introduction to fourier transform and signal analysis

Continuous Fourier transformDiscrete Fourier transform

References

Introduction to Fourier transform and signalanalysis

Zong-han, Xieicbm0926@gmail.com

January 7, 2015

Zong-han, Xieicbm0926@gmail.com Introduction to Fourier transform and signal analysis

References

License of this document

Introduction to Fourier transform and signal analysis by Zong-han,Xie (icbm0926@gmail.com) is licensed under a Creative CommonsAttribution-NonCommercial 4.0 International License.

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Outline

1 Continuous Fourier transform

2 Discrete Fourier transform

3 References

References

Orthogonal condition

Any two vectors a, b satisfying following condition aremutually orthogonal.

a∗ · b = 0 (1)

Any two functions a(x), b(x) satisfying the followingcondition are mutually orthogonal.

∫a∗(x) · b(x)dx = 0 (2)

* means complex conjugate.

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Complete and orthogonal basis

cos nx and sinmx are mutually orthogonal in which n and mare integers. ∫ π

−πcos nx · sinmxdx = 0∫ π

−πcos nx · cosmxdx = πδnm∫ π

−πsin nx · sinmxdx = πδnm (3)

δnm is Dirac-delta symbol. It means δnn = 1 and δnm = 0when n 6= m.

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Fourier series

Since cos nx and sinmx are mutually orthogonal, we can expandan arbitrary periodic function f (x) by them. we shall have a seriesexpansion of f (x) which has 2π period.

f (x) = a0 +∞∑k=1

(ak cos kx + bk sin kx)

∫ π

−πf (x)dx

∫ π

−πf (x) cos kxdx

∫ π

−πf (x) sin kxdx (4)

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Fourier series

If f (x) has L period instead of 2π, x is replaced with πx/L.

f (x) = a0 +∞∑k=1

(ak cos

L+ bk sin

∫ L2

− L2

f (x)dx

∫ L2

− L2

f (x) cos2kπx

Ldx , k = 1, 2, ...

∫ L2

− L2

f (x) sin2kπx

Ldx , k = 1, 2, ... (5)

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Fourier series of step function

f (x) is a periodic function with 2π period and it’s defined asfollows.

f (x) = 0,−π < x < 0

f (x) = h, 0 < x < π (6)

Fourier series expansion of f (x) is

f (x) =h

(sin x

sin 3x

sin 5x

5+ ...

f (x) is piecewise continuous within the periodic region. Fourierseries of f (x) converges at speed of 1/n.

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Fourier series of step function

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Fourier series of triangular function

f (x) is a periodic function with 2π period and it’s defined asfollows.

f (x) = −x ,−π < x < 0

f (x) = x , 0 < x < π (8)

f (x) =π

2− 4

∑n=1,3,5...

(cos nx

f (x) is continuous and its derivative is piecewise continuous withinthe periodic region. Fourier series of f (x) converges at speed of1/n2.

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Fourier series of triangular function

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Fourier series of full wave rectifier

f (t) is a periodic function with 2π period and it’s defined asfollows.

f (t) = − sinωt,−π < t < 0

f (t) = sinωt, 0 < t < π (10)

f (t) =2

π− 4

∑n=2,4,6...

(cos nωt

n2 − 1

f (x) is continuous and its derivative is piecewise continuous withinthe periodic region. Fourier series of f (x) converges at speed of1/n2.

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Fourier series of full wave rectifier

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Complex Fourier series

Using Euler’s formula, Eq. 4 becomes

f (x) = a0 +∞∑k=1

(ak − ibk

2e ikx +

ak + ibk2

e−ikx)

Let c0 ≡ a0, ck ≡ ak−ibk2 and c−k ≡ ak+ibk

2 , we have

f (x) =∞∑

m=−∞cme

∫ π

−πf (x)e−imxdx (12)

e imx and e inx are also mutually orthogonal provided n 6= m and itforms a complete set. Therfore, it can be used as orthogonal basis.

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Complex Fourier series

If f (x) has T period instead of 2π, x is replaced with 2πx/T .

f (x) =∞∑

m=−∞cme

i 2πmxT

∫ T2

f (x)e−i2πmxT dx ,m = 0, 1, 2... (13)

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Fourier transform

from Eq. 13, we define variables k ≡ 2πmT , f (k) ≡ cmT√

2πand

4k ≡ 2π(m+1)T − 2πm

T = 2πT .

We can have

f (x) =1√2π

∞∑m=−∞

f (k)e ikx 4 k

f (k) =1√2π

∫ T2

f (x)e−ikxdx

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Fourier transform

Let T −→∞

f (x) =1√2π

∫ ∞−∞

f (k)e ikxdk (14)

f (k) =1√2π

∫ ∞−∞

f (x)e−ikxdx (15)

Eq.15 is the Fourier transform of f (x) and Eq.14 is the inverseFourier transform of f (k).

