Image transforms

DIGITALIMAGEPROCESSING
Filtering in the
Frequency Domain
by
Dr. K. M. Bhurchandi

The DFT of One - Variable
• Obtaining the DFT from the Continuous Transform of a Sampled
Function
sampled, continuous,
band-limited function FT periodic function
extending from -∞ to ∞ extending from -∞ to ∞
• Equation of transform of sampled data in terms of sampled function is
given by:
• And we know 2

• Although fn is a discrete function, its Fourier is continuous and
infinitely periodic with period 1/ ΔT.
• Sampling one period is the basis for the DFT.
3

• To have M equally spaced samples of the transform taken over the
period μ = 0 to 1/ ΔT, consider following frequencies,
where, m = 0, 1, 2, ….., M-1
Substituting it in previous equation, we get DFT equation:
Conversely, IDFT can be obtained by:
where, n = 0, 1, 2, …, N-1
4

• In a more intuitive approach, notations used will be
x & y - image coordinates variables
u & v – frequency variables
Thus, the above equations become:
where, u = 0, 1, 2, …., M-1
where, x = 0, 1, 2, …, M-1
• Both the forward & inverse discrete transforms are infinitely
periodic with period M.
5

Extension to Functions of Two
Variables
• The 2-D Impulse and Its Sifting Property:
• The impulse of 2 continuous variables t & z is given by:
• &
• As in the 1D case, the 2D impulse exhibits the sifting property
under integration,
• More generally for an impulse located at coordinates (to, zo)
6

Variables
• For discrete variables x & y, the 2D discrete impulse is defined
as:
• Its sifting property is:
• For an impulse located at coordinates (xo, yo)
7

Variables
• Two-dimensional unit discrete impulse.
8

The 2-D Continuous Fourier
Transform Pair
• f(t, z) – continuous function of two variables, t & z.
• μ & γ – frequency variables define continuous frequency
domain.
9

• A 2-D function & its spectrum
10

2-D Sampling and the 2-D
Sampling Theorem
• 2-D impulse train is given by:
where, ΔT & ΔZ are separations between samples along t
& z axis.
11

2-D Sampling and the 2-D
Sampling Theorem
• Function f(t, z) is said to be band-limited if its Fourier
Transform is 0 outside a rectangle established by the intervals
[-μmax, μmax] and [-γmax, γmax].
• 2-Dimesional sampling theorem states that a continuous,
band limited function f(t, z) can be recovered with no error
from a set of its samples if the sampling intervals are
and
12

Aliasing in Images
• 2-D Fourier Transform of oversampled & under-sampled band
limited function
13

Aliasing in Images
• Two manifestations of aliasing in images:
• Spatial Aliasing & Temporal Aliasing
• Spatial Aliasing: It is due to under-sampling
• Temporal Aliasing: It is related to time intervals between
images in a sequence of images.
• E.g. wagon wheel effect (wheel appear to rotate backwards)
14

The 2-D DFT and Its Inverse
• 2-D Discrete Fourier Transform Pairs:
where, f(x, y) – digital image of size M x N.
u – 0, 1, 2, ….., M-1
v – 0, 1, 2, ….., N-1
where, F(u, v) – transform of f(x, y).
x – 0, 1, 2, ….., M-1
y – 0, 1, 2, …., N-1
16

Properties of 2-D DFT
1) Relationships between Spatial & Frequency Intervals
• If a continuous function f(t, z) is sampled to form a digital
image, f(x, y) consisting of M x N samples in t & z resp.
• If ΔT & ΔZ denote the separations between samples, then,
separations between the corresponding discrete, frequency
domain variables are given by:
Δ𝑢 =
1
𝑀 Δ𝑇
& Δ𝑣 =
1
𝑁 Δ𝑍
17

2) Translation and Rotation
• Fourier transform pair satisfies translation properties
Multiplying f(x,y) by exponential shown shifts the origin of DFT
to (u0,v0) and conversely, multiplying F(u,v) by negative of that
exponential shifts the origin of f(x,y) to (x0,y0).
18

