Robert Sobot
Engineering
Mathematics
by Example
Vol. III: Special Functions
and Transformations
Second Edition
Engineering Mathematics by Example
Robert Sobot
Engineering Mathematics
by Example
Vol. III: Special Functions and Transformations
Second Edition
Robert Sobot
ETIS-ENSEA
Cergy-Pontoise, France
ISBN 978-3-031-41202-8
ISBN 978-3-031-41203-5
https://doi.org/10.1007/978-3-031-41203-5
(eBook)
1st edition: © Springer Nature Switzerland AG 2021
2nd edition: © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG
2024
This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part
of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting,
reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic
adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed.
The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply,
even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and
therefore free for general use.
The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be
true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or
implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher
remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
This Springer imprint is published by the registered company Springer Nature Switzerland AG
The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
Paper in this product is recyclable.
To my math teacher Mr. Miloš Belušević
Preface
Preface to the Second Edition
It is inevitable that first edition of any type of textbook, and especially textbooks for mathematics,
includes quite a few errors that slipped by all preproduction reviews. In this second edition, errors are
thoroughly reviewed and corrected, with hopes that not many new ones are created.
As the original volume doubled, this second edition is split into three separate books, Vol. I, II,
III. In order to reinforce natural development in the study, best effort is put to logically organize the
presented examples and techniques so that subsequent problems reference the once already solved.
Île-de France, France
June 27, 2023
Robert Sobot
Preface to the First Edition
This tutorial book resulted from my lecture notes developed for undergraduate engineering courses in
mathematics that I teach over the last several years at l’École Nationale Supérieure de l’Électronique
et de ses Applications (ENSEA), Cergy in Val d’Oise department, France.
My main inspiration to write this tutorial type collection of solved problems came from my
students who would often ask “How do I solve this? It is impossible to find the solution,” while
struggling to logically connect all the little steps and techniques that are required to combine together
before reaching solution. In the traditional classical school systems mathematics used to be thought
with the help of systematically organized volumes of problems that help us develop “the way of
thinking.” In other words, to learn how to apply the abstract mathematical concepts to everyday
engineering problems. Same as for music, it is also true for mathematics that in order to reach high
level of competence one must put daily effort into studying of typical forms over long period of time.
In this tutorial book I choose to give not only the complete solutions to the given problems, but
also guided hints to techniques being used at the given moment. Therefore, problems presented in this
book do not provide review of the rigorous mathematical theory, instead the theoretical background
is assumed, while this set of classic problems provides a playground to play and to adopt some of the
main problem-solving techniques.
vii
viii
Preface
The intended audience of this book are primarily undergraduate students in science and engineering. At the same time, my hope is that students of mathematics at any level will find this book to be
useful source of practical problems to practice.
Île-de France, France
November 30, 2020
Robert Sobot
Acknowledgments
I would like to acknowledge all those classic wonderful collections of mathematical problems that
I grew up with and used as the source of my knowledge, and I would like to say thank you to
their authors for providing me with thousands of problems to work on. Specifically, I would like to
acknowledge classic collections written in ex–Yugoslavian, Russian, English and French languages,
some are listed in the bibliography. Hence, I do want to acknowledge their contributions that are
clearly visible throughout this book, which are now being passed on to my readers.
I would like to thank all of my former and current students who I had opportunity to tutor in mathematics since my student years. Their relentless stream of questions posed with unconstrained curiosity
forced me to open my mind, to broaden my horizons, and to improve my own understandings.
I want to acknowledge Mr. Allen Sobot for reviewing and verifying some of the problems, for very
useful discussions and suggestions, as well as his work on technical redaction in first edition of this
book.
Sincere gratitude goes to my publisher and editors for their support and making this book possible.
ix
Contents
1
Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Sequence Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Finite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
3
4
4
4
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Sequence Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Finite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
6
9
11
14
17
2
Elementary Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 Basic Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Derivatives and Integrals with Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Special Functions Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
23
24
25
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 Basic Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Derivatives and Integrals with Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Special Functions Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
26
32
42
3
Continuous Time Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Basic Function Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Function Synthesis and Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Continuous Time Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
46
47
48
49
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Basic Function Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Function Synthesis and Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Continuous Time Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
50
54
59
82
xi
xii
4
Contents
Discrete Time Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1 Discrete Signal Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Discrete Signal Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
92
92
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.1 Discrete Signal Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.2 Discrete Signal Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5
Continuous Time Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.1 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.2 Composite Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
5.1 Special Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
5.2 Composite Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
5.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
6
Discrete-Time Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
6.1 Simple Seqences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
6.2 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
6.3 Inverse Discrete Time FT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.4 Fast Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
6.1 Simple Seqences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
6.2 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
6.3 Inverse Discrete Time FT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
6.4 Fast Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
7
Laplace Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
7.1 Basic Laplace Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
7.2 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
7.3 Inverse Laplace Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
7.4 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
7.1 Basic Laplace Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
7.2 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
7.3 Inverse Laplace Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
7.4 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
Acronyms
f (x), g(x)
f (x), g (x)
f (g(x))
sin(x)
cos(x)
tan(x)
arctan(x)
sinc (t)
log(x)
ln(x)
logn (x)
x∗y
i
j
ℑ(z)
ℜ(z)
F (x(t))
X(ω)
L −1 {F (s)}
F −1 (X(ω))
F
−
→
LT
−→
l’H
=
DF T
FFT
δ(x)
(x)
(x)
XT
(t)
u(t)
r(t)
sign (t)
(a, b)
(a, b)
≡
Functions of x
First derivatives of functions of f (x), g(x)
Composite function
Sine of x
Cosine of x
Tangent of x
Arctangent of x
Sine cardinal of t
Logarithm of x, base 10
Natural logarithm of x, base e
Logarithm of x, base n
Convolution product
Imaginary unit, i 2 = −1
Imaginary unit, j 2 = −1
Imaginary part of a complex number z
Real part of a complex number z
Fourier transform of x(t)
Fourier transform of x(t)
Inverse Laplace transform of F (s)
Inverse Fourier transform of X(ω)
Apply Fourier transform
Apply Laplace transform
Apply l’Hôpital rule
Discrete Fourier transform
Fast Fourier transform
Dirac distribution
Triangular distribution
Square (rectangular) distribution
Dirac comb whose period is T
Heaviside step function of t
Heaviside step function of t
Ramp function of t
Sign function of t
Pair of numbers
Open interval
Equivalence
xiii
xiv
Acronyms
{a1 , a2 , a3 , . . . }
{an }
R
C
N
Q
a∈C
a∈R
∀x
i=0 ai
dB
dBm
H (j x)
x→ a
List of elements, vector, series
List of elements, vector, series
The set of real numbers
The set of complex numbers
The set of natural numbers
The set of rational numbers
Number a is included in the set of complex numbers
Number a is included in the set of real numbers
For all values of x
Angle, argument
It follows that
Derivative of y relative to x
Partial derivative of u(x, y, z, . . . ) relative to x
Second partial derivative of u(x, y, z, . . . ), relative to x then to y
Complex conjugate of number z
Absolute value of x
Matrix A whose size is m × n
The identity square matrix whose size is n × n
Determinant of matrix A
Transpose of matrix A
Difference between two variables
Main determinant of a matrix
Cramer’s sub-determinant relative to variable x
Vector a
Sum of elements {a0 , a1 , . . . , a∞ }
Decibel
Decibel normalized to “10−3 ”, i.e. “milli”
Transfer function
Limiting to a from its right side (x-axis)
x→ a
Limiting to a from its left side (x-axis)
x → a
x →
a
·
Limiting to a from above (y-axis)
Limiting to a from below (y-axis)
in-line hints
⇒
dy
dx
ux
ux,y
∗
z
|x|
Am,n
In
|A|
AT
x
a
∞
1
Series
Infinite series: an infinite ordered list of elements (terms) to be added
∞
ai = a1 + a2 + a3 + · · ·
i=1
Limit of infinite series: first, the sum of finite number n of elements ai , and then limit n to infinity
∞
ai = lim
n→∞
i=1
n
ai
i=1
Arithmetic progression: a sequence of numbers where the difference d between any two
subsequent terms is constant. That being case, given the first term a1 , a general term an is found
as
an = a1 + d (n − 1)
The sum of the first n terms in arithmetic progression
Sn =
n
ak =
k=1
n
(a1 + an )
2
Geometric progression: a sequence of numbers where the ratio r between any two subsequent
terms is constant. Given the first term a1 , a general term an is found as
a1 , a2 = a1 r, a3 = a2 r, . . .
an = a1 r n−1
The sum of geometric progression: if a1 = a, (a = 0) then the sum of first n terms is found as
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_1
1
2
1 Series
Sn =
n
n
ak =
a r k−1 = ar 0 + ar 1 + ar 2 + · · · + ar n−1
k=1
n terms
k=1
Assuming a = 0, there are two main cases:
1. r = 1 : the sum of first n elements is
Sn =
n
k=1
ak =
n
*1
a
= na
r k−1
k=1
thus, the sum of infinite series is
n→∞
n=1
a r n−1 =
n→∞
n→∞
*1 1n−1
a
=
a = a + a + a + ··· → ∞
n=1
n=1
infinitely terms
In conclusion, if r = 1, infinite geometric series is divergent.
2. r = 1 : The sum of first n elements is derived as follows:
n−1
+
ar
ar2 + · · · + ar
∴
Sn = a + n−1
+
ar
ar2 + · · · + ar
+ ar n
r Sn = (Sn − r Sn ) = a − ar n ⇒ Sn =
a (1 − r n )
1−r
Convergence of the geometric sum: depending upon the value of |r| = 0, where a is the first term
in the series, then the infinite sum may be calculated as
(a) |r| > 1: in this case, each power term r n of the geometric series is greater and greater; therefore
∞
n=1
∞
a (1 − r n )
−anr n−1
l’H
(=)
→∞
= lim
n→∞
n→∞
1−r
∞
−1
a r n−1 = lim
In conclusion, for |r| >, 1 infinite geometric series is divergent.
(b) |r| < 1: in this case, recall that powers r n of number r inferior to one are smaller and smaller
tending to zero; thus
∞
n=1
ar
n−1
n
>
a 1−
r
= lim
n→∞
1−r
0
=
a
1−r
In conclusion, for |r| < 1, infinite geometric series is convergent.
Mathematical induction: given identity F (n) in function of n can be proven by the mathematical
induction technique that consists of three steps:
1.2 Finite Series
3
1. F (n) must be proven to be true for n = 1.
2. Assumption is that F (k) is also true, where k is an arbitrary value of n = k.
3. F (n) must be proven to be true for the next value of n = k + 1
and then, the conclusion is that given identity F (n) is true for all n.
Problems
Finite or infinite series are summed in accordance with their respective forms. In general, it is
important to first determine if the series is in the arithmetic or geometric form and then to apply
various algebraic methods.
1.1
Sequence Notation
Determine general term an of infinite series in P.1.7 to P.1.9.
1.1. 1,3,7,15,31,63,127 · · · 1.2. 1, −1, 1, −1, 1, · · ·
2 3 4
1.4. 1, , , , · · ·
3 5 7
1.7.
3 4 5 6
, , , ,···
2 3 4 5
1.5.
1.3. S = 23 + 33 + · · · + n3
1
1
1
1
1 1
1
,
,
,
, · · · 1.6. 1, − , , − , · · ·
1·2 2·3 3·4 4·5
2 3
4
1.8. S =
1 2 3
+ + + ···
2 3 4
1.9. 1,
2
3
4
,
,
,···
1·2 1·2·3 1·2·3·4
Calculate sums of finite series P.1.10 to P.1.15.
1.10.
4
i
1.11.
i=1
1.13.
5
2
1.12.
i=1
2j
1.14.
j =0
1.2
4
i2
i=1
4
1
k=1
3
k
1.15.
3
n−1
n=1
n2 + 3
Finite Series
Derive sums in P.1.16 to P.1.21.
1.16.
n
i=1
1
1.17.
n
i=1
i
1.18.
5
2n+2
n=1
3n
4
1.19.
1 Series
n
i(4i − 3)
1.20.
25
(5 − 2i)
1.23.
i=1
1.22.
i=1
1.3
n
1
1.21.
2k
i=1
10
(−2)n−1
5 11 23 47
, , , ,···
6 12 24 48
1.24.
n
(−1)k−1 x 2k
k=1
n=1
Infinite Series
Derive infinite sums in P.1.25 to P.1.30.
1.25.
∞
√
n−3
1.26.
n=3
1.27.
lim
n→∞
∞
3
i=1
n
1.29.
1.4
1−
lim
n→∞
i
n
1
4
2
1.28.
+1
1−
1
1
··· 1 − 2
9
n
11
lim
k→∞
n=0
lim
n→∞
1.30.
lim
cos
k nπ
6
2
n
1
+ 2 + ··· + 2
2
n
n
n
∞
n→∞
1
n(n + 1)(n + 2)
n=1
Geometric Series
Deduce general terms of series in P.1.31 to P.1.38 and then calculate the sums.
1.31.
∞
−2
n=1
3
4
n−1
1.32.
3 1 1
1
1
+ + +
+
+ ···
2 2 6 18 54
∞
22n 31−n
1.35. S =
1.34. 5 −
1.36.
10 20 40
+
−
+ ···
3
9
27
10
(4, −8, 16, −32, 64, . . . )
n=1
k=0
∞
∞
(−1)n−1
x n−1 ; |x| < 1
1.38.
1
n=1
1.5
5(−5)n
1
1.33.
1.37. S =
5
Mathematical Induction
Prove identities in P.1.39 to P.1.45.
1.39. 1 + 2 + 3 + · · · + n =
n(n + 1)
2
3n−1
1.5
Mathematical Induction
5
1.40. 12 + 22 + 32 + · · · + n2 =
n(n + 1)(2n + 1)
6
1.41. 13 + 23 + 33 + · · · + n3 =
n(n + 1)
2
2
1.42. 1 + 2 + 22 + · · · + 2n−1 = 2n − 1
1.43. 1 + 3 + 5 + · · · + (2n − 1) = n2
1.44. 1 · 2 + 2 · 3 + 3 · 4 + · · · + n(n + 1) =
1.45.
n(n + 1)(n + 2)
3
1
1
1
n
1
+
+
+ ··· +
=
1·3 3·5 5·7
(2n − 1)(2n + 1)
2n + 1
6
1 Series
Answers
1.1
Sequence Notation
1.1. One possible set of solutions may be derived by noticing that all terms of given series are close
to powers of two series, i.e. 2n = 2, 4, 8, 16, 32, . . . , thus
1, 3, 7, 15, 31, 63, 127 · · · ⇒ ak = 2k − 1, k ∈ (1, ∞)
or,
ak = 2k+1 − 1, k ∈ (0, ∞), etc.
1.2. Series whose elements are alternating ±1 may be written as
1, −1, 1, −1, 1, · · · ⇒ am = (−1)m , m ∈ (2, ∞),
or,
am = (−1)m−1 , m ∈ (1, ∞),
or,
am = cos(mπ ), m ∈ (1, ∞),
or,
am = ej mπ , (j 2 = −1) and m ∈ (1, ∞), etc.
1.3. Probably most obvious response is to write,
S = 23 + 3 3 + · · · + n 3 =
n
k3
k=2
However, all indexes may be reduced, for example, by one without changing the total number of terms
nor the total sum, as
S=
n−1
( j + 1)3 = ( 1 + 1)3 + ( 2 + 1)3 + ( 3 + 1)3 + · · · + ( (n − 1) + 1)3
j =1
= 23 + 33 + · · · + n3
Or, if for any reason it is preferred that the index counter starts with zero, then
S=
n−2
( m + 2)3 = ( 0 + 2)3 + ( 1 + 2)3 + ( 2 + 2)3 + · · · + ( (n − 2) + 2)3
m=0
= 23 + 33 + · · · + n3
1.4. One possible set of relations between numerator and denominator may be 2n − 1, that is
2 3 4
n
1, , , , · · · ⇒ an =
, n ∈ (1, ∞),
3 5 7
2n − 1
or,
1.1 Sequence Notation
7
an =
n+1
, n ∈ (0, ∞), etc.
2n + 1
1.5. Denominator is product of two subsequent numbers, thus
1
1
1
1
1
,
,
,
, · · · ⇒ ak =
, k ∈ (1, ∞) or,
1·2 2·3 3·4 4·5
k(k + 1)
ak =
1
, k ∈ (0, ∞) or,
(k + 1)(k + 2)
ak =
1
, k ∈ (2, ∞) etc.
(k − 1) k
1.6. Given series may be written as
1 1
1
+1 −1 +1 −1
1, − , , − , · · · =
,
,
,
, · · · = see A.1.2
2 3
4
1
2
3
4
⇒ an =
(−1)n+1
, n ∈ (1, ∞)
n
It is worth mentioning that this infinite sum converges to ln 2.
1.7. Note that general term of any given series may be written in infinite number of ways, there
is no unique solution. Some forms seem more evident than the others, nevertheless there are many
possibilities. For example, it may seem natural to say that numerator of any term equal denominator
plus one, i.e.
3 4 5 6
i+1
, , , , · · · ⇒ ai =
, i ∈ (2, ∞)
2 3 4 5
i
or,
ai =
i+2
i
, i ∈ (1, ∞) or, ai =
, i ∈ (3, ∞), · · ·
i+1
i−1
1.8. By inspection, denominator equal numerator plus one, thus general form of this sum may be
written in any of the following forms
S=
∞
∞
∞
1 2 3
n
k+1
m+2
+ + + ··· =
=
=
2 3 4
n+1
k + 2 m=−1 m + 3
n=1
k=0
or relative to any other index starting at arbitrary value.
1.9. Denominator is expanding in the form of factorial, thus
1,
k
3
4
2
,
,
, · · · ⇒ ak = , k ∈ (1, ∞)
1·2 1·2·3 1·2·3·4
k!
1.10. Finite number of terms k = 4 in series can be added in a finite amount of time. Thus, a four
element sequence where each term equal to its position in the series, here a general term ni = i, i.e.
8
1 Series
n1 = 1, n2 = 2, n3 = 3, n4 = 4, adds up as
4
i = 1 + 2 + 3 + 4 = 10
i=1
This general form of a sequence is better known as “arithmetic progression” or, in this case, the sum
of first m integers, where the number of terms to be added m = 4. Note that indexing of terms starts
with “1” and finishes with m = 4, by consequence there the total of m terms to be added.
1.11. A finite sequence of k identical terms, here a general term ni = 2 (that is to say, not a function
of i), i.e. n1 = 2, n2 = 2, n3 = 2, n4 = 2, is added as
4
i=1
2 = 2 + 2 + 2 + 2 = 4·2 = 8
4 times
where, the indexing range is i ∈ (1, 4). As index of the last element in the series is ‘4’ by consequence
there is the total of k = 4 terms to be added.
1.12. Sum of the first k = 3 integer squares ni = i 2 , i.e. n1 = 12 , n2 = 22 , n3 = 32 , is
3
i 2 = 12 + 22 + 32 = 1 + 4 + 9 = 14
i=1
where, the indexing range is i ∈ (1, 3). As index of the last element in the series is ‘3’ by consequence
there is the total of k = 3 terms (i.e. 1, 4, and 9) to be added.
1.13. Finite series of “powers of two” aj = 2j , i.e. a0 = 20 , a1 = 21 , a2 = 22 , a3 = 23 , a4 =
24 , a5 = 25 , is added as
5
2j = 20 + 21 + 22 + 23 + 24 + 25 = 1 + 2 + 4 + 8 + 16 + 32 = 63
j =0
where, the indexing range is j ∈ (0, 5). As index of the last element in the series is ‘n = 5’ and index
of the first element is ‘0’, by consequence there is the total of k = n + 1 = 6 terms (i.e. 1, 2, 4, 8, 16
and 32) to be added. That is to say, elements indexed one to five, plus the one with the index zero.
1.14. Finite series of the first n = 4 integer fractions, a1 = 1/1, a2 = 1/2, a3 = 1/3, a4 = 1/4, sums as
4
1
k=1
k
=
1 1 1 1
25
+ + + =
1 2 3 4
12
where, the indexing range is k ∈ (1, 4). As index of the last element in the series is ‘n = 4’ and index
of the first element is ‘1’, by consequence there is the total of n = 4 terms to be added.
1.2 Finite Series
9
1.15. Given finite series of three elements in the form
ai =
n−1
n2 + 3
∴ its sum is
3
n=1
1.2
ai =
3
n−1
n=1
n2 + 3
=
2 1
1−1
2−1
3−1
0 1
13
+ 2
+ 2
= + +
=
2
6
1 +3
2 +3
3 +3
3 7 12
42
Finite Series
1.16. In the general case of the last index value n, the sum S is derived in function of n as
S=
n
i=1
1 = 1 + 1 + 1 + ··· + 1 = n
n×1
because, a general term of this series is ai = 1, that is to say, a1 = 1, a2 = 1, a3 = 1, . . . repeated n
times. Note that this solution is general because n may be finite or infinite. In the case that this series
is infinite, then the sum is found as limit
lim S = lim n = ∞
n→∞
n→∞
1.17. In comparison with P.1.10, general solution (a.k.a. Gauss’ formula) of this sum is derived
in function of n. The legend has it that this classic problem of the sum of an arithmetic sequence
was solved by Gauss when he was still in the elementary school. His proof takes advantage of the
commutative property of addition by writing the same sequence in both ascending and descending
order and creating n identical summing terms as
S=
S=
∴
1
n
+2
+3
+(n − 1) + · · ·
+···
+3
2S = (n + 1) +(n + 1) +(n + 1) + · · ·
+(n − 1) +n
+2
+1
+(n + 1) +(n + 1)
n×(n+1)
because vertical sum of each pair results in (n + 1) repeated n times: (n + 1), 2 + (n − 1) = (n + 1),
etc. Thus, double sum is
2S = n(n + 1) ⇒ S =
n(n + 1)
2
In conclusion, sum of the first n integers is
S=
n
i=1
i = 1 + 2 + 3 + ··· + n =
n(n + 1)
2
10
1 Series
5
2n+2
1.18.
n=1
3n
5
2 1
2
2 n
=
=
4
=
4
+
n
3
3
3
3
n=1
n=1
8
16
32
422
2 4
+ +
+
+
=4
=4
3 9 27 81 243
243
5
2n 22
2
+
2
3
3
+
2
3
4
+
2
3
5
1.19. This sum may be split into two already known sums as
n
i(4i − 3) =
i=1
n
4i −
2
i=1
n
3i = 4
i=1
n
i −3
2
i=1
n
i = see P.1.40 and P.1.17
i=1
n(n + 1)(2n + 1)
n(n + 1)
4n(n + 1)(2n + 1) − 9n(n + 1)
=4
−3
=
6
2
6
= (n + 1)
n(n + 1)(8n − 5)
8n2 − 5n
=
6
6
1.20. When n → ∞ this series is known as description of Zeno’s paradox. Noninfinite sum may be
solved as
Sn =
1
1
1
1
1
+ 3 + · · · + n−1 + n
+
2 22
2
2
2
∴
2Sn =
2
2
2
2
1
1
1
1
1
2
1
+ 2 + 3 + · · · + n−1 + n = 1 + + 2 + 3 + · · · + n−1 + n − n
2 2
2
2
2
2 2
2 2
2 2
Sn
= 1 + Sn −
1
2n
⇒ Sn = 1 −
1
2n
Note that for n → ∞ this sum converges to one.
1.21. General term of the given series may be deduced as
5 11 23 47
, , , ,··· ∴
6 12 24 48
5
6−1
1
1
=
=1− =1−
6
6
6
3 · 21
12 − 1
1
1
11
=
=1−
=1−
12
12
12
3 · 22
24 − 1
1
1
23
=
=1−
=1−
24
24
24
3 · 23
48 − 1
1
1
27
=
=1−
=1−
48
48
48
3 · 24
···
ak = 1 −
so that,
1 1
3 2k
1.3 Infinite Series
n
ak =
k=1
11
n
1−
k=1
1 1
3 2k
=
n
k=1
25
25
25
1.22.
(5 − 2i) =
5−2
i=
i=1
i=1
i=1
= 25 · 5 − 2
1.23.
1 1
1
1
= see A.1.20 = n −
1− n
k
3 k=1 2
3
2
n
1−
n
n
ak = (a1 + an )
2
k=1
25
(1 + 25) = 25 · 5 − 25 − 25 · 25 = 25(5 − 1 − 25) = −525
2
10
10
10
1 (−2)n−1 =
(−1)n−1 2n 2−1 =
(−1)n−1 2n
2
n=1
n=1
n=1
1
1−1 1
=
(−1) 2 + (−1)2−1 22 + (−1)3−1 23 · · · + (−1)10−1 210
2
1
2 − 4 + 8 − 16 + 32 − 64 + 128 − 256 + 512 − 1024
=
2
= 1 − 2 + 4 − 8 + 16 − 32 + 64 − 128 + 256 − 512 = −341
n
(−1)k−1 x 2k = (−1)1−1 x 2(1) + (−1)2−1 x 2(2) + (−1)3−1 x 2(3) + · · · + (−1)n−1 x 2n
1.24.
k=1
= (−1)0 x 2 + (−1)1 x 4 + (−1)2 x 6 + · · · + (−1)n−1 x 2n
= x 2 − x 4 + x 6 − · · · + (−1)n−1 x 2n
1.3
Infinite Series
1.25. S =
∞
√
√
√
√
n − 3 = 0 + 1 + 2 + 3 + ··· + n − 3 + ···
n=3
Obviously, this infinite sum is divergent because the last term in the sum is infinite,
√
∞−3=∞
1.26. Terms of this series in the first period are
S=
11
n=0
cos
0π
1π
2π
3π
4π
5π
nπ
= cos
+ cos
+ cos
+ cos
+ cos
+ cos
6
6
6
6
6
6
6
7π
8π
9π
10π
11π
6π
+ cos
+ cos
+ cos
+ cos
+ cos
6
6
6
6
6
6
√
√
√
√
1
1
3 1
3
3 1
3
=1+
+ +0− −
−1−
− −0+ +
2
2
2
2 2
2
2
2
+ cos
=0
=0
12
1 Series
Therefore, when k → ∞ this sum is extended to infinite number of full periods, thus the total sum is
zero (i.e. series is convergent). In any other case, the infinite sum of this series is divergent.
1.27. Note which variable is the sum counter, and which one is the limit counter,
lim
n→∞
∞
3
i=1
n
i
n
2
∞ 3 2 3
3
= lim
i
+
n→∞
n→∞
n n
n
n3
n
i=1
i=1
∞
∞
3 2 3 = lim
i
+
1
n→∞ n3
n i=1
i=1
= see P.1.40 and P.1.16
3 n(n
+ 1)(2n + 1) + 3 n
= lim
n→∞ n3 2
n
6 2
1 n + 1 2n + 1
= lim
+3
n→∞ 2
n
n
1
1
1 n
1+ n n
2+ n
= lim
+3
n→∞ 2
n
n
⎡
⎤
0
0
⎢1
⎥
1
1
1+ 2 + + 3⎥
= lim ⎢
⎣
⎦
n→∞ 2
n
n
+ 1 = lim
=
∞ 3 i2
+
2
1
(1 + 0)(2 + 0) + 3
2
=4
1.28. Given series may be rewritten so that
lim
n→∞
1
2
n
+ 2 + ··· + 2
2
n
n
n
1 + 2 + ··· + n
n(n + 1)
= see P.1.17 = lim
2
n→∞
n→∞
n
2 n2
= lim
0
n
= lim
2
1
1+ n
n→∞
2n
2
=
1
2
1.29. Given series may be factorized so that
1
1
1
1−
··· 1 − 2
1−
lim
n→∞
4
9
n
2
2
= a − b = (a − b)(a + b)
= lim
n→∞
= lim
n→∞
n2 − 1
4−1 9−1
···
4
9
n2
(2 − 1)(2 + 1) (3 − 1)(3 + 1)
(n − 1)(n + 1)
···
22
32
n2
1.3 Infinite Series
13
(n
−
2) n
(n
−
1) (n + 1)
1 · 3A 2 · 4A 3 · 5A 4 · 6A
A
··· X
X
n→∞
(n −X
1) · (n − 1)
n
2 · 2 3A · 3 4A · 4 5A · 5
X
A· n
∞ l’H
1
n+1
1
= lim
= lim =
=
n→∞ 2n
n→∞ 2
∞
2
= lim
1.30. One possible method to resolve limits of series similar to
lim
n→∞
∞
1
n(n
+
1)(n
+ 2)
n=1
is to use partial fractions decomposition method in such a way that sum of the resulting repetitive
pattern cancels with possible exception of a constant, e.g. ‘1’. The repetitive pattern may be deduced
by decomposing a few last terms in the series, which (if exists) may be extended back to the leading
terms as
1
1
1
1
=
−
+
n(n + 1)(n + 2)
2 n (n + 1)
2 (n + 2)
1
1
1
1
=
− +
(n − 1)n(n + 1)
2(n − 1) n
2 (n + 1)
1
1
1
1
=
−
+
(n − 2)(n − 1)n
2(n − 2) (n − 1)
2n
···
The emerging pattern may be applied to the leading terms (and verified if still correct) as
···
1
1
1
1
=
− +
6·7·8
2·6 7
2·8
1
1
1
1
=
− +
5·6·7
2·5 6
2·7
1
1
1
1
=
− +
4·5·6
2·4 5
2·6
1
1
1
1
=
− +
3·4·5
2·3 4
2·5
1
1
1
1
=
− +
2·3·4
2·2 3
2·4
1
1
1
1
=
− +
1·2·3
2·1 2
2·3
As a consequence, the sum simplifies as
14
1 Series
∞
1
1
1
1
= −
+
n(n
+
1)(n
+
2)
2
n
(n
+
1)
2
(n
+ 2)
n=1
1
1 1
− 1) − n + 2 (n + 1)
2(n
1 1 1
+ − + 2(n − 2) (n − 1) 2 n
···
+
1
1
1
+ A − C + AA 7CC 16
12
1
1
1
+A −C +A
A
C
AA
10A 6C 14
1 1
1
+ = − C + A
C
AA
8 5C 12
1 1
1
+ − + A
AA
6 4 10
1 1 1
+ − + 4 3 8
1 1 1
+ − + 2 2 6
That is to say, the series sum consists of non–canceled terms and converges as
0
>
*0
*0
1
1 1
1 1
1
lim
= lim
+ − + =
n→∞
n→∞ 4
n(n
+
1)(n
+
2)
2
(n
+
1)
(n
+
1)
2
(n
+
2)
4
n=1
∞
In conclusion, this series is convergent, to 1/4.
1.4
Geometric Series
1.31. First few terms of this infinite series where n ∈ (1, ∞) are
9
27
3
3
an : −2, − , − , − , · · · − 2
2
8
32
4
n−1
···
Therefore, the sum of infinite series where first term in the series is a1 = −2, r = 3/4 < 1 is
∞
n=1
Note that indexing starts with n = 1.
−2
3
4
n−1
=
−2
= −8
1 − 34
1.4 Geometric Series
15
1.32. Property of geometric series is that ratio r of any two subsequent terms is constant, i.e.
5(−5)n+1
an+1
=
= −5n+1−n = −5 (const.)
an
5(−5)n
r=
First few terms of this infinite series where n ∈ (1, ∞) are
an : −25, 125, · · · 5(−5)n · · ·
then, the first term in the series is a1 = 5(−5)1 = −25 and | − 5| = 5 > 1. Sum of the first n = 5
terms of geometric progression is
S5 =
5
5(−5)n =
n=1
a1 (1 − r n )
1−r
=
−25 1 − (−5)5
−78 150
=
= −13 025
1 − (−5)
6
1.33. Given
3 1 1
1
1
+ + +
+
+ ···
2 2 6 18 54
Property of geometric series is that ratio r of any two subsequent terms is constant, i.e.
r=
1/2
1/6
1/18
1
=
=
= ··· =
(const.)
3/2
1/2
1/6
3
then, the first term in the series is for n = 0, i.e. a0 = 3/2. Therefore, the first few terms are
a0 =
3
2
1
3
0
=
3
3
; a1 =
2
2
1
3
1
=
1
3
; a2 =
2
2
1
3
2
=
1
3
; · · · an =
6
2
Infinite geometric progression is convergent under condition
1
<1 3
|r| < 1 ⇒
therefore, sum of infinite geometric progression is
S∞ =
∞
ar n =
n=0
∞
3
n=0
2
1
3
n
=
! a0 "
3/2
9
=
=
1−r
1 − 1/3
4
Note that indexing starts with n = 0.
1.34. Given infinite series,
S =5−
it follows that
10 20 40
+
−
+ ···
3
9
27
1
3
n
;
16
1 Series
1. ratio of subsequent terms is
−
2
20
2
2
10
40
−
3 = · · · = − 2 = r ⇒ |r| = 2 < 1
3 = 9 3 = 27
10
20
3
3
5
−
9
3
1
2. then given series may be written as
S =5−
10 20 40
+
−
+ ··· = 5
3
9
27
−
2
3
0
+5
−
2
3
1
+5
−
2
3
2
+ ··· + 5
−
2
3
n
3. therefore, sum of infinitely long series where r = −2/3 and the first term in the series is a0 = 5 is
calculated as
5−
∞
10 20 40
+
−
+ ··· =
5
3
9
27
n=0
−
2
3
n
=
! a0 "
5
5
=
=3
=
2
1−r
5
1 − −3
3
Note that indexing starts at n = 0.
1.35. Given expression may be converted into the general form of geometric series as
S=
∞
22n 31−n =
n=1
=
∞
n=1
4
4
3
∞
∞
∞
4n
4 × 4n−1
(22 )n 3−(n−1) =
=
3n−1
3n−1
n=1
n=1
n=1
n−1
∴ |r| =
4
>1
3
Therefore, since |r| > 1 the conclusion is that this series is divergent, i.e. S → ∞.
1.36. Given series 4, −8, 16, −32, 64, . . . Property of geometric series is that ratio r of any two
subsequent terms is constant, i.e.
r=
−8
16
−32
=
=
= −2 (const.)
4
−8
16
then, given series may be written as
4 − 8 + 16 − 32 + 64 + · · · = 4(−2)0 + 4(−2)1 + 4(−2)2 + 4(−2)3 + 4(−2)4 + · · · + 4(−2)n
therefore, r = −2 and the first term is for n = 0, i.e. a0 = 4. Sum of the first n = 1 to n = 10 terms
of geometric progression is
S10 =
10
n=0
Note that indexing starts at n = 0.
4(−2)n =
−2 1 − (−2)10
= 2 732
1 − (−2)
1.5 Mathematical Induction
17
1.37. This expression is in the general form of geometric series, then given condition that |x| < 1,
it follows that
S=
∞
x
n−1
=
n=1
∞
1 x n−1 = a1 = 1, r = x =
n=1
1
1−x
1.38. Given infinite geometric series may be written as
∞
(−1)n−1
3n−1
1
=
∞
−1
3
1
n−1
first few terms of this infinite series where n ∈ (1, ∞) are
1
(−1)n−1
1 1
···
an : 1, − , , − · · ·
3 9
27
3n−1
where, r = −1/3, that is to say |r| < 1, and the first term in the series a1 = 1 so that
∞
(−1)n−1
3n−1
1
1.5
=
1
3
=
1
4
1 − −3
Mathematical Induction
1.39. Mathematical induction technique is based on the three step procedure:
1. given identity must be true for n = 1, as
1 + 2 + 3 + ··· + n =
n( n + 1)
⇒
2
1=
1( 1 + 1)
2
= =1 2
2
2. given that the identity is true for n = 1, it should be true for other n = k, i.e.
1 + 2 + 3 + ··· + k =
k(k + 1)
2
3. assuming that the identity is true for n = k, it must be proven that it is true for the next natural
number, i.e. n = k + 1. That is to say,
k( k + 1)
2
∴ add the next term (k + 1) to both sides
1 + 2 + 3 + ··· + k =
1 + 2 + 3 + · · · + k +(k + 1) =
k(k + 1)
2
(1.1)
k(k + 1)
k(k + 1) + 2 (k + 1)
(k + 1)(k + 2)
+(k + 1) =
=
2
2
2
18
1 Series
=
(k + 1) (k + 1) + 1
2
(1.2)
Obviously, (1.2) is in the same form as (1.1) after ‘ k → (k + 1)’ replacement. In conclusion, if
(1.1) is true then (1.2) is also true for (k + 1).
Final argument: for n = 1 (1.1) is true and, that being the case, it is true for the next n = 2.
Consequently, being true for n = 2 it is also true for the next n = 3, etc.. By the “domino effect”
(1.1) is therefore true for all subsequent n.
1.40. General term of this series is ai = i 2 , and the sum of first n squares is found to be
S=
n
i 2 = 12 + 22 + 32 + · · · + (n − 1)2 + n2 =
i=1
n( n + 1)(2 n + 1)
6
(1.3)
Proof and derivation of sum (1.3) is based on the three step procedure mathematical induction
technique
1. Case: n = 1 given identity (1.3) must be true.
S1 = 12 =
1 × ( 1 + 1)(2 × 1 + 1)
1×2×3
6
=
= =1 6
6
6
which simply proves that (1.3) is true for n = 1.
2. Assumption: given that the identity is true for n = 1, it should be true for other n = k, i.e.
Sk =
k( k + 1)(2 k + 1)
6
(1.4)
3. Case: n = k + 1 assuming that the identity (1.4) is true for n = k, it must be proven that it is true
for the next natural number, i.e. n = k + 1. That is to say, when the following term is added as
Sk+1 = 12 + 22 + · · · + k 2 +(k + 1)2 = Sk + (k + 1)2 =
Sk
=
k(2k + 1) + 6(k + 1)
k(k + 1)(2k + 1) + 6(k + 1)2
= (k + 1)
6
6
= (k + 1)
=
k(k + 1)(2k + 1)
+ (k + 1)2
6
2k 2 + 4k + 3k + 6
(k + 1) (k + 2)(2k + 3)
2k 2 + 7k + 6
= (k + 1)
=
6
6
6
(k + 1) (k + 1) + 1
6
2 (k + 1) + 1
(1.5)
Obviously, (1.5) is in the same form as (1.4) after ‘ k → (k + 1)’ replacement. In conclusion, if
(1.4) is true then (1.5) is also true for (k + 1).
1.5 Mathematical Induction
19
Final argument: for n = 1 (1.3) is true, and in the previous step it is proven that the next sum of
n = 2 elements is also true. Then, by induction, it must be that the next sum of n = 3 elements is
also correct, etc., for any subsequent sum of n elements.
1.41. Given sum of n cubes,
13 + 2 3 + 3 3 + · · · + n 3 =
n(n + 1)
2
2
(1.6)
Proof and derivation of sum (1.6) is based on the three step procedure mathematical induction
technique
1. Case: n = 1 given identity (1.6) must be true.
S1 = 13 =
n( n + 1)
2
2
=
1( 1 + 1)
2
2
=
4
=1 4
which simply proves that (1.6) is true for n = 1.
2. Assumption: given that the identity is true for n = 1, it should be true for other n = k, i.e.
Sk =
k( k + 1)
2
2
(1.7)
3. Case: n = k + 1 assuming that the identity (1.7) is true for n = k, it must be proven that it is true
for the next natural number, i.e. n = k + 1. That is to say, when the following term is added as
Sk+1 = 13 + 23 + 33 + · · · + k 3 +(k + 1)3 = Sk + (k + 1)3 =
Sk
k( k + 1)
2
2
+ (k + 1)3
2
2
2
k ( k + 1) + 22 (k + 1)3
2 k + 2 (k + 1)
2 k + 4k + 4
=
(k
+
1)
=
(k
+
1)
22
22
22
2
(k + 1) (k + 1) + 1
(k + 1) (k + 2) 2
=
=
(1.8)
2
2
=
2
2
Obviously, (1.7) is in the same form as (1.8) after ‘ k → (k + 1)’ replacement. In conclusion, if
(1.7) is true then (1.8) is also true for (k + 1).
Final argument: for n = 1 (1.6) is true, and in the previous step it is proven that the next sum of
n = 2 elements is also true. Then, by induction, it must be that the next sum of n = 3 elements is
also correct, etc., for any subsequent sum of n elements.
1.42. Given sum of n powers of two,
1 + 2 + 22 + · · · + 2n−1 = 2n − 1
(1.9)
20
1 Series
Proof and derivation of sum (1.9) is based on the three step procedure mathematical induction
technique
1. Case: n = 1 given identity (1.9) must be true.
S1 = 21−1 = 2 1 − 1 ∴ 20 = 1 which simply proves that (1.9) is true for n = 1.
2. Assumption: given that the identity is true for n = 1, it should be true for other n = k, i.e.
Sk = 2 k − 1
(1.10)
3. Case: n = k + 1 assuming that the identity (1.10) is true for n = k, it must be proven that it is
true for the next natural number, i.e. n = k + 1. That is to say, when the following term is added as
Sk+1 = 1 + 2 + 22 + · · · + 2k−1 +2k = Sk + 2k = 2 k − 1 + 2k
Sk
= 2k + 2k − 1 = 2 · 2k − 1 = 2 k+1 − 1 (1.11)
Obviously, (1.10) is in the same form as (1.11) after ‘ k → (k + 1)’ replacement. In conclusion,
if (1.10) is true then (1.11) is also true for (k + 1).
Final argument: for n = 1 (1.9) is true, and in the previous step it is proven that the next sum, i.e. of
n = 2 elements, is also correct. Then, by induction, it must be that the next sum of n = 3 elements is
also correct, etc., for any subsequent sum of n elements.
1.43. Given sum of n odd numbers
1 + 3 + 5 + · · · + (2n − 1) = n2
(1.12)
Proof and derivation of sum (1.12) is based on the three step procedure mathematical induction
technique
1. Case: n = 1 given identity (1.12) must be true.
S1 = (2 · 1 − 1) = 12 ∴ 1 = 1 which simply proves that (1.12) is true for n = 1.
2. Assumption: given that the identity is true for n = 1, it should be true for other n = k, i.e.
Sk = 1 + 3 + 5 + · · · + (2 k − 1) = k 2
(1.13)
3. Case: n = k + 1 assuming that the identity (1.13) is true for n = k, it must be proven that it is
true for the next natural number, i.e. n = k + 1. That is to say, when the following term is added as
1.5 Mathematical Induction
21
Sk+1 = 1 + 3 + 5 + · · · + (2k − 1) +(2k + 1) = Sk + (2k + 1) = k 2 + (2k + 1)
Sk
= (k + 1) 2
(1.14)
Note that given odd number (2k − 1) the next odd number is (2k − 1) + 2 = (2k + 1) because
all odd numbers (as well as even) are separated by ±2. Obviously, (1.13) is in the same form as
(1.14) after ‘ k → (k + 1)’ replacement. In conclusion, if (1.13) is true then (1.14) is also true
for (k + 1).
Final argument: for n = 1 (1.12) is true, and in the previous step it is proven that the next sum, i.e. of
n = 2 elements, is also correct. Then, by induction, it must be that the next sum of n = 3 elements is
also correct, etc., for any subsequent sum of n elements.
1.44. Given sum of n binary products
1 · 2 + 2 · 3 + 3 · 4 + · · · + n(n + 1) =
n(n + 1)(n + 2)
3
(1.15)
Proof and derivation of sum (1.15) is based on the three step procedure mathematical induction
technique
1. Case: n = 1 given identity (1.15) must be true.
S1 = 1 ( 1 + 1) =
6
1( 1 + 1)( 1 + 2)
∴ 2=
3
3
which simply proves that (1.15) is true for n = 1.
2. Assumption: given that the identity is true for n = 1, it should be true for other n = k, i.e.
Sk = 1 · 2 + 2 · 3 + 3 · 4 + · · · + k(k + 1) =
k( k + 1)( k + 2)
3
(1.16)
3. Case: n = k + 1 assuming that the identity (1.16) is true for n = k, it must be proven that it is
true for the next natural number, i.e. n = k + 1. That is to say, when the following term is added as
Sk+1 = 1 · 2 + 2 · 3 + 3 · 4 + · · · + k(k + 1) +(k + 1)(k + 2) = Sk + (k + 1)(k + 2)
Sk
=
k(k + 1)(k + 2)
k(k + 1)(k + 2) + 3(k + 1)(k + 2)
+ (k + 1)(k + 2) =
3
3
=
(k + 1) (k + 1) + 1 (k + 1) + 2
(k + 1)(k + 2)(k + 3)
=
3
3
(1.17)
Obviously, (1.16) is in the same form as (1.17) after ‘ k → (k + 1)’ replacement. In conclusion,
if (1.16) is true then (1.17) is also true for (k + 1).
