal and initial output (or initial stored
energy).
24. The convolution operation is performed to find the response y(n) of an LTI discrete time system from the
input x(n) and impulse response h(n).
25. The deconvolution operation is performed to extract the input x(n) of an LTI system from the response y(n) of
the system.
26. A point-s1 on left half of s-plane (LHP), will map as a point-z1 inside the unit circle in z-plane.
27. A point-s1 on imaginary axis of s-plane, will map as a point-z1 on the unit circle in z-plane.
28. A point-s1 on the right half of s-plane (RHP), will map as a point-z1 outside the unit circle in z-plane.
29. The mapping of s-plane to z-plane, using the transformation, esT = z is not one-to-one.
30. The mapping of the continuous time frequency W to the discrete time frequency w is many-to-one.
7. 73
Signals & Systems
7.10 Short Questions and Answers
Q7.1
Infinite geometric series sum formula
¥
Find the Z-transform of an u(n).
å
By the definition of Z-transform,
¥
Z{an u(n)} =
n -n
å
=
n= 0
Q7.2
n = 0
¥
åaz
Cn =
(a z -1)n =
n =0
1
1- C
; if, 0 < |C| < 1
1
z
1
=
=
1 - az -1 1 - a / z z - a
Find the Z-transform of e- anT u(n).
By the definition of Z-transform,
¥
Z{e - anT u(n)} =
¥
åe
- anT -n
z
=
n= 0
Q7.3
å (e
- aT -1 n
z )
=
n= 0
z
1
1
=
=
1 - e - aT z -1 1- e -aT / z z - e -aT
Find the Z-transform of x(n) defined as,
x(n) = b n
=0
;
0 £ n£ N -1
;
otherwise
Finite geometric series
Solution
sum formula.
By the definition of Z-transform,
l q å
x(n) z -n =
å
n
=
n= 0
d
z -N zN - bN
1 - (b z -1)N
1 - bN z -N
=
=
-1
-1
1 - bz
1- b z
z -1(z - b)
N- 1
å (bz
-1 n
) =
n= 0
Q7.4
bn z -n
1 - cN
1- c
n =0
n =-¥
=
N-1
åc
N -1
+¥
Z x(n) =
i
d
z -N +1 zN - bN
=
i
z -b
Find the Z-transform of x(n) = an+1 u(n+1).
Solution
Given that, x(n) = an+1 u(n + 1) = an+1
;
n ³ –1
By the definition of Z-transform,
+¥
+¥
l q å x(n) z = å a
-n
Z x(n) =
n =-¥
å da z
-1 n
i
åa
+¥
n +1
z -n = a 0 z +
n =0
=z+a
n= 0
Q7.5
z - n = a -1+1 z +
n =-1
+¥
=z+a
+¥
n +1
;
By differentiation property of Z-transform we get,
d
X(z)
dz
l
q
= -z
d
1
a z -1
log (1+ a z -1) = - z
(- a z -2 ) =
-1
1+ a z
1 + a z -1
dz
=
RS
T
z
z - (- a )
UV
W
= a(-a )n -1 u(n - 1)
x(n) =
;
|z| > |a|
Since ROC is exterior of a
circle of radius (a), the x(n)
should be a causal signal
a z -1
a
z
= a z -1
=
z -1(z + a) z + a
z - ( -a )
\ n x(n) = Z -1 a z -1
\
a z -n
|z| > |a|
Let, x(n) = Z–1 {X(z)}
Z nx(n) = -z
n
n= 0
1
az
z(z - a ) + az
z2
=
= z+
=
-1
1- a z
z-a
z-a
z-a
Determine the inverse Z-transform of X(z) = log (1 + az –1)
Solution
Given that, X(z) = log (1 + az–1)
åa
a
(-a )n - 1 u(n - 1)
n
If Z {x(n)} = X(z)
then by shifting property
Z {x(n – m} = z–m X(z)
Chapter 7 - Z - Transform
Q7.6
7. 74
Determine x(0) if the Z-transform of x(n) is X(z) =
2 z2
.
(z + 3) (z - 4)
Solution
By initial value theorem of Z-transform,
2 z2
x(0) = Lt X(z) = Lt
z ®¥
z ®¥ (z + 3) (z - 4 )
= Lt
z ®¥
Q7.7
F
z G
H
2
2 z2
3
4
1+
1z
z
IJ FG
KH
2
IJ = Lt FG1+ 3 IJ FG1- 4 IJ
K
H zK H zK
z ®¥
=
2
=2
(1+ 0) (1 - 0)
Determine the Z-transform of x(n) = (n – 3) u(n).
Solution
l q z z- 1
d
Zln x(n)q = -z
X(z)
dz
Z u(n) =
l q l
q l
q
= Zln u(n)q - 3 Zlu(n)q
d F z I
= -z G
J - 3 z z- 1 = -z z(z--11-) z - z3z- 1
dz H z - 1K
Z x(n) = Z (n - 3) u(n) = Z n u(n) - 3 u(n)
d
2
=
Q7.8
u
= v du - u dv
v
z - 3z(z - 1) z - 3z2 + 3 z -3z2 + 4 z z(4 - 3z)
3z
z
=
=
=
=
2
z -1
(z - 1)
(z - 1)2
(z - 1)2
(z - 1)2
(z - 1)2
Determine the transfer function of the LTI system defined by the equation,
y(n) - 0.5 y(n - 1) = x(n) + 0.4 x(n - 1)
Solution
Given that, y(n) - 0.5 y(n - 1) = x(n) + 0.4 x(n - 1)
On taking Z-transform we get,
Y(z) - 0.5 z -1Y(z) = X(z ) + 0.4 z -1X (z)
\ Transfer function,
Q7.9
Þ
Y (z ) 1 - 0.5 z -1 = X(z) 1 + 0.4 z -1
Y(z) 1 + 0.4 z -1
=
X(z) 1 - 0.5 z -1
The transfer function of a system is given by, H(z) = 1 – z–1. Find the response of the system for
any input, x(n).
