Chapter 5 FOURIER TRANSFORMS
5.1
INTRODUCTION
Fourier series are powerful tools for problems
involving functions that are periodic or are of
interest on a finite interval only.
Since many problems involve functions that are non
periodic and are of interest on either semi-infinite
interval or the whole x -axis, we ask what can be
done to extend the method of Fourier series to such
functions.
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FOURIER TRANSFORMS
Chapter 5 FOURIER TRANSFORMS
5.1
INTRODUCTION
The Fourier integral can be regarded as the limiting
case of a Fourier series representation of a function
f (x ) defined over an interval −L < x < L as
L → ∞.
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Chapter 5 FOURIER TRANSFORMS
5.2
FOURIER INTEGRAL
5.2.1 Fourier Integral
We know that the expansion of a function f (x )
defined in an interval [−L, L] into Fourier series has
the form
∞
a0 X
+
(an cos ωn x + bn sin ωn x )
f (x ) =
2
n=1
where ωn =
nπ
L ,
Z
1 L
an =
f (t) cos(ωn t)dt and
L −L
Z
1 L
bn =
f (t) sin(ωn t)dt.
L −L
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
The substitution of the expressions of an and bn into
Fourier series leads to
1
f (x ) =
2L
+
Z
L
f (t)dt
1
L
−L
∞ Z L
X
n=1
f (t) cos ωn t cos ωn x + sin ωn t sin ωn x dt.
−L
Thus,
1
f (x ) =
2L
Z
L
∞
1X
f (t)dt+
L n=1
−L
Z
L
f (t) cos ωn (t−x )dt
−L
(1)
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5.2.1
FOURIER INTEGRAL
Now we set
∆ω = ωn+1 − ωn =
Then
1
L
=
π
(n + 1)π nπ
−
= .
L
L
L
∆ω
π
and we may write (1) in the form
Z L
1
f (x ) =
f (t)dt
2L −L
Z L
(2)
∞
1X
f (t) cos ωn (t − x )dt.
+
∆ω
π n=1
−L
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
Suppose f (x ) is defined throughout the x -axis and
piecewise smooth in any finite interval [−L, L] and
thus can be expanded into Fourier series in every
such interval.
Then the representation is valid for any fixed L,
arbitrarily large, but finite.
We shall assume that f is absolutely integrable
on (−∞, ∞), that is the improper integral
Z ∞
|f (t)|dt = Q
−∞
is convergent (and so Q is finite).
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
Letting L → ∞ we find that
Z L
Z L
1
1
f (t)dt ≤
|f (t)|dt
2L −L
2L −L
Z ∞
Q
1
→ 0.
|f (t)|dt =
≤
2L −∞
2L
Hence
1
lim
L→∞ 2L
Z
L
f (t)dt = 0.
−L
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5.2.1
FOURIER INTEGRAL
Note that the improper integral
Z ∞
φ(ω) :=
f (t) cos ω(t − x )dt
−∞
is absolutely convergent since
Z
Z ∞
f (t) cos ω(t−x ) dt ≤
−∞
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∞
|f (t)|dt = Q < ∞.
−∞
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
Moreover, the improper integral φ(ω) is absolutely
convergent, and hence, if f is a continuous function,
the function φ is continuous as well.
For sufficiently large values of L the integral
Z L
f (t) cos ωn (t − x )dt
−L
can be regarded as being almost equal to the value
of the function φ(ω) assumed for ω = ωn , that is,
Z L
f (t) cos ωn (t − x )dt ≈ φ(ωn ).
−L
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
The second term on the right side of (2) therefore
can be rewritten as
Z L
∞
∞
1X
1X
∆ω
φ(ωn )∆ω.
f (t) cos ωn (t−x )dt ≈
π n=1
π n=1
−L
P
π
In the sum ∞
n=1 φ(ωn )∆ω the quantity ∆ω = L
tends to zero as L → ∞ while, in the limit, the
variable ωn run through all the values of ω from 0
to ∞. Therefore it is natural to expect that this
sum tends to the integral
Z ∞
Z ∞
Z ∞
φ(ω)dω =
dω
f (t) cos ω(t − x )dt.
0
0
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CALCULUS 3 Chapter 5
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
On substituting the expression obtained into the
formula of f (x ) we obtain from (2),
Z ∞
Z
1 ∞
dω
f (t) cos ω(t − x )dt (3)
f (x ) =
π 0
−∞
The integral on the right is called Fourier integral
and the formula itself is Fourier integral formula.
It is clear that our naive approach merely suggest
the representation (3).
Sufficient conditions for validity of (3) are as
follows.
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
Theorem 5.1 (Fourier integral theorem)