2) Translation and Rotation
Using Polar coordinates
Results in transform pair
Rotating f(x,y) with rotates F(u,v) by same angle and viceversa.
19

3) Periodicity
2D Fourier transform and its inverse are infinitely periodic in u
and v directions
and
where k1 and k2 are integers
20

3) Periodicity
21

Problems on DTFT
1) Let f(x, y) =
1 1 1
1 1 1
1 1 1
is image matrix. Then DFT(F) = ?
Sol:- y 0 1 2
f(x, y) = x
0
1
2
1 1 1
1 1 1
1 1 1
=>
𝑓 0, 𝑦 = [1 1 1]
𝑓 1, 𝑦 = [1 1 1]
𝑓 2, 𝑦 = [1 1 1]
Generally, F(v) = 𝑓(𝑦)𝑒−𝑗2𝜋𝑣𝑦/𝑁
𝑁−1
𝑦=0
Thus, F1(0, v) = 𝑓(0, 𝑦)𝑒−𝑗2𝜋𝑣𝑦/𝑁
𝑁−1
𝑦=0 22

F1(0, v) = 𝑓(0, 𝑦)𝑒−𝑗2𝜋𝑣𝑦/𝑁
𝑁−1
𝑦=0
F1(0, v) = 𝑓(0, 0)𝑒−𝑗2𝜋𝑣0/3 + 𝑓(0, 1)𝑒−𝑗2𝜋𝑣1/3 + 𝑓(0, 2)𝑒−𝑗2𝜋𝑣2/3
For, v = 0: F1(0, 0) = 𝑓(0, 0)𝑒−𝑗2𝜋0∗0/3 + 𝑓(0, 1)𝑒−𝑗2𝜋0∗1/3 + 𝑓(0, 2)𝑒−𝑗2𝜋0∗2/3
= 1 x 1 + 1 x 1 + 1 x 1 = 3
For, v = 1: F1(0, 1) = 𝑓(0, 0)𝑒−𝑗2𝜋1∗0/3
+ 𝑓(0, 1)𝑒−𝑗2𝜋1∗1/3
+ 𝑓(0, 2)𝑒−𝑗2𝜋1∗2/3
= 1 x 1 + 1 x (-0.5000 - 0.8660i) + 1 x (-0.5000 + 0.8660i)
= 0
For, v = 2: F1(0, 2) = 𝑓(0, 0)𝑒−𝑗2𝜋2∗0/3 + 𝑓(0, 1)𝑒−𝑗2𝜋2∗1/3 + 𝑓(0, 2)𝑒−𝑗2𝜋2∗2/3
= 1 x 1 + 1 x (-0.5000 + 0.8660i) + 1 x (-0.5000 - 0.8660i)
= 0
So, F1(0, v) = [3 0 0];
Similarly, F1(1, v) = [3 0 0]; & F1(2, v) = [3 0 0];
Thus, F1 =
3 0 0
3 0 0
3 0 0
Now, applying transformation column-wise.
23

Generally, F(u) = 𝑓(𝑥)𝑒−𝑗2𝜋𝑢𝑥/𝑀
𝑀−1
𝑥=0
Thus, F(u, 0) = 𝐹1(𝑥, 0)𝑒−𝑗2𝜋𝑢𝑥/𝑀
𝑀−1
𝑥=0
F(u, 0) = F1 (0, 0)𝑒−𝑗2𝜋𝑢0/3 + F1(1, 0)𝑒−𝑗2𝜋𝑢1/3 +F1(2, 0)𝑒−𝑗2𝜋𝑢2/3
For, u = 0: F(0, 0) = F1(0, 0)𝑒−𝑗2𝜋0∗0/3
+F1(1, 0)𝑒−𝑗2𝜋0∗1/3
+F1(2, 0)𝑒−𝑗2𝜋0∗2/3
= 3 x 1 + 3 x 1 + 3 x 1
= 9
For, u = 1: F(1, 0) = F1(0, 0)𝑒−𝑗2𝜋1∗0/3
+F1(1, 0)𝑒−𝑗2𝜋1∗1/3
+F1(2, 0)𝑒−𝑗2𝜋1∗2/3
= 0 x 1 + 0 x (-0.5000 - 0.8660i) + 0 x (-0.5000 + 0.8660i)
= 0
For, u = 2: F(2, 0) = F1(0, 0)𝑒−𝑗2𝜋2∗0/3 +F1(1, 0)𝑒−𝑗2𝜋2∗1/3 +F1(2, 0)𝑒−𝑗2𝜋2∗2/3
= 0 x 1 + 0 x (-0.5000 + 0.8660i) + 0 x (-0.5000 - 0.8660i)
= 0
So, F(u, 0) = [9 0 0]; Similarly, F(u, 1) = [0 0 0]; & F(u, 2) = [0 0 0];
Thus, F =
9 0 0
0 0 0
0 0 0
24

