Chapter 1

Mathematical tools
Fourier analysis in two dimensions

1. Introduction

In this chapter we review some of the mathematical tools that are useful in describing linear
phenomena, and discuss some of the mathematical decompositions that are often employed in
their analysis.

Learning outcomes
After this course the student is able to:
-Calculate analytically the expression of the 2D Fourier transforms of all function combinations
commonly used in optics
-Explain the properties of a 2D or 1D function based on the analysis of its 2D or 1D spectrum,
respectively, and vice versa.
-Interpret what is a spatial or a temporal frequency

1
2. Fourier transform
2.1. Definition in one dimension

The Fourier transform is a mathematical operator that, when applied to a function 𝑓 𝑥

real or complex, leads to a real or complex function 𝐹 𝑢 = 𝐹𝑇𝑢 𝑓 𝑥

𝐹 𝑢 = 𝐹𝑇𝑢 𝑓 𝑥 = න 𝑓 𝑥 𝑒 −𝑗2𝜋𝑢𝑥 𝑑𝑥 (1-0)

−∞

Where 𝑥 is a coordinate in the space domain and 𝑢 the corresponding coordinate in

the dual Fourier domain

If we write 𝐹 𝑢 in a polar form 𝐹 𝑢 = 𝐴 𝑢 𝑒 −𝑗𝜑 𝑢

with 𝐴 𝑢 > 0 and 𝜑 𝑢 defined
modulo 2𝜋, we can use the following terminology
• 𝐴 𝑢 is the spectrum of 𝑓 𝑥
• 𝐴² 𝑢 is the spectral energy of 𝑓
• 𝜑 𝑢 is the spectral phase of 𝑓

2
2.2. Definition in two dimensions

The Fourier transform of a function 𝑔 𝑥, 𝑦 of two independent variables x and y, will

be represented here by a function 𝐺 𝑢, 𝑣 = 𝐹𝑇𝑢,𝑣 𝑔 𝑥, 𝑦 and defined by

𝐺 𝑢, 𝑣 = 𝐹𝑇𝑢,𝑣 𝑔 𝑥, 𝑦 = ඵ 𝑔 𝑥, 𝑦 𝑒 −𝑗2𝜋 𝑢𝑥+𝑣𝑦

𝑑𝑥𝑑𝑦 (1-1)
−∞

The transform so defined is itself a complex-valued function of two independent

variables u and v, which we generally refer to as frequencies. Similarly, the
inverse Fourier transform of a function 𝐺 𝑢, 𝑣 will be represented by 𝐹𝑇𝑥,𝑦
−1
𝐺 𝑢, 𝑣
and is defined as

∞
−1
𝑔 𝑥, 𝑦 = 𝐹𝑇𝑥,𝑦 𝐺 𝑢, 𝑣 = ඵ 𝐺 𝑢, 𝑣 𝑒 𝑗2𝜋 𝑢𝑥+𝑣𝑦
𝑑𝑢𝑑𝑣 (1-2)
−∞

3
2.3. Existence conditions

While a variety of sets of sufficient conditions for the existence of (1-1) are possible,
perhaps the most common set is the following:

1. g must be absolutely integrable over the infinite (x,y) plane.

2. g must have only a finite number of discontinuities and a finite number of maxima
and minima in any finite rectangle.
3. g must have no infinite discontinuities.

Usually "physical possibility is a valid sufficient condition for the existence of

a transform."

However, it is often convenient in the analysis of systems to represent true physical

waveforms by idealized mathematical functions, and for such functions one or more of
the above existence conditions may be violated.
For example, it is common to represent an idealized point source of light by the two-
dimensional Dirac delta function often represented by

𝛿 𝑥, 𝑦 = lim 𝑁 2 𝑒𝑥𝑝 −𝑁 2 𝜋 𝑥 2 + 𝑦² (1-3)

𝑛→∞

The Dirac delta function being infinite at the origin and zero elsewhere, has an infinite
discontinuity and therefore fails to satisfy existence condition 3. But, each member
function of the defining sequence (1-3) does satisfy the existence requirements and
each, in fact, has a Fourier transform given by
4
 2 2
𝜋 𝑢2 + 𝑣²
𝐹𝑇𝑢,𝑣 𝑁²𝑒𝑥𝑝 −𝑁 𝜋 𝑥 + 𝑦² = 𝑒𝑥𝑝 − (1-4)
𝑁²

