image processing - lesson 6 discrete fourier...

Fourier Transform - Part II

Image Processing - Lesson 6

• Discrete Fourier Transform - 1D

• Discrete Fourier Transform - 2D

• Fourier Properties

• Convolution Theorem

• FFT

• Examples

Discrete Fourier Transform

Move from f(x) (x R) to f(x) (x Z) by sampling at equal intervals.

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

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

f(x) = f(x0+x∆x) x = 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+2∆x) f(x

0+3∆x)

f(xn) = f(x0 + x∆x)

∈ ∈

Discrete Fourier Transform

The Discrete Fourier Transform (DFT) is defined as:

u = 0, 1, 2, ..., N-1∑−

=

−

=1

0

2

)(1)(N

x

Niux

exfN

uFπ

∑−

=

=1

0

2

)()(N

u

Niux

euFxfπ

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

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

Matlab: F=fft(f);

Matlab: F=ifft(f);

4

3

2

1

Discrete Fourier Transform - Example

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+2∆x) f(x

0+3∆x)

f(x) = f(x0 + x∆x)

F(0) = 1/4 Σ f(x) e = 1/4 Σ f(x) 13

x=0

-2πi0x4

= 1/4(f(0) + f(1) + f(2) + f(3)) = 1/4(2+3+4+4) = 3.25

3

x=0

F(1) = 1/4 Σ f(x) e = 1/4 [2e0+3e-iπ/2+4e−πi 4e-i3π/2] = [-2+i]3

x=0

-2πix4 1

4

F(3) = 1/4 Σ f(x) e = 1/4 [2e0+3e-i3π/2+4e−3π i 4e-i9π/2] = [-2-i]3

x=0

-6πix4 1

4

F(2) = 1/4 Σ f(x) e = 1/4 [2e0+3e-iπ+4e−2π i 4e-3πi] = [-1-0i]=3

x=0

-4πix4 -1

4-14

Fourier Spectrum:

|F(0)| = 3.25|F(1)| = [(-1/2)2 + (1/4)2]0.5

|F(2)| = [(-1/4)2 + (0)2]0.5

|F(3)| = [(-1/2)2 + (-1/4)2]0.5

Discrete Fourier Transform -2D

Image f(x,y) x = 0,1,...,N-1 y= 0,1,...,M-1

The Discrete Fourier Transform (DFT) is defined as:

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

x = 0, 1, 2, ..., N-1∑−

=∑

−

=

+=

1

0

1

0

)(2),(),(

N

u

Mvy

NuxM

v

ievuFyxf

π

y = 0, 1, 2, ..., M-1

u = 0, 1, 2, ..., N-1v = 0, 1, 2, ..., M-1

∑−

=

−

=∑

+−=

1

0

1

0

)(2),(1),(F

N

x

M

y

Mvy

Nu xi

eyxfM N

vuπ

Matlab: F=fft2(f);

Matlab: F=ifft2(f);

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

v

u

Fourier Transform - ImageVisualizing the Fourier Transform

Image using Matlab Routines• F(u,v) is a Fourier transform of f(x,y) and it has

complex entries.

F = fft2(f);• In order to display the Fourier Spectrum |F(u,v)|

– Cyclically rotate the image so that F(0,0) is in the center:

F = fftshift(F);– Reduce dynamic range of |F(u,v)| by displaying

the log:

D = log(1+abs(F));

Example:

|F(u)| = 100 4 2 1 0 0 1 2 4

|F(u)| = 0 1 2 4 100 4 2 1 0Cyclic

|F(u)|/10 = 0 0 0 0 10 0 0 0 0

log(1+|F(u)|) = 0 0.69 1.01 1.61 4.62 1.61 1.01 0.69 0

0 1 2 4 10 4 2 1 0

Display in Range([0..10]):

log(1+|F(u)|)/0.462 =

Original

Visualizing the Fourier Image - Example

|F(u,v)|

|F(0,0)| at center log(1 + |F(u,v)|)

Original

Shifted Fourier Image Shifted Log Fourier Image = log(1+ |F(u,v)|)