Numerical on DFT
2) Let f(x, y) =
1 2 3
4 5 6
7 8 9
Sol:- y 0 1 2
f(x, y) = x
0
1
2
1 2 3
4 5 6
7 8 9
=>
𝑓 0, 𝑦 = [1 2 3]
𝑓 1, 𝑦 = [4 5 6]
𝑓 2, 𝑦 = [7 8 9]
𝑁−1
𝑦=0
Thus, F1(0, v) = 𝑓(0, 𝑦)𝑒−𝑗2𝜋𝑣𝑦/𝑁
𝑁−1
𝑦=0 25

F1(0, v) = 𝑓(0, 𝑦)𝑒−𝑗2𝜋𝑣𝑦/𝑁
𝑁−1
𝑦=0
F1(0, v) = 𝑓(0, 0)𝑒−𝑗2𝜋𝑣0/3
+ 𝑓(0, 1)𝑒−𝑗2𝜋𝑣1/3
+ 𝑓(0, 2)𝑒−𝑗2𝜋𝑣2/3
For, v = 0: F1(0, 0) = 𝑓(0, 0)𝑒−𝑗2𝜋0∗0/3
+ 𝑓(0, 1)𝑒−𝑗2𝜋0∗1/3
+ 𝑓(0, 2)𝑒−𝑗2𝜋0∗2/3
= 1 x 1 + 2 x 1 + 3 x 1 = 6
For, v = 1: F1(0, 1) = 𝑓(0, 0)𝑒−𝑗2𝜋1∗0/3
+ 𝑓(0, 1)𝑒−𝑗2𝜋1∗1/3
+ 𝑓(0, 2)𝑒−𝑗2𝜋1∗2/3
= 1 x 1 + 2 x (-0.5000 - 0.8660i) + 3 x (-0.5000 + 0.8660i)
= -1.5000 + 0.8660i
For, v = 2: F1(0, 2) = 𝑓(0, 0)𝑒−𝑗2𝜋2∗0/3
+ 𝑓(0, 1)𝑒−𝑗2𝜋2∗1/3
+ 𝑓(0, 2)𝑒−𝑗2𝜋2∗2/3
= 1 x 1 + 2 x (-0.5000 + 0.8660i) + 3 x (-0.5000 - 0.8660i)
= -1.5000 - 0.8660i
So, F1(0, v) = [6 -1.5000 + 0.8660i -1.5000 - 0.8660i];
Similarly, F1(1, v) = [15 -1.5000 + 0.8660i -1.5000 - 0.8660i];
& F1(2, v) = [24 -1.5000 + 0.8660i -1.5000 - 0.8660i];
Thus, F1 =
6 −1.5000 + 0.8660i −1.5000 − 0.8660i
15 −1.5000 + 0.8660i −1.5000 − 0.8660i
24 −1.5000 + 0.8660i −1.5000 − 0.8660i
Now, applying transformation column-wise. 26