22
1 Series
Final argument: for n = 1 (1.15) is true, and in the previous step it is proven that the next sum, i.e. of
n = 2 elements, is also correct. Then, by induction, it must be that the next sum of n = 3 elements is
also correct, etc., for any subsequent sum of n elements.
1.45. Given sum of n inverse binary odd number products
1
1
1
1
n
+
+
+ ··· +
=
1·3 3·5 5·7
(2n − 1)(2n + 1)
2n + 1
(1.18)
Proof and derivation of sum (1.18) is based on the three step procedure mathematical induction
technique
1. Case: n = 1 given identity (1.18) must be true.
S1 =
1
1
1
1
=
∴
=
(2 · 1 − 1)(2 · 1 + 1)
2· 1 + 1
3
3
which simply proves that (1.18) is true for n = 1.
2. Assumption: given that the identity is true for n = 1, it should be true for other n = k, i.e.
Sk =
1
1
1
k
1
+
+
+ ··· +
=
1·3 3·5 5·7
(2k − 1)(2k + 1)
2k + 1
(1.19)
3. Case: n = k + 1 assuming that the identity (1.19) is true for n = k, it must be proven that it is
true for the next natural number, i.e. n = k + 1. That is to say, when the following term is added as
Sk+1 =
1
1
1
1
1
+
+
+
+ ··· +
1·3 3·5 5·7
(2k − 1)(2k + 1) (2k + 1)(2k + 3)
Sk
1
k
1
k(2k + 1)(2k + 3) + (2k + 1)
=
+
=
(2k + 1)(2k + 3)
2k + 1 (2k + 1)(2k + 3)
(2k + 1)(2k + 1)(2k + 3)
(2k+
1) k(2k + 3) + 1
(2k+
1)
2k 2 + 3k + 1
(k + 1)
k+1
=
=
=
=
(2k
+
1)(2k
+
1)(2k
+
3)
(2k
+
1)(2k
+
3)
(2k
+
1)(2k
+
3)
2
(k
+ 1) + 1
(1.20)
= Sk +
Obviously, (1.19) is in the same form as (1.20) after ‘ k → (k + 1)’ replacement. In conclusion,
if (1.19) is true then (1.20) is also true for (k + 1).
Final argument: for n = 1 (1.18) is true, and in the previous step it is proven that the next sum, i.e. of
n = 2 elements, is also correct. Then, by induction, it must be that the next sum of n = 3 elements is
also correct, etc., for any subsequent sum of n elements.
2
Elementary Special Functions
Delta function is defined as
∞
−∞
δ(t) = 0; (t = 0)
δ(t) dt = 1
Geometrical interpretation of Dirac function δ(t) is that its area is normalized to one (as definite
integral), even though its height tends to infinity while at the same time its length tends to zero (thus,
strictly speaking, its area A = 0 · ∞ is not determined).
The main properties of Dirac delta function δ(t) are the following:
1. It is an even function, i.e., δ(t) = δ(−t)
∞
x(t)δ(t) dt = x(0)
2. Sampling property,
∞−∞
x(t)δ(t − t0 ) dt = x(t0 )
−∞
b
x(t)δ(t − t0 ) dt =
a
x(t0 ), t0 ∈ (a, b)
0, otherwise
Problems
2.1
Basic Special Functions
Sketch graph and give definition of special functions in P.2.1 to P.2.9.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_2
23
24
2 Special Functions
2.1. e(σ +j ω ) t , (σ, ω ∈ R)
2.2. sinc (t)
2.3. Heaviside u(t).
2.4. sign (t)
2.5. r(t)
2.6. (t)
2.7. (t)
2.8. δ(t)
2.9. XT (t)
2.2
Derivatives and Integrals with Special Functions
Calculate derivatives and integrals in P.2.10 to P.2.29.
2.10. u (t)
2.11. sign (t)
2.12. r (t)
2.13. (t/2)
t
2.14. (t)
2.15.
sign (τ ) dτ
−t
t
t
2.16.
u(τ ) dt
2.17.
−∞
∞
2.18.
1/4
(t) dt
−∞
2.19.
1/4
(t) dt
−∞
∞
2.22.
−∞
∞
2.24.
−∞
∞
2.26.
−∞
δ(t − 4) dt
(t) dt
3
2.23.
−∞
t 2 δ(t + 2) dt
sin(2.5 π t) δ(t − 1) dt
e
0
2.21.
−∞
∞
2.28.
(t) dt
−∞
∞
2.20.
r(τ ) dt
−∞
−at 2
δ(t + 1) dt, (a > 0)
∞
2.25.
−3/2
∞
2.27.
−∞
∞
2.29.
−∞
δ(t − 4) dt
t 2 δ(t + 2) dt
e−at δ(t + 1) dt, (a > 0)
2
x(t) δ(t − a) dt, a ∈ ℜ
2.3 Special Functions Relations
2.3
25
Special Functions Relations
Derive relationship of functions in P.2.30 to P.2.33.
2.30. u(t) as function of sign (t)
2.31. (t/2T ) as function of δ(t)
2.32. (t/2τ ) as function of u(t/2τ )
2.33. (t) as function of sign (t)
26
2 Special Functions
Answers
2.1
Basic Special Functions
2.1. Complex exponential function
x(t) = es t
where, s = σ + j ω
and σ, ω ∈ R
may be written as
x(t) = e(σ +j ω )t = eσ t ej ω t = eσ t
cos(ω t) + j sin(ω t)
ℜ(x(t))
ℑ(x(t))
x ∗ (t) = e(σ −j ω )t = eσ t e−j ω t = eσ t cos(ω t) − j sin(ω t)
where, eσ t is a real exponential function, while e±j ω t is a strictly complex exponential function.
Then, after applying Euler’s formula, it is evident that complex exponential term by itself consists of
cosine (along real axis) and sine (along the perpendicular imaginary axis) functions, Fig. 2.1 (left).
As the argument t progresses, amplitudes of the two trigonometric functions is multiplied by the real
exponential term.
Case 1: σ = 0, ω = 0 : complex exponential function becomes a constant.
x(t) = e(σ +j ω )t = e(0+j 0)t = e0 = 1
a real constant
Case 2: σ > 0, ω = 0 : real part is positive and imaginary part is zero. Therefore, only the real
exponential term is left.
x(t) = e(σ +j ω )t = e(σ +j 0)t = eσ t
a real exponential function
Case 3: σ = 0, ω = 0 : real part is zero and imaginary part is non–zero, therefore only the strictly
complex exponential term is left. By consequence, amplitudes of both sine and cosine functions are
constant ±1, Fig. 2.1 (right).
Fig. 2.1 Example P.2.1
2.1 Basic Special Functions
27
x(t) = e(σ +j ω )t = e(0+j ω )t = ej ω t = cos(ω t) + j sin(ω t)
a strictly complex exponential function
In addition, the sum and difference of complex conjugate exponents give
x(t) + x ∗ (t) = ej ω t + e−j ω t
= cos(ω t) + j sin(ω t) + cos(ω t) − j sin(ω t)
= 2 cos(ω t)
cos(ω t) =
∴
ej ω t + e−j ω t
2
x(t) − x ∗ (t) = ej ω t − e−j ω t
= cos(ω t) + j sin(ω t) − cos(ω t) − j sin(ω t)
= 2j sin(ω t)
sin(ω t) =
∴
ej ω t − e−j ω t
2j
Case 4: σ > 0, ω = 0 : increasing real exponential function multiplies amplitude of both cosine
function along the real axis, and sine function along the imaginary axis, Fig. 2.2 (left).
Case 5: σ < 0, ω = 0 : decreasing real exponential function multiplies amplitude of both cosine
function along the real axis, and sine function along the imaginary axis, Fig. 2.2 (right).
Reminder: Depending on specific values of (σ, ω ∈ R) complex exponential function e(σ +j ω ) t
transforms to a constant or other fundamental engineering functions.
Fig. 2.2 Example P.2.1
28
2 Special Functions
Fig. 2.3 Example P.2.2
sinc (t)
1
t
0
−2p
Fig. 2.4 Example P.2.3
0
2p
u(t)
1
t
0
0
2.2. Cardinal sine function sinc(t), see Fig. 2.3, is analytically defined as
sinc (t) =
sin t
t
or, sinc (π t) =
sin(π t)
πt
where the last definition is normalized to π . Cardinal sinus function is not defined for t = 0, however
when t → 0 it limits as sinc (0) → 1 (see the limits chapter in Calculus book). To indicate that point
(0, 1) is not included, it is marked with the crossed circle symbol.
2.3. Heaviside u(t), see Fig. 2.4, is a piecewise linear function that is analytically defined as
u(t) =
⎧
⎨ 0, t < 0 i.e.
limt→ 0 u(t) = 0
⎩ 1, t ≥ 0 i.e.
limt→ 0 u(t) = 1
Step function u(t) is not continuous at t = 0, where by agreement u(t) = 1 for t ≥ 0. To indicate
that point (0, 0) is not defined, it is marked with the crossed circle symbol. Note that limit at t = 0
does not exist because the left and right side limits are different.
2.4. Piecewise linear sign(t) function, see Fig. 2.5, is analytically defined as,
2.1 Basic Special Functions
29
Fig. 2.5 Example P.2.4
sign (t)
1
t
0
−1
0
Fig. 2.6 Example P.2.5
r(t)
1
t
0
0
⎧
⎪
−1, t < 0
⎪
⎨
0, t = 0 i.e.
sign (t) =
⎪
⎪
⎩ 1, t > 0 i.e.
1
limt→ 0 sign (t) = −1
limt→ 0 sign (t) = 1
or, alternatively, relative to absolute function
sign (t) =
|t|
t
= , ∀x = 0
|t|
t
By agreement, point of discontinuity at t = 0 is declared as sign (0) = 0.
2.5. Piecewise linear ramp function r(t), see Fig. 2.6, is analytically defined as follows.
r(t) =
0, t ≤ 0
t, t ≥ 0
Note that ramp function is well defined for t = 0, however the left and right side derivatives are not
the same (r (t) = 0, t ≤ 0, and r (t) = 1, t ≥ 0). Alternative analytical definition of ramp function
may be
r(t) = t u(t)
because, for t < 0 the product of any function thus including linear function y(t) = t with u(t)
equal zero.
30
2 Special Functions
Fig. 2.7 Example P.2.6
Fig. 2.8 Example P.2.7
2.6. Piecewise rectangular function (t), Fig. 2.7, is analytically defined as
t
T
⎧
⎨ 1, − T ≥ t ≥ T
=
2
2
⎩ 0, otherwise
Where, T is the length of non–zero interval centred around the origin. By agreement, the two points
of discontinuities at t = ±T/2 are declared as (±T/2) = 1. It should be evident that limits at two
discontinuity points t = ±T/2 are not existent. As shown in Ch. 5, P.5.10 and P.5.15, (t) and sinc (ω)
are dual functions in respect to Fourier transformation.
2.7. Triangle function (t), Fig. 2.8, is analytically defined as
t
T
⎧
t
T
⎪
⎪
+ 1, − ≤ t ≤ 0
⎪
⎪
2
⎨ T /2
= 1− t , 0≤t ≤ T
⎪
⎪
T /2
2
⎪
⎪
⎩ 0, otherwise
Where, parameter T is the length of non–zero interval centred around the origin. Note that triangle
function is well defined for ∀t, however the left and right side derivatives at t = ±T/2 and t = 0 are
not the same.
2.1 Basic Special Functions
31
Fig. 2.9 Example P.2.8
d (t)
1
t
0
0
Note that (t) = 0, |t| ≥ ±T/2, but (t) = 2/T , −T/2 ≤ t ≤ 0, (t) = −2/T , 0 ≤ t ≤ T/2. As
shown in Ch. 5, P.5.11, (t) and sinc2 (ω) are dual functions in respect to Fourier transformation.
2.8. Dirac delta function δ(t), Fig. 2.9, is defined by the means of definite integral as
∞
−∞
δ(t) dt = 1
δ(t) = 0, t = 0
Dirac delta function therefore equal zero for all t, except at t = 0 where, because being defined
by definite integral, its area is normalized to one, as indicated at y–axis (its height is infinite). By
consequence, the product of any continuous function and Dirac delta results in only one non–zero
point, assuming that Dirac delta is found in the given interval.
Reminder: Main properties of δ(t) are:
1. it is an even function, i.e. δ(t) = δ(−t)
∞
x(t)δ(t) dt = x(0)
2. sampling property,
−∞
∞
x(t)δ(t − t0 ) dt = x(t0 )
−∞
b
x(t)δ(t − t0 ) dt =
a
x(t0 ), t0 ∈ (a, b)
0, otherwise
2.9. Dirac comb (a.k.a. “sampling”) periodic function XT (t) is created by a sum of equidistant
Dirac functions, Fig. 2.10, analytically defined as
XT (t) =
∞
k=−∞
δ(t − kT )
32
2 Special Functions
Fig. 2.10 Example P.2.9
where, T is sampling period, thus sampling frequency is fs = 1/T . Using periodically placed
Dirac delta functions enables sampling of arbitrary number of sampling points—each product of the
particular Delta function with any continuous function results in one non–zero product. Dirac comb
is the key function for “analog to digital” (A/D) signal conversion.
2.2
Derivatives and Integrals with Special Functions
2.10. Derivative u (t) may be deduced from graphical interpretation, Fig. 2.11 (left). Obviously in
the two intervals where t = 0 the step function is constant, therefore by definition, its derivative is
1. Case t < 0: then u(t) = 0 ⇒ u (t) = 0,
2. Case t > 0: then u(t) = 1 ⇒ u (t) = 0, and
3. Case t = 0: then as the interval between u(t) = 0 and u(t) = 1 tends to zero, i.e. t → 0, the
rate of change between ‘0’ and ‘1’ tends to infinity, which by definition creates Dirac function. In
other words, as t → 0,
d u(t)
= δ(t)
dt
In summary, the zero time change between between ‘0’ and ‘1’ levels shows as Delta function.
Physical interpretation may be that zero time change requires an infinite amount of energy.
2.11. Derivative sign (t) may be deduced from graphical interpretation, Fig. 2.11 (right). In the two
intervals where t = 0 sign function is constant, by definition, its derivative is
1. Case t > 0: then sign (t) = +1 ⇒ sign (t) = 0,
2. Case t < 0: then sign (t) = −1 ⇒ sign (t) = 0,
3. Case t = 0: then as the interval between sign (t) = −1 and sign (t) = 1 tends to zero, i.e.
t → 0, the rate of change between ‘−1’ and ‘1’ tends to infinity, which by definition describes
Dirac function. However, the vertical change covers two units, each associated with “one Dirac”,
in other words, as t → 0,
d sign (t)
= 2δ(t)
dt
2.2 Derivatives and Integrals with Special Functions
0
const.
u(t)
1
const.
u (t) = d (t)
u (t) = 0
u (t) = 0
Δt → 0
const.
t
0
Δt → 0
const.
−1
Δt → 0
0
sign (t)
1
t
1
33
sign (t) = 2d (t)
2
t
u (t) = 0
0
u (t) = 0
t
Δt → 0
0
0
Fig. 2.11 Examples P.2.10, P.2.11
Fig. 2.12 Example P.2.12
r(t)
1
0
r(t) = t
const.
t
0
1
0
1
r (t) = u(t)
r (t) = 1
r (t) = 0
t
0
In summary, the zero time change between between ‘±1’ levels shows as two times Delta function.
2.12. Derivative r (t) may be deduced from graphical interpretation, Fig. 2.12. Obviously for t < 0
ramp function is constant and it is linear for t ≥ 0, therefore by definition its derivative is
1. Case t < 0: then r(t) = 0 ⇒ r (t) = 0,
2. Case t ≥ 0: then r(t) = t ⇒ r (t) = 1
In conclusion, for t < 0 derivative r (t) = 0 and for t ≥ 0 derivative r (t) = 1, which is identical to
definition of Heaviside step function u(t).
2.13. Derivative (t/2) may be deduced from graphical interpretation, Fig. 2.13. Obviously, except
at the two discontinuity points t = ±1 rectangular function is constant, therefore by definition, its
derivative is
1. Case t < −1: then (t/2) = 0 ⇒ (t/2) = 0,
2. Case −1 ≤ t ≤ 1: then (t/2) = 1 ⇒ (t/2) = 0,
34
2 Special Functions
Fig. 2.13 Examples P.2.13, P.2.14
3. Case t > 1: then (t/2) = 0 ⇒ (t/2) = 0,
4. Case t = −1: then as the interval between (t/2) = 0 and (t/2) = 1 tends to zero, i.e. t →
0, the rate of change between ‘0’ and ‘1’ tends to infinity, which by definition describes Dirac
function. In other words, as t → −1 (see A.2.22),
d (t/2)
= δ(t + 1), (t = −1)
dt
5. Case t = 1: then as the interval between (t/2) = 0 and (t/2) = −1 tends to zero, i.e. t → 0,
the rate of change between ‘0’ and ‘−1’ tends to negative infinity, which by definition describes
negative Dirac function. In other words, as t → 1,
d (t/2)
= −δ(t − 1), (t = 1)
dt
In summary, at each discontinuity point of (t/2) the rate of function’s change tends to infinity; to
positive infinity at t = −1 and negative infinity at t = 1, as indicated with the two Dirac functions in
Fig. 2.13 (left).
2.14. Derivative (t) may be deduced from graphical interpretation, Fig. 2.13 (right). There are
four distinct intervals, thus its derivative is
1. Case t < −1: then (t) = 0 ⇒ (t) = 0,
2. Case −1 ≤ t ≤ 0: then (t) = 1 + t ⇒ (t) = 1,
3. Case 0 ≤ t ≤ 1: then (t) = 1 − t ⇒ (t) = −1,
4. Case t > 1: then (t) = 0 ⇒ (t) = 0,
In summary, even though (t) is continuous, its derivative is a piecewise linear non–continuous
function.
2.15. Sign function equal ‘−1’ for negative t, and ‘+1’ for t ≥ 0, thus (by employing temporarily
variable τ )
2.2 Derivatives and Integrals with Special Functions
35
Case 1: τ ≤ 0
0
−t
sign (τ ) dτ =
0
−t
sign (τ ) dτ =
0
−t
0
(−1) dτ = − τ = − 0 − (−t) = −t
−t
Case 2: τ ≥ 0
t
t
sign (τ ) dτ =
0
sign (τ ) dτ =
t
0
0
t
(1) dτ = τ = (t − 0) = t
0
In summary, see Fig. 2.14
sign (t) dt =
−t
t
(t < 0) def
= |t|
(t ≥ 0)
2.16. Step function equal zero for t < 0, and equal one (i.e. a constant) for t ≥ 0, thus (by employing
temporarily variable τ )
Case 1: τ ≤ 0
0
−∞
u(τ ) dτ =
0
−∞
u(τ ) dτ =
0
>
0 dτ = 0
0
−∞
Case 2: τ ≥ 0
t
0
u(τ ) dτ =
t
0
u(τ ) dτ =
t
1 dτ =
0
t
0
t
dτ = τ = t
0
Fig. 2.14 Example P.2.15
f (t)
1
t
0
−1
sign
(t)
sign (t) dt
−1
0
1
36
2 Special Functions
In summary,
t
−∞
u(τ ) dτ =
t; t ≥ 0
0; t < 0
= r(t)
which should be easy to visualize, for t ≥ 0, as area of a rectangle whose height equal one and its
length, i.e. t, increases linearly increases.
2.17. Ramp function equal zero for t < 0, and t for t ≥ 0, thus (by using temporarily variable τ )
Case 1: τ ≤ 0
0
−∞
r(τ ) dτ =
0
−∞
r(τ ) dτ =
0
>
0 dτ = 0
0
−∞
Case 2: τ ≥ 0
t
r(τ ) dτ =
t
0
r(τ ) dτ =
t
0
τ dτ =
0
1 2 t
t2
τ =
2
2
0
In conclusion,
⎧ 2
⎨t
; t ≥0
r(τ ) dτ = 2
⎩
−∞
0; t < 0
t
=
t2
u(t)
2
2.18. By definition, see P.2.6, rectangular function is
t
T
⎧
⎨ 1, − T ≥ t ≥ T
=
2
2
⎩ 0, otherwise
that is to say, T = 1, see Fig. 2.15 (left). Defined integral results in area between function and the
horizontal axis, in this case calculated over three distinct intervals
∞
−∞
(t) dt =
=
−1/2
−∞
1/2
−1/2
(t) dt +
1/2
−1/2
1/2
dt = t =1
(t) dt +
∞
1/2
(t) dt =
−1/2
−∞
0 dt +
1/2
−1/2
1 dt +
∞
0 dt
1/2
−1/2
The only non–zero area is within the (−1/2, 1/2) interval. From the elementary geometry, it is evident
that rectangular area is height times length, which in this case equal one square unit despite infinite
bounds of the integral (because, elsewhere port function equal zero).
2.19. By definition, see P.2.6, rectangular function is
2.2 Derivatives and Integrals with Special Functions
37
Fig. 2.15 Examples P.2.18, P.2.19
t
T
⎧
⎨ 1, − T ≥ t ≥ T
=
2
2
⎩ 0, otherwise
i.e. T = 1, see Fig. 2.15 (right). Resulting non–zero area between function and horizontal axis, is
∞
−∞
(t) dt =
=
−1/2
−∞
(t) dt +
1/4
−1/2
(t) dt =
1/4
1
1
dt = t = − −
4
2
1
− /2
−1/2
1/4
−1/2
−∞
=
0 dt +
1/4
1 dt
−1/2
3
4
Evidently rectangular area is height times length, i.e. three quarters of a square unit.
2.20. By definition, see P.2.7, triangular function is
t
T
⎧ t
⎪
+ 1, −T/2 ≤ t ≤ 0
⎪
⎪
⎨ T /2
= 1 − t , 0 ≤ t ≤ T/2
⎪
⎪
T /2
⎪
⎩
0, otherwise
that is to say, T = 1, see Fig. 2.16 (left). That being the case, this triangular function is
⎧
1
⎪
⎨ 2t + 1, − /2 ≤ t ≤ 0
(t) = 1 − 2t, 0 ≤ t ≤ 1/2
⎪
⎩
0, otherwise
Result of a defined integral is area between function and horizontal axis, as
∞
−∞
(t) dt =
−1/2
−∞
(t) dt +
0
−1/2
(t) dt +
1/2
0
(t) dt +
∞
(t) dt
1/2
38
2 Special Functions
Fig. 2.16 Examples P.2.20, P.2.21
=
−1/2
−∞
0 dt +
0
= t 0
−1/2
(1 + 2t) dt +
1/2
(1 − 2t) dt +
0
∞
0 dt
1/2
1/2
1/2
1 1 1 1
1
+ t − t 2 = − + − =
2
4
2
4
2
0
0
−1/2
0
+ t 2 −1/2
Evidently, triangular area is height times length divided by two, i.e. a half square unit.
2.21. By definition, see P.2.7, triangular function is
t
T
⎧
t
T
⎪
⎪
+ 1, − ≤ t ≤ 0
⎪
⎪
2
⎨ T /2
= 1− t , 0≤t ≤ T
⎪
⎪
T /2
2
⎪
⎪
⎩ 0, otherwise
that is to say, T = 1, see Fig. 2.16 (right). That being the case, this triangular function is
⎧
1
⎪
⎨ 2t + 1, − /2 ≤ t ≤ 0
(t) = 1 − 2t, 0 ≤ t ≤ 1/2
⎪
⎩
0, otherwise
Result of a defined integral is area between function and horizontal axis, in this case calculated
over three distinct intervals
∞
−∞
(t) dt =
=
−1/2
−∞
−1/2
−∞
(t) dt +
0 dt +
0
0
−1/2
−1/2
(t) dt +
(1 + 2t) dt +
1/4
(t) dt
0
1/4
0
(1 − 2t) dt
2.2 Derivatives and Integrals with Special Functions
0
= t −1/2
39
1/4
1/4
1 1 1
1
7
+ t − t 2 = − + −
=
2 4 4 16
16
0
0
−1/2
0
+ t 2 The only non–zero area is within the (−1/2, 1/4) interval, which in this case equal 7/16 square units
because of the integral bounds that do not cover complete non–zero interval.
2.22. By definition, see P.2.8, Dirac delta function is
∞
−∞
δ(t) dt = 1
δ(t) = 0, t = 0
Another way of reading this definition is to say that Dirac delta is located at coordinate where its
argument equal zero. That is to say, δ(t) is located at t = 0 and equal zero for all t = 0.
So where is δ(t − 4) located then? Its argument is ‘t − 4’, therefore this Dirac delta is located at
the coordinate where its argument equal zero, that is
t −4=0⇒t =4
as illustrated in Fig. 2.17 (left). That being the case, the coordinate of Dirac delta t = 4 is still within
the integration interval (−∞, ∞), see Fig. 2.17 (left). By consequence, the calculated integral equal
zero for all t = 4 except in single point t = 4 that is covered by Dirac delta, thus
∞
−∞
δ(t − 4) dt = 1
2.23. As already shown in P.2.22, argument of this Dirac delta is ‘t − 4’, therefore it is located at the
coordinate where its argument equal zero, that is
t −4=0⇒t =4
d (t − 4)
d (t − 4)
1
1
0
t
0
Fig. 2.17 Examples P.2.22, P.2.23
4
0
t
0
3
4
40
2 Special Functions
as illustrated in Fig. 2.17 (right). That being the case, the coordinate t = 4 is not anymore within the
integration interval (−∞, 3), which is to say that the non–zero area of Dirac delta (equal to one) is
not included in the total calculated area whose interval is illustrated by shading in Fig. 2.17 (right).
By consequence, within given integration interval Dirac delta function always equal zero, thus
3
−∞
:0
δ(t−4) dt = 0
2.24. As already shown in P.2.22, argument of this Dirac delta is ‘t + 2’, therefore it is located at the
coordinate where its argument equal zero, that is
t + 2 = 0 ⇒ t = −2
Given f (t) = t 2 , product f (t) δ(t + 2) equal zero within the shaded interval, see Fig. 2.18 (left),
because δ(t + 2) = 0 for all t = −2, by consequence ‘0 · f (t) = 0’. However, at single point t = −2
Dirac delta is not equal zero, and f (−2) = (−2)2 = 4. That being the case, their non–zero product,
and therefore integral, is calculated as
Case 1: t = −2
∞
t δ(t + 2) dt =
2
−∞
∞
−∞
t 0 dt =
2
∞
−∞
0 dt = 0
Case 2: t = −2
∞
−∞
t δ(t + 2) dt =
2
∞
∞
:1
(−2) δ(t + 2) dt = 4 δ(t+2) dt = 4
−∞
−∞
2
because, see A.2.22, Dirac delta at δ(t + 2) is still within the integration interval, thus its integral
equal one. This is known as “sampling property” of ‘f (t) δ(t)’ product, see Fig. 2.18 (left).
f (t) d (t + 1)
f (t) d (t + 1)
4
f (−2)
1
1
f (t) d (t + 1) = 0
f (t) d (t + 1) = 0
0
t
−2
Fig. 2.18 Examples P.2.24, P.2.25
0
0
t
−2 −3/2
0
2.2 Derivatives and Integrals with Special Functions
41
2.25. As already shown in P.2.22, argument of this Dirac delta is ‘t + 2’, therefore it is located at the
coordinate where its argument equal zero, that is
t + 2 = 0 ⇒ t = −2
that is to say, outside of the given interval (−3/2, +∞). Given f (t) = t 2 , product f (t) δ(t + 2) equal
zero within the integration interval (shaded), see Fig. 2.18 (right), because δ(t +2) = 0 for all t = −2
and by consequence ‘0 · f (t) = 0’. As Dirac delta is outside of the integration interval,
∞
−3/2
t 2 δ(t + 2) dt =
∞
−3/2
t 2 0 dt =
∞
−3/2
0 dt = 0
which is expected, see A.2.23, as Dirac delta at δ(t + 2) is not within the integration interval.
2.26. Given,
∞
−∞
sin(2.5 π t) δ(t − 1) dt
it follows that Dirac delta is located at ‘t − 1 = 0 ⇒ t = 1’. Consequently, the product
‘sin(2.5 π t) δ(t − 1)’ is not equal zero only at ‘t = 1’, see Fig. 2.19. Thus,
∞
−∞
sin(2.5 π t) δ(t − 1) dt =
∞
∞
:1
sin(2.5 π × 1) δ(t − 1) dt = sin(2.5 π ) δ(t−1) dt
−∞
−∞ const.
= sin(2.5 π ) = sin(π/2)
=1
2.27. Given
∞
−∞
e−at δ(t + 1) dt, (a > 0)
2
Fig. 2.19 Example P.2.26
f (t) d (t − 1)
1
0
t
f (t) d (t + 1) = 0
−1
0
1
42
2 Special Functions
f (t) d (t + 1)
f (t) d (t + 1)
1
1
f (t) d (t + 1) = 0
f (−1)
1/ea
0
f (t) d (t + 1) = 0
t
−1
0
0
t
−1
0
Fig. 2.20 Examples P.2.27, P.2.28
position of δ(t + 1) is at t + 1 = 0 ∴ t = −1, which is inside the bonds of given definite integral,
see Fig. 2.20 (left). Then,
∞
−∞
e−at δ(t + 1) dt =
2
∞
:1
2
e−a(−1) δ(t + 1) dt = e−a δ(t+1) dt = e−a
−∞
−∞
∞
2.28. As oppose to P.2.27, as δ(t) is not found within the bonds of definite integral, then for ‘t = −1’
all products equal zero because δ(t) = 0, see Fig. 2.20 (right), and,
∞
e−at δ(t + 1) dt =
2
0
∞
0
2
:0
δ(t+1) dt =
e−at ∞
0 dt = 0
0
2.29. Considering that δ(t − a) is located at t = a, which is certainly within interval t ∈ (−∞, ∞),
then
∞
−∞
2.3
x(t) δ(t − a) dt =
∞
−∞
x(a) δ(t − a) dt = x(a)
∞
:1
−a)
δ(t
dt
= x(a)
−∞
Special Functions Relations
2.30. One way to derive sign (t) as function of u(t) may be by a couple of simple transformations as
illustrated in Fig. 2.21. In the first step, see Fig. 2.21 (left), u(t) is multiplied by factor two to create
2u(t) whose amplitude equal two when t ≥ 0, see Fig. 2.21 (centre). In the second step 2u(t) is
shifted down to create 2u(t) − 1, which is identical to sign (t) . Thus,
sign (t) = 2u(t) − 1
As a side note, derivative
d
d
d
2u(t) − 1 = 2 u(t) − (1) = see A.2.10 = 2δ(t)
dt
dt
dt
2.3 Special Functions Relations
2
43
u(t)
2
1
2u(t)
1
t
0
−1
1
t
0
t
0
−1
0
2u(t) − 1
2
−1
0
0
Fig. 2.21 Example P.2.30
Fig. 2.22 Example P.2.32
1
0
u(t + t )
1
t
u(t − t )
t
0
1
0
Π (t/2t )
−t
t
0
t
which is same result as already derived in A.2.11.
2.31. As already shown in P.2.13 and Fig. 2.13, derivative of rectangular function (t/2T ) may be
expressed as function of Dirac delta as
d
(t/2T ) = δ (t + T ) − δ (t − T )
dt
2.32. One possible relation between (t/2τ ) and u(t) is illustrated in Fig. 2.22, where one step
function may be advanced and one delayed in time. The difference between these two u(t) functions
produces (t/2τ ) as
(t/2τ ) = u (t + τ ) − u (t − τ )
which is deduced by calculating the difference within each piecewise interval. For example, within
the interval t ∈ (−τ, τ ),
u(t + τ ) = 1 and
u(t − τ ) = 0
∴
u(t + τ ) − u(t − τ ) = 1 − 0 = 1
2.33. One possible relation between (t) and sign (t) is illustrated in Fig. 2.23. On the left side,
it is shown how to create two sign (t) functions: one advanced by τ , Fig. 2.23 (left, top), one
44
2 Special Functions
1
0
−1
sign(t + t )
t
1
0
−1
sign(t − t )
t
1
0
−1 −sign(t − t )
−t
t
1
0
−1
sign(t +t )
t
1
0
−1 −sign(t − t )
t
1
Π (t/2t )
t
0
0
−t
t
0
t
Fig. 2.23 Example P.2.33
delayed by τ , Fig. 2.23 (left, centre), and then inverted, Fig. 2.23 (left, bottom). Then, two functions
sign (t + τ ) and sign (t − τ ) are added and divided by two (because 1 + 1 = 2) to create (t/2τ ), see
Fig. 2.23 (right), as summarized by equation
t
2τ
=
1
1
sign (t + τ ) − sign (t − τ )
2
2
3
Continuous Time Convolution
Convolution integral is defined as
z(t) = x(t) ∗ y(t) =
∞
−∞
x(τ ) y(t − τ ) dτ
(3.1)
where τ is the provisional integration variable. That is to say, at any given instance of time t, the
products of the two functions x(τ ) and y(t − τ ) are integrated relative to the parametric variable
τ . It is important to realize that, being definite integral, the result of convolution integral may
be interpreted as time progression of “area under the curve,” where in this case “the curve” is the
product x(τ )h(t −τ ) functions. Therefore, in the preparation part, it is necessary to perform a series of
function transformations to realize the two required functions, i.e., x(t) → x(τ ) and y(t) → y(t −τ ).
Energy and power: the basic relationship between energy and power within a time interval t is
E(t) =
t
P (τ ) dτ
0
Given a signal x(t), its energy Ex and average power Px may be calculated in multiple ways:
2
Reminder: Note that x(t) = x(t) x ∗ (t)
(a) Energy within a time interval (t1 , t2 )
Ex =
t2
x(t)2 dt
t1
(b) Total energy
Ex =
∞
x(t)2 dt
−∞
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_3
45
46
3 Continuous Time Convolution
(c) Average power within a time interval (t1 , t2 )
Px =
1
t2 − t1
t2
x(t)2 dt
t1
(d) Average of total power
1
Px = lim
τ →∞ τ
τ/2
x(t)2 dt
−τ/2
(e) Average power of a periodic signal within a period T0
Px =
1
T0
T0/2
x(t)2 dt
−T0/2
Problems
3.1
Basic Function Transformations
Given x(t) function in Fig. 3.1, sketch graphs of functions as defined in P.3.1 to P.3.4.
3.1. x(t − 2)
3.2. x(−t)
3.3. x(2t)
3.4. x(0.2t)
3.5. Given x(t) function in Fig. 3.2, deduce and sketch graph of y(t) = x(2t + 3).
3.6. Given x(t) function in Fig. 3.3, show graph of y(t) = x(−t/2 − 4).
Fig. 3.1 Examples P.3.1
to P.3.4
x(t)
1
0
t
−1
Fig. 3.2 Example P.3.5
0
2
x(t)
1
0
t
−1
−2
0
2
3.2 Function Synthesis and Decomposition
47
Fig. 3.3 Example P.3.6
2
x(t)
1
0
t
−1
−1
Fig. 3.4 Example P.3.7
1
0
1
2
x(t)
0
3.2
t
Function Synthesis and Decomposition
Reminder: given arbitrary x(t) function may be decomposed into sum of its even xp (t) and
odd xi (t) functions that are defined as
xp (t) =
x(t) + x(−t)
2
(3.2)
xi (t) =
x(t) − x(−t)
2
(3.3)
Verification:
xp (t) + xi (t) =
x(t) x(t) x(−t) x(−t)
x(t) + x(−t) x(t) − x(−t)
+
=
+
+
−
2
2
2
2
2
2
= x(t)
Given functions in P.3.7 to P.3.11, do the required syntheses/decomposition operations.
3.7. Given x(t) function in Fig. 3.4, decompose x(t) into a sum of one even xp (t) and one odd xi (t)
function so that x(t) = xp (t) + xi (t), and then show their respective graphs.
3.8. Show the graphical representation of the following function:
x(t) = 3u(t) + r(t) − r(t − 1) − 5u(t − 2)
3.9. Derive the analytical form of function x(t) given in Fig. 3.5.
48
3 Continuous Time Convolution
Fig. 3.5 Example P.3.9
x(t)
2
1
0
0
Fig. 3.6 Example P.3.10
2
t
4
x(t)
2
1
0
t
−2
−5
Fig. 3.7 Example P.3.11
−3
0
3
x(t)
2
1
0
t
−2
−5
x(t)
1
t
−1
0
1
3
0
t
−1
0
Fig. 3.8 Example P.3.12
3.10. Derive the analytical form of function x(t) given in Fig. 3.6.
3.11. Derive the analytical form of function x(t) given in Fig. 3.7.
3.3
0
h(t)
1
0
−3
Continuous Time Convolution
Resolve convolutions in P.3.12 to P.3.18.
3.12. Calculate y(t) = x(t) ∗ h(t), where x(t) and h(t) are given in Fig. 3.8.
3.13. Calculate z(t) = x(t) ∗ y(t), where x(t) and y(t) are given in Fig. 3.9.
3
3.4 Energy and Power
49
x(t)
1
y(t)
1
0
t
0
0
t
0
1
3
Fig. 3.9 Example P.3.13
3.14. Calculate z(t) = u(t) ∗ u(t).
3.15. Calculate y(t) = x(t) ∗ h(t), given
x(t) = u(t + 1) − u(t − 1) t
h(t) = u(t + 2) − u(t)
3.16. Calculate x(t) = (t/2) ∗ (t/2).
3.17. Calculate z(t) = x(t) ∗ y(t), where x(t) = t 3 u(t) and y(t) = t 2 u(t).
3.18. Calculate w(t) = x(t) ∗ v(t), given
x(t) =
3.4
1 − t;
0;
0≤t ≤1
otherwise
and,
v(t) =
e−t ;
0;
t ≥0
otherwise
Energy and Power
Given x(t) functions, calculate their respective energies and powers in P.3.19 to P.3.33.
3.19. x(t) = 2(t − 1)
3.20. x(t) = (t)
3.21. x(t) = 1/4 − t 2 (t)
3.22. x(t) = a (a = const.)
3.23. x(t) = u(t)
3.24. x(t) = u(−t)
3.25. x(t) = sin(t)
3.26. x(t) = a ej ω 0 t
3.27. x(t) = 1 + cos(2t)
3.28. x(t) = e−t u(t)
3.29. x(t) = e−t
3.30. x(t) = e−|t|
3.31. x(t) = r(t)
3.32. x(t) = et u(t)
2
3.33. x(t) = (1 − x 3 ) u(−t)
50
3 Continuous Time Convolution
Answers
3.1
Basic Function Transformations
3.1. After adding a constant ±a to its argument, the function does not change its original form while
it translates along the horizontal axis by distance ∓a.
In the case of piecewise linear functions, it is sufficient to identify coordinates of some typical
points, such as endpoints of each line segment, and to recalculate their new positions by replacing
the original argument t with the new argument t ± a. Given x(t) in Fig. 3.1, one can notice changes
(discontinuities) in the function’s shape happening at t = −1, t = 0, and t = 2. Thus, to apply
x(t) → x(to − 2) transformation, where to is the new t coordinate of each calculated point, is to say
that by equalizing the original “(t)” and the new “(to − 2)” arguments, it follows that
t
−1
0
2
t = to − 2
−1 = to − 2
0 = to − 2
2 = to − 2
⇒
to
1
2
4
that is to say, all points are shifted in the positive direction by two units, see Fig. 3.10. Note the left
and right side difference in limits at each point of discontinuity.
3.2. Following same reasoning as in P.3.1, given x(t) in Fig. 3.1, it follows that after applying
x(t) → x(−to ) transformation the new coordinates of x(t) are calculated as
t
−1
0
2
t = −to
−1 = −to
0 = −to
2 = −to
⇒
to
1
0
−2
that is to say, after multiplying function’s argument by “−1” all points are mirrored (“time reversed”)
around the vertical axis, see Fig. 3.11.
3.3. Given x(t) in Fig. 3.1 it follows that after applying x(t) → x(2 to ) transformation (“time
compression”) the new coordinates of x(t) are calculated as
Fig. 3.10 Example P.3.1
1
x(t) x(t − 2)
+2
0
t
0
1
2
4
3.1 Basic Function Transformations
51
Fig. 3.11 Example P.3.2
x(−t) x(t)
1
0
t
−2
0
Fig. 3.12 Example P.3.3
1
x(t) x(2t)
1
0
t
−1/2 0
Fig. 3.13 Example P.3.4
1
x(t) x(0.2t)
1
0
t
0
−5
t
−1
0
2
t = 2to
−1 = 2to
0 = 2to
2 = 2to
⇒
10
to
−1/2
0
1
that is to say, after multiplying function’s argument by number greater than one, all coordinates are
“time compressed” by the same factor, see Fig. 3.12. Note the left and right side limits at each point
of discontinuity.
3.4. Given x(t) in Fig. 3.1 it follows that after applying x(t) → x(0.2 to ) transformation (“time
expansion”) the new coordinates of x(t) are calculated as
t
−1
0
2
t = 0.2to
−1 = 0.2to
0 = 0.2to
2 = 0.2to
⇒
to
−5
0
10
that is to say, after multiplying function’s argument by number inferior to one (or, equivalently,
dividing by a number superior to one), all coordinates are “time expanded” by the same factor, see
Fig. 3.13.
52
3 Continuous Time Convolution
3.5. Subsequent multiple transformations may be resolved by using various methods. Note the
points of discontinuities at t = −1, 0, 1, 2 that are not difficult to follow visually. For example,
Method 1: Given x(t) in Fig. 3.2 it follows that after applying x(t) → x(2 to + 3) transformation all
x(t) points are moved to the new coordinates, that may be calculated directly as
t
−1
0
1
2
t = 2to + 3
−1 = 2to + 3
0 = 2to + 3
1 = 2to + 3
2 = 2to + 3
⇒
to
−2
−3/2
−1
−1/2
that is to say, all coordinates are “time compressed” and “shifted”, see Fig. 3.14.
Method 2: Transformation y(t) = x(2t + 3) consists of one compression (i.e 2t) and one delay (i.e.
+3). If the delay is applied before the compression, then
delay
compression
x(t) −−−−−→ a(t) = x(t + 3) −−−−−−−−→ b(t) = a(2t)
In that case, the two step transformation is as follows
1. Applying delay x(t + 3) results in (see Fig. 3.15)
t
−1
0
1
2
t = ta + 3
−1 = ta + 3
0 = ta + 3
1 = ta + 3
2 = ta + 3
⇒
Fig. 3.14 Example P.3.5
ta
−4
−3
−2
−1
x(2t + 3)
1
x(t)
0
t
−1
−2
Fig. 3.15 Example P.3.5
−1
0
a(t)
1
0
t
−1
−4
−2
0
3.1 Basic Function Transformations
53
2. Applying compression a(2tb ) results in
ta
−4
−3
−2
−1
ta = 2tb
−4 = 2tb
−3 = 2tb
−2 = 2tb
−1 = 2tb
⇒
tb
−2
−3/2
−1
−1/2
which is same result x(2t + 3) as already shown in Fig. 3.14.
Method 3: If compression is applied before delay, then the original delay must be compressed as
well,
compression
delay
3
x(t) −−−−−−−−→ a(t) = x(2t) −−−−−→ b(t) = a(t + 3) = x 2 t +
2
In that case, the two step transformation is as follows,
1. Applying compression x(2t) results in (see Fig. 3.16.)
t
−1
0
1
2
t = 2ta
−1 = 2ta
0 = 2ta
1 = 2ta
2 = 2ta
⇒
ta
−1/2
0
1/2
1
ta = tb + 3/2
1
− /2 = tb + 3/2
0 = tb + 3/2
1/2 = t + 3/2
b
1 = tb + 3/2
⇒
tb
−2
−3/2
−1
−1/2
2. Applying delay a(t + 3/2) results in
ta
1
− /2
0
1/2
1
which is in agreement with the previous two methods, as shown in Fig. 3.14.
