image enhancement in the frequency domain (chapter...

Digital Image Processing

Image Enhancement in the

Frequency Domain

(Chapter 5)

Any function that periodically repeats itself can be expressedas the sum of sines and/or cosines of different frequencies, each multiplied by a different coefficients.This sum is called a Fourierseries.

Fourier Series

00

0

00

0

00

0

0

0

0

0

0

0

1

000

sin)(2

cos)(2

)(1

sincos)(

Tt

t

n

Tt

t

n

Tt

t

n

nn

dttnwtgT

b

dttnwtgT

a

dttgT

a

tnwbtnwaatg

Fourier Series

A function that is not periodic but the area under its curve is finite can be expressed as the integral of sines and/or cosines multiplied by a weighing function. The formulation in this case is Fourier transform.

Fourier Transform

dueuFxf

dxexfuF

uxj

uxj

2

2

)()(

)()(

Continuous One-Dimensional Fourier Transform and Its Inverse

Where 1j

• (u) is the frequency variable.

• F(u) is composed of an infinite sum of sine and cosine terms and…

• Each value of u determines the frequency of its corresponding sine-cosine pair.

2

10

2

11

)(

x

xx

2 sinc )(

uuF

0.5- 0.5

x

2

4

u

Continuous One-Dimensional Fourier Transform and Its Inverse

ExampleFind the Fourier transform of a gate function (t) defined by

Discrete One-Dimensional Fourier Transform and Its Inverse

• A continuous function f(x) is discretized into a sequence:

)}]1[(),...,2(),(),({ 0000 xNxfxxfxxfxf

by taking N or M samples x units apart.


• Where x assumes the discrete values (0,1,2,3,…,M-1) then

)()( 0 xxxfxf

• The sequence {f(0),f(1),f(2),…f(M-1)} denotes any

M uniformly spaced samples from a corresponding

continuous function.

1

0

1

0

2

2sin2cos)(1

)(

)(1

)(

M

x

M

x

M

xuj

M

xuj

M

xuxf

MuF

exfM

uF

1

0

2

)()(M

u

xM

uj

euFxf

u =[0,1,2, …, M-1]

x =[0,1,2, …, M-1]



• The values u = 0, 1, 2, …, M-1 correspond to

samples of the continuous transform at values 0, u, 2u, …, (M-1)u.

i.e. F(u) represents F(uu), where:

u 1

Mx


• The Fourier transform of a real function is generally complex and we use polar coordinates:

• Its phase angle

)(

)(tan)(

)]()([)(

)()(

)()()(

1

2/122

)(

uR

uIu

uIuRuF

euFuF

ujIuRuF

uj

• The square of the spectrum

)()()()( 222uIuRuFuP

is referred to as the Power Spectrum of f(x)

(spectral density).


• Fourier spectrum: 2/122 ),(),(),( vuIvuRvuF

• Phase:

),(

),(tan),( 1

vuR

vuIvu

• Power spectrum: ),(),(),(),( 222vuIvuRvuFvuP

Discrete 2-Dimensional Fourier Transform

1 second

Fmax = 100 HzWhat is the sampling rate (Nyquist)?What is the time resolution?What is the frequency resolution?What if we take samples for two seconds with the Nyquist sampling rate?

Time and Frequency Resolution and Sampling

2/122

1

0

1

0

)(2

1

0

1

0

)(2

),(),(),(

),(),(

),(1

),(

vuIvuRvuF

evuFyxf

eyxfMN

vuF

M

u

N

v

yN

vx

M

uj

M

x

N

y

N

yv

M

xuj

Fourier Spectrum

Discrete Two-Dimensional Fourier Transform and Its Inverse

1

0

1

0

),(1

)0,0(M

x

N

y

yxfMN

F

F(0,0) is the average intensity of an image


Use Matlab to generate the above figures


),(),(

)()(

then

)()(

If

00

)(2

0

00

0

vvuuFeyxf

Getg

Gtg

N

yv

M

xuj

tj

Frequency Shifting Property of the Fourier Transform

Frequency Shifting Property of the Fourier Transform

function Normalized_DFT = Img_DFT(img)

img=double(img); % So mathematical operations can be conducted on

% the image pixels.