Accordingly the generalized transform of 𝛿 𝑥, 𝑦 is found to be

𝜋 𝑢2 + 𝑣²
𝐹𝑇𝑢,𝑣 𝛿 𝑥, 𝑦 = lim 𝑒𝑥𝑝 − =1 (1-5)
𝑛→∞ 𝑁²

The spectrum of a delta function extends uniformly over the entire frequency
domain. We come back to the Delta function in section 3

5
2.4. The Fourier transform as a decomposition

We may regard the two-dimensional Fourier transform as a decomposition of a

function 𝑔 𝑥, 𝑦 into a linear combination of elementary functions of the form
𝒆𝒋𝟐𝝅 𝒖𝒙+𝒗𝒚 according to equation (1-2). From this it is clear that the complex number
𝑮 𝒖, 𝒗 is simply a weighting factor that must be applied to the elementary function
of frequencies u,v in order to synthesize the desired 𝑔 𝑥, 𝑦 .

The great advantage afforded by linearity is the ability to express the response (be it
light amplitude, or light intensity) to a complicated stimulus in terms of the responses
to certain "elementary“ stimuli. Thus if a stimulus is decomposed into a linear
combination of elementary stimuli, each of which produces a known response of
convenient form, then by virtue of linearity, the total response can be found as a
corresponding linear combination of the responses to the elementary stimuli.

6
2.5. Fourier Transform Theorems and properties

We now consider a few of the basic mathematical properties of the transform,

properties that will find wide use in later material. These properties are presented as
mathematical theorems, followed by brief statements of their physical significance.
These theorems are direct extensions of the analogous one-dimensional statements,
whose proofs can be found in Appendix A of Reference [1].

 Linearity theorem

𝐹𝑇𝑢,𝑣 𝛼𝑔 𝑥, 𝑦 + 𝛽ℎ 𝑥, 𝑦 = 𝛼𝐹𝑇𝑢,𝑣 𝑔 𝑥, 𝑦 + 𝛽𝐹𝑇𝑢,𝑣 ℎ 𝑥, 𝑦 (1-6)

The transform of a weighted sum of two (or more) functions is simply the identically
weighted sum of their individual transforms. α and β are two scalar coefficients

 Conjugate

∗
𝐹𝑇𝑢,𝑣 𝑔∗ 𝑥, 𝑦 = 𝐹𝑇−𝑢,−𝑣 𝑔 𝑥, 𝑦

The transform of a conjugate is the conjugate of the transform taken in opposite values

7
  Similarity theorem.

1 𝑢 𝑣 (1-7)
a and b are two scalar coefficients 𝐹𝑇𝑢,𝑣 𝑔 𝑎𝑥, 𝑏𝑦 = 𝐺 ,
𝑎𝑏 𝑎 𝑏

A "stretch" of the coordinates in the space domain (x, y) results in a contraction of

the coordinates in the frequency domain (u,v), plus a change in the overall amplitude
of the spectrum

 Shift theorem.

𝐹𝑇𝑢,𝑣 𝑔 𝑥 − 𝑎, 𝑦 − 𝑏 = 𝐺 𝑢, 𝑣 𝑒 −𝑗2𝜋 𝑢𝑎+𝑣𝑏 (1-8)

Translation in the space domain introduces a linear phase shift in the frequency
domain

 Rayleigh's theorem (Parseval's theorem).

∞ ∞

ඵ 𝑔 𝑥, 𝑦 2
𝑑𝑥𝑑𝑦 = ඵ 𝐺 𝑢, 𝑣 2
𝑑𝑢𝑑𝑣 (1-9)
−∞ −∞

The integral on the left-hand side of this theorem can be interpreted as the energy
contained in the waveform 𝑔 𝑥, 𝑦 . This in turn leads us to the idea that the
quantity 𝐺 𝑢, 𝑣 2 can be interpreted as an energy density in the frequency
domain.
8
 Space domain Fourier dual domain = frequency domain
Similarity theorem (illustration)
𝑓 𝑥 𝐹 𝑢

𝐹 𝑢 = 𝐹𝑇𝑢 𝑓 𝑥

x u

x u
 0 
0
Initial function Corresponding Fourier transform
 


𝑢
𝑓 𝑘𝑥 𝐹
𝑘

k  x u k

x u
0
 0
Outstretched function Corresponding Fourier transform
 
 Space domain Fourier dual domain = frequency domain
Shift theorem (illustration if 𝑓 𝑥 is real)
𝑓 𝑥 𝐹 𝑢