Generally, F(u) = 𝑓(𝑥)𝑒−𝑗2𝜋𝑢𝑥/𝑀
𝑀−1
𝑥=0
Thus, F(u, 0) = 𝐹1(𝑥, 0)𝑒−𝑗2𝜋𝑢𝑥/𝑀
𝑀−1
𝑥=0
F(u, 0) = F1 (0, 0)𝑒−𝑗2𝜋𝑢0/3 + F1(1, 0)𝑒−𝑗2𝜋𝑢1/3 +F1(2, 0)𝑒−𝑗2𝜋𝑢2/3
= 6 x 1 + 15 x 1 + 24 x 1 = 45
For, u = 1: F(0, 1) = F1(0, 1)𝑒−𝑗2𝜋1∗0/3
+F1(1, 1)𝑒−𝑗2𝜋1∗1/3
+F1(2, 1)𝑒−𝑗2𝜋1∗2/3
= 6 x 1 + 15 x (-0.5000 - 0.8660i) + 24 x (-0.5000 + 0.8660i)
= -13.5000 + 7.7940i
= 6 x 1 + 15 x (-0.5000 + 0.8660i) + 24 x (-0.5000 - 0.8660i)
= -13.5000 - 7.7940i
Thus, F =
45 −4.5 + 2.59𝑖 −4.5 − 2.59𝑖
−13.5000 + 7.7940i 0 0
−13.5000 − 7.7940i 0 0
27

3) Let f(x, y) =
0 1 0
1 0 1
0 1 0
Sol:- y −1 0 1
f(x, y) = x
−1
0
1
0 1 0
1 0 1
0 1 0
𝑁/2
𝑦=−𝑁/2
Thus, F1(-1, v) = 𝑓(−1, 𝑦)𝑒−𝑗2𝜋𝑣𝑦/𝑁
𝑁/2
𝑦=−𝑁/2 28

F1(-1, v) = 𝑓(−1, −1)𝑒−𝑗2𝜋𝑣(−1)/3 + 𝑓(−1, 0)𝑒−𝑗2𝜋𝑣0/3 + 𝑓 −1, 1 𝑒−
𝑗2𝜋𝑣1
3
&
F(u, -1) = F1 (−1, −1)𝑒−𝑗2𝜋𝑢(−1)/3
+ F1(0, −1)𝑒−𝑗2𝜋𝑢0/3
+F1(1, −1)𝑒−𝑗2𝜋𝑢1/3
Then,
F =
4 −0.5 − 0.866𝑖 −0.5 + 0.866𝑖
−0.5 − 0.866𝑖 1 − 1.73𝑖 −2
−0.5 + 0.866𝑖 −2 1 + 1.73𝑖
𝑢𝑠𝑖𝑛𝑔 𝑟𝑒𝑓. (−1, −1)
F =
−2 1 −2
1 4 1
−2 1 −2
𝑢𝑠𝑖𝑛𝑔 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 (0, 0)
29

Symmetry Properties of 2D-DFT
30

Spectrumis insensitive to translation.But
rotatesby the same angle as the rotated
image
35

Translation affects the phase
36

Reconstruction using either
spectrum or phase
37

Frequency Domain Filtering
Fundamentals
• Given a digital image f(x,y) of size MxN, the basic filtering
equation is
39
Result of Filtering an image
Input image Frequency
Spectrum

More Results on Filtering
40
Low Pass Filter
High Pass Filter
Top row: Frequency domain Filters, Bottom Row: Corresponding Filtered Images

Gaussian Low pass Filtering
41
Gaussian Low
Pass Filter
without padding
Gaussian Low
Pass Filter with
padding

Summary of steps for filtering in
Frequency domain
1) Given an input image f(x, y) of size M x N, obtain the padding parameters
P & Q as 2M & 2N respectively.
2) Form a padded image, fp(x, y), of size P x Q be appending the necessary
number of zeros to f(x, y).
3) Multiply fp(x, y) by (-1)x+y to center its transform.
4) Compute the DFT, F(u, v), of the image from step 3.
5) Generate a real, symmetric filter function, H(u, v), of size P x Q with
center at coordinates (P/2, Q/2). Form the product G(u, v) = H(u, v)F(u, v)
using array multiplication; that is, G(i, k) = H(i, k)F(i, k).
6) Obtain the processed image: gp(x, y) = {real[Ṫ-1[G(u, v)]]}(-1)x+y where
the real part is selected in order to ignore parasitic complex components
resulting from computational inaccuracies, and the subscript p indicates
that we are dealing with padded arrays.
7) Obtain the final processed result, g(x, y), by extracting the M x N region
from the top, left quadrant of gp(x, y).
42