Fig. 3.16 Example P.3.5
a(t)
1
0
t
−1
−1
0
1
54
3 Continuous Time Convolution
Fig. 3.17 Example P.3.6
x(−t/2 − 4)
2
1
0
t
−1
−12
−10
−8
−6
3.6. By following same reasoning as in A.3.5, subsequent multiple transformations may be resolved
using various methods. For example,
Method 1: Given x(t) in Fig. 3.3 note points of discontinuity at t = −1, 0, 1, 2, it follows that after
applying x(t) → x(− to /2 − 4) transformation all x(t) points are moved to the new coordinates, that
may be calculated directly as
t = −to /2 − 4
−1 = −to /2 − 4
0 = −to /2 − 4
1 = −to /2 − 4
2 = −to /2 − 4
t
−1
0
1
2
⇒
to
−6
−8
−10
−12
that is to say, all coordinates are time “inverted”, “expanded” and “shifted”, see Fig. 3.17. As an
exercise, this example may be solved using the other two methods in A.3.5.
3.2
Function Synthesis and Decomposition
3.7. Given function in Fig. 3.4 is piecewise linear, its even and odd functions (3.2) and (3.3) may
be deduced by using graphical technique. That is to say, by adding or subtracting all line segments
within each individual interval.
Two summing terms, x(t) and x(−t), that are needed to derive even function xp (t), are shown
Fig. 3.18 (left). Summation of linear segments making these two terms is done interval by interval as
follows,
t
x(t)
0
x(−t)
0
xp (t)
0
−2
0, 0
0
0, 1
1
0 , 1/2
1/2
−1
0, 1
1
1, 1
−t
1/2, 1
(1−t)/2
0
1, 0
t
0, 1
1
1/2, 1/2
(1+t)/2
1
1, 1
1
1, 0
0
1 , 1/2
1/2
2
1, 0
0
0, 0
0
1/2, 0
0
For example, at t = 0 the left/right side summations x(t) + x(−t) (see Fig. 3.18 (left)) are
t= 0 :
x(t) + x(−t)
1+0
1
=
=
2
2
2
3.2 Function Synthesis and Decomposition
55
x(t)
1
0
0
t
x(−t)
1
t
x(−t)
1
0
0
t
x p (t)
1
t
1/2
0
−2
x(t)
1
0
2
xi (t)
0
−1/2
t
t
−2
0
2
Fig. 3.18 Example P.3.7
0+1
1
x(t) + x(−t)
=
=
2
2
2
t= 0 :
The difference (3.3) between x(t) and x(−t) (see Fig. 3.18 (right)) is
t
x(t)
0
x(−t)
0
xi (t)
0
−2
0, 0
0
0, 1
1
0 , − 1/2
−1/2
−1
0, 1
1
1, 1
−t
− 1/2, 0 (1+t)/2
0
1, 0
t
0, 1
1
1/2, − 1/2 (t−1)/2
1
1, 1
1
1, 0
0
0 , 1/2
1/2
2
1, 0
0
0, 0
0
1/2, 0
0
Verification: sum x(t) = xp (t) + xi (t) is analytically confirmed as (see Fig. 3.19)
t
xp (t)
0
xi (t)
0
x(t)
0
−2
0, 1/2
−1
1/2
0, − 1/2 −1/2
0, 0
0
0
(1−t)/2
1/2, 1
− 1/2, 0 (1+t)/2
0, 1
1
1/2, 1/2
(1+t)/2
1/2, − 1/2 (t−1)/2
1, 0
t
1
1 , 1/2
0 , 1/2
1, 1
2
1/2
1/2
1
1/2, 0
1/2, 0
0
1, 0
0
0
3.8. Function x(t) may be synthesized by summation of the basic terms in the form of u(t) and r(t)
functions. The individual terms needed to derive function x(t) are shown Fig. 3.20 (left). Summation
of all four terms is done interval by interval as follows,
56
3 Continuous Time Convolution
Fig. 3.19 Example P.3.7
x p (t)
1
0
−2
0
1/2
2
t
xi (t)
0
−1/2
t
−2
0
2
x(t)
1
0
t
3u(t) + r(t) − r(t − 1) − 5u(t − 2)
4
3
4
3
0
−1
−5
0
−1
t
r(t)
3u(t)
−r(t − 1)
−5u(t − 2)
t
−5
0
1
t
3u(t)
0
r(t)
0
−r(t − 1)
0
−5u(t − 2)
0
x(t)
0
2
0
1
2
Fig. 3.20 Example P.3.8
0
0, 3
3
0, 0
t
0, 0
0
0, 0
0
0, 3
3+t
1
3, 3
3
1, 1
t
0, 0
1−t
0, 0
0
4, 4
4
2
3, 3
3
2, 2
t
− 1, − 1
1−t
0, − 5
−5
4, − 1
−1
and drafted in Fig. 3.20 (right). Note that the left and right side limits are not same at discontinuities.
3.9. There are infinite ways to synthesize function x(t) given in Fig. 3.5, thus any possible analytical
form that corresponds to x(t) is not unique. Nevertheless, arguably simplest analytical form may be
deduced by following changes interval by interval. For example:
1. Interval from minus infinity to minus one: x(t) starts as 2u(t + 2), see Fig. 3.21 (top). Then,
at t = −1 its amplitude is reduced by one, which may be forced by adding −u(t + 1), see
Fig. 3.21 (bottom).
3.2 Function Synthesis and Decomposition
57
Fig. 3.21 Example P.3.9
2
1
0
−1
2u(t + 2)
−u(t + 1)
2
2u(t + 2) − u(t + 1)
1
t
0
−2
0
2u(t + 2) − u(t + 1)
t
−r(t − 1)
2
1
0
−1
Fig. 3.23 Example P.3.9
−1
2
1
0
−1
Fig. 3.22 Example P.3.9
t
2u(t + 2) − u(t + 1) − r(t − 1)
−2
−1
0
1
2
1
0
−1
2u(t + 2) − u(t + 1) − r(t − 1)
t
r(t − 2)
2
x(t)
1
t
0
−2
Fig. 3.24 Example P.3.10
t
1
−1
u(t)
0
1
r(t − 1)
0
2
2
t
u(t) + r(t − 1)
1
t
0
0
1
2
2. The following change of x(t) form happens at t = 1, which may be forced by adding ramp
−r(t − 1), see Fig. 3.22 (top). If not cancelled, downward line segment of ramp of −r(t − 1)
would continue to negative infinity, see Fig. 3.22 (bottom).
3. In order to cancel −r(t − 1) ramp after t = 2 point, it is necessary to add positive ramp with the
same slope, i.e
−r(t − 1) + r(t − 2) = 0, (x ≥ 2)
as is evident in Fig. 3.23 (bottom) so that compete synthesis of x(t) is
x(t) = 2u(t + 2) − u(t + 1) − r(t − 1) + r(t − 2)
3.10. There are infinite ways to synthesize function x(t) given in Fig. 3.6, for example:
58
3 Continuous Time Convolution
2
Fig. 3.25 Example P.3.10
u(t) + r(t − 1)
0
−2
u(t) + r(t − 1) − 2r(t − 2)
2
1
0
t
0
Fig. 3.26 Example P.3.10
1
2
3
2
4
u(t) + r(t − 1) − 2r(t − 2)
r(t − 3)
0
2
1
0
u(t) + r(t − 1) − 2r(t − 2) + r(t − 3)
0
Fig. 3.27 Example P.3.10
t
−2r(t − 2)
1
2
3
2
4
t
t
(3)
u(t − 4)
0
2
1
0
(3) + u(t − 4)
0
1
2
3
4
t
t
1. Interval from minus infinity to minus one: x(t) starts as u(t) function followed by delayed ramp
r(t − 1), see Fig. 3.24 (top). If not cancelled, upward line segment of ramp of r(t − 1) would
continue to negative infinity, see Fig. 3.24 (bottom).
2. The following change of x(t) form is at t = 2, which may be forced by adding delayed ramp whose
slope is two times negative, i.e. −2r(t − 2), see Fig. 3.25 (top). Note that it is necessary to add two
ramp functions −r(t − 2) so that ramp r(t − 1) for x ≤ 2 is replaced by r(t − 1) − 2r(t − 2) =
−r(t − 2), t ≥ 2, see Fig. 3.25 (bottom).
3. If not cancelled, downward line segment of ramp of −r(t − 2) would continue to negative infinity.
In order to cancel it after t = 3 point, it is necessary to add positive ramp with the same slope (see
Fig. 3.26 (top)), i.e
−r(t − 2) + r(t − 3) = 0, (x ≥ 3)
as is evident in Fig. 3.26 (bottom), thereafter x(t) = 1 is constant.
4. As the next summation term, delayed positive step function u(t − 4) is added after t = 4 point to
the form derived in (3), as is evident in Fig. 3.27.
5. Finally, form derived in (4) is added to negative step −u(t − 5) at t = 5 to complete x(t), see
Fig. 3.28.
In conclusion, one possible analytical form is
x(t) = u(t) + r(t − 1) − 2r(t − 2) + r(t − 3) + u(t − 4) − 2u(t − 5)
Note how x(t) is formed in t ∈ [1, 3] interval by adding “r(t − 1) − 2r(t − 2) + r(t − 3)” ramp
functions.
3.3 Continuous Time Convolution
59
2
Fig. 3.28 Example P.3.10
(4)
t
0
−2u(t − 5)
−2
2
1
0
x(t)
0
1
2
3
4
5
t
3.11. Following the same reasoning as in P.3.10, by inspection of graph in Fig. 3.7, it may be deduced
that,
2
2
f (t) = u(t + 5) + u(t + 3) − r(t + 3) + r(t) − 2u(t) + 3u(t − 3)
3
3
It may be verified, for example at x = 0 , i.e. from the left side, as
2
2
f ( 0 ) = u( 0 + 5) + u( 0 + 3) − r( 0 + 3) + r( 0 ) − 2u( 0 ) + 3u( 0 − 3)
3 3 2
2
= 1 + 1 − (3) + (0) − 2 (0) + 3 (0) = 0
3
3
which is correct coordinate of the point ( 0 , 0) in Fig. 3.7. Similarly, from the right side of t = 0 it is
calculated that
2
2
f ( 0 ) = u( 0 + 5) + u( 0 + 3) − r( 0 + 3) + r( 0 ) − 2u( 0 ) + 3u( 0 − 3)
3 3 2
2
= 1 + 1 − (3) + (0) − 2 (1) + 3 (0) = −2
3
3
Note left and right side values of −2u(0) term, as underlined above.
3.3
Continuous Time Convolution
Convolution integral (3.1) consists of the x(τ )h(t − τ ) product. Therefore, in the preparation part,
it is necessary to create these two functions by the change of variables method and then to resolve
their product. First, the change of variable x(t) → x(τ ) does not change the x(t) function’s form.
Then, it is necessary to perform a sequence of h(t) transformations, that is to say, to convert h(t) →
h(τ ) → h(−τ ) → h(t − τ ) function. As the change of variable t into provisional integration variable
τ does not change the original forms of neither x(t) nor h(t) functions, only transformations of
h(t) → h(t − τ ) function are illustrated.
3.12. Graphical method of resolving convolution integral consist the following steps, where the
original integral form is decomposed into several simpler integrals.
60
3 Continuous Time Convolution
Fig. 3.29 Example P.3.12
1
h(t,t )
0
t,t
1
h(−t )
0
−3
−1
0
1
1
3
h(−t + t)
0
−3 + t
Fig. 3.30 Example P.3.12
0
t
1+t
x(t )
h(t − t )
1
t
t −3
t +1
t
−1
0
1
Preparatory work : h(t) → h(τ ) → h(−τ ) → h(t − τ ) transformations.
(a) h(t) → h(τ ) transformation does not change form, only the variable t → τ , Fig. 3.29 (top).
(b) h(τ ) → h(−τ ) time inversion transformation results in the mirrored form, Fig. 3.29 (middle).
Note that τ = 0 reference point is still known.
(c) h(−τ ) → h(t − τ ) transformation shifts each point of h(−τ ) by t, Fig. 3.29 (bottom). Note that
τ = 0 reference point is not known, it is relative to t, thus is resolved case by case.
Case 1: (t + 1) ≤ −1:
The leading edge of h(t − τ ) is inferior to (τ = −1), in other words, there is no overlapping area
between h(t − τ ) and x(τ )
t + 1 ≤ −1 ∴ t ≤ −2
That being the case, it is irrelevant where exactly (t + 1) point is relative to the trailing edge of
x(τ ) at (τ = −1), see Fig. 3.30. What is important is that under (t ≤ −2) condition, due to either
h(t − τ ) or x(τ ) equal zero, their product always equal zero
x(τ ) h(t − τ ) = 0, ∀τ
and by consequence, the convolution integral also equal zero,
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
∞
−∞
0 dτ = 0 (t ≤ −2)
Case 2: −1 < (t + 1) ≤ 1:
The leading edge of h(t − τ ) is found within τ ∈ (−1,
1) interval bound by x(τ ), however the
trailing edge of h(t − τ ) is still outside of this interval note that h(t − τ ) rectangle is longer than
3.3 Continuous Time Convolution
61
x(τ ) , i.e.
t + 1 > −1
∴ t > −2
t +1≤1
∴ t ≤0
⇒
−2<t ≤0
Therefore, within interval τ ∈ (−1, t + 1), see Fig. 3.31, there is the overlapping non–zero area
because that is where h(t − τ ) = 1 and x(τ ) = 1, therefore the convolution integral is
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
t+1
−1
t+1
(1) (1) dτ = τ = t + 1 − (−1)
−1
= t + 2 (−2 ≤ t ≤ 0)
Case 3 : (t + 1) > 1 and (t − 3) ≤ −1:
The leading edge of h(t − τ ) is found at (t + 1 > 1) while its trailing edge is still (t − 3 ≤ −1),
i.e.
t +1>1
∴ t >0
t − 3 ≤ −1
∴ t ≤2
⇒ 0<t ≤2
Note that h(t − τ ) rectangle is longer than x(τ ). Under this condition, see Fig. 3.32, the overlapping
area is found in the interval τ ∈ (−1, +1) where h(t − τ ) = 1 and x(τ ) = 1, in other words it is
maximal possible overlapping rectangular surface whose area equals two,
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
1
−1
1
(1) (1) dτ = τ −1
= 2 (0 ≤ t ≤ 2)
Case 4 : (t − 3) ≥ −1 and (t − 3) ≤ 1:
The trailing edge at (t − 3) of h(t − τ ) entered the τ ∈ (−1, 1) interval bound by x(τ ), which is
to say that the leading edge at (t + 1) of h(t − τ ) must be somewhere at t > 1, see Fig. 3.33, i.e.
t − 3 ≥ −1
∴ t ≤2
t −3<1
∴ t <4
⇒ 2≤t <4
therefore, the overlapping area is found in the interval τ ∈ (t − 3, +1) where both h(t − τ ) = 1 and
x(τ ) = 1, i.e.
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
= 4 − t (2 ≤ t < 4)
1
t−3
1
(1) (1) dτ = τ t−3
62
3 Continuous Time Convolution
Fig. 3.31 Example P.3.12
x(t )
h(t − t )
1
0
t −3
t +1
−1
Fig. 3.32 Example P.3.12
0
Fig. 3.33 Example P.3.12
t +1
x(t )
1
0
h(t − t )
t +1
0
t
1
t −3
−1
1
h(t − t )
t −3
−1
0
0
x(t )
1
t
t
1
Case 5 : (t − 3) ≥ 1:
The trailing edge at (t − 3) of h(t − τ ) passed completely the x(τ ) region, i.e. (t − 3 ≥ 1), see
Fig. 3.34, i.e.
t −3≥1 ∴ t ≥4
That being the case, it is irrelevant where exactly (t − 3) is relative to the leading edge of x(τ ) at
τ = 1, see Fig. 3.34. What is important is that under this condition, due to either h(t − τ ) or x(τ )
equals zero, their product is therefore zero
x(τ ) h(t − τ ) = 0, ∀τ
∞
∞
x(τ ) h(t − τ ) dτ =
0 dτ = 0 (t ≥ 4)
y(t) =
−∞
−∞
∴
3.3 Continuous Time Convolution
63
Fig. 3.34 Example P.3.12
x(t )
1
0
h(t − t )
t −3
−1
0
t
1
Fig. 3.35 Example P.3.12
y(t)
2
0
t +1
(1)
(2)
−2
(3)
0
(4)
2
(5)
4
t
Summary: The total convolution integral y(t) shows the progression of the overlapping surface area
between x(τ ) and h(t − τ ) in function of t, see Fig. 3.35, where within each interval the solution
function is different, as resolved in cases (1) to (5). Note that the use of ‘≤’ and ‘<’ symbols to
bound the intervals is not strict because y(t) is continuous, thus the interval bound points may be
assigned either on the left or right side of the inequalities, as
⎧
⎪
0
⎪
⎪
⎪
⎪
⎪
⎨2 + t
y(t) = 2
⎪
⎪
⎪
4−t
⎪
⎪
⎪
⎩
0
t ≤ −2
−2 ≤ t ≤ 0
0≤t ≤2
2≤t ≤4
t ≥4
3.13. Graphical method of resolving convolution integral consist the following steps, where the
original integral form is decomposed into several simpler integrals.
Preparatory work : y(t) → y(τ ) → y(−τ ) → y(t − τ ) transformations.
(a) y(t) → y(τ ) transformation does not change form, only the variable t → τ , Fig. 3.36 (top).
(b) y(τ ) → y(−τ ) time inversion transformation results in the mirrored form, Fig. 3.36 (middle).
Note that τ = 0 reference point is still known.
(c) y(−τ ) → y(t − τ ) transformation shifts each point of y(−τ ) by t, Fig. 3.36 (bottom). Note that
τ = 0 reference point is not known, it is relative to t, thus is resolved case by case.
Case 1 : (t − 1) ≤ 0:
The leading edge (t − 1) of y(t − τ ) is inferior to τ = 0 i.e.
t −1≤0 ∴ t ≤1
64
3 Continuous Time Convolution
Fig. 3.36 Example P.3.13
1
y(t, t )
0
t, t
1
y(−t )
0
−3
−1
0
1
−3 + t
−1 + t
0
t
t −3
t
x(t )
y(t − t )
1
3
y(−t + t)
0
Fig. 3.37 Example P.3.13
1
t −1
t
0
That being the case, it is irrelevant where exactly t is relative to the origin, see Fig. 3.37. What
is important is that under this condition, because always either of the two function equals zero, their
product equal zero, as
x(τ ) y(t − τ ) = 0, ∀τ
and by consequence the convolution integral equal zero, as
z(t) =
∞
−∞
x(τ ) y(t − τ ) dτ =
∞
0 dτ
−∞
= 0 (t ≤ 1)
Case 2 : (t − 1) > 0 and (t − 3) ≤ 0:
The leading edge at (t − 1) of y(t − τ ) passed into the positive side, however the trailing edge at
(t − 3) of y(t − τ ) is still in the negative side, i.e.
t −1>0
∴ t >1
t −3≤0
∴ t ≤3
⇒ 1<t ≤3
Under this condition, the overlapping area is found in the interval τ ∈ (0, t − 1), see Fig. 3.38, where
both y(t − τ ) = 1 and x(τ ) = 1, therefore the convolution integral (i.e. the overlapping surface area)
equal
z(t) =
∞
−∞
x(τ ) y(t − τ ) dτ =
= t − 1 (1 ≤ t ≤ 3)
t−1
0
t−1
(1) (1) dτ = τ 0
3.3 Continuous Time Convolution
65
Fig. 3.38 Example P.3.13
x(t )
1
0
y(t − t )
t −3
t −1
t
0
Fig. 3.39 Example P.3.13
x(t )
1
y(t − t )
0
t −3
t −1
t
0
Fig. 3.40 Example P.3.13
z(t)
2
(1)
0
0
(2)
1
(3)
3
t
Case 3 : (t − 3) > 0:
The trailing edge (t − 3) of y(t − τ ) shifted into the positive side, see Fig. 3.39, i.e.
t −3>0 ∴ t >3
Therefore, the overlapping surface area is at its maximum (i.e. the full area of rectangular function,
and henceforth stays constant)
z(t) =
∞
−∞
x(τ ) y(t − τ ) dτ =
t−1
t−3
t−1
(1) (1) dτ = τ = (t − 1) − (t − 3)
t−3
= 2 (t > 3)
Summary:
The total convolution integral shows progression of the overlapping area in function of t as
resolved in the cases (1) to (3), Fig. 3.40, as
⎧
⎪
⎨0
z(t) = t − 1
⎪
⎩
2
t ≤1
1≤t ≤3
t >3
66
3 Continuous Time Convolution
Fig. 3.41 Example P.3.14
1
u(t, t )
0
t,t
1
u(−t )
0
t
0
1
u(−t + t)
0
t
t
3.14. Graphical method of resolving convolution integral consist the following steps, where the
original integral form is decomposed into several simpler integrals.
Preparatory work : u(t) → u(τ ) → u(−τ ) → u(t − τ ) transformations.
(a) u(t) → u(τ ) transformation does not change form, only the variable t → τ , Fig. 3.41 (top).
(b) u(τ ) → u(−τ ) time inversion transformation results in the mirrored form, Fig. 3.41 (middle).
Note that τ = 0 reference point is still known.
(c) u(−τ ) → u(t − τ ) transformation shifts each point of u(−τ ) by t, Fig. 3.41 (bottom). Note that
τ = 0 reference point is not known, it is relative to t, thus is resolved case by case.
There are two distinct cases to solve.
Case 1 : t ≤ 0:
The leading edge of u(t − τ ) is still in the negative side, i.e.
t ≤0
It is irrelevant where exactly t is relative to the origin, see Fig. 3.42. What is important is that under
this condition, because always either of the two function equals zero, their product equal zero as
u(τ ) u(t − τ ) = 0, ∀τ
by consequence, convolution integral equal zero as,
z(t) =
∞
−∞
u(τ ) u(t − τ ) dτ =
∞
0 dτ
−∞
= 0 (t ≤ 0)
Case 2 : t > 0:
The leading edge of u(t − τ ) is shifted into the positive side, Fig. 3.43, i.e.
t >0
3.3 Continuous Time Convolution
67
Fig. 3.42 Example P.3.14
1
0
u(t )
u(t − t )
t
t
0
Fig. 3.43 Example P.3.14
u(t )
u(t − t )
1
0
t
t
0
Fig. 3.44 Example P.3.14
z(t) = r(t)
1
0
(1)
(2)
t
0
1
thus,
z(t) =
∞
−∞
u(τ ) u(t − τ ) dτ =
t
0
t
(1) (1) dτ = τ 0
= t (t > 0)
(Rectangular area is calculated as height (equal one) times its length (equal t).)
Summary:
The total convolution integral is therefore r(t) function, see Fig. 3.44, or written in equally elegant
product form of linear and step functions as
z(t) =
0
t ≤0
t
t >0
= r(t) or, z(t) = t u(t)
3.15. Graphical method of resolving convolution integral consist the following steps, where the
original integral form is decomposed into several simpler integrals.
68
3 Continuous Time Convolution
1
0
−1
1
0
−1
1
u(t + 2)
−u(t)
u(t + 2) − u(t)
−2
t
t
0
h(t − t )
t
0
t +2
t
1 u(t + 1)
0
−u(t − 1)
−1
1
0
−1
t
t
u(t + 1) − u(t − 1)
t
1 (u(t + 1) − u(t − 1))t
0
−1
−1
0
t
1
Fig. 3.45 Example P.3.15
Fig. 3.46 Example P.3.15
h(t − t )
1
0
x(t )
t +2
t
t
−1
−1
0
1
Preparatory work :
(a) the syntheses of h(t) = u(t + 2) − u(t) and h(t − τ ) is illustrated in Fig. 3.45 (left).
(b) the syntheses of x(t) = u(t + 1) − u(t − 1) t is illustrated in Fig. 3.45 (right)
There are four distinct cases to solve.
Case 1 : t + 2 ≤ −1:
The leading edge of h(t − τ ) is inferior to τ = −1, that is to say
t + 2 ≤ −1 ∴ t ≤ −3
That being the case, it is irrelevant where exactly t is relative to τ = −1, see Fig. 3.46. What is
important is that under this condition, because either of the two function equals zero, their product
equal zero as
x(τ ) h(t − τ ) = 0, ∀τ
and by consequence the convolution integral equal zero as
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
= 0 (t ≤ −3)
∞
0 dτ
−∞
3.3 Continuous Time Convolution
69
Fig. 3.47 Example P.3.15
x(t )
h(t − t )
1
0
−1
t
t +2
t
−1
0
1
Case 2 : t + 2 > −1 and t + 2 ≤ 1
The leading edge of h(t − τ ) is found within the interval τ ∈ (−1, 1), that is to say
t + 2 > −1
∴ t > −3
t +2≤1
∴ t ≤ −1
⇒
− 3 < t ≤ −1
Under this condition, see Fig. 3.47, the overlapping surface is found in the interval τ ∈ (−1, t + 2)
where h(t − τ ) = 1 and x(τ ) = t, consequently
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
t+2
−1
1 2 t+2
(τ ) (1) dτ = τ 2
−1
12
1
(t + 2)2 − (−1)2 =
t + 4t + 3
2
2
1
= (t + 1)(t + 3), (−3 < t ≤ −1)
2
=
Note that both h(t − τ ) and x(τ ) non–zero intervals are equal, both lengths equal two.
Case 3 : t > −1 and t ≤ 1
The trailing edge of h(t − τ ) is found within the interval τ ∈ (−1, 1), that is to say
−1<t ≤1
Under this condition, see Fig. 3.48, the overlapping surface is found in the interval τ ∈ (t, 1) where
h(t − τ ) = 1 and x(τ ) = t, consequently
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
1
t
1 2 1
(τ ) (1) dτ = τ 2
t
1
= (1 − t 2 ) = (a 2 − b2 ) = (a − b)(a + b)
2
1
= − (t − 1)(t + 1), (−1 < t ≤ 1)
2
70
3 Continuous Time Convolution
Fig. 3.48 Example P.3.15
x(t )
1
h(t − t )
0
t
−1
t +2
t
−1
Fig. 3.49 Example P.3.15
0
1
h(t − t )
x(t )
1
0
t
−1
t
−1
0
t +2
1
Case 4 : t > 1
The trailing edge of h(t − τ ) is superior to τ = 1, that is to say
t >1
That being the case, it is irrelevant where exactly t is relative to τ = 1, see Fig. 3.49. What is important
is that under this condition either of the two function equal zero, therefore their product equal zero as,
x(τ ) h(t − τ ) = 0, ∀τ
and by consequence the convolution integral equal to zero as
y(t) =
∞
−∞
x(τ ) h(t − τ ) dτ =
∞
0 dτ
−∞
= 0 (t > 1)
Summary:
The total convolution integral is therefore y(t) function, see Fig. 3.50, as
y(t) =
⎧
⎪
0
⎪
⎪
⎪
⎨ 1 (t + 1)(t + 3)
t < −3
⎪
− 21 (t − 1)(t + 1)
⎪
⎪
⎪
⎩0
(−1 ≤ t ≤ 1)
t >1
2
(−3 ≤ t ≤ −1)
3.3 Continuous Time Convolution
71
Fig. 3.50 Example P.3.15
1/2
0
y(t)
(1)
(3)
(2)
(4)
t
−1/2
−3
−1
0
1
Fig. 3.51 Example P.3.16
1
x(t,
x(t t , −t )
1 + tτ
1 + tτ
1 − tτ
1 − tτ
0
tt
t,
−1
1
0
1
x(t − t )
1 + tτ − t
1 − tτ + t
0
t
t −1
t
t +1
3.16. Graphical method of resolving convolution integral consist the following steps, where the
original integral form is decomposed into several simpler integrals.
Preparatory work : x(t) → x(τ ) → x(−τ ) → x(t − τ ) transformations.
(a) x(t) → x(τ ) transformation does not change form, only the variable t → τ . However, triangular
function is even and by consequence x(τ ) → x(−τ ) time inversion transformation also does not
change its form.
Even after x(−τ ) → x(t − τ ) transformation, visually, triangular form does not change, see
Fig. 3.51.
Although changes are not evident visually, note how time inversion alters linear equations on
(τ ≥ 0) and (τ ≤ 0) sides
x(τ ) = 1 + τ ⇒
x(−τ ) = 1 − τ
x(τ ) = 1 − τ ⇒
x(−τ ) = 1 − (−τ ) = 1 + τ
(b) It is important, however, to explicitly write linear equations that correspond to each linear segment
after x(−τ ) → x(t − τ ) transformation as:
⎫
⎧
1 + τ, −1 ≥ τ ≥ 0⎪
⎪
⎪ x(t − τ ) =
⎨
⎬
x(−τ ) = 1 − τ, 0 ≤ τ ≤ 1
∴
x(t − τ ) =
⎪
⎪
⎩
⎪
⎭
x(t − τ ) =
x(−τ ) = 0, |τ | ≥ 1
x(−τ ) =
1 + τ − t, t − 1 ≤ τ ≤ t
1 − τ + t, t ≤ τ ≤ t + 1
0, otherwise
72
3 Continuous Time Convolution
Fig. 3.52 Example P.3.16
0
x(t )
x(t − t )
1
t −1
t +1
−1
t
0
1
Reminder: Note that for τ ≥ 0 shifting by ‘+t’ means adding ‘+t’ so that (1−τ ) → (1−τ +t),
while for τ ≤ 0 shifting by ‘+t’ means adding ‘−t’ so that (1 + τ ) → (1 + τ − t).
There are six distinct cases to solve.
Case 1 : t + 1 < −1
The leading (t + 1) point of x(t − τ ) is inferior to τ = −1, i.e.
t + 1 < −1 ∴
t < −2
That being the case, it is irrelevant where exactly t is relative to the origin, see Fig. 3.52. What is
important is that under this condition at least one of the two functions equal zero, therefore their
product equal zero as
x(τ ) x(t − τ ) = 0, ∀τ
and by consequence, convolution integral equal zero,
z(t) =
∞
−∞
x(τ ) x(t − τ ) dτ =
∞
0 dτ
−∞
= 0 (t < −2)
Case 2 : t + 1 ≥ −1, t + 1 ≤ 0
As long as the leading (t + 1) point of x(t − τ ) is in τ ∈ (−1, 0) interval, i.e.
t + 1 ≥ −1
∴ t ≥ −2 and
t +1≤0
∴ t < −1
∴ − 2 ≤ t < −1
the overlapping surface is bound by ( 1 − τ + t) and ( 1 + τ ) linear segments, and (−1, t +1) interval,
see Fig. 3.53. Thus, product of these two linear segments defines convolution integral as:
3.3 Continuous Time Convolution
73
Fig. 3.53 Example P.3.16
0
t −1
t +1
−1
z(t) =
=
∞
−∞
t+1
−1
x(τ ) x(t − τ ) dτ =
t+1
−1
x(t )
x(t − t )
1
t
0
1
(1 + τ ) (1 − τ + t) dτ
(1 − τ + t + τ + tτ − τ 2 ) dτ
t+1
t+1
t+1
t
1
= (1 + t) τ + τ 2 − τ 3 2
3
−1
−1
−1
t
1
= (1 + t) (t + 1) − (−1) + (t + 1)2 − (−1)2 − (t + 1)3 − (−1)3
2
3
= use Pascal triangle for binomial power development
=
(t + 2)3
6
after factorizing third order polynomial.
Case 3 : t + 1 ≥ 0, t + 1 ≤ 1
The leading (t + 1) point of x(t − τ ) is found in τ ∈ (0, 1) interval, i.e.
t + 1 > 0 ∴ t > −1 and
t +1≤1 ∴ t <
0
∴ −1<t ≤0
where, overlapping surface is found in τ ∈ (−1, t +1) interval, Fig. 3.54. However, this time, actually
there are three distinct sub–intervals (a), (b), (c), as
(a) :(−1, t), (1 + τ ) (1 + τ − t)
(b) :(t, 0), (1 + τ ) (1 − τ + t)
(c) :(0, t + 1), (1 − τ ) (1 − τ + t)
where the convolution products found in each of sub–intervals are different, see Fig. 3.54, and the
convolution integral is split into the sum of three integrals so that the overlapping areas are correctly
calculated as
74
3 Continuous Time Convolution
Fig. 3.54 Example P.3.16
1
0
(a)
(b)
x(t − t )
x(t )
t −1
Fig. 3.55 Example P.3.16
z(t) =
+
−∞
x(τ ) x(t − τ ) dτ =
0
−1
(b)
t
1
(c)
x(t )
x(t − t )
t −1
−1
t
0
(a)
0
∞
t +1
t
−1
1
(c)
t +1 t
t
0
1
(1 + τ ) (1 + τ − t) dτ
(1 + τ ) (1 − τ + t) dτ
t
+
t+1
(1 − τ ) (1 − τ + t) dτ
0
= ··· =
1
(−t 3 + 3t + 2)
6
1
t (t 2 + 6t + 6)
6
1
+ (−t 3 + 3t + 2)
6
1
1
= (−t 3 + 3t + 2) − t (t 2 + 6t + 6), (−1 < t ≤ 0)
3
6
−
Case 4 : t − 1 ≥ −1, t − 1 ≤ 0
As long as the trailing (t − 1) point of x(t − τ ) is in τ ∈ (−1, 0) interval, i.e.
t − 1 ≥ −1
∴ t >0
t −1<0
∴ t <1
∴ 0<t ≤1
where, overlapping surface is found in τ ∈ (t − 1, 1) interval, Fig. 3.55. There are three distinct
sub–intervals (a), (b), (c),
3.3 Continuous Time Convolution
75
(a) :(t − 1, 0), (1 + τ ) (1 + τ − t)
(b) :(0, t), (1 − τ ) (1 + τ − t)
(c) :(t, 1), (1 − τ ) (1 − τ + t)
and the convolution integral is split into the sum of three integrals so that the overlapping areas are
correctly calculated as
z(t) =
∞
−∞
+
t
x(τ ) x(t − τ ) dτ =
0
(1 + τ ) (1 + τ − t) dτ
t−1
(1 − τ ) (1 + τ − t) dτ
0
+
1
(1 − τ ) (1 − τ + t) dτ
t
= after polynomial development, integration, and factorization
=
1
1 3
(t − 3t + 2) + t (t 2 − 6t + 6), (0 < t ≤ 1)
3
6
Case 5 : t − 1 ≥ 0, t − 1 ≤ 1
As long as trailing point (t − 1) of x(t − τ ) is in τ ∈ (0, 1), i.e.
t −1≥0
∴ t >1
t −1<1
∴ t <2
∴ 1≤t ≤2
the overlapping surface is bound by (1 − τ ) and (1 + τ − t) linear segments, and (t − 1, 1) interval,
see Fig. 3.56. Thus, product of these two linear segments defines convolution integral as:
z(t) =
1
(1 − τ ) (1 + τ − t) dτ
t−1
= similar to the case 2
1
= − (t − 2)3
6
Fig. 3.56 Example P.3.16
x(t )
1
x(t − t )
0
t −1
−1
0
t +1
1
t
76
3 Continuous Time Convolution
Fig. 3.57 Example P.3.16
x(t )
1
x(t − t )
0
t −1
−1
0
t
t +1
1
Case 6 : t − 1 > 1
The trailing point (t − 1) of x(t − τ ) is superior to τ = 1, i.e.
t −1>1 ∴ t >2
It is irrelevant how much exactly t is superior to τ = 1, see Fig. 3.57. What is important is that under
this condition at least one of the two functions equal zero, therefore their product equal zero as
x(τ ) x(t − τ ) = 0, ∀τ
and by consequence, convolution integral equal zero as
z(t) =
∞
−∞
x(τ ) x(t − τ ) dτ =
∞
0 dτ
−∞
= 0 (t ≥ 2)
Summary:
The total convolution integral is therefore the sum of these six cases, see also Fig. 3.58.
z(t) =
⎧
⎪
0
⎪
⎪
⎪
⎪
1/6(t + 2)3
⎪
⎪
⎪
⎪
⎨ 1/3(−t 3 + 3t + 2) − 1 t (t 2 + 6t + 6)
6
⎪ 1/3(t 3 − 3t + 2) + 1 t (t 2 − 6t + 6)
⎪
6
⎪
⎪
⎪
3
⎪
⎪ −1/6(t − 2)
⎪
⎪
⎩0
t < −2
−2 ≤ t ≤ −1
−1 ≤ t ≤ 0
0≤t ≤1
1≤t ≤2
t >2
3.17. Graphical method of resolving convolution integral consist the following steps, where the
original integral form is decomposed into several simpler integrals.
Preparatory work : y(t) → y(τ ) → y(−τ ) → y(t − τ ) transformations.
(a) y(t) → y(τ ) transformation does not change form, only the variable t → τ , Fig. 3.59 (top).
(b) y(τ ) → y(−τ ) time inversion transformation results in the mirrored form, Fig. 3.59 (middle).
Note that τ = 0 reference point is still known.
3.3 Continuous Time Convolution
77
Fig. 3.58 Example P.3.16
z(t)
1/2
0
t
(1)
(2)
(3)
−2
(4)
(5)
0
(6)
2
Fig. 3.59 Example P.3.17
y(t, t )
t, t
0
y(−t )
0
t
0
y(t − t )
t
0
t
(c) y(−τ ) → y(t − τ ) transformation shifts each point of y(−τ ) by t, Fig. 3.59 (bottom). Note that
τ = 0 reference point is not known, it is relative to t, thus is resolved case by case.
Note that,
y(t) = t 2 u(t)
∴
y(t − τ ) = (t − τ )2 u(t − τ ) = (t 2 − 2t τ + τ 2 ) u(t − τ )
There are two distinct cases to solve.
Case 1 : t ≤ 0:
The leading t point of y(t − τ ) is inferior to zero, i.e.
t ≤0
It is irrelevant where exactly is t < 0 point, see Fig. 3.60. What is important is that under this
condition, either x(τ ) or y(t − τ ) function equal zero, as
x(τ ) y(t − τ ) = 0, ∀τ
and by consequence convolution integral equal zero, as
78
3 Continuous Time Convolution
Fig. 3.60 Example P.3.17
y(t − t )
x(t )
t
0
t
0
Fig. 3.61 Example P.3.17
y(t )
y(t − t )
0
t
0
z(t) =
=
∞
−∞
∞
−∞
x(τ ) y(t − τ ) dτ
0 dτ = 0
Case 2 : t > 0:
The leading t point of y(t − τ ) is superiour to zero, i.e.
t >1
It is irrelevant where exactly is t > 0 point, see Fig. 3.61, as
z(t) =
=
∞
−∞
t
0
x(τ ) y(t − τ ) dτ =
τ 3 (t − τ )2 dτ =
t
∞
−∞
τ 3 u(τ ) (t − τ )2 u(t − τ ) dτ
(τ 5 − 2tτ 4 + t 2 τ 3 ) dτ
0
=
t
1 6 t
2t 6
t6
2t t
t2 t6
τ − τ 5 + τ 4 =
−
+
6
5
4
6
5
4
0
0
0
=
t6
60
Summary:
The total integral is therefore the sum of these two cases, see Fig. 3.62,
t
3.3 Continuous Time Convolution
79
Fig. 3.62 Example P.3.17
z(t)
1
t
0
0
x(t)
1
t
0
v(t)
1
Π (t )
0
1
u(t)
0
1
t
0
Fig. 3.63 Example P.3.18
Fig. 3.64 Example P.3.18
v(t,t )
t, t
0
v(−t )
t
0
0
v(t − t )
t
0
t
⎧
⎨0
z(t) = t 6
⎩
60
t ≤0
t >0
=
t6
u(t)
60
3.18. Given analytical forms of two functions
x(t) =
1 − t;
0≤t ≤1
0;
otherwise
and,
v(t) =
e−t ;
t ≥0
0;
otherwise
where there are well defined boundaries, these two functions x(t) and v(t) may also be defined by
products with step and port functions, see Fig. 3.63.
x(t) = (1 − t) (t) and v(t) = e−t u(t)
Graphical method of resolving convolution integral consist the following steps, where the original
integral form is decomposed into several simpler integrals.
Preparatory work : v(t) → v(τ ) → v(−τ ) → v(t − τ ) transformations.
(a) v(t) → v(τ ) transformation does not change form, only the variable t → τ , Fig. 3.64 (top).
80
3 Continuous Time Convolution
v(τ ) → v(−τ ) time inversion transformation results in the mirrored form, Fig. 3.64 (top). Note
that τ = 0 reference point is still known.
(b) v(−τ ) → v(t − τ ) transformation shifts each point of v(−τ ) by t, Fig. 3.64 (bottom). Note that
τ = 0 reference point is not known, it is relative to t, thus is resolved case by case.
There are three distinct cases to be solved.
Case 1 : t < 0
Within this interval, either one of the two function equal zero, see Fig. 3.65, thus their product
equal zero, as
x(τ ) v(t − τ ) = 0, ∀τ
and by consequence convolution integral equal zero, as
w(t) =
∞
−∞
x(τ ) v(t − τ ) dτ =
t−1
(0) (0) dτ
0
= 0 (t < 0)
Case 2 : 0 ≤ t ≤ 1 :
The leading edge point t of v(t − τ ) is found in interval τ ∈ (0, 1), i.e.
t
≥0
t
≤1
∴ 0≤t ≤1
where the overlapping surface is bound by τ ∈ (0, t) inteval, see Fig. 3.66, as
w(t) =
∞
−∞
= e−t
x(τ ) v(t − τ ) dτ =
t
0
eτ dτ −
t
t
0
(1 − τ ) e
τ eτ dτ = e
−(t−τ )
dτ =
t
−t τ
e e − τ e−t eτ dτ
0
−t
I1 − I2
0
Where,
Fig. 3.65 Example P.3.18
1
0
x(t )
v(t − t )
t
t
0
1
3.3 Continuous Time Convolution
81
Fig. 3.66 Example P.3.18
x(t )
v(t − t )
t
0
t
0
Fig. 3.67 Example P.3.18
x(t )
1
v(t − t )
t
0
0
I1 =
t
0
1
1
t
t
eτ dτ = eτ = et − 1
0
and,
I2 =
t
τ eτ dτ = partial integration = t et − et + 1
0
Therefore,
w(t) = e−t et − 1 − t et + et − 1
= 2 − t − 2 e−t (0 ≤ t ≤ 1)
Case 3 : t > 1
The leading edge of v(t − τ ) is in (t > 1) interval, thus the overlapping surface is bound in
τ ∈ (0, 1) interval, see Fig. 3.67, as
w(t) =
∞
=e
−∞
−t
x(τ ) v(t − τ ) dτ =
1
1
(1 − τ ) e−(t−τ ) dτ
0
(eτ − τ eτ ) dτ = etc. as in Case 2.
0
−t
= e (e − 2) (t > 1)
Summary:
The total convolution integral is therefore, see Fig. 3.68,
82
3 Continuous Time Convolution
Fig. 3.68 Example P.3.18
w(t)
(1)
0
(2)
(3)
t
0
1
2
Fig. 3.69 Example P.3.19
2Π(t − 1)
t
0
0
1
Λ (t)
1
t
0
−1/2
0
1/4
(1/4 − t 2 )Π (t)
t
0
−1/2
⎧
⎪
⎨ 0;
w(t) = 2 − t − 2 e−t ;
⎪
⎩ −t
e (e − 2);
3.4
1/2
0
1/2
t <0
0≤ t ≤1
t >1
Energy and Power
3.19. Amplitude of a delayed port signal x(t) = 2(t − 1) equal two for t ∈ (0, 1), otherwise it
equal zero. Thus,
Ex =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
|2(t − 1)|2 dt =
1
(2)2 dt = 4
0
Therefore, given a finite source of energy,
Px = lim
def
t→∞
Ex
4
= lim = 0
t→∞
t
t
That is to say, having a finite source of energy that must be distributed over infinitely long period of
time, the average power reduces to zero.
Summary : this signal x(t) belongs to the category of signals that are contained within a finite time
interval (t1 , t2 ) and whose amplitude is finite, such as three forms in Fig. 3.69. By consequence,
Ex < ∞ and Px = 0 .
3.4 Energy and Power
83
3.20. Triangular signal x(t) = (t) is bound in t ∈ (−1/2, 1/2) interval, thus
Ex =
def
∞
−∞
|x(t)| dt =
2
∞
−∞
|(t)| dt =
0
2
−1/2
(t + 1/2)2 dt +
1/2
(−t + 1/2)2 dt
0
=
1
1
1
1 3 0
1 0
1 0
1 /2 1 /2 1 /2
+ t 2 + t + t 3 − t 2 + t t 3
2
4 −1/2 3
2
4 0
0
0
−1/2
−1/2
=
1
1
1
+
=
24 24
12
Therefore, given a finite source of energy,
1/12
Ex
= lim
=0
t→∞ t
t→∞ t
Px = lim
def
That is to say, having a finite source of energy that must be distributed over infinitely long period of
time, the average power reduces to zero.