𝐹 𝑢 = 𝐹𝑇𝑢 𝑓 𝑥

u

x u
0 
0
Initial function Corresponding Fourier transform
 


𝑓 𝑥−𝑎 ℜ𝑒 𝐹𝑇𝑢 𝑓 𝑥 − 𝑎
= 𝑐𝑜𝑠 2𝜋𝑢𝑎 𝐹 𝑢

𝑎 u
0
x 1

0
 𝑎
Translated function Corresponding Fourier transform

The convolution of two funtions is defined by:
∞

ඵ 𝑔 𝜉, 𝜂 ℎ 𝑥 − 𝜉, 𝑦 − 𝜂 𝑑𝜉𝑑𝜂 = 𝑔 ⋆ ℎ 𝑥, 𝑦 (1-10)
−∞

 Convolution theorem.

𝐹𝑇𝑢,𝑣 ඵ 𝑔 𝜉, 𝜂 ℎ 𝑥 − 𝜉, 𝑦 − 𝜂 𝑑𝜉𝑑𝜂 = 𝐺 𝑢, 𝑣 𝐻 𝑢, 𝑣 (1-11)

−∞

Or
𝐹𝑇𝑢,𝑣 𝑔 ⋆ ℎ 𝑥, 𝑦 = 𝐺 𝑢, 𝑣 𝐻 𝑢, 𝑣 (1-12)

The convolution of two functions in the space domain is entirely equivalent to the
more simple operation of multiplying their individual transforms and inverse
transforming.
Similarly
∞

𝐹𝑇𝑢,𝑣 𝑔 𝑢, 𝑣 ℎ 𝑢, 𝑣 = ඵ 𝐺 𝜉, 𝜂 𝐻 𝑢 − 𝜉, 𝑣 − 𝜂 𝑑𝜉𝑑𝜂 = 𝐺 ⋆ 𝐻 𝑢, 𝑣 (1-13)

−∞

The Fourier transform of the multiplication of two functions in the space domain is
11
equivalent to the convolution of their individual Fourier transforms.
Geometric interpretation of the convolution in one dimension

𝑔 𝜉 ℎ 𝜉

⋆
𝑥

𝜉 0
𝜉
0

ℎ 𝜉  ℎ 𝑥−𝜉


0 𝜉 0 𝜉
𝑥

 

𝑔⋆ℎ 𝑥 =
∞

ඵ 𝑔 𝜉 ℎ 𝑥 − 𝜉 𝑑𝜉
−∞ 𝜉
0 𝑥
 Autocorrelation theorem.

𝐹𝑇𝑢,𝑣 ඵ 𝑔 𝜉, 𝜂 𝑔∗ 𝜉 − 𝑥, 𝜂 − 𝑦 𝑑𝜉𝑑𝜂 = 𝐺 𝑢, 𝑣 2 (1-14)

−∞

Similarly
∞
2
𝐹𝑇𝑢,𝑣 𝑔 𝑢, 𝑣 = ඵ 𝐺 𝜉, 𝜂 𝐺 ∗ 𝜉 − 𝑢, 𝜂 − 𝑣 𝑑𝜉𝑑𝜂 (1-15)
−∞

This theorem may be regarded as a special case of the convolution theorem in which
we convolve 𝑔 𝑥, 𝑦 with 𝑔∗ −𝑥, −𝑦 where 𝑔∗ 𝑥, 𝑦 is the complex conjugate of 𝑔 𝑥, 𝑦

The correlation between two functions is defined by

ඵ 𝑔 𝜉, 𝜂 ℎ∗ 𝜉 − 𝑥, 𝜂 − 𝑦 𝑑𝜉𝑑𝜂 = 𝑔⨂ℎ 𝑥, 𝑦
−∞
(1-16)
∞

ඵ 𝑔 𝜉 + 𝑥, 𝜂 + 𝑦 ℎ∗ 𝜉, 𝜂 𝑑𝜉𝑑𝜂 = 𝑔⨂ℎ 𝑥, 𝑦
−∞
13
 Fourier integral theorem.

