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IMPLEMENTATIONIMPLEMENTATION

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Introduction to Image ProcessingIntroduction to Image Processing

Dr. Kourosh KianiEmail: kkiani2004@yahoo.comEmail: Kourosh.kiani@aut.ac.irEmail: Kourosh.kiani@semnan.ac.irWeb: www.kouroshkiani.com

Present to:Amirkabir University of Technology (Tehran

Polytechnic) & Semnan University

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Lecture 09

Discrete Fourier Transform

(DFT )of 1D signal

Discrete Time Periodic Signals

A discrete time signal x[n] is periodic with period N if and only if

][][ Nnxnx for all n .

Definition:

N

][nx

n

Meaning: a periodic signal keeps repeating itself forever!

Example: a Sinusoid

[ ] 2cos 0.2 0.9x n n

Consider the Sinusoid:

It is periodic with period since 10N

][29.02.0cos2

9.0)10(2.0cos2]10[

nxn

nnx

for all n.

General Periodic Sinusoid

n

N

kAnx 2cos][

Consider a Sinusoid of the form:

It is periodic with period N since

][22cos

)(2cos][

nxknN

kA

NnN

kANnx

for all n.

with k, N integers.

1.03.0cos5][ nnxConsider the sinusoid:

It is periodic with period since 20N

][231.03.0cos5

1.0)20(3.0cos5]20[

nxn

nnx

for all n.

We can write it as:

1.0

20

32cos5][ nnx

Example of Periodic Sinusoid

nN

kj

Aenx

2

][

Consider a Complex Exponential of the form:

for all n.

It is periodic with period N since

Periodic Complex Exponentials

][

][

22

)(2

nxeAe

AeNnx

jkn

Nkj

NnN

kj

1

njejnx 1.0)21(][

Consider the Complex Exponential:

We can write it as

Example of a Periodic Complex Exponential

nj

ejnx

20

12

)21(][

and it is periodic with period N = 20.

Goal:

We want to write all discrete time periodic signals in terms of a common set of “reference signals”.

Reference Frames

It is like the problem of representing a vector in a reference frame defined by

• an origin “0”

• reference vectors

x

,..., 21 ee

x

01e

2e

Reference Frame

Reference Frames in the Plane and in Space

For example in the plane we need two reference vectors

x

01e

2e21,ee

Reference Frame

… while in space we need three reference vectors 321 ,, eee

0

1e

2e

Reference Frame

x

3e

A Reference Frame in the Plane

If the reference vectors have unit length and they are perpendicular (orthogonal) to each other, then it is very simple:

2211 eaeax

0

11ea

22ea

Where projection of along

projection of along1a

2a 2e1e

x

x

The plane is a 2 dimensional space.

A Reference Frame in the Space

If the reference vectors have unit length and they are perpendicular (orthogonal) to each other, then it is very simple:

332211 eaeaeax

0

11ea

22ea

Where projection of along

projection of along

projection of along

1a

2a 2e1e

x

x

The “space” is a 3 dimensional space.

3a x

3e

33ea

Example: where am I ?

N

E0

1e

2e

x

m300

m200

Point “x” is 300m East and 200m North of point “0”.

Reference Frames for Signals

We want to expand a generic signal into the sum of reference signals.

The reference signals can be, for example, sinusoids or complex exponentials

n

][nx

reference signals

Back to Periodic Signals

A periodic signal x[n] with period N can be expanded in terms of N complex exponentials

1,...,0 ,][2

Nkenen

N

kj

k

as

1

0

][][N

kkk neanx

A Simple Example

Take the periodic signal x[n] shown below:

n0

1

2

Notice that it is periodic with period N=2.

Then the reference signals are

nnj

nnj

ene

ene

)1(][

11][

2

12

1

2

02

0

We can easily verify that (try to believe!):

nn

nenenx

)1(5.015.1

][5.0][5.1][ 10

for all n.

Another Simple Example

Take another periodic signal x[n] with the same period (N=2):

n0

3.0

3.1

Then the reference signals are the same0

22

0

12

21

[ ] 1 1

[ ] ( 1)

j n n

j n n

e n e

e n e

We can easily verify that (again try to believe!):

nn

nenenx

)1(8.015.0

][8.0][5.0][ 10

for all n.