Let f be absolutely integrable over the interval
−∞ < x < ∞ and suppose that f (x ) satisfies
Dirichlet conditions:
In any finite interval, f is piecewise continuous
and has only a finite number of maxima and
minima.
Then the Fourier integral representation of f (x ),
given in (3), converges to f (x ) at all points where f
is continuous and to 12 [f (x −) + f (x +)] where f is
discontinuous.
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
Since
cos ω(t − x ) = cos ωt cos ωx + sin ωt sin ωx ,
formula (3) can be rewritten in the form
Z ∞
Z
1 ∞
f (t) cos ωtdt
cos ωxdω
f (x ) =
π 0
−∞
Z ∞
Z
1 ∞
f (t) sin ωtdt.
+
sin ωxdω
π 0
−∞
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5.2 FOURIER INTEGRAL
5.2.1
FOURIER INTEGRAL
By defining the functions A(ω) and B(ω) as
Z
1 ∞
f (t) cos ωtdt and
A(ω) =
π −∞
Z
1 ∞
B(ω) =
f (t) sin ωtdt
π −∞
(4)
the Fourier integral representation in (3) can be
written in the simpler form
Z ∞
f (x ) =
A(ω) cos ωx + B(ω) sin ωx dω
0
(5)
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5.2.1
FOURIER INTEGRAL
Relation (5) is analogous to an expansion of a
function into a Fourier series, and expressions (4)
are similar to the formulae for the Fourier
coefficients.
Example 1.1 Find the Fourier integral
representation of f (x ) = e −|x | .
2
, B(ω) = 0,
π(1 + ω 2 )
Z
2 ∞ cos ωx
dω.
=
π 0 1 + ω2
A(ω) =
e −|x |
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5.2.2
FOURIER INTEGRAL FOR EVEN AND ODD FUNCTIONS
For an even or odd function the Fourier integral
becomes simpler.
Just as in the case of Fourier series, this is of
practical interest in saving work and avoiding errors.
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5.2.1
FOURIER INTEGRAL
If f is an even function, then f (t) sin ωt is odd, so
B(ω) = 0 in (4) and
Z
2 ∞
f (t) cos ωtdt.
A(ω) =
π 0
The Fourier integral (5) then reduces to the
Fourier cosine integral
Z ∞
f (x ) =
A(ω) cos ωxdω
(6)
0
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5.2.2
FOURIER INTEGRAL FOR EVEN AND ODD FUNCTIONS
Similarly, if f is an odd function, then f (t) cos ωt is
odd, so A(ω) = 0 in (4) and
Z
2 ∞
f (t) sin ωtdt.
B(ω) =
π 0
The Fourier integral (5) then reduces to the
Fourier sine integral
Z ∞
f (x ) =
B(ω) sin ωxdω
(7)
0
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5.2 FOURIER INTEGRAL
5.2.2
FOURIER INTEGRAL FOR EVEN AND ODD FUNCTIONS
Example 1.2 Find Fourier cosine and Fourier
sine integrals of
f (x ) = e −βx
where x > 0 and β > 0.
ANS.
f (x ) = e
−βx
2β
=
π
Z
0
∞
cos ωx
2
dω
=
ω2 + β 2
π
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Z
0
∞
ω sin ωx
dω
ω2 + β 2
FOURIER TRANSFORMS
5.2 FOURIER INTEGRAL
5.2.2
FOURIER INTEGRAL FOR EVEN AND ODD FUNCTIONS
Summary of Fourier integral representations
Let f (x ) satisfy the conditions of Theorem 5.1.
(a) f (x ) has the general Fourier integral
representation
Z ∞
1
f (x +)+f (x −) = [A(ω) cos ωx+B(ω) sin ωx ]dω,
2
0
where
Z
1 ∞
f (t) cos ωtdt
and
A(ω) =
π −∞
Z
1 ∞
B(ω) =
f (t) sin ωtdt.
π −∞
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5.2.2
FOURIER INTEGRAL FOR EVEN AND ODD FUNCTIONS
(b) If f (x ) is even, it has the Fourier cosine integral
representation
Z ∞
1
f (x +) + f (x −) =
A(ω) cos ωxdω.
2
0
(c) If f (x ) is odd, it has the Fourier sine integral
representation
Z ∞
1
B(ω) sin ωxdω.
f (x +) + f (x −) =
2
0
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5.3 FOURIER TRANSFORM
An integral transform is a transformation in the form
of an integral that produces from given functions
new functions depending on a different variable.
The Laplace transform is of this kind and is by far
the most important integral transform in
engineering.
The next in order of importance are Fourier
transforms.
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5.3 FOURIER TRANSFORM
5.3.1
FOURIER INTEGRAL IN COMPLEX FORMS
The starting point for the development of the
Fourier transform is the complex form of the Fourier
integral representation of a function f (x ).
To derive this representation in which f (x ) is
defined over the interval (−∞, ∞), we write
