A Tutorial on Fourier Analysis - Université de Montréalpift6080/H09/documents/eck_fft.pdfA...

A Tutorial on Fourier Analysis


Douglas Eck

University of Montreal

NYU March 2006


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−0.5

0

0.5

1A fundamental and three odd harmonics (3,5,7)

fund (freq 100)3rd harm5th harm7th harmm

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1Sum of odd harmonics approximate square wave

fund (freq 100)fund+3rd harmfund+3rd+5thfund+3rd+5th+7th


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1Sum of odd harmonics from 1 to 127


Linear Combination

In the interval [u1, u2] a function Θ(u) can be written as a linearcombination:

Θ(u) =∞∑i=0

αiψi (u)

where functions ψi (u) make up a set of simple elementaryfunctions. If functions are orthogonal (roughly, perpindicular; innerproduct is zero)then coefficients αi are independant from oneanother.


Continuous Fourier Transform

The most commonly used set of orthogonal functions is the Fourierseries. Here is the analog version of the Fourier and Inverse Fourier:

X (w) =

∫ +∞

−∞x(t)e(−2πjwt)dt

x(t) =

∫ +∞

−∞X (w)e(2πjwt)dw


Discrete Fourier and Inverse Fourier Transform

X (n) =N−1∑k=0

x(k)e−2πjnk/N

x(k) =1

N

N−1∑n=0

X (n)e2πjnk/N


Taylor Series Expansion

f (x) = f (x0)+f ′(x0)

1(x−x0)+

f ′′(x0)

1 ∗ 2(x−x0)

2+f ′′′(x0)

1 ∗ 2 ∗ 3(x−x0)

3+...

or more compactly as

f (x) =∞∑

n=0

f (n)(x0)

n!(x − x0)

n


Taylor series expansion of e jΘ

Since ex is its own derivative, the Taylor series expansion forf (x) = ex is very simple:

ex =∞∑

n=0

xn

n!= 1 + x +

x2

2+

x3

3!+ ...

We can define:

e jΘ ==∞∑

n=0

(jΘ)n

n!= 1 + jΘ− Θ2

2− j

Θ3

3!+ ...


Splitting out real and imaginary parts

All even order terms are real; all odd-order terms are imaginary:

ree jΘ = 1−Θ2/2 + Θ4/4!− ...

ime jΘ = Θ−Θ3/3! + Θ5/5!− ...


Fourier Transform as sum of sines and cosines

Observe that:

cos(Θ) =∞∑

n>=0;n is even

(−1)n/2

n!Θn

sin(Θ) =∞∑

n>=0;n is odd

(−1)(n−1)/2

n!Θn

Thus yielding Euler’s formula:

e jθ = cos(Θ) + j sin(Θ)


Fourier transform as kernel matrix


Example

Sum of cosines with frequencies 12 and 9, sampling rate = 120

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0

0.01

0.02

signal

0 20 40 60 80 100 120−0.5

0

0.5

1

real part

0 20 40 60 80 100 120−0.2

−0.1

0

0.1

0.2

imag part

two cosines (freqs=9, 12)


Example

FFT coefficients mapped onto unit circle

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

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0

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1FFT projected onto unit circle


Impulse response

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1

signal

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0.01

0.01

magnitude

impulse response

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0

0.5

1

phase


Impulse response

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0.5

1

signal

0 10 20 30 40 50 60 70 80 90 1000.01

0.01

0.01

magnitude

delayed impulse response

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−2

0

2

4

phase


A look at phase shifting

Sinusoid frequency=5 phase shifted multiple times.

0 10 20 30 40 50 60 70 80 90 10049.5361

49.5361

49.5361

49.5361

mag

nitu

de

number of points shifted

sinusoid freq=5 phase shifted repeatedly

0 10 20 30 40 50 60 70 80 90 100−3

−2

−1

0

1

2

3

angl

e fr

eq=

6

number of points shifted


Sin period 10 + period 30

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.50

100

200

freq (mag)

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5−5

0

5

phase

0 100 200 300 400 500 600−1

0

1

component sinusoids

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0

5

reconstructed signal using component sinusoids vs original

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0

2

reconstructed signal using ifft vs original


Aliasing

The useful range is the “Nyquist frequency” (fs/2)

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0

1cos(21)

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0

1cos(21) sampled at 24 Hz

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

0

1cos(45)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

0

1cos(45) sampled at 24 Hz

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

0

1cos(−3)

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0

1cos(−3) sampled at 24 Hz


Leakage

Even below Nyquist, when frequencies in the signal do not alignwell with sampling rate of signal, there can be “leakage”. Firstconsider a well-aligned exampl (freq = .25 sampling rate)

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1Sinusoid at 1/4 the Sampling Rate

Time (samples)

Am

plitu

de

a)

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10

20

30

40

Normalized Frequency (cycles per sample))

Mag

nitu

de (

Line

ar)b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−300

−200

−100

0


Mag

nitu

de (

dB)

c)


Leakage

Now consider a poorly-aligned example (freq = (.25 + .5/N) *sampling rate)

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0

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1Sinusoid NEAR 1/4 the Sampling Rate

Time (samples)

Am

plitu

de

a)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30


Mag

nitu

de (

Line

ar)

b)

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0

10

20

30


Mag

nitu

de (

dB)

c)


Leakage

Comparison:

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0

0.5


Time (samples)

Am

plitu

de

a)

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10

20

30

40


Mag

nitu

de (

Line

ar)b)