• Steps for Filtering in the Frequency Domain
43

Image smoothing using Frequency
domain filters
• Ideal Lowpass Filters:
• A 2-D lowpass filter that passes without attenuation all
frequencies within a circle of radius D0 from the origin and
“cut off” all frequencies outside this circle is called an ideal
lowpass filter (ILPF).
H 𝑢, 𝑣 =
1, 𝑖𝑓 𝐷(𝑢, 𝑣) ≤ 𝐷0
0, 𝑖𝑓 𝐷(𝑢, 𝑣) > 𝐷0
Where, D0 - a positive constant
D(u, v) – distance between (u, v) in the frequency domain
and the center of the frequency rectangle; i.e.
D(u, v) = [(u – P/2)2 + (v – Q/2)2]1/2
Where P & Q are 2M & 2N resp. 44

domain filters Perspective plot
of an Ideal LPF
Filter as an image
Filter radial cross section
45

• Ideal Low Pass Filtering
46

47
R=30
R=10
R=160
R=60
R=460
Input Image
Results of Ideal LPF

The blurring and “ringing”
properties of Ideal LPFs
48
Representation in the spatial
domain od an ILPF of radius 5
and size 1000x1000.
Intensity profile of a horizontal
line passing through the
center of the image.

domain filters
• Butterworth Lowpass Filters:
• The transfer function of a Butterworth lowpass filter (BLPF) of
order n, and with cutoff frequency at a distance D0 from the
origin, is defined as,
H 𝑢, 𝑣 =
1
1+
𝐷 𝑢,𝑣
𝐷0
2𝑛
where, D(u, v) = [(u – P/2)2 + (v – Q/2)2]1/2
49

domain filters
from order 1 - 4
Perspective
plot of BLPF
Radial cross section of filter
Filter as an image
50

domain filters
• Spatial representation of BLPF of order 1, 2, 5 & 20 &
corresponding intensity profiles through the center of the
filter.
• Ringing increases as a function of filter order.
51

52
Original Image Results of filtering
using BLPF of order
2 with various cut
off frequencies

domain filters
• Gaussian Lowpass Filters:
• Gaussian lowpass filters(GLPF) in two dimensions is given by
𝐻 𝑢, 𝑣 = 𝑒−𝐷2
(𝑢,𝑣)/2σ2
where, σ – is measure of spread about the center.
• Let σ = D0, thus the filter can be expressed as:
𝐻 𝑢, 𝑣 = 𝑒−𝐷2
(𝑢,𝑣)/2𝐷0
2
where D0 is the cutoff frequency 53

domain filters
Perspective
Plot of GLPF
Filter as an image Filter radial cross section
for diff D0
54

Gaussian Low pass Filters
55
Input Image Results of
filtering using
GLPF with
various cut off
frequencies

Applications of LPF
• Character Recognition
57
Sample text of low resolution
Note: broken characters
Result of filtering with GLPF
Note: broken character
segments are joined

Contd..
• Printing and publishing industry: “cosmetic” processing
58

Image sharpening using Frequency
domain filter
• A highpass filter is obtained from a given lowpass filter using
the equation
HHP(u, v) = 1 – HLP(u, v)
Where, HLP(u, v) – transfer function of the LPF
• Ideal Highpass Filter
• A 2-D highpass filter is given by
H 𝑢, 𝑣 =
0, 𝑖𝑓 𝐷(𝑢, 𝑣) ≤ 𝐷0
1, 𝑖𝑓 𝐷(𝑢, 𝑣) > 𝐷0
59

domain filter
IHPF
BHPF
GHPF
60

Spatial Representation of High
Pass Filter
61

domain filter
• Butterworth Highpass Filter
• A 2-D Butterworth highpass filter (BHPF) of order n and cutoff
frequency D0 is defined as
H 𝑢, 𝑣 =
1
1+
𝐷0
𝐷 𝑢,𝑣
2𝑛
• Gaussian Highpass Filter
• The transfer function of the Gaussian highpass filter (GHPF)
with cutoff frequency locus of at a distance D0 from the center
of the frequency rectangle is given by
𝐻 𝑢, 𝑣 = 1 − 𝑒−𝐷2
(𝑢,𝑣)/2𝐷0
2
63