Fourier Image = |F(u,v)|

Properties of The Fourier Transform

Distributive (addition)

LinearityF[a f(x,y)] = aF [f(x,y)]

a f(x,y) a F(u,v)

CyclicF(u,v) = F(u+N,v) = F(u,v+N) = F(u+N,v+N)

F(x,y) = F(x+N,y+N)

SymmetricF(u,v) = F*(-u,-v)

thus:

|F(u,v)| = |F(-u,-v)| Fourier Spectrum is symmetric

DC (Average)1MF(0,0) = Σ Σ f(x) e0

N-1

x=0

1N

M-1

y=0

if f(x) is real:

F[ f1(x,y) + f2(x,y)] = [f1(x,y)] + F [f2(x,y)]F~

F~

F~

F~

F~ x

f(x)

ω

|F(ω)|

x

g(x)

ω

|G(ω)|

x

f+g

ω

|F(ω)+G( ω)|

Distributive: { } { } { }gFfFgfF ~~~ +=+

1 cycle

0 N/2 N-1N

-N/2

Cyclic and Symmetry of the Fourier Transform - 1D Example

|F(u)|

Translation

Image Transformations

f(x-x0,y-y0) F(u,v)e-2πi(ux

0+vy

0)

N

f(x,y)e F(u-u0,v-v0)2πi(u

0x+v

0y)

N

The Fourier Spectrum remains unchanged under translation:

|F(u,v)| = |F(u,v)e |-2πi(ux0+vy0)

N

Rotation

Rotation of f(x,y) by θ

Rotation of F(u,v)by θ

Scale

f(ax,by) F(u/a,v/b)1|ab|

x

f(x)

ω

|F(ω)|

x

f(x)

ω

|F(ω)|

Change of Scale- 1D:

x

f(x)

ω

|F(ω)|

( ){ } ( ) ( ){ }

==

aF

a1

axfF~thenFxfF~ifω

ω

x

f(x)

ω

F(ω)

x

f(2x)

ω

0.5 F(ω/2)

Change of Scale

x

f(x/2)

ω

2 F(2ω)

2D Image 2D Image - Rotated

Fourier Spectrum Fourier Spectrum

Example - Rotation

Image Domain Frequency Domain

Fourier Transform Examples

Separabitity

F(u,v) = Σ Σ f(x,y) e-2πi(ux+vy)/nN-1

x=0

1N

N-1

y=0

F(u,v) = Σ (Σ f(x,y) e-2πivy/n ) e-2πiux/nN-1

x=0

1N

N-1

y=0

F(u,v) = Σ F(x,v) e-2πiux/nN-1

x=0

1N

Thus, to perform a 2D Fourier Transform is equivalent to performing 2 1D transforms:

1) Perform 1D transform on EACH column of image f(x,y).Obtain F(x,v).

2) Perform 1D transform on EACH row of F(x,v).

Higher Dimensions:Fourier in any dimension can be performed by applying 1D transform on each dimension.





Linear Systems and Responses

SpatialDomain

FrequencyDomain

Input f F

Output g G

ImpulseResponse h

Freq.Response H

Relationship g=f*h G=FH

The Convolution Theorem

g = f * h g = f h

implies implies

G = F H G = F * H

Convolution in one domain is multiplication in the other and vice

versa

The Convolution Theorem

Convolution in one domain is multiplication in the other and vice

versa

( ) ( ){ } ( ){ } ( ){ }xgFxfFxgxfF ~~~ =∗

and likewise

( ) ( ){ } ( ){ } ( ){ }xgFxfFxgxfF ~~~ ∗=

f(x,y) * g(x,y) F(u,v) G(u,v)

f(x,y) g(x,y) F(u,v) * G(u,v)

f(x)

-0.5 0.5

x

h(x)Example:

-1 1

f(x)

-0.5 0.5

*h(x) =

F(ω) F(ω)

H(ω ) = .