Summary : this signal x(t) belongs to the category of signals that are contained within a finite time
interval (t1 , t2 ) and whose amplitude is finite, see Fig. 3.69. By consequence, Ex < ∞ and Px = 0 .
3.21. Non–zero part of quadratic signal x(t) = 1/4 − t 2 (t) is bound in t ∈ (−1/2, 1/2) interval,
and its magnitude is finite. Thus, non–zero integrals are
Ex =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
| 1/4 − t 2 (t)|2 dt =
1/2
−1/2
| 1/4 − t 2 (1)|2 dt
1
1
1/2
1 5 /2
1 3 /2
1
1
= x − x +
=
x 5
6
16
30
−1/2
−1/2
−1/2
Therefore, given a finite source of energy,
Px = lim
def
t→∞
1/30
Ex
= lim
=0
t→∞
t
t
That is to say, having a finite source of energy that must be distributed over infinitely long period of
time, the average power reduces to zero.
Summary : this signal x(t) belongs to the category of signals that are contained within a finite time
interval (t1 , t2 ) and whose amplitude is finite, see Fig. 3.69. By consequence, Ex < ∞ and Px = 0 .
3.22. Given a continuous non–periodic infinitely long signal, its average power is calculated as
1 τ/2 2
1 τ/2
|a| dt = a 2 lim
dt
τ →∞ τ −τ/2
τ →∞ τ −τ/2
−τ/2
−τ
1 τ/2
1 τ
τ 1
2
2
t
−
= a 2 lim = a 2
= a lim
= a lim
τ →∞ τ
τ →∞ τ
τ
→∞
2
2
τ
−τ/2
1
τ →∞ τ
P (t) = lim
def
τ/2
|x(t)|2 dt = lim
84
3 Continuous Time Convolution
And,
E(t) =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
|a|2 dt = a 2
∞
= a t = a 2 (∞ + ∞)
∞
dt
−∞
2
−∞
=∞
That is to say, in order to sustain the non-zero finite average power over infinitely long period of time,
evidently, it is necessary to have infinite source of energy.
Summary : this signal x(t) belongs to the category of signals that are not contained within a
finite time interval and whose amplitude is constant, such as forms in Fig. 3.70. By consequence,
Ex = ∞ and 0 < Px < ∞ .
3.23. Given an infinitely long u(t) signal, its energy is
Ex =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
|u(t)|2 dt =
∞
0
∞
(1)2 dt = t = ∞
0
And its average power is
1 τ/2
|x(t)| dt = lim
|u(t)|2 dt
τ →∞ τ
−τ/2
−τ/2
1 0
1 τ/2 2
1 τ/2
1 τ
|0|2 dt + lim
|1| dt = lim
t = lim
= lim
τ →∞ τ
τ →∞ τ
τ →∞ τ
τ →∞ τ 2
−τ/2
0
0
1
Px = lim
τ →∞ τ
def
=
τ/2
2
1
2
which makes sense because half of the time u(t) signal amplitude equals zero (for t < 0) and the
other half of time (i..e. t ≥ 0) it equal one, thus average of ‘0’ and ‘1’ is one half.
Fig. 3.70 Example P.3.22
a
t
0
0
1
0
u(t)
0
1
0
t
u(−t)
0
t
3.4 Energy and Power
85
That is to say, in order to sustain the non-zero finite average power over infinitely long period of
time, evidently, it is necessary to have infinite source of energy.
Summary : this signal x(t) belongs to the category of signals that are not bound within a finite time
interval and whose amplitude is constant, see Fig. 3.70. By consequence, Ex = ∞ and 0<Px < ∞ .
3.24. Given an infinitely long u(−t) signal, its energy is
Ex =
def
∞
−∞
|x(t)| dt =
2
∞
−∞
|u(−t)| dt =
0
2
0
(1) dt = t 2
−∞
−∞
=∞
And its average power is
1 τ/2
|u(−t)|2 dt
τ →∞ τ
−τ/2
−τ/2
1 0
1 τ/2 2
1 0
1 τ
2
= lim
|1| dt + lim
|0| dt = lim
= lim
t
τ →∞ τ
τ
→∞
τ
→∞
τ
→∞
τ
τ
τ
−τ/2
0
−τ/2
2
1
τ →∞ τ
Px = lim
def
=
τ/2
|x(t)|2 dt = lim
1
2
which makes sense because half of the time u(t) signal amplitude equals zero (for t < 0) and the
other half of time (i..e. t ≥ 0) it equal one, thus average of ‘0’ and ‘1’ is one half.
That is to say, in order to sustain the non–zero finite average power over infinitely long period of
time, evidently, it is necessary to have infinite source of energy.
Summary : this signal x(t) belongs to the category of signals that are not bound within a finite time
interval and whose amplitude is constant, see Fig. 3.70. By consequence, Ex = ∞ and 0 < Px <∞ .
3.25. Given a continuous periodic infinitely long signal its average power over one period T is
T /2
T /2
1 − cos 2x
sin (x) =
2
−T /2
−T /2
T /2
T /2 1 − cos 2x
1
1 1
1
=
dt =
t − sin(2t)
T −T /2
2
T 2
4
−T /2
T
1
1 1 T
T
T
+
−
sin 2
=
− sin −2
T 2 2
2
4
2
2
1
1 1
0
0
:
:
sin
sin
T −
=
(T
) +
(T
)
@
4
T 2
@
Px =
1
T
=
1
2
def
| sin(t)|2 dt =
1
T
sin2 (t) dt =
2
86
3 Continuous Time Convolution
That is to say, average power within a singe period is non–zero. By consequence, in order to sustain
the same non–zero average power over infinitely many periods there must be an infinite source of
energy, as
Ex = lim Px t =
def
t→∞
1
lim t = ∞
2 t→∞
Summary : this signal x(t) belongs to the category of periodic signals, i.e. that are not bound within
a finite time interval and whose amplitude is finite. By consequence, Ex = ∞ and 0 < Px < ∞ .
3.26. Given a complex exponential signal, by Euler’s formula it is converted as
x(t) = a ej ω 0 t = a [cos(ω 0 t) − j sin(ω 0 t)] ∴ ω 0 T0 = 2π ∴ T0 =
2π
ω0
which is to say that period of x(t) is T0 . Average power of a periodic signal is calculated as
Px =
1
T0
=
a2
T0
def
T0 /2
1
|a ej ω 0 t |2 dt = ej x = 1 =
T0
−T0 /2
T0 /2
a 2 T0
T0
t =
+
= a2
T
2
2
0
−T0 /2
T0 /2
−T0 /2
:1
ω 0
t
|ej
a2 |2 dt
This is periodic signal, therefore,
Ex = lim Px t = ∞
def
t→∞
That is to say, in order to sustain a non–zero average power over infinitely many periods it must be
that Ex = ∞.
Summary : this signal x(t) belongs to the category of periodic signals, i.e. that are not bound within
a finite time interval and whose amplitude is finite. By consequence, Ex = ∞ and 0 < Px < ∞ .
3.27. Given a continuous periodic infinitely long signal its average power over one period T is
1 T /2
1 T /2 2
2
cos (2t) + 2 cos(2t) + 1 dt
Px =
|1 + cos(2t)| dt =
T −T /2
T −T /2
1 + cos 2x
2
= cos (x) =
2
T /2 1 1
1
=
+ cos(4t) + 2 cos(2t) + 1 dt
T −T /2 2 2
3 T /2
1 1
sin(4t) + sin(2t) + t
=
T 8
2 −T /2
def
3.4 Energy and Power
87
3 T
T
T
1
T
+ sin 2
+
sin 4 2
+
8
2 2
2
2
2
1 1
0 3
:0
:
sin
sin
=
(2T
) +
(T
) + T
8
2
T
@
3
=
2
1
=
T
That is to say, average power within a singe period is non–zero. By consequence, in order to sustain
the same non–zero average power over infinitely many periods there must be an infinite source of
energy, as
Ex = lim Px t =
def
t→∞
3
lim t = ∞
2 t→∞
Summary : this signal x(t) belongs to the category of periodic signals, i.e. that are not bound within
a finite time interval and whose amplitude is finite. By consequence, Ex = ∞ and 0 < Px < ∞ .
3.28. Given exponential signal converges, i.e. x(t) → 0 for t → ±∞, thus
Ex =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
−t
e u(t)2 dt =
∞
e−2t (1)2 dt = −
0
1 −2t ∞
e
2
0
1
1
= − [0 − 1] =
2
2
Therefore,
Ex
=0
t→∞ t
Px = lim
def
That is to say, having a finite source of energy that must be distributed over infinitely long period of
time, the average power reduces to zero.
Summary : this signal x(t) belongs to the category of signals where if (t → ±∞) then (x(t) → 0),
such as forms in Fig. 3.71. By consequence, 0 < Ex < ∞ and Px = 0 .
Fig. 3.71 Example P.3.28
1
e−t u(t)
t
0
0
1
−t 2
e
t
0
1
0
e−|t|
t
0
0
88
3 Continuous Time Convolution
3.29. Given exponential signal converges, i.e. x(t) → 0 for t → ±∞, thus
∞
∞
2
π
−t 2 2
−t 2
|x(t)| dt =
dt = see chapters on integrals =
Ex =
e
e dt =
2
−∞
−∞
−∞
def
∞
2
Therefore,
Px = lim
def
t→∞
Ex
=0
t
That is to say, having a finite source of energy that must be distributed over infinitely long period of
time, the average power reduces to zero.
Summary : this signal x(t) belongs to the category of signals where if (t → ±∞) then (x(t) → 0),
see Fig. 3.71. By consequence, 0 < Ex < ∞ and Px = 0 .
3.30. Given exponential signal converges, i.e. x(t) → 0 for t → ±∞, thus
Ex =
def
∞
−∞
|x(t)| dt =
2
1 2t 0
= e 2
−∞
∞
−∞
−|t| 2
e dt =
1 −2t ∞
+ e
=1
2
0
−∞
t 2
dt +
e
∞
−t 2
dt = see A.3.28
e
0
0
Therefore,
Px = lim
def
t→∞
Ex
=0
t
That is to say, having a finite source of energy that must be distributed over infinitely long period of
time, the average power reduces to zero.
Summary : this signal x(t) belongs to the category of signals where if (t → ±∞) then (x(t) → 0),
see Fig. 3.71. By consequence, 0 < Ex < ∞ and Px = 0 .
3.31. Given an r(t) signal, its energy is
Ex =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
|r(t)|2 dt =
∞
r 2 dt =
0
1 3 ∞
t =∞
3
0
And its average power is
1
Px = lim
τ →∞ τ
def
τ/2
1
|x(t)| dt = lim
τ
→∞
τ
−τ/2
1
= 0 + lim
τ →∞ τ
τ/2
2
τ/2
|r(t)|2 dt
2
1 3 τ/2
1
τ 3
1
1
lim
t =
lim
r dt =
3 τ →∞ τ
3 τ →∞ τ 8
0
2
0
−τ/2
3.4 Energy and Power
89
Fig. 3.72 Example P.3.31
1
r(t)
t
0
0
1
0
t
0
(1 − t 3 ) u(−t)
1
0
=
1
e t u(t)
t
0
τ2
1
lim
=∞
3 τ →∞ 8
This is an example of a signal whose both energy and average power equal infinity.
Summary : this signal x(t) belongs to the category of signals where if (t → ∞) or (t → −∞) then
(x(t) → ∞), such as forms in Fig. 3.72. By consequence, Ex = ∞ and Px = ∞ .
3.32. Given exponential et u(t) signal, its energy is
Ex =
def
∞
−∞
|x(t)| dt =
2
∞
−∞
|e u(t)| dt =
t
2
∞
0
1 2t ∞
e (1) dt = e = ∞
2
0
2t
2
1 τ/2 t
|x(t)| dt = lim
|e u(t)|2 dt
τ →∞ τ
−τ/2
−τ/2
1
1
1 τ/2 2t
1 2t τ/2
1 τ
lim
e =
lim
= 0 + lim
e dt =
(e − 1)
τ →∞ τ
2 τ →∞ τ
2 τ →∞ τ
0
0
1
Px = lim
τ →∞ τ
τ/2
def
2
∞
0
τ
>
>
1
1
e
1
= lim
− lim
=∞
τ →∞ τ
τ →∞ τ
2
2
Summary : this signal x(t) belongs to the category of signals where if (t → ∞) or (t → −∞) then
(x(t) → ∞), see Fig. 3.72. By consequence, Ex = ∞ and Px = ∞ .
3.33. Given cubic x(t) signal, its energy is
Ex =
def
∞
−∞
|x(t)|2 dt =
∞
−∞
|(1 − x 3 ) u(−t)|2 dt =
0
1 4 0
1 7 0
− t + t =∞
= t 7
2
−∞
−∞
−∞
0
−∞
(1 − x 3 )2 (1)2 dt
90
3 Continuous Time Convolution
And its average power is
1
Px = lim
τ →∞ τ
def
1
= lim
τ →∞ τ
τ/2
1
|x(t)| dt = lim
τ
→∞
τ
−τ/2
τ/2
2
−τ/2
(1 − x 3 ) u(−t)2 dt
1 7 0
1 4 0
1 0
1
1
3 2
dt + 0 =
−
+ lim
1−x
lim
t lim
t t
τ →∞ τ
7 τ →∞ τ
2 τ →∞ τ
−τ/2
−τ/2
−τ/2
−τ/2
0
=∞
Summary : this signal x(t) belongs to the category of signals where if (t → ∞) or (t → −∞) then
(x(t) → ∞), see Fig. 3.72. By consequence, Ex = ∞ and Px = ∞ .
4
Discrete Time Convolution
Convolution sum is defined as
y[ n] = x[k] ∗ y[n] =
∞
x[ k] y[ n − k]
(4.1)
k=−∞
where k is the provisional summation variable.
Energy and power definitions: by definition, E(t) =
Calculations of sampled functions are the following:
t
0 P (τ )dτ
for continuous functions.
1. Energy within an interval
Ex =
m
x[k]2
k=n
2. Total energy
Ex =
∞
x[k]2
k=−∞
3. Average power within an interval
Px =
n
2
1 x[k]
2n + 1 k=n
4. Total power
n
2
1 x[k]
n→∞ 2n + 1
k=n
Px = lim
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_4
91
92
4 Convolution D
Problems
4.1
Discrete Signal Convolution
Given short sequences in P.4.1 to P.4.3, calculate their convolution sums.
4.1. x[n] = [1 , 1, 1], y[n] = [1 , 1, 1]
4.2. x[n] = [1, 2, 3], y[n] = [1, 1, 1]
4.3. f [k], g[k], see Fig. 4.1
Given two short sequences h[n] = [1, 2 , 0, −3] and x[n] in P.4.4 to P.4.7, calculate their convolution
sums. (Note: number whose index n = 0 is underlined.)
4.4. x[n] = δ[n]
4.5. x[n] = δ[n + 1] + δ[n − 2]
4.6. x[n] = [1 , 1, 1]
4.7. x[n] = [2, 1, − 1 , −2, −3]
Given analytical form of sequences in P.4.8 to P.4.10, calculate their convolution sums.
4.8. x[k] = α u[k]; (0 < α < 1)
k
h[k] = u[k]
1; (0 ≤ k ≤ 4)
4.9. x[k] =
0; otherwise
α k ; (0 ≤ k ≤ 6)
h[k] =
0; otherwise
4.10. x[n] = sin(ω 0 n) u(n)
y[n] = u[n]
4.2
Discrete Signal Energy and Power
Calculate energy and power of sequences x[k] in P.4.11 to P.4.16.
4.11. x[k] = δ[k]
αk ;
4.14. x[k] =
0;
Fig. 4.1 Example P.4.3
k≥0
k<0
k
1
4.12. x[k] =
u[k]
4
4.13. x[k] = ej 10k u[k]
4.15. x[k] = u[k]
4.16. x[k] = cos
2
π
6
k
f [k] g[k]
1
0
−3 −2 −1
k
0
1
2
3
4
5
4.1 Discrete Signal Convolution
93
Answers
4.1
Discrete Signal Convolution
There are many ways to tabulate convolution sums, mostly the question of preference and style. Some
of the typical methods are illustrated here.
4.1. There is a number of ways to tabulate convolution of discrete series, which is question of style
and preference.
Method 1: Given two short sequences x[n] = [1 , 1, 1] and y[n] = [1 , 1, 1], that is to say
with a clearly defined finite interval of non–zero samples, the interval of convolution z[n] series
is determined as
x[0] = 1, x[1] = 1, x[2] = 1 ∴
finite non–zero interval begins at n = 0 and ends at n = 2
y[0] = 1, y[1] = 1, y[2] = 1 ∴
finite non–zero interval begins at n = 0 and ends at n = 2
∴
z[n] ∴
∴
finite non–zero interval begins at n = 0 + 0 = 0
finite non–zero interval ends at n = 2 + 2 = 4
that is to say, z[n], 0 ≤ n ≤ 4. By consequence, it is necessary to evaluate z[n] in five cases
n = [0, 1, 2, 3, 4], as
z[n] = x[n] ∗ y[n] =
def
∞
m=−∞
x[m] y[k − m] =
2
x[m] y[n − m]
m=0
Note that “fixed” series is x[m] and y[k − m] is the “sliding” sequence relative to parameter m,
therefore there are only three simples in the interval m = (0, −1) as,
Case 1: n = 0 z[ 0] =
2
x[m] y[ 0 − m] = x[0] y[ 0 − (0)] + x[1] y[ 0 − (1)] + x[2] y[ 0 − (2)]
m=0
= x[0] y[0] + x[1] y[−1] + x[2] y[−2] = (1)(1) + (1)(0) + (1)(0)
=1
Case 2: n = 0 z[ 1] =
2
x[m] y[ 1 − m] = x[0] y[ 1 − (0)] + x[1] y[ 1 − (1)] + x[2] y[ 1 − (2)]
m=0
= x[0] y[1] + x[1] y[0] + x[2] y[−1] = (1)(1) + (1)(1) + (1)(0)
=2
94
4 Convolution D
Case 3: n = 0 z[ 2] =
2
x[m] y[ 2 − m] = x[0] y[ 2 − (0)] + x[1] y[ 2 − (1)] + x[2] y[ 2 − (2)]
m=0
= x[0] y[2] + x[1] y[1] + x[2] y[0] = (1)(1) + (1)(1) + (1)(1)
=3
Case 4: n = 0 z[ 3] =
2
x[m] y[ 3 − m] = x[0] y[ 3 − (0)] + x[1] y[ 3 − (1)] + x[2] y[ 3 − (2)]
m=0
= x[0] y[3] + x[1] y[2] + x[2] y[1] = (1)(0) + (1)(1) + (1)(1)
=2
Case 5: n = 0 z[ 4] =
2
x[m] y[ 4 − m] = x[0] y[ 4 − (0)] + x[1] y[ 4 − (1)] + x[2] y[ 4 − (2)]
m=0
= x[0] y[4] + x[1] y[3] + x[2] y[2] = (1)(0) + (1)(0) + (1)(1)
=1
In conclusion, z[n] = [1 , 2, 3, 2, 1], see Fig. 4.2. (Compare with the continuous time example
P.3.12). Note that, if given as a list, sample z[0] is underlined.
Method 2: Given two short sequences x[n] = [1 , 1, 1] and y[n] = [1 , 1, 1], sequences x[n] may be
written, for example, as
x[n] = δ[n] + δ[n − 1] + δ[n − 2]
By definition, convolution sum is
z[n] = x[n] ∗ y[n] = (δ[n] + δ[n − 1] + δ[n − 2]) ∗ y[n] = y[n] + y[n − 1] + y[n − 2]
where, y[n − 1] and y[n − 2] are shifted y[n] as
y[n − 1 = [ , 1, 1, 1]
y[n − 2] = [ , , 1, 1, 1]
so that ‘ ’ syntax is used to mark position of n = 0 sample. The convolution sum
Fig. 4.2 Example P.4.1
3
z[n]
n
0
−1
0
1
2
3
4
5
4.1 Discrete Signal Convolution
95
z[n] = y[n] + y[n − 1] + y[n − 2]
may be done by the tabular method as
n
y[n]
y[n − 1]
y[n − 2]
z[n]
0
1
1
1
1
+
1
2
2
1
1
1
3
3
4
1
1
2
1
1
which is in agreement with the Method 1: result.
4.2. There is a number of ways to tabulate convolution of discrete series, which is question of style
and preference.
Method 1: Given two short sequences f [k] and g[k] in Fig. 4.1, that is to say with a clearly defined
finite interval of non–zero samples, the interval of convolution h[k] series is determined as
f [−2] = 2, f [−1] = 2 ∴
finite non–zero interval begins at
k = −2 and ends at
g[1] = 2, f [3] = 1 ∴
finite non–zero interval begins at
k = 2 and ends at
finite non–zero interval begins at
k = −2 + 2 = 0
k = −1
k=3
∴
h[k] ∴
∴
finite non–zero interval ends at
k = −1 + 3 = 2
that is to say, h[k] = 0, 0 ≤ k ≤ 2. By consequence, it is necessary to evaluate h[k] in three cases
k = [0, 1, 2], as
h[k] = f [k] ∗ g[k] =
def
∞
m=−∞
f [m] g[k − m] =
−1
f [m] g[k − m]
m=−2
Note that “fixed” series is f [m] and g[k − m] is the “sliding” sequence relative to parameter m,
therefore there are only two simples in the interval m = (−2, −1) as,
Case 1: k = 0 h[ 0] =
−1
f [m] g[ 0 − m] = f [−2] g[0 − (−2)] + f [−1] g[0 − (−1)]
m=−2
= f [−2] g[2] + f [−1] g[1] = (2)(2) + (2)(0) = 4
Case 2: k = 1 h[ 1] =
−1
f [m] g[ 1 − m] = f [−2] g[1 − (−2)] + f [−1] g[1 − (−1)]
m=−2
= f [−2] g[3] + f [−1] g[2] = (2)(1) + (2)(2) = 6
96
4 Convolution D
6
Fig. 4.3 Example P.4.3
h[k]
4
2
0
Case 3: k = 2 h[ 2] =
−1
k
−1
0
1
2
3
f [m] g[ 2 − m] = f [−2] g[2 − (−2)] + f [−1] g[2 − (−1)]
m=−2
= f [−2] g[4] + f [−1] g[3] = (2)(0) + (2)(1) = 2
In conclusion, h[k] = [4 , 6, 2], see Fig. 4.3. Note that, if given as a list, sample h[0] is underlined.
Method 2: Given graphs in Fig. 4.1, two sequences f [k] and g[k] may be written, for example, as
f [k] = 2δ[k + 2] + 2δ[k + 1]
g[k] = [ , , 2, 1]
where the leading empty spaces in g[k] keep the sequence index so that g[2] = 2 and g[3] = 1.
Underlined space ‘ ’ is used to indicate the index 0 position. By definition, convolution sum is
h[k] = f [k] ∗ g[k] = (2δ[k + 2] + 2δ[k + 1]) ∗ g[k] = 2 g[k + 2] + 2 g[k + 1]
where, g[k + 2] and g[k + 1] are shifted g[k] as
2 g[k + 2] = 2 [2 , 1] = [4 , 2]
2 g[k + 1] = 2 [ , 2, 1] = [ , 4, 2]
so that 2 syntax is used to mark k = 0 index in g[k + 2] sequence. The convolution sum
h[k] = 2 g[k + 2] + 2 g[k + 1] may be done by the tabular method as
k
2 g[k + 2]
2 g[k + 1]
h[k]
−2
−1
0
4
+
4
1
2
4
6
2
3
2
2
which is in agreement with the Method 1: result.
4.3. There is a number of ways to tabulate convolution of discrete series, which is question of style
and preference.
Method 1: Given two short sequences f [k] and g[k] in Fig. 4.1, that is to say with a clearly defined
finite interval of non–zero samples, the interval of convolution h[k] series is determined as
4.1 Discrete Signal Convolution
97
f [−2] = 2, f [−1] = 2 ∴
finite non–zero interval begins at
k = −2 and ends at
g[1] = 2, f [3] = 1 ∴
finite non–zero interval begins at
k = 2 and ends at
finite non–zero interval begins at
k = −2 + 2 = 0
k = −1
k=3
∴
h[k] ∴
∴
k = −1 + 3 = 2
finite non–zero interval ends at
that is to say, h[k] = 0, 0 ≤ k ≤ 2. By consequence, it is necessary to evaluate h[k] in three cases
k = [0, 1, 2], as
h[k] = f [k] ∗ g[k] =
def
∞
f [m] g[k − m] =
m=−∞
−1
f [m] g[k − m]
m=−2
Note that “fixed” series is f [m] and g[k − m] is the “sliding” sequence relative to parameter m,
therefore there are only two simples in the interval m = (−2, −1) as,
Case 1: k = 0 h[ 0] =
−1
f [m] g[ 0 − m] = f [−2] g[0 − (−2)] + f [−1] g[0 − (−1)]
m=−2
= f [−2] g[2] + f [−1] g[1] = (2)(2) + (2)(0) = 4
Case 2: k = 1 h[ 1] =
−1
f [m] g[ 1 − m] = f [−2] g[1 − (−2)] + f [−1] g[1 − (−1)]
m=−2
= f [−2] g[3] + f [−1] g[2] = (2)(1) + (2)(2) = 6
Case 3: k = 2 h[ 2] =
−1
f [m] g[ 2 − m] = f [−2] g[2 − (−2)] + f [−1] g[2 − (−1)]
m=−2
= f [−2] g[4] + f [−1] g[3] = (2)(0) + (2)(1) = 2
In conclusion, h[k] = [4 , 6, 2], see Fig. 4.4. Note that, if given as a list, sample h[0] is underlined.
Method 2: Given graphs in Fig. 4.1, two sequences f [k] and g[k] may be written, for example, as
f [k] = 2δ[k + 2] + 2δ[k + 1]
g[k] = [ , , 2, 1]
where the leading empty spaces in g[k] keep the sequence index so that g[2] = 2 and g[3] = 1.
Underlined space ‘ ’ is used to indicate the index 0 position. By definition, convolution sum is
Fig. 4.4 Example P.4.3
6
h[k]
4
2
0
k
−1
0
1
2
3
98
4 Convolution D
h[k] = f [k] ∗ g[k] = (2δ[k + 2] + 2δ[k + 1]) ∗ g[k] = 2 g[k + 2] + 2 g[k + 1]
where, g[k + 2] and g[k + 1] are shifted g[k] as
2 g[k + 2] = 2 [2 , 1] = [4 , 2]
2 g[k + 1] = 2 [ , 2, 1] = [ , 4, 2]
so that 2 syntax is used to mark k = 0 index in g[k + 2] sequence. The convolution sum
h[k] = 2 g[k + 2] + 2 g[k + 1] may be done by the tabular method as
k
2 g[k + 2]
2 g[k + 1]
h[k]
−2
−1
0
4
+
4
1
2
4
6
2
3
2
2
which is in agreement with the Method 1: result.
4.4. Given sequence h[n] = [1, 2 , 0, −3], and Dirac delta function δ[n], convolution is equivalent
to multiplication by one because Dirac delta is the identity function, as
y[n] = δ[n] ∗ h[n] = h[n] ∗ δ[n] = h[n]
4.5. Given h[n] = [1, 2 , 0, −3] and sequence of two Dirac delta, as
x[n] = δ[n + 1] + δ[n − 2]
it follows that
y[n] = x[n] ∗ h[n] = (δ[n + 1] + δ[n − 2]) ∗ h[n]
= δ[n + 1] ∗ h[n] + δ[n − 2] ∗ h[n]
= h[n + 1] + h[n − 2]
Where, two shifted sequences h[n + 1] and h[n − 2] are
h[n] = [1, 2 , 0, −3]
h[n + 1] = [1, 2, 0 , −3] shifted one places to right
h[n − 2] = [ , 1, 2, 0, −3] shifted two places to left
One way to calculate the convolution sum y[n] = h[n + 1] + h[n − 2] is by tabular method as
4.1 Discrete Signal Convolution
99
Fig. 4.5 Example P.4.4
y[n]
2
0
n
−2
−4
n
h[n + 1]
h[n − 2]
y[n]
−2
1
−1
2
0
0
1
2
0
−2
1
−3
1
−2
+
0
2
2
3
4
2
2
0
0
−3
−3
That is to say, (see also Fig. 4.5)
y[n] = [1, 2, 0 , −2, 2, 0, −3]
4.6. Given h[n] = [1, 2 , 0, −3] and sequence of three Dirac delta, as
x[n] = [1 , 1, 1] = δ[n] + δ[n − 1] + δ[n − 2]
it follows that,
y[n] = x[n] ∗ h[n] = (δ[n] + δ[n − 1] + δ[n − 2]) ∗ h[n]
= h[n] + h[n − 1] + h[n − 2]
Knowing h[n] = [1, 2 , 0, −3] it follows that two shifted sequences are
h[n − 1] = [1 , 2, 0, −3]
h[n − 2] = [ , 1, 2, 0, −3]
so that the convolution sum y[n] = h[n] + h[n − 1] + h[n − 2] is
n
h[n]
h[n − 1]
h[n − 2]
y[n]
−1
1
0
2
1
+
1
3
1
0
2
1
3
2
−3
0
2
−1
That is to say, (see also Fig. 4.6)
y[n] = [1, 3 , 3, −1, −3, −3]
3
4
−3
0
−3
−3
−3
4
100
4 Convolution D
4
Fig. 4.6 Example P.4.6
y[n]
2
0
n
−2
−4
−2
0
2
4.7. Given two sequeces
x[n] = [2, 1, − 1 , −2, −3]
h[n] = [1, 2 , 0, −3]
the tabular method may be used to calculate convolution
y[n] = x[n] ∗ h[n]
Noting the index reference for x[n], it may be written as
x[n] = 2δ[n + 2] + δ[n + 1] − δ[n] − 2δ[n − 1] − 3δ[n − 2]
so that
y[n] = x[n] ∗ h[n]
= (2δ[n + 2] + δ[n + 1] − δ[n] − 2δ[n − 1] − 3δ[n − 2]) ∗ h[n]
= 2h[n + 2] + h[n + 1] − h[n] − 2h[n − 1] − 3h[n − 2]
where the required shifted sequences of h[n] are
2 h[n + 2] = 2 [1, 2, 0, − 3 ] = [2, 4, 0, − 6 ]
h[n + 1] = [1, 2, 0 , −3]
−h[n] = [−1, − 2 , 0, 3]
−2 h[n − 1] = −2 [1 , 2, 0, −3] = [ − 2 , −4, 0, 6]
−3 h[n − 2] = −3 [ , 1, 2, 0, −3] = [
, −3, −6, 0, 9]
Above intermediate results are tabulated and added in tabular form as follows
y[n] =
2
k=−2
x[k] h[n − k]
4
4.1 Discrete Signal Convolution
n
x[−2] h[n + 2]
x[−1] h[n + 1]
x[0] h[n]
x[1] h[n − 1]
x[2] h[n − 2]
y[n]
101
−3
2
−2
4
1
−1
0
2
−1
0
−6
0
−2
−2
+
2
5
1
− 10
1
2
3
4
−3
0
−4
−3
−10
3
0
−6
−3
6
0
6
9
9
Visually, tabular method does the same “sweep” of h[k − n] as already illustrated in the graphics
method. In conclusion, (see Fig. 4.7)
2
y[n] = x[n] ∗ h[n] =
x[k] h[n − k] = [2, 5, 1, −10 , −10, −3, 6, 9]
k=−2
4.8. Given two long sequences x[k] and h[k] in analytical form, that is to say with the infinite
non–zero interval, the convolution is resolved as same as in the case of continuous signals. Function
transformations may be used to derive x[m] and h[k − m], see Fig. 4.8.
x[k] = α k u[k]; (0 < α < 1)
h[k] = u[k]
Case 1 : k < 0
Given that k < 0, all products inside the convolution sum equal zero, see Fig. 4.9, as
x[m] h[k − m] = 0; ∀m
and by consequence, convolution sum equal zero, as
10
Fig. 4.7 Example P.4.7
y[n]
5
0
n
−5
−10
−4
x[m]
1
···
m
0
−2 −1
Fig. 4.8 Example P.4.8
0
1
2
3
4
5
0
2
4
h[k − m]
1
···
−2
···
···
m
0
··· k −1 k k +1
102
4 Convolution D
Fig. 4.9 Example P.4.8
h[k − m] x[m]
1
···
···
m
0
···
k −1 k ··· 0 1 2 3 4 5
Fig. 4.10 Example P.4.8
h[k − m] x[m]
1
···
···
m
0
···
0
k k + 1· · ·
y[k] = x[k] ∗ h[k] = 0; (k < 0)
Case 2 : k > 0
Given that k ≥ 0, there are non–zero products inside the convolution sum interval, see Fig. 4.10,
as
x[m] h[k − m] = 0; m = [0, 1, . . . , k]
and by consequence, convolution sum is
y[k] = x[k] ∗ h[k] =
k
x[m] h[k − m]
0
which leads into geometric sum as
=
k
α m (1)
0
=
N −1
αn =
n=0
N; α = 1
N
1−α
;
1−α
α=1
=
1 − α k+1
1−α
geometric sum
Summary: the two cases together may be written in the compact form as
y[k] =
1 − α k+1
u[k]
1−α
Note that due to α < 1 then (see Fig. 4.11)
lim α k+1 → 0 ∴
k→∞
1
1 − α k+1
→
k→∞ 1 − α
1−α
lim
4.9. Given port x[k] and exponential h[k] functions in the analytical form, as
(k ≥ 0)
4.1 Discrete Signal Convolution
103
Fig. 4.11 Example P.4.8
1
1−a
y[k]
1
···
0
···
· · · −1 0
1
2
3
···
···
0
−2
0
k
h[k − m]
x[m]
1
4 ···
2
4
6
m
1
0
···
···
k ···
k−6
···
m
Fig. 4.12 Example P.4.9
x[k] =
1; (0 ≤ k ≤ 4)
0; otherwise
and
h[k] =
∞
by definition, convolution sum is y[k] = x[k] ∗ h[k] =
α k ; (0 ≤ k ≤ 6)
0; otherwise
x[m] h[k − m]. After function trans-
m=−∞
formations x[m] and h[k − m] are as in Fig. 4.12.
Case 1 : k < 0
For k < 0 all products inside the convolution sum equal zero, see Fig. 4.13, as
x[m] h[k − m] = 0; ∀m
and by consequence, convolution sum equal zero, as
y[k] = x[k] ∗ h[k] = 0; (k < 0)
Case 2 : 0 ≤ k ≤ 4
As the leading sample hk of h[k −m] enters interval [0, 4] defined by port function x[m], non–zero
products are in [0, k] interval, see Fig. 4.14, thus convolution sum is
y[k] = x[k] ∗ h[k] =
=
k
m=0
r =k−m ⇒
x[m] h[k − m] =
k
(1) α k−m =
m=0
m=0 ∴ r=k
m=k ∴ r=0
k
α k−m
m=0
=
0
r=k
αr =
k
r=0
αr =
1 − α k+1
1−α
104
4 Convolution D
Fig. 4.13 Example P.4.9
h[k − m] x[m]
1
0
···
··· k −6
k ··· 0
Fig. 4.14 Example P.4.9
4 ···
···
m
x[m]
h[k − m]
1
0
···
0
k−6
Fig. 4.15 Example P.4.9
4 ···
k
···
m
x[m] h[k − m]
1
0
···
··· k −6 0
4 5 ···
···
m
Case 3 : 5 ≤ k ≤ 6
As the leading sample hk of h[k − m] leaves interval [0, 4] defined by port function x[m], but hk−4
is inferior to zero, (note that exponential suite has six and port has four samples) i.e.
k
k−6
>4
≤0 ∴ k≤6
⇒ k = (5, 6)
Therefore, non–zero products are in interval [0, 4], see Fig. 4.15, and convolution sum is
y[k] = x[k] ∗ h[k] =
4
m=0
x[m] h[k − m] =
4
(1) α k−m =
m=0
−1 5
4
−1 m
α k − α k−5
k
k1− α
=α
=α
=
α
1 − α −1
1 − α −1
m=0
4
α k α −m
m=0
α k+1 − α k−4
α
=
α
α−1
Case 4 : 7 ≤ k ≤ 10
As the trailing sample hk of h[k − m] enters interval [0, 4] defined by port function x[m],
4.1 Discrete Signal Convolution
105
Fig. 4.16 Example P.4.9
h[k − m]
x[m]
1
0
···
···
0
Fig. 4.17 Example P.4.9
k−6
4
0
k
h[k − m]
x[m]
1
···
m
···
···
··· 0
4··· k − 6
k
···
m
···
k−6
>0 ∴ k>6
k−6
≤ 4 ∴ k ≤ 10
⇒ k = (7, 8, 9, 10)
and non–zero products are in interval [k − 6, 4], see Fig. 4.16, thus convolution sum is
4
y[k] = x[k] ∗ h[k] =
x[m] h[k − m] =
m=k−6
=
=
⎧
⎪
⎪r = m − k + 6;
⎨
⎪
⎪
⎩
10−k
r=0
4
(1) α k−m
m=k−6
k−m=6−r
m=k−6 ∴ r =0
m = 4 ∴ r = 10 − k
α 6−r = α 6
⎫
⎪
⎪
⎬
⎪
⎪
⎭
10−k
−1 r
α 7 − α k−4
1 − α k−11
= α6
=
α
−1
1−α
α−1
r=0
Case 5 : k > 10
As the trailing sample hk−6 of h[k − m] leaves interval [0, 4] defined by port function x[m], that
is k − 6 > 4 ∴ k > 10, see Fig. 4.17, thus all products equal zero, as
x[m] h[k − m] = 0; ∀m
and by consequence, convolution sum equal zero, as
y[k] = x[k] ∗ h[k] = 0; (k > 10)
106
4 Convolution D
Fig. 4.18 Example P.4.9
y[k]
1 ···
(1)
(2)
(3)
(4) (5)
4 5 6 7
10
0
···
k
Summary: see Fig. 4.18
⎧
0;
⎪
⎪
⎪
⎪
⎪
1 − α k+1
⎪
⎪
;
⎪
⎪
1−α
⎪
⎪
⎪
⎪
⎪
⎪
⎨ k+1
α
− α k−4
y[k] =
;
⎪
α−1
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
α 7 − α k−4
⎪
⎪
⎪
;
⎪
⎪
⎪
⎩ α−1
0;
k<0
0≤k≤4
5≤k≤6
7 ≤ k ≤ 10
k > 10
4.10. Given two sequences x[n] and y[n] in analytical form and assuming finite n, see Fig. 4.19,
convolution
h[n] = x[n] ∗ y[n]
may be resolved analytically as follows.
For n < 0 all convolution products equal zero, thus (assuming finite m) for n ≥ 0
h[n] = x[n] ∗ y[n] =
m
x[m] y[n − m] = 0 +
−m
m
sin(ω 0 m) (1) =
0
m
sin(ω 0 m)
0
m
m
1 −j ω 0 m
1 j ω0 m
ej x − e−j x
ej x + e−j x
= sin(j x) =
e
−
e
=
, cos(j x) =
2j 0
2j 0
2j
2
m
m
1 j ω 0 m
1 −j ω 0 m
e
e
= eab = (ea )b , ea+b = ea eb =
−
2j 0
2j 0
m
1 − a m+1
1 1 − ej ω 0 (m+1)
1 1 − e−j ω 0 (m+1)
n
a =
=
=
−
1−a
2j
1 − ej ω 0
2j
1 − e−j ω 0
0
m+1
m+1
m+1
m+1
= 1 = e0 = ej ω 0 ( 2 )−j ω 0 ( 2 ) = ej ω 0 ( 2 ) e−j ω 0 ( 2 )
1 ej ω 0 ( 2 ) e−j ω 0 ( 2 ) − ej ω 0 (m+1)
1 ej ω 0 ( 2 ) e−j ω 0 ( 2 ) − e−j ω 0 (m+1)
−
j ω0
−j ω 0
j ω0
−j ω 0
2j
2j
e 2 e 2 − ej ω 0
e 2 e 2 − e−j ω 0
m+1
=
m+1
m+1
m+1
4.2 Discrete Signal Energy and Power
1
107
x[n]
···
···
0
−1
y[n]
1
···
n
···
0
−2
0
2
4
6
8
10
12
−2
0
3
h[n]
2
4
6
8
10
n
12
Fig. 4.19 Example P.4.10
Fig. 4.20 Example P.4.10
2
1
0
···
−2
0
2
4
6
8
10
···
n
12
m+1
m+1
j ω 0 ( m+1
−j ω 0 ( m+1
−j ω 0 ( m+1
j ω 0 ( m+1
2 )
2 ) − ej ω 0 ( 2 )
2 )
2 ) − e −j ω 0 ( 2 )
e
e
e
e
1
1
jω
jω
−
=
j ω0
j ω0
j ω0
j ω0
0
0
−
2j
2j
e 2 e 2 −e 2
e− 2 e 2 − e− 2
j ω0 m
j
ω0
j ω0 m
j
ω0
ω 0 (m+1)
ω 0 (m+1)
−2j
2j
e 2 H
e−2 @
1 e 2 1 e− 2 H sin
2
2
@ sin
=
−
ω
ω0
j
ω
0
j
ω0
H
0
2j
−2j
sin
2j
−
2j
sin
@
H
2
2
2
e
2
@
e
sin ω 0 (m+1) 1
j ω0 m
sin ω 0 (m+1)
ω0 m
1 j ω0 m
2
2
=
2j
=
e 2 − e− 2
@
ω0
@ sin 2
sin 2
2j
sin ω20
2j
@
@
=
sin ω 0 (m+1)
ω0 m
2
; (0 ≤ m ≤ n)
sin
ω0
sin 2
2
The convolution sequence h[n] is shown in Fig. 4.20
4.2
Discrete Signal Energy and Power
4.11. Given Dirac function, x[k] = δ[k], by definition
E[k] =
∞
|x[k]|2 =
k=−∞
∞
|δ[k]|2 = · · · + 0 + 0 + · · · + 12 + 0 + 0 + · · ·
k=−∞
=1<∞
which is to say that since the energy is distributed over infinitely long time interval k = ∞, therefore
at any given moment the available average power must be
108
4 Convolution D
P [k] = lim
k→∞
E[k]
1
= lim = 0
k→∞ k
k
as same as for continuous functions where t → ∞.
4.12. Given infinitely long series x[k] =
k
1
u[k], it follows that
4
2
∞ k
∞ 2k
∞ 1
1
1 k
2
|x[k]| =
u[k] =
=
E[k] =
4
4
16
k=−∞
k=−∞
k=0
k=0
∞
1
k
where, |r| < 1
r =
geometric series,
1−r
k=0
∞
a
2k
as well as,
ar =
1 − r2
k=0
∞
ar
2k+1
or,
ar
=
1 − r2
k=0
1
∴ a = 1 and r =
16
∞
=
1
1
1−
16
=
16
<∞
15
which is to say that since the energy is distributed over infinitely long time interval k = ∞, therefore
P [k] = lim
k→∞
E[k]
16/15
= lim
=0
k→∞
k
k
as same as for continuous functions where t → ∞.
4.13. Samples of complex exponential series, x[k] = ej 10k u[k], are added for k ∈ [−n, n],
consequently there are in total 2n + 1 samples to add. That is, k = 1, 2, 3, . . . , n (therefore n points
in total) on the positive side, plus k = −1, −2, −3, . . . , −n (i.e. n points in total) on the negative
side, plus k = 0 (i.e. one additional point) in the middle, thus 2n + 1 points in total. Accordingly, by
definition for power we find
!
!
n
n
j 10k
2
1
1
2
e
|x[k]| = lim
u[k]
n→∞ 2n + 1
2n + 1 k=−n
k=−n
!
!
n
n
1 j 10k 2
1
2
|1k |
= lim
e
n→∞
2n + 1 k=0
2n + 1 k=0
P [k] = lim
n→∞
= lim
n→∞
"
= lim
n→∞
#
n+1
∞
n+1
1
1
l’H
=
= lim = < ∞
n→∞ 2
2n + 1
∞
2
4.2 Discrete Signal Energy and Power
109
which is to say that in order to support non–zero average power over infinite long interval k = ∞,
there must be E[k] = P [k] k = ∞.