Solution
Given that, H(z) = 1 – z –1
Y( z )
X( z )
\ Response in z - domain,
We know that, H(z) =
Y(z) = H(z) X(z) = (1 - z -1) X(z) = X(z) - z -1 X(z)
l q
n
s
\ Response in time domain, y(n) = Z -1 Y(z) = Z -1 X(z) - z -1 X(z ) = x(n) - x(n - 1)
Q7.10
An LTI system is governed by the equation, y(n) = –2 y(n – 2) – 0.5 y(n – 1) + 3 x(n – 1) + 5 x(n).
Determine the transfer function of the system.
Solution
Given that, y(n) = –2 y(n – 2) – 0.5 y(n – 1) + 3 x(n – 1) + 5 x(n)
On taking Z-transform of above equation we get,
Y (z) = -2 z-2 Y(z) - 0.5z -1 Y(z) + 3 z -1 X(z) + 5 X(z)
Y (z) + 2 z-2 Y(z) + 0.5 z -1 Y(z) = 3 z -1 X(z) + 5 X(z)
Y (z) 1 + 2 z -2 + 0.5z -1 = 3 z -1 + 5 X(z)
\ Transfer function, H(z) =
Y(z)
3z -1 + 5
5 z2 + 3 z
=
= 2
-1
-2
X(z) 1 + 0.5 z + 2 z
z + 0.5 z + 2
7. 75
Q7.11
Signals & Systems
The transfer function of an LTI system is H(z) =
z-1
. Determine the impulse response.
(z - 2) (z + 3)
Solution
H(z) =
A=
A
B
z -1
=
+
(z - 2) (z + 3) z - 2 z + 3
z -1
´ (z - 2)
(z - 2) (z + 3)
z -1
´ (z + 3)
(z - 2) (z + 3)
4
1
1
1
+
H(z) =
5 z-2 5 z+3
B=
z=2
z -1
z+3
=
z =-3
=
z -1
z-2
2 -1 1
=
2+3
5
=
z=2
=
z =-3
-3 - 1 -4 4
=
=
-3 - 2
-5 5
l q RST 51 z -1 2 + 54 z +1 3 UVW
RS 1 z z + 4 z z UV
T 5 z - 2 5 z - (-3) W
Impulse response, h(n) = Z -1 H(z) = Z -1
= Z -1
=
Q7.12
-1
-1
1 (n-1)
1 (n -1)
4
+ 4 (-3)(n-1) u(n - 1)
2
u(n - 1) + (-3)(n -1) u(n - 1) =
2
5
5
5
Determine the response of LTI system governed by the difference equation,y(n) – 0.5 y(n – 1) = x(n),
for input x(n) = 5n u(n), and initial condition y(–1) = 2.
Solution
Given that, x(n) = 5n u(n)
l q
;
\ X(z) = Z u(n) =
z
z-5
Given that, y(n) – 0.5 y(n – 1) = x(n),
On taking Z-transform of above equation we get,
Y(z) - 0.5 z -1Y(z ) + y (-1) = X(z )
Y(z) - 0.5 z -1Y(z ) + 2 =
Y(z) - 0.5 z -1Y (z) - 1 =
\ Y( z ) =
Let,
\
z
z-5
z
z-5
z(2 z - 5)
(z - 0.5) (z - 5)
LM
N
0.5
z
OP
Q
z
+1
z-5
Þ
Y ( z) 1 -
Þ
Y (z)
2z - 5
=
z
(z - 0.5) (z - 5)
=
Þ
2z - 5
A
B
Y(z)
=
=
+
(z - 0.5) (z - 5) z - 0.5 z - 5
z
A=
2z - 5
´ (z - 0.5)
(z - 0.5) (z - 5)
B=
2z - 5
´ (z - 5)
(z - 0.5) (z - 5)
1
10
1
Y(z) 8
=
+
9 z - 0.5 9 z - 5
z
l q
\ Response, y(n) = Z -1 Y(z) = Z -1
=
=
z = 0.5
z=5
=
2´ 5 - 5
5
50 10
=
=
=
5 - 0.5
4.5 45
9
8
z
10
z
+
9 z - 0.5 9 z - 5
8
z
10
z
+
9 z - 0.5 9 z - 5
Þ
RS
T
-4
2 ´ 0.5 - 5
40 8
=
=
=
-4.5 45 9
0.5 - 5
Y(z) =
LM
N
UV
W
OP
Q
8
10 n
8
10 n
0.5n u(n) +
5 u(n) =
0.5 n +
5 u(n)
9
9
9
9
Y(z)
LM z - 0.5 OP = z + z - 5
N z Q z-5
Chapter 7 - Z - Transform
Q7.13
7. 76
t
A signal x(t) = a is sampled at a frequency of 1/T Hz in the range –¥ < t < 0. Find the
Z-transform of the sampled version of the signal.
Solution
Given that, x(t) = at
;
–¥ < t < 0
The sampled version of the signal x(nT) is given by, x(nT) = anT ; –¥ < nT < 0
Now the Z - transform of x(n T) is,
+¥
l
q å x(n T) z
Z x(n T) =
0
-n
n =-¥
¥
=
å da
n =0
Q7.14
-T
z
n
i
=
=
åa
nT
z -n
n =-¥
1
1
aT
=
= T
-T
T
1- a z 1- z a
a -z
The transfer function of a system is given by, H(z) =
1
1
. Determine the stability
+
1 - 0.5 z -1 1 - 2 z -1
and causality of the system for a) ROC : |z| > 2 ; b) ROC : |z| < 0.5.
