Transformada Fourier

Compiled from:S. NarasimhanNjegos NincicOther

Frequency domain analysis and Fourier Transform

How to Represent Signals?

• Option 1: Taylor series represents any function using polynomials.

• Polynomials are not the best - unstable and not very physically meaningful.

• Easier to talk about “signals” in terms of its “frequencies”

(how fast/often signals change, etc).

Jean Baptiste Joseph Fourier (1768-1830)

• Had crazy idea (1807):• Any periodic function

can be rewritten as a weighted sum of Sines and Cosines of different frequencies.

• Don’t believe it? – Neither did Lagrange,

Laplace, Poisson and other big wigs

– Not translated into English until 1878!

• But it’s true!– called Fourier Series

– Possibly the greatest tool

used in Engineering

Time and Frequency

• example : g(t) = sin(2pi f t) + (1/3)sin(2pi (3f) t)

Time and Frequency

= +


Frequency Spectra


= +

Frequency Spectra

• Usually, frequency is more interesting than the phase

= +

=

Frequency Spectra

Source: http://mathworld.wolfram.com/FourierSeries.html

Gibbs phenomenon: ringing near discontinuity

f ( t )=4π (sin ω0 t+

sin3 ω0 t

3+

sin5ω0 t

5+

sin7 ω0 t

7+…)

= 1

1sin(2 )

k

A ktk

Frequency Spectra

DIRICHLET CONDITIONS

Suppose that

1.f(x) is defined and single valued except possibly at finite number of points in (-l,+l)

2.f(x) is periodic outside (-l,+l) with period 2l

3.f(x) and f’(x) are piecewise continuous in(-l,+l)

Then the Fourier series of f(x) converges to

a)f(x) if x is a point of continuity

b)[f(x+0)+f(x-0)]/2 if x is a point of discontinuity

METHOD OF OBTAINING FOURIER SERIES OF

f ( x )=1.

a0=1l∫−l

+l

f ( x )dx

an=1l∫−l

+l

f ( x )cosnπxl

dx

bn=1l∫−l

l

f ( x )sinnπxl

dx

n=1,2,3, . .. .

2.

4.

3.

f ( x )a0

2+∑

n=1

∞

(an cosnπx

l+bn sin

nπxl )

Expressing cos nx and sin nx in exponential form, we may rewrite Eq.(7.1) as

f ( x )= ∑n=−∞

∞

cn e inx(7.4)

in which

cn=12

(an−ibn ) ,

c−n=12

(an +ibn ) , n> 0,(7.5)

and c0=

12

a0 .

f ( x )=a0

2+∑

n=1

∞

( an cosnx+bn sin nx ) .

cos nx=12

( einx +e−inx ) , sin nx=12 i

( einx−e−inx )

(7.6)

(7.3)

f ( x )= ∑n=−∞

∞

cn e inx

in which 1( ),

21

( ), 0,2

n n n

n n n

c a ib

c a ib n

and c0=12

a0 .

COMPLEX FORM OF FOURIER SERIES

Recall from earlier that we can write a Fourier series in the complex form

From the earlier definitions of we can show that ,n na b

0

1( ) , 2 / ,

T in xnc f x e dx T

T

Real and Complex Sinusoids

cos x ejx e jx

2

sin x ejx e jx

j2

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 20

Fourier Transform

• We want to understand the frequency of our signal. So, let’s reparametrize the signal by instead of x:

A sin( ωx+φ )

f(x) F()Fourier Transform

F() f(x)Inverse Fourier Transform

• For every from 0 to inf, F() holds the amplitude A and phase of the corresponding sine

– How can F hold both? Complex number trick!

