fourier transforms for dip

8/22/2019 Fourier Transforms for DIP

1/40

Chapter 4Image Enhancement in the

Frequency Domain

Fourier s idea :Any periodic function can be represented

as a sum of sine and cosine functions, with appropriate

amplitudes and phases.

Since we represent an image by a two-dimensional

function f(x,y), we can represent it as a sum of sines and

cosines, if it is periodic.

If the function is not periodic, we have to use a

generalization of the Fourier series representation, known

as the Fourier Transform.


2/40

Development of Fourier Series


3/40

+

+=

=

=

xL

nbxL

naaxf nn

n

n

2sin2cos2

)(11

0

Any function f(x) which is periodic and is integrable can be

represented (expanded) as:

This can be conveniently written in exponential form as:

=

= n n Ljnx

cxf

2

exp)(


4/40

The Cns are known as the Fourier coefficients.

The original function can be reconstructed from a

knowledge of all the Fourier coefficients.

If the function is non-periodic, i.e., L , n/L can

take continuous values. We set n/L = u (called spatial

frequency).

Therefore:

= dujuxuFxf )2exp()()(


5/40

This relation can be inverted to give:

= dxjuxxfuF )2exp()()(

F(u) is known as the Fourier Transform of f(x). It is

analogous to cn .

This can be easily generalized to two-dimensions:

+= dxdyvyuxjyxfvuF ))(2exp(),(),(


6/40

The Meaning of the Fourier Transform:

Let f(x) = sin(2uo

x). The representation of this function

in the Fourier Space, would be:

= dxjuxxuuF o )2exp()sin()(


7/40

function (also called impulse function):

Property :

0for x0

0for x)(

====x

Also area under the delta function = 1

=1)( dxx


8/40

= )0()()( fdxxxf

Rectangular or box function:

f(x) = 1 for a < x < a

= 0 for |x| > a

f(x)

-a 0 a


9/40

Fourier Transform of rect function:

)2(sin2)2sin(2

)]2exp()2[exp()2exp( uacauua

jujuajuadxjux

a

a

===


10/40


11/40

Does this remind you of something?

Clue: Optical phenomena which you studied in Physics-

I/MT-I

Ans: The diffraction pattern of a single slit!

In fact, this is true in general. Diffraction patterns are the

FT of the aperture.


12/40

Points to ponder:

Reciprocal nature of F(u) and f(x). Here the width of the

central peak is inversely proportional to a (the width of

f(x)). This is a general property of the Fourier transform.

F(u) is complex (in general).

Sharp changes in the image (f(x)) give rise to peaks at

large values of u.

The phase part of F(u) gives us information about where

these sharp changes occur.


13/40

DISCRETE FOURIER TRANSFORM (DFT)

1-.M0,1,2,....u

2exp)(

1)(

1

0

=

=

=

for

M

juxxf

MuF

M

x

1-.M0,1,2,....x

2exp)()(1

0

=

=

=

for

MjuxuFxf

M

u

Inverse DFT


14/40

Proof:

Substituting for F(u) in the definition of inverse DFT, we

get:

=

=

=

1

0

1

0

'2exp

2exp)(

1)(

M

u

M

x M

jux

M

juxxf

Mxf

=

=

=

1

0

1

0

)'(2exp)(

1)(

M

x

M

u M

xxjuxf

Mxf


15/40

If x = x, then the summand = 1, otherwise it is zero.

How?

Phasor addition:

Here M=6; let (x-x)=1,

therefore

Angle between successive

phasors = 2/6

Therefore the summation over

u equals M if (x-x)=0 and the

other summation has only one

term. Hence the proof.


16/40

Alternatively

The sum for f(x) is equal to:

0)'(2exp1

))'(2exp(1)'(2

exp

1

0 =

=

=

M

xxju

xx

M

xxjuM

u


17/40

ADDITIONAL PROPERTIES OF THE DFT:

PERIODICITY: F(u+M)=F(u)

This follows from the definition of the DFT.

=

=

+

=+

=

=

=

M

juxxf

Mjux

M

juxxf

M

M

xMuj

xfMMuF

M

x

M

x

M

x

2exp)(

1)2exp(

2exp)(

1

)(2

exp)(

1

)(

1

0

1

0

1

0


18/40

Due to the periodic nature of the DFT, the Fourier

spectrum may appear scrambled if we apply it directly

on an image.