At each point of continuity of g

−1
𝐹𝑇𝑥,𝑦 𝐹𝑇𝑢,𝑣 𝑔 𝑥, 𝑦 = 𝑔 𝑥, 𝑦 (1-17)

At each point of discontinuity of g, the two successive transforms yield the angular
average of the values of g in a small neighborhood of that point. That is, the
successive transformation and inverse transformation of a function yields that function
again, except at points of discontinuity

𝐹𝑇𝑥,𝑦 𝐹𝑇𝑢,𝑣 𝑔 𝑥, 𝑦 = 𝑔 −𝑥, −𝑦

 Derivative theorem.

𝑑𝑛 𝑓 𝑥
𝐹𝑇𝑢 = 𝑖2𝜋𝑢 𝑛
𝐹𝑇𝑢 𝑓 𝑥 (1-18)
𝑑𝑥 𝑛

The above transform theorems are of far more than just theoretical interest. They will
be used frequently, since they provide the basic tools for the manipulation of Fourier
transforms and can save enormous amounts of work in the solution of Fourier analysis
14
problems.
 3. Dirac function
 x 

Absence of event 𝛿 𝑥, 𝑦 = 0 𝑖𝑓 𝑥 𝑎𝑛𝑑 𝑦 ≠ 0



Localization of the event

𝛿 𝑥, 𝑦 → ∞ 𝑖𝑓 𝑥𝑎𝑛𝑑 𝑦 = 0
∞

Condition of normalisation
න 𝛿 𝑥, 𝑦 𝑑𝑥 𝑑𝑦 = 1 x

0
−∞


 𝛿 𝑥 − 𝑥0 , 𝑦
𝛿 𝑥 − 𝑥0 , 𝑦 = 0 𝑖𝑓 𝑥 ≠ 𝑥0 𝑎𝑛𝑑 𝑦 ≠ 0
𝛿 𝑥 − 𝑥0 , 𝑦 → ∞ 𝑖𝑓 𝑥 = 𝑥0 𝑎𝑛𝑑 𝑦 = 0
∞

න 𝛿 𝑥 − 𝑥0 , 𝑦 𝑑𝑥 𝑑𝑦 = 1
x
−∞ 0 0


The convolution of a function 𝑔(𝑥,𝑦) by the Dirac function centered on 𝒙𝟎 , 𝛿 𝑥− 𝑥0 , 𝑦 , corresponds to a
translation of the initial function. 
∞

ඵ 𝑔 𝜉, 𝜂 𝛿 𝑥 − 𝜉 − 𝑥0 , 𝑦 − 𝜂 𝑑𝜉𝑑𝜂 = 𝑔 𝑥 − 𝑥0 , 𝑦
−∞
𝑔(𝑥,𝑦) 𝛿 𝑥 − 𝑥0 , 𝑦 𝑔 𝑥 − 𝑥0 , 𝑦
Translated Dirac function Translation effect
Initial function
⋆

x0 x0 x0
15
0 0 x0 0 x0
The product of a function 𝑔(𝑥,𝑦) by the Dirac function centered on 𝑥0 , 𝛿 𝑥 − 𝑥0 , 𝑦 , corresponds to a
sampling of the initial function.

𝑔(𝑥,𝑦) 𝛿 𝑥 − 𝑥0 , 𝑦 − 𝑦0 = 𝑔(𝑥0 ,𝑦0 ) 𝛿 𝑥 − 𝑥0 , 𝑦 − 𝑦0

𝑔(𝑥,𝑦) 𝛿 𝑥, 𝑦 = 𝑔(0,0) 𝛿 𝑥, 𝑦

f x   x  x 0 
f x 0 
Multiplied by
 

x x x
0 0 x0 0 x0

 
Initial function TranslatedDirac function Sampling effect
   


16
4. Separable Functions

A function of two independent variables is called separable with respect to a specific

coordinate system if it can be written as a product of two functions, each of which
depends on only one of the independent variables. Thus the function g is separable in
rectangular coordinates 𝑥, 𝑦 if

𝑔 𝑥, 𝑦 = 𝑔𝑋 𝑥 𝑔𝑌 𝑦 (1-19)

while it is separable in polar coordinates 𝑟, 𝜃 if

𝑔 𝑟, 𝜃 = 𝑔𝑅 𝑟 𝑔Θ 𝜃 (1-20)

a function separable in rectangular coordinates has the particularly simple

property that its two-dimensional Fourier transform can be found as a product of one-
dimensional Fourier transforms