Solution
a) ROC is |z| > 2
When ROC is |z| > 2, the impulse response h(n) should be right sided signal.
l q
\ Impulse response, h(n) = Z -1 H(z) = Z -1
RS 1
T1- 0.5 z
+
-1
UV d
W
1
= 0.5 n + 2n u(n)
1 - 2 z -1
i
1. The ROC does not include unit circle. Hence the system is unstable.
2. The impulse response is right sided signal. Hence the system is causal.
b) ROC is |z| < 0.5
When ROC is |z| < 0.5, the impulse response h(n) should be left sided signal.
l q
\ Impulse response, h(n) = Z -1 H(z) = Z -1
RS 1
T1- 0.5 z
-1
+
UV d
W
1
= -0.5 n - 2n u( -n - 1)
1 - 2 z -1
i
1. The ROC does not include unit circle. Hence the system is unstable.
2. The impulse response is left sided sequence. Hence the system is anticausal.
Q7.15
Determine the stability and causality of the system described by the transfer function,
1
1
+
for ROC : 0.25 < |z| < 2.
1 - 0.25 z -1 1 - 2 z -1
Solution
H(z) =
Given that, ROC is 0.25 < |z| < 2
When ROC is 0.25 < |z| < 2, the impulse response h(n) is two sided signal. Since |z| > 0.25, the term with
pole z = 0.25 corresponds to right sided signal. Since |z| < 2, the term with pole z = 2 corresponds to left
sided signal.
l q
\ Impulse response, h(n) = Z -1 H(z) = Z -1
RS 1
T1- 0.25 z
-1
+
UV
W
1
= 0.25 n u(n) - 2n u( -n - 1)
1 - 2 z -1
1. The ROC includes the unit circle. Hence the system is stable.
2. The impulse response is two sided noncausal signal. Hence the system is noncausal.
7. 77
Q7.16
Signals & Systems
U s i n g Z -transform, determine the response of the LTI system with impulse response,
l
q
l
q
h(n) = 1, -1, 1 , for an input x(n) = -2, 3, 1 .
Solution
l
q
\ X(z) = Zlx(n)q = å x(n) z = å x(n) z = x(0) + x(1) z + x(2) z = -2 + 3 z + z
Given that, h(n) = l1, -1, 1q
\ H(z) = Zlh(n)q = å h(n) z = å h(n) z = h(0) + h(1) z + h(2) z = 1 - z + z
Given that, x(n) = -2, 3, 1
+¥
2
-n
-n
n= -¥
+¥
-2
-1
-2
2
-n
-n
n= -¥
We know that, H(z) =
-1
n=0
-1
-2
-1
-2
n=0
Y(z)
X(z)
d
i d
\ Y(z) = X(z) H(z) = -2 + 3 z -1 + z -2 ´ 1 - z -1 + z -2
-1
-2
+ 3z - 3z
-1
-2
-3
= -2 + 2 z - 2 z
= -2 + 5 z - 4 z
-1
+ 2z
+z
-2
i
+ 3 z -3 + z -2 - z -3 + z -4
-4
.....(1)
By definition of Z - transform,
+¥
l q å y(n) z
Y(z) = Z y (n) =
-n
n= -¥
On expanding the above summation we get,
.....(2)
Y(z) =......+ y(0) + y(1) z -1 + y(2) z -2 + y(3) z -3 + y(4) z -4 +.......
On comparing equations (1) and (2) we get,
y(0) = -2 ; y(1) = 5
;
y(2) = -4
l
; y(3) = 2
; y(4) = 1
q
\ Response, y(n) = -2, 5, -4, 2, 1
Q7.17
l
q
Using Z-transform, perform deconvolution of response y(n) = -2, 5, -4, 2, 1 and impulse
l
q
response h(n) = 1, -1, 1 , to extract the input x(n).
Solution
l
q
Given that, y(n) = -2, 5, -4, 2, 1
+¥
4
l q å y(n) z = å y(n) z
-n
Y(z) = Z y(n) =
n= -¥
-n
n=0
= y(0) + y(1) z-1 + y(2) z-2 + y(3) z -3 + y(4) z -4 = -2 + 5 z -1 - 4z -2 + 2 z -3 + z -4
l
q
Given that, h(n) = 1, -1, 1
+¥
2
l q å h(n) z = å h(n) z
-n
H (z) = Z h(n) =
n= -¥
We know that, H(z) =
-n
= h(0) + h(1) z-1 + h(2) z -2 = 1 - z -1 + z -2
n=0
Y(z)
X(z)
- 2 + 3 z - 1 + z -2
Y(z) -2 + 5 z -1 - 4 z -2 + 2 z -3 + z -4
\ X(z) =
=
1 - z -1 + z -2
H(z)
1 - z -1 + z -2
- 2 (–)
+ 2 z -1 (+)
- 2 z -2
(+)
3 z -1 - 2 z -2 + 2 z -3
= -2 + 3 z -1 + z -2
. ....(1)
By definition of Z - transform,
- 2 + 5 z -1 - 4 z -2 + 2 z -3 + z -4
- 3 z -2 (–)
3 z -1 (+)
+ 3 z -3
(–)
z - 2 - z - 3 + z -4
+¥
l q å x(n) z
X(z) = Z x(n) =
n= -¥
-n
(–)
- z -3 (–)
+ z -4
z-2 (+)
0
Chapter 7 - Z - Transform
7. 78
On expanding the above summation we get,
X(z) =......+ x(0) + x(1) z -1 + x(2) z -2 + x(3) z -3 +.......