Same reference signals, just different coefficients

Orthogonal Reference Signals

Notice that, given any N, the reference signals are all orthogonal to each other, in the sense

kmN

kmnene

N

nmk if

if 0][][

1

0

*

1

0 2

)(221

0

21

0

*

1

1][][

N

n N

kmj

kmjn

N

kmjN

n

nN

kmjN

nmk

e

eeenene

Since

by the geometric sum

… apply it to the signal representation …

k

N

m

N

nmkm

N

nk

nx

N

mmm

N

nk

Nanenea

neneanenx

1

0

1

0

*

1

0

*

][

1

0

1

0

*

][][

][][][][

and we can compute the coefficients. Call then kNakX ][

1,...,0 ,][][1

0

2

NkenxNakX

N

n

knNj

k

Discrete Fourier Series

1,...,0 ,][][1

0

2

NkenxkX

N

n

knNj

Given a periodic signal x[n] with period N we define the

Discrete Fourier Series (DFS) as

Since x[n] is periodic, we can sum over any period. The general definition of Discrete Fourier Series (DFS) is

1,...,0 ,][][][1 20

0

NkenxnxDFSkX

Nn

nn

knNj

for any0n

Inverse Discrete Fourier Series

1

0

2

][1

][][N

k

knNjekX

NkXIDFSnx

The inverse operation is called Inverse Discrete Fourier Series (IDFS), defined as

Revisit the Simple Example

Recall the periodic signal x[n] shown below, with period N=2:

n0

1

2

1,0,)1(21)1](1[]0[][][1

0

2

2

kxxenxkX kk

n

nkj

Then 1]1[,3]0[ XX

Therefore we can write the sequence as

n

k

knjekXkXIDFSnx

)1(5.05.1

][2

1][][

1

0

2

2

Example of Discrete Fourier Series

Consider this periodic signal

The period is N=10. We compute the Discrete Fourier Series

25

102 29 4

210 10

100 0

1 if 1, 2,...,9

[ ] [ ]15 if 0

j k

j kn j kn

j k

n n

ek

X k x n e ee

k

][nx

n010

1

… now plot the values …

0 2 4 6 8 100

5magnitude

0 2 4 6 8 10-2

0

2phase (rad)

k

k

|][| kX

][kX

Example of DFS

Compute the DFS of the periodic signal

)5.0cos(2][ nnx

Compute a few values of the sequence

,...0]3[,2]2[,0]1[,2]0[ xxxx

and we see the period is N=2. Then

k

n

knjxxenxkX )1(]1[]0[][][

1

0

2

2

which yields

2]1[]0[ XX

Signals of Finite Length

All signals we collect in experiments have finite length

)(tx )(][ snTxnx

ss TF

1

MAXTMAX SN T F

Example: we have 30ms of data sampled at 20kHz (ie 20,000 samples/sec). Then we have

points data 60010201030 33 N

Series Expansion of Finite Data

We want to determine a series expansion of a data set of length N.

Very easy: just look at the data as one period of a periodic sequence with period N and use the DFS:

n

1N0

Discrete Fourier Transform (DFT)

Given a finite interval of a data set of length N, we define the Discrete Fourier Transform (DFT) with the same expression as the Discrete Fourier Series (DFS):

21

0

[ ] [ ] [ ] , 0,..., 1N j kn

N

n

X k DFT x n x n e k N

And its inverse

21

0

1[ ] [ ] [ ] , 0,..., 1

N j knN

n

x n IDFT X k X k e n NN

Signals of Finite Length

All signals we collect in experiments have finite length in time

)(tx )(][ snTxnx

ss TF

1

MAXTMAX SN T F

Example: we have 30ms of data sampled at 20kHz (ie 20,000 samples/sec). Then we have

points data 60010201030 33 N

Series Expansion of Finite Data

We want to determine a series expansion of a data set of length N.