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−300

−200

−100

0


Mag

nitu

de (

dB)

c)

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−0.5

0

0.5


Time (samples)

Am

plitu

de

a)

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10

20

30


Mag

nitu

de (

Line

ar)

b)

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0

10

20

30


Mag

nitu

de (

dB)

c)


Windowing can help

Can minimize effects by multiplying time series by a window thatdiminishes magnitude of points near signal edge:

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1Blackman Window

Time (samples)

Am

plitu

de

a)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−100

−50

0


Mag

nitu

de (

dB)

b)

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−50

0


Mag

nitu

de (

dB)

c)


Leakage Reduced

Comparison:

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0

0.5


Time (samples)

Am

plitu

de

a)

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10

20

30


Mag

nitu

de (

Line

ar)

b)

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0

10

20

30


Mag

nitu

de (

dB)

c)

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−0.5

0

0.5


Time (samples)

Am

plitu

de

a)

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5

10

15


Mag

nitu

de (

Line

ar)

b)

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−60

−40

−20

0

20


Mag

nitu

de (

dB)

c)


Convolution theorem

This can be understood in terms of the Convolution Theorem.Convolution in the time domain is multiplication in the frequencydomain via the Fourier transform (F).

F(f ∗ g) = F(f )×F(g)


Computing impulse response

The impulse response h[n] is the response of a system to the unitimpulse function.


Using the impulse response

Once computed, the impulse response can be used to filter anysignal x [n] yielding y [n].


Examples


Filtering using DFT

Goal is to choose good impulse response h[n]

Transform signal into frequency domain

Modify frequency properties of signal via multiplication

Transform back into time domain


Difficulties (Why not a perfect filter?)

You can have a perfect filter(!)

Need long impulse response function in both directions

Very non causal

In generating causal version, challenges arise


Gibbs Phenomenon

−500 −400 −300 −200 −100 0 100 200 300 400 5000

0.5

1

ideal lopass filter in frequency domain

−500 −400 −300 −200 −100 0 100 200 300 400 500−0.1

0

0.1

ideal filter coeffs in time domain

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0

0.1

truncated causal filter

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0.5

1

1.5

Gibbs phenomenon


Spectral Analysis

Often we want to see spectral energy as a signal evolves overtime

Compute Fourier Transform over evenly-spaced frames of data

Apply window to minimize edge effects


Short-Timescale Fourier Transform (STFT)

X (m, n) =N−1∑k=0

x(k)w(k −m)e−2πjnk/N

Where w is some windowing function such as Hanning or gaussiancentered around zero.The spectrogram is simply the squared magnitude of these STFTvalues


Trumpet (G4)

Time

Fre

quen

cy

0 1 2 3 4 5 60

1000

2000

3000

4000

5000

[Listen]


Violin (G4)

Time

Fre

quen

cy

0 0.5 1 1.5 2 2.5 30

1000

2000

3000

4000

5000

[Listen]


Flute (G4)

Time

Fre

quen

cy

0 0.5 1 1.5 2 2.50

1000

2000

3000

4000

5000

[Listen]


Piano (G4)

Time

Fre

quen

cy

0 0.1 0.2 0.3 0.4 0.5 0.6 0.70

1000

2000

3000

4000

5000

[Listen]


Voice

Time

Fre

quen

cy

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1000

2000

3000

4000

5000

6000

7000

8000

[Listen]


C Major Scale (Piano)

Time

Fre

quen

cy

0 1 2 3 4 5 6 7 80

2000

4000

6000

8000

10000

12000

14000

16000

[Listen]


C Major Scale (Piano)

Log Spectrogram (Constant-Q Transform) reveals low-frequencystructure

Time

Fre

quen

cy

0 1 2 3 4 5 6 7 8

135

240

427

761

1356

2416

4305

7671

13669


Time-Space Tradeoff

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Am

p.

spoken "Steven Usma"


Time-Space Tradeoff

Time

Fre

quen

cy

Narrowband Spectrogram overlap=152 timepts=1633

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

1000

2000

3000

4000

Time

Fre

quen

cy

Wideband Spectrogram overlap=30 timepts=6593

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

1000

2000

3000

4000


Auto-correlation and meter

Autocorrelation long used to find meter in music (Brown 1993)Lag k auto-correlation a(k) is a special case of cross-correlationwhere a signal x is correlated with itself:

a(k) =1

N

N−1∑n=k

x(n) x(n − k)

Autocorrelation can also be defined in terms of Fourier analysis

a = F−1(|F(x)|)

where F is the Fourier transform, F−1 is the inverse Fouriertransform and || indicates taking magnitude from a complex value.


0 5 10 15 20 25 30time (seconds)

0 5 10 15 20 25 30time (seconds)

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000lags (ms)

Time series (top), envelope (middle) and autocorrelation (bottom) of

excerpt from ISMIR 2004 Tempo Induction contest

(Albums-Cafe Paradiso-08.wav). A vertical line marks the actual tempo

(484 msec, 124bpm). Stars mark the tempo and its integer multiples.

Triangles mark levels in the metrical hierarchy.


Fast Fourier Transform

Fourier Transform O(N2)

Fast version possible O(NlogN)

Size must be a power of two

Strategy is decimation in time or frequency

Divide and conquer

Rearrange the inputs in bit reversed orderOutput transformation (Decimation in Time)

A Tutorial on Fourier Analysis - Université de Montréalpift6080/H09/documents/eck_fft.pdfA...

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