Image Processing - Lesson 5

• Introduction to Fourier Transform

• Image Transforms

• Basis to Basis

• Fourier Basis Functions

• Fourier Coefficients

• Fourier Transform - 1D

• Fourier Transform - 2D

Fourier Transform - Part I

The Fourier Transform

Jean Baptiste Joseph Fourier

Efficient Data Representation

• Data can be represented in many ways.

• There is a great advantage using an appropriate representation.

• It is often appropriate to view images as combinations of waves.

How can we enhance such an image?

Solution: Image Representation

2 1 3

5 8 7

0 3 5

= 1 0 0

0 0 0

0 0 0

2 + 1

0 1 0

0 0 0

0 0 0

0 0 1

0 0 0

0 0 0

+ 3 + 5

0 0 0

1 0 0

0 0 0

+

+ ...

= 3 + 5 +

+ 10 + 23 + ...

The inverse Fourier Transform

• For linear-systems we saw that it is convenient to represent a signal f(x) as a sum of scaled and shifted sinusoids.

( ) ωωω

πω deFxf xi∫= 2)(

How is this done?

X Coordinate

Grayscale Image

[ a1 a2 a3 a4 ] =

a1 [ 1 0 0 0 ] + a2 [ 0 1 0 0 ] + a3 [ 0 0 1 0 ] + a4 [ 0 0 0 1 ]

Hadamard Transform:

1. Basis Functions.2. Method for finding the image given the transform coefficients. 3. Method for finding the transform coefficients given the image.

Standard Basis:

Transforms: Change of Basis

Frequency Coordinate

Fourier Image

Standard Basis New Basis

[ 2 1 0 1 ] =

= 1 [ 1 1 1 1 ] + 1/2 [ 1 1 -1 -1 ] - 1/2 [ -1 1 1 -1 ] + 0 [ -1 1 -1 1 ]

= [ 1 1/2 -1/2 0 ]Hadamard

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Wave Number

Hadamard Basis Functions - 1D

N = 16

X = [ 2 1 0 1 ] standard

New Basis:

Finding the transform coefficients

T0 = [ 1 1 1 1 ]T1 = [ 1 1 -1 -1 ]T2 = [ -1 1 1 -1 ]T3 = [ -1 1 -1 1 ]

Signal:

New Coefficients:

a0 = <X,T0 > =< [ 2 1 0 1 ] , [ 1 1 1 1 ] > /4 = 1a1 = <X,T1 > =< [ 2 1 0 1 ] , [ 1 1 -1 -1 ] > /4 = 1/2

a2 = <X,T2 > =< [ 2 1 0 1 ] , [ -1 1 1 -1 ] > /4 = -1/2

a3 = <X,T3 > =< [ 2 1 0 1 ] , [ -1 1 -1 1 ] > /4 = 0

X = [ 1 1/2 -1/2 0 ] newSignal:

1. Basis Functions.2. Method for finding the image given the transform coefficients. 3. Method for finding the transform coefficients given the image.

X Frequency

Y F

requ

ency

X Coordinate

Y C

oord

inat

e

Grayscale Image

FourierImage

1 00 0[ ]a11 a12

a21 a22[ ] 0 1

0 0[ ] 0 01 0[ ] 0 0

0 1[ ]+ a22+ a21+ a12= a11

Hadamard Transform:

1 11 1[ ]2 1

1 0[ ] 1 -11 -1[ ] -1 -1

1 1[ ] -1 11 -1[ ]+ 0- 1/2+ 1/2= 1

1 1/2-1/2 0[ ]=

Standard Basis:

Hadamard

Transforms: Change of Basis - 2D

[1 11 1[ ] 1 -1

1 -1[ ]-1 -11 1[ ] -1 1

1 -1[ ]1 1/2

-1/2 0 ][

Hadamard Transform:

1 11 1[ ]1 1/2

-1/2 0 ] 1 -11 -1[ ] -1 -1

1 1[ ] -1 11 -1[ ]+ 0- 1/2+ 1/2= 1

1 00 0[ ]2 1

1 0[ ] 0 10 0[ ] 0 0

1 0[ ] 0 00 1[ ]+ 0+ 1+ 1= 2

Standard Basis:

[[

[1 00 0[ ] 0 1

0 0]0 01 0] 0 0

0 1]2 1

1 0[ ]

Hadamard

Hadamard

Standardcoefficients

Basis Elements

Basis Elements

coefficients

Hadamard Basis Functions

size = 8x8 Black = +1 White = -1

For continuous images/signals f(x):

1) The number of Basis Elements Bi is ∞.

<f(x),Bi(x)> = dx

x

(x)iBf(x)∫

diBxf ii∫=i

(x)a)(

2) The dot product:

Fourier Transform

Basis Functions are sines and cosines

sin(x)

cos(2x)

sin(4x)

The transform coefficients determine the amplitude:

a sin(2x) 2a sin(2x) -a sin(2x)

3 sin(x)

+ 1 sin(3x)

+ 0.8 sin(5x)

+ 0.4 sin(7x)

=

A

B

C

D

A+B

A+B+C

A+B+C+D

Every function equals a sum of sines and cosines

The Fourier Transform

• The inverse Fourier Transform composes a signal f(x) given ( )ωF

( ) ωωω

πω deFxf xi∫= 2)(

• The Fourier Transform finds the given the signal f(x):

( ) dx∫ −=x

xi2f(x)eF πωω

( )ωF

• F(ω) is the Fourier transform of f(x):

• f(x) is the inverse Fourier transform of F(ω):

• f(x) and F(ω) are a Fourier transform pair.