.....(2)
On comparing equations (1) and (2) we get,
x(0) = -2 ; x(1) = 3
l
;
x(2) = 1
q
\ Input, x(n) = -2, 3, 1
Q7.18
In an LTI system the impulse response h(n) = C
for which the system is stable.
n
for n £ 0. Determine the range of values of C,
Solution
Given that, h(n) = Cn for n £ 0.
+¥
\
+¥
¥
0
å h(n) = å C =å C
n
n =-¥
n = -¥
-n
=
å (C
+¥
If, 0 < C -1 < 1, then
å (C
-1 n
)
n=0
n= 0
-1 n
) =
n= 0
1
1 - C -1
+¥
If, C
-1
å (C
> 1, then
-1 n
) =¥
n =0
\ For stability, C -1 < 1
Q7.19
1
<1
C
Þ
Þ
C >1
Using Z-transform, determine the response of the LTI system with impulse response h(n) = 0.4 n u(n),
for an input x(n) = 0.2n u(n).
Solution
Given that, x(n) = 0.2n u(n).
l q n
s
z
z - 0.2
s
z
z - 0.4
\ X(z) = Z x(n) = Z 0.2n u(n) =
Given that, h(n) = 0.4n u(n)
l q n
\ H(z) = Z h(n) = Z 0.4n u(n) =
We know that, H(z) =
Y(z)
X(z)
\Y (z) = X(z) H(z) =
Let,
z
z
z2
´
=
z - 0.2 z - 0.4 (z - 0.2) (z - 0.4 )
Y(z)
z
A
B
=
=
+
z
(z - 0.2) (z - 0.4) z - 0.2 z - 0.4
A=
B=
\
z
´ (z - 0.2)
(z - 0.2) (z - 0.4 )
z
´ (z - 0.4)
(z - 0.2) (z - 0.4 )
2
Y(z)
-1
=
+
z - 0.2 z - 0.4
z
l q
RS
T
Response, y(n) = Z-1 Y(z) = Z-1 -
z = 0.2
z = 0. 4
Þ
=
0.2
0.2
=
= -1
0.2 - 0.4 -0.2
=
0.4
0.4
=
=2
0.4 - 0.2 0.2
Y(z) = -
z
z
+2
z - 0.2
z - 0.4
z
z
+2
z - 0.2
z - 0.4
UV
W
= -(0.2)n u(n) + 2 (0.4)n u(n) = 2 (0.4)n - (0.2)n u(n)
7. 79
Signals & Systems
Using Z-transform perform deconvolution of response, y(n) = 2 (0.4)n u(n) – (0.2)n u(n) and impulse response, h(n) = 0.4n u(n), to extract the input x(n).
Solution
Given that, y(n) = 2 (0.4)n u(n) – (0.2)n u(n)
Q7.20
l q n
s
\ Y(z) = Z y(n) = Z 2 (0.4)n u(n) - (0.2)n u(n)
=
2z
2 z (z - 0.2) - z (z - 0.4 ) 2 z 2 - 0.4 z - z2 + 0.4 z
z
z2
=
=
=
( z - 0.4 ) (z - 0.2)
( z - 0.4) (z - 0.2)
( z - 0.4 ) (z - 0.2)
z - 0.4 z - 0.2
Given that, h(n) = 0.4 n u(n)
l q n
s
\ H(z) = Z h(n) = Z 0.4 n u(n) =
We know that, H(z) =
\ X(z) =
z
z - 0.4
Y(z)
X(z)
Y(z)
1
z - 0.4
z
z2
=
= Y(z) ´
´
=
H(z) ( z - 0.4 ) (z - 0.2)
z
z - 0.2
H(z)
l q
\ Input, x(n) = Z -1 X(z) = Z -1
RS z UV = 0.2
T z - 0.2 W
n
u(n)
7.11 MATLAB Programs
Program
7.1
Write a MATLAB program to find one sided Z-transform of the following
standard causal signals.
a) n
b) an c) nan
d) e-anT
%Program
to
find
the
Z-transform
clea r a ll
syms n T a real;
sym s z co m plex;
%(a)
x = n;
disp(‘(a)
ztrans(x)
%(b)
x = a^n;
disp(‘(b)
ztrans(x)
%Let
%Let
of
of
z-transform
of
“a^n”
of
“n(a^n)”
%(d)
x=exp(-a*n*T);
disp(‘(d) z-transform
ztrans(x)
OUTPUT
(a) z-transform of “n” is
ans =
z/(z-1)^2
of
standard
signals
n, T, a be real variable
z be co m plex va r ia ble
z-transform
%(c)
x=n*(a^n);
disp(‘(c) z-transform
ztrans(x)
some
“n”
is’);
is’);
is’);
“exp(-a*n*T)”
is’);
Chapter 7 - Z - Transform
7. 80
(b) z-transform of
“a^n” is
ans =
z/a/(z/a-1)
(c) z-transform of
“n(a^n)” is
ans =
z*a/(z-a)^2
(d) z-transform of “exp(-a*n*T)” is
ans =
z/exp(-a*T)/(z/exp(-a*T)-1)
Program
7.2
Write a MATLAB program to find Z-transform of the following causal
signals.