4.14. Infinite sum of power series where |α| < 1 is convergent, thus by definition
E[k] =
∞
|x[k]|2 =
k=−∞
∞
∞
k
k 2
α u[k]2 =
α =
∞
2 k
α
k=−∞
k=0
k=0
geometric series
=
1
<∞
1 − α2
Finite amount of energy must be distributed over infinitely long interval, thus the average power equal
1
2
E[k]
P [k] = lim
= lim 1 − α = 0
k→∞ k
k→∞
k
4.15. Infinitely long step function has average power calculated as
P [k] = lim
n→∞
= lim
n→∞
!
!
n
n
1
1
2
2
|u[k]|
|x[k]| = lim
n→∞ 2n + 1
2n + 1 k=−n
k=−n
!
# "
n
1
∞
1
n+1
2
|1k |
= lim
=
= LH lim
n→∞ 2n + 1
n→∞ 2
2n + 1 k=0
∞
n+1
=
1
<∞
2
which is to say that in order to support non–zero average power over infinite long interval k = ∞,
there must be E[k] = P [k] k = ∞.
4.16. Average power of an infinitely long periodic series is calculated over one period as
!
!
n
n π 2
1
1
2
k P [k] = lim
|x[k]| = lim
cos
n→∞ 2n + 1
n→∞ 2n + 1
6
k=−n
k=−n
1 + cos(2x)
= cos2 (x) =
2
"
##!
"
n
1 1
1
π
+ cos 2
k
= lim
n→∞ 2n + 1
2 2
6 3
k=−n
⎡
⎤
⎢
⎢ 1
1
= lim ⎢
n→∞ ⎢ 2n + 1 2
⎣
n
1k
k=−n
2n+1
1
1
+
2n + 1 2
n
cos
π
k=−n
=0
3
k
⎥
⎥
⎥
⎥
⎦
110
4 Convolution D
=
in total, there are (2n+1) terms of 1/2 each;
the sum of sin or cos over one period equal zero ;
"
#
1
1
1
= lim
2n
+
1)
= <∞
(
n→∞ 2n
+1 2
2
which is to say that in order to support non–zero average power over infinite long interval k = ∞,
there must be E[k] = P [k] k = ∞.
5
Continuous Time Fourier Transform
Classification of Fourier transformations, sums and series:
Continuous time (CT) periodic, Fig. 5.1 (left), and nonperiodic, Fig. 5.1 (right), x(t) functions.
Periodic functions are transformed by continuous-time Fourier series (CT FS), while continuous time
nonperiodic functions are transformed by continuous-time Fourier transformation (CT FT).
Similarly, discrete-time (DT) periodic functions are mapped by discrete-time Fourier series (DT FS),
while discrete-time nonperiodic functions are transformed by discrete-time Fourier transformation
(DT FS).
In addition, in order to carry out Fourier transformation (FT) of continuous signals on numerical
processors, CT signals must be sampled before applying discrete Fourier transformation (DFT)
algorithm. Version of DFT algorithm that is optimized for fast execution is referred to as fast Fourier
transformation (FFT).
Fourier and inverse Fourier integrals are defined as
def
F x(t) = X(ω) =
∞
x(t) e−j ω t dt
(5.1)
−∞
def
1
F −1 X(ω) = x(t) =
2π
∞
X(ω) ej ω t dω
(5.2)
−∞
Fourier series may be defined in various forms. The most common form is a periodic function f (x)
development, where the resulting series is in the form of an infinite sum of sines and cosines terms,
as
f (t) =
∞
∞
a0 +
an cos(nt) +
bn sin(nt)
2
n=1
n=1
(5.3)
where coefficients an and bn are defined as
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_5
111
112
5 CT Fourier Transformation
x(t)
x(t)
t
T
t
CT FT
CT FS
|X (w )|
|C[n]|
n
w
Fig. 5.1 Comparison between CT FS and CT FT
a0 =
an =
1
π
1
bn =
π
π
−π
π
−π
1
π
π
f (t) dt
(5.4)
f (t) cos(nt) dt (n ≥ 1)
(5.5)
f (t) sin(nt) dt (n ≥ 1)
(5.6)
−π
Note: the resulting infinite series may or may not be convergent.
Properties of Fourier Transformation
FT
FT
FT
1. Linearity: given three functions and their FT as f (t) → F (ω), x(t) → X(ω), and y(t) → Y (ω),
and constants a, b, then
f (t) = ax(t) + b(y(t) ⇒ F (ω) = aX(ω) + bY (ω)
FT
2. Time shifting: given x(t) → X(ω) and a constant τ , then
x(t − τ ) → e−j τ ω X(ω)
FT
That is to say, time delay (t − τ ) translates into phase shift e−j τ ω in frequency domain.
FT
3. Frequency shifting: given x(t) → X(ω) and a constant ω 0 , then
e−j ω 0 t x(t) → X(ω − ω 0 )
FT
That is to say, phase shift e−j ω 0 t in time domain translates into frequency shift (ω − ω 0 ).
FT
4. Time scaling: given x(t) → X(ω) and a constant a, then
FT
x(at) →
1 ω
X
|a|
a
5.2 Composite Functions
113
FT
5. Derivative property: given x(t) → X(ω), then FT of (n)–th derivative is
FT
x (n) (t) → (−j ω )n X(ω)
FT
6. Multiplication by t n : given x(t) → X(ω), then
dn
X (ω)
dωn
FT
t n x(at) → (−j )n
FT
FT
7. Convolution property: given x(t) → X(ω) and y(t) → Y (ω), then
FT
x(t) ∗ y(t) → X(ω) Y (ω)
Problems
5.1
Special Functions
Given analytical functions x(t) in P.5.1 to P.5.15 (where applies, assume a, τ > 0), derive Fourier
transformations X(ω).
5.1. x(t) = δ(t)
5.2. x(t) = δ(t − τ )
5.3. x(t) = a
5.4. x(t) = ej ω 0 t
5.5. x(t) = cos(ω 0 t)
5.6. x(t) = sin(ω 0 t)
5.7. x(t) = e−at u(t)
5.8. x(t) = eat u(−t)
5.9. x(t) = e−a|t|
t
5.10. x(t) = a τ
5.11. x(t) = a 5.13. x(t) = u(t)
5.14. x(t) = e−t
5.2
t
2τ
2
5.12. x(t) = sign (t)
5.15. x(t) = sinc (t)
Composite Functions
Given graphs of composite functions in P.5.16 to P.5.21, derive FT.
5.16. Given graph of function x(t) in Fig. 5.2, derive X(ω).
5.17. Given graph of function x(t) in Fig. 5.3 derive X(ω).
5.18. Given graph of RF pulse h(t) where ω 0 = 2πf0 is the RF frequency, a is sine amplitude, and
τ is duration of RF pulse, see Fig. 5.4, derive its Fourier transform.
114
5 CT Fourier Transformation
Fig. 5.2 Example P.5.16
x(t)
2
0
−2
0
Fig. 5.3 Example P.5.17
2
x(t)
2
0
−2
t
t
−2
0
Fig. 5.4 Example P.5.18
2
h(t)
a
0
t
−a
(1/ f0 )
−t/2
Fig. 5.5 Example P.5.19
0
t /2
x(t)
2
0
0
2
4
t
5.19. Given graph of x(t) in Fig. 5.5, derive its Fourier transform.
5.20. Given graph of x(t) in Fig. 5.6, derive its Fourier transform.
5.21. Given graph of x(t) in Fig. 5.7, derive its Fourier transform.
5.22. Given graph of x(t) in Fig. 5.8, derive its Fourier transform.
5.3
Fourier Series
Unless specified otherwise, derive Fourier series within (−π ≤ t ≤ π ) of functions x(t) in P.5.23 to
P.5.33.
5.23. a
5.24.
a
t
π
5.25. | sin t|
5.3
Fourier Series
115
Fig. 5.6 Example P.5.20
x(t)
1
0
t
−4
−2
0
Fig. 5.7 Example P.5.21
2
x(t)
1
0
t
−4
−2
0
Fig. 5.8 Example P.5.22
2
x(t)
1
0
t
−2
−1
0
1
5.26. sin3 t
5.27. et
5.28. a ((t−π/2)/π ) , x ∈ (0, 2π )
5.29. a (t/π )
5.30. t 2 /pi 2
5.31. π 2 /16 π 2 − t 2
5.32. −
2a
|t| + a
π
5.33.
a
t, (0 ≤ t ≤ π )
π
a
− (t − π ) + a, (π ≤ t ≤ 2π )
π
116
5
CT Fourier Transformation
Answers
5.1
Special Functions
5.1. Dirac functions equal zero for all arguments t, except in t = 0 point. Its Fourier transformation
is derived by definition (5.1) as
X(ω) =
def
=
∞
−∞
∞
−∞
x(t) e−j ω t dt = δ(t) = 0 ⇒ t = 0 ∴ ej ω 0 = 1
:1
e−jω t dt =
δ(t)
∞
−∞
δ(t) dt = 1
def
These two functions, Dirac delta and a constant, Fig. 5.9, are said to be “dual” because Fourier
transform of Dirac function is a constant function, and vice versa, i.e. Fourier transform of a constant
function is Dirac function. Physical interpretation is that in order to synthesize Dirac function it is
necessary to add infinitely sin(ω t), and/or cos(ω t), functions of equal powers whose frequencies are
in the range ω ∈ [−∞, ∞].
5.2. Fourier transformation of a delayed Dirac function δ(t − τ ) is derived by definition as
X(ω) =
def
=
∞
−∞
∞
−∞
x(t) e−j ω t dt = δ(t − τ ) = 0 ⇒ t = τ
δ(t − τ ) e−j ω τ dt = e−j ω τ
∞
−∞
δ(t − τ ) dt
def
=1
= (1) e
−j ω τ
That is, see A.5.1, delaying function x(t) by τ in time domain is equivalent to multiplying Fourier
transformation X(ω) by exp(−j ω τ ) in frequency domain (i.e. it adds phase term).
Reminder: Fourier transformation of a delayed
function f (t − τ ) is equivalent to Fourier
transformation of its non–delayed version F f (t) multiplied by the phase delayed term, as
F f (t − τ ) = F f (t) e−j ω τ
x(t)
1
t
0
0
Fig. 5.9 Example P.5.1
X (w )
1
w
0
0
5.1 Special Functions
117
5.3. Fourier transformation of a constant function may be derived, for example, by using “reverse
engineering” technique, i.e by starting with the inverse Fourier transformation integral (5.2)
1
2π
F −1 (X(ω)) = x(t) =
def
∞
X(ω) ej ω t dω
−∞
Given that x(t) = a, its transformation function X(ω) is therefore found as
a=
def
1
2π
∞
X(ω) ej ω t dω
(5.7)
−∞
The idea is to gradually deduce the form of X(ω) until the inverse Fourier equality (5.7) is satisfied,
i.e. its right side of the equation is reduced to a. Evidently, it must be that X(ω) = 2π ×? so that the
already existing 1/2π term may be canceled, as
a=
1
H
2π
H
∞
H
2π
H×? ej ω t dω
−∞
Also, it must be that X(ω) = 2π a×?, i.e.
a=
1
2π
∞
2π a×? ej ω t dω
−∞
∴
1
2π a
a=
2π
∞
?
−∞
ej ω t dω
(5.8)
=1
Note that after 2π terms cancel and (5.8) is reduced to a = a× integral, it is possible only if the
integral equal one. As already seen in A.5.1 the exponential term is cancelled by exploiting the
sampling property of Dirac function because δ(ω) = 0 for all ω = 0, so that the product
δ(ω) e
−j ω t
=
0
δ(ω) e0
∀ω = 0
ω=0
and therefore, by definition
∞
−∞
δ(ω) e−j ω t dω =
∞
−∞
1
def
> dω =
δ(ω) 1
e0
That is to say, it has to be that
1
a=
2π
In conclusion,
∞
2π a δ(ω)
−∞
ej ω t dω
118
5
1
a=
2π
def
∴
a=
1
2π
∞
CT Fourier Transformation
X(ω) ej ω t dω
−∞
∞
2π a δ(ω) ej ω t dω
−∞
∴
X(ω) = 2π a δ(ω)
Thus, Fourier transformation of a constant ‘a’ is indeed proportional to Dirac function (the 2π a term
is the multiplying constant), as derived in A.5.1 and illustrated in Fig. 5.9.
5.4. Amplitude of complex exponential function x(t) = ej ω 0 t is constant (i.e. equal one, as the
radius of a unit circle). Note that ω 0 is a constant as oppose to ω that is variable, thus (see A.5.3)
F −1 (X(ω)) = x(t) =
def
1
2π
∴
e
j ω0 t
1
=
2π
∴
ej ω 0 t =
1
2π
∞
X(ω) ej ω t dω
−∞
∞
X(ω) ej ω t dω
−∞
∞
−∞
2π ×?
ej ω t dω
In order to factor out exp(j ω 0 t) from the right side integral, it is necessary to sample exp(j ω t) at
ω = ω 0 (note that exp(j ω 0 t) is not function of ω). That is equivalent of multiplying exp(j ω t) by
delayed Dirac function δ(ω − ω 0 ), as
e
j ω0 t
1
=
2π
∴
ej ω 0 t =
∞
−∞
∞
1
2π
δ(ω − ω 0 ) ej ω 0 t dω
2π
−∞
∴
e
j ω0 t
2π × δ(ω − ω 0 ) ej ω t dω
=e
j ω0 t
∞
−∞
δ(ω − ω 0 ) dω
def
=1
∴
ej ω 0 t = ej ω 0 t In conclusion, given x(t) = ej ω 0 t ,
5.1 Special Functions
119
X(ω) = F (x(t)) = 2π δ(ω − ω 0 )
5.5. Application of Euler’s formula transforms cosine functions into the sum of complex exponential
functions, as
x(t) = cos(ω 0 t) =
ej ω 0 t + e−j ω 0 t
2
Therefore,
j ω0 t
j ω0 t
+ e−j ω 0 t
e
e
X(ω) = F (x(t)) = F
=F
2
2
−j ω 0 t
e
+F
2
Fourier transformation of a complex exponential function is already solved in A.5.4, thus
j ω0 t
e
F
2
= 2π δ(ω − ω 0 )
∴
−j ω 0 t
e
F
= 2π δ(ω + ω 0 )
2
In conclusion,
X(ω) =
1
2π δ(ω − ω 0 ) + 2π δ(ω + ω 0 ) = π [δ(ω + ω 0 ) + δ(ω − ω 0 )]
2
That is to say, Fourier transform X(ω) of x(t) = cos(ω 0 t) function is the sum of two Dirac functions,
one delayed and the other advanced by ω 0 , Fig. 5.10. Note that, X(ω) ∈ R. This symmetrical form
is known as “bilateral”, where ω 0 argument is mathematically abstract therefore can be both positive
and negative. In engineering applications, specifically if ω 0 argument is used to represent physical
frequency (that can not be negative), bilateral form is transformed into its unilateral equivalent so that
ω 0 ≥ 0.
5.6. Application of Euler formula transforms sine function to the difference of complex exponential
functions.
Fig. 5.10 Example P.5.5
p
t
0
−1
X(w )
x(t)
1
w
0
0
−w 0
0
w0
120
5
x(t) = sin(ω 0 t) =
CT Fourier Transformation
ej ω 0 t − e−j ω 0 t
2j
Therefore,
j ω0 t
j ω0 t
− e−j ω 0 t
e
e
=F
X(ω) = F (x(t)) = F
2j
2j
−j ω 0 t
e
−F
2j
In A.5.4 Fourier transformation of a complex exponential function is already derived, thus
j ω0 t
e
F
2
= 2π δ(ω − ω 0 )
∴
−j ω 0 t
e
F
= 2π δ(ω + ω 0 )
2
In conclusion,
X(ω) =
=
1 2π δ(ω − ω 0 ) − 2π δ(ω + ω 0 )
2j
π j
[δ(ω − ω 0 ) − δ(ω + ω 0 )]
j j
= j π [δ(ω + ω 0 ) − δ(ω − ω 0 )]
That is to say, Fourier transform of sin(ω 0 t) function is also the sum of two Dirac functions, one
delayed and the other advanced by ω 0 . However, there is also phase rotation by π/2 (which is due to
j term) as well as the negative δ(ω − ω 0 ) term. Note that, X(ω) ∈ C.
5.7. Sketch of exponential e−at and u(t) functions product is in Fig. 5.11. By definition, Fourier
integral of x(t), a > 0, is
X(ω) =
def
=
=
∞
x(t) e−j ω t dt
−∞
∞
e−at u(t) e−j ω t dt
−∞
∞
e−at (1) e−j ω t dt
0
Fig. 5.11 Example P.5.7
x(t)
e−at
1
0
u(t)
e−at u(t)
0
t
5.1 Special Functions
121
=
∞
e−(a+j ω )t dt
0
= integration by change of variable
∞
1
=−
e−(a+j ω )t a + jω
0
=−
1
1
0
>
(0 − )
e
a + jω
∴
X(ω) =
1
, ∈C
a + jω
Modulus |X(ω)| and argument θ(ω) = arg X(ω) plots of complex function X(ω) (i.e. Bode plot)
are derived as,
(a) |X(ω)| : real and imaginary parts of an inverse complex function may be derived as
|X(ω)| = 1
a + jω
ω a − j ω a − j ω a
−
j
=
=
a − j ω a 2 + ω2 a 2 + ω2
a 2 + ω2 note negative ℜ(X(ω)) > 0 and ℑ(X(ω)) < 0 parts
ω2
1
a2
a 2
+
ω2
+
=
=
=
2
2
2
2
(a 2 + ω2 )2
(a 2 + ω2 )2
a + ω2 )
(a + ω ) or, alternatively
1 1
1
=
|X(ω)| = =√
2
|a + j ω |
a + jω
a + ω2
It is sufficient to evaluate this even function in only a few important points as
ω = 0 ∴ |X(ω)| =
1
a
1
ω = ±a ∴ |X(ω)| = √
a 2
lim |X(ω)| = 0
ω→±∞
then, its graph is sketched in Fig. 5.12 (top).
(b) θ(ω) = arg X(ω) : as ℜ(X(ω)) > 0 and ℑ(X(ω)) < 0 the argument θ(ω) is in the fourth
quadrant, thus negative.
ℑ(X(ω))
def
θ(ω) = arctan
= arctan
ℜ(X(ω))
ω
− 2 ω
a + ω2 = atan2
(−ω, a) = − arctan
a
a
a 2
+
ω2
122
5
CT Fourier Transformation
Fig. 5.12 Example P.5.7
√
|X (w )|
1/a
2/2a
0
w
p/2
q (w )
p/4
0
−p/4
−p/2
w
−a 0
Fig. 5.13 Example P.5.8
a
x(t)
eat
u(−t)
1
eat u(−t)
0
t
0
It is sufficient to evaluate this odd function in only a few important points as
ω = 0 ∴ θ(0) = 0
ω = +a ∴ θ(a) = − arctan(1) = −
π
4
ω = −a ∴ θ(−a) = − arctan(−1) =
lim θ(ω) = −
ω→∞
lim θ(ω) =
ω→−∞
π
4
π
2
π
2
then, its graph is sketched in Fig. 5.12 (bottom).
5.8. Sketch of exponential x(t) = eat u(−t) functions product is in Fig. 5.13. By definition, Fourier
integral of x(t), a > 0, is
X(ω) =
def
∞
−∞
x(t) e−j ω t dt =
∞
−∞
eat u(−t) e−j ω t dt =
= integration by change of variable
0
1
1
1
(a−j ω )t 0
>
e
e
(
=
−
0)
=
a − jω
a − jω
−∞
∴
X(ω) =
1
, ∈C
a − jω
0
−∞
eat e−j ω t dt =
0
e(a−j ω )t dt
−∞
5.1 Special Functions
123
Fig. 5.14 Example P.5.8
|X (w )|
1/a
√
2/2a
0
w
p/2
q (w )
p/4
0
−p/4
−p/2
w
−a 0
a
Modulus |X(ω)| and argument θ(ω) = arg X(ω) plots of complex function X(ω) (i.e. Bode plot)
are derived as,
(a) |X(ω)| : real and imaginary parts of an inverse complex function may be derived as
|X(ω)| = 1
a − jω
a + j ω a + j ω a
ω =
=
+
j
a + j ω a 2 + ω2 a 2 + ω2
a 2 + ω2 note negative ℜ(X(ω)) > 0 and ℑ(X(ω)) > 0 parts
a2
a 2
+
ω2
ω2
1
=
+
=
=
2
2
2
2
(a 2 + ω2 )2
(a 2 + ω2 )2
a
+ ω2 )
(a + ω ) or, alternatively
1 1
1
=
=√
|X(ω)| = 2
|a − j ω |
a − jω
a + ω2
Module of a complex and its conjugate number are same, compare |X(ω)| in Fig. 5.12 (top) and
Fig. 5.14 (top).
(b) θ(ω) = arg X(ω) : as ℜ(X(ω)) > 0 and ℑ(X(ω)) > 0 the argument θ(ω) is in the first quadrant,
thus positive.
ω
2
ℑ(X(ω))
ω
a
+
ω2 = atan2
= arctan
θ(ω) = arctan
(ω, a) = arctan
a
ℜ(X(ω))
a
a 2
+
ω2
def
It is sufficient to evaluate this odd function in only a few important points as
ω = 0 ∴ θ(0) = 0
ω = +a ∴ θ(a) = arctan(1) =
π
4
ω = −a ∴ θ(−a) = arctan(−1) = −
lim θ(ω) =
ω→∞
π
2
π
4
124
5
lim θ(ω) = −
ω→−∞
CT Fourier Transformation
π
2
then, its graph is sketched in Fig. 5.14 (bottom).
5.9. Given function x(t) = e−a|t| may be written as
⎧
−at
⎪
⎨e ,
x(t) = e−a|t| = 1,
⎪
⎩ at
e ,
t >0
t =0
t <0
as illustrated in Fig. 5.15 (left). Fourier transformation is derived by definition as
X(ω) =
def
∞
x(t) e
−j ω t
−∞
dt =
∞
e
−a|t|
e
−j ω t
−∞
dt =
0
at
e e
−∞
−j ω t
dt +
∞
e−at e−j ω t dt
0
= see P.5.8 and P.5.7
=
1
a +
j
ω + a −
j
ω
1
+
=
a − jω
a + jω
a 2 + ω2
∴
2a
X(ω) = 2
a + ω2
∴ X(ω) ∈ R ⇒
|X(ω)|
θ(ω)
= X(ω)
= 0◦
Being a real function X(ω) its argument θ(ω) = arg X(ω) = 0◦ , i.e. it is a constant. It is sufficient to
evaluate X(ω) function in only a few important points as
2
a
1
ω = ±a ∴ |X(ω)| =
a
ω = 0 ∴ |X(ω)| =
lim |X(ω)| = 0
ω→±∞
then, its graph is sketched in Fig. 5.15 (right).
1
x(t)
e−at
1/a
0
t
0
Fig. 5.15 Example P.5.9
X (w )
2/a
eat
0
−a 0
w
a
5.1 Special Functions
125
5.10. Fourier transformation of (t) function x(t) = a (t/τ ) where t ∈ (−τ/2, τ/2) is
X(ω) =
def
∞
x(t) e
−j ω t
−∞
∞
t
dt =
τ
−∞
e
−j ω t
dt =
τ/2
a e−j ω t dt
−τ/2
= integration by change of variable
a −j ω t τ/2
a −j ω τ/2
=−
=−
− ej ω τ/2
e
e
jω
jω
−τ/2
2a ej ω τ/2 − e−j ω τ/2
a 2 j ω τ/2
− e−j ω τ/2 =
e
jω 2
ω
2j
ωτ ωτ τ sin 2
2a τ/2
sin
= 2a
=
ωτ
ω τ/2
2
2
2
ωτ = a τ sinc
2
=
In conclusion, (t) and sinc (ω) are dual functions in respect to Fourier transformation, Fig. 5.16.
5.11. Fourier transform of the triangle function x(t) = (t/2τ ) where t ∈ (−τ, τ ) may be
performed by exploring derivative properties of Fourier transform of function derivatives (Fig. 5.17).
Reminder: Given f (t) and its Fourier transformation F (ω), derivative property of FT is
F f (x) = j ω F (ω)
F f (x) = (j ω )2 F (ω) = −ω2 F (ω)
F f (3) (x) = (j ω )3 F (ω) = −j ω3 F (ω)
···
or, in general
FT
f (n) (t) → (−j ω )n F (ω)
Π(t /t )
a
t
0
−t/2
Fig. 5.16 Example P.5.10
0
at sinc wt
2
at
t/2
w
0
0
126
5
Fig. 5.17 Example P.5.11
CT Fourier Transformation
x(t)
a
0
t
x (t)
a/t
0
− a/t
t
x (t)
a/t
0
t
−2a/t
−t
0
t
Derivatives of a triangle function may be deduced by graph inspection method.
(a) First derivative x (t) of a piecewise linear triangle function x(t) may be deduced for each interval
separately, Fig. 5.17 (top) and (middle). Horizontal intervals of x(t) are constant, thus their
derivatives equal zero. Two slope intervals are constant, thus by definition of tangent function
in right triangle, as a/(±τ ), i.e.
a
dx
= ; x ∈ (−τ, 0)
dt
τ
dx
a
= − ; x ∈ (0, τ )
dt
τ
(b) Second derivatives x (t) indicate points where x (t) change. Evidently, see Fig. 5.17 (middle),
there are instantaneous changes of x (t) at t = (−τ, 0, τ ). An instantaneous change is modelled
by Dirac function. Sharp step from zero to a/τ at (t = −τ ) generates positive Delta function,
sharp step from zero to −a/τ at t = τ generates negative Delta function, and sharp step from
a/τ to −a/τ at t = 0 generates negative −2a/τ Delta function.
By inspection of Fig. 5.17 (bottom) it follows that
2a
a
d 2x
a
δ(t) + δ(t − τ )
= δ(t + τ ) −
2
dt
τ
τ
τ
a
= (δ(t + τ ) − 2 δ(t) + δ(t − τ ))
τ
(5.9)
Fourier transformations of Dirac delta functions are already derived in A.5.1 and A.5.2. Then, by
knowing FT of each Dirac term in (5.9) and by applying derivative property of FT it follows that
(j ω ) X(ω) = F
2
d 2x
dt 2
=
a jω τ
e
− 2 + e−j ω τ
τ
∴
X(ω) =
2a a jω τ
e
1 − cos ωτ
+ e−j ω τ −2 = 2
2
−ω τ
ω τ
2 cos(ωτ )
2 sin2 (ωτ/2)
5.1 Special Functions
127
Λ 2tt
a
0
−t
0
at sinc2
at
t
t
wt
2
w
0
0
Fig. 5.18 Example P.5.11
Fig. 5.19 Example P.5.12
x(t)
1
u(t)
e−at u(t)
lima→
0
t
0
x(t)
0
t
lima→−
−eat u(−t)
−1
0
ωτ τ 2a
sin (ωτ/2) sin (ωτ/2)
2 sin2
= aτ
τ ωωτ
2
ωτ/2
ωτ/2
ωτ ωτ sinc
= aτ sinc
2
2
2 ωτ
= aτ sinc
2
=
In conclusion, (t) and sinc2 [ω] are dual functions in respect to Fourier transformation, Fig. 5.18.
5.12. Sign function sign (t) is non–convergent for t → ∞, and by consequence direct application
of Fourier integral also produces non–converging result. Some of the possible techniques that may be
used are as follows (Fig. 5.19).
Method 1: sign function may be expressed as
sign (t) = u(t) − u(−t)
In addition, e0 = 1, which can be also written as
lim e±at = 1
a→0
and therefore it is also valid to write
u(t) = (1) u(t) = lim e−at u(t)
a=0
−u(−t) = (−1) u(−t) = − lim eat u(−t)
a=0
128
5
CT Fourier Transformation
By doing so, see Fig. 5.19, sign (t) is replaced with the equivalent x(t) that is convergent, as
x(t) = sign (t) = u(t) − u(−t) = lim e−at u(t) − eat u(−t)
a→0
= see A.5.7 and A.5.8
∴
X(ω) = F x(t) = lim
* + jω
a
0
a→0
=
1
−
1
* − jω
a
0
=
1
1
+
jω
jω
2
jω
Method 2: exploit the derivative property of Fourier transform as
x(t) = sign (t) = 2u(t) − 1
see A.2.11 and A.5.1
∴
dx(t)
2
FT
= 2 δ(t) −→ j ω X(ω) = 2 ∴ X(ω) =
dt
jω
5.13. Step function u(t) is nom–convergent for t → ∞, and by consequence direct application of
Fourier integral also produces non–converging result. However, see A.2.30 and A.5.12,
x(t) = u(t) =
∴
1 + sign (t)
2
1 sign (t)
+
= F ( constant ) + F (sign (t) )
2
2
1
1 1
1 2
=F
+ F sign (t) = 2π δ(ω) +
2
2
2
2 j ω
X(ω) = F
= π δ(t) +
1
jω
5.14. Derivative of Gaussian function is
x(t) = e−t
2
∴ x (t) = −2t e−t = −2t x(t)
2
(5.10)
therefore, FT of left and right sides in (5.10) is
F x (t) = −2F tx(t)
(5.11)
after substituting FT derivative property F x (t) = j ω X(ω) in (5.11) it follows that
j ω X(ω) = −2F tx(t)
(5.12)
5.1 Special Functions
129
thus it is necessary to derive F tx(t) . By definition,
X(ω) = F x(t) =
∞
x(t) e−j ω t dt
−∞
∴
left and right side derivative, note that x(t) = f (ω)
∞
∞
d
d −j ω t
d
−j ω t
x(t) e
dt =
x(t)
dt
X(ω) =
e
dω
dω −∞
dω
−∞
∞
def
= −j ω
t x(t) e−j ω t dt = −j ω F t x(t)
−∞
∴
F t x(t) =
1 j
j
X (ω) = X (ω)
−j ω j
ω
(5.13)
Substituting (5.13) into (5.12) results in a first order differential equation with constant coefficients
relative to X(ω) as
j ω X(ω) = −2
X (ω) = −
j
X (ω)
ω
ω2
X(ω)
2
X(ω) = y
This differential equation is solved by the separation of variables method as
dy
ω
=− y ⇒
dω
2
dy
1
=−
y
2
1
ω dω ⇒ ln y − ln C = − ω2 ∴
4
Return to X(ω),
X(ω) =
def
∞
−∞
ω2
e−x e−j ω t dt = C e− 4
2
Simplest initial condition is X(0) = C, so that e0 = 1 and
C=
∞
−∞
e−x dt = see Vol. II
2
=
√
π
In conclusion,
X(ω) =
√
π e−ω /4
2
5.15. Fourier transform of sinc (t) function may be derived as as follows
X(ω) =
def
=
∞
−∞
∞
1
−∞ ω 0 t
∞
sin(ω 0 t) −j ω t
dt
e
ω0 t
−∞
sin(ω 0 t) cos(ω t) − j sin(ω t) dt
x(t) e−j ω t dt =
y = C e−ω /4
2
130
5
=
∞
1
−∞ ω 0 t
sin(ω 0 t) cos(ω t) dt − j
CT Fourier Transformation
∞
1
sin(ω 0 t) sin(ω t) dt
ω
0 t
−∞
= I1 − j I2
See Vol. II where it is shown that
∞
0
e−ω 0 t y dy =
1
ω0 t
(5.14)
That is to say, the two integrals I1 and I2 are solved separately as follows. First, solution to I1
I1 =
∞
−∞
1
ω0 t
sin(ω 0 t) cos(ω t) dt
= substitute (5.14) and use trig identity for product, as
∞ ∞
1
−ω 0 t y
sin (ω 0 + ω) t] + sin[(ω 0 − ω) t dt
e
dy
=
2
−∞ 0
∞ ∞
1
=
e−ω 0 t y dy sin (ω 0 + ω) t] + sin[(ω 0 − ω) t dt
2 −∞ 0
∞
∞
∞
∞
1
1
=
dy
e−ω 0 y t sin (ω + ω 0 ) t dt −
dy
e−ω 0 y t sin (ω − ω 0 ) t dt
2 −∞
2 −∞
0
0
= see: Vol. II =
ω + ω0
1 ∞
ω − ω0
1 ∞
dy −
dy
=
2 −∞ (ω + ω 0 )2 + (ω 0 y)2
2 −∞ (ω − ω 0 )2 + (ω 0 y)2
see: Vol. II, as well as ω, ω 0 = f (y) i.e. they are constants relative to y
π
=
sign (ω + ω 0 ) − sign (ω − ω 0 )
2ω 0
The last result defines (ω/2ω 0 ) as illustrated in Fig. 5.20, which is expected as dual function of
sinc (t) function.
Integral I2 is solved by following the same idea as for I1 ,
∞
1
sin(ω 0 t) sin(ω t) dt
ω
0 t
−∞
∞ ∞
1
e−ω 0 t y dy cos (ω − ω 0 ) t − cos (ω + ω 0 ) t dt
=
2 −∞ 0
∞
∞
∞
∞
1
−ω 0 y t
−ω 0 y t
=
dy
e
cos (ω − ω 0 ) t dt −
dy
e
cos (ω + ω 0 ) t dt
2
−∞
0
−∞
0
∞
∞
ω0 y
ω0 y
1
dy −
dy
=
2
2
2
2
2
−∞ (ω − ω 0 ) + (ω 0 y)
−∞ (ω + ω 0 ) + (ω 0 y)
I2 =
5.1 Special Functions
131
(a)
x(w )
1
sign(w + w 0 )
0
x(w )
sign (w + w 0 ) − sign (w − w 0 )
0
−w 0
−w 0
w0
0
(b)
0
(c)
w
−1
1
2
x(w )
p
w0
−sign(w − w 0 )
0
(d)
w0
w
x(w )
p
2w 0 (sign (w + w 0 ) − sign (w − w 0 ))
w
−1
0
−w 0
0
w0
−w 0
0
w0
w
Fig. 5.20 Example P.5.15
⎧
⎪
⎪
⎪
⎨x
= (ω ± ω 0 )2 + (ω 0 y)2 ∴
dx = 2ω20 y dy ∴ ω 0 y dy =
⎫
1
⎪
dx ⎪
⎪
⎬
2ω 0
∞
∞
⎪
1
dx
1
⎪
⎪
ln x =
∴
⎭
2ω 0 −∞ x
2ω 0
−∞
∞
∞ 1
2
2 2
2 − ln (ω + ω 0 ) + (ω 0 y) ln (ω − ω 0 ) + (ω 0 y) =
4ω 0
−∞
−∞
∞
1
=
ln (ω − ω 0 )2 + (ω 0 y)2 − ln (ω − ω 0 )2 + (ω 0 y)2 4ω 0
−∞
∞
− ln (ω + ω 0 )2 + (ω 0 y)2 + ln (ω + ω 0 )2 + (ω 0 y)2 ⎪
⎪
⎪
⎩
−∞
#
$
a
ln a + ln b = ln(ab); ln a − ln b = ln ; ln a b = b ln a
b
%
&
∞
(ω − ω 0 )2 + (ω 0 y)2 (ω − ω 0 )2 + (ω 0 y)2 1
ln
− ln
=
4ω 0
(ω + ω 0 )2 + (ω 0 y)2 (ω + ω 0 )2 + (ω 0 y)2 −∞
'
(
(ω − ω 0 )2 + (ω 0 y)2
(ω − ω 0 )2 + (ω 0 y)2
1
lim ln
− lim ln
=
y→−∞
4ω 0 y→∞
(ω + ω 0 )2 + (ω 0 y)2
(ω + ω 0 )2 + (ω 0 y)2
' ( ∞
1
(ω − ω 0 )2 + (ω 0 y)2
(ω − ω 0 )2 + (ω 0 y)2
ln lim
− ln lim
=
=
2
2
2
2
y→∞
y→−∞
4ω 0
(ω + ω 0 ) + (ω 0 y)
(ω + ω 0 ) + (ω 0 y)
∞
' 2 2 (
2ω 0 y
2ω 0 y
∞
1
l’H
− ln lim
=
=
ln lim
2
2
y→∞
y→−∞
4ω 0
∞
2ω 0 y
2ω 0 y
%
&
%
&)
2
2
2ω
2ω
1
l’H
=
ln lim 0 − ln lim 0
2
2
y→∞ 2ω
y→−∞ 2ω
4ω 0
0
0
=
l’H
1
[ln(1) − ln(1)] = 0
4ω 0
132
5
CT Fourier Transformation
In conclusion,
X(ω) =
5.2
0
sin(ω 0 t) −j ω t
π
>
e
sign
+
ω
I
−
sign
−
ω
dt = I1 + j =
I
=
(ω
)
(ω
)
2
1
0
0
ω0 t
2ω 0
−∞
∞
Composite Functions
5.16. Given x(t) may be decomposed into the sum of two (t) functions, x1 (t) and x2 (t) as
illustrated in Fig. 5.21 (left), where
t
x1 (t) = 4
∴
x(t) = t
4
and x2 (t) = +
t
2
t
2
the total Fourier transformation is therefore the sum of two FT, see P.5.10, as
t
X(ω) = F 4
t
+F 2
= 4 sinc (2ω) + 2 sinc (ω)
5.17. Using graphical derivation method and derivative properties of FT it follows that,
x(t) = −2 sign (t − 2) ∴ x (t) = −4δ(t − 2) ∴ j ω X(ω) = −4e−2j ω
∴
X(ω) = −
4 −2j ω
e
jω
2
0
x(t)
t
−2
0
2
0
0
t
−a
x1 (t)
t
−2
0
2
0
2
x(t)
a
Π (t/t )
1
2
x2 (t)
t
−2
0
Fig. 5.21 Examples P.5.16, P.5.18
2
0
−t/2
0
t/2
t
5.2 Composite Functions
133
5.18. Given RF pulse h(t) may be decomposed into the product of x(t) = a cos(j ω 0 t) and (t/τ )
functions, Fig. 5.21 (right), i.e.
h(t) = a cos(j ω 0 t) (t/τ )
where, ω 0 = 2π f0 is the RF sine frequency, and τ is duration of RF pulse.
By using Euler’s formula it follows that,
x(t) = a cos(ω 0 t) =
a
exp(j ω 0 t) + exp(−j ω 0 t)
2
then, Fourier transformation is derived by applying the frequency shifting and time scaling properties,
Reminder:
Frequency shifting property: e−j ω 0 t x(t) → X(ω − ω 0 )
FT
1 ω
X
|a|
a
FT
Time scaling property: x(at) →
that is to say
FT
FT
(t) → sinc (t) ∴ (t) exp(j ω 0 t) → sinc (ω − ω 0 )
∴
h(t) =
a
a
(t/τ ) exp(j ω 0 t) + (t/τ ) exp(−j ω 0 t)
2
2
∴
aτ
H (ω) = F h(t) =
2
sinc (ω) τ (ω − ω 0 ) + sinc (ω) τ (ω + ω 0 )
That is to say, if ω 0 τ 1 then, see Fig. 5.22
⎧ aτ
⎪
sinc (ω) τ (ω − ω 0 )
⎪
⎨ 2
H (ω) = 0
⎪
⎪
⎩ aτ sinc (ω) τ (ω + ω )
0
2
; ω>0
; ω=0
; ω<0
Fig. 5.22 Example P.5.18
H (w )
at
2
0
w
w0
0
w0
134
5
Fig. 5.23 Example P.5.19
CT Fourier Transformation
Λ(t − 2)
1
0
t
Π (t − 2)
1
0
0
2
4
t
5.19. Given function x(t) may be decomposed into the sum of port and triangle funcitons functions,
Fig. 5.23, i.e.
x(t) = (t/4 − 2) + (t/2 − 2)
then, Fourier transformation is derived by applying the time shifting and scaling properties as
F x(t) = F ( (t/4 − 2)) + F ((t/2 − 2))
= 4e−j 2ω sinc (4ω) + 2e−j 2ω sinc2 (2ω)
5.20. Given function x(t) in Fig. 5.6 may be decomposed into the sum of two port functions, as
x(t) = (t/2 − 1) + (t/2 + 3)
that is to say
F x(t) = F ( (t/2 − 1)) + F ( (t/2 + 3))
= 2e−j ω sinc (2ω) + 2ej 3ω sinc (3ω)
5.21. Given x(t) in Fig. 5.7, it may be decomposed into the sum of two triangle functions, as
x(t) = t/2 − 1 + t/2 + 3
that is to say, after applying the time scaling and delay properties of FT,
F x(t) = F t/2 − 1 + F t/2 + 3
= 2e−j ω sinc2 (2ω) + 2ej 3ω sinc2 (3ω)
5.22. Given function x(t) in Fig. 5.8 may be decomposed into the sum of two triangle functions, see
Fig. 5.24, as
x(t) = t/2 − 1 + t/2
5.3 Fourier Series
135
Fig. 5.24 Example P.5.22
x(t)
1
0
t
−2
−1
0
1
that is to say, after applying the time scaling and delay properties of FT,
F x(t) = F t/2 − 1 + F t/2
= 2e−j ω sinc2 (2ω) + 2sinc2 (2ω)
= 2sinc2 (2ω) e−j ω + 1
5.3
Fourier Series
5.23. Given f (t) = a, in (−π, π ) interval, coefficients of Fourier series are
a π
a
a dt =
π = 2a
t = 2
π −π π
−π
π
1 π
1 π
a
sin(nt) f (t) cos(nt) dt =
a cos(nt) dt =
an =
π −π
π −π
nπ
−π
1
a0 =
π
π
1
f (t) dt =
π
−π
π
a : 0
:0
sin(−nπ
=0
sin(nπ
) −
)
nπ
π
1 π
1 π
a
f (t) sin(nt) dt =
a sin(nt) dt = −
bn =
cos(nt) π −π
π −π
nπ
−π
a ) − ) = 0
=−
cos(nπ
cos(nπ
nπ
=
Therefore, Fourier series of f (t) = a is
∞
∞
∞
∞
*0 *0
2A a a0 an cos(nt) +
bn sin(nt) =
0cos(nt) +
0 sin(nt)
f (t) =
+
+
2
2A
n=1
n=1
n=1
n=1
=a
That is to say, Fourier series of a constant is simply the constant itself.
5.24. Given sawtooth f (t) = (a/π ) t, in (−π, π ) interval, coefficients of Fourier series are
136
5
CT Fourier Transformation
a 2
a 2 π
a
t =
π − (−π )2 = 0
t dt =
2
2
2π
2π
−π π
−π
π
π
a
1
1
t cos(nt) dt = see Vol. II
an =
f (t) cos(nt) dt =
π −π
π −π π
π
a = 2 2 nt sin(nt) + cos(nt) π n
−π
1
a0 =
π
=
π
1
f (t) dt =
π
−π
π
a :0
:0
) − n(−π )
cos(nπ
cos(−nπ
nπ sin(−nπ
sin(nπ
) +
) − )
π 2 n2
=0
1
π
π
1
π
π
a
t sin(nt) dt = see Vol. II
π
−π
−π
π
a = 2 2 −nt cos(nt) + sin(nt) nπ
bn =
f (t) sin(nt) dt =
−π
a : 0
:0
sin(−nπ
= 2 2 −nπ cos(nπ ) + sin(nπ
) + n(−π ) cos(−nπ ) − )
nπ
2n
2a
πa
=− cos(nπ ) = −
(−1)n
2
2
nπ
n π (−1)n
Therefore, Fourier series of sawtooth f (t) = (a/π ) t is
f (t) =
∞
∞
∞
∞
a0 0 2a
an cos(nt) +
bn sin(nt) = +
0 cos(nt) +
−
+
(−1)n sin(nt)
2
2
nπ
n=1
n=1
n=1
n=1
∞
=−
=
2a (−1)n sin(nt)
nπ n=1
2a
2a
2a
sin(t) −
sin(2t) +
sin(3t) − · · ·
π
2π
3π
Note that sawtooth f (t) = (a/π ) tfunction is odd, i.e. f (x) = −f (−x), and therefore all an = 0.
This is a general property of all odd functions. The sum of first five terms of the Fourier series f (t)
is already close approximation of ideal sawtooth function, see Fig. 5.25.