[R,C]=size(img);

for r = 1:R % To phase shift the image so the DFT will be

for c=1:C % centered on the display monitor

phased_img(r,c)=(img(r,c))*(-1)^(r+c);

end

end

fourier_img = fft2(phased_img); %Discrete Fourier Transform

mag_fourier_img = abs(fourier_img ); % Magnitude of DFT

Log_mag_fourier_img = log10(mag_fourier_img +1);

Max = max(max(Log_mag_fourier_img ));

Normalized_DFT = (Log_ mag_fourier_img )*(255/Max);

imshow(uint8(Normalized_DFT))

Basic Filtering in the Frequency Domain using Matlab

1. Multiply the input image by (-1)x+y to center the transform2. Compute F(u,v), the DFT of the image from (1)3. Multiply F(u,v) by a filter function H(u,v)4. Compute the inverse DFT of the result in (3)5. Obtain the real part of the result in (4)6. Multiply the result in (5) by (-1)x+y

Basic Filtering in the Frequency Domain

An image and its Frequency information

1

)2/,2/(),( if 0),(

NMvuvuH

otherwise

Notch Filter

Filtering out the DC Frequency Component

Low Pass Filter attenuate high frequencies while “passing” low frequencies.

High Pass Filterattenuate low frequencies while “passing” high frequencies.

Low-pass and High-pass Filters

High-pass Filtering

Low-pass and High-pass Filters

2/122

0

0

)2/()2/(),(

),( if 0

),( if 1),(

NvMuvuD

DvuD

DvuDvuH

Smoothing Frequency Domain, Ideal Low-pass Filters

u v

T

M

u

N

v

T

PvuF

vuFP

/),(100

),(1

0

1

0

2

Total Power

The remained percentagepower after filtration

Smoothing Frequency Domain, Ideal Low-pass Filters

For circle radius fc =5

Enclose image power = 92%

fc =15

= 94.6%

fc =30

= 96.4%

fc =80

= 98%fc =230

= 99.5%

Smoothing Frequency Domain,Ideal Low-pass Filters

Cause of Ringing

nDvuD

vuH2

0/),(1

1),(

Smoothing Frequency Domain, Butterworth Low-pass Filters

Smoothing Frequency Domain, Butterworth Low-pass Filters

Radii= 5

Radii= 15Radii= 30

Radii= 80 Radii= 230

Butterworth Low-pass Filter: n=2

Smoothing Frequency Domain,Butterworth Low-pass Filters

20

2 2/),(),(

DvuDevuH

Smoothing Frequency Domain, Gaussian Low-pass Filters

Radii= 5

Radii= 15Radii= 30

Radii= 80 Radii= 230

Gaussian Low-pass


Hhp(u,v) = 1 - Hlp(u,v)

Ideal HPF

Butterworth HPF

Gaussian HPF

Sharpening Frequency Domain Filters

Sharpening Frequency Domain Filters

),( if 1

),( if 0),(

0

0

DvuD

DvuDvuH

Sharpening Frequency Domain, Ideal High-pass Filters

nvuDD

vuH2

0 ),(/1

1),(

Sharpening Frequency Domain,Butterworth High-pass Filters

20

2 2/),(1),(

DvuDevuH

Sharpening Frequency Domain,Gaussian High-pass Filters

Homomorphic Filtering

g t k t h tM

k t m h mm

M

( ) ( ) * ( ) ( ) ( )

1

0

1

h(t) g(t)k(t)g(t) = k(t)*h(t)

G(f) = K(f)H(f)

-1

1

t

h(t)

1

2

3

t

k(t)

1 2 3

* is a convolution operator and not multiplication

Convolution

gM

k m h mm

M

( ) ( ) ( )01

00

1

m

k(-m)

-1

1

m

h(m)

g(t)

Convolution

-1

1

m

h(m)

Convolution

m

k(1-m)

g(t)

-1

1

m

h(m)

Convolution

m

k(2-m)

g(t)

-1

1

m

h(m)

Convolution

m

k(3-m)

g(t)

-1

1

m

h(m)

Convolution

m

k(4-m)

g(t)

-1

1

m

h(m)

Convolution

m

k(5-m)

g(t)

-1

1

m

h(m)

Convolution

m

k(6-m)

g(t)

f x y h x yMN

f m n h x m y nn

N

m

M

( , ) ( , ) ( , ) ( , )

1

0

1

0

1

2-Dimensions Convolution

g t k t h tM

k t m h mm

M

( ) ( ) ( ) ( ) ( )

1

0

1

h(t) g(t)k(t)g(t) = k(t)h(t)

G(f) = K(f)* H(f)

-1

1 1

2

t t

k(t)

h(t)

g(t)

2

Correlation

Correlation

Convolution

image enhancement in the frequency domain (chapter...

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