F (ω )=R (ω )+iI (ω )

A=±√R (ω )2 +I (ω )2 φ= tan−1 I ( ω )

R(ω )

Transforms

Transform: In mathematics, a function that results when

a given function is multiplied by a so-called kernel function, and the product is integrated between suitable limits. (Britannica)

Can be thought of as a substitution

Transforms

Example of a substitution: Original equation: x + 4x² – 8 = 0 Familiar form: ax² + bx + c = 0 Let: y = x² Solve for y x = ±√y

4

Transforms

Transforms are used in mathematics to solve differential equations: Original equation: Apply Laplace Transform:

Take inverse Transform: y = Lˉ¹(y)

y'' − 9y =15e− 2t

s 2 L y − 9 L y =1 5

s 2

Ly=15

s2s2−9

Fourier Transform

Property of transforms: They convert a function from one domain to

another with no loss of information

Fourier Transform:

converts a function from the time (or spatial) domain to the frequency domain

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Fourier Series to Fourier Transform

For periodic signals, we can represent them as linear combinations of harmonically related complex exponentials

To extend this to non-periodic signals, we need to consider aperiodic signals as periodic signals with infinite period.

As the period becomes infinite, the corresponding frequency components form a continuum and the Fourier series sum becomes an integral (like the derivation of CT convolution)

Instead of looking at the coefficients a harmonically –related Fourier series, we’ll now look at the Fourier transform which is a complex valued function in the frequency domain

27

Fourier Series to Fourier Transform

Effect of increasing period T

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0 50 100 150

frequency

co

eff

icie

nt

va

lue

, T

=2

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0 50 100 150

frequency

coeff

icie

nt valu

e, T=5

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0 50 100 150

frequency

co

eff

icie

nt

va

lue

, T

=1

0

a/T a/T

a/T

Frequency Spectra

FT: Just a change of basis

.

.

.

* =

M * f(x) = F()

IFT: Just a change of basis

.

.

.

* =

M-1 * F() = f(x)

Fourier Transform – more formally

Arbitrary function Single Analytic Expression

Spatial Domain (x) Frequency Domain (u)

Represent the signal as an infinite weighted sum of an infinite number of sinusoids

F (u )=∫ f ( x ) e−i 2 πuxdx

(Frequency Spectrum F(u))

e ik=cos k+isin k i=√−1Note:

Inverse Fourier Transform (IFT)

f ( x )=∫ F (u ) ei 2 πuxdx

• Also, defined as:

F (u )=∫ f ( x ) e−iux dxe ik=cos k+isin k i=√−1Note:

• Inverse Fourier Transform (IFT)

f ( x )=1

2 π∫F (u ) e iux dx

Fourier Transform

Review of Cauchy Principal Review of Cauchy Principal Value IntegralsValue Integrals

2 0 2 0 2

1 01 1 0ln ln

x x

dx dx dxI x x

x x x

Recall for real integrals,

but a finite result is obtained if the integral interpreted as

2 2 2

11 10 0

0

lim lim ln ln

lim ln

x x

dx dx dxI x x

x x x

ln1 ln 2 ln ln 2

because the infinite contributions from the two symmetrical shadedparts shown exactly cancel. Integrals evaluated in this way are said to be (Cauchy) principal value integrals (or “deleted” integrals) and are often written as 2 2

1 1

dx dxI PV

x x

or

x

1/x

Delta function

Definition

Area for any > 0

Sifting property

since

δ ( x )={0 x≠0∞ x=0 }

∫ δ( x )dx= 1=∫ δ ( x )dx

∫ δ( x ) f ( x )dx =f (0 )

∫ δ( x−x0 ) f ( x )dx=f ( x0 )

36

Dirac delta Function

This allows an arbitrary sequence x(n) or continuous-time function f(t) to be expressed as:

x (n )= ∑k=−∞

∞

x (k )δ (n−k )

f ( t )=∫ f ( x )δ( x−t )dt

37

Example 1: Decaying Exponential

Consider the (non-periodic) signal

Then the Fourier transform is:

x ( t )=e−at u( t ) a>0

X ( jω)=∫e−at u( t )e− jωt dt=∫e−(a+jω)t dt1−(a+jω )

e−(a+jω ) t|0∞

1(a+jω )

a = 1

38

Example 2: Single Rectangular Pulse

Consider the non-periodic rectangular pulse at zero

The Fourier transform is:

x ( t )={1 |t|<T 1

0 |t|≥T1}

X ( jω)=∫ x ( t )e− jωt dt =∫ e− jωt dt1− jω

e− jωt|−T 1

T1

2sin(ωT 1 )