=

f(x,y) g(x,y) f * g (x,y)

F(u,v) G(u,v) F(u,v) G(u,v)

F[F(u,v) G(u,v)]

Convolution Theorem -2D Example

F-1ℑ

1−ℑ

Sampling the Image

x

f(x)

ω

|F(ω)|

F̂

x

g(x)

ω

F̂|G(ω)|

ω

T 1/T

x

*

F̂

-1/2T 1/2T

Undersampling the Image

x

g(x)

F̂|G(ω)|

ω

T 1/T

*

ω-1/2T 1/2Tx

F̂

x

f(x)

ω

|F(ω)|

F̂

Critical Sampling• If the maximal frequency of f(x) is ωmax , it

is clear from the above replicas that ωmax should be smaller that 1/2T.

• Alternatively:

• Nyquist Theorem : If the maximal frequency of f(x) is ωmax the sampling rate should be larger than 2 ωmax in order to fully reconstruct f(x) from its samples.

• If the sampling rate is smaller than 2ωmax overlapping replicas produce aliasing .

max21

ω>T

ω

|F(ω)|

-1/2T 1/2T

Optimal Interpolation

• It is possible to fully reconstruct f(x) from its samples:

ω

|F(ω)|

x

f(x)F-1^

-1/2T 1/2T

ω

|F(ω)|

x

f(x)F-1^

-1/2T 1/2T

ω

|F(ω)|

x

f(x)F-1^

-1/2T 1/2T

1

*

Note, that only N/2 different transform values are obtainedfor the N/2 point transforms.

Fast Fourier Transform - FFT

u = 0, 1, 2, ..., N-1∑−

=

−

=1

0

2

)(1)(N

x

Niux

exfN

uFπ

O(n2) operations

∑∑−

=

−

=

+−−

++=12/

0

12/

0

)12(222

)12(1)2(1)(N

x

N

x

Nxiu

Nxiu

exfN

exfN

uFππ

The Fourier transform of N inputs, can be performed as2 Fourier Transforms of N/2 inputs each + one complex multiplication and addition for each value i.e. O(N).

even x odd x

−−

∑ ++∑=−

=

−−

=

12/

0

212/

0

2/2

2/2

2/)12(

2/1)2(1

21 N

x

NiuN

x

Niux

Niux

Nexf

Neexf

πππ

Fourier Transform ofof N/2 even points

Fourier Transform ofof N/2 odd points

0 1 2 3 4 5 6 7 0 2 4 6 1 3 5 7

All sampling pointsOdd

sampling pointsEven

sampling points

For u = 0, 1, 2, ..., N/2-1

−−−

∑ ++∑=−

=

−

=

12/

0

12/

0

2/22

2/2

2/)12(

2/1)2(1

21)(

N

x

N

xN

Niux

Niu

Niux

Nexf

NeexfuF

πππ

−

+= )()(21)(

2/2/

2

uFeuFuF o

N

e

NNN

iuπ

For u' = u+N/2 :N

iuiN

iuN

NuiNiu

eeeeeπ

ππππ 22)2/(2'2 −

−=−−

=

+−

=

−

obtain :

Thus: only one complex multiplication is needed for two terms.

−

+= )()(21)(

2/

2

2/uFeuFuF o

NN

iueNN

π

−

−=+ )()(21)(

2/

2

2/2 uFeuFuF oN

Niu

eN

NN

π

)u(e2/NFCalculating and is done recursivelyby

calculating and .)u(o

2/NF)u(e

4/NF )u(o4/NF

F(0) F(1) F(2) F(3) F(4) F(5) F(6) F(7)

F(0) F(2) F(4) F(6) F(1) F(3) F(5) F(7)

F(0) F(4) F(2) F(6) F(1) F(5) F(3) F(7)

F(0) F(1) F(2) F(3) F(4) F(5) F(6) F(7)

2-pointtransform

4-pointtransform

FFT

FFT : O(nlog(n)) operations

FFT of NxN Image: O(n2log(n)) operations

FilteredImage

Image Transform

Filtered

FFT

FFT -1NewImage

Frequency Enhancement

image processing - lesson 6 discrete fourier...

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