Image Processing Fourier Transform 1D Efficient Data Representation Discrete Fourier Transform - 1D...

Image Processing

Fourier Transform 1D

• Efficient Data Representation

• Discrete Fourier Transform - 1D

• Continuous Fourier Transform - 1D

• Examples

2

The Fourier Transform

Jean Baptiste Joseph Fourier

3

Efficient Data Representation

• Data can be represented in many ways.

• There is a great advantage using an appropriate representation.

• Examples: – Equalizers in Stereo systems.

– Noisy points along a line

– Color space red/green/blue v.s. Hue/Brightness

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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?

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How can we enhance such an image?

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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 + 23 +

+ 10 + 7 + ...

7

Transforms

1. Basis Functions.

2. Method for finding the image given the transform coefficients.

3. Method for finding the transform coefficients given the image.

U Coordinates

V C

oord

inat

es

X Coordinate

Y C

oord

inat

e

Grayscale Image

TransformedImage

8

Representation in different bases

• It is possible to go back and forth between representations:

v1

v2

a

u1

u2

ivu u,aia

ii

uv uiaa

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The Inner Product

• Discrete vectors (real numbers):

• Discrete vectors (complex numbers):

The vector aH denotes the conjugate

transpose of a.

• Continuous functions:

ibiabab,ai

T

ibiabab,ai

*H

x

* dxxgxfxg,xf

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The Fourier basis functions

Basis Functions are sines and cosines

sin(x)

cos(2x)

sin(4x)

The transform coefficients determine the amplitude and phase:

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

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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

12

Sum of cosines only symmetric functions

Sum of sines only antisymmetric functions

{ {

13

Using Complex Numbers:

cos(kx) , sin(kx)

Ckcos(kx) + Sksin(kx) = Rk sin(kx + k)

f(x) = C0 + C1cos(x) + S1sin(x) + ... + Ckcos(kx) + Sksin(kx) + ...

Fourier Coefficients

eikx

Terms are considered in pairs:

k

kkkkk C

SSCR 122 tanandwhere

Ckcos(kx) + Sksin(kx) R ei eikx

Amplitude+phase

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• The inverse Continuous Fourier Transform composes a signal f(x) given F():

deF)x(fk

xi2

The 1D Continues Fourier Transform

• The Continuous Fourier Transform finds the F() given the (cont.) signal f(x):

B(x)=ei2x is a complex wave function for

each (continues) given

dxxi x

2ef(x)F

• F() and f(x) are continues.

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Continuous vs sampled Signals

Sampling: Move from f(x) (x R) to f(xj) (j Z) by sampling at equal intervals.

f(x0), f(x0+x), f(x0+2x), .... , f(x0+[n-1]x),

Given N samples at equal intervals, we redefine f as:

f(j) = f(x0+jx) j = 0, 1, 2, ... , N-1

4

3

2

1

4

3

2

1

0 0.25 0.5 0.75 1.0 1.25 0 1 2 3

f(x)

f(x0)

f(x0+x)

f(x0+2x)f(x0+3x)

f(j) = f(x0 + jx)

f(0)

f(3)f(2)

f(1)

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• The discrete basis functions are

• For frequency k the Fourier coefficient is:

1..0

21 NkN

ikx

k eN

xb

N/kx2iikkk eeR

Nkx2

sinSNkx2

cosC k

kikeRkF

1..0 Nx

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k = 0, 1, 2, ..., N-1

1

0

2

)(,)(N

x

N

ikx

exfxbxfkFk

1

0

21 )()(N

k

N

ikx

NekFxf

x = 0, 1, 2, ..., N-1

The Inverse Discrete Fourier Transform (IDFT) is defined as:

Matlab: F=fft(f);

Matlab: F=ifft(f);

The Discrete Fourier Transform (DFT)

Remark: Normalization constant might be different!

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Discrete Fourier Transform - Example

F(0) = f(x) e = f(x) 13

x=0

-2i0x4

= (f(0) + f(1) + f(2) + f(3)) = (2+3+4+4) = 13

3

x=0

F(1) = f(x) e = [2e0+3e-i/2+4ei+4e-i3/2] = [-2+i]3

x=0

-2ix4

F(3) = f(x) e = [2e0+3e-i3/2+4ei+4e-i9/2] = [-2-i]3

x=0

-6ix4

F(2) = f(x) e = [2e0+3e-i+4ei+4e-3i] = [-1-0i]= -13

x=0

-4ix4

DFT of [2 3 4 4] is [ 13 (-2+i) -1 (-2-i) ]

f(x) = [2 3 4 4]

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

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

• f(x) and F(k) are a Fourier pair.