( ){ } ( )ωFxfF =~

( ){ } ( )xfFF =− ω1~

• The Fourier transform F(ω) is a function over the complex numbers:

– Rω tells us how much of frequency ωis needed.

– θω tells us the shift of the Sine wave with frequency ω.

• Alternatively:

– aω tells us how much of cos with frequency ω is needed.

– bω tells us how much of sin with frequency ω is needed.

( ) ωθωω ieRF =

( ) ωωω ibaF +=

• Rω - is the amplitude of F(ω).

• θω - is the phase of F(ω).

• |Rω|2=F*(ω) F(ω) - is the power

spectrum of F(ω) .

• If a signal f(x) has a lot of fine details

F(ω) will be high for high ω.

• If the signal f(x) is "smooth" F(ω) will

be low for high ω.

Problem in Frequency

Space

Original Problem

Solution in Frequency

Space

Solution of Original Problem

Relatively easy solution

Difficult solution

Fourier Transform

Inverse Fourier

Transform

Why do we need representation in the frequency domain?

Examples:

• Let ( )xxf δ=)(

( ) ( ) 12 =⋅= −∞

∞−∫ xiexF πωδω

Fourier

x

f(x)

The Delta Function:

0

• Let 1)( =xf

( ) ( )ωδω πω == ∫∞

∞−

− xieF 2

Fourier

ω

x

f(x)

Rω

ω

θω

ω

Real

ω

Imag

The Constant Function:

0

0

0

• Let ( )xxf 02cos)( πω=

( ) ( ) =⋅+= −∞

∞−

−∫ dxeeeF xixixi πωπωπωω 222 00

21

x

ω

Rω

ω

θω

( ) ( )[ ]0021

ωωδωωδ ++−=

f(x)

Fourier

ω0-ω0ω

Real

ω

Imag

ω0-ω0

The Cosine wave:

• Let

<=

otherwise021if1)(

21

xxrect

( ) ( ) ( )πωπω

πωω πω sincsin5.0

5.0

2 === ∫−

− dxeF xi

x

f(x)

Fourier

Rω

ω

The Window Function (rect):

-0.5 0.5

Proof:

f(x) = rect1/2(x) ={1 |x| ≤ 1/2

0 otherwise1

-1/2 1/2

dxedxexfuF iuxiux ∫∫−

−∞

∞−

− ==2/1

2/1

22)()( ππ

[ ] 2/12/1

2

21

−−

−= iuxe

iuπ

π[ ]iuiu ee

iuππ

π −−

= −

21

[ ])sin()cos()sin()cos(21

uiuuiuiu

πππππ

−−−−

=

uu

ππ )sin(

= = sinc(u)

sinc(u)

u

F(u)

• Let2

)( xexf π−=

( ) 2πωω −=eF

x

f(x)

Fourier

Rω

ω

The Gaussian:

x

f(x)

Fourier

Rω

ω

k

1/k

The bed of nails function:

ck(x)

C1/k(ω)

{ } dxdyeyxfvuFyxfF vyuxi∫ ∫∞

∞−

∞

∞−

+−== )(2),(),(),(~ π

Fourier Transform - 2D

Given a continuous real function f(x,y),its Fourier transform F(u,v) is defined as:

The Inverse Fourier Transform:

F(u,v) = a(u,v) + ib(u,v) =|F(u,v)|eiφ(u,v)

Phase =

Spectrum (Amplitude) =

φ(u,v) = tg-1(b(u,v)/a(u,v))

|F(u,v)| = a2(u,v) + b2(u,v)

|F(u,v)|2 = a2(u,v) + b2(u,v)

√Power Spectrum =

{ } dudvevuFyxfvuFF vyuxi∫ ∫∞

∞−

∞

∞−

+− == )(21 ),(),(),(~ π

Fourier Wave Functions - 2D

F(u,v) is the coefficient of the sine wave e2πi(ux+vy)

1/u

1v

x

y

lines of (real) value 1

e2πi(ux+vy) = cos(2π(ux+vy)) + isin(2π(ux+vy))

The ratio determines the Direction.uv

The size of u,v determines the Frequency.

u = 0 direction of waves

v = 0

direction of waves

22

1

vu +

u=0, v=0 u=1, v=0 u=2, v=0u=-2, v=0 u=-1, v=0

u=0, v=1 u=1, v=1 u=2, v=1u=-2, v=1 u=-1, v=1

u=0, v=2 u=1, v=2 u=2, v=2u=-2, v=2 u=-1, v=2

u=0, v=-1 u=1, v=-1 u=2, v=-1u=-2, v=-1 u=-1, v=-1

u=0, v=-2 u=1, v=-2 u=2, v=-2u=-2, v=-2 u=-1, v=-2

U

V

Fourier Transform 2D - Example

2D Function

2D Fourier Transform

Fourier Transform 2D - Example

x

y

f(x,y)

1/2

1/2

1

f(x,y) = rect(x,y) = {1 |x| ≤ 1/2, |y| ≤ 1/2

0 otherwise

|F(u,v)|

F(u,v) = sinc(u) ⋅ sinc(v) = sinc(u,v)

Proof of Fourier of Rect = sinc in 2D

),()sin()sin(

vuSincv

vu

u==

ππ

ππ

∫ ∫∫ ∫− −

+−∞

∞−

∞

∞−

+− ==2/1

2/1

2/1

2/1

)(2)(2),()( dxdyedxdyeyxfuF vyuxivyuxi ππ

dyedxe ivyiux ∫∫−

−

−

−=2/1

2/1

22/1

2/1

2 ππ

Image Domain Frequency Domain

Fourier Transform Examples