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Math for CS Lecture 11 1 Lecture 11 Fourier Transforms

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Page 1: Fourier

Math for CS Lecture 11 1

Lecture 11

Fourier Transforms

Page 2: Fourier

Math for CS Lecture 11 2

Fourier Series in exponential form

Consider the Fourier series of the 2T periodic function:

Due to the Euler formula

It can be rewritten as

With the decomposition coefficients calculated as:

(1)

1

0 )sincos(2

)(~

nnn T

nxb

T

nxa

axf

sincos ie i

n

inxnecxf )(

~

T

T

tT

in

n tfeT

c )(2

1

(2)

Page 3: Fourier

Math for CS Lecture 11 3

Fourier transform

The frequencies are and

Therefore (1) and (2) are represented as

Since, on one hand the function with period T has also the periods kT for any integer k, and

on the other hand any non-periodic function can be considered as a function with infinite

period, we can run the T to infinity, and obtain the Riemann sum with ∆w→∞, converging to

the integral:

(3)

T

nwn

wdttfeexfn

T

T

iwtiwx

)(2

1)(

~

Tw

dwdttfeexf

T

T

iwtiwx )(2

1)(

~

(4)

Page 4: Fourier

Math for CS Lecture 11 4

Fourier transform definition

The integral (4) suggests the formal definition:

The funciotn F(w) is called a Fourier Transform of function f(x) if:

The function

Is called an inverse Fourier transform of F(w).

dttfetfFwF iwt )(:)}({:)(

dwwFewFF iwx )(2

1:)}({1

(6)

(5)

Page 5: Fourier

Math for CS Lecture 11 5

Example 1

The Fourier transform of

is

The inverse Fourier transform is

1||

1||

0

1:)(

t

ttf

02

0sin

2)}({1

1 w

ww

wdtetfF iwt

00

0

10

12/1

1||1))1(sin(1))1(sin(1

sincos

2sin

2

2

x

x

x

dww

xwdw

w

xw

dww

wwxdw

w

we iwx

Page 6: Fourier

Math for CS Lecture 11 6

Fourier Integral

If f(x) and f’(x) are piecewise continuous in every finite interval, and f(x) is absolutely

integrable on R, i.e.

converges, then

Remark: the above conditions are sufficient, but not necessary.

dwdttfeexfxf iwtiwx )(

2

1)]()([

2

1

Page 7: Fourier

Math for CS Lecture 11 7

Properties of Fourier transform

1 Linearity:

For any constants a, b the following equality holds:

Proof is by substitution into (5).

2 Scaling:

For any constant c, the following equality holds:

)()()}({)}({)}()({ wbGwaFtgbFtfaFtbgtafF

)(||

1)}({

c

wF

cctfF

Page 8: Fourier

Math for CS Lecture 11 8

Properties of Fourier transform 2

3 Time shifting:

Proof:

4. Frequency shifting:

Proof:

)()}({ 00 wFettfF iwt

dueufedtettfttfF iwuiwtiwt

)()()}({ 0

00

)()}({ 00 wwFtfeF iwt

)()()}({ 000 wwFdtetfetfeF iwtiwttiw

Page 9: Fourier

Math for CS Lecture 11 9

Properties of Fourier transform 3

5. Symmetry:

Proof:

The inverse Fourier transform is

therefore

dwewFwfFtf iwt)(2

1)}({)( 1

)(2)}({ wftFF

)}({)(2

1)(2 tFFdtetFwf itw

Page 10: Fourier

Math for CS Lecture 11 10

Properties of Fourier transform 4

6. Modulation:

Proof:

Using Euler formula, properties 1 (linearity) and 4 (frequency shifting):

)]()([2

1)}sin()({

)]()([2

1)}cos()({

000

000

wwFwwFtwtfF

wwFwwFtwtfF

)]()([2

1

)}]({)}({[2

1)}cos()({

00

000

wwFwwF

tfeFtfeFtwtfF tiwtiw

Page 11: Fourier

Math for CS Lecture 11 11

Differentiation in time

7. Transform of derivatives

Suppose that f(n) is piecewise continuous, and absolutely integrable on R. Then

In particular

and

Proof:

From the definition of F{f(n)(t)} via integrating by parts.

)()()}({ )( wFiwtfF nn

)()}({ ' wiwFtfF )()}({ 2'' wFwtfF

Page 12: Fourier

Math for CS Lecture 11 12

Example 2

The property of Fourier transform of derivatives can be used for solution of differential

equations:

Setting F{y(t)}=Y(w), we have

tetHyy 4)(4

01

00)(

t

ttH

iwetHFyFyF t

4

1})({}{4}{ 4

iwwYwiwY

4

1)(4)(

Page 13: Fourier

Math for CS Lecture 11 13

Example 2

Then

Therefore

216

1

)4)(4(

1)(

wiwiwwY

||41

8

1)}({)( tewYFwy

Page 14: Fourier

Math for CS Lecture 11 14

Frequency Differentiation

In particular and

Which can be proved from the definition of F{f(t)}.

)()}({ )( wFitftF nnn

)()}({ wFittfF )()}({ 2 wFtftF

Page 15: Fourier

Math for CS Lecture 11 15

ConvolutionThe convolution of two functions f(t) and g(t) is defined as:

Theorem:

Proof:

duutgufgf )()(*

)](*[2

1)}()({

}{}{}*{

wGFtgtfF

gFfFgfF

}{}{)()()(

)()(}*{

)( gFfFduutdutgeufe

dtduutgufegfF

utiwiwu

iwt