Fig. 5.25 Example P.5.24
a
f (t)
0
t
−a
−2p
−p
0
p
2p
5.3 Fourier Series
137
5.25. Given absolute sines f (t) = | sin t|, in (−π, π ) interval, note that it is an even function and
always positive as | sin t| ≥ 0 (a cosines is also even function). Thus, integrals may be calculated only
in the positive (0, π ) interval then multiplied by two so that coefficients of Fourier series are
1
a0 =
π
π
1
f (t) dt =
π
−π
π
2
| sin t| dt =
π
−π
π
0
π
2
sin t dt =
(− cos t) π
0
2
4
(− cos π + cos 0) =
π
π
1 π
1 π
2 π
f (t) cos(nt) dt =
| sin t| cos(nt) dt =
sin t cos(nt) dt
an =
π −π
π −π
π 0
(
'
1 π
1 π
1
= sin x cos y = sin(x + y) + sin(x − y) =
sin(t + nt) dt +
sin(t − nt) dt
π 0
π 0
2
(
'
1
= t ± nt = x ∴ (1 ± n) dt = dx ∴ dt =
dx
1±n
π
π
1 1
1 1
=−
cos(t + nt) −
cos(t − nt) π 1+n
π 1−n
0
0
1 1
1 1 cos (1 + n)π − cos (1 + n) · 0 −
cos (1 − n)π − cos (1 − n) · 0
=−
π 1+n
π 1−n
=
=1
(−1)n
=1
(−1)n
1 (−1) + 1
2 (−1) + 1
1 (−1) + 1
−
=
, (n ≥ 2)
π 1+n
π 1−n
π (1 + n)(1 − n)
1 π
1 π
2 π
f (t) sin(nt) dt =
| sin t| sin(nt) dt =
sin t sin(nt) dt
bn =
π −π
π −π
π 0
(
'
1 π
1 π
1
1
cos(t − nt) dt −
cos(t + nt) dt
= sin x sin y = cos(x − y) − cos(x + y) =
π 0
π 0
2
2
'
(
1
= t ± nt = x ∴ (1 ± n) dt = dx ∴ dt =
dx
1±n
π
π
1 1
1 1
=
sin(t − nt) −
sin(t + nt) π 1−n
π 1+n
0
0
1 1 1 1
sin (1 − n)π − sin (1 − n) · 0 −
sin (1 + n)π − sin (1 + n) · 0
=
π 1−n
π 1+n
=−
n
n
=0
n
=0
= 0, (n ≥ 2)
Therefore, Fourier series of absolute sines f (t) = | sin t| is
∞
∞
2 2 (−1)n + 1
cos(nt) +
f (t) = +
0 sin(nt)
π
π (1 + n)(1 − n)
n=2
n=2
=
∞
2 2 (−1)n + 1
+
cos(nt)
π
π (1 + n)(1 − n)
n=2
=0
=0
138
5
Fig. 5.26 Example P.5.25
CT Fourier Transformation
f (t)
a
0
−p
0
p
t
:0
3 2
(−1)
+1
4
2 4
4
= −
cos(2t) +
cos(3t) −
cos(4t) + 0 −
cos(6t) + 0 − · · ·
π
3π
π (1 + 3)(1 − 3)
15π
35π
Note that absolute sines f (t) = | sin t| function is even, i.e. f (x) = f (−x), and therefore all bn = 0.
This is a general property of all even functions. The sum of first four terms of the Fourier series f (t)
is already close approximation of ideal absolute sines function, see Fig. 5.26.
5.26. Given sines cube f (t) = sin3 t, in (−π, π ) interval. An attempt to calculate coefficients of
Fourier series by definition may turn out to be very long and tedious, as an and bn integrals are
π
π
1 π
1
1
3
3 f (t) dt =
sin t dt = see Vol. II =
(− cos t) + cos t π −π
π
3
−π
−π
−π
1
1 3 ) + =
cos
π +
cos(−π
cos3 π − cos
(−π )
−
π
3
1
a0 =
π
=0
an =
1
π
1
bn =
π
π
π
−π
f (t) cos(nt) dt =
π
1
π
1
f (t) sin(nt) dt =
π
−π
π
sin3 t cos(nt) dt
−π
π
sin3 t sin(nt) dt
−π
which is doable by multiple methods but not always trivial to resolve.
Instead, by noting that sin3 t is already trigonometric function, first it may be transformed by
applying identities, as
sin t sin t =
1
1
cos(t − t) − cos(t + t) = 1 − cos(2t)
2
2
∴
1
1
1
sin3 t = sin2 t sin t = 1 − cos(2t) sin t = sin t − cos(2t) sin t
2
2
2
1 1
1
sin(2t + t) − sin(2t − t)
= sin t −
2
2 2
=
1
1
1
sin t − sin 3t + sin t
2
4
4
5.3 Fourier Series
1
139
f (t)
f (t)
t
0
−1
t
0
−p
0
−p
p
0
p
Fig. 5.27 Examples P.5.26, P.5.27
Therefore, Fourier series of sines cube f (t) = sin3 t consists of only two non–zero terms , as
f (t) =
3
1
sin t − sin 3t
4
4
and all the other terms are eliminated by zero coefficients. Which is to say that Fourier series does
not have to be infinitely long. In addition, this finite series contains the only terms of sines that are
needed to exactly create sin3 t waveform, see Fig. 5.27 (left).
5.27. Given exponential function f (t) = et , in (−π, π ) interval, note that exponential function is
neither odd nor even. Coefficients of Fourier series are
1 π
1 π t
1 t π
1
a0 =
(e ) = (eπ − e−π )
f (t) dt =
e dt =
π −π
π −π
π
π
−π
π
π
1
1
f (t) cos(nt) dt =
et cos(nt) dt = see Vol. II
an =
π −π
π −π
π
π
1
1 n
−π
−π
e
+
e
sin(nπ
)
cos(nπ
)
=
+
e
−
e
π n2 + 1
n2 + 1
=0
(−1)n
1 (−1)n π
−π
e
−
e
π n2 + 1
1 π
1 π t
bn =
f (t) sin(nt) dt =
e sin(nt) dt = see Vol. II
π −π
π −π
1
n
1
− 2
cos(nπ ) eπ − e−π + 2
sin(nπ ) eπ + e−π
=
π
n +1
n +1
=
(−1)n
=−
=0
1
n
(−1)n eπ − e−π
2
π n +1
Therefore, Fourier series of exponential function f (t) = et is
∞
∞
n
1 π
1 (−1)n π
1
−π
(e − e−π ) +
e
(−1)n eπ − e−π sin(nt)
−
e
cos(nt)
−
2
2
2π
π n +1
π n +1
n=1
n=1
∞
∞
1
n
1
1 π
+
(−1)n cos(nt) −
(−1)n sin(nt)
= (e − e−π )
2+1
π
2 n=1 n2 + 1
n
n=1
f (t) =
140
5
CT Fourier Transformation
Note that exponential function f (t) = et function is neither even nor odd, and therefore an = 0 and
bn = 0. This is a general property of non–even non–odd functions. The sum of first several terms of
the Fourier series f (t) is shown in Fig. 5.27 (right). Note that many more terms are needed for good
approximation close to the interval’s ends.
5.28. Given port function a ((t−π/2)/π ), in (0, 2π ) interval, (it is not even). Thus,
a0 =
π
1
an =
π
1
π
π
−π
f (t) dt =
1
f (t) cos(nt) dt =
π
−π
1
π
π
0
π
a dt =
0
a
π =a
Z
π
Z
π
1a
sin(nt) a cos(nt) dt =
πn
0
1a
2a
=
sin(0) − sin(−nπ ) =
sin(nπ ) = 0
πn
πn
π
1 π
1 π
1a
f (t) sin(nt) dt =
a sin(nt) dt = −
cos(nt) bn =
π −π
π 0
πn
0
=−
1a
1a
cos(nπ ) − cos(0) = −
(−1)n − 1
πn
πn
Therefore, Fourier series of port function a ((t−π/2)/π ) is
f (t) =
∞
∞
1 a
a 0 cos(nt) +
+
1 − (−1)n sin(nt)
2 n=1
π n=1 n
∞
a 1
a
1 − (−1)n sin(nt)
= +
2 π n=1 n
=
a
2a
2a
2a
+
sin(t) + 0 +
sin(3t) + 0 +
sin(5t) + · · ·
2
π
3π
5π
Note that port function a ((t−π/2)/π ) function is not even. The sum of first few terms of the Fourier
series f (t) is close approximation of an ideal port function, see Fig. 5.28.
Fig. 5.28 Example P.5.28
a
f (t)
0
0
p
2p
t
5.3 Fourier Series
141
5.29. Given port function a (t/π ), in (−π, π ) interval, note that it is an even function. Thus,
coefficients of Fourier series are
a0 =
1
π
1
an =
π
π
−π
f (t) dt =
π
1
π
π/2
−π/2
a dt =
1
f (t) cos(nt) dt =
π
−π
a
π =a
Z
π
Z
π/2
1a
a cos(nt) dt =
sin(nt) πn
−π/2
−π/2
π/2
1a
2a
sin(nπ/2) − sin(−nπ/2) =
sin(nπ/2)
πn
πn
π/2
1 π
1 π/2
1a
f (t) sin(nt) dt =
a sin(nt) dt = −
bn =
cos(nt) π −π
π −π/2
πn
−π/2
=
=−
1a
1a
cos(nπ/2) − cos(−nπ/2) = −
cos(nπ/2)
cos(nπ/2)
− = 0
πn
πn
Therefore, Fourier series of port function a (t/π ) is
f (t) =
∞
∞
a 2a
0 sin(nt)
+
sin(nπ/2) cos(nt) +
2 n=1 π n
n=1
=
∞
a 2a
+
sin(nπ/2) cos(nt)
2 n=1 π n
=
2a
2a
2a
a
+
cos(t) + 0 −
cos(3t) + 0 +
cos(5t) + · · ·
2
π
3π
5π
Note that port function a (t/π ) is even, i.e. f (x) = f (−x), and therefore all bn = 0. This is a
general property of all even functions. The sum of first few terms of the Fourier series f (t) is close
approximation of an ideal port function, see Fig. 5.29.
5.30. Given square function t 2/π 2 , in (−π, π ) interval, note that it is an even function. Thus,
coefficients of Fourier series are
2
1 π
1 π t2
1 3 π
1 3
2
π − (−π )3 =
f (t) dt =
dt =
t =
π3 =
a0 =
2
3
3
3
π −π
π −π π
3π
3π
3
3
π
−π
π
π
1
1
an =
f (t) cos(nt) dt = 3
t 2 cos(nt) dt = see Vol. II
π −π
π −π
Fig. 5.29 Example P.5.29
f (t)
a
0
−p
0
p
t
142
5
CT Fourier Transformation
Fig. 5.30 Example P.5.30
f (t)
1
0
−p
0
p
t
⎛
⎞
1 4
1 ⎝2 2
4
4
π sin(nπ ) − 3 sin(nπ ) + 2 π cos(nπ ) ⎠ = 2 2 (−1)n
= 3
n
n
π
n
π n
bn =
1
π
π
−π
=0
f (t) sin(nt) dt =
=0
1
π
π
−π
(−1)n
t 2 sin(nt) dt = see Vol. II = 0
Therefore, Fourier series of square function t 2/π 2 is
∞
∞
1 1 4
n
(−1) cos(nt) +
0 sin(nt)
f (t) = +
3 n=1 π 2 n2
n=1
=
∞
1 1 4
+
(−1)n cos(nt)
3 n=1 π 2 n2
=
4
4 1
4 1
1
4 1
−
cos(4t) + · · ·
cos(t) + 2 cos(2t) − 2 cos(3t) + 2
3 π2
π 4
π 9
π 16
Note that square function t 2/π 2 is even, i.e. f (x) = f (−x), and therefore all bn = 0. This is a
general property of all even functions. The sum of first few terms of the Fourier series f (t) is close
approximation of an ideal square function, see Fig. 5.30.
5.31. Given square function π 2/16 π 2 − x 2 , in (−π, π ) interval, note that it is an even function.
Thus, coefficients of Fourier series are
1
a0 =
π
π
1
f (t) dt =
π
−π
π
π3
π2 2
π − t 2 dt =
16
−π 16
π
π
dt −
16
−π
π4
π4
π4
−
=
8
24
12
1 π
1 π π2 2
π − t 2 cos(nt) dt
f (t) cos(nt) dt =
an =
π −π
π −π 16
π
:0 π
π3 π
=
dt −
t 2 cos(nt) dt = see P.5.30
cos(nt)
16−π
16 −π
=
=−
π2 1
(−1)n
4 n2
π
t 2 dt
−π
5.3 Fourier Series
143
Fig. 5.31 Example P.5.31
f (t)
2p
p
t
0
−p
1
π
bn =
π
−π
p
π
π2 2
π − t 2 sin(nt) dt
−π 16
1
π
f (t) sin(nt) dt =
0
*0
:0
π 3 π π π 2 =
sin(nt)
dt
−
tsin(nt)
dt = see P.5.30 = 0
16 −π
16−π
Therefore, Fourier series of square function π 2/16 π 2 − x 2 is
f (t) =
=
∞
∞
π4 π2 1
n
(−1)
cos(nt)
+
0 sin(nt)
−
24 n=1 4 n2
n=1
∞
π4 π2 1
(−1)n cos(nt)
−
24 n=1 4 n2
π4
π2
π2 1
π2 1
π2 1
+
cos(t) −
cos(2t) +
cos(3t) −
cos(4t) + · · ·
24
4
4 4
4 9
4 16
Note that square function π 2/16 π 2 − x 2 is even, i.e. f (x) = f (−x), and therefore all bn = 0. This is
a general property of all even functions. The sum of first few terms of the Fourier series f (t) is close
approximation of an ideal square function, see Fig. 5.31.
=
5.32. Given triangle function −2a/π |t| + a, in (−π, π ) interval. Thus, coefficients of Fourier series
are
1 π
1 π
2a
2a
2 π
f (t) dt =
a0 =
− |t| + a dt = even function =
− t + a dt
π −π
π −π
π
π 0
π
π
π
π
4 2 a 1 2 2a
2a
2a
4a
t +
t dt +
dt = − 2
π = − 2 π 2 + 2a = 0
=− 2
π 0
π 0
π 2
π
π
0
Integral of absolute linear function and cosine product should be split in two intervals, as
an =
1
π
π
−π
2a
=− 2
π
2a
=− 2
π
f (t) cos(nt) dt =
π
1
π
a
|t| cos(nt) dt +
π
−π
0
π
−
−π
π
2a
|t| + a
π
cos(nt) dt
cos(nt) dt
−π
2a
(−t) cos(nt) dt − 2
π
−π
π
0
a
t cos(nt) dt +
π
π
−π
cos(nt) dt = see Vol. II
144
5
0
π
CT Fourier Transformation
a ) − sin(−nπ
sin(nπ
)
nπ
−π
0
0
π
1
2a
2a 1
1
1
= − 2 − t sin(nt) − 2 cos(nt) − 2
t sin(nt) + 2 cos(nt) π
n
n
π
n
n
−π
0
1
1
2a
1
2a
1
−
−
=− 2 − 0 sin(0)
cos(0)
+
sin(−nπ
)
−
cos(−nπ
)
(−π
)
π
n
n2
π2
n
n2
=−
2a
π2
2a
π2
(−t) cos(nt) dt −
t cos(nt) dt +
(−1)n
2a 1
+ 1 cos(0)
) + 1 cos(nπ ) + 2a 1 0 sin(0)
sin(nπ
π
π2 n
n2
π2 n
n2
−
(−1)n
2a
2a
2a
4a
4a
2a
− 2 2 (−1)n − 2 2 (−1)n + 2 2 = − 2 2 (−1)n + 2 2
2
2
nπ
nπ
nπ
n π
nπ
n π
4a = 2 2 1 − (−1)n
nπ
=
Integral of absolute linear function and sine product should be split in two intervals, as
bn =
1
π
=−
π
−π
2a
π2
2a
=− 2
π
f (t) sin(nt) dt =
π
−π
1
π
|t| sin(nt) dt +
0
−π
a
π
π
−π
−
π
2a
|t| + a
π
sin(nt) dt
−π
(−t) sin(nt) dt +
π
0
π
a t sin(nt) dt +
− cos(nt) nπ
−π
= see Vol. II
2a
1
1
=− 2
t cos(nt) − 2 sin(nt)
π
n
n
+
sin(nt) dt
0
+ − 1 t cos(nt) + 1 sin(nt)
n
n2
−π
π 0
a 1
− cos(nπ ) − − cos(−nπ )
πn
=0
⎡⎛
⎞ ⎛
⎞
2a ⎣⎝ 1
1
1
1
− 2 sin(0) ⎠ − ⎝ (−π ) cos(−nπ ) − 2 sin(−nπ ) ⎠
=− 2
0 cos(0)
π
n
n
n
n
=0
2a
=− 2
π
=0
⎞
=0
⎛
=0
⎞⎤
1
1
1
1
+ 2 sin(0) ⎠⎦
0 cos(0)
+ ⎝− π cos(nπ ) + 2 sin(nπ ) ⎠ − ⎝− n
n
n
n
⎛
=0
1
1
2a ) − )
− (−π ) cos(−nπ ) − π cos(nπ ) = −
cos(nπ
cos(nπ
n
n
nπ
5.3 Fourier Series
145
Fig. 5.32 Example P.5.32
f (t)
a
t
0
−a
−p
0
p
Therefore, Fourier series of triangle function −2a/π |t| + a is
f (t) =
=
∞
4a 1 − (−1)n cos(nt)
2
2
nπ
n=1
8a
8a
8a
cos(t) + 0 + 2 2 cos(3t) + 0 + 2 2 cos(5t) + · · ·
π2
3 π
5π
Note that triangle function −2a/π |t| + a is even, i.e. f (x) = f (−x), and therefore all bn = 0. This is
a general property of all even functions. The sum of first few terms of the Fourier series f (t) is close
approximation of an ideal triangle function, see Fig. 5.32.
5.33. Given triangle function
x(t) =
⎧a
⎪
t,
⎪
⎨π
(0 ≤ t ≤ π )
⎪
⎪
⎩ − a (t − π ) + a,
π
(π ≤ t ≤ 2π )
in (0, 2π ) interval. Thus, coefficients of Fourier series are
1 2π a
1 π a
t dt +
f (t) dt =
− (t − π ) + a dt
π 0 π
π π
π
−π
π
2π π
2π
a
1
1
a
a
a
2a 2π
− (t − π ) + a dt = 2
=
t dt +
t dt − 2
t dt +
dt
π 0 π
π π
π
π 0
π π
π π
a 4π 2
π2
2a
a
3a
a π2
− 2
−
+
π= −
+ 2a
= 2
π 2
π
2
2
π
2
2
1
a0 =
π
π
=a
Integrals an of piecewise linear segments may be resolved separately, as
an =
1
π
π
−π
f (t) cos(nt) dt =
a
π2
π
t cos(nt) dt −
0
= see Vol. II and P.5.32
a a = − 2 2 1 − (−1)n − 2 2 1 − (−1)n + 0
nπ
nπ
2a = − 2 2 1 − (−1)n
nπ
a
π2
2π
π
t cos(nt) dt +
2a
π
2π
cos(nt) dt
π
146
5
Fig. 5.33 Example P.5.33
CT Fourier Transformation
f (t)
a
t
0
0
2p
p
Integrals bn of piecewise linear segments may also be resolved separately, as
bn =
1
π
π
−π
f (t) sin(nt) dt =
a
π2
π
t sin(nt) dt −
0
a
π2
2π
t sin(nt) dt +
π
2a
π
2π
sin(nt) dt
π
2a a
a (−1)n −
(−1)n − 2 +
(−1)n − 1
nπ
nπ
nπ
2a
a
a
2a 2a
n
n − (−1)
n + S + (−1)
−S
= − (−1)
S
S
nπ
nπ
nπS nπ
nπ
S
=−
=0
Therefore, Fourier series of this triangle function is
f (t) =
=
∞
∞
a 2a n
0 sin(nt)
1
−
(−1)
cos(nt)
−
−
2 n=1 n2 π 2
n=1
4a
4a
a
4a
cos(t) − 2 2 cos(3t) − 2 2 cos(5t) + · · ·
−
2 π2
3π
5 π
The sum of first few terms of the Fourier series f (t) is close approximation of an ideal triangle
function, see Fig. 5.33.
6
Discrete-Time Fourier Transform
Fourier sum is defined as
F (xn ) = Xk =
def
N
−1
xn e
−j
2π k
n
N
n=0
F −1 (X[k]) = x[n] =
∞
1 2π k
X[k] exp j
n
T −∞
N
Fourier series is defined as
x(t) =
∞
Ck e j ω 0 t , ω 0 =
k=−∞
Ck =
1
T
t0 +T
2π
T
x(t) e−j ω 0 t dt
t0
Problems
6.1
Simple Seqences
Given functions x[n] in P.6.1 to P.6.6 (assume (|α| < 1, τ > 0), derive Fourier transformations X(ω).
6.1. x[n] = {1, 2, 3, 4}
6.2. x[n] = {2, 3, −1, 1}
6.3. x[n] = δ[n], N = 16
6.4. x[n] = δ[n − n0 ]
6.5. x[n] = 1, N = 16
6.6. x[n] = α n u[n − 1]
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_6
147
148
6.2
6 DT Fourier Transformation
Fourier Series
Given functions x(t) in P.6.7 to P.6.12, derive their Fourier series.
6.7. Derive discrete Fourier transformations X[k] of the following functions x(t).
x(t) = sin(1 Hz), sampling frequency fs = 8 Hz.
6.8. Given f (t) = 1 − (t/π ), derive its Fourier series if t ∈ [−π, π ].
6.9. Given square waveform in Fig. 6.1, derive Fourier series assuming a = T /4.
6.10. Given square waveform in Fig. 6.1, derive Fourier series assuming a = T /8.
6.11. Given square waveform in Fig. 6.2, derive its Fourier series.
6.12. Given x(t) waveform in Fig. 6.3 derive its Fourier series.
Fig. 6.1 Example P.6.9
x(t)
1
t
0
−T
−a 0
Fig. 6.2 Example P.6.11
T
a
x(t)
1
0
−T
Fig. 6.3 Example P.6.12
0
2a
t
T
x(t)
a
0
t
−a
−2p
0
2p
6.4 Fast Fourier Transform
6.3
149
Inverse Discrete Time FT
Given discrete sequences X[k] in P.6.13 to P.6.14, derive their inverse Fourier transforms x[n].
6.13. X[k] = {10, −2 + 2j, −2, −2 − 2j }
6.4
6.14. X[k] = {5, 3 − j 2, −3, 3 + j 2}
Fast Fourier Transform
Resolve questions in P.6.15 to P.6.21.
6.15. Assuming the audio spectrum signal, [20 Hz, 20 kHz], estimate the time needed to calculate
discrete Fourier transformation (DFF) by hands, if the desired resolution is 1 Hz and it takes 1min to
perform one calculation (extremely fast, but for the sake of argument). Then, compare with the time
needed to do the same calculations by using fast Fourier transformation (FFT).
6.16. Given a general discrete series xn = x[n], briefly develop the FFT equations and explain the
main idea behind it.
6.17. Given a general discrete series xn = x[n] and N = 2, apply FFT algorithm.
6.18. Given a general discrete series xn = x[n] and N = 4, apply FFT algorithm.
6.19. By using FFT graph algorithm in P.6.18, calculate Fourier transformation of x[n] = {1, 2, 3, 4}
(it is already solved using DFT in P.6.1).
6.20. Further to discussion in P.6.16, deduce FFT graph diagram if number of transformation points
is N = 8.
6.21. By using FFT graph algorithm in P.6.20, calculate F (xn ) if xn = [1, 1, −1, −1, 1, 1, −1, −1].
150
6 DT Fourier Transformation
Answers
6.1
Simple Seqences
6.1. Given x[n] = {1, 2, 3, 4}, DT FT in N = 4 points is derived by definition
3
2π k
2π k
X[k] =
x[n] exp −j
x[n] exp −j
n =
n
N
4
n=0
n=0
N
−1
In discrete sequence interval, both n, k ∈ (0, 3), thus for each case k it follows that,
3
1
2π 0
>
x[n] exp −j
x[n] n =
e0
k = 0 : X[0] =
4
n=0
n=0
3
= x[0] + x[1] + x[2] + x[3] = 1 + 2 + 3 + 4
= 10
3
π 1 2 π 1
n =
x[n] exp −j
x[n] exp −j n
k = 1 : X[1] =
2
2 4
n=0
n=0
3
(−j )n
if needed, review calculations of complex number powers
= x[0] (−j )0 + x[1] (−j )1 + x[2] (−j )2 + x[3] (−j )3
= 1(1) + 2(−j ) + 3(−1) + 4(j ) = 1 − 2j − 3 + 4j
= −2 + 2j
3
2π 2
n =
x[n] exp −j
x[n] exp (−j π n)
k = 2 : X[2] =
1 4
n=0
n=0
3
(−1)n
= x[0] (−1)0 + x[1] (−1)1 + x[2] (−1)2 + x[3] (−1)3
= 1(1) + 2(−1) + 3(1) + 4(−1) = 1 − 2 + 3 − 4
= −2
3
1 2 π 3
3π
n =
n
x[n] exp −j
x[n] exp −j
k = 3 : X[3] =
2
2 4
n=0
n=0
3
(j )n
= x[0] (j )0 + x[1] (j )1 + x[2] (j )2 + x[3] (j )3
= 1(1) + 2(j ) + 3(−1) + 4(−j ) = 1 + 2j − 3 − 4j
= −2 − 2j = x[1]∗
Recall that as = −π ≡ π and = −(3π/4) ≡ (5π/4), then the complete solution may be written
as,
6.1 Simple Seqences
151
Fig. 6.4 Example P.6.1
|X[k]|
10
√
2 2
2
0
k
Θ[k]
p
p
2
k
0
0
1
2
X[k] = {10, −2 + 2j, −2, −2 − 2j }
√
√ |X[k]| = 10, 2 2, 2, 2 2
5π
3π
, π,
[k] = 0,
4
4
where |X[k]| and phase [k] diagrams are in Fig. 6.4
6.2. Given x[n] = {2, 3, −1, 1}, DT FT in N = 4 points is derived by definition
X[k] =
3
2 πk
2π k
n =
x[n] exp −j
x[n] exp −j n
N
2 4
n=0
n=0
N
−1
As both n, k ∈ (0, 3) interval, thus for each case k it follows that,
3
3
π 0
:=1
(0)
exp
x[n] exp −j
x[n]
x[n]
k = 0 : X[0] =
n =
2
n=0
n=0
n=0
3
=2+3−1+1=5
π 1
x[n] exp −j
k = 1 : X[1] =
n
2
n=0
1π
2π
3π
0π
+ 3 exp −j
− 1 exp −j
+ 1 exp −j
= 2 exp −j
2
2
2
2
3
= 2 + 3(−j ) − 1(−1) + 1(j )
= 3 − 2j
k = 2 : X[2] =
π 2
n
x[n] exp −j
2
n=0
3
= 2 exp (−j (0π )) + 3 exp (−j (1π )) − 1 exp (−j (2π )) + 1 exp (−j (3π ))
3
152
6 DT Fourier Transformation
Fig. 6.5 Example P.6.2
|X[k]|
5
k
0
Θ [k]
p
p
2
k
0
0
1
2
3
= 2 + 3(−1) − 1(1) + 1(−1)
= −3
π 3
n
x[n] exp −j
k = 3 : X[3] =
2
n=0
π3
π3
π3
π3
= 2 exp −j
0 + 3 exp −j
1 − 1 exp −j
2 + 1 exp −j
3
2
2
2
2
3
= 2 + 3(j ) − 1(−1) + 1(−j )
= 3 + 2j
Recall that as π ≡ 180◦ , then the complete solution may be written as,
X[k] = {5, 3 − 2j, −3, 3 + 2j }
√
√ |X[k]| = 5, 13, 3, 13
[k] = {0, −33.7◦ , π, 33.7◦ }
where |X[k]| and phase [k] diagrams are in Fig. 6.5
6.3. Given Delta function x[n] = δ[n] in N = 16 points, that is to say x[0] = 1 otherwise x[n] =
0, n = 0, or
x[n] = {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
by definition it follows that,
15
2π k
2π k
n =
n
x[n] exp −j
x[n] exp −j
X[k] =
8
N
16
n=0
n=0
N
−1
Calculations for each k are as follows.
6.1 Simple Seqences
153
15
π 0
:1
(0)
k = 0 : X[0] =
x[n] exp −j
x[n]
n =
exp
8
n=0
n=0
15
= x[0] + x[1] + x[2] + · · · + x[15] = 1 + 0 + 0 + · · · + 0
=1
15
π π 1
n =
k = 1 : X[1] =
x[n] exp −j
x[n] exp −j n
8
8
n=0
n=0
π = x[0] exp −j 0 + 0 + 0 + · · · + 0
8
15
=1
15
π 2
2π
k = 2 : X[2] =
n =
n
x[n] exp −j
x[n] exp −j
8
8
n=0
n=0
2π
= x[0] exp −j
0 + 0 + 0 + ··· + 0
8
15
=1
..
.
etc.
In conclusion, given x[k] = δ[k], then X[k] = 1, ∀k, which is a constant function that is well known
to be dual function to Dirac impulse.
6.4. Method 1: DT FT of delayed impulse x[n] = δ[n − n0 ] = {· · · 0 0 · · · 1 · · · 0 0 · · · } in N points,
where n = n0 is indice of the only non–zero sample may be derived by definition as
∞
2π k
2π k
x[n] exp −j
δ[n − n0 ] exp −j
X[k] =
n =
n
N
N
n=−∞
n=−∞
2π k
n0 +0 + 0 + · · ·
= · · · + 0 + 0 + δ[n0 − n0 ] exp −j
N
∞
= 1 exp −j
2π k
n0
N
n=n0
Therefore, |X[k]| = 1, ∀k and its argument is (2π k n0 /N), which illustrates the property of delayed
functions in general where the delay transforms into the equivalent argument (i.e. angle or phase).
FT
Method 2: knowing time domain delay property of FT and that δ[n] → 1, then it follows directly
that
2π k
2π k
n0
n0
F (δ[n − n0 ]) = F (δ[n]) exp −j
⇒ X[k] = 1 exp −j
N
N
154
6 DT Fourier Transformation
6.5. Given constant function x[n] = 1 in N = 16 points may be transformed by definition as
X[k] =
15
2π k
2π k
n =
n
x[n] exp −j
1 exp −j
8
N
16
n=0
n=0
N
−1
As both n, k ∈ (0, 15) interval, thus for each case k it follows that,
k = 0 : X[0] =
15
15
π 0
:1 =
exp
exp −j
1n
n =
(0)
8
n=0
n=0
n=0
15
= 16
15
π π 1
exp −j
exp −j n
n =
k = 1 : X[1] =
8
8
n=0
n=0
π π π = exp −j 0 + exp −j 1 + · · · + exp −j 15
8
8
8
15
The question is: how to add these sixteen exponential terms? Recall that modules of all complex
exponential terms equal one (the unit circle radius). See Vol. I to recall the exponential form of
complex numbers. Finally, sampled complex terms are symmetrically and periodically distributed
over one period, T = 2π i.e. one circle circumference. Given N = 16 points, each term in the sum
is a complex number that is shifted along the circle at the multiples of 2π/N = 2π/16 = π/8, see
Fig. 6.6. Therefore, the summation may be done by considering symmetrical pairs of numbers.
For example, see Fig. 6.6, x[6] and x[14] are at the opposite sides and by consequence their real
parts ±Re have opposite signs, as well as their imaginary parts ±Im, therefore their sum equals zero
as
x[14] + x[6] = ℜ(x[6]) + j ℑ(x[14]) − ℜ(x[6]) − ℑ(x[6]) = 0
Or, more formally, this conclusion may be confirmed as
π π x[6] + x[14] = exp −j 6 + exp −j 14
8
8
2
π
2π
= exp −j
3 + exp −j
7
4 8
4 8
π
3π
+ exp j
= exp −j
4
4
π π 3π
3π
= cos −
+ j sin −
+ cos
+ j sin
4
4
4
4
√
√
√
√
2
2
2
2
−j
+
+j
=−
2
2
2
2
=0
Similarly, it may be shown that all the remaining sums of the opposite complex pairs also equal zero,
i.e.
6.1 Simple Seqences
155
Fig. 6.6 Example P.6.5
x10
x11 x12
j
x9
x8
x13
Im
x14
x15
x0
-Re
−1
1
Re
x7
x6
-Im
x5
x1
−j
x4
x[0] + x[8] = 0
x[4] + x[12] = 0
x[1] + x[9] = 0
x[5] + x[13] = 0
x[2] + x[10] = 0
x[6] + x[14] = 0
x[3] + x[11] = 0
x[7] + x[15] = 0
x3
x2
In conclusion, X[1] = 0
k = 2, 3, ... 15 :
After following the same reasoning, it may be shown that for all the remaining values of k = 2, 3, ...15
it is true that X[k], therefore
X[0] = 16
π π π X[1] = exp −j 0 + exp −j 1 + · · · + exp −j 15 = 0
8
8
8
X[2] = X[3] = · · · = X[15] = 0
In summary, given N = 16
F (x[n] = 1) = 16 δ[n]
and X[k] being real function, the argument θ[k] = 0.
6.6. Given infinitely long exponential series x[n] = α n u[n − 1], where |α| < 1, its DT FT may be
derived by definition as
∞
2π k
2π k
n
x[n] exp −j
α u[n − 1] exp −j
n =
n
X[k] =
N
N
n=−∞
n=−∞
∞
delayed step function u[n − 1] = 0, ∀n < 1 ∴ u[n − 1] = 1, ∀n ≥ 1
156
6 DT Fourier Transformation
∞ 2π k
2π k n
α n ( 1) exp −j
n =
α exp −j
N
N
n=1
n=1
b x a y a = (xy)a ; as well as x ab = x a
N
a N +1 − a m
k
, (|a| < 1)
a =
a−1
k=m
=
∞
:0
∞+1 2π k 1
2π
k
−
α
exp
−j
α exp
−j
N
N
=
2π k
α exp −j
−1
N
exp −j 2π k = 1, α < 1 ∴ α exp −j 2π k < 1 see Vol. I
N N
2π k
α exp −j
N
=
2π k
1 − α exp −j
N
6.2
Fourier Series
6.7. Given a CT sine function x(t) = sin(1 Hz), before performing DT FT it is necessary to convert
x(t) → x[n], which is done in the following steps.
1. Sampling frequency normalization : good practical choice for N is to use the powers of two, here
N = 23 = 8 points. Given the sampling frequency fs = 8 Hz and the sine function’s period
T (1 Hz) = 1s, it follows that
f = 1 Hz ⇒ T =
1
= 1s ⇔ 2π
f
fs = 8 Hz ⇒ N = 8
∴
2π
π
1
s≡
=
8
8
4
that is to say, each sample is found at the multiples of π/4, see Fig. 6.7. Note that first sample
x[0] = 0, as is the first sample of the next period.
2. Calculations of each sample : N = 8 samples are evenly distributed over one period, i.e. full circle,
therefore separated by π/4 angle. Each sample is then calculated as,
6.2 Fourier Series
157
0
Fig. 6.7 Example P.6.7
√
1
2/2
2p
p
f (t)
f [k]
t
k
0
√
− 2/2
−1
0
1
2
3
4
π
=0
x[4] = (1) sin 4
4
π
x[0] = (1) sin 0 ) = 0
4
√
π
2
x[1] = (1) sin 1 ) =
4
2
π
x[2] = (1) sin 2
=1
4
√
π
2
=
x[3] = (1) sin 3
4
2
√
π
2
=−
x[5] = (1) sin 5
4
2
π
x[6] = (1) sin 6
= −1
4
√
π
2
=−
x[7] = (1) sin 7
4
2
Therefore,
√
√
√
√ 2
2
2
2
x[n] = 0,
, 1,
, 0, −
, −1, −
2
2
2
2
Note that simple sum of samples is
√
√
√
√
2
2
2
2
x[n] = 0 +
+1+
+0−
−1−
=0
2
2
2
2
n=1
7
FT
3. x(t) → X[k] : given N = 8 by definition it follows that
X[k] =
7
2 πk
2π k
n
n =
x[n] exp −j
x[n] exp −j N
4 8
n=0
n=0
N
−1
so that calculations for each case k are as follows.
1 7
7
π 0
0
7
k = 0 : X[0] =
e
n =
x[n] exp −j
x[n] =
x[n] = 0
4
n=0
n=0
n=0
7
π 1
n
x[n] exp −j
k = 1 : X[1] =
4
n=0
√
π π 2
= (0) exp(0) +
exp −j 1 + (1) exp −j 2
2
4
4
√
√
π π π 2
2
+
exp −j 3 + (0) exp −j 4 −
exp −j 5
2
4
4
2
4
π π √2
exp −j 7
− (1) exp −j 6 −
4
2
4
7
5
6
7
158
6 DT Fourier Transformation
cos(−x) = cos(x) sin(−x) = − sin(x)
√ π π 2π
2π
2
=0+
cos
− j sin
+ (1) cos
− j sin
+ ···
2
4
4
4
4
− j 0.5) + (−j ) + (−
− j 0.5)
= 0 + (
0.5
0.5
− j 0.5) + (−j ) + (−
− j 0.5)
0.5
0.5
+ 0 + (
= −4j
2
π
k = 2 : X[2] =
n
x[n] exp −j
4 2
n=0
√
π
π 2
exp −j
+ (1) exp −j 2
= (0) exp(0) +
2
2
2
√
√
π 2
2
π
π
+
exp −j 3 + (0) exp −j 4 −
exp −j 5
2
2
2
2
2
π π √2
exp −j 7
− (1) exp −j 6 −
2
2
2
√ π
π
2π
2π
2
cos
− j sin
+ (1) cos
− j sin
+ ···
=
2
2
2
2
2
√
√
√
√
2
2
2
2
= −j
−1+j
+j
+1−j
2
2
2
2
7
=0
Similarly, X[2] = X[3] = X[4] = X[5] = X[6] = 0.
However,
π 7
k = 7 : X[7] =
x[n] exp −j
n
4
n=0
√
7π
7π
2
exp −j
1 + (1) exp −j
2
= (0) exp(0) +
2
4
4
√
√
7π
7π
7π
2
2
+
exp −j
3 + (0) exp −j
4 −
exp −j
5
2
4
4
2
4
√
7π
2
7π
6 −
exp −j
7
− (1) exp −j
4
2
4
7
cos(−x) = cos(x) sin(−x) = − sin(x)
π
7π
≡ , etc.
use unit circle to convert angles, for example −
4
4
√ π π π π 2
cos
+ j sin
+ (1) cos
+ j sin
+ ···
=0+
2
4
4
2
2
6.2 Fourier Series
159
+ j 0.5) + (+j ) + (−
+ j 0.5)
= 0 + (
0.5
0.5
+ j 0.5) + (+j ) + (−
+ j 0.5)
+ 0 + (
0.5
0.5
= 4j
In summary, given N = 8, fs = 8 Hz, and x(t) = sin(1 Hz), DT FT is
X[k] = {0, −4j, 0, 0, 0, 0, 0, 4j }
|X[k]| = {0, 4, 0, 0, 0, 0, 0, 4}
Note that Fourier transformation result X[k] is “bilateral”, see Fig. 6.8 (top), there are two non–
zero impulses in |X[k]| sequence. At this stage of calculations, FT result is relative strictly to an
abstract variable k.
4. In order to express the amplitude |X[k]| relative to the physical units of frequency, i.e. [Hz],
resolution of each frequency “bin” is calculated as
fs
8 Hz
=
= 1 Hz/bin
N
8
In this case, because N and sampling frequency fs are numerically equal it happens that k
maps directly into Hz. That is to say, by selecting different either N or fs the resolution of
Fourier transformation may be set arbitrarily. In addition, impulses found above half of sampling
frequency, known as Nyquist frequency, are folded over to create “unilateral” view of Fourier
transformation in 0 to fs /2 range (as commonly shown by spectrum analyzers or simulated
graphs), see Fig. 6.8 (bottom).
Consequently, in unilateral view amplitudes at each frequency are calculated as the sum of samples
already in place and the ones “folded over”.
For example amplitude of sample at x(1 Hz) is increased by amplitude of x(7 Hz) that after folding
over falls on top, thus |H [1 Hz]| = 4 + 4 = 8. This amplitude is the consequence of N = 8, thus
in order to show correct amplitude, i.e |x(t)| = 1 the unilateral graph must be renormalized by
dividing the amplitudes by N, see Fig. 6.9.
Fig. 6.8 Example P.6.7
4 |X[k]|
k
0
4 |X[k]|
folding
0
[Hz]
fs /2
0
1
2
3
4
5
6
7
160
6 DT Fourier Transformation
(1) 8 |X[k]|
Fig. 6.9 Example P.6.7
fs /2
0
f [Hz]
0
1
2
3
4
6.8. Given a linear function f (t) = 1− t/π its Fourier series within the periodic interval x ∈ [−π, π ]
is calculated by definition as,
∞
∞
a0 f (t) =
an cos (n t) +
bn sin (n t)
+
2
n=1
n=1
where,
a0 =
1
π
1
an =
π
bn =
1
π
π
f (t) dt
−π
π
f (t) cos (n t) dt
−π
π
f (t) sin (n t) dt
−π
Therefore, given period T = 2π ,
1
a0 =
π
=2
an =
1
π
π
t
1−
π
−π
π
−π
1−
t
π
1
dt =
π
π
1 2
1
1 2 π
2
t −
t 2π −
=
π −π
2π
π
2π
−π
−π
cos (n t) dt = see Vol. II
π
:0
1
cos (n t) dt − 2
x cos
(n t) dt
π −π
−π
π
1
1
0
:0
:
) −
sin(nπ
sin(nt) =
=
sin(−nπ
)
nπ
nπ
−π
1
=
π
π
=0
bn =
1
π
π
−π
1 1
=− 2
π n
1−
t
π
sin (n t) dt =
π
:0 1 π
1
sin
t)
dt
− 2
t sin (n t) dt = see Vol. II
(n
π −π
π −π
π
− t cos(nt) + sin(nt)
sin(nπ )=0
cos(nπ )−cos(nπ )=0
1
=
nπ 2
−π
π cos(nπ ) + π cos(nπ )
6.2 Fourier Series
161
Fig. 6.10 Example P.6.8
2
f (t)
1
0
−p
=
1
nπ 2C
0
p
t
π 2 cos(nπ )
Z
= (−1)n
(−1)n
2
nπ
Therefore,
∞
2
2
2
2
f (t) = 1 +
(−1)n
sin(nt) = 1 − sin(t) +
sin(2t) −
sin(3t) + · · ·
nπ
π
2π
3π
n=1
This series is superimposed to f (t), see Fig. 6.10, which illustrates the series approximation. To
achieve “perfect” approximation, there should be infinite number of terms. The largest discrepancy
between the continuous function f (t) and its approximation is at the ends of given interval. Over
wider interval Fourier series is a periodic function.
