04 1 - frequency domain filtering fundamentals

4/28/2008
1
Frequency domain filtering
fundamentals
Spring 2008 ELEN 4304/5365 DIP 1
by Gleb V. Tcheslavski: gleb@ee.lamar.edu
http://ee.lamar.edu/gleb/dip/index.htm
Preliminaries
For a digital image f(x,y) the basic filtering equation is
{ }1
{ }1
( , ) ( , ) ( , )g x y H u v F u v−
= ℑ
Where H(u,v) and F(u,v) are DFTs of the image and of the filter.
Their product is defined by array (element-by-element)
multiplication. Specifications of H(u,v) are simplified considerably
when using functions symmetric about their centers This requires
when using functions symmetric about their centers. This requires
that F(u,v) is also centered, which is accomplished by multiplying
the input image by (-1)x+y before computing its transform.

4/28/2008
2
Simple filters
Considering the following image with centered (and scaled) DFT
Simple filters
One of the simplest filters would be H(u,v) having 0 at the center
and 1 elsewhere. Such filter will reject the DC (constant) term and
l thi l h dleave everything else unchanged.
Since the DC term
represents an average
intensity, setting it to zero
results in reduction of
average intensity to zero.
Therefore, the resulting
image appears darker.
Actually, aero average
implies negative intensities.

4/28/2008
3
Simple filters
Low frequencies in the transform are related to slowly varying
intensity components of image. High frequencies are caused by
sharp transitions in intensity (edges, noise).
Therefore, LPF will blur an image reducing details (and noise).
HPF will enhance sharp details but cause reduction in contrast
since it eliminates DC component. Adding a small constant to a
HPF d t ff t h i ti b t t li i ti
HPF does not affect sharpening properties but prevents elimination
of DC term and, thus, preserves contrast.
Simple filters
a=0.85

4/28/2008
4
Zero‐padding
Periodicity implied by DFT
leads to a “wraparound”
error: data from adjacent
periods aliases leading to
incorrect results of
convolution.
To avoid “wraparounds”,
convolved functions must
be zero padded
Spring 2008 ELEN 4304/5365 DIP 7Correct result
be zero‐padded.
More on zero‐padding
Sample image Blurred without padding Blurred with padding
p g p g p g
Blurred images are not uniform: top white edges are blurred but
side edges are not (without zero‐padding). The zero‐padding leads
to an expected result.

4/28/2008
5
More on zero‐padding
Image
Image
No padding With padding
Due to periodicity implicit while using DFT, a spatial filter passing
through the top edge of the image encompasses a part of the image
and also a part of the bottom of the periodic image right above it.
Padding helps to avoid this…
Zero‐padding and ringing
Considering an ideal LPF
and its spatial representation
obtained via multiplication
by (-1)u and IDFT, we
notice that zero-padding
will lead to discontinuities.
If we compete DFT of zero-
padded filter (to obtain its
f h t i ti )
Spring 2008 ELEN 4304/5365 DIP 10
frequency characteristic),
ringing (Gibbs?) will occur.
Ideal – infinite sinc –
truncation – ringing!

4/28/2008
6
Appearance of ringing
Results of filtering
with ILPF… Notice
ringing
We cannot use ideal filters AND zero-padding at the same time!
ringing
appearance.
Spring 2008 ELEN 4304/5365 DIP 11
We cannot use ideal filters AND zero padding at the same time!
One approach would be to zero-pad images and then create filters
(of the same zero-padded size) in frequency domain. However, this
will produce (insignificant though) wraparounds since the filter
won’t be zero-padded.
On phase angle
Since DFT of an image is a complex array, say
( , ) ( , ) ( , )F u v R u v jI u v= +
The filtered image would be
{ }1
( , ) ( , ) ( , ) ( , ) ( , )g x y H u v R u v jH u v I u v−
= ℑ +
We are interested in filters
affecting the real and
imaginary parts equally –
Spring 2008 ELEN 4304/5365 DIP 12
ag a y pa ts equa y
zero-phase-shift filters.
Even small changes in the
phase angle may have
drastic effect!
Phase multiplied by 0.5 Phase multiplied by 0.25

4/28/2008
7
Summary of steps
1. For the input image f(x,y) of size MxN, form the zero-padded
image fp(x,y) of size PxQ (typically P = 2M, Q = 2N), where
2 1 2 1P M Q N≥ ≥2 1; 2 1P M Q N≥ − ≥ −
2. Obtain fc(x,y) multiplying fp(x,y) by (-1)x+y to center its transform;
3. Compute the DFT of the fc(x,y);
4. Generate a real, symmetric filter function H(u,v) of size PxQ with
center at coordinates (P/2, Q/2);
5. Form the product G(u,v) = H(u,v) F(u,v) via array multiplication;
6 Ob i h d i
Spring 2008 ELEN 4304/5365 DIP 13
6. Obtain the processed image
7. Finally, extract g(x,y) – the MxN region from the top left quadrant
of gp(x,y).
{ }( )( )1
( , ) ( , ) 1
x y
pg x y real G u v
+−
⎡ ⎤= ℑ −⎣ ⎦
Summary of steps
The original image
– zero-padded –
scaled by (-1)x+y –scaled by ( 1)
DFT – DFT of LPF
– DFT of filtering
result – IDFT –
extracted filtered
image.
Spring 2008 ELEN 4304/5365 DIP 14

4/28/2008
8
Spatial vs. Frequency domains
The link between filtering in spatial and in frequency domains is the
convolution theorem...
( ) ( )h H⇔( , ) ( , )h x y H u v⇔
However, DFT implements a circular convolution.
Considering
next the image
and its spectrum
Spring 2008 ELEN 4304/5365 DIP 15
(magnitude of
its DFT) of size
600 x 600
Spatial vs. Frequency domains
We attempt to use a 3x3
Sobel mask…
Therefore, we need to
zero-pad both the image
and the mask to size of
602 x 602 (keeping the
mask at the center to
preserve symmetry) to
avoid wraparounds.
h fil f
Spring 2008 ELEN 4304/5365 DIP 16
The filter I n frequency
domain…
Results of frequency-
domain filtering and
spatial filtering are same.

04 1 - frequency domain filtering fundamentals

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04 1 - frequency domain filtering fundamentals