a) 0.5n
b) 1+n(0.4)(n-1)
%**********
clear all
syms n real;
program
%Let
to
n
determine
be
%(a)
x1=0.5^n;
disp(‘(a) z-transform
X1=ztrans(x1)
%(b)
x2=1+n*(0.4^(n-1));
disp(‘(b) z-transform
X2=ztrans(x2)
real
z-transform
of
the
given
signals
variable
of
“0.5^n”
is’);
of
“1+n*(0.4^(n-1))”
is’);
OUTPUT
(a) z-transform of “0.5^n” is
X1 =
2*z/(2*z-1)
(b) z-transform of “1+n*(0.4^(n-1))”
X2 =
z/(z-1)+25*z/(5*z-2)^2
Program
is
7.3
Write a MATLAB program to find inverse Z-transform of the following
z-domain signals.
b) 1/((1+z-1)(1-z-1)2)
a) 1/(1-1.5z-1+0.5z-2)
%***********Program
syms
n
to
determine
the
inverse
z-transform
z
X=1/(1-1.5*(z^(-1))+0.5*(z^(-2)));
disp(‘Inverse z-transform of 1/(1-1.5z^-1+0.5z^-2)is’);
x=iztrans(X,z,n);
simplify(x)
X=1/((1+(z^(-1)))*((1-(z^(-1))^2)));
disp(‘Inverse z-transform of 1/((1+z^-1)*(1-z^-1)^2))is’);
x=iztrans(X,z,n);
simplify(x)
OUTPUT
Inverse
z-transform of 1/(1-1.5z^-1+0.5z^-2) is
ans =
2-2^(-n)
Inverse z-transform of 1/((1+z^-1)*(1-z^-1)^2)) is
ans =
3/4*(-1)^n+1/2*(-1)^n*n+1/4
7. 81
Signals & Systems
Program
7.4
Write a MATLAB program to perform convolution of signals,x1(n)= (0.4)nu(n)
and x2(n)=(0.5)nu(n),using Z-transform, and then to perform deconvolution
using the result of convolution to extract
x1(n) and x2(n).
%***
Program
to
perform
convolution
and
deconvolution
using
z-transform
clear all;
syms n z
x1n=0.4^n;
x2n=0.5^n;
X1z=ztrans(x1n);
X2z=ztrans(x2n);
X3z=X1z*X2z;
con12=iztrans(X3z);
disp(‘Convolution of
simplify(con12)
%product
x1(n)
of
z-transform
of
inputs
and x2(n) is’);
% convolution output
decon_X1z=X3z/X1z;
decon_x1n=iztrans(decon_X1z);
disp(‘The signal x1(n) obtained
simplify(decon_x1n)
decon_X2z=X3z/X2z;
decon_x2n=iztrans(decon_X2z);
disp(‘The signal x2(n) obtained
simplify(decon_x2n)
by
deconvolution
is’);
by
deconvolution
is’);
OUTPUT
Convolution of x1(n) and x2(n) is
ans =
5*2^(-n)-4*2^n*5^(-n)
The signal x1(n) obtained by deconvolution is
ans =
2^(-n)
The signal x2(n) obtained by deconvolution is
ans =
2^n*5^(-n)
Program
7.5
Write a MATLAB program to find residues and poles of z-domain signal,
(3z 2+2z+1)/(z2-3z+2)
%*** Program to
% function of z
find
partial
fraction
expansion
of
rational
clear all
H=tf(‘z’);
Ts=0.1;
b=[3
a=[1
2 1 ];
-3 2];
disp(‘The
H=tf([b],
%Numerator coefficients
%Denominator coefficients
given transfer
[a],Ts)
function
is,’);
disp(‘The residues, poles and direct terms of given TF
disp(‘(r - residue ; p - poles ; k - direct terms)’);
[r,p,k]=residue(b,a)
disp(‘The num. and den.
[b,a]=residue(r,p,k)
coefficients
extracted
from
are,’);
r,p,k,’);
Chapter 7 - Z - Transform
7. 82
OUTPUT
The given transfer function is,
Transfer function:
3 z^2 + 2 z + 1
--------------z^2 - 3 z + 2
Sampling time: 0.1
The residues, poles and direct terms of given TF are,
(r - residue ; p - poles ; k - direct terms)
r =
17
-6
p =
2
1
k =
3
The num. and den. coefficients extracted from r,p,k are,
b =
3
2
1
a =
1
-3
2
Program
7.6
Write a MATLAB program to find poles and zeros of z-domain signal,
(z2+0.8z+0.8)/(z2+0.49), and sketch the pole zero plot.
%
%
Program to determine poles and zeros of rational
to plot the poles and zeros in z-plane
function
of
Z
and
clear all
syms z
num_coeff=[1 0.8 0.8];
disp(‘Roots of numerator
zeros=roots(num_coeff)
%find
polynomial
den_coeff=[1 0 0.49];
disp(‘Roots of denominator
poles=roots(den_coeff)
the factors of z^2+0.8z+0.8
z^2+0.8z+0.8 are zeros.’);
%find the factors of
polynomial z^2+0.49 are
H=tf(‘z’);
Ts=0.1;
H=tf([num_coeff],[den_coeff],Ts);
zgrid on;
pzmap(H);
%Pole-zero
plot
OUTPUT
Roots of numerator polynomial z^2+0.8z+0.8 are zeros.
zeros =
-0.4000 + 0.8000i
-0.4000 - 0.8000i
Roots of denominator polynomial
poles =
0 + 0.7000i
0 - 0.7000i
z^2+4z+13
The pole-zero plot is shown in fig P7.6.
are
poles.
z^2+0.49
poles.’);
7. 83
Signals & Systems
Fig P7.6 : Pole-Zero plot of program 7.6.
Program
7.7
Write a MATLAB program to compute and sketch the impulse response of
discrete time system governed by transfer function, H(z)=1/(1-0.8z-1+0.16z2).