𝐹𝑇𝑢,𝑣 𝑔 𝑥, 𝑦 = 𝐹𝑇𝑢 𝑔𝑋 𝑥 𝐹𝑇𝑣 𝑔𝑌 𝑦 (1-21)

The Fourier transform of a general function separable in polar coordinates can

be expressed as an infinite sum of weighted Hankel transforms

𝐹𝑇𝜌,𝜙 𝑔 𝑟, 𝜃 = σ∞
𝑘=−∞ 𝑐𝑘 −𝑗
𝑘
𝑒𝑥𝑝 𝑗𝑘𝜙 ℋ𝑘 𝑔𝑅 𝑟 (1-22)
1 2𝜋
Where 𝑐𝑘 = ‫𝑔 ׬‬Θ 𝜃 𝑒𝑥𝑝 −𝑗𝑘𝜃 𝑑𝜃 (1-23)
17
2𝜋 0
And ℋ𝑘 is the Hankel transform operator of order k, defined by
∞
ℋ𝑘 𝑔𝑅 𝑟 = 2𝜋 න 𝑟 𝑔𝑅 𝑟 𝐽𝑘 2𝜋𝑟𝜌 𝑑𝑟 (1-24)
0

Here the function 𝑱𝒌 is the kth-order Bessel function of the first kind.

The Bessel function of the first kind, zeroth order, 𝑱𝟎 can be written as:

2𝜋
1
𝐽0 𝑎 = න 𝑒 −𝑗𝑎 𝑐𝑜𝑠 𝜃−𝜙
𝑑𝜃 (1-25)
2𝜋
0

The Bessel function of the first kind, kth order, 𝑱𝒌 can be written as:

2𝜋
−𝑘
𝑗 (1-26)
𝐽𝑘 𝑎 = න 𝑒 −𝑗 𝑘𝜃+𝑎𝑐𝑜𝑠𝜃
𝑑𝜃
2𝜋
0

The kth-order and (k-1)th-order Bessel function of the first kind are related by the
following recurrence relation:

𝑑 𝑘
𝑎 𝐽𝑘 𝑎 = 𝑎𝑘 𝐽𝑘−1 𝑎 (1-27)
𝑑𝑎

18
5. Functions with Circular Symmetry: Fourier-Bessel transforms

The function g is said to be circularly symmetric if it can be written as a function of r

alone, that is,

𝑔 𝑟, 𝜃 = 𝑔𝑅 𝑟 (1-28)

Such functions play an important role in the problems of interest here, since most
optical systems have precisely this type of symmetry. We accordingly devote special
attention to the problem of Fourier transforming a circularly symmetric function.
The Fourier transform of g in a system of rectangular coordinates is, of course, given
by
∞

𝐺 𝑢, 𝑣 = ඵ 𝑔 𝑥, 𝑦 𝑒 −𝑗2𝜋 𝑢𝑥+𝑣𝑦
𝑑𝑥𝑑𝑦 (1-1)
−∞

To fully exploit the circular symmetry of g, we make a transformation to polar

coordinates in both the 𝒙, 𝒚 and the 𝒖, 𝒗 planes as follows:

𝑟= 𝑥 2 + 𝑦2 𝑥 = 𝑟 𝑐𝑜𝑠𝜃
𝑦
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛
𝑥
𝑦 = 𝑟 𝑠𝑖𝑛𝜃 (1-29)
𝜌 = 𝑢2 + 𝑣 2 𝑢 = 𝜌 𝑐𝑜𝑠𝜙
𝑣
𝜙 = 𝑎𝑟𝑐𝑡𝑎𝑛 v = 𝜌 𝑠𝑖𝑛𝜙
𝑢

19
Applying the coordinate transformations (1-1) to Eq. (1-28), the Fourier transform
of 𝑔 can be written
∞ 2𝜋

𝐺0 𝜌, 𝜙 = න න 𝑟𝑔𝑅 𝑟 𝑑𝑟𝑒 −𝑗2𝜋𝑟𝜌 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜙+𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜙

𝑑𝜃 (1-30)
0 0
∞ 2𝜋

𝐺0 𝜌, 𝜙 = න 𝑟𝑔𝑅 𝑟 𝑑𝑟 න 𝑒 −𝑗2𝜋𝑟𝜌 𝑐𝑜𝑠 𝜃−𝜙

𝑑𝜃 (1-31)
0 0

Note the subscript in 𝐺0 is added simply because the functional form of the expression
for the transform in polar coordinates is in general different than the functional form
for the same transform in rectangular coordinates.