Z ∞
Z ∞
(3) 1
dω
f (t) cos ω(t − x )dt
f (x ) =
π 0
−∞
Z
Z ∞
1 ∞
=
dω
f (t) cos ω(x − t)dt
π 0
−∞
where we have used the result
cos ω(t − x ) = cos ω(x − t).
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5.3.1
FOURIER INTEGRAL IN COMPLEX FORMS
As the integrand in the last integral is an even
function of ω,
Z ∞
Z ∞
1
dω
f (t) cos ω(x − t)dt
f (x ) =
2π −∞
−∞
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(8)
FOURIER TRANSFORMS
5.3 FOURIER TRANSFORM
5.3.1
FOURIER INTEGRAL IN COMPLEX FORMS
The function sin ω(x − t) is an odd function of ω, so
Z ∞
Z ∞
1
dω
f (t) sin ω(x − t)dt (9)
0=
2π −∞
−∞
Multiplying equation (9) by j, adding the result to
equation (8), and using the Euler formula
e jθ = cos θ + j sin θ,
we arrive at the complex Fourier integral
representation
Z ∞
Z ∞
1
dω
f (t)e jω(x −t) dt
f (x ) =
2π −∞
−∞
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(10)
FOURIER TRANSFORMS
5.3 FOURIER TRANSFORM
5.3.2
FOURIER TRANSFORM PAIR
We write the factor e jω(x −t) in (10) as the product
e jω(x −t) = e jωx · e −jωt and have
1
f (x ) =
2π
Z
∞
e jωx
Z
−∞
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∞
f (t)e −jωt dt dω
(11)
−∞
CALCULUS 3 Chapter 5
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5.3 FOURIER TRANSFORM
5.3.2
FOURIER TRANSFORM PAIR
Definition 3.1
For a given function f : R → C the function F (ω)
(for ω ∈ R) is defined by
Z ∞
(12)
F (ω) =
f (t)e −jωt dt
−∞
provided the integral exists as an improper Riemann
integral. The function F (ω) is called the Fourier
transform or spectrum of f (t).
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5.3.2
FOURIER TRANSFORM PAIR
F (ω) exists when f (t) is absolutely integrable on
the interval (−∞, ∞).
The transformation from f (t) to F (ω) is called the
Fourier transformation.
We will also write F{f }(ω) instead of F (ω).
The symbol F denotes the Fourier transformation.
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5.3.2
FOURIER TRANSFORM PAIR
• f (t) will represent a function depending on
time, while F (ω) usually depends on frequency.
• Thus we say that f (t) is defined in the time
domain and F (ω) is defined in the frequency
domain.
Note that F (ω) will in general be a complex-valued
function, so F : R → C.
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5.3.2
FOURIER TRANSFORM PAIR
It follows from (11) that
Z ∞
1
F (ω)e jωt dω
f (t) =
2π −∞
(13)
Since formula (13) recovers f (t) from its Fourier
transform F (ω), f (t) is termed the Fourier inverse
transform of F (ω).
As with the Laplace transform, the pair
{f (t), F (ω)}
is called a Fourier transform pair.
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5.3.2
FOURIER TRANSFORM PAIR
There are a number of variations of definition of
F (ω) and you need to be aware that different
authors may use slightly different formulae.
This can cause confusion when the Fourier
transform is first met.
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5.3.2
FOURIER TRANSFORM PAIR
Although the convention that we have adopted here
is fairly standard, some authors associate the factor
1
with (12) rather than (13), while others
2π
1
associate a factor of √ with each of (12) and
2π
(13).
We stress that this is a choice that we have made,
but you should have no difficulty in using any form.
All that is required of the factors is that their
1
product be
and so, the pair combine to give the
2π
Fourier integral (10).
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5.3.2
FOURIER TRANSFORM PAIR
Example 3.1 Find the Fourier transform of the
one-sided exponential function
f (t) = H(t)e −at ,
(a > 0)
where H(t) is the Heaviside unit step function.
ANS.
F{f (t)} =
1
.
a + jω
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5.3.2
FOURIER TRANSFORM PAIR
Example 3.2 Calculate the Fourier transform of
the rectangular pulse
(
A
if |t| ≤ T ,
f (t) =
0
if |t| > T .
Solution
F{f (t)} =
Z
T
Ae −jωt dt = 2AT sinc ωT .
−T
where sinc x is defined by
(
sinc x =
sin x
x
1
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CALCULUS 3 Chapter 5
