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Introduction to Fourier Transforms
Overview and Motivation : Fourier transform theory is the extension of Fourier
series theory to functions that are defined for all values of x. Thus, we will be able to
represent a function defined for − ∞≤ x ≤∞ as a linear combination of harmonic
functions.
Key Mathematics : Fourier transforms and more vector-space theory.
I. Fourier Series vs the Fourier Transform
By now you should be intimately familiar with the Fourier series representation of a
function f ( ) x on the interval − L ≤ x ≤ L. A representation that uses the normalized
harmonic functions
inxL
L
e
π
2
1
(introduced in Lecture 14) is
( )
∞
=−∞
n
inxL
n
ce
L
f x
π
, (1a)
( )
−
−
L
L
inxL
n
f x e dx
L
c
π
. (1b)
As we know the Fourier series representation is useful for any function that we only
need to define within the bounds − L ≤ x ≤ L. Outside that interval, the
representation is periodic with period 2 L because the rhs of Eq. (1a) has a period of
2 L.
There are many times, however, when we wish to represent a (nonperiodic) function
on the entire real line as a linear combination of harmonic functions. To do this we
can take the L →∞ limit of Eq. (1). This limit (which we will not go through, but is
well defined) yields the following pair of relationships
( ) ( )
∞
−∞
f x = hk e dk
ikx
, (2a)
( ) ( )
∞
−∞
−
h k = f x e dx
ikx
, (2b)
There are several things to notice about Eq. (2). First, we have traded in the discrete
index n in Eq. (1) for the continuous variable k = ( n π L ) ( = 2 π λ), which is already
familiar as the wave vector. Second, if we compare Eqs. (1) and (2), we might
conclude that
ikx
e
2 π
1
are now our normalized harmonic functions. That is correct, as
we discuss in further detail below. With that, we can then interpret the function h ( k )
as the coefficient (or component) of the harmonic function
ikx
e
2 π
1
. This function
h ( ) k has a special name: it is known as the Fourier transform of the function f ( x ).
The Fourier transform h ( ) k is thus analogous to the Fourier coefficients
n
c that
appear in the Fourier series. The other feature of Eq. (2) that you undoubtedly
noticed is that f ( ) x is expressed as a continuous sum (integral) over basis functions
rather than a discrete sum over basis functions. This is a necessary consequence of
the L →∞limit.
Because of the resultant symmetry in the two relationships in Eq. (2), the function
f ( ) x is also known as the inverse Fourier transform of h ( k ). In fact, because f ( x )
and h ( ) k are obtainable from each other, they each contain the same information, just
in a different form.
1
We remark that there are technical criteria that the function f ( x )must meet for Eq.
(2) to be valid. A sufficient condition is that f ( x )be square integrable,
( ) <∞
∞
−∞
f x dx
2
If Eq. (3) is true, then h ( ) k is also square integrable and it can be shown that
( ) ( )
∞
−∞
∞
−∞
f x dx = hk dk
2 2
The proof of Eq. (4) is left as an exercise.
II. The Fourier Transform and Vector Space Theory
As we also discussed in Lecture 14, the Fourier series [Eq. (1)] can be thought of as a
pair of vector-space relationships
1
Of course, the same is true about the function f ( x )and the coefficients
n
c in the Fourier series.
Knowing the
n
c 's is equivalent to knowing the function f ( x )itself.
A comparison of Eqs. (5) and (9) then suggests that we define the inner product on
this vector space as
( g ( ) x f ( ) x ) g ( ) x f ( ) x dx
∞
−∞
Let's see what this gives us if we calculate ( u ˆ( k , x ), u ˆ( k , x ))
for this vector space. If life
is good then we expect to get ( ( ) ( ))
kk
u k x uk x
′
ˆ , ,ˆ , δ , similar to Eq. (7). Let's see what
happens. Using Eq. (10) we have
( ( ) ( ))
( )
∞
−∞
−′
u k x uk x e dx
ikk x
We consider two cases separately, k = k ′and k ≠ k ′.