Finally, we use of the Bessel function of the first kind, zero order, 𝐽0 Eq. 1.25
allows to simplify the expression for the transform

𝐺0 𝜌, 𝜙 = 𝐺0 𝜌 = 2𝜋 න 𝑟𝑔𝑅 𝑟 𝐽0 2𝜋𝑟𝜌 𝑑𝑟 (1-32)

0

Thus the Fourier transform of a circularly symmetric function is itself

circularly symmetric and can be found by performing the 1D manipulation of (1-
31).
This particular form of the Fourier transform occurs frequently enough to warrant
a special designation; it is accordingly referred to as the Fourier-Bessel transform,
or alternatively as the Hankel transform of zero order (cf. Eq. (1-24)).
20
By means of arguments identical with those used above, the inverse Fourier
transform of a circularly symmetric spectrum 𝐺0 𝜌 can be expressed as
∞

𝑔𝑅 𝑟 = 2𝜋 න 𝜌𝐺0 𝜌 𝐽0 2𝜋𝑟𝜌 𝑑𝜌 (1-33)

0

Thus for circularly symmetric functions there is no difference between the transform
and the inverse-transform operations.

21
6. Some frequently used functions

1
1 𝑖𝑓 𝑥 <
2
Rectangle function 𝒓𝒆𝒄𝒕 𝒙 = 1
𝑖𝑓 𝑥 =2
1
2
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

𝑠𝑖𝑛 𝑥
Sinc function 𝑠𝑖𝑛𝑐 𝑥 =
𝑥

1 𝑖𝑓 𝑥 > 0
Signum function 𝑠𝑔𝑛 𝒙 = ቐ 0 𝑖𝑓 𝑥=0
−1 𝑥<0

22
 1− 𝑥 𝑖𝑓 𝑥 ≤1
Triangle function 𝚲 𝒙 =ቊ
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Comb function 𝑐𝑜𝑚𝑏 𝑥 = σ∞

−∞ 𝛿 𝑥 − 𝑛

1 𝑖𝑓 𝑟 < 1
Circle function 𝐂𝐢𝐫𝐜𝟏 𝒓 = ൞ 1
2
𝑖𝑓 𝑟=1
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
23
 7. Some useful 1D Fourier transform pairs

Initial function FOURIER transform

f x  F u  T.F.u f x 
f x  F u 
 u 
1
1 x u
  0  0
 
x 
f  F 
u
1
  x   x  1  u
0 0

 
f x  F u

 x  x 0   u
expi2ux 0

 x   0
0 x0
 
f x  F u

H x  1  1 1
x  u   
   2 i2u u
0
  0
 f x  F u 
x, x 0 
1  sin2ux 0  
 x 2x 0 u
x 0 0 
x0 2ux 0 
  0

24  
 Initial function FOURIER transform
f x  F u  T.F.u f x 
f x 
x, x 1, x 2  1 x 2  x1   sinux 2  x1   exp iux 2  x1 

x
 u x 2  x1 
0 x1 x2

 f x  F u 
1  sinux 0 2
x, x 0     x x0  u
x 0 0 x0  ux 0 
   0
f x  F u 
expi 2u0 x
 
   0
x
 u  u0  u

0 u0
  
Fu  F u 
 1  u  u0   
cos2u0x  u
2 
 u  u0    u
0   u0 u0
0
 F u  
F u
 i  u  u0  
sin2u0x  
u  u u
2 
 u  u0 
0
0   u 
0
  
0


 
25 
 
 Initial function FOURIER transform
f x  F u  T.F.u f x 
1 f x  1 F u 

expx 2 x expu 2
     
u
0 0
1 f x  F u 

   2u 2  1

expkx 2 exp  


  x k  k   u
   0
0
f x  F u
expix 2 
x   
expi   expiu 2  u
  4  

0   0
Fu  F u 
1 
H x   expkx  u
  i2u  k  u
0 0
  F u 

  x  kx 
f x 


 1  k 
0 x u  
 u

0 x0 
x0  x 0   0 1 x0
k k 

 
26
  
  

27
Exercises

Calculate the Fourier transform using the integral definition

Π 𝑥, 𝑥0
1.1 Calculate the Fourier transform of the door function Π 𝑥, 𝑥0 defined by: 1
x