%*******
Program
to
find
impulse
response
of
a
discrete
time
clear all
syms z n
H=1/(1-0.8*(z^(-1))+0.16*(z^(-2)));
disp(‘Impulse response h(n) is’);
h=iztrans(H);
%compute impulse
simplify(h)
N=15;
b=[0 0 1];
a=[1 -0.8 0.16];
[H,n]=impz(b,a,N);
stem(n,H);
xlabel(‘n’);
ylabel(‘h(n)’);
response
%numerator coefficients
%denominator coefficients
%compute N samples of impulse
%sketch
impulse
response
OUTPUT
Impulse response h(n) is
ans =
2^n*5^(-n)+2^n*5^(-n)*n
The sketch of impulse response is shown in fig P7.7.
response
system
Chapter 7 - Z - Transform
7. 84
Fig P7.7 : Impulse response of program 7.7.
7.12 Exercises
I. Fill in the blanks with appropriate words
1. The –––––––– of X(z) is the set of all values of z, for which X(z) attains a finite value.
2. The transformation –––––– maps the s-plane into z-plane.
3. The ––––– of s-plane can be mapped into the ––––– of the unit circle in z-plane.
4. The ratio of Z-transform of output to Z-transform of input is called _______ of the system.
5. In the mapping z =esT, the _______ poles of s-plane are mapped into _______ of unit circle in z-plane.
6. In impulse invariant mapping the _______ poles of s-plane are mapped into _______ of unit circle in z-plane.
7. In impulse invariant mapping the poles on the imaginary axis in s-plane are mapped on the _______ in
z-plane.
8. In _______ transformation any strip of width 2p/T in s-plane is mapped into the entire z-plane.
9. The phenomena of high frequency components acquiring the identity of low frequency components is
called _______.
10. For a causal LTI discrete time system the ROC should be _______ the circle of radius whose value corresponds
to pole with _______ magnitude.
11. If X(z) is rational, then the ROC does not include _______ of X(z).
12. The sequences multiplied by u(–n –1) are _______ and defined for _______.
13. The inverse Z-transform of transfer function is _______ of the system.
14. If Z-transform of x(n) is X(z), then Z-transform of x*(n) is _______.
15. The Z-transform of a shifted signal, shifted by 'q' units of time is obtained by _______ to Z-transform of
unshifted signal.
Answers
1. region of convergence
2. s = (1/T) ln z
6. right half, exterior
7. unit circle
11. poles
12. anticausal sequences, n £ 0
3. left half, interior
4. transfer function
5. left half, interior
8. impulse invariant
9. aliasing
10. outside, largest
13. impulse response
14. X*(z*)
15. multiplying z q
7. 85
Signals & Systems
II. State whether the following statements are True/False.
1. The Z-transform exists only for those values of z for which X(z) is finite.
2. When the input is an impulse sampled signal, the z-domain transfer function can be directly obtained from
s-domain transfer function.
3. The jW axis in s-plane maps into the unit circle of z-plane in the clockwise direction.
4. The left half of s-plane maps into the interior of the unit circle in z-plane.
5. The system is unstable if all the poles of transfer function lies inside the unit circle in z-plane.
6. The Z-transform of impulse response gives the transfer function of LTI system.
7. If X(z) and H(z) are Z-transform of input and impulse response respectively, then the response of LTI system
is given by inverse Z-transform of the product X(z) H(z).
8 . For a stable LTI continuous time system the poles should lie on the right half of s-plane.
9. For a stable LTI discrete time system the poles should lie on the unit circle.
o
FG d IJ
H dz K
t
m
10. If Z{x(n)} = X(z), then Z n mx(n) = - z
X(z) .
Answers
1. True
2. True
3. False
4. True
5. False
6. True
7. True
8. False
9. False
10. False
III. Choose the right answer for the following questions
1. The impulse response, h(n) = 1
; n = 0 , can be represented as,
= -(1 - b) b
n- 1
; n³1
a) d(n)
c) d(n) – (1–b) bn–1 u(n – 1)
b) u(n) – (1–b) bn–1 u(n – 1)
d) u(n) – (1–b) bn–1 u(n)
2. The Z-transform of a–n u(– n – 1) is,
a)
-z
z -1 a
b)
c)
z
z -1 a
z
z-a
d)
-z
z-a
3. The ROC of the sequence x(n) = u(–n) is,
a) |z| > 1
4. The inverse Z-transform of
a) 3(4)n u(n–1)
b) |z| < 1
c) no ROC
d) –1 < |z| <1
3
, |z| > 4 is,
z-4
b) 3(4)n–1 u(n)
c) 3(4)n–1 u(n+1)
d) 3(4)n–1 u(n–1)
b) zeros
c) no poles
d) no zeros
5. ROC of x(n) contains,
a) poles
6. The inverse Z-transform of X(z) = e a z , z > 0 is,
a) x(n) =
-a n
u(n)
n!
b) x(n) =
an
u(n)
n!
c) x(n) =
a n -1
u(n - 1)
n!
