the fourier series edited

7/28/2019 The Fourier Series Edited

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1

The Fourier Series

A Fourier series is an expansion of a

periodicfunctionf(t) in terms of an infinite sum

ofcosines and sines

Introduction

=

++=

1

0 )sincos(2

)(n

nn tnbtnaa

tf

In other words, any periodicfunction can be

resolved as a summation of

constant value and cosine and sine functions:

=

++=

1

0 )sincos(2

)(n

nn tnbtnaa

tf

)sincos(11

tbta ++

2

0a

=

)2sin2cos( 22 tbta ++

)3sin3cos( 33 tbta ++ +

The computation and study of Fourier series isknown as harmonic analysis and is extremely

useful as a way to break up an arbitrary

periodic function into a set of simple terms thatcan be plugged in, solved individually, and

then recombinedto obtain the solution to the

original problem or an approximation to it towhatever accuracy is desired or practical.

=

+ +

+ + +

Periodic Function2

0a

ta cos1

ta 2cos2

tb sin1

tb 2sin2

f(t)

t

=

++=

1

0 )sincos(2

)(n

nn tnbtnaa

tf

where

=T

dttfT

a0

0 )(2

frequencylFundementa2

==

T

=T

n tdtntfT

a0

cos)(2

=T

n tdtntfT

b0

sin)(2

*we can also use the integrals limit .

=T

dttfT

a0

0 )(2

2/

2/

T

T


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2

Example 1

Determine the Fourier series representation of thefollowing waveform.

Solution

First, determine the period & describe the one periodof the function:

T= 2


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3

Some helpful identities

Forn integers,n

n )1(cos =0sin =n

02sin =n 12cos =n

xx sin)sin( = xx cos)cos( =

[Supplementary]

The sum of the Fourier series terms canevolve (progress) into the original waveform

From Example 1, we obtain++++= ttttf

5sin5

23sin

3

2sin

2

2

1)(

It can be demonstrated that the sum will leadto the square wave:

tttt

7sin7

25sin

5

23sin

3

2sin

2+++ttt

5sin5

23sin

3

2sin

2++

tt

3sin3

2sin

2+t

sin2

(a) (b)

(c) (d)

ttttt

9sin9

27sin

7

25sin

5

23sin

3

2sin

2++++

ttt

23sin23

23sin

3

2sin

2

2

1++++

(e)

(f)

Example 2

Given,)( ttf =

11 t

)()2( tftf =+

Sketch the graph off(t) such that .33 t

Then compute the Fourier series expansion off(t).


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4

Solution

The function is described by the following graph:

T= 2

==

T

2We find that

Then we compute the coefficients:

02

11

22

2

)(2

1

1

21

1

1

1

0

=

=

==

=

ttdt

dttfT

a

0coscos

)cos(cos0

cos)]sin([sin

sinsin

coscos)(2

22

22

1

1

22

1

1

1

1

1

1

1

1

=

=

+=

+

=

=

==

n

nn

nnn

n

tn

n

nn

dtn

tn

n

tnt

tdtnttdtntfT

an

since xx cos)cos( =

nnn

n

n

nn

n

n

n

tn

n

nn

dtn

tn

n

tnt

tdtnttdtntfT

b

nn

n

1

22

1

1

22

1

1

1

1

1

1

1

1

)1(2)1(2cos2

)sin(sincos2

sin)]cos([cos

coscos

sinsin)(2

+

=

==

+=

+

+=

+

=

==

+=

=

++=

=

+

=

ttt

tnn

tnbtnaa

tf

n

n

n

nn

3sin3

22sin

2

2sin

2

sin)1(2

)sincos(2

)(

1

1

1

0

Finally,

Example 3

Given


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5

Solution

The function is described by the following graph:

T= 4

2

2 ==

TWe find that

0 2 4 6 8 10 12t

v (t)

2

Then we compute the coefficients:

12

22

1)2(

2

1

0)2(4

2

)(2

2

0

22

0

4

2

2

0

4

0

0

=

==

+=

=

ttdtt

dtdtt

dttvT

a

222222

2

0

22

2

0

2

0

4

2

2

0

4

0

])1(1[2)cos1(2

2

2cos1

cos

2

10

sin

2

1sin)2(

2

1

0cos)2(2

1cos)(

2

nn

n

n

n

n

tn

dtn

tn

n

tnt

tdtnttdtntvT

a

n

n

=

=

=

+=

+

=

+==

nnn

n

n

n

tn

n

dtn

tn

n

tnt

tdtnttdtntvT

bn

21

2

2sin1

sin

2

11

cos

2

1cos)2(

2

1

0sin)2(2

1sin)(

2

22

2

0

22

2

0

2

0

4

2

2

0

4

0

===

=

=

+==

since 0sin2sin == nn

=

=

+

+=

++=

122

1

0

2sin

2

2cos

])1(1[2

2

1

)sincos(2

)(

n

n

n

nn

tn

n

tn

n

tnbtnaa

tv

Finally,

Symmetry Considerations

Symmetry functions:(i) even symmetry

(ii) odd symmetry


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6

Even symmetry

Any functionf(t) is even if its plot issymmetrical about the vertical axis, i.e.

)()( tftf =

Even symmetry (cont.)

The examples ofeven functions are:2)( ttf =

t t

t

||)( ttf =

ttf cos)( =

Even symmetry (cont.)

The integral of an even function from A to +Ais twice the integral from 0 to +A

t ++

=

AA

A

dttfdttf0

eveneven )(2)(

A +A

)(even tf

Odd symmetry

Any functionf(t) is odd if its plot isantisymmetrical about the vertical axis, i.e.

)()( tftf =

Odd symmetry (cont.) The examples ofodd functions are:

3)( ttf =

tt

t

ttf =)(

ttf sin)( =

Odd symmetry (cont.) The integral of an odd function from A to +A

is zero

t 0)(odd =+

A

A

dttfA +A

)(odd tf


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7

Even and odd functions

(even)(even) = (even) (odd)(odd) = (even) (even)(odd) = (odd) (odd)(even) = (odd)

The product properties ofeven and odd

functions are:

Symmetry consideration

From the properties ofeven and odd

functions, we can show that:

foreven periodic function;=

2