( i ) If k = k ′, then Eq. (11) integrates to
( ( ) ( ))
∞
−∞
uk x uk x = x
, (12a)
which is undefined. Hum… Not too good. Let's look at the other case.
( ii ) If k ≠ k ′, then Eq. (11) integrates to
( ( ) ( ))
( )
( )
∞
−∞
−′
ikk x
e
ik k
u k x uk x
(12b)
Unfortunately, this is not defined either! So it looks like either the basis functions or
the inner product is unsuitable.
As it turns out, we can fix this dilemma by defining the inner product slightly
differently as
( g ( ) x f ( ) x ) e g ( ) x f ( ) x dx
x n
n
∞
−∞
−
→∞
2
, lim. (13)
Notice what the function
x n
e
2
−
does for us. For any finite n this Gaussian function
cuts off the integrand fast enough to make the integral converge. Furthermore in the
limit
n →∞
the function itself simply approaches 1.
3
Let's see what happens with this definition of the inner product. We now have
( ( ) ( ))
( )
∞
−∞
− −′
→∞
u k ′ x uk x = e e dx
x n ik kx
n
2
lim
To take advantage of the symmetry of the Gaussian we rewrite this as
( ( ) ( )) { [( ) ] [( ) ]}
∞
−∞
−
→∞
u k x uk x e k k x i k k x dx
x n
n
lim cos sin
2
Because
x n
e
2
−
is even, the integral involving the sine function is zero, so this simplifies
to
( ( ) ( )) [( ) ]
∞
−∞
−
→∞
u k x uk x e k k xdx
x n
n
lim cos
2
We have seen this integral before (see Lecture 12). Calculating the integral, Eq. (15)
becomes
( ( ) ( ))
( )
2
2
lim
n nkk
n
u k x uk x e
− −′
→∞
π
Now you may remember from the last lecture that the limit of a similar sequence of
Gaussian functions is the Dirac delta function. If you closely compare Eq. (16) with
Eq. (8) from the Lecture 15 notes, you will see that Eq. (16) can be expressed as
( ( ) ( ))
k k
uk x uk x δ (17)
And using the relationship δ ( xa ) = a δ( ) x , this simplifies to
3
In fact,
x n
n
e
2
lim
−
→∞
is one definition of the unit distribution.
( ( ) ( ))
( )
u k x f x e e dx h ( k ) dk
x n ik kx
n
∞
−∞
∞
−∞
−
′ −
→∞
2
lim
But from Eq. (20) we see that Eq. (22) is simply
( u ( k x ) f ( ) x ) = ( k ′− k ) ( h k ′) dk ′
∞
−∞
which gives us the result that we want,
( u ( k , x ), f ( ) x ) = h ( ) k
Thus, as with the
n
c 's in the Fourier series representation of a function, the Fourier
transform h ( ) k can be though of as the inner product of the normalized basis
function with the original function
f ( x ).
Exercises
- 16.1 Show that Eq. (4), ( ) ( )
∞
−∞
∞
−∞
f x dx = hk dk
2 2
, is true.
* 16.2 Calculate the Fourier transform of the function
( )
x
f x e
−
=. Plot the resulting
function vs k.
** 16.3 As the notes discuss, the original attempt at defining the inner product as
( g ( ) x f ( ) x ) g ( ) x f ( ) x dx
∞
−∞
, needs to be slightly modified. We chose one particular way
that this can be done. Another choice that we could have made is
( g ( ) x f ( ) x ) g ( ) x f ( ) xdx
n
n
n
−
→∞
, lim. Show that this definition of the inner product also
gives the result
( u ˆ ( k ′, x ), u ˆ( k , x )) =δ( k − k ′)for the inner product of two basis functions.
{Hint: you will need to use the second definition of the delta function from Lecture
15 [Eq. (9) on p. 4]}.