d) None of the above
z - 4z -2 + 3 z -3
(z - 1) 2
d) X(z) =
7. The Z-transform of x(n) = [u(n) – u(n – 3)], for ROC |z| > 1 is,
a) X(z) =
z - z -2
z -1
b) X(z) =
z -2
(z - 1)2
c) X(z) =
z - z -2
z -1
Chapter 7 - Z - Transform
7. 86
8. The system function H(z) =
z3 - 2 z2 + z
is,
z 2 + 0.25 z + 0 .125
a) causal
b) noncausal
c) unstable but causal
d) can not be defined
9. If all the poles of the system function H(z) have magnitude smaller than 1, then the system will be,
a) stable
b) unstable
c) BIBO stable
d) a and c
l
q
10. If x(n) = 0.5, -0.25, 1 , then Z-transform of the signal is,
a)
z2
0.5 z - 0.25z + 1
b)
2
z2
z - 0.5 z + 0.25
2
c)
0.5 z 2 - 0.25 z + 1
z2
d)
2 z2 + 4 z + 1
z2
11. The ROC of the signal x(n) = an for –5 < n < 5 is,
a) Entire z-plane
c) Entire z-plane except z = 0
b) Entire z-plane except z = 0 and z = ¥
d) Entire z-plane except z = ¥
12. If Z-transform of x(n) is X(z) then Z-transform of x(–n) is,
a) –X(z)
b) X(–z)
13. The inverse Z-transform of X(z) can be defined as,
1
a) x(n) =
X( z) z n -1 dz
2p c
c) x(n) =
z
z
1
X( z) z n -1 dz
2pj c
14. The Z-transform is a,
a) finite series
c) –X(z–1)
d) X(z–1)
z
b) x(n) =
1
X( z) z n -1 dz
2j c
d) x(n) =
1
X( z) z - n dz
2pj c
z
b) infinite power series c) geometric series
d) both a and c
n
15. If the Z-transform of x(n) is X(z), then Z-transform of (0.5) x(n) is,
a) X(0.5 z)
b) X(0.5–1 z)
c) X(2–1 z)
d) X(2z)
16. The Z-transform of correlation of the sequences x(n) and y(n) is,
a) X*(z) Y*(z–1)
b) X(z) Y( z–1)
d) X(z–1) Y( z–1)
c) X( z) * Y( z)
+¥
17. The parseval's relation states that if Z{x1 (n)} = X1(z) and Z{x2 (n)} = X2(z) then
å x (n) x (n) is,
1
*
2
n =-¥
z
z
FG IJ
HK
F 1I
X ( z) X G J z
Hz K
z
z
FG IJ
H K
F 1I
X ( z) X G J z
Hz K
a)
1
1 -1
X1 ( z) X*2
z dz
2p c
z
b)
1
1
X1 ( z) X2 * z -1 dz
2p c
z
c)
1
2pj c
d)
1
2p
1
*
2
*
-1
dz
c
1
*
2
*
-1
dz
18. For a stable LTI discrete time system poles should lie ––––– and unit circle should be ––––––– .
a) outside unit circle, included in ROC
b) inside unit circle, outside of ROC
c) inside unit circle, included in ROC
d) outside unit circle, outside of ROC
19. An LTI system with impulse response, h(n) = (–a)n u(n) and –a < –1 will be,
a) stable system
b) unstable system
c) anticausal system
d) neither stable nor causal
20. If X(z) has a single pole on the unit circle, on negative real axis then, x(n) is,
a) signed constant sequence
b) signed decaying sequence
c) signed growing sequence
d) constant sequence
7. 87
Signals & Systems
n
21. The Z-transform of x(n) = –na u(–n – 1) is,
a) X(z) =
az
( z - a)2
22. The ROC for x(n) ¬
a)
b)
Z
Z -1
a z(z + a)
( z - a )3
® X(z)
R1
a
c) X(z) =
is R1, then ROC of an x(n) ¬
b) aR1
a z -1
(1 - a z -1 ) 2
Z
Z-1
d) both a and c
® X e az j is,
d)
c) R 1
1
R1
23. The Z-transform of a ramp function x(n) = n u(n) is,
a) X(z) =
z
; ROC is z > 1
(z - 1) 2
b) X( z) =
-z
; ROC is z > 1
( z - 1) 2
c) X( z) =
z
; ROC is z < 1
( z - 1) 2
d) X( z) =
-z
; ROC is z < 1
( z - 1) 2
24. By impulse invariant transformation, if x(nT) is sampled version of x(t), then Z{x(nT)} is,
l
q
a) L x(n T)
l
LM
N
25. The Z-transform of x(n) = sin
a)
z
z +1
q
b) L-1 x(n T)
z = e sT
b)
z = e - sT
l
q
c) L x(n T)
z = e - sT
l
q
d) L-1 x(n T)
z = e sT
OP
Q
p
n u(n) is,
2
z2
z +1
c)
2
1
z +1
d)
z
z2 + 1
Answers
1. c
6. b
11. b
16. b
21. d
2. a
7. d
12. d
17. c
22. a
3. b
8. b
13. c
18. c
23. a
4. d
9. a
14. b
19. b
24. a
5. c
10. c
15. b
20. a
25. d
IV. Answer the following questions
1. Define one sided and two-sided Z-transform.
2. What is region of convergence (ROC)?
3. State the final value theorem with regard to Z-transform.
4. State the initial value theorem with regard to Z-transform.
5. Define Z-transform of unit step signal.
6. What are the different methods available for inverse Z-transform?
7. When the z-domain transfer function of the system can be directly obtained from s-domain transfer function?
8. Define the transfer function of an LTI system.
9. Write the transfer function of N th order LTI system.
10. What is impulse invariant transformation?
11. How a-point in s-plane is mapped to z-plane in impulse invariant transformation?
12. Why an impulse invariant transformation is not considered to be one-to-one?
13. Give the importance of convolution and deconvolution operations using Z-transform.
14. Give the conditions for stability of an LTI discrete time system in z-plane.
Chapter 7 - Z - Transform
7. 88
15. Explain when an LTI dicrete time system will be causal.
16. Define ROC for various finite and infinite discrete time signals.
17. Explain the shifting property of a discrete time signal defined in the range 0 < n < ¥ with an example.
18. What are all the properties of ROC of a rational function of z?
19. List the various elements that can be used to realise the structure of an LTI discrete time system.
20. State and prove the linearity property of Z-transform.
V. Solve the Following Problems
E7.1
Determine the Z-transform and their ROC of the following discrete time signals.