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5.3 FOURIER TRANSFORM
5.3.2
FOURIER TRANSFORM PAIR
Example 3.3 Show that the Fourier transform of
the unit step function H(t) does not exist.
Solution
Z ∞
Z
H(t)e −jωt dt =
−∞
∞
e
0
−jωt ′
Since e
Z ∞
−jωt
Z
dt = lim
b→∞
b
e −jωt dt.
0
= −jωe −jωt , it follows that
b
1
lim e −jωt
0
−jω b→∞
−∞
j
=
lim e −jωb − 1 .
ω b→∞
However, this limit does not exists.
H(t)e −jωt dt =
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5.3.2
FOURIER TRANSFORM PAIR
Table of Fourier transforms
f (t)
e −at H(t) (a > 0)
F{f }
1
a + jω
1
(a + jω)2
te −at H(t) (a> 0)
(
A (|t| ≤ T )
2AT sinc ωT
0 (|t| > T )
2a
e −a|t| (a > 0)
a2 + ω 2
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5.3.3
CONTINUOUS FOURIER SPECTRA
Fourier transforms are generally complex-valued
functions of the real frequency variable ω.
Writing F (ω) in the exponential form
F (ω) = |F (ω)|e j arg F (ω) ,
plots of |F (ω)| and arg F (ω), which are both
real-valued functions of ω, are called the amplitude
and phase spectra respectively of the signal f (t).
The two spectra represent the frequency-domain
portrait of the signal f (t).
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5.3 FOURIER TRANSFORM
5.3.3
CONTINUOUS FOURIER SPECTRA
Example 3.4 Determine the amplitude and
phase spectra of the causal signal
f (t) = e −at H(t)
(a > 0)
and plot their graphs.
ANS.
|F (ω)| = √
1
a2 + ω 2
−1
arg F (ω) = tan (1) − tan
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ω CALCULUS 3 Chapter 5
a
FOURIER TRANSFORMS
5.3 FOURIER TRANSFORM
5.3.3
CONTINUOUS FOURIER SPECTRA
The Fourier transform discussed so far have two
properties in common.
First, the amplitude spectra are even functions of
the frequency variable ω. This is always the case
when the time signal f (t) is real.
The second common feature is that all the
amplitude spectra decrease rapidly as ω increases.
This means that most of information concerning the
“shape” of the signal f (t) is contained in a fairly
small interval of frequency axis around ω = 0.
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5.4 PROPERTIES OF FOURIER TRANSFORM
5.4.1
LINEARLY PROPERTY
Theorem 5.2 (Linearity of the
Fourier Transform)
If f (t) and g(t) are functions having Fourier
transforms and if α and β are constants, then
F{αf (t) + βg(t)} = αF{f (t)} + βF{g(t)}
As a consequence of this, we say that the Fourier
transform operator F is a linear operator.
Clearly the linearity property also applies to the
inverse transform operator F−1 .
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5.4.2
TIME-DIFFERENTIATION PROPERTY
Theorem 5.3 (Fourier Transform of
Derivatives)
Let f (t) be continuous on the t-axis and f (t) → 0
as |t| → ∞. Furthermore, let f ′ (t) be absolutely
integrable on the t-axis. Then
F{f ′ (t)} = jωF{f (t)}
(14)
Two successive applications of (14) give the
transform of the second derivative of f
F{f ′′ (t)} = −ω 2 F{f (t)}
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5.4.2
TIME-DIFFERENTIATION PROPERTY
Repeating the argument n times, it follows that
n
o
(n)
(15)
F f (t) = (jω)n F{f (t)}
The result (15) is referred to as the
time-differentiation property, and may be used
to obtained frequency-domain representation of
differential equations.
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5.4.2
TIME-DIFFERENTIATION PROPERTY
Example 4.1
Given
2
√
2
F e −t (ω) = π e −ω /4 ,
2
find F{te −t }(ω).
ANS. F{te −t
2
r
π
2
}(ω) = −
jωe −ω /4 .
2
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5.4 PROPERTIES OF FOURIER TRANSFORM
5.4.3
TIME-SHIFT PROPERTY
Theorem 5.4
(Time-shift Property)
If f (t) has a Fourier transform F (ω), then
F{f (t − t0 )} = e −jωt0 F (ω)
(16)
The result (16) is known as the time-shift
property, and implies that delaying a signal by a
time t0 causes its Fourier transform to be multiplied
by e −jωt0 . (16) implies
F−1 e −jωt0 F (ω) = f (t − t0 ) = f (t) t→t−t0
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.4 PROPERTIES OF FOURIER TRANSFORM
5.4.3
TIME-SHIFT PROPERTY