1 𝑖𝑓 𝑥 < 𝑥0 −𝑥0 𝑥0
Π 𝑥, 𝑥0 = ቊ 
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒 
by using equation 1.0 (result given in the table of section 7)

Use known Fourier transforms and the mathematical properties of the Fourier transform to
calculate the FT of new functions

1.2 Calculate the Fourier transform of the door function Π 𝑥, 𝑥1 , 𝑥2 defined by: Π 𝑥, 𝑥1 , 𝑥2
1 𝑖𝑓 𝑥1 < 𝑥 < 𝑥2 1
Π 𝑥, 𝑥1 , 𝑥2 = ቊ x
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
x x1 x2
by using the result of exercise 1.1, and the shift theorem
(result given in the table of section 7) 

𝛬 𝑥, 𝑥0
 
1
𝑥
1− 𝑖𝑓 𝑥 ≤ 𝑥0  x
1.3 Let’s consider the Triangle function 𝛬 𝑥, 𝑥0 = ቐ 𝑥0
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 −𝑥0 𝑥0

Give the expression of its derivative Λ′ 𝑥, 𝑥0 using a difference of door functions 
Using the result of exercise 1.1 and the shift theorem calculate the Fourier transform of Λ′ 𝑥, 𝑥0
Using the derivative theorem, calculate the Fourier transform of 𝛬 𝑥, 𝑥0 .
(result given in the table of section 7) 28
Use the properties of separable functions to calculate the Fourier transform of two-
dimensional functions from the Fourier transform of one-dimensional functions

∆𝑥 ∆𝑦
1.4 Calculate the Fourier transform of the 2D door function Π 𝑥, ; 𝑦, defined by:
2 2

∆𝑥 ∆𝑦 ∆𝑥 ∆𝑥 ∆𝑦 ∆𝑦
Π 𝑥, ; 𝑦, =ቐ 1 𝑖𝑓 − < 𝑥 < 𝑎𝑛𝑑 − < 𝑦 <
2 2 2 2 2 2
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

1.5 Let’s consider the Circle function with a radius R in Cartesian coordinates
1 𝑖𝑓 𝑥 2 + 𝑦² < 𝑅
Circ𝑅 𝑥, 𝑦 = 1
2 𝑖𝑓 𝑥 2 +𝑦²=𝑅

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Give the expression of its Fourier transform in Cartesian coordinates (eq 1.2) and make the transformation
to polar coordinates in both the 𝑥, 𝑦 and the 𝑢, 𝑣 planes using the following equations:
𝑟 = 𝑥 2 + 𝑦2 𝜌 = 𝑢2 + 𝑣 2 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑢 = 𝜌 𝑐𝑜𝑠𝜙
𝑦 𝑣
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝜙 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 = 𝑟 𝑠𝑖𝑛𝜃 𝑣 = 𝜌 𝑠𝑖𝑛𝜙
𝑥 𝑢

Simplify the expression of the Fourier transform by using 𝐽0 , the Bessel function of the first kind, zero order,
given in Eq 1.30.
Then, using the recurrence relation 1.27 applied to the Bessel functions of the first kind and orders 0 and 1
deduce that the Fourier transform of the Circle function is

𝑅
𝐹𝑇𝜌 Circ𝑅 𝑟 = 𝐽 2𝜋𝑅𝜌
𝜌 1 29
Create a periodic 2D function and calculate its Fourier transform

∆𝑥 ∆𝑦
1.6 Use the Dirac comb to periodize the 2D door function Π 𝑥, ; 𝑦, with a period of 2∆𝑥 along the 𝑥
2 2
direction and a period of 2∆𝑦 along the 𝑦 direction
Calculate the Fourier transform of this grating (periodic function)
We limit the width of the grating to 𝐿𝑋 along the 𝑥 direction and to 𝐿𝑌 along the 𝑦 direction. Write the new
expression for the grating function and its Fourier transform.