• f(x) is a representation of the signal in the Spatial Domain and F(k) is a representation in the Frequency Domain.

kFxfF~

xfkFF~ 1

The Fourier Transform - Summary

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

– Rk tells us how much of frequency k is needed.

k tells us the shift of the Sine wave with frequency k.

• Alternatively:

– ak tells us how much of cos with frequency k is needed.

– bk tells us how much of sin with frequency k is needed.

kikeRkF

kk ibakF

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x

f(x)

k

Rk

k

k

kikeRkF

The signal f(x)

k

Real

k

Imag

kk ibakF

Amplitude (spectrum) and Phase

Real and Imaginary

The Frequency Domain

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• Rk - is the amplitude of F(k).

k - is the phase of F(k).

• |Rk|2=F*(k) F(k) - is the power spectrum of

F(k) .

• If a signal f(x) has a lot of fine details |Rk|2

will be high for high k.

• If the signal f(x) is "smooth" |Rk|2 will be

low for high k.

3 sin(x)

+ 1 sin(3x)

+ 0.8 sin(5x)

+ 0.4 sin(7x)

=

DEMO

23

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Examples:

• Let xxf δ)(

1dxexF x2i

Fourier

x

f(x)

R

Real

Imag

The Delta Function:

0

00

0

1

xgdxxxxg

dxxxx

)(

)(;)(lim

25

• Let 1)( xf

dxeF x2i

Fourier

x

f(x)

R

Real

Imag

The Constant Function:

0

0

0

26

• Letx2i 0e)x(f

dxeeF x2ix2i 0

Fourier

x

f(x)

R

Real

Imag

A Basis Function:

0

0dxe x)(2i 0

00

27

• Let xxf 02cos)(

dxeeeF xixixi 222 00

2

1

x

R

002

1

f(x)

Fourier

0-0

Real

Imag

0-0

The Cosine wave:

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• Let xxf 02sin)(

dxeeei

F xixixi 222 00

2

x

R

002

i

f(x)

Fourier

0-0

/2

Real

Imag

0

-0

The Sine wave:

-/2

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• Let

otherwise021if1)(

21

xxrect

sinc

sin5.0

5.0

2

dxeF xi

x

f(x)

Fourier

R

The Window Function (rect):

-0.5 0.5

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Proof:

f(x) = rect1/2(x) ={ 1 |x| 1/2

0 otherwise1

-1/2 1/2

dxedxexfF xixi

2/1

2/1

22)()(

2/1

2/12

2

1

xie

i

ii ee

i

2

1

)sin(i)cos()sin(i)cos(i

21

)(SINC)sin(

sinc()

F()

31

• Let2

)( xexf

2 eF

x

f(x)

Fourier

R

The Gaussian:

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• Let kmodx)x(ck

ω1modωδ~1

kk C

kcF

x

f(x)

Fourier

R

k

1/k

The Comb Function:

ck(x)

C1/k()

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Properties of The Fourier Transform

• Linearity:

• Distributive (additivity):

• DC (average):

• Symmetric: If f(x) is real then,

2121 fF~

fF~

ffF~

fF~

fF~

FFthusFF *

dxexfF 00

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x

f(x)

|F()|

x

g(x)

|G()|

x

f+g

|F()+G()|

Distributive: gFfFgfF ~~~

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Transformations

• Translation:

The Fourier Spectrum remains unchanged under translation:

• Scaling:

020

xieFxxfF ~

02 xieFF

a

Fa

xafF1~

0 50 1000

0.2

0.4

0.6

0.8

1

0 50 1000

0.2

0.4

0.6

0.8

1

0 50 100-15

-10

-5

0

5

10

0 50 100-15

-10

-5

0

5

10

0 50 100-15

-10

-5

0

5

10

0 50 100-15

-10

-5

0

5

10

0 50 1000

2

4

6

8

10

0 50 1000

2

4

6

8

10

0 50 100-15

-10

-5

0

5

10

0 50 100-15

-10

-5

0

5

10

0 50 100-15

-10

-5

0

5

10

Example - Translation

1D Image

Translated

real(F(u)) imag(F(u)) |F(u)|

Differences:

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x

f(x)

|F()|

x

f(x)

|F()|

Change of Scale- 1D:

x

f(x)

|F()|

a

ωF

a

1axfF

~thenωFxfF

~if

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x

f(x)

F()

x

f(2x)

0.5 F(/2)

Change of Scale

x

f(x/2)

2 F(2)