ω

Note, the values are real

T1 = 1

40

Example 3: Impulse SignalThe Fourier transform of the impulse signal can be calculated

as follows:

Therefore, the Fourier transform of the impulse function has a constant contribution for all frequencies

x ( t )=δ ( t )

X ( jω)=∫ δ ( t )e− jωt dt=1

X(j)

41

Example 4: Periodic SignalsA periodic signal violates condition 1 of the Dirichlet conditions for the

Fourier transform to exist

However, lets consider a Fourier transform which is a single impulse of area 2 at a particular (harmonic) frequency =0.

The corresponding signal can be obtained by:

which is a (complex) sinusoidal signal of frequency 0. More generally, when

Then the corresponding (periodic) signal is

The Fourier transform of a periodic signal is a train of impulses at the harmonic frequencies with amplitude 2ak

x ( t )=1

2 π∫2 πδ ( ω−ω0 )e jωt dω=e

jω0 t

X ( jω)=2 πδ (ω−ω0 )

X ( jω)= ∑k=−∞

∞

2 πak δ (ω−kω0 )

x ( t )= ∑k=−∞

∞

ak ejkω0 t

Fourier Transform Pairs (I)

angular frequency ( )e−iuxNote that these are derived using

angular frequency ( )e−iuxNote that these are derived using

Fourier Transform Pairs (I)

Fourier Transform and Convolution

g=f∗h

G (u )=∫ g ( x ) e−i 2 πuxdx

=∫∫ f ( τ )h ( x−τ ) e−i2 πux dτdx

=∫∫ [ f ( τ ) e−i 2 πuτ dτ ] [h ( x−τ ) e−i2 πu ( x−τ ) dx ]

=∫ [ f ( τ ) e−i 2 πuτ dτ ]∫ [h ( x' ) e−i 2 πux' dx' ]

Let

Then

=F (u ) H (u )

Convolution in spatial domain

Multiplication in frequency domain⇔

Fourier Transform and Convolution

g=f∗h G=FH

g=fh G=F∗H


So, we can find g(x) by Fourier transform

g = f h

G = F H

FT FTIFT

Properties of Fourier Transform


Linearity c1 f ( x ) +c2 g ( x ) c1 F (u )+c2 G (u )

Scaling f ( ax )1|a|

F ( ua )

Shifting f ( x−x0 ) e−i2 πux0 F (u )

Symmetry F ( x ) f (−u )

Conjugation f ( x ) F (−u )Convolution f ( x )∗g ( x ) F (u )G (u )

Differentiationdn f ( x )

dxn (i2 πu )n F (u )

frequency ( )e−i2 πuxNote that these are derived using

Properties of Fourier Transform

Example: Human ears do not hear wave-like

oscilations, but constant tone

Often it is easier to work in the frequency domain

Example use: Hearing mechanism

Example use: Smoothing/Blurring

• We want a smoothed function of f(x)

g ( x )=f ( x )∗h ( x )

H(u) attenuates high frequencies in F(u) (Low-pass Filter)!

• Then

H (u )=exp [−12

(2 πu )2 σ2 ]G (u ) =F (u ) H (u )

12πσ

u

H (u )

h ( x )=1

√2 π σexp[−1

2x2

σ 2 ]• Let us use a Gaussian kernel

σ

h ( x )

x

Image Processing in the Fourier Domain

Low-pass Filtering

Let the low frequencies pass and eliminating the high frequencies.

Generates image with overall shading, but not much detail

High-pass Filtering

Lets through the high frequencies (the detail), but eliminates the low frequencies (the overall shape). It acts like an edge enhancer.

Exercícios

Transformada Fourier

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