6.9. Note that this square pulse waveform is even as f (−t) = f (t) (i.e. symmetric around t = 0).
Exponential form of Fourier series, by definition, is
∞
x(t) =
Cn exp (j nω 0 t) ;
where,
ω0 =
k=−∞
Cn =
1
T
2π
T
and
t0 +T
x(t) exp (−j nω 0 t) dt
t0
Case n = 0 : note that x(t) = 1, t ∈ [−a, a], thus
C0 =
1
T
t0 +T
t0
:1
1 a
2a
0
ω
x(t) exp
t)
dt
=
(1) dt =
(−j
0
T −a
T
Case n ≥ 1 :
Cn =
=
2
T
t0 +T
x(t) exp (−j nω 0 t) dt =
t0
2
1
−j nω 0 T
2
T
a
−a
(1) exp (−j nω 0 t) dt
exp (−j nω 0 a) − exp (j nω 0 a)
162
6 DT Fourier Transformation
2T 1 exp (j nω 0 a) − exp (−j nω 0 a)
nπ T
2j
2
2nπ
=
sin
a
nπ
T
=
Then, given specific high/low ratio,
a = T /4 : therefore,
1
2a
2T
=
=
T
2
T 4 2
2
sin (nπ/2)
2nπ
2
2nπ T
=
sin
a =
sin
= sinc (nπ/2)
Cn =
nπ
T
nπ
nπ/2
T 4 2
C0 =
∴
|Cn | = |sinc (nπ/2) |
A few calculated samples in the series are, see Fig. 6.11
⎧
⎪
⎪
⎪ C1
⎪
⎪
⎪
⎪C
2
⎪
π ⎪
⎪
⎪
⎪
⎨C
sin n
3
2 =
π
⎪
⎪
n
C4
⎪
⎪
2
⎪
⎪
⎪
⎪
C5
⎪
⎪
⎪
⎪
⎩
2
π
= 2/2π sin (2π/2) = 0
= 2/π sin (π/2) =
= 2/3π sin (3π/2) = −
2
3π
= 2/4π sin (4π/2) = 0
2
= 2/5π sin (5π/2) =
5π
···
That is to say, function x(t) can be synthesized as the sum of infinite sine !
series, where all even
terms equal zero. Being a complex number Cn = an + j bn , recall that |Cn | = an2 + bn2 , thus
x(t) =
∞
k=−∞
Cn exp (j nω 0 t) = a0 +
∞
an cos(nω 0 t) + bn sin(nω 0 t)
n=1
for even functions bn = 0 ∴ |an | = |Cn |
∞
nπ 1 2
T
2π
sin
cos(n ω 0 t) = ω 0 =
= +
, a=
2 n=1 nπ
2
T
4
1
2
2π
1
2π
1
2π
= +
cos
t − cos 3 ×
t + cos 5 ×
t − ···
2 π
T
3
T
5
T
As a comparison, this series is overlaid on x(t), see Fig. 6.12 where only the first four terms are used,
and spectrum is illustrated in Fig. 6.11. Note that all even terms equal zero.
6.10. Note that this square pulse waveform is even as f (−t) = f (t) (i.e. symmetric around t = 0).
6.2 Fourier Series
163
Fig. 6.11 Example P.6.9
1
2
|C[n]|
0
n
0
1
2
3
Fig. 6.12 Example P.6.9
4
5
6
f (t)
1
0
−a
0
t
a
See A.6.9, where after setting a = T /8 it follows that
2 T
1
2a
=
=
T
4
T 8 4
2nπ
2 2
1
2nπ T
1 sin (nπ/4)
2
Cn =
sin
a =
sin
= sinc (nπ/4)
=
nπ
T
2 nπ
2 nπ/4
2
T 8 4
C0 =
∴
|Cn | =
1
|sinc (nπ/4) |
2
A few calculated samples in the series are, see Fig. 6.13
⎧
⎪
⎪
⎪
C1
⎪
⎪
⎪
⎪
⎪
⎪ C2
⎪
π ⎪
⎪
⎪
⎪
⎨
sin
n
1
4 = C3
nπ
⎪
2
⎪
⎪
C4
⎪
4 2
⎪
⎪
⎪
⎪
⎪
⎪
C5
⎪
⎪
⎪
⎪
⎩
√
2
π
1
= 2/2π sin (2π/4) =
π
√
2
2
3π
= /3π sin ( /4) =
3π
= 2/4π sin (4π/4) = 0
√
2
= 2/5π sin (5π/4) = −
5π
···
= 2/π sin (π/4) =
That is to say, function x(t) can be synthesized
as the sum of infinite sine series. Being a complex
!
number Cn = an + j bn , recall that |Cn | = an2 + bn2 , thus
x(t) =
∞
k=−∞
Cn exp (j nω 0 t) = a0 +
∞
an cos(nω 0 t) + bn sin(nω 0 t)
n=1
for even functions bn = 0 ∴ |an | = |Cn |
164
6 DT Fourier Transformation
Fig. 6.13 Example P.6.10
|C[n]|
1
4
0
n
0
1
Fig. 6.14 Example P.6.10
2
3
4
5
6
f (t)
1
0
−T
−a 0 a
t
T
∞
nπ 1 2
2π
T
sin
cos(n ω 0 t) = ω 0 =
= +
, a=
2 n=1 nπ
2
T
4
1
2
2π
1
2π
1
2π
= +
cos
t − cos 3 ×
t + cos 5 ×
t − ···
2 π
T
3
T
5
T
As a comparison, this series is overlaid on x(t), see Fig. 6.14 where only the first four terms are used,
and spectrum is illustrated in Fig. 6.13.
6.11. Method 1: Given delayed (odd) pulse waveform, in (0, T ) period, coefficients of Fourier series
may be calculated in sine/cosine form as
a0 =
2
T
2
an =
T
T
0
T
0
f (t) dt =
2
T
2a
(1) dt =
0
2
f (t) cos(nt) dt =
T
2a
0
4a
T
2a
2
sin(nt) (1) cos(nt) dt =
nT
0
2
2 sin(2a n) − sin(0) =
sin(2a n)
=
nT
nT
2a
2 T
2 2a
2
bn =
cos(nt) f (t) sin(nt) dt =
(1) sin(nt) dt = −
T 0
T 0
nT
0
=−
2 2 cos(2a n) − cos(0) =
1 − cos(2a n)
nT
nT
Then, given specific high/low ratio and period T = 2π ,
a = T /4 : therefore,
6.2 Fourier Series
a0 =
an =
165
4a
4T
=
=1
T
T 4
2
T
2
1
2
T
2 1
2π
:0
sin
n =
sin
n =
sin(2a n) =
sin 2
sin
n =
(nπ )
nT
nT
nT
2
nπ 4 2
2nπ
2
=0
2
2 1 − cos(2a n) =
bn =
nT
nT
1 T
1 − cos 2
1 − cos (nπ )
n
=
nπ
4 2
A few calculated samples in the series are,
⎧
⎪
⎪
b1
⎪
⎪
⎪
⎪
⎪
⎪
b2
⎪
⎪
⎪
⎪
⎨ b3
1 bn =
1 − cos (nπ ) =
⎪
nπ
⎪
b4
⎪
⎪
⎪
⎪
⎪
⎪
b5
⎪
⎪
⎪
⎪
⎩
Therefore, Fourier series of port function a
f (t) =
=
2
= 1/π 1 − cos (π ) =
π
= 1/2π 1 − cos (2π ) = 0
2
= 1/3π 1 − cos (3π ) =
3π
= 1/4π 1 − cos (4π ) = 0
2
= 1/5π 1 − cos (5π ) =
5π
···
((t−π/2)/π ) is
∞
∞
1 1 1 +
1 − cos (nπ ) sin(nt)
0 cos(nt) +
2 n=1
π n=1 nπ
1
2
2
2
+ sin(t) + 0 +
sin(3t) + 0 +
sin(5t) + · · ·
2 π
3π
5π
Note that delayed port function is not even and that even coefficients equal zero. The sum of first few
terms of the Fourier series f (t) is close approximation of an ideal square wave function, see Fig. 6.15.
Method 2: Relative to square waveform in P.6.9 function f (t) in this example is time delayed, that is
to say y(t) = f (t − a). Knowing the time delay property of Fourier transform it can be written that
2nπ
1
sin
a
Cn = exp(−j nω 0 a)
nπ
T
and derive the same Fourier series as per Method 1: but faster.
6.12. Fourier series of delayed triangular function with period T = 2π may be derived, for example,
in sine/cosine form by definition.
f (t) =
∞
∞
a0 an cos (n t) +
bn sin (n t)
+
2
n=1
n=1
166
6 DT Fourier Transformation
Fig. 6.15 Example P.6.11
x(t)
1
0
−T
0
2a
t
T
where,
a0 =
1
π
1
an =
π
bn =
1
π
π
f (t) dt
−π
π
f (t) cos (n t) dt
−π
π
f (t) sin (n t) dt
−π
Method 1: Given triangular function is periodic with period T = 2π . Note equations of linear
sections within the chosen intervals. Without loosing generality of T , it may be written that
1 π/2 2a
2a
1 3π/2
f (t) dt =
−
t dt +
(t − π ) dt
π −π/2 π
π π/2
π
−π/2
π/2
3π/2
π/2
3π/2
2a
2a
2a 3π/2
2a
2a
2a 3π/2
= 2
t
t dt − 2
t dt +
dt = 2 t 2 − 2 t 2 +
π −π/2
π π/2
π π/2
π
π
π
−π/2
π/2
π/2
"
#
2
2
2
2
)
2a 1 9π
2a 1 π (−π
π
2a 3π
π
− = 2
− 2
−
+
−
π 2 4
π 2
4
4
π
2
2
4
3π/2
1
a0 =
π
=−
2a
π2
a 8 2 π = −a + a
+
2
π
π 4
=0
As well, given period T = 2π ,
1 3π/2 2a
1 π/2 2a
−
t cos (n t) dt +
(t−π ) cos (n t) dt
f (t) cos (n t) dt =
π −π/2 π
π π/2
π
−π/2
π/2
3π/2
2a
2a 3π/2
2a
t cos (n t) dt − 2
t cos (n t) dt +
cos (n t) dt
= 2
π −π/2
π π/2
π π/2
1
an =
π
3π/2
= see Vol. II
π/2
π/2
3π/2
3π/2
2a
1
2a
1
=
t
sin(n
t)
+
cos(n
t)
−
t
sin(n
t)
−
cos(n
t)
nπ 2
n2 π 2
nπ 2
n2 π 2
−π/2
−π/2
π/2
π/2
6.2 Fourier Series
167
3π/2
2a
sin(n t) nπ
π/2
π
2a π
π
π
π
π
1
=
cos
n
cos
−n
sin
n
+
sin
−n
+
−
nπ 2 2
2
2
2
n2 π 2 2
2
π
π
3π
π
1
3π
3π
2a
sin
n
−
sin
n
−
−
cos
n
cos
n
−
2
nπ 2
2
2
2
2
n2 π 2 2
π 2a
3π
+
sin n
− sin n
nπ
2
2
3π
π
=
= − , cos(−x) = cos(x), sin(−x) = − sin(x)
2
2
π 2a
π
π
a
π 2a
3a
sin −n
+
sin n
+
sin −n
−
sin n
=−
nπ
2
nπ
2
nπ
2
nπ
2
3a
2a
π
2a
a
π
=
−
sin n
−
−
sin n
nπ
nπ
2
nπ
nπ
2
+
=0
and,
1 π/2 2a
1 3π/2 2a
f (t) sin (n t) dt =
t sin (n t) dt +
(t−π ) sin (n t) dt
−
π −π/2 π
π π/2
π
−π/2
π/2
3π/2
2a
2a
2a 3π/2
= 2
t sin (n t) dt − 2
t sin (n t) dt +
sin (n t) dt
π −π/2
π π/2
π π/2
1
bn =
π
3π/2
= see Vol. II
π/2
3π/2
2a
1
1
t
t
2a
= 2 − cos(nt) + 2 sin(nt) − 2 − cos(nt) + 2 sin(nt) π
n
n
π
n
n
−π/2
π/2
3π/2
2a
cos(nt) −
nπ
π/2
3π
π
=
= − , cos(−x) = cos(x), sin(−x) = − sin(x)
2
2
=−
π 2a
2a
π 2a
2a
cos(nπ/2) + 2 2 sin(nπ/2) −
cos(nπ/2) + 2 2 sin(nπ/2)
2
2
nπ
n π
2 nπ
2 nπ
2a
2a
π 2 a
π 2 a
3
cos(3nπ/2) − 2 2 sin(3nπ/2) −
cos(nπ/2) + 2 2 sin(nπ/2)
nπ
nπ
2 nπ 2
2 nπ 2
2a
+ 2acos(nπ/2)
− cos(3nπ/2)
nπ
nπ
3a
3a
8a
− cos(nπ/2)
+ 2 2 sin(nπ/2)
= cos(nπ/2)
nπ
nπ
nπ
+
168
6 DT Fourier Transformation
=
8a
sin(nπ/2)
n2 π 2
A few calculated terms in the series are
⎧
⎪
⎪
⎪ b1
⎪
⎪
⎪
⎪
⎪
b2
⎪
⎪
⎪
⎪
⎪
⎪
b3
⎪
⎪
⎪
⎪
⎨b
8a
4
sin(nπ/2) =
2
2
⎪
n π
⎪
b5
⎪
⎪
⎪
⎪
⎪
⎪
b6
⎪
⎪
⎪
⎪
⎪
⎪
b7
⎪
⎪
⎪
⎪
⎩
8a
π2
2 2
8a
= /2 π sin(2π/2) = 0
= 8a/12 π 2 sin(π/2) =
8a
= 8a/32 π 2 sin(3π/2) = − 2 2
3π
= 8a/42 π 2 sin(4π/2) = 0
8a
= 8a/52 π 2 sin(5π/2) = 2 2
5π
= 8a/62 π 2 sin(6π/2) = 0
8a
= 8a/62 π 2 sin(7π/2) = − 2 2
7π
···
which can be generalized as,
⎧
⎪
0;
⎪
⎪
⎪
⎨
8a
8a
;
sin(nπ/2)
=
=
(nπ )2
⎪
n2 π 2
⎪
8a
⎪
⎪
⎩−
;
(nπ )2
n even
n = 1, 5, 9, . . .
n = 3, 7, 11, . . .
In summary,
∞
8a
f (t) =
bn sin (n t) = 2
π
n=1
1
1
sin (t) − 2 sin (3t) + 2 sin (5t) − · · ·
3
5
A first few terms are superimposed over the ideal delayed triangle function x(t) in Fig. 6.16.
Method 2: With a bit of practice and experience, one can take advantage of function’s symmetry and
summation of signed areas by definite integrals (see Calculus).
(a) By itself, delayed triangular function x(t) is odd, see Fig. 6.17. Therefore, within a symmetric
interval definite interval of x(t) must equal zero. For example, an easy choice of symmetric
interval is −π/2, π/2, so that
1
a0 =
π
π
1
f (t) dt =
π
−π
a 2
a 2 π
2a
2
(π/2) − (−π/2)
=0
t dt = 2 t = 2 π
π
−π/2 π
−π
π/2
This result is general for odd functions over symmetric interval
(b) The product of x(t) cos(nt) is also odd, see Fig. 6.17 (top). Definite integral of an odd function
over symmetric interval must equal zero, which may be shown, for example as
1
an =
π
π
1
f (t) cos (n t) dt =
π
−π
π/2
2a
t cos (n t) dt
−π/2 π
6.2 Fourier Series
169
Fig. 6.16 Example P.6.12
x(t)
a
0
t
−a
−2p
0
2p
= see Vol. II
π/2
π/2
2a
1
t
sin(n
t)
+
cos(n
t)
2
2
2
nπ
nπ
−π/2
−π/2
π
2a π
π
π
1
π
π
=
cos
n
cos
−n
sin
n
+
sin
−n
+
−
nπ 2 2
2
2
2
n2 π 2 2
2
=
=0
(c) The product of x(t) sin(nt) is always positive, see Fig. 6.17 (bottom). By consequence, definite
integral may be calculated as two times the integral on half the period. Thus,
1
bn =
π
π
f (t) sin (n t) dt
−π
2 π 2a
2a
t sin (n t) dt −
(t − π ) sin (n t) dt
π
π π/2 π
0
4a π
4a π
4a π/2
t sin (n t) dt − 2
t sin (n t) dt +
sin (n t) dt = see Vol. II
= 2
π 0
π π/2
π π/2
π
π
2
π
4a 4a 4a
= 2 2 sin(n t) − n t cos(n t) − 2 2 sin(n t) − n t cos(n t) −
cos(n t) π
π
nπ
nπ
nπ
0
2
2
4a − = 2 2 sin(n π/2) − n π/2 cos(n π/2) − sin(0)
0 cos(0)
nπ
4a π) − sin(n π/2) − n π/2 cos(n π/2)
− 2 2 sin(n
π) − n π cos(n
n π
4a π) − cos(n π/2)
cos(n
−
nπ
4a
2a
4a
2a
4a
π/2) +
π/2) −
π/2)
= 2 2 sin(n π/2) + 2 2 sin(n π/2) − cos(n
cos(n
cos(n
nπ
nπ
nπ
nπ
nπ
2
=
π
=
π/2
8a
sin(n π/2)
n2 π 2
Which is the same result as by definition in Method 1: but with a bit less calculations.
170
6 DT Fourier Transformation
Fig. 6.17 Example P.6.12
x(t)
a ( +) × (−) = (−)
t
0
−a
( +) × (+) = (+)
−p
0
p
x(t)
a
( −) × (−) = (+)
0
−a
6.3
t
( +) × (+) = (+)
0
1
2
Inverse Discrete Time FT
6.13. Given a simple sequence X[k] = {10, −2 + 2j, −2, −2 − 2j } it follows that N = 4 and
periodicity T = 4, therefore by definition
x[n] = F −1 (X[k]) =
=
1
4
1
:
2π0
2π 1
j
n
10 exp
n
+
(−2
+
j
2)
exp
j
4
4 2
2π 2
2π 3
− 2 exp j
n + (−2 − j 2) exp j
n
4
4 2
1
Case n = 0 : x[0] =
4
=
∞
3
1 2π k
2π k
1
X[k] exp j
X[k] exp j
n =
n
T −∞
N
4 n=0
4
π 3π
10+(−2 + j 2) exp j 0 − 2 exp (j π 0) + (−2 − j 2) exp j
0
2
2
%
1$
10 − 2 + j 2 − 2 − 2 − j 2 = 1
4
−j
*
π j
−1
1
3π
:
:
π1)+(−2
exp
− j 2)exp Case n = 1 : x[1] =
10+(−2 + j 2)
j 1
exp
j 1 −2
(j
2
4
2
1
10 + (−2 + j 2)(j ) − 2(−1) + (−2 − j 2)(−j )
=
4
1
H
2j
=
10 + −2j
+2
+@
H− 2
@− 2 = 2
4
6.3 Inverse Discrete Time FT
1
Case n = 2 : x[2] =
4
171
*−1
π −1
1
3π
:
:
π 2)+(−2 − j 2)exp exp
j 2 − 2
j 2
exp
10+(−2+j 2)
(j
2
2
1
10 + (−2 + j 2)(−1) − 2(1) + (−2 − j 2)(−1)
4
1
=
10 + 2 − 2j − 2 + 2 + 2j = 3
4
=
j
*
−j
−1
π 1
3π
:
:
π3)+(−2 − j 2)exp Case n = 3 : x[3] =
exp
j 3 −2
exp
10+(−2 + j 2)
j 3
(j
4
2
2
1
10 + (−2 + j 2)(−j ) − 2(−1) + (−2 − j 2)(j )
=
4
1
10 + 2j + 2 + 2 − 2j + 2 = 4
=
4
In conclusion, x[n] = {1, 2, 3, 4}.
6.14. Given a simple sequence X[k] = {5, 3−j 2, −3, 3+j 2}, it follows that N = 4 and periodicity
T = 4, therefore by definition
x[n] = F −1 (X[k]) =
1
=
4
∞
3
1
2π k
1 2π k
n =
n
X[k] exp j
X[k] exp j
T −∞
N
4 n=0
4 2
1
:
2π 1
2π0
n
5 exp
n + (3 − j 2) exp j
j
4
4 2
2π 2
2π 3
− 3 exp j
n + (3 + j 2) exp j
n
4
4 2
Case n = 0 : x[0] =
=
1
4
π 3π
5 + (3 − j 2) exp j 0 − 3 exp (j π 0) + (3 + j 2) exp j
0
2
2
%
1$
5 + 3 −@
j@
2 − 3 + 3 + @
j@
2
4
=2
−j
*
j
−1
π 1
3π
:
:
π1)
5 + (3 − j 2)
exp
+ (3 + j 2) exp exp
j 1 − 3
Case n = 1 : x[1] =
j 1
(j
4
2
2
1
=
5 + (3 − j 2)(j ) − 3(−1) + (3 + j 2)(−j )
4
172
6 DT Fourier Transformation
1
=
3j
3j
5 +@
@+ 2 + 3 −@
@+ 2
4
=3
1
Case n = 2 : x[2] =
4
*−1
−1
π 1
3π
:
:
π2) + (3 + j 2) exp 5 + (3 − j 2)
exp
exp
j 2 − 3
j 2
(j
2
2
1
5 + (3 − j 2)(−1) − 3(1) + (3 + j 2)(−1)
4
1
=
5 − 3 + 2j − 3 − 3 − 2j = −1
4
=
1
Case n = 3 : x[3] =
4
j
*
−j
π −1
3π
:
:
π3) + (3 + j 2) exp exp
j 3
5 + (3 − j 2)
exp
j 3 − 3
(j
2
2
1
5 + (3 − j 2)(−j ) − 3(−1) + (3 + j 2)(j )
4
1
5 − 3j − 2 + 3 + 3j − 2 = 1
=
4
=
In conclusion, x[n] = {2, 3, −1, 1}.
6.4
Fast Fourier Transform
6.15. Discrete Fourier Transformation (DFT) of series x[n] is by definition
N
−1
2π k
n
Fk (x[n]) =
x[n] exp −j
N
n=0
def
where, k = 0, 1, 2, . . .
In order to simplify the long equations writing, often exp() is used to denote the necessary exponential
term in the above definition. As n, k, N changes, the number of necessary calculations increase as
follows
N = 1 F0 = x[0] exp() (one calculation)
N = 2 F0 = x[0] exp() + x[1] exp() (two calculations)
F1 = x[0] exp() + x[1] exp() (two calculations)
(total: four calculations)
N = 3 F0 = x[0] exp() + x[1] exp() + x[2] exp() (three calculations)
F1 = x[0] exp() + x[1] exp() + x[2] exp() (three calculations)
6.4 Fast Fourier Transform
173
F2 = x[0] exp() + x[1] exp() + x[2] exp() (three calculations)
(total: nine calculations)
..
.
N ⇒
(total: N 2 calculations)
That is to say, for example, the resolution of 1 Hz within 20 kHz bandwidth means that it is
necessary to do N = 20,000 points in the transformation. Knowing that DFT calculation does N 2
operations, it follows that
2
t = N 2 × 1min = 20 × 103 min = 400 × 106 min ≈ 278 × 103
days
≈ 760 years
By comparison, FFT algorithm does N log2 N calculations, that is to say
ln N
t = (N log2 N) × 1min = N
ln 2
× 1min = 286 × 103 min ≈ 198 days
The advantage of FFT over DFT increases as N becomes larger.
6.16. The main idea of FFT is to split the grand sum into two separate sums, in the first iteration to
one with odd and one with even coefficients. In the subsequent iteration, again every second term is
grouped as “odd’ and the other half as “even” group, see Fig. 6.18. The process is repeated until each
group consists of two terms only. This process is highly symmetric, as a consequence the binary form
of aligned term indexes in the final iteration is binary inverted relative to the binary form of aligned
term indexes at the beginning. Formally, FFT equations are rearranged as follows,
Fig. 6.18 Example P.6.16
bit inversion
[0 0 0]
[1 0 0]
x0
x4
x0
x2
x0
x1
[0 0 0]
[0 0 1]
[0 1 0]
[1 1 0]
x2
x6
x4
x6
x2
x3
[0 1 0]
[0 1 1]
[0 0 1]
[1 0 1]
x1
x5
x1
x3
x4
x5
[1 0 0]
[1 0 1]
[0 1 1]
[1 1 1]
x3
x7
x5
x7
x6
x7
[1 1 0]
[1 1 1]
N =2
N =4
N =8
···
174
6 DT Fourier Transformation
N
−1
2π k
Fk (x[n]) =
x[n] exp −j
n
N
n=0
def
N/2−1
2π k 2m
2π k 2m + 1
x[2m] exp −j
x[2m + 1] exp −j
+
N/2 2
N/2
2
m=0
m=0
even index
odd index
2π k
Ck = exp −j
N
N/2−1
N/2−1
2π k
2π k
x[2m] exp −j
x[2m + 1] exp −j
=
m + Ck
m
N/2
N/2
m=0
m=0
N/2−1
=
Where,
2π k
2π k
2π k
m = cos −
m + j sin −
m ; (k = 0, 1, . . . , N)
exp −j
N/2
N/2
N/2
However, when k > N/2, that is to say k = N/2 + r and r = 1, 2, . . . , N/2, then
2π (N/2 + r)
−2π (N/2) m − 2π r m
cos −
m = cos
N/2
N/2
2π r m
= cos −2π m −
N/2
m = 0, 1, . . . , N/2 ∴ −2π m = 0
2π r m
2π (N/2 + r) m
= cos −
= cos −
N/2
N/2
known as “symmetry identity”
This identity is valid for sine terms as well, which is to say that
2π (N/2 + k)
2π k
m = exp −j
m ; (k = 0, 1, 2, . . . , N)
exp −j
N/2
N/2
As a consequence, it is sufficient to cary out only N/2 calculations. However, each of the summing
groups can be split again and again, thus the total number of calculations is N log2 N.
6.17. Given N = 2 and a general discrete series xn = x[n], Fourier transform Fk (x[n]) = Xk
equations are written by using following syntax,
Xk =
N
−1
2π
nk =
xn exp −j
xn wNn · k = N = 2
N
n=0
n=0
N
−1
6.4 Fast Fourier Transform
=
175
1
xn w2n · k
n=0
This syntax simplifies the repetitive writing of complex exponential terms, keep in mind that wNn · k
represents the following calculations
2π
wN ≡ exp −j
N
2π
2π n · k def n · k
= wN
∴ exp −j
n k = exp −j
N
N
where the power term is calculated as the product nk, for example, if N = 2 it follows that
0 · 0
$
%0 · 0
2π
w20 · 0 = exp −j
= exp (−j π )
= (−1)0 = 1
2
or,
w21 · 1 =
2π
exp −j
2
1 · 1
$
%1 · 1
= exp (−j π )
= (−1)1 = −1
For the moment, let us keep n > 0 exponential terms in place, and also use the identity w21 · 1 ≡
−w20 · 0 , even if their values are as simple as +1 or −1. Reason for this choice becomes evident as
N = 4, N = 8, etc.. In this case,
k=0: X[0] = x w0 · 0 + x w1 · 0 = x + x ≡ x + x w0
0
k=1:
2
1
2
0
1
0
1
2
X[1] = x0 w20 · 1 + x1 w21 · 1 = x0 − x1 ≡ x0 − x1 w20
This calculation process is illustrated by “butterfly diagram”, Fig. 6.19. The solutions Xk on the right
side are formed by the graph inspection. For example, x0 is first multiplied by “1” then added to x1 w20
at the summing node to produce X[0] = x0 + x1 w20 . Similarly, graph illustrates the formation of X[1]
equation. This basic butterfly diagram is then reused for any N = 2 FFT. As illustrated by the colour
scheme, note paths that carry ‘ + 1 multiplication factor, paths that carry positive exponential factor,
and paths that carry negative exponential factor.
6.18. In the case of N = 4, the four Xk DFT transformation are reorganized into symmetric form
that can be directly translated into butterfly diagram as follows. As a reminder, shorthand syntax for
N = 4 is
π
π n k
2π
w4 ≡ exp −j
= exp −j
∴ exp −j
≡ w4n · k
4
2
2
Fig. 6.19 Example P.6.17
x0
1
1
2π·0
X0 = x0 + x1 e−j 2
w20 = 1
x1
−w20 = −1
2π·0
X1 = x0 − x1 e−j 2
176
6 DT Fourier Transformation
∴
π 0
w40 = exp −j
=1
2
π 1
= −j
w41 = exp −j
2
π 2
w42 = exp −j
= exp (−j π ) = −1
2
π 3
π
3π
= exp −j
= exp j
=j
w43 = exp −j
2
2
2
so that,
k = 0 : X0 = x0 w40 · 0 + x1 w41 · 0 + x2 w42 · 0 + x3 w43 · 0
= x0 + x1 + x2 + x3 (“DC component”)
k = 1 : X1 = x0 w40 · 1 + x1 w41 · 1 + x2 w42 · 1 + x3 w43 · 1
π
3π
1
2
3
= x0 + x1 w4 + x2 w4 + x3 w4 =
≡−
2
2
= x0 + x1 w41 − x2 − x3 w41
k = 2 : X2 = x0 w40 · 2 + x1 w41 · 2 + x2 w42 · 2 + x3 w43 · 2
= x0 + x1 w42 + x2 w44 + x3 w46
= x0 − x1 + x2 − x3
k = 3 : X3 = x0 w40 · 3 + x1 w41 · 3 + x2 w42 · 3 + x3 w43 · 3
3π
9π
= x0 + x1 w43 + x2 w46 + x3 w49 =
≡−
2
2
= x0 + x1 w41 − x2 − x3 w41
This set of equation may be written in a more symmetric form as,
X0 = (x0 + x2 ) + w40 (x1 + x3 )
X1 = (x0 − x2 ) + w41 (x1 − x3 )
X2 = (x0 + x2 ) + w42 (x1 + x3 )
X3 = (x0 − x2 ) + w43 (x1 − x3 )
which can be further structured to explicitly include the form of two-bit butterfly (i.e. N = 2) as
X0 = (x0 + w20 x2 ) + w40 (x1 + w20 x3 )
6.4 Fast Fourier Transform
177
bit inversion
bit inversion
[0 0] x0
even
X0 [0 0]
+w20
−w20
[1 0] x2
X1 [0 1]
[0 0] x0 = [1]
[1 0] x2 = [3]
4
+w20
−w20
−2
−w40
odd
+w41
[1 1] x3
+w20
−w20
[0 0]
X1 = [−2 − 2j]
[0 1]
X2 = [−2]
[1 0]
X3 = [−2 + 2j]
[1 1]
+w40
+w40
[0 1] x1
X0 = [10]
X2 [1 0]
−w41
X3 [1 1]
[0 1] x1 = [2]
[1 1] x3 = [4]
6
−w40
+w41
+w20
−w20
−2
−w41
Fig. 6.20 Examples P.6.18, P.6.19
X1 = (x0 − w20 x2 ) + w41 (x1 − w20 x3 )
X2 = (x0 + w20 x2 ) − w40 (x1 + w20 x3 )
X3 = (x0 − w20 x2 ) − w41 (x1 − w20 x3 )
Direct graph implementation of these equations is illustrated in Fig. 6.20 (left), where starting from
the left side of the graph and following xn terms along the associated paths (the addition operations
are done at the merging nodes) the four Xk equations are composed simply by inspection. Note the
index order of xn and Xk indexes in this graph structure.
6.19. In order to use FFT graph diagram, in the first step xn terms are reorganized on the left side by
using the index bit inversion relative to Xk , see Fig. 6.20 (right). The intermediate sums are calculated
using the two-bit grouping on the left side, then the second group of addition/multiplication operations
is done by following paths in the right side of the graph.
6.20. In the case of N = 8, shorthand syntax is
π
π n k
2π
≡ w8n · k
w8 ≡ exp −j
= exp −j
∴ exp −j
8
4
4
∴
π 0
w80 = exp −j
=1
4
√
π 1
2
1
w8 = exp −j
(1 − j )
=
4
2
π 2
π
= −j
= exp −j
w82 = exp −j
4
2
√
π 3
3π
2
= exp −j
w83 = exp −j
=−
(1 + j )
4
4
2
π 4
w84 = exp −j
= exp (−j π ) = −1 ≡ −w80
4
178
6 DT Fourier Transformation
Fig. 6.21 Example P.6.20
x0
X0
x4
X1
x2
X2
w82
x6
−w82
X3
x1
X4
w81
x5
−w81
X5
−w82
X6
−w83
X7
w82
x3
w83
w82
x7
−w82
√
2
π 5
5π
−j
= exp −j
=−
(1 − j ) ≡ −w81
4
4
2
π 6
6π
w86 = exp −j
= exp −j
= j ≡ −w82
4
4
√
π 7
7π
2
7
w8 = exp −j
= exp −j
=
(1 + j ) ≡ −w83
4
4
2
w85 = exp
In reference to Fig. 6.18, FFT graph flow is derived for N = 8 as in Fig. 6.21, where horizontal red
lines imply multiplication by ‘−1’, dark lines imply multiplication by ‘1’, and other multiplication
factors are written explicitly.
6.21. Given sequence xn = [1, 1, −1 − 1, 1, 1, −1 − 1], by inspection of graph in Fig. 6.21, where
horizontal red lines imply multiplication by ‘−1’ it follows that,
X0 = x0 + x4 + x2 + x6 + x1 + x5 + x3 + x7 = 1 + 1 − 1 − 1 + 1 + 1 − 1 − 1
=0
X1 = x0 − x4 + (x2 − x6 )(−j ) + x1 − x5 + (x3 − x7 )(−j ) w82
= 1 − 1 − 1 + 1 + (1 − 1 + (−1 + 1)(−j ))w82
=0
X2 = x0 + x4 + (x2 + x6 )(−1) + x1 + x5 + (x3 + x7 )(−1) (−j )
= 1 + 1 + 1 + 1 + (1 + 1 + 1 + 1)(−j )
= 4 − 4j
X3 = x0 − x4 + (x2 − x6 )(−w82 ) + x1 − x5 + (x3 − x7 )(−w82 )
6.4 Fast Fourier Transform
179
= 1 − 1 + (−1 + 1)(−w82 ) + 1 − 1 + (−1 + 1)(−w82 )
=0
X4 = x0 + x4 + x2 + x6 + (x1 + x5 + x3 + x7 )(−1)
=1+1−1−1−1−1+1+1
=0
X5 = x0 − x4 + (x2 − x6 )(−j ) + x1 − x5 + (x3 − x7 )(−j ) (−w81 )
= 1 − 1 + (−1 + 1)(−j ) + 1 − 1 + (−1 + 1)(−j ) (−w81 )
=0
X6 = x0 + x4 + (x2 + x6 )(−1) + x1 + x5 + (x3 + x7 )(−1) (j )
= 1 + 1 + 1 + 1 + (1 + 1 + 1 + 1)(j )
= 4 + 4j
X7 = x0 + x4 + (x2 − x6 )(j ) + x1 − x5 + (x3 − x7 )(j ) (−w83 )
= 1 + 1 + (−1 + 1)(j ) + 1 − 1 + (−1 + 1)(j ) (−w83 )
=0
The total number of sum/product calculations is 8 log2 8 = 24 in comparison with DFT where the
total number of calculations equals 82 = 64.
7
Laplace Transformation
Laplace transformation (LT) is a tool to convert time domain function f (t) into its s–domain
equivalent F (s). Virtually, all books on Laplace transformation first present table with f (t) → F (s)
LT pairs of basic functions. However, it is very important that, instead of memorizing these tabulated
pairs, these f (t) → F (s) pairs are derived from definition. After some practice and experience, the
lookup tables are used to speed up the calculations.
Laplace and inverse Laplace integrals: given complex variable s = σ + j ω, (σ, ω ∈ R), then
unilateral LT and inverse LT integrals are
F (s) = L f (t) =
+∞
f (t) e−st dt
(7.1)
c+j τ
1
lim
F (s) est ds
j 2π τ →∞ c−j τ
(7.2)
0
f (t) = L −1 F (s) =
where c ∈ R and “contour integral” (7.2) is solved with the help of the residue theorem. Note that
solution to inverse LT my not be unique. Traditionally, the inverse LT integral is not solved at the
introductory level courses; instead, “the table of simple inverse transforms” and LT properties are
used.
Properties of Laplace Transformation
1. Linearity: given three functions f (t), x(t), and y(t) and constants a, b, then
f (t) = ax(t) + by(t) ⇒ L {f (t)} = a L {x(t)} + bL {y(t)}
LT
2. Time shifting: given x(t) → X(s), a constant a > 0, and step function u(t), then
L {x(t − a)u(t − a)} = e−as X(s)
LT
3. Frequency shifting: given x(t) → X(s) and a constant a > 0, then
L eat x(t) = X(s − a)
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5_7
181
182
7 Laplace Transformation
LT
4. Time scaling: given x(t) → X(s) and a constant a > 0, then
L {x(at)} =
1 s X
a
a
LT
5. Derivative property: given n times differentiable x(t) → X(s), then
n
L x (n) = s n X(s) −
x k−1 (0)
k=1
LT
6. Multiplication by t n : given x(t) → X(s) and a positive integer n > 0, then
L t n x(t) = (−1)n X(n) (s)
LT
LT
7. Convolution property: given functions x(t) → X(s) and y(t) → Y (s), then
L {x(t) ∗ y(t)} = X(s) Y (s)
Problems
7.1
Basic Laplace Transformations
Given functions in P.7.1 to P.7.15, derive their respective LT by using definition integral (7.1). Assume
a > 0, a ∈ R.
7.1. f (t) = 1
7.2. f (t) = u(t)
7.3. f (t) = u(t − a)
7.4. f (t) = δ(t)
7.5. f (t) = δ(t − a)
7.6. f (t) = t u(t)
7.7. f (t) = t 2 u(t)
7.8. f (t) = t a u(t)
7.9. f (t) = eat u(t)
7.10. f (t) = e−at u(t)
7.11. f (t) = eat + e−at u(t) 7.12. f (t)= u(t) − u(t − a)
7.13. f (t) = sin(at) u(t)
7.14. f (t) = cos(at) u(t)
7.2
7.15. ∀t ∈ {R} : f (t) = f (t + T )
Basic Properties
It is educative, instead of memorizing the properties table, to prove LT properties starting from the
definition integral. After some practice, these properties may be applied directly and speed up the
calculations.
Given functions in P.7.16 to P.7.30, where applicable, derive properties of Laplace transformation
assuming t ≥ 0 and a > 0.
7.4 Differential Equations
183
7.16. g(t) = f (t)
7.17. g(t) = f (t)
7.18. g(t) = f (t)
7.19. f (t) = sin(2t)
7.20. f (t) = cos(2t)
7.21. f (t) = t sin(2t)
7.22. f (t) = t 2 cos t
7.23. g(t) = eat f (t)
7.24. g(t) = f (at)
7.25. g(t) =
7.28. f (t) =
7.3
f (t)
t
e4t − e−3t
t
7.26. g(t) =
t
f (z) dz
7.27. f (t) =
t
0
(p2 + p − ep ) dp
0
7.29. f (t) = (t − 2)3 u(t − 2) 7.30. f (t) = sin t u(t)
t
Inverse Laplace Transformation
Given in P.7.31 to P.7.42, derive their respective inverse LT by taking advantage of already derived
transformation pairs in the previous examples of this chapter.
7.31.
1
s
7.34. e−s
7.37.
s
s2 + 9
7.40. arctan
7.4
1
s
7.32.
1 −2s
e
s
7.35.
1
s2
7.36.
7.38.
1
1 − e−2s
s
7.39.
7.41.
2s
s 2 − a2
7.33. 1
7.42.
2
s3
4s
(s 2 + 4)2
s
s 2 + 4s + 5
Differential Equations
Differential equations include the solution function as well as its derivatives. Very efficient method
for solving some classes of differential equations is to apply “derivative property” of LT and convert
these equations into ordinary algebraic polynomials. Even without formal (inductive) proof, it should
not be difficult to adopt this property and use it within specific problems.
Reminder:: “derivative property” of LT is
Fn (s) = L f (n) (t) = s n · L {f (t)} − s n−1 f (0) − · · · − f (n−1) (0)
where s n is nth power of the polynomial variable s, f (n) (t) stands for nth derivative of f (t),
and f (0) stands for the initial condition of f (t) at t = 0, i.e., before the input signal is injected.
184
7 Laplace Transformation
By applying LT derivative property, solve differential equations given in P.7.43 to P.7.51. As already
known, ordinary integrals deliver general solution, i.e., there is an infinity of solution functions
separated by a constant. Given sufficient number (equal to the order of differential equation) of the
initial conditions, general solution is resolved into specific solution.
7.43. y + 1 = 0, y(0) = 0
7.44. −2y + y = 0, y(0) = 1
7.45. y − 1 = 0, y (0) = 0, y(0) = 0
7.46. y − y + 1 = 0, y (0) = 1, y(0) = 1
7.47. y + 2y + 2y = 0, y (0) = 1, y(0) = −1 7.48. y + y = sin(2t), y (0) = 0, y(0) = 0
7.49. y + y = 2e−t , y (0) = 1, y(0) = 1
7.50. y − y = cos(t), y(0) = 0, y (0) = 0
7.51. y − y = 0, y(0) = −3, y (0) = 1, y (0) = 4
7.1 Basic Laplace Transformations
185
Answers
Tutorial-type solutions to given examples are aligned so that subsequent forms are, as much as
possible, logical development from the preceding ones. Thus, it is highly recommended, before
looking into the proposed solutions, to make every effort and to explore own ideas. What is more, as
they are backward referenced in the subsequent examples, it is assumed that the previously introduced
ideas are already adopted. After reading the solutions, it is important to independently try again and
even to explore other possible methods. Finally, as each problem could be solved with several different
techniques, the methods presented in each of the problems are by no means the only ones possible.
7.1
Basic Laplace Transformations
7.1. Given time domain function
f (t) = 1
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
+∞
1 e−st dt = change variable
0
dz
1
= −s ∴ dt = − dz
dt
s
1
1
=−
ez dz = − ez back to the original variable s and its limits
s
s
+∞
1
1
1 −∞
1
*0 0
= − e−st e −
e7
=− =
s
s
s
0
∴
z = −st
Note that, given t ≥ 0, a constant “1” is indistinguishable from step function u(t).
7.2. Given time domain function of unit step
f (t) = u(t)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e
−st
dt =
0
+∞
u(t) e
−st
dt = u(t) = 1 for t ≥ 0 =
0
1
= see P.7.1 =
s
which is expected, considering that for x ≥ 0 function u(t) = 1.
+∞
0
1 e−st dt
186
7 Laplace Transformation
7.3. Given time domain function of delayed unit step
f (t) = u(t − a)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
+∞
u(t − a) e−st dt = u(t − a) = 1 for t ≥ a
0
a
* 0 +∞
1
−st
ez dz
=
0e dt +
1 e−st dt = −
s
0
a
dz
1
change variable: z = −st ∴
= −s ∴ dt = − dz
dt
s
+∞
1 −∞
1 −as
1
*0
e
= − e−st = − e − e−as =
s
s
s
a
Compare this result with P.7.2 and note that time–delaying a function by a translates into multiplying
its Laplace transformation with the exponential term exp(−as). In general, this case is known as
“time shifting” property of Laplace transformation.
Reminder: Given that L {f (t)} = F (s), time shifting property of Laplace transformation is
generalized as
L {f (t − a)} = e−as F (s)
7.4. Given time domain Dirac delta function
f (t) = δ(t)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e
0
−st
dt =
+∞
δ(t) e
0
−st
1 +∞
*1
0
7
dt = t = 0 = δ(t) dt
e
0
=1
because, at t = 0 Dirac delta function is included within the (0, +∞) interval.
7.5. Given time–delayed Dirac delta function
f (t) = δ(t − a)
by definition, Laplace transformation F (s) = L f (t) is
7.1 Basic Laplace Transformations
F (s) =
+∞
f (t) e
−st
dt =
187
+∞
0
δ(t − a) e
−st
dt = t = a, a > 0 = e−as
0
+∞
:1
δ(t
−
a)
dt
0
= e−as
compare with the result in P.7.4 (as expected and already noted in P.7.3).
7.6. Given time domain ramp function
f (t) = t u(t)
Method 1: by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
+∞
t u(t) e−st dt = u(t) = 1 for t ≥ 0 =
0
+∞
t e−st dt
0
dz
1
1
= −s ∴ dt = − dz, and t = − z
dt
s
s
1
= 2 z ez dz = integration by parts: u = z, dv = ez dz ∴ du = dz, v = ez
s
1
1 1
= 2 z ez − ez dz = 2 z ez − ez = 2 ez z − 1
s
s
s
return to the original variable
+∞
1
1
1
= 2 e−st −st − 1 = 2 0 − e0 −(0)s − 1 = 2
s
s
s
0
=
z = −st ∴
Method 2: Compare with the result in P.7.2 and note the consequence of multiplying u(t) by t. In
general, this case is known as “multiplying with power of t” property of Laplace transformation.
Reminder: Given that L {f (t)} = F (s), multiplying with power of t property of Laplace
transformation is generalized as
dn
L t n f (t) = (−1)n n F (s) = (−1)n F (n) (s)
ds
where, F (n) (s) stands for nth derivative of F (s).
