discrete fourier transform (dft)

Discrete Fourier Transform (DFT)

Jyh-Shing Roger Jang ( 張智星 )MIR Lab ( 多媒體資訊檢索實驗室 )

CSIE, NTU ( 台灣大學資訊工程系 )[email protected], http://mirlab.org/jang

mailto:[email protected]

http://mirlab.org/jang

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Discrete Fourier Transform

Goal Decompose a given vector of discrete signals into sinusoidal components whose amplitudes can be view as energy distribution over a range of frequencies

Applications Speech recognition Digital filtering Many many more...

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Decomposition

Given xn=x(n/fs), n=0~N-1, we want to decompose x(t) into a linear combination of sinusoidal functions.

Magnitude spectrum Phase spectrum

Remaining problems To determine no. of terms and their frequencies

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4321 ,,, mmmm

4321 ,,,

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Bin Frequencies

Frequencies of these sinusoidal basis functions (aka “bin frequencies”):

/ / 2/ 2 / 3: , , , , ,

/ 22 3: 0, , , , ,

s s s s

ss s s

N NN N NPeriods

f f f f

N ff f fFrequencies

N N N N

fsNfsfsfstime

Nindex

/)1(/2/1/0:

1210:

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Sinusoidal Basis Functions (1/2)

1

2

3

4

5

6

7

sin 2

2 2/ 2sin 2

3 3/ 3sin 2

4 4/ 4sin 2

5 5/ 5sin 2

6 6/ 6sin 2

7 7/ 7sin 2

s s

s

s s

s

s s

s

s s

s

s s

s

s s

s

s s

s

T f Basis

f fNt

f N N

f fNt

f N N

f fNt

f N N

f fNt

f N N

f fNt

f N N

f fNt

f N N

f fNt

f N N

N

8

8 8/ 8sin 2s s

s

f ft

f N N

1,16,8~1,2sin

ss fNktN

kf

0 2 4 6 8 10 12 14 16-1

0

1

k=1

N=16, fs=1

0 2 4 6 8 10 12 14 16-1

0

1

k=2

0 2 4 6 8 10 12 14 16-1

0

1

k=3

0 2 4 6 8 10 12 14 16-1

0

1

k=4

0 2 4 6 8 10 12 14 16-1

0

1

k=5

0 2 4 6 8 10 12 14 16-1

0

1

k=6

0 2 4 6 8 10 12 14 16-1

0

1

k=7

0 2 4 6 8 10 12 14 16-1

0

1

k=8

Time (sec)

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Sinusoidal Basis Functions (2/2)

, 0 1

, cos 2 ,sin 2

2 2 2/ 2, cos 2 ,sin 2

3 3 3/ 3, cos 2 ,sin 2

/, cos 2 ,sin 2

s s s

s

s s s

s

s s s

s

s s s

s

T f

f f fNT f t t

f N N N

f f fNT f t t

f N N N

f f fNT f t t

f N N N

kf kf kfN kT f t t

f N N N

T

/ ( / 2), cos 2 ,sin 2

2 2 2s s s

s

f f fN Nf t t

f

0 2 4 6 8 10 12 14 16-1

0

1

k=

1

N=16, fs=1

0 2 4 6 8 10 12 14 16-1

0

1

k=

2

0 2 4 6 8 10 12 14 16-1

0

1

k=

3

0 2 4 6 8 10 12 14 16-1

0

1

k=

4

0 2 4 6 8 10 12 14 16-1

0

1

k=

5

0 2 4 6 8 10 12 14 16-1

0

1

k=

6

0 2 4 6 8 10 12 14 16-1

0

1

k=

7

0 2 4 6 8 10 12 14 16-1

0

1

k=

8

Time (sec)

1,16,8~1,2cos

ss fNktN

kf

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Decomposition into Basis Functions

0

1 1

2 2

3 3

/2 /2

( )

cos 2 sin 2

2 2cos 2 sin 2

3 3cos 2 sin 2

cos 2 sin 2

cos 2 sin 22 2

s s

s s

s s

s sk k

s sN N

x t a

f fa t b t

N N

f fa t b t

N N

f fa t b t

N N

kf kfa t b t

N N

f fa t b t

0

1 1

2 2

3 3

2

( / )

2 2cos 1 sin 1

2 2cos 2 sin 2

2 2cos 3 sin 3

2 2cos sin

cos

n s

k k

N

x x t n f a

n na b

N N

n na b

N N

n na b

N N

n na k b k

N N

a n

No. of modifiable parameters = N

-8-

Decomposition into Basis Functions

Overall expression:

No. of parameters is 1+2*(N/2-1)+1=N, which is equal to the no. of data points Exact solution is likely to exist.

12

01 2

12

2 20

1 2

2 2cos sin cos , 1 ~ 1

2cos cos

N

n k k Nk

N

k k k Nk

n nx a a k b k a n n N

N N

na a b k a n

N

Amplitude PhaseFreq= /skf N

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Parameter Identification

How to identify coefficients of the basis functions: Fast Fourier transform (FFT)

A fast algorithm with a complexity of O(n log n)

Solving simultaneous linear equations Integration (which take advantage of the

orthogonality of the basis functions)

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Frequencies of Basis Functions

Due to Euler identity, we can express the k-th term shown on the right.

