Image Enhancement in the

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Chapter 4 Image 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.
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Development of Fourier Series
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Any function f(x) which is periodic and is
integrable can be represented (expanded) as
This can be conveniently written in exponential
form as
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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
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This relation can be inverted to give
F(u) is known as the Fourier Transform of f(x).
It is analogous to cn . This can be easily
generalized to two-dimensions
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The Meaning of the Fourier Transform Let f(x)
sin(2puox). The representation of this function
in the Fourier Space, would be
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• function (also called impulse function)
• Property

Also area under the delta function 1
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Rectangular or box function f(x) 1 for a lt
x lt a 0 for x gt a
f(x)
-a 0 a
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Fourier Transform of rect function
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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.
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• 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.

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DISCRETE FOURIER TRANSFORM (DFT)
Inverse DFT
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Proof Substituting for F(u) in the definition of
inverse DFT, we get
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If x x, then the summand 1, otherwise it is
zero. How?
Phasor addition Here M6 let (x-x)1,
therefore Angle between successive phasors
2p/6 Therefore the summation over u equals M if
(x-x)0 and the other summation has only one
term. Hence the proof.
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Alternatively The sum for f(x) is equal to
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• ADDITIONAL PROPERTIES OF THE DFT
• PERIODICITY F(uM)F(u)
• This follows from the definition of the DFT.

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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.
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Fourier Spectrum of a white rectangle
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Centering the Fourier Transform can be done by
multiplying the Image by (-1)(xy) before taking
the Fourier Transform. We can show that this
centers the Transfrom. From the definition of
DFT
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• Significance of Fourier components
• The zero frequency component , F(0,0) is equal to
  the average gray level (intensity) of the image.
• Proof

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Fourier Spectrum of typical images
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• BASIC STEPS FOR FILTERING IN THE FREQUENCY DOMAIN
• Multiply the input image by (-1)(xy) to center
  the transform
• Compute F(u,v)
• Multiply F(u,v) by a filter function H(u,v)
  (component wise)
• Compute the inverse DFT
• Obtain the real part of the result in (4)
• Multiply the result by (-1)(xy)

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NOTCH FILTER Sets average to zero Display is
scaled by setting most negative to zero
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Low pass filtering Result is similar to averaging
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IDEAL LOW PASS FILTER
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EFFECT OF APPLYING AN IDEAL LOW PASS FILTER
RINGING
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What is the origin of ringing? To understand
ringing we need to understand an operation
called convolution.
Convolution, is usually defined as
This is similar to averaging
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If the domain of the functions is continuous,
then convolution is defined as
Let us now consider convolution in 1-d and look
at its Fourier transform
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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
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RINGING
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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.
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Gaussian Filters
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Why does the Gaussian filter not suffer from
ringing? To understand this, let us look at
its spatial representation.
Converting this integral to polar co-ordinates,
we have
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Butterworth Low pass filters
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Butterworth Low pass filters
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 are obtained with higher
values of n but at the cost of increase in
ringing effect.
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EFFECT OF FILTERING WITH BUTTERWORTH FILTERS
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