Very easy: just look at the data as one period of a periodic sequence with period N and use the DFS:

n

1N0

Discrete Fourier Transform (DFT)

Given a finite of a data set of length N we define the Discrete Fourier Transform (DFT) with the same expression as the Discrete Fourier Series (DFS):

21

0

[ ] [ ] [ ] , 0,..., 1N j kn

N

n

X k DFT x n x n e k N

and its inverse

21

0

1[ ] [ ] [ ] , 0,..., 1

N j knN

n

x n IDFT X k X k e n NN

Example of Discrete Fourier Transform

Consider this signal

The length is N=10. We compute the Discrete Fourier Transform

25

102 29 4

210 10

100 0

1 if 1, 2,...,9

[ ] [ ]15 if 0

j k

j kn j kn

j k

n n

ek

X k x n e ee

k

][nx

n0

9

1

…now plot the values…

0 2 4 6 8 100

5magnitude

0 2 4 6 8 10-2

0

2phase (rad)

k

k

|][| kX

][kX

Fs = 1000; % Sampling frequencyT = 1/Fs; % Sample timeL = 1000; % Length of signalt = (0:L-1)*T; % Time vector% Sum of a 50 Hz sinusoid and a 120 Hz sinusoidx = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t); y = x + 2*randn(size(t)); % Sinusoids plus noisefigure (1);plot(Fs*t(1:50),y(1:50))title('Signal Corrupted with Zero-Mean Random Noise')xlabel('time (milliseconds)')

0 5 10 15 20 25 30 35 40 45 50-6

-4

-2

0

2

4

6Signal Corrupted with Zero-Mean Random Noise

time (milliseconds)

NFFT = 2^nextpow2(L); % Next power of 2 from length of yY = fft(y,NFFT)/L;f = Fs/2*linspace(0,1,NFFT/2+1);figure (2);% Plot single-sided amplitude spectrum.plot(f,2*abs(Y(1:NFFT/2+1))) title('Single-Sided Amplitude Spectrum of y(t)')xlabel('Frequency (Hz)')ylabel('|Y(f)|')

0 50 100 150 200 250 300 350 400 450 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Single-Sided Amplitude Spectrum of y(t)

Frequency (Hz)

|Y(f

)|

Relation Between Dx and Du

For a signal f(x) with M points, let spatial resolution Dx be space between samples in f(x) and let frequency resolution Du be space between frequencies components in F(u), we have

xMu

DD

1

Example: for a signal f(x) with sampling period 0.5 sec, 100 point, we will get frequency resolution equal to

Hz02.05.0100

1

Du

This means that in F(u) we can distinguish 2 frequencies that are apart by 0.02 Hertz or more.

m = 4;n = 2^m-1;x = gf(randi([0 2^m-1],n,1),m); % Random vectory = fft(x); % Transform of xz = ifft(y); % Inverse transform of yck = isequal(z,x) % Check that ifft(fft(x)) recovers x.

[x z]= 5 5 2 2 3 3 0 014 1412 12 5 513 1312 12 9 9 6 6 5 5 1 1 5 5 8 8

>> a=[1 2 3 4 5 6]a = 1 2 3 4 5 6

>> b=fft(a')b= 21.0000 -3.0000 + 5.1962i -3.0000 + 1.7321i -3.0000 -3.0000 - 1.7321i -3.0000 - 5.1962i

Example in Matlab

>> c=ifft(b);>> a=c'

a =

1.0000 2.0000 3.0000 4.0000 5.0000 6.0000

Periodicity of 1-D DFT

0 N 2N-N

DFT repeats itself every N points (Period = N) but we usually display it for n = 0 ,…, N-1

We display only in this range

1

0

/2)(1

)(M

x

MuxjexfM

uF From DFT:

Conventional Display for 1-D DFT

0 N-1

Time Domain Signal

DFT

f(x))(uF

0 N-1

Low frequencyarea

High frequencyarea

The graph F(u) is not easy to understand !

)(uF

0-N/2 N/2-1

)(uF

0 N-1

Conventional Display for DFT : FFT Shift

FFT Shift: Shift center of thegraph F(u) to 0 to get betterDisplay which is easier to understand.

High frequency area

Low frequency area

0 200 400 600 800 1000 12000

200

400

600FFT

0 200 400 600 800 1000 12000

200

400

600FFTHIFT

y=rand(1024,1);z=fft(y)subplot(2,1,1)plot(abs(z))title('FFT')w=fftshift(z)subplot(2,1,2)plot(abs(w))title('FFTHIFT')

Questions? Discussion? Suggestions?

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