Example 4.2 Determine the Fourier transform of
the rectangular pulse
(
A, 0 ≤ t ≤ 2T
f (t) =
0, elswhere.
ANS.
F (ω) = 2ATe −jωT sinc ωT .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.4 PROPERTIES OF FOURIER TRANSFORM
5.4.4
FREQUENCY-SHIFT PROPERTY
Theorem 5.5 (Frequency-shift Property)
If f (t) has a Fourier transform F (ω), then
F e jω0 t f (t) = F (ω − ω0 )
(17)
The result (17) is known as the frequency-shift
property, and indicates that multiplication by e jω0 t
simply shifts the spectrum of f (t) so that it is
centered on the point ω = ω0 in the frequency
domain.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.4 PROPERTIES OF FOURIER TRANSFORM
5.4.4
FREQUENCY-SHIFT PROPERTY
Example 4.3
of the signal
Determine the frequency spectrum
f (t) cos ωc t
if F (ω) = F{f (t)} is known.
ANS.
1
1
F{f (t) cos ωc t} = F (ω − ωc ) + F (ω + ωc ).
2
2
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
In this section we define two quantities associated
with time signals, namely signal energy and signal
power. These quantities play an important role in
characterizing signal types.
The energy associated with the signal f (t) is
defined as
Z ∞
2
E=
f (t) dt
−∞
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
If f has a Fourier transform F (ω),
Z ∞
1
f (t) =
F (ω)e jωt dω,
2π −∞
then
Z ∞
1
jωt
E=
f (t)f (t)dt =
f (t)
F (ω)e dω dt
2π −∞
−∞
−∞
Z ∞
Z ∞
Z ∞
1
1
jωt
F (ω)
f (t)e dt dω =
F (ω)F (−ω)dω
=
2π −∞
2π −∞
−∞
Z ∞
Z ∞
1
1
2
=
F (ω)F (ω)dω =
F (ω) dω.
2π −∞
2π −∞
Z
∞
Z
∞
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
Theorem 5.6 (The Parseval relation for
the Fourier transform)
If f (t) has the Fourier transform F (ω), then
Z ∞
Z ∞
2
1
2
f (t) dt =
F (ω) dω
2π −∞
−∞
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
(18)
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
Equation (18) relates the total energy of the signal
2
f (t) to the integral over all frequencies of F (ω) .
2
For this reason, F (ω) is called the energy
2
spectral density, and a plot of F (ω) versus ω is
called the energy spectrum of the signal f (t).
The result (18) is called Parserval’s theorem.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
Example 5.1
densities of
Determine the energy spectral
(a) the one-sided exponential function
f (t) = e −at H(t),
(a > 0);
(
A if |t| ≤ T
(b) the rectangular pulse f (t) =
0 if |t| > T .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
Solution (a) By Example 3.1,
F (ω) =
1
.
a + jω
The energy spectral density of the function f (t) is
therefore
1
1
2
F (ω) =
=
.
|a + jω|2
a2 + ω 2
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
(b) By Example 3.2,
F (ω) = 2AT sinc ωT .
Thus the energy spectral density of the pulse is
2
F (ω) = 4A2 T 2 (sinc ωT )2 .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.1
ENERGY AND POWER
There are important signals f (t), defined in general
for
<2t < ∞, for which the integral
R ∞ −∞
−∞ f (t) dt is unbounded.
For such signals, instead of considering energy, we
consider the average power P, frequently referred to
as the power of the signal:
Z
2
1 T
P = lim
f (t) dt
T →∞ T −T
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.2
TRANSFORMS OF IMPULSE FUNCTIONS
We now consider the Fourier transform of the Dirac
delta function δ(t). Recall that
(
Z b
g(c)
if a < c < b
g(t)δ(t − c)dt =
0
otherwise
a
Thus we readily obtain the following two Fourier
transforms:
Z ∞
F{δ(t)} =
δ(t)e −jωt dt = 1
Z−∞
∞
F{δ(t)} =
δ(t − t0 )e −jωt dt = e −jωt0
−∞
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.3
FOURIER TRANSFORMS OF A CONVOLUTION
Now we examine how the Fourier transform behaves
under convolutions.
2
2
R∞ R∞ Suppose that −∞ f (t) dt and −∞ g(t) dt are
finite. The convolution of f and g, denoted f ∗ g, is
defined by
Z ∞
f (t − x )g(x )dx ,
f ∗ g (t) =
−∞
or, equivalently,
Z
f ∗ g (t) =
∞
f (x )g(t − x )dx
−∞
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.3
FOURIER TRANSFORMS OF A CONVOLUTION
Theorem 5.7 (Convolution theorem)
Let F (ω) and G(ω) be the Fourier transforms of f
and g. Then
F{f ∗ g}(ω) = F (ω)G(ω)