30
 Bibliography

[1] Introduction to Fourier Optics, Joseph W. Goodman, The McGraw Hill

Companies, reissued 1988

31
 ∆𝑥 ∆𝑦
∆𝑥 ∆𝑦 𝐹𝑇𝑢,𝑣 Π 𝑥, ; 𝑦,
2D door function Π 𝑥, 2 ; 𝑦, 2 2 2
= ∆𝑥 ∆𝑦 𝑠𝑖𝑛𝑐 𝜋𝑢∆𝑥 𝑠𝑖𝑛𝑐 𝜋𝑣∆𝑦
v

f x, y 
 1 y
 u
x
 T.F.u f x 

 
 1 x
 y F u,v 
0

v
 

x 0

y
u


33

1.5 Solution with more calculation details
Let’s consider the Circle function with a radius R in Cartesian coordinates
1 𝑖𝑓 𝑥 2 + 𝑦² < 𝑅
Circ𝑅 𝑥, 𝑦 = 1
2 𝑖𝑓 𝑥 2 +𝑦²=𝑅

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Give the expression of its Fourier transform in Cartesian coordinates (eq 1.2)
∞

𝐹𝑇𝑢,𝑣 Circ𝑅 𝑥, 𝑦 = ඵ Circ𝑅 𝑥, 𝑦 𝑒 −𝑗2𝜋 𝑢𝑥+𝑣𝑦

𝑑𝑥𝑑𝑦
−∞

and make the transformation to polar coordinates in both the 𝑥, 𝑦 and the 𝑢, 𝑣 planes using the
following equations:
𝑟 = 𝑥 2 + 𝑦2 𝜌 = 𝑢2 + 𝑣 2 𝑥 = 𝑟 𝑐𝑜𝑠𝜃 𝑢 = 𝜌 𝑐𝑜𝑠𝜙
𝑦 𝑣
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝜙 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 = 𝑟 𝑠𝑖𝑛𝜃 𝑣 = 𝜌 𝑠𝑖𝑛𝜙
𝑥 𝑢

∞ 2𝜋

𝐹𝑇𝜌 Circ𝑅 𝑟 = න න 𝑟Circ𝑅 𝑟 𝑑𝑟𝑒 −𝑗2𝜋𝑟𝜌 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜙+𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜙

𝑑𝜃
0 0
∞ 2𝜋

𝐹𝑇𝜌 Circ𝑅 𝑟 = න 𝑟Circ𝑅 𝑟 𝑑𝑟 න 𝑒 −𝑗2𝜋𝑟𝜌 𝑐𝑜𝑠 𝜃−𝜙

𝑑𝜃
0 0
Finally, we use of the Bessel function of the first kind, zero order, 𝐽0 Eq. 1.25 to simplify the expression
35
 ∞

𝐹𝑇𝜌 Circ𝑅 𝑟 = 2𝜋 න 𝑟Circ𝑅 𝑟 𝐽0 2𝜋𝑟𝜌 𝑑𝑟

0
𝑅

𝐹𝑇𝜌 Circ𝑅 𝑟 = 2𝜋 න 𝑟 𝐽0 2𝜋𝑟𝜌 𝑑𝑟

0
And by using the integral form of Eq 1-27 with k = 1
𝑅
𝑅
𝑎𝐽1 𝑎 0 = න 𝑎𝐽0 𝑎 𝑑𝑎
0
With 2𝜋𝑟𝜌 = 𝑎 we get and 2𝜋𝜌𝑑𝑟 = 𝑑𝑎
𝑅
1
𝐹𝑇𝜌 Circ𝑅 𝑟 = න 2𝜋𝑟𝜌 𝐽0 2𝜋𝑟𝜌 2𝜋𝜌𝑑𝑟
2𝜋𝜌²
0
1
𝐹𝑇𝜌 Circ𝑅 𝑟 = 2𝜋𝑟𝜌𝐽1 2𝜋𝑟𝜌 𝑅0
2𝜋𝜌²
1
𝐹𝑇𝜌 Circ𝑅 𝑟 = 2𝜋𝑅𝜌𝐽1 2𝜋𝑅𝜌 − 0
2𝜋𝜌²
𝑅
𝐹𝑇𝜌 Circ𝑅 𝑟 = 𝐽1 2𝜋𝑅𝜌
𝜌

36
 𝑅
1 𝑖𝑓 𝑥 2 + 𝑦² < 𝑅 𝐹𝑇𝜌 Circ𝑅 𝑟 = 𝐽 2𝜋𝑅𝜌
𝜌 1
Circ𝑅 𝑥, 𝑦 = 1
2 𝑖𝑓 𝑥 2 +𝑦²=𝑅 v
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

 1 2R 
u


f x, y  
T.F.u f x 
2R
F u,v 


 
y 
v
0

x 
37
 u