Results of BHPF of order 2
• Results are much smoother than ideal HPF
64

Laplacian in the frequency domain
• The Laplacian can be implemented in the frequency domain
using the filter
H(u, v) = -4π2(u2 + v2)
Or with respect to the center of the frequency rectangle, using
the filter
H(u, v) = -4π2D2(u, v)
Where, D(u, v) is the distance function.
65

Laplacian in the frequency domain
The Laplacian image is obtained as:
𝛻2𝑓 𝑥, 𝑦 = T
− 1{H u, v F(u, v)}
Enhancement is achieved using the equation
𝑔 𝑥, 𝑦 = 𝑓 𝑥, 𝑦 + 𝑐𝛻2𝑓(𝑥, 𝑦)
• Here, c = -1, because, H(u, v) is negative
• In frequency domain the above equation is written as:
𝑔 𝑥, 𝑦 = T
− 1{[1 + 4π2D2 u, v ]F(u, v)}
66

Homomorphic Filtering
• According to Illumination-Reflectance model:
• An image f(x, y) can be expressed as the product of its illumination,
i(x, y), and reflectance, r(x, y), components:
f(x, y) = i(x, y)r(x, y)
• Since the Fourier Transform of a product is not the product of he
transforms:
• So, F[f(x, y)] ≠ F[i(x, y)] F[r(x, y)]
• However, if,
z(x, y) = ln f(x, y)
= ln i(x, y) + ln r(x, y)
Then,
F{z(x, y)} = F{ln f(x, y)}
= F{ln i(x, y)} + F{ln r(x, y)}
67

or
Z(u, v) = Fi(u, v) + Fr(u, v)
Where, Fi(u, v) and Fr(u, v) are the Fourier transforms of ln i(x, y) & ln
r(x, y) resp.
We can filter Z(u, v) using a filter H(u, v)
S(u, v) = H(u, v) Fi(u, v) + H(u, v) Fr(u, v)
The filtered image in spatial domain is
s(x, y) = T-1{S(u, v)}
= T-1{H(u, v) Fi(u, v)} + T-1{H(u, v) Fr(u, v)}
= i’(x, y) + r’(x, y)
where, i’(x, y) = T-1{H(u, v) Fi(u, v)}
r’(x, y) = T-1{H(u, v) Fr(u, v)}
68

• In order to get the original image, we reverse the process,
g(x, y) = es(x, y)
= ei’(x, y)er’(x, y)
= i0(x, y)r0(x, y)
where, i0(x, y) = ei’(x, y) & r0(x, y) = er’(x, y) are the illumination and
reflectance components of the output (processed) image.
69

Summary of steps in homomorphic
filtering
• The illumination component of an image is characterized by
slow spatial variations, while the reflectance components tend
to vary abruptly.
• The low frequencies of DFT of Logarithm of an image are
associated with illumination and high frequencies with
reluctance.
70

• Shows cross section of such a homomorphic filter H(u, v) that
affects low and high frequency components differently.
71

• If parameters γL & γH are chosen such that γL < 1 & γH > 1, the
filter attenuates low frequencies and amplifies high
frequencies.
• It results in simultaneous dynamic range compression and
contrast enhancement. (Highpass filter)
• Slightly modified Gaussian highpass filter yields the function
𝐻 𝑢, 𝑣 = γH − γL [1 − 𝑒−𝑐[𝐷2
(𝑢,𝑣)/𝐷0
2]
] + γL
where, controls the sharpness of the slope of the
function as it transitions between γL & γH
72

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