F {F ∗ G}(t) = 2πf (t)g(t)
−1
The Fourier transform of the convolution of two
functions equals the product of the Fourier
transforms of each of the functions.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.5 TRANSFORMS OF THE STEP
AND IMPULSE FUNCTIONS
5.5.3
FOURIER TRANSFORMS OF A CONVOLUTION
Example 5.2
Use the convolution theorem to
evaluate F f (t) ∗ g(t) , where
(
1, −1 ≤ t ≤ 1
f (t) =
and g(t) = e −t H(t).
0, otherwise
Solution
According to Examples 3.1 and 3.2,
sin ω
1
F (ω) = 2
and G(ω) =
.
ω
1 + jω
Thus, by Theorem 5.7,
sin ω
F f (t) ∗ g(t) = 2
.
ω(1 + jω)
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.1
DEFINITION OF THE DISCRETE FOURIER TRANSFORM
In this section we show how a function f (t), with
Fourier transform F (ω), can be sampled at intervals
of T to give a sequence of values
f (0), f (T ), f (2T ) . . ..
The discrete Fourier transform takes such a
sequence and processes it to produce a new
sequence which can be thought of as a sampled
version of F (ω).
The discrete Fourier transform is important as it is
the basis of most signal and image processing
methods.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.1
DEFINITION OF THE DISCRETE FOURIER TRANSFORM
Definition 6.1
Let N be a positive integer and let {fn }N−1
n=0 be a
sequence of N complex numbers. The discrete
Fourier transform (DFT) of {fn }N−1
n=0 is another
sequence Fk , also having N terms, defined by
Fk =
N−1
X
fn e −j2nkπ/N ,
k = 0, 1, . . . , N − 1.
n=0
We also write {Fk }N−1
k=0 by F{fn } to denote the DFT
of the sequence {fn }N−1
n=0 .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.1
DEFINITION OF THE DISCRETE FOURIER TRANSFORM
Example 6.1 Find the DFT of the sequence
{fn }30 = {1, 2, −5, 3}.
Solution Here N = 4, so
3
3
X
X
−j2nkπ/4
Fk =
fn e
=
fn e −jnkπ/2 ,
n=0
k = 0, 1, 2, 3.
n=0
So,
F0 =
F1 =
3
X
n=0
3
X
fn e 0 = 1 + 2 + (−5) + 3 = 1,
fn e −jnπ/2 = · · · = 6 + j,
n=0
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.1
DEFINITION OF THE DISCRETE FOURIER TRANSFORM
F2 =
F3 =
3
X
n=0
3
X
fn e
−j2nπ/2
=
3
X
fn e −jnπ · · · = −9
n=0
fn e −j3nπ/2 = · · · = 6 − j.
n=0
So, the DFT of the sequence 1, 2, −5, 3 is the
sequence 1, 6 + j, −9, 6 − j.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.2
THE INVERSE DISCRETE FOURIER TRANSFORM
The Inverse Discrete Fourier Transform
Suppose we know the numbers Fk , the N-point
DFT of some sequence {fn }N−1
n=0 . We would like to
retrieve {fn }.
We can prove that
N−1
1X
Fk e j2nkπ/N ,
fn =
N
n = 0, 1, . . . , N − 1.
k=0
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CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.2
THE INVERSE DISCRETE FOURIER TRANSFORM
Definition 6.2
The inverse DFT of the sequence {Fk }N−1
k=0 is the
sequence {fn }, also having N terms, given by
N−1
1X
Fk e j2nkπ/N , n = 0, 1, . . . , N−1.
F {Fk } = fn =
N
−1
k=0
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 DISCRETE FOURIER TRANSFORM
5.6.2
THE INVERSE DISCRETE FOURIER TRANSFORM
Example 6.2 Using the definition, find the
inverse DFT of the sequence {−4, 1, 0, 1}.
Solution In this sequence N = 4 and
F0 = −4, F1 = 1, F2 = 0, F3 = 1.
From the definition, for n = 0, 1, 2, 3,
3
3
k=0
k=0
1X
1X
Fk e j2nkπ/4 =
Fk e jnkπ/2
fn =
4
4
1
− 4 + 1 · e jnπ/2 + 0 · e jnπ + 1 · e j3nπ .
4
Then, taking n = 0, 1, 2, 3 gives the sequence
{fn }3n=0 = {−1/2, −1, −3/2, −1}.
=
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.3
MATRIX REPRESENTATION OF THE DFT
The DFT of a sequence {fn }N−1
n=0 is given by
Fk =
N−1
X
fn e −j2nkπ/N
n=0
Define
w = e −j2π/N
so that
Fk =
N−1
X
fn w nk
n=0
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.3
MATRIX REPRESENTATION OF THE DFT
Writing out this sum explicitly for each k we find
F0 = f0 w 0 + f1 w 0 + f2 w 0 + · · · + fN−1 w 0
F1 = f0 w 0 + f1 w 1 + f2 w 2 + · · · + fN−1 w N−1
F2 = f0 w 0 + f1 w 2 + f2 w 4 + · · · + fN−1 w 2(N−1)
..
.
FN−1 = f0 w 0 + f1 w N−1 + f2 w 2(N−1) + · · · + fN−1 w (N−1)(N−1) .
These equations can be written in matrix form as follows:
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.3