Consider the Fourier spectrum of a 2-D rectangular (or

box) function.

Direct application of the DFT would give rise to a

scrambled Fourier Transform.

F i S t f hit t l


19/40

Fourier Spectrum of a white rectangle


20/40

Centering the Fourier Transform can be done by

multiplying the Image by (-1)(x+y) before taking the

Fourier Transform.

We can show that this centers the Transfrom.

From the definition of DFT:

22exp)()1(1),(1

0

)(=

+

=

M

x

yx

Njvy

Mjuxxf

MvuF

=

+=

1

0

22exp)()exp(

1 M

x N

jvy

M

juxxfjyjx

M


21/40

Significance of Fourier components:

The zero frequency component , F(0,0) is equal to the

average gray level (intensity) of the image.

Proof:

=

=

= Njvy

Mjuxyxf

MNvuF

M

x

N

y

2exp2exp),(1),(1

0

1

0

y)f(x,ofAverage),(1

)0,0(1

0

1

0

==

=

=

M

x

N

y

yxfMN

F


22/40

Fourier Spectrum of typical images


23/40

BASIC STEPS FOR FILTERING IN THE FREQUENCY

DOMAIN

1. Multiply the input image by (-1)(x+y)

to center thetransform

2. Compute F(u,v)

3. Multiply F(u,v) by a filter function H(u,v) (componentwise)

4. Compute the inverse DFT

5. Obtain the real part of the result in (4)

6. Multiply the result by (-1)(x+y)


24/40

NOTCH FILTER : Sets average to zero:


25/40

NOTCH FILTER : Sets average to zero:Display is scaled by setting most

negative to zero


26/40

Low pass filtering: Result is similar to averaging


27/40

IDEAL LOW PASS FILTER


28/40

EFFECT OF APPLYING AN IDEAL LOW PASS FILTER:


29/40

EFFECT OF APPLYING AN IDEAL LOW PASS FILTER:RINGING

What is the origin of ringing?


30/40

What is the origin of ringing?

To understand ringing we need to understand an

operation called convolution.

===

=++=mn

i

ii

b

bt

a

as

zwtysxftswyxg1

),(),(),(

Convolution, is usually defined as:

=

= ==

1

0

1

0),(),(

1

),(*),(),(

N

n

M

mnymxhnmfMNyxhyxfyxg

This is similar to averaging


31/40

If the domain of the functions is continuous, then

convolution is defined as:

= ddyxhfyxg ),(),(),(

Let us now consider convolution in 1-d and look at itsFourier transform:

= dxjuxdxhfuG )2exp()()()(


32/40

)()(

)2exp())(2exp()()()(

uHuF

dxjuxjudxhfuG

=

=

Therefore, convolution in the spatial domain is equivalent

to multiplication in the Fourier Domain and vice-versa.

Let us consider a simple point image (impulse function).

The Fourier transform of this would be a constant. We

multiply this by the filter function. Hence in the real

domain we have to convolve the impulse function with the

spatial representation of the filter function.

The filter function in the spatial domain will have

circular symmetry. It has a central peak and it reduces in

height as we move away from the origin


33/40

RINGING


34/40

In order to avoid ringing Ideal Low pass filter is not used.

Other filters are available: Gaussian and Butterworth

Filters.For ringing to be absent, the filter function should have a

spatial representation which varies monotnically. This is

true for Gaussian function and the I order Butterworth

filter.

= 22

2

),(

exp),(

vuD

vuH


35/40

Gaussian Filters


36/40

Why does the Gaussian filter not suffer from ringing?

To understand this, let us look at its spatial representation.

++= dudvvyuxjvuvuH ))(2exp()2/)(exp(),( 222

( )( )dudvyxjyv

jx

u 22222222

22exp22exp22exp

+

Converting this integral to polar co-ordinates, we have

( )( ) +

0

2

0

2222222 22exp)exp( ddyx


37/40

Butterworth Low pass filters


38/40

Butterworth Low pass filters:

nDvuDvuH

2

0 ]/),([11),(

+=

Here D(u,v) is the distance from the origin , D0 is the

distance of the cut-off frequency from the origin and n is

the order of the Butterworth filter. Sharper cutoffs areobtained with higher values of n but at the cost of increase

in ringing effect.


39/40

EFFECT OF FILTERING WITH BUTTERWORTH FILTERS


40/40

fourier transforms for dip

Documents