2 2

2 2

2 2

2 2cos sin

2,

2 2

1 1

2 2

1 1

2 2

1 1

2 2

k k

j j j j

k k

j jk k k k

n nj k j kN N

k k k k

n nj k j N kN N

k k k k

n nj k j N kN N

k N k

n nk th term a k b k

N N

e e e e na b with k

j N

a b j e a b j e

a b j e a b j e

a b j e a b j e

c e c e

:

cos sin

cos sin

cos / 2

sin / 2

j

j

j j

j j

Euler Identity

e j

e j

e e

e e j

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Frequencies of Basis Functions

Plug in the simplified k-th term:

Include the first and last terms:

2 21

0 /21 0

0 0

/2 /2

cos

1, 1 ~ / 2 1

2

1, / 2 1 ~ 1

2

n nN Nj k j kN N

n k N kk k

k k k

N N

k k k

x a c e a n c e

where

c a

c a b j k N

c a

c a b j k N N

0

2 21 1

1 1

2 22 2

2 2

2 23 3

3 3

2 2

/2

21

0 /21

cos

cos

n

n nj j NN N

N

n nj j NN N

N

n nj j NN N

N

n nj k j N kN N

k N k

N

nN j kN

k Nk

x a

c e c e

c e c e

c e c e

c e c e

a n

a c e a n

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Representations of DFT: Two-side

Characteristics General representation If x is complex, then c is not conjugate symmetric MATLAB command: c=fft(x)

0 1 2 1 0 1 2 1[ , , , ] [ , , , ]fft

N Nx x x x x c c c c c

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Representations of DFT: One-side

Characteristics When x is real, c is conjugate symmetric and we

only have to look at one-side of FFT result. MATLAB command: magSpec=fftOneside(x)

0 1 2 3 /2 1 /2 /2 1 3 2 1[ , , , , , , , , , , , ]N N N N N Nc c c c c c c c c c

Real Real

Complexconjugate

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Representations of DFT: One-side

Formulas

0 1 2 1

0 1 2 /2 1 /2

0 1 2 /2 1 /2

2 2 2 2 2 20 1 1 2 2 /2 1 /2 1 /2

, , ,

/ 22 3,2 ,2 , 2 , at bin frequencies 0, , , ,...,

Magnitude spectrum , 2 , 2 , 2 ,

, , , ,

Power spectru

N

ss s sN N

N N

N N N

x x x x x

N ff f fc c c c c c

N N N N

c c c c c

a a b a b a b a

10 0 1 2 /2 1 /2

2 2 2 2 2 210 0 10 1 1 10 2 2 10 /2 1 /2 1 10 /2

m 20log , 2 , 2 , 2 ,

20log ,10log ,10log , 10log ,20log

N N

N N N

c c c c c

a a b a b a b a

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Example: Conjugate Symmetric

Conjugate symmetric of DFT for real x fftSym01.m

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Example: Two-side FFT

Two-side FFT of a pure sinusoid at one of the bin freqency fft01.m

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Example: Two-side FFT

Two-side FFT of a pure sinusoid NOT at one of the bin freqency fft02.m

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Example: One-side FFT

One-side FFT of a pure sinusoid NOT at one of the bin freqency fft03.m

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Example: One-side FFT

One-side FFT of a frame of audio signals fft04.m

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Example: FFT for Data Compression

Use partial coefficients to reconstruct the original signals fftApproximate01.m High-frequency

components are not so important

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Example: FFT for Data Compression

Use low-freq components to reconstruct a frame fftApproximate01.m

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Example: FFT for Periodic Signals

FFT on periodic signals fftRepeat01.m Useful for pitch tracking Intepretation of

harmonics from viewpoints ofIntegration for obtaining

the coefficientsApproximation by basis

functions

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Example: Zero-padding for FFT

Zero-padding Purpose

Easier for FFTInterpolation

fftRepeat01.m

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Example: Down-sampling

Down sampling High-frequency

components are missing fftResample01.m

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Harmonics

Why do we have harmonic structures in a power spectrum? Since the original frame is quasi-periodic…

If we want to use the spectrum for pitch tracking, we need to enhance the harmonics. How?

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How to Enhance Harmonics?

Goal Enhance harmonics for pitch tracking

Approach Take an integer number of fundamental periods

Hard! Use windowing Use zero-padding for better resolution

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Example: Harmonics Enhancement (1)

fftHarmonics01.m

0 100 200 300-1

-0.5

0

0.5

1Frame 1 (length=256)

Am

plitu

de

0 100 200 300-1

-0.5

0

0.5


Am

plitu

de

0 1000 2000 3000 4000-50

0

50

100

Frequency (Hz)

Pow

er (

db)

Power spectrum 1

0 1000 2000 3000 4000-50

0

50

100

Frequency (Hz)

Pow

er (

db)

Power spectrum 2

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fftHarmonics02.m

0 100 200 300-1

-0.5

0

0.5


Am

plitu

de

0 100 200 300-1

-0.5

0

0.5


Am

plitu

de

0 1000 2000 3000 4000-50

0

50

100

Frequency (Hz)

Pow

er (

db)

Power spectrum 1

0 1000 2000 3000 4000-100

-50

0

50

100

Frequency (Hz)

Pow

er (

db)

Power spectrum 2

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fftHarmonics03.m

0 100 200 300-1

-0.5

0

0.5


Am

plitu

de

0 1000 2000 3000-1

-0.5

0

0.5


Am

plitu

de

0 1000 2000 3000 4000-100

-50

0

50

100

Frequency (Hz)

Pow

er (

db)

Power spectrum 1

0 1000 2000 3000 4000-100

-50

0

50

100

Frequency (Hz)

Pow

er (

db)

Power spectrum 2

discrete fourier transform (dft)

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