F0
F1
F2
..
.
MATRIX REPRESENTATION OF THE DFT


 
 
=
 
 
FN−1






Likewise,

f0
f1 

f2  = 1
N
.. 
. 
fN−1
w0
w0
w0
..
.
w0
w1
w2
..
.
w0
w2
w4
..
.
···
···
···
..
.
w 0 w N−1 w 2(N−1) · · ·







w0
w0
w0
..
.
w0
w1
w2
..
.
w0
w2
w4
..
.
···
···
···
..
.
w 0 w N−1 w 2(N−1) · · ·
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
w0
w N−1
w 2(N−1)
..
.





w (N−1)(N−1)
w0
w N−1
w 2(N−1)
..
.
w (N−1)(N−1)
CALCULUS 3 Chapter 5
f0
f1
f2
..
.




.


fN−1
(19)







F0
F1
F2
..
.
FN−1
FOURIER TRANSFORMS



.


5.6 FAST FOURIER TRANSFORM
5.6.3
MATRIX REPRESENTATION OF THE DFT
Example 6.3
(a) Find the matrix representing a three-point DFT.
(b) Use the matrix to find the DFT of the sequence
{fn }2n=0 = {4, −7, 11}.
Solution
(a) Here N = 3 and so
√
3
−1
−
j
.
w = e −j2π/3 =
2
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.3
MATRIX REPRESENTATION OF THE DFT
The matrix representing a three-point DFT is
 0
 

1
1
1
w w0 w0
 w 0 w 1 w 2  =  1 e j2π/3 e j4π/3 
w0 w2 w4
1 e j4π/3 e j8π/3


1
1√
1√


=  1 −1−j2 √3 −1+j2 √3  .
1 −1+j2 3 −1−j2 3
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.3
MATRIX REPRESENTATION OF THE DFT
(b)

1
F0
 F1  = 
1
F2
1



1√
−1−j 3
2√
−1+j 3
2
1√
−1+j 3
2√
−1−j 3
2


4

 −7 
11

8
=  2 + j15.5885  .
2 − j15.5885
So, the required DFT is the sequence
8, 2 + j15.5885, 2 − j15.5885.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
The fast Fourier transform (FFT) is an
algorithm that has been developed to compute the
DFT in an extremely economical fashion.
This algorithm exploits the periodicity and
symmetry of trigonometric functions to compute
the transform.
It makes the DFT a practical tool for large N.
Here one chooses N = 2p (p integer) and uses the
special form of the Fourier matrix to break down the
given problem into smaller problems.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
We consider the case where N = 4 = 22
measurements.
Then
w = e −j2π/4 = e −jπ/2 = −j
and thus,
w nk = (−j)nk .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Thus the matrix representing a four-point DFT is
w0
 w0

 w0
w0

w0
w1
w2
w3
w0
w2
w4
w6
 
w0
1 1 1 1
3 
j
w  
 1 −j −1
6 =
1 −1 1 −1
w
9
w
1
j −1 −j
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5


.

FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Let the sample values {gn }3n=0 be given. We find
F{gk }.
Step 1. Determine the matrix representing the DFT.
By (19), Gk = F{gk }
representation as

  0
G0
w
 G1 (19)  w 0

 
 G2  =  w 0
G3
w0
is given in the matrix
w0
w1
w2
w3
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
w0
w2
w4
w6

w0
g0
3 
w   g1
w 6   g2
w9
g3
CALCULUS 3 Chapter 5


.

FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Since
w 4 = 1, w 5 = w , w 6 = w 2 , w 7 = w 3 ,
w 8 = w 4 = 1, w 9 = w 5 = w ,
we have

G0
 G1

 G2
G3



1 1 1 1
g0
  1 w 1 w 2 w 3   g1
=

  1 w 2 w 0 w 2   g2
1 w3 w2 w1
g3
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5


.

(20)
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Step 2.
Factorize the matrix:

 
1 1
1
1
1 w0
 1 w2 w0 w2   1 w2

 
 1 w1 w2 w3  =  0 0
1 w3 w2 w1
0 0


0 0
1 0 w0 0
 0 1 0 w0 
0 0 


.
1 w1   1 0 w2 0 
1 w3
0 1 0 w2
(21)
The matrix on the left-hand side of (21) is the
coefficient matrix of (20), but with rows 2 and 3
interchanged.
Thus we can write (20) as
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
 
G0
1 w0
 G2   1 w 2
 

 G1  =  0 0
G3
0 0




0 0
1 0 w0 0
g0
0 


0 0 
  0 1 0 w   g1  .
1 w 1   1 0 w 2 0   g2 
1 w3
0 1 0 w2
g3
(22)
We now define a vector
 ′  
1 0
g0
 g1′   0 1
 

 g2′  =  1 0
0 1
g3′

g0
w0 0
0 
0 w   g1
w 2 0   g2
g3
0 w2
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5


.

(23)
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
It then follows from (23) that
g0′ = g0 + w 0 g2
g1′ = g1 + w 0 g3
and
g2′ = g0 + w 2 g2 = g0 − w 0 g2
g3′ = g1 + w 2 g3 = g1 − w 0 g3 .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
(22) can be rewritten in the
 

1 w0 0
G0
 G2   1 w 2 0
 

 G1  =  0 0 1
0 0 1
G3
Hence,
G0
G2
G1
G3
form

0

0 

w1  
w3

g0′
g1′ 
.
g2′ 
g3′
(24)
= g0′ + w 0 g1′
= g0′ + w 2 g1′
= g2′ + w 1 g3′
= g2′ + w 3 g3′ .
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Example 6.4 Use the method of the FFT
algorithm to compute the Fourier transform of the
sequence
{gn }3n=0 = {1, 2, 1, 0}.
In this case N = 4 = 22 , so
n
n
w n = e −j2π/4 = e −jπ/2 = (−j)n .
Solution.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
We computing the vector
 


g0′
1 0 w0 0
g0
 g1′ (23)  0 1 0 w 0   g1 

 


 g3′  =  1 0 w 2 0   g2 
0 1 0 w2
g3
g3′

  
1 0 1 0
1
 0 1 0 1  2  
  
=
 1 0 −1 0   1  = 
0 1 0 −1
0

Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5

2
2
.
0
2
FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Set
[G0′ G1′ G2′ G3′ ]T = [G0 G2 G1 G3 ]T .
From (24) we find that this vector is given by
 ′  
 ′ 
G0
1 w0 0 0
g0
 G1′   1 w 2 0 0   g1′ 

 


 G2′  =  0 0 1 w 1   g2′ 
G3′
0 0 1 w3
g3′

  
2
4
1 1 0 0
 1 −1 0 0   2   0
  
=
 0 0 1 −j   0  =  −2j
2
2j
0 0 1 j
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5


.

FOURIER TRANSFORMS
5.6 FAST FOURIER TRANSFORM
5.6.4
THE FAST FOURIER TRANSFORM
Finally, since
G0′ = G0 , G1′ = G2 , G2′ = G1 ,
and G3′ = G3 ,
we recover the transform vector [G0 G1 G2 G3 ]T as
  ′  


G0
G0
4
 G1   G2′   −2j 
 
 


 G2  =  G1′  =  0  .
G3
G3′
2j
Therefore the Fourier transform of the sequence
1, 2, 1, 0 is the sequence 4, −2j, 0, 2j.
Assoc. Prof. Dr. Nguyen Dinh & Dr. Nguyen Ngoc Hai
CALCULUS 3 Chapter 5
FOURIER TRANSFORMS