In this example, power of t 1 is n = 1, so multiplying with power of t property is applied directly
as
f (t) = u(t) ∴ F (s) =
∴
1
1
dF (s)
∴ F (1) (s) =
=− 2
s
ds
s
and (−1)1 = −1
188
7 Laplace Transformation
L {t u(t)} = (−1)1
dF (s)
= −1
ds
1
− 2
s
=
1
s2
7.7. Given u(t) multiplied by t 2 , as
f (t) = t 2 u(t)
Method 1: by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
+∞
t 2 u(t) e−st dt = u(t) = 1 for t ≥ 0 =
0
+∞
t 2 e−st dt
0
1
1
dz
= −s ∴ dt = − dz, and t 2 = 2 z2
dt
s
s
1
= − 3 z2 ez dz = integration by parts: u = z2 , dv = ez dz ∴ du = 2z dz, v = ez
s
1 2 z
z
see P.7.6 : z ez dz = ez z − 1
= − 3 z e − 2 z e dz
s
1 1
= − 3 z2 ez − 2ez z − 1 = − 3 ez z2 − 2z + 2 return to the original variable
s
s
+∞
1
1
2
= − 3 e−st (−st)2 − 2(−st) + 2
= − 3 0 − 1(0 + 0 + 2) = 3
s
s
s
0
=
z = −st ∴
Method 2: By using the “multiplying with power of t” property of Laplace transformation, in this
example, power of t 2 is n = 2, so multiplying with power of t property is applied directly as
f (t) = u(t) ∴ F (s) =
1
2
d 2 F (s 2 )
∴ F (2) (s) =
= 3
s
ds
s
and (−1)2 = 1
∴
L {t u(t)} = (−1)2
d 2 F (s)
=1
ds 2
2
s3
=
2
s3
Evidently, with the experience, the properties of Laplace transformation are used as a shortcut.
7.8. Given differential equation with constant coefficients
f (t) = t a u(t)
without knowing value of a, evidently, solving by definition is not much practical because it requires
a bit longer proof by the induction method.
7.1 Basic Laplace Transformations
189
Reminder: Positive integer powers of t a are generalized by the induction method as
a!
L t a u(t) = (−1)a u(a) (s) = a+1
s
where, u(a) (s) stands for ath derivative of u(s).
Therefore,
F (s) =
+∞
t a e−st dt =
0
a!
s a+1
7.9. Given time domain function
f (t) = eat u(t)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
+∞
eat e−st dt =
0
+∞
e−(s−a)t dt = change variable
0
1
dz
= −(s − a) ∴ dt = −
dz
z = −(s − a)t ∴
dt
s−a
1
1
=−
ez dz = −
ez return to the original variable
s−a
s−a
=−
1
+∞
1
1 a 0 1
0
e−(s−a)t e7
=−
=
e − 0
s−a
s−a
s−a
Note, if
a = 1 : F (s) =
+∞
et e−st dt =
0
a = −1 : F (s) =
+∞
1
s−1
e−t e−st dt =
0
1
s+1
7.10. Given time domain function
f (t) = e−at u(t)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e
−st
dt =
0
z = −(s + a)t
+∞
e
0
∴
−at
e
−st
dt =
+∞
e−(s+a)t dt = change variable
0
dz
1
= −(s + a) ∴ dt = −
dz
dt
s+a
190
7 Laplace Transformation
1
=−
s+a
ez dz = −
1
ez return to the original variable
s+a
+∞
1
1 a 0 1
1
−(s+a)t 0
7
=−
e
=−
e −
e
=
s+a
s
+
a
s
+
a
0
7.11. Given time domain function
f (t) = eat u(t) + e−at u(t)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e
−st
0
dt =
+∞
at
e + e−at e−st dt =
+∞
0
dt +
0
see P.7.9 and P.7.10
=
e
−(s−a)t
+∞
e−(s+a)t dt
0
1
1
(s + a) + (s − a)
2s
+
=
= 2
s−a
s+a
(s − a)(s + a)
s − a2
7.12. Given time domain function of rectangular impulse
f (t) = u(t) − u(t − a)
by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
0
−
f (t) e−st dt =
+∞
+∞
u(t) − u(t − a) e−st dt =
+∞
0
u(t) e−st dt
0
u(t − a) e−st dt
0
1 1 −as
1
= see P.7.2 and P.7.3 = −
= 1 − e−as
e
s
s
s
7.13. Given time domain function sinusoidal function
f (t) = sin(at) u(t) =
ej at − e−j at
u(t)
2j
and u(t) = 1, (t ≥ 0) then, by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
0
−
1
2j
f (t) e−st dt =
+∞
+∞
0
1
ej at − e−j at
u(t) e−st dt =
2j
2j
+∞
e(j a−s)t dt
0
e−(j a+s)t dt
0
see P.7.11
=
1
2j
1
1
−
s − ja
s + ja
=
1
2j
(s + j a) − (s − j a)
(s − j a)(s + j a)
=
1 @
2j
a
@a =
2
2
2
s
+
a
s
+
a2
2j
@
@
7.2 Basic Properties
191
7.14. Given time domain function
f (t) = cos(at) u(t) =
ej at + e−j at
u(t)
2
and u(t) = 1, (t ≥ 0) then, by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
see P.7.11
=
1
2
+∞
0
1
1
+
s − ja
s + ja
=
ej at + e−j at −st
1
e dt =
2
2
+∞
0
(s + j a) + (s − j a)
(s − j a)(s + j a)
1
2
e(j a−s)t dt +
=
1
2
+∞
e−(j a+s)t dt
0
1 2A s
s
= 2
2
2
s + a2
2A s + a
7.15. Given time–domain periodic function
∀t ∈ R : f (t) = f (t + T )
by taking advantage of the time–shifting property, it follows that
F (s) =
+∞
f (t) e−st dt =
T
0
e−st f (t) dt +
2T
0
e−st f (t) dt +
3T
e−st f (t) dt + · · ·
2T
T
change variable: t = z, t = z + T , t = z + 2T , . . . ∴ dt = dz, . . .
T
2T
3T
−sz
−s(z+T )
=
e f (z) dz +
e
f (z + t) dz +
e−s(z+2T ) f (z + 2T ) dz + · · ·
0
2T
T
L {f (t − a)} = e−as F (s)
T
T
T
e−sz f (z) dz + e−sT
e−sz f (z) dz + e−s2T
e−sz f (z) dz + · · ·
=
0
0
= 1 + e−sT + e−s2T + · · ·
T
0
e−sz f (z) dz
0
∞
1
sum of infinite geometric series:
z =
1
−
z
n=0
return to the original variable
T
1
e−st f (t) dt
=
1 − e−sT 0
7.2
n
Basic Properties
7.16. Given first derivative f (x) as
g(t) = f (t) =
df (t)
dt
(see Ch. 1)
192
7 Laplace Transformation
and by recalling that condition for existence of the transform is that f (t) is finite, i.e. f (t) < ∞
for all t, then by definition, Laplace transformation G(s) = L g(t) is
G(s) =
+∞
g(t) e−st dt =
+∞
0
f (t) e−st dt =
0
+∞
e−st
0
df (t)
Z
dZt
Z
dZt
integration by parts: u = e−st , dv = df (t) ∴ du = −s e−st , v =
df (t) = f (t)
+∞
1
+∞
:0
−∞
0
7
e
+ s F (s)
f (+∞)
= f (t) e−st +s
f (t) e−st dt = − f (0) e
0
0
F (s)=L f (t)
= s F (s) − f (0)
because f (t) < ∞ ∴ f (+∞) · 0 = 0 . In addition, f (0) stands for the initial condition of f (t)
at t = 0 (i.e. just before the signal is injected into system).
Reminder: Laplace transformation of a function’s first derivative is
L
df (t)
dt
= s F (s) − f (0)
This general case is known as “first derivative property” of Laplace transformation.
7.17. Given second derivative f (x) as
d 2 f (t)
df (t)
=
dt 2
dt
by definition, Laplace transformation G(s) = L g(t) is
g(t) = f (t) =
G(s) =
+∞
g(t) e
−st
dt =
0
+∞
f (t) e
0
−st
dt =
+∞
0
e−st
df (t)
Z
dZt
Z
dZt
integration by parts: u = e−st , dv = df (t) ∴ du = −s e−st , v =
= f (t) e
+s
+∞
1
+∞
:0
−st
−∞
0
7
f
e
+s
f (t) e dt = (+∞) e
− f (0) −st +∞
0
0
0
f (t) e−st dt
see P.7.16
= s s F (s) − f (0) − f (0) = s 2 F (s) − s f (0) − f (0)
df (t) = f (t)
7.2 Basic Properties
193
7.18. Given third derivative f (x) as
d 3 f (t)
df (t)
=
dt 3
dt
by definition, Laplace transformation G(s) = L g(t) is
g(t) = f (t) =
G(s) =
+∞
g(t) e−st dt =
+∞
0
0
integration by parts: u = e
= f (t) e
+s
f (t) e−st dt =
e−st
0
−st
df (t)
Z
dZt
Z
dZt
, dv = df (t) ∴ du = −s e
−st
, v=
df (t) = f (t)
+∞
1
+∞
:0
−st
−∞
0
7
e
e
+s
f (t) e dt = (+∞)
− f (0) f
−st +∞
+∞
0
0
f (t) e−st dt
see P.7.17
0
= s s 2 F (s) − s f (0) − f (0) − f (0) = s 3 F (s) − s 2 f (0) − s f (0) − f (0)
Reminder: For the reference, this derivative property of Laplace transformation is generalized
for nth derivative as
L f (n) (t) = s n L {f (t)} − s n−1 f (0) − s n−2 f (0) − · · · − f (n−1) (0)
7.19. Given composite sinusoidal function
f (t) = sin(2t) =
e2j t − e−2j t
2j
by definition (see P.7.13), Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e
0
=
1
2j
+∞
0
see P.7.11
=
1
2j
−st
dt =
+∞
0
e(2j −s)t dt −
1
1
−
s − 2j
s + 2j
1
2j
e2j t − e−2j t −st
e dt
2j
+∞
e−(2j +s)t dt
0
=
2j
1 (s + 2j ) − (s − 2j )
1 2@
@ = 2
=
2
2
2j (s − 2j )(s + 2j )
s2 + 4
2j
@
@ s +2
194
7 Laplace Transformation
7.20. Given sinusoidal function
e2j t + e−2j t
2
by definition (see P.7.14), Laplace transformation F (s) = L f (t) is
f (t) = cos(2t) =
F (s) =
+∞
f (t) e−st dt =
0
see P.7.11
=
1
2
+∞
0
1
1
+
s − 2j
s + 2j
=
e2j t + e−2j t −st
1
e dt =
2
2
+∞
e(2j −s)t dt +
0
1
2
+∞
e−(2j +s)t dt
0
s
1 (s + 2j ) + (s − 2j )
1 2A s
= 2
=
2
2
2 (s − 2j )(s + 2j )
s +4
2A s + 2
7.21. Given sinusoidal function multiplied by power of t
f (t) = t sin(2t)
Method 1: by definition, Laplace transformation F (s) = L f (t) is
F (s) =
+∞
f (t) e−st dt =
0
+∞
t sin(2t) e−st dt
0
1
integration by parts: u = te−st , dv = sin(2t) dt ∴ du = e−st (1−st) dt, v = − cos(2t)
2
+∞
+∞
1
1
= − cos(2t) t e−st +
cos(2t) e−st (1 − st) dt
0
2
2 0
+∞
1 +∞
1
cos(2t) e−st dt − s
cos(2t) e−st t dt
=0+
2 0
2
0
L {cos(2t)}, P.7.20
+∞
1 s
1
=
−
s
cos(2t) e−st t dt
2 s2 + 4 2
0
because
:0
+∞
:0
−st
−s(0)
t cos(2t)
e
−
cos(2t) t e−st = lim cos(2(0))
(0)
e
t→∞
0
where the above limit may be proved, for example, by the Squeeze theorem. In addition,
+∞
cos(2t) e−st t dt =
0
integration by parts: u = te−st , dv = cos(2t) dt ∴ du = e−st (1 − st)dt, v =
=
+∞ 1 +∞
1 −st
te
sin(2t)
−
sin(2t)e−st (1 − st) dt
0
2
2 0
1
sin(2t)
2
7.2 Basic Properties
1
=0−
2
195
+∞
1
sin(2t) e dt + s
2
0
L {sin(2t)}, P.7.19
+∞
1
+ s
t sin(2t) e−st dt
2 0
−st
+∞
t sin(2t) e−st dt = −
0
1 2
2 s2 + 4
F (s)
=−
1 2A 1
1
+ s F (s)
2
s
+
4
2
2A
because
:0
+∞
:0
−st
−s(0)
= lim sin(2(0))
(0)
e
sin(2t)
t
e
−
sin(2t) t e−st t→∞
0
Altogether,
1 s
1
F (s) = 2
− s
2 s +4 2
=
F (s)
+∞
cos(2t) e
−st
0
1 s
1
1
1
t dt =
− s − 2
+ s F (s)
2 s2 + 4 2
s +4 2
1 s
1
1 s
+
− s 2 F (s)
2 s2 + 4 2 s2 + 4 4
s2 + 4
s
4s
= 2
∴ F (s) = 2
4
s +4
(s + 4)2
Method 2: After some practice, the property of multiplication of a function by powers of t (see P.7.6)
may be adapted directly, as
dn
L t n f (t) = (−1)n n F (s) = (−1)n F (n) (s)
ds
In this example, power of t 1 is n = 1, so multiplying with power of t property is applied directly as
f (t) = sin(2t) ∴ F (s) =
4s
dF (s)
2
∴ F (1) (s) =
=− 2
s2 + 4
ds
(s + 4)2
and (−1)1 = −1
∴
L {t sin(2t)} = (−1)1
4s
dF (s)
= 2
ds
(s + 4)2
7.22. Given sinusoidal function multiplied by power of t as
f (t) = t 2 cos t
derivation by definition, Laplace transformation F (s) = L f (t) is a bit longer than in P.7.21.
Taking advantage of the property of multiplication of a function by powers of t,
196
7 Laplace Transformation
dn
L t n f (t) = (−1)n n F (s) = (−1)n F (n) (s)
ds
Here, power of t 2 is n = 2 thus (−1)2 = 1, so “multiplying with power of t” property leads to
f (t) = cos t ∴ F (s) =
s
s2 + 1
∴ F (2) (s) =
d 2 F (s)
d d
=
2
ds
ds ds
s
s2 + 1
=
d
ds
−s 2 + 1
(s 2 + 1)2
=
2s(s 2 − 3)
(s 2 + 1)3
∴
d 2 F (s)
2s(s 2 − 3)
L t 2 cos t = (−1)2
= 2
2
ds
(s + 1)3
which is, evidently, much faster than derivation by definition.
7.23. Given function multiplied by exponent
g(t) = eat f (t)
by definition, Laplace transformation is
L {f (t)} = F (s) =
+∞
f (t) e−st dt
0
∴
+∞
+∞
+∞
F (s − a) =
f (t) e−(s−a)t dt =
f (t) eat e−st dt = eat
f (t) e−st dt
0
0
0
= e L {f (t)}
at
Reminder: This general relation is known as “frequency shifting” property
F (s − a) = eat L {f (t)}
Note that the shifting of s argument by a generates exponential term eat that multiples the
non–shifted function’s transformation.
7.24. Given function f (t) whose argument is time–scaled by a as
g(t) = f (at)
by definition, Laplace transformation F (s) = L f (t) is
7.2 Basic Properties
L {f (t)} =
197
+∞
f (t) e−st dt
0
∴
+∞
+∞
+∞
a
1
L {f (at} =
f (at) e−st dt =
f (at) e−st dt =
f (at) e−st d(at)
a
a
0
0
0
s
1 +∞
∴ s = az
f (at) e−st d(at) = change variable: z =
=
a
a 0
1 +∞
1
=
f (at) e−z(at) d(at) = L {f (z)}
a 0
a
s
1
s
1
= F
= L f
a
a
a
a
Note that basic idea in this proof is to replace all arguments t with arguments at. By doing so, form
of Laplace transformation relative to variable z is achieved.
Reminder: This general relation is known as “time scaling” property of Laplace transformation
L {f (at)} =
1 s F
a
a
7.25. Given function f (t) that is scaled by t as
g(t) =
f (t)
t
assuming that limt→0 g(t) exists. Note that 1/t = t −1 , then by definition and the multiplying with
power of t property, Laplace transformation is derived as
f (t) = t g(t) ∴ L {t g(t)} = L {f (t)}
∴
d
L {g(t)} = F (s)
(−1)1
ds
d L {g(t)} = − F (s) ds
∴
change to temporary variable s = z
s
∞
L {g(t)} = −
F (z) dz =
F (z) dz
∞
L
f (t)
t
∴
∞
F (z) dz
=
s
s
198
7 Laplace Transformation
Reminder: This general relation of division by t is also known as “frequency–domain
integration” property of Laplace transformation
f (t)
t
L
∞
=
F (z) dz
s
where z is temporary variable that helps to preserve s as the final result variable.
7.26. Given time–domain function g(t) as
g(t) =
t
f (z) dz
0
Method 1: by definition, Laplace transformation F (s) = L f (t) is
G(s) =
+∞
t
g(t) e−st dt = L
f (z) dz
0
=
0
+∞ t
0
f (z) dz e−st dt
0
t
1
f (z) dz, v = e−st ∴ u = f (t), v = − e−st
s
0
∞ ∞
t
1
1
f (z) dz
−
− e−st f (t) dt
= − e−st
s
s
0
0
0
∞
+∞
1 −∞
1 ∞
*o
e = − f (z) dz
+
f (t)e−st dt
s
s 0
0
0
integration by parts: u =
F (s)=L f (t)
∴
L
t
f (z) dz =
0
F (s)
s
Method 2: taking advantage of derivative property of Laplace transformation,
g(t) =
t
0
f (z) dz ∴ g (t) = f (t),
and, g(0) = 0
L g (t) = L {f (t)} = F (s)
where,
L g (t) = s L {g(t)} − g(0) = s L {g(t)}
∴
F (s) = s L
t
f (z) dz
0
∴
7.2 Basic Properties
199
L
t
F (s)
s
f (z) dz =
0
Reminder: This general relation also known as “time-domain integration” property of Laplace
transformation
t
F (s)
L
f (z) dz =
s
0
where z is temporary variable that helps to preserve t as the final result variable.
7.27. Given time domain function f (t) as
f (t) =
t
(z2 + z − ez ) dz
0
by taking advantage of the transformation properties as
L {f (t)} = L
t
t
(z + z − e ) dz = L
2
z dz + L
t
2
z
0
0
z dz − L
0
t
ez dz
0
it follows that,
2 L p2 = 3 see P.7.7 ∴ L
s
L {p} =
L ep =
1 see P.7.6 ∴ L
2
s
t
z2 dz =
0
t
0
1 see P.7.9 ∴ L
s−1
z dz =
t
2/s 3
2
= 4
s
s
1
1/s 2
= 3
s
s
ez dz =
0
1
1/(s − 1)
=
s
s(s − 1)
∴
L {f (t)} =
2
−s 3 + s 2 + s − 2
1
1
=
+
−
s4
s3
s(s − 1)
s 4 (s − 1)
7.28. Given time–domain function
f (t) =
et − e−3t
t
by taking advantage of the transformation properties as
L
it follows that
et − e−3t
t
=L
et
t
−L
e−3t
t
200
7 Laplace Transformation
1 see P.7.9
s−1
1 L e−3t =
see P.7.9
s+3
L et =
L
e −e
t
t
−3t
∴
∞
=
∞
1
1
dz −
dz see P.7.25
z−1
z+3
s
s
∞
∞
∞
∞
z − 1 z − 1 z − 1 =
ln
−
ln
= ln(z − 1) − ln(z + 3) = ln
s
s
z + 3 s
z + 3
z + 3 s
=
∞ ∞
= ln
*0
s − 1 l’H
s−1
s−1
s−1
1
− ln
− ln
= − ln
= ln
= lim
ln
s→∞
s+3
1
s+3
s+3
s+3
−1
s+3
s−1
7.29. Given time–domain function
f (t) = (t − 2)3 u(t − 2)
Reminder:
n!
L t n u(t) = n+1
s
L {f (t − a)} = e−as F (s)
Method 1: by taking advantage of transformation properties, first, transformation of the the non–
delayed version of f (t) simply f (t) = t 3 ∴ (n = 3), thus
f (t) = t 3 ∴ F (s) =
3!
s 3+1
=
6
s4
then, delaying f (t) by a = 2 results in
6
f (t) = (t − 1)3 ∴ F (s) = e−2s 4
s
Method 2: by definition,
7.2 Basic Properties
F (s) =
201
+∞
f (t) e−st dt
0
=
2
3 −st
(t − 2) e
dt +
+∞
0
2
0
L z3 (see P.7.8)
3 −st
(t − 2) e
dt = 0 +
+∞
(t − 2)3 e−st dt
2
change variable: z = t − 2, dz = dt, t = z + 2
adjust the limits: t → 2 ⇒ z → 0, t → ∞ ⇒ z → ∞
+∞
+∞
+∞
z3 e−s(z+2) dz =
z3 e−2s e−sz dz = e−2s
z3 e−sz dz
=
=e
−2s
= e−2s
0
0
6
s4
7.30. Given time–domain function a.k.a sinc (t)
sin t
t
Reminder:
L
f (t)
t
=
∞
F (z) dz
s
L {sin(t)} =
1
s2 + 1
by taking advantage of transformation’s frequency–domain integration properties, first, transformation of f (t) version that is not divided by t,
F (s) = L {sin(t)} =
L
sin t
t
1
s2 + 1
∴
∞
∞
=
F (z) dz =
s
s
∞
1
dz
=
arctan(z)
= arctan(∞) − arctan(s)
s
z2 + 1
π
= − arctan(s)
2
trig identity: arctan(s) + arctan
= arctan
1
s
1
s
=
π
, (s > 0)
2
202
7.3
7 Laplace Transformation
Inverse Laplace Transformation
7.31. For example,
Method 1: Inverse of 1/s may be derived, for example, by using result in P.7.8 where it is found that
a!
L t a u(t) = a+1
s
which after setting a = 0 reduces to L t 0 u(t) = L {u(t)} (because t 0 = 1), so that
L {u(t)} =
0!
1
1
=
⇒ f (t) = L −1
s 0+1
s
s
= u(t)
Method 2: See P.7.1 and P.7.2, as t ≥ 0, then f (t) = L −1 {1/s} = u(t).
1 −2s
= u(t − 2)
e
s
7.32. See P.7.3, as a = 2, then f (t) = L −1
7.33. See P.7.4, f (t) = L −1 {1} = δ(t)
7.34. See P.7.5, as a = 1, then f (t) = L −1 e−s = δ(t − 1)
7.35. See P.7.6, f (t) = L −1
1
s2
= t u(t)
7.36. See P.7.7, f (t) = L −1
2
s3
= t 2 u(t).
s
7.37. See P.7.14, as a = 3 then f (t) = L −1
s 2 + 32
1
1 − e−2s
s
7.38. See P.7.12, as a = 2 then f (t) = L −1
7.39. See P.7.21, f (t) = L −1
4s
(s 2 + 4)2
7.40. See P.7.30, L −1 arctan
1
s
7.41. See P.7.11, L −1
2s
s 2 − a2
=
= cos(3 t) u(t)
= u(t) − u(t − 2)
= t sin(2t)
sin t
t
= eat + e−at u(t)
7.42. Inverse LT of rational function are found by partial fractions and decomposition of rational
fractions into already known rational terms. For example,
7.4 Differential Equations
f (t) =
=
203
s
2
2
s
=
+
−
s 2 + 4s + 5
(s + 2)2 + 1 (s + 2)2 + 1 (s + 2)2 + 1
s+2
1
−2
2
(s + 2) + 1
(s + 2)2 + 1
Therefore,
L −1 {f (t)} = L −1
1
s+2
−2
(s + 2)2 + 1
(s + 2)2 + 1
− 2L −1
= L −1
s+2
(s + 2)2 + 1
1
(s + 2)2 + 1
= see P.7.13, P.7.14 and P.7.23
= e−2t cos(t) − 2 e−2t sin(t) = e−2t (cos(t) − 2 sin(t))
7.4
Differential Equations
7.43. Given first order differential equation with constant coefficients
y +1=0 ∴
dy(t)
+1=0
dt
Method 1: postulate that Y (s) = L {y(t)}, then by taking advantage of LT derivative properties as
y + 1 = 0 ∴ L y + 1 = L {0} ∴ L y + L {1} = L {0}
sY (s) − y(0) +
1
1
1
= 0 ∴ sY (s) = −
∴ Y (s) = − 2
s
s
s
Reminder:
L {t} =
1
s2
1
s2
⇒ L −1
=t
Inverse Laplace, see P.7.6, is then
y(t) = −t
Method 2: direct integration results in
dy(t)
dy(t)
+1=0 ∴
= −1 ∴
dt
dt
∴
dy(t) = −
dt ∴ y(t) = −t + C
y(0) = 0
204
7 Laplace Transformation
y(0) = −(0) + C = 0 ∴ C = 0
∴
y(t) = −t
Verification:
y(t) = −t
y (t) = −1
∴
−1 + 1 = 0 7.44. Given differential equation with constant coefficients
−2y + y = 0
Method 1: assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
−2y + y = 0 ∴ L −2y + y = L {0} ∴ −2L y + L {y} = L {0}
−2 sY (s) − y(0) +Y (s) = 0 ∴ −2sY (s) + Y (s) + 2 = 0 ∴ Y (s) =
H
−2
1
−2
H
=
H
−2s + 1
−2
H s − 1/2
Reminder:
L et/2 =
1
1
⇒ L −1
s − 1/2
s − 1/2
= et/2
Inverse transformation results in (see P.7.9)
y(t) = et/2
Method 2: direct integration of differential equation with separable variables
dy
dy
1
−2
+y =0 ∴
= y ∴
dt
dt
2
dy
1
=
y
2
dt
∴
ln y(t) + ln C =
t
t
*1
∴ ln C y(t) =
∴ y(t) = C et/2 ∴ y(0) = C =C=1
e0/2
2
2
∴
y(t) = et/2
7.4 Differential Equations
205
Verification:
y(t) = et/2
y (t) =
1 t/2
e
2
∴
−2A
1 t/2
e + et/2 = 0 2A
7.45. Method 1: Given differential equation with constant coefficients
y −1=0
assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
y − 1 = 0 ∴ L y − 1 = L {0} ∴ L y − L {1} = L {0}
1
1
1
s 2 Y (s) − s y(0) − y (0) − = 0 ∴ s 2 Y (s) − = 0 ∴ Y (s) = 3
s
s
s
Reminder:
2
L t2 = 3
s
⇒ L −1
2
s3
= t2
Inverse transformation (see P.7.7) results in
Y (s) =
1 2
1
1 2
=
∴ y(t) = t 2
3
3
s 2
2 s
2
Method 2: direct integration results in
y −1=0
dy
−1=0 ∴
dt
dy =
dt ∴ y (t) = t + C1 ∴ y (0) = 0 + C1 = 0 ∴ C1 = 0
∴ y (t) = t
∴
dy
=t ∴
y =
dt
y(t) =
1 2
t
2
dy =
t dt ∴ y(t) =
02
t2
+ C2 ∴ y(0) =
+ C2 = 0 ∴ C2 = 0
2
2
206
7 Laplace Transformation
Verification:
y(t) =
t2
2
y (t) = t
y (t) = 1
∴
y −1=1−1=0 7.46. Given differential equation with constant coefficients
y −y +1=0
assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
y − y + 1 = 0 ∴ L y − y + 1 = L {0}
L y − L y + L {1} = L {0}
2
1
s Y (s) − s y(0) − y (0) − sY (s) − y(0) + = 0
s
2
1
s Y (s) − s − 1) − sY (s) − 1) + = 0
s
s 2 Y (s) − sY (s) = s −
1
s
s2 − 1
Y (s) s(s − 1) =
s
s+1
(s −
1)(s + 1)
=
Y (s) =
2
s
(s − 1)
s2
Rational function Y (s) must be decomposed by partial fraction method, so that it is possible to find
its inverse Laplace transformation.
Y (s) =
A
As + B
B
s+1
= + 2 =
s2
s
s
s2
which is to say (by equality of the numerators)
A = 1 and B = 1
∴
Y (s) =
1
1
+ 2
s
s
7.4 Differential Equations
207
Reminder:
L {1} =
1
1
⇒ L −1
s
s
=1
L {t} =
1
s2
=t
⇒ L −1
1
s2
Inverse Laplace transformation is
Y (s) =
1
1
+ 2 ∴ y(t) = 1 + t
s
s
Verification:
y(t) = 1 + t
y (t) = 1
y (t) = 0
∴
y −y +1=0−1+1=0 7.47. Given differential equation with constant coefficients
y + 2y + 2y = 0
assume Y (s) = L {y(t)}, then by taking advantage of LT properties as
y + 2y + 2y = 0 ∴ L y + 2y + 2y = L {0}
L y + 2L y + 2L {y} = L {0}
2
s Y (s) − s y(0) − y (0) + 2 sY (s) − y(0) +2Y (s) = 0
2
s Y (s) + s − 1) + 2 sY (s) + 1) +2Y (s) = 0
s 2 Y (s) + 2sY (s) + 2Y (s) = −(s + 1)
Y (s) s 2 + 2s + 2 = −(s + 1)
Y (s) =
−s + −1
s 2 + 2s + 2
Rational function Y (s) must be decomposed by partial fraction method, so that it is possible to find its
inverse Laplace transformation. Denominator of Y (s) has complex roots, s1 = −1+i and s2 = −1−i.
It follows that
208
7 Laplace Transformation
Y (s) =
=
−s − 1
B
A(s − s2 ) + B(s − s1 )
−s − 1
A
+
=
=
=
s 2 + 2s + 2
(s − s1 )(s − s2 )
s − s1
s − s2
(s − s1 )(s − s2 )
s(A + B) − As2 − Bs1
(s − s1 )(s − s2 )
which is to say (by equality of the left and right side numerators)
A + B = −1 ∴ B = −1 − A
−As2 − Bs1 = −1 ∴ As2 + Bs1 = 1
∴
As2 + (−1 − A)s1 = 1 ∴ A =
i
1
1 + s1
1 + (−1 + i)
=−
=
=
s2 − s1
−1 − i − (−1 + i)
2
−2i
∴ B = −1 +
1
1
=−
2
2
Reminder:
L e−st =
1
1
⇒ L −1
s+1
s+1
= e−st
Inverse Laplace transformation is
Y (s) = −
1 1
1 1
−
2 s − s1
2 s − s2
1
1
1
1
1
1
y(t) = − es1 t − es2 t = − e(−1+i)t − e(−1−i)t = − t eit − t e−it
2
2
2
2
2e
2e
1 1 = − t cos t + i sin t − t cos t − i sin t = −e−t cos t
2e
2e
Verification:
y(t) = −e−t
y (t) = e−t
y (t) = −e−t
∴
−e−t + 2e−t − 2e−t = 2e−t − 2e−t = 0 7.48. Given non–homogenous differential equation with constant coefficients
y + y = sin(2t)
7.4 Differential Equations
209
assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
y + y = sin(2t) ∴ L y + y = L {sin(2t)}
L y + L {y} = L {sin(2t)}
2
s Y (s) − s y(0) − y (0) + Y (s) =
s 2 Y (s) + Y (s) =
Y (s)(s 2 + 1) =
Y (s) =
2
s2 + 4
2
s2 + 4
2
s2 + 4
2
(s 2 + 1)(s 2 + 4)
Rational function Y (s) must be decomposed by partial fraction method, so that it is possible to find
its inverse Laplace transformation. Denominator of Y (s) has complex roots, s1 = +i, s2 = −i,
s3 = +2i, and s4 = −2i. It follows that
Y (s) =
=
As + B
Cs + D
(As + B)(s 2 + 4) + (Cs + D)(s 2 + 1)
2
=
+
=
(s 2 + 1)(s 2 + 4)
s2 + 1
s2 + 4
(s 2 + 1)(s 2 + 4)
s 3 (A + C) + s 2 (B + D) + s(4A + C) + 4B + D
(s 2 + 1)(s 2 + 4)
which is to say (by equality of the numerators)
A+C
B +D
4A + C
4B + D
⎫
= 0⎪
⎪
⎪
⎪
⎪
= 0⎬
= 0⎪
⎪
⎪
⎪
⎪
⎭
=2
∴ A = 0, B =
2
2
, C = 0, D = −
3
3
Reminder:
L {sin(at)} =
a
s 2 + a2
⇒ L −1
a
s 2 + a2
Inverse Laplace transformation is (see P.7.13)
1 2
2 1
−
2
3 s + 1 3 s2 + 4
2
1
y(t) = sin t − sin(2t)
3
3
Y (s) =
= sin(at)
210
7 Laplace Transformation
Verification:
⎫
= 23 sin t − 13 sin(2t) ⎪
⎪
⎬
2
2
4
1
∴ − sint + sin(2t) + sin t − sin(2t) = sin(2t) y (t) = 23 cos t − 23 cos(2t)
⎪
3
3
3
3
⎪
⎭
y (t) = − 32 sin t + 43 sin(2t)
y(t)
7.49. Given non–homogenous differential equation with constant coefficients
y + y = 2e−t
assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
y + y = 2e−t ∴ L y + y = L 2e−t
L y + L y = L 2e−t
2
s Y (s) − s y(0) − y (0) + sY (s) − y(0) =
2
s+1
2
s 2 Y (s) − s − 1 + sY (s) − 1 =
s+1
2
Y (s)(s 2 + s) − s − 2 =
s+1
Y (s) =
s 2 + 3s + 4
(s + 2)(s + 1) + 2
=
(s 2 + s)(s + 1)
s(s + 1)(s + 1)
Rational function Y (s) must be decomposed by partial fraction method, so that it is possible to find
its inverse Laplace transformation. It follows that
Y (s) =
=
s 2 + 3s + 4
A
B
C
A(s 2 + 2s + 1) + Bs(s + 1) + Cs
= +
+
=
s(s + 1)(s + 1)
s
s + 1 (s + 1)2
s(s + 1)2
s 2 (A + B) + s(2A + B + C) + A
s(s + 1)(s + 1)
which is to say (by equality of the numerators)
A+B
2A + B + C
A
⎫
= 1⎪
⎪
⎬
=3
∴ A = 4, B = −3, C = −2
⎪
⎪
⎭
=4
so that inverse Laplace transformation is derived as
Y (s) =
4
3
2
−
−
s
s + 1 (s + 1)2
y(t) = 4 − 3e−t − 2te−t
7.4 Differential Equations
211
Verification:
y(t)
y (t)
y (t)
⎫
= 4 − 3e−t − 2te−t ⎪
⎪
⎬
−t
−t
−t
−t
∴ −
= 2te + e
2te
+ e−t + 2te
+ e−t = 2e−t ⎪
⎪
⎭
= −2te−t + e−t
7.50. Given non–homogenous differential equation with constant coefficients
y − y = cos(t)
assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
y − y = sin(t) ∴ L y − y = L {cos(t)}
L y − L y = L {cos(t)}
2
s
s Y (s) − s y(0) − y (0) − sY (s) − y(0) = 2
s +1
s
s 2 Y (s) − sY (s) = 2
s +1
1
s 1
=
Y (s) =
2 + 1)
2 + 1)
s(s
−
1)(s
(s
−
1)(s
Rational function Y (s) must be decomposed by partial fraction method, so that it is possible to find
its inverse Laplace transformation. Denominator of Y (s) has complex roots, s1 = +i and s2 = −i. It
follows that
Y (s) =
=
1
A
Bs + C
A(s 2 + 1) + (Bs + C)(s − 1)
=
+ 2
=
2
(s − 1)(s + 1)
s−1
s +1
(s − 1)(s 2 + 1)
s 2 (A + B) + s(C − B) + A − C
(s − 1)(s 2 + 1)
which is to say (by equality of the numerators)
A+B
C−B
A−C
⎫
= 0⎪
⎪
⎬
1
1
1
=0
∴ A = ,B = − ,C = −
⎪
2
2
2
⎪
⎭
=1
Reminder:
L {cos(at)} =
s
s 2 + a2
⇒ L −1
Inverse Laplace transformation is (see P.7.14)
s
s 2 + a2
= cos(at)
212
7 Laplace Transformation
1 s
1 1
1 1
−
−
2 s − 1 2 s2 + 1 2 s2 + 1
1
1
1
y(t) = et − cos(t) − sin(t)
2
2
2
Y (s) =
Verification:
y(t)
y (t)
y (t)
⎫
= 12 et − 12 cos(t) − 12 sin(t) ⎪
⎪
⎬
1 x
1
1
= 2 e + 2 sin(x) − 2 cos(x)
⎪
⎪
⎭
= 12 ex + 12 cos(x) + 12 sin(x)
∴
1 x 1
1 1 x 1 1
e + cos(x) + sin(x) − e − sin(x) + cos(x) = cos(t) 2
2
2
2
2
2
7.51. Given non–homogenous differential equation with constant coefficients
y −y =0
assuming Y (s) = L {y(t)}, then by taking advantage of LT properties as
y − y = 0 ∴ L y − y = L {0}
L y − L y = L {0}
s 3 Y (s) − s 2 y(0) − sy (0) − y (0) − s 2 Y (s) − s y(0) − y (0) = 0
s 3 Y (s) + 3s 2 − s − 4 − s 2 Y (s) − 3s + 1 = 0
Y (s) =
−3s 2 + 4s + 3
s 2 (s − 1)
Rational function Y (s) must be decomposed by partial fraction method, so that it is possible to find
its inverse Laplace transformation. It follows that
Y (s) =
=
A
B
As(s − 1) + B(s − 1) + Cs 2
C
−3s 2 + 4s + 3
= + 2+
=
2
s (s − 1)
s
s
s−1
s 2 (s − 1)
s 2 (A + C) + s(B − A) − B
s 2 (s − 1)
which is to say (by equality of the numerators)
⎫
= −3⎪
⎪
⎬
B −A =4
∴ A = −7, B = −3, C = 4
⎪
⎪
⎭
−B
=3
A+C
so that inverse Laplace transformation is derived as
7.4 Differential Equations
213
3
7
4
Y (s) = − − 2 +
s
s
s−1
y(t) = −7 − 3t + 4et
Verification:
y (t)
⎫
= −7 − 3t + 4et ⎪
⎪
⎪
⎪
⎪
t
⎬
= 4e − 3
y (t)
= 4e
y (t)
= 4et
y(t)
t
⎪
⎪
⎪
⎪
⎪
⎭
∴ 4et − 4et = 0 Bibliography
[Ap17] W. Appel, Mathématiques pour la physique et les physiciens, H&K editions, 5 edn. (2017). Number 978–
2–35141–339–5.
[MIT20b] MIT Open Courseware, Multivariable Calculus (2020). https://ocw.mit.edu/courses/mathematics/1802sc-multivariable-calculus-fall-2010/. on–line
[MIT20] MIT Open Courseware, Signals and Systems (2020). https://ocw.mit.edu/resources/res-6-007-signalsand-systems-spring-2011. on–line
[MIT20a] MIT Open Courseware, Single Variable Calculus (2020). https://ocw.mit.edu/courses/mathematics/1801sc-single-variable-calculus-fall-2010/. on–line
[Dem63] B.P. Demidovich, Collection of problems and exercises for mathematical analysis (in Russian), 5 edn.
(1963). Number 517.2 D30. National Printing House for Literature in Mathematics and Physics
[Ler09] Y. Leroyer, P. Tessen, Mathématique pour l’ingénieur: Exercises et problémes, 1 edn. (2009). Number
978–2–10–052186–9. Dunod
[Mit67] D.S. Mitrinović, Matematika I: u obliku metodičke zbirke zadataka sa rešenjima, 3 edn. (1967). Number
2043. Gradjevinska Knjiga
[Mit77] D.S. Mitrinović, Kompleksna analiza, 4 edn. (1977). Gradjevinska Knjiga
[Mit78] D.S. Mitrinović, Matematika I: u obliku metodičke zbirke zadataka sa rešenjima, 5 edn. (1978). Number
06–1785/1. Gradjevinska Knjiga
[Mit79] D.S. Mitrinović, Kompleksna analiza: zbornik zadataka i problema, vol. 3, 2 edn. (1979). Number 06–
1431/1. Naučna Knjiga
[Mo64] P.S. Modenov, Collection of problems in special program (in Russian) (1964). National Printing House
for Literature in Mathematics and Physics
[Mu20] D. Müller, Mathématiques (2020). http://www.apprendre-en-ligne.net/MADIMU2/INDEX.HTM. on–
line
[Sobo20] R. Sobot, Private Collection of Course Notes in Mathematics (2020)
[Sobo21] R. Sobot, Engineering Mathematics by Example, 1 edn. (Springer, Berlin, 2021). Number 978–3–030–
79544–3
[Spi80] M.R. Spiegel, Analyse de Fourier et Application aux Problémes de Valeurs aux Limites, 1 edn. (McGraw–
Hill, New York, 1980). Number France 2–7042–1019–5
[St11] Stewart, Analyse concepts et contextes. Fonctions d’une variable, vol. 1, 3 edn. (De Boeck Supérieur,
New York, 2011). Number 978–2–8041–6306–8
[Ste11a] Stewart, Analyse concepts et contextes. Fonctions de plusieurs variables, vol. 2, 3 edn. (De Boeck
Supérieur, New York, 2011). Number 978–2–8041–6327–3
[Sym20] Symbolab, Online Calculator (2020). https://www.symbolab.com/. on–line
[Tkz20] TikZ. graphs Generated by TikZ (2020). https://en.wikipedia.org/wiki/PGF/TikZ. on–line
[Ven96] T.B. Vene, Zbirka rešenih zadataka iz matematike, vol. 2, 22 edn. (Zavod za Udžbenike i Nastavna
Sredstva, New York, 1996). Number 86–17–09617–9
[Ven01] T.B. Vene, Zbirka rešenih zadataka iz matematike, vol. 1, 28 edn. (Zavod za Udžbenike i Nastavna
Sredstva, New York, 2001). Number 86–17–09031–6
[Ven03] T.B. Vene, Zbirka rešenih zadataka iz matematike, vol. 3, 27 edn. (Zavod za Udšbenike i Nastavna
Sredstva, New York, 2003). Number 86–17–10786–3
[Ven04] T.B. Vene, Zbirka rešenih zadataka iz matematike, vol. 4, 36 edn. (Zavod za Udšbenike i Nastavna
Sredstva, New York, 2004). Number 86–17–11156–9
[Wik20a] Wikipedia, Fourier Transform (2020). https://en.wikipedia.org/wiki/Fourier_transform. on–line
[Wik20b] Wikipedia, Linear Algebra (2020). https://en.wikipedia.org/wiki/Linear_algebra. on–line
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5
215
Index
A
Arithmetic progression, 1, 8
C
Calculus, 28, 168
Continuous functions, 31, 32, 91, 108, 161
Convergence, 2
Convolution integral, 45, 59–61, 63–68, 70–73, 75–77,
79–81
Convolution sum, 91–94, 96–99, 101–105
D
Differential equations, 129, 183–184, 188, 203–213
Dirac delta, 23, 31, 32, 39–41, 43, 98, 99, 116, 126, 185,
186
DT Fourier transformation, 147–179
E
Energy, 32, 45, 49, 82–92, 107–110
F
Fourier series, 111, 114–115, 135–148, 156–170
Fourier transformation, 30, 31, 111–114, 116–120,
124–129, 132–134, 147–149, 159, 165, 174
Function, 2, 23, 45, 98, 111, 147, 181
Function transformations, 45–47, 50–54
G
Geometric progression, 1, 15, 16
Geometric series, 2, 4, 14–17
H
Heaviside step, 24, 28, 33
L
Laplace transformation, 181–212
P
Power, 2, 6, 8, 19, 45, 46, 49, 82–92, 107–110, 116, 156,
175, 183, 186–188, 193–195, 197
R
Rectangular, 30, 33, 36, 37, 43, 61, 65, 67, 189
S
Sampled functions, 91
Sequence, 1, 3, 6–9, 59, 92–101, 106, 107, 149, 150, 159,
170, 171, 178
Series, 1–22, 45, 93, 95–97, 108, 109, 111, 112, 139,
149, 155, 161–165, 168, 172, 174
Sinus cardinal, 28
Special functions, 23–44, 113, 116–132
Sum, 1, 27, 47, 102, 111, 154
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature
Switzerland AG 2024
R. Sobot, Engineering Mathematics by Example,
https://doi.org/10.1007/978-3-031-41203-5
217