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Properties of Fourier transform

f (x), g(x) and h(x) are functions and their Fourier transforms aref (k), g(k) and h(k). a, b x0 and k0 are real numbers.

Linearity: If h(x) = af (x) + bg(x), then Fourier transform ofh(x) equals to h(k) = af (k) + bg(k).

Translation: If h(x) = f (x − x0), then h(k) = f (k)e−ikx0

Modulation: If h(x) = e ik0x f (x), then h(k) = f (k − k0)

Scaling: If h(x) = f (ax), then h(k) = 1a f (ka )

Conjugation: If h(x) = f ∗(x), then h(k) = f ∗(−k). With thisproperty, one can know that if f (x) is real and thenf ∗(−k) = f (k). One can also find that if f (x) is real and then|f (k)| = |f (−k)|.

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Properties of Fourier transform

If f (x) is even, then f (−k) = f (k).

If f (x) is odd, then f (−k) = −f (k).

If f (x) is real and even, then f (k) is real and even.

If f (x) is real and odd, then f (k) is imaginary and odd.

If f (x) is imaginary and even, then f (k) is imaginary and even.

If f (x) is imaginary and odd, then f (k) is real and odd.

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Dirac delta function

Dirac delta function is a generalized function defined as thefollowing equation.

f (0) =

∫ ∞−∞

f (x)δ(x)dx∫ ∞−∞

δ(x)dx = 1 (16)

The Dirac delta function can be loosely thought as a functionwhich equals to infinite at x = 0 and to zero else where.

δ(x) =

{+∞, x = 0

0, x 6= 0

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Dirac delta function

From Eq.15 and Eq.14

f (k) =1

∫ ∞−∞

f (k ′)e ik′xdk ′e−ikxdx

∫ ∞−∞

f (k ′)e i(k′−k)xdxdk ′

Comparing to ”Dirac delta function”, we have

f (k) =

∫ ∞−∞

f (k ′)δ(k ′ − k)dk ′

δ(k ′ − k) =1

∫ ∞−∞

e i(k′−k)xdx (17)

Eq.17 doesn’t converge by itself, it is only well defined as part ofan integrand.

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Convolution theory

Considering two functions f (x) and g(x) with their Fouriertransform F (k) and G (k). We define an operation

f ∗ g =

∫ ∞−∞

g(y)f (x − y)dy (18)

as the convolution of the two functions f (x) and g(x) over theinterval {−∞ ∼ ∞}. It satisfies the following relation:

f ∗ g =

∫ ∞−∞

F (k)G (k)e ikxdt (19)

Let h(x) be f ∗ g and h(k) be the Fourier transform of h(x), wehave

h(k) =√

2πF (k)G (k) (20)

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Parseval relation

∫ ∞−∞

f (x)g(x)∗dx =

∫ ∞−∞

1√2π

∫ ∞−∞

F (k)e ikxdk1√2π∫ ∞

−∞G ∗(k ′)e−ik

′xdk ′dx

∫ ∞−∞

F (k)G ∗(k ′)e i(k−k′)xdkdk ′

By using Eq. 17, we have the Parseval’s relation.∫ ∞−∞

f (x)g∗(x)dx =

∫ ∞−∞

F (k)G ∗(k)dk (21)

Calculating inner product of two fuctions gets same result as theinner product of their Fourier transform.

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Cross-correlation

Considering two functions f (x) and g(x) with their Fouriertransform F (k) and G (k). We define cross-correlation as

(f ? g)(x) =

∫ ∞−∞

f ∗(x + y)g(x)dy (22)

as the cross-correlation of the two functions f (x) and g(x) overthe interval {−∞ ∼ ∞}. It satisfies the following relation: Leth(x) be f ? g and h(k) be the Fourier transform of h(x), we have

h(k) =√

2πF ∗(k)G (k) (23)

Autocorrelation is the cross-correlation of the signal with itself.

(f ? f )(x) =

∫ ∞−∞

f ∗(x + y)f (x)dy (24)

References

Uncertainty principle

One important properties of Fourier transform is the uncertaintyprinciple. It states that the more concentrated f (x) is, the morespread its Fourier transform f (k) is.Without loss of generality, we consider f (x) as a normalizedfunction which means

∫∞−∞ |f (x)|2dx = 1, we have uncertainty

relation:(∫ ∞−∞

(x − x0)2|f (x)|2dx)(∫ ∞

−∞(k − k0)2|f (k)|2dk

16π2(25)

for any x0 and k0 ∈ R. [3]

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Fourier transform of a Gaussian pulse

f (x) = f0e−x2

2σ2 e ik0x

f (k) =1√2π

∫ ∞−∞

f (x)e−ikxdx

1/σ2e−(k0−k)2

|f (k)|2 ∝ e−(k0−k)2

Wider the f (x) spread, the more concentrated f (k) is.

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Signals with different width.

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The bandwidth of the signals are different as well.

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Outline

3 References

References

Nyquist critical frequency

Critical sampling of a sine wave is two sample points per cycle.This leads to Nyquist critical frequency fc .

2∆(26)

In above equation, ∆ is the sampling interval.Sampling theorem: If a continuos signal h(t) sampled with interval∆ happens to be bandwidth limited to frequencies smaller than fc .h(t) is completely determined by its samples hn. In fact, h(t) isgiven by

h(t) = ∆∞∑

n=−∞hn

sin[2πfc(t − n∆)]

π(t − n∆)(27)

It’s known as Whittaker - Shannon interpolation formula.

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Discrete Fourier transform

Signal h(t) is sampled with N consecutive values and samplinginterval ∆. We have hk ≡ h(tk) and tk ≡ k ∗∆,k = 0, 1, 2, ...,N − 1.With N discrete input, we evidently can only output independentvalues no more than N. Therefore, we seek for frequencies withvalues

fn ≡n

N∆, n = −N

2, ...,

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Discrete Fourier transform

Fourier transform of Signal h(t) is H(f ). We have discrete Fouriertransform Hn.

H(fn) =

∫ ∞−∞

h(t)e−i2πfntdt ≈ ∆N−1∑k=0

hke−i2πfntk

= ∆N−1∑k=0

hke−i2πkn/N

Hn ≡N−1∑k=0

hke−i2πkn/N (29)

Inverse Fourier transform is

hk ≡ 1

N−1∑n=0

Hnei2πkn/N (30)

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Periodicity of discrete Fourier transform

From Eq.29, if we substitute n with n + N, we have Hn = Hn+N .Therefore, discrete Fourier transform has periodicity of N.

Hn+N =N−1∑k=0

hke−i2πk(n+N)/N

=N−1∑k=0

hke−i2πk(n)/Ne−i2πkN/N

= Hn (31)

Critical frequency fc corresponds to 12∆ .

We can see that discrete Fourier transform has fs period wherefs = 1/∆ = 2 ∗ fc is the sampling frequency.

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Aliasing

If we have a signal with its bandlimit larger than fc , we havefollowing spectrum due to periodicity of DFT.

Aliased frequency is f − N ∗ fs where N is an integer.

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Aliasing example: original signal

Let’s say we have a sinusoidal sig-nal of frequency 0.05. The sampling interval is 1. We have the signal

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Aliasing example: spectrum of original signal

and we have its spectrum

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Aliasing example: critical sampling of original signal

The critical sampling interval of the original signal is 10 which ishalf of the signal period.

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Aliasing example: under sampling of original signal

If we sampled the original sinusiodal signal with period 12, aliasinghappens.

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Aliasing example: DFT of under sampled signal

fc of downsampled signal is 12∗12 , aliased frequency is

f − 2 ∗ fc = −0.03333 and it has symmetric spectrum due to realsignal.

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Aliasing example: two frequency signal

Let’s say we have a signal containing two sinusoidal signal offrequency 0.05 and 0.0125. The sampling interval is 1. We havethe signal

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Aliasing example: spectrum of two frequency signal

and we have its spectrum

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Aliasing example: downsampled two frequency signal

Doing same undersampling with interval 12.

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Aliasing example: DFT of downsampled signal

We have the spectrum of downsampled signal.

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Filtering

We now want to get one of the two frequency out of the signal.Wewill adapt a proper rectangular window to the spectrum.Assuming we have a filter function w(f ) and a multi-frequencysignal f (t), we simply do following steps to get the frequency bandwe want.

F−1{w(f )F{f (t)}} (32)

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Filtering example: filtering window and signal spectrum

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Filtering example: filtered signal

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Outline

3 References

References

Supplementary Notes of General Physics by Jyhpyng Wang,http:

//idv.sinica.edu.tw/jwang/SNGP/SNGP20090621.pdf

http://en.wikipedia.org/wiki/Fourier_series

http://en.wikipedia.org/wiki/Fourier_transform

http://en.wikipedia.org/wiki/Aliasing

http://en.wikipedia.org/wiki/Nyquist-Shannon_

sampling_theorem

MATHEMATICAL METHODS FOR PHYSICISTS by GeorgeB. Arfken and Hans J. Weber. ISBN-13: 978-0120598762

Numerical Recipes 3rd Edition: The Art of ScientificComputing by William H. Press (Author), Saul A. Teukolsky.ISBN-13: 978-0521880688Zong-han, Xieicbm0926@gmail.com Introduction to Fourier transform and signal analysis

References

Chapter 12 and 13 inhttp://www.nrbook.com/a/bookcpdf.php

http://docs.scipy.org/doc/scipy-0.14.0/reference/fftpack.html

http://docs.scipy.org/doc/scipy-0.14.0/reference/signal.html

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