l
A
c) x(n) = l2,
q
l
a) x(n) = 1, 3, 5, 6
q
1.0, 1, 2, 5, 7, 2
A
e) x(n) = (0.1) n u(n) + (0.3) n u( - n - 1)
E7.2
q
b) x(n) = 3, 0, 9, 0, 27, 2
A
d) x(n) = -0.4 n u(n - 1)
f) x(n) = (0.4) n
Find the one sided Z-transform of the following discrete time signals.
b) x(n) = n(0.5) n +2
a) x(n) = n 2 2 n u(n)
c) x(n) = (0.5) n u(n) - u( n - 2)
E7.3
Find the one sided Z-transform of the discrete signals generated by mathematically sampling the
following continuous time signals.
a) x(n) = 3 t e -0.5t u(t)
b) x(t) = 2 t 3 u(t)
E7.4
Find the time domain initial value x(0) and final value x(¥) of the following z-domain functions.
2
z2
a) X(z) =
2
b) X(z) =
-1
-1
1+ z
1- z
z - 1 z - 0.2
d
E7.5
i
b gb
g
Determine the inverse Z-transform of the following functions by partial fraction method.
5 z2
z ( z2 - 1)
4 z2 - 2 z
a) X(z) =
c) X(z) = 2
b) X(z) = 3
2
2
( z + 1) 2
( z + 1) ( z + 2)
z - 5z + 8 z - 4
3E7.7
Determine the inverse Z-transform of the function, X(z) =
E7.9
5 -1
z
6
LM1 - 1 z OP LM1 - 1 z OP
N 4 QN 3 Q
-1
E7.8
g
Determine the inverse Z-transform of the following functions using contour integral method.
2z - 1 z
z2 + z
(1 - e -a ) z
a) X(z) =
b) X(z) =
c) X(z) =
2
3
4 ( z - 1)
( z - 2)
( z - 1) ( z - e -a )
b
E7.6
id
-1
1
1
1
1
b) ROC : z < ,
a) ROC : z > ,
c) ROC : < z < .
4
3
4
3
Determine the inverse Z-transform for the following function using power series method.
z
X(z) = 2
2 z - 3z + 1
a) ROC : z < 0.5,
b) ROC : z > 1
Determine the inverse Z-transform for the following functions using power series method.
z2 + z
1 - 0.5 z -1
a) X(z) = 2
; ROC : z > 1
b) X(z) =
;
ROC : z > 0.5
z - 2z + 1
1 + 0.5 z -1
7. 89
Signals & Systems
E7.10
Determine the transfer function and impulse response for the systems described by the following
equations.
3
1
b) y(n) - y( n - 1) + y( n - 2) = 2 x( n)
a) y(n) - 2 y( n - 1) - 3 y( n - 2) = x( n - 1)
4
8
c) y(n) = 0.2 x(n) - 0.5 x( n - 1) + 0.6 y( n - 1) - 0.08 y( n - 2)
1
1
d) y(n) - y(n - 1) = x( n) + x( n - 1)
2
3
E7.11
A discrete time LTI system is characterized by a transfer function, H ( z) =
z ( 3 z - 4)
FG z - 1 IJ (z - 3) . Specify the
H 2K
ROC of H(z) and determine h(n) for the system to be (i) stable (ii) causal.
E7.12
E7.13
Determine the unit step response of a discrete time LTI system, whose input and output relation is
described by the differential equation, y(n) + 3 y(n–1) = x(n), where the initial condition
is y(–1) = 1.
Determine the response of discrete time LTI system governed by the following differential equation,
3 y(n) – 4 y(n –1) + y(n – 2) = x(n) ; with initial conditions, y(–2) = –2 and y(–1) = 1 for the input
x( n) =
E7.14
FG 1 IJ
H 2K
n
u(n).
An LTI system has the impulse response h(n) defined by h( n) = x1 (n - 1) * x2 (n) .The Z-transform of
the two signals x1(n) and x2(n) are X1(z) = 1– 3z–1 and X2(z) = 1 + 2z–2. Determine the output of the
system for the input d(n – 1).
Answers
E7.1
3 5
6
+ +
b) X(z) = 3 z5 + 9 z3 + 27 z + 2
z z2 z 3
ROC is entire z - plane except at z = ¥.
ROC is entire z - plane except at z = 0.
5 7 2
c) X(z) = 2 z3 + 10 z 2 + z + 2 + + 2 + 3
z z
z
ROC is entire z - plane except at z = 0 and z = ¥.
-0.4
d) X(z) =
;
ROC is exterior of the circle of radius 0.4 in z - plane.
z - 0.4
a) X(z) = 1 +
e) X(z) =
-0.2 z
(z - 0.1) (z - 0.3)
;
ROC is 0.1 < z < 0.3
f) X(z) =
-2.1 z
(z - 0.4) (z - 2.5)
;
ROC is 0.4 < z < 2.5
E7.2
a) X(z) =
E7.3
a) X(z) =
E7.4
2 z(z + 2)
( z - 2)3
3zT e-0.5T
( z - e -0.5T ) 2
a) Initial value, x(0) = 2
Final value, x(¥) = 0.5
b) X(z) =
0.53 z
( z - 0.5) 2
b) X(z) =
c) X(z) =
z2 - 0.25
z ( z - 0.5)
d
i
2 T 3 z z2 + 4 z + 1
( z - 1) 4
b) Initial value, x(0) = 1
Final value, x( ¥ ) = 1.25
Chapter 7 - Z - Transform
E7.5
a) x(n) = 0.12