Introduction to Fourier transform and signal analysis

Continuous Fourier transform
Discrete Fourier transform
References
Introduction to Fourier transform and signal
analysis
Zong-han, Xie
icbm0926@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 Commons
Attribution-NonCommercial 4.0 International License.

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Outline
1 Continuous Fourier transform
2 Discrete Fourier transform
3 References

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Orthogonal condition
Any two vectors a, b satisfying following condition are
mutually orthogonal.
a∗
· b = 0 (1)
Any two functions a(x), b(x) satisfying the following
condition are mutually orthogonal.
a∗
(x) · b(x)dx = 0 (2)
* means complex conjugate.

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Complete and orthogonal basis
cos nx and sin mx are mutually orthogonal in which n and m
are integers.
π
−π
cos nx · sin mxdx = 0
π
−π
cos nx · cos mxdx = πδnm
π
−π
sin nx · sin mxdx = πδnm (3)
δnm is Dirac-delta symbol. It means δnn = 1 and δnm = 0
when n = m.

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Fourier series
Since cos nx and sin mx are mutually orthogonal, we can expand
an arbitrary periodic function f (x) by them. we shall have a series
expansion of f (x) which has 2π period.
f (x) = a0 +
∞
k=1
(ak cos kx + bk sin kx)
a0 =
1
2π
π
−π
f (x)dx
ak =
1
π
π
−π
f (x) cos kxdx
bk =
1
π
π
−π
f (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
2kπx
L
+ bk sin
2kπx
L
a0 =
1
L
L
2
− L
2
f (x)dx
ak =
2
L
L
2
− L
2
f (x) cos
2kπx
L
dx, k = 1, 2, ...
bk =
2
L
L
2
− L
2
f (x) sin
2kπx
L
dx, 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 as
follows.
f (x) = 0, −π < x < 0
f (x) = h, 0 < x < π (6)
Fourier series expansion of f (x) is
f (x) =
h
2
+
2h
π
sin x
1
+
sin 3x
3
+
sin 5x
5
+ ... (7)
f (x) is piecewise continuous within the periodic region. Fourier
series 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 as
follows.
f (x) = −x, −π < x < 0
f (x) = x, 0 < x < π (8)
f (x) =
π
2
−
4
π
n=1,3,5...
cos nx
n2
(9)
f (x) is continuous and its derivative is piecewise continuous within
the periodic region. Fourier series of f (x) converges at speed of
1/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 as
follows.
f (t) = − sin ωt, −π < t < 0
f (t) = sin ωt, 0 < t < π (10)
f (t) =
2
π
−
4
π
n=2,4,6...
cos nωt
n2 − 1
(11)
f (x) is continuous and its derivative is piecewise continuous within
the periodic region. Fourier series of f (x) converges at speed of
1/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
2
eikx
+
ak + ibk
2
e−ikx
Let c0 ≡ a0, ck ≡ ak −ibk
2 and c−k ≡ ak +ibk
2 , we have
f (x) =
∞
m=−∞
cmeimx
cm =
1
2π
π
−π
f (x)e−imx
dx (12)
eimx and einx are also mutually orthogonal provided n = m and it
forms 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=−∞
cmei 2πmx
T
cm =
1
T
T
2
−T
2
f (x)e−i 2πmx
T dx, m = 0, 1, 2... (13)

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Fourier transform
from Eq. 13, we deﬁne variables k ≡ 2πm
T , ˆf (k) ≡ cmT√
2π
and
k ≡ 2π(m+1)
T − 2πm
T = 2π
T .
We can have
f (x) =
1
√
2π
∞
m=−∞
ˆf (k)eikx
k
ˆf (k) =
1
√
2π
T
2
−T
2
f (x)e−ikx
dx

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Fourier transform
Let T −→ ∞
f (x) =
1
√
2π
∞
−∞
ˆf (k)eikx
dk (14)
ˆf (k) =
1
√
2π
∞
−∞
f (x)e−ikx
dx (15)
Eq.15 is the Fourier transform of f (x) and Eq.14 is the inverse
Fourier transform of ˆf (k).

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Properties of Fourier transform
f (x), g(x) and h(x) are functions and their Fourier transforms are
ˆf (k), ˆg(k) and ˆh(k). a, b x0 and k0 are real numbers.
Linearity: If h(x) = af (x) + bg(x), then Fourier transform of
h(x) equals to ˆh(k) = aˆf (k) + bˆg(k).
Translation: If h(x) = f (x − x0), then ˆh(k) = ˆf (k)e−ikx0
Modulation: If h(x) = eik0x f (x), then ˆh(k) = ˆf (k − k0)
Scaling: If h(x) = f (ax), then ˆh(k) = 1
a
ˆf (k
a )
Conjugation: If h(x) = f ∗(x), then ˆh(k) = ˆf ∗(−k). With this
property, one can know that if f (x) is real and then
ˆf ∗(−k) = ˆf (k). One can also ﬁnd that if f (x) is real and then
|ˆf (k)| = |ˆf (−k)|.

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Properties of Fourier transform
If f (x) is even, then ˆf (−k) = ˆf (k).
If f (x) is odd, then ˆf (−k) = −ˆf (k).
If f (x) is real and even, then ˆf (k) is real and even.
If f (x) is real and odd, then ˆf (k) is imaginary and odd.
If f (x) is imaginary and even, then ˆf (k) is imaginary and even.
If f (x) is imaginary and odd, then ˆf (k) is real and odd.

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Dirac delta function
Dirac delta function is a generalized function deﬁned as the
following equation.
f (0) =
∞
−∞
f (x)δ(x)dx
∞
−∞
δ(x)dx = 1 (16)
The Dirac delta function can be loosely thought as a function
which equals to inﬁnite at x = 0 and to zero else where.
δ(x) =
+∞, x = 0
0, x = 0

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Dirac delta function
From Eq.15 and Eq.14
ˆf (k) =
1
2π
∞
−∞
∞
−∞
ˆf (k )eik x
dk e−ikx
dx
=
1
2π
∞
−∞
∞
−∞
ˆf (k )ei(k −k)x
dxdk
Comparing to ”Dirac delta function”, we have
ˆf (k) =
∞
−∞
ˆf (k )δ(k − k)dk
δ(k − k) =
1
2π
∞
−∞
ei(k −k)x
dx (17)
Eq.17 doesn’t converge by itself, it is only well deﬁned as part of
an integrand.

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Convolution theory
Considering two functions f (x) and g(x) with their Fourier
transform 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 the
interval {−∞ ∼ ∞}. It satisﬁes the following relation:
f ∗ g =
∞
−∞
F(k)G(k)eikx
dt (19)
Let h(x) be f ∗ g and ˆh(k) be the Fourier transform of h(x), we
have
ˆh(k) =
√
2πF(k)G(k) (20)

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Parseval relation
∞
−∞
f (x)g(x)∗
dx =
∞
−∞
1
√
2π
∞
−∞
F(k)eikx
dk
1
√
2π
∞
−∞
G∗
(k )e−ik x
dk dx
=
∞
−∞
1
2π
∞
−∞
F(k)G∗
(k )ei(k−k )x
dkdk
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 the
inner product of their Fourier transform.

References
Cross-correlation
Considering two functions f (x) and g(x) with their Fourier
transform 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) over
the interval {−∞ ∼ ∞}. It satisﬁes the following relation: Let
h(x) be f g and ˆh(k) be the Fourier transform of h(x), we have
ˆh(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)

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Uncertainty principle
One important properties of Fourier transform is the uncertainty
principle. It states that the more concentrated f (x) is, the more
spread its Fourier transform ˆf (k) is.
Without loss of generality, we consider f (x) as a normalized
function which means
∞
−∞ |f (x)|2dx = 1, we have uncertainty
relation:
∞
−∞
(x − x0)2
|f (x)|2
dx
∞
−∞
(k − k0)2
|ˆf (k)|2
dk
1
16π2
(25)
for any x0 and k0 ∈ R. [3]

References
Fourier transform of a Gaussian pulse
f (x) = f0e
−x2
2σ2 eik0x
ˆf (k) =
1
√
2π
∞
−∞
f (x)e−ikx
dx
=
f0
1/σ2
e
−(k0−k)2
2/σ2
|ˆf (k)|2
∝ e
−(k0−k)2
1/σ2
Wider the f (x) spread, the more concentrated ˆf (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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Nyquist critical frequency
Critical sampling of a sine wave is two sample points per cycle.
This leads to Nyquist critical frequency fc.
fc =
1
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) is
given 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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Signal h(t) is sampled with N consecutive values and sampling
interval ∆. We have hk ≡ h(tk) and tk ≡ k ∗ ∆,
k = 0, 1, 2, ..., N − 1.
With N discrete input, we evidently can only output independent
values no more than N. Therefore, we seek for frequencies with
values
fn ≡
n
N∆
, n = −
N
2
, ...,
N
2
(28)

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Fourier transform of Signal h(t) is H(f ). We have discrete Fourier
transform Hn.
H(fn) =
∞
−∞
h(t)e−i2πfnt
dt ≈ ∆
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
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)/N
e−i2πkN/N
= Hn (31)
Critical frequency fc corresponds to 1
2∆ .
We can see that discrete Fourier transform has fs period where
fs = 1/∆ = 2 ∗ fc is the sampling frequency.

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Aliasing
If we have a signal with its bandlimit larger than fc, we have
following 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 is
half 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, aliasing
happens.

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Aliasing example: DFT of under sampled signal
fc of downsampled signal is 1
2∗12, aliased frequency is
f − 2 ∗ fc = −0.03333 and it has symmetric spectrum due to real
signal.

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Aliasing example: two frequency signal
Let’s say we have a signal containing two sinusoidal signal of
frequency 0.05 and 0.0125. The sampling interval is 1. We have
the 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.We
will adapt a proper rectangular window to the spectrum.
Assuming we have a ﬁlter function w(f ) and a multi-frequency
signal f (t), we simply do following steps to get the frequency band
we 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

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 George
B. Arfken and Hans J. Weber. ISBN-13: 978-0120598762
Numerical Recipes 3rd Edition: The Art of Scientiﬁc
Computing by William H. Press (Author), Saul A. Teukolsky.
ISBN-13: 978-0521880688Zong-han, Xieicbm0926@gmail.com Introduction to Fourier transform and signal analysis

References
References
Chapter 12 and 13 in
http://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

Introduction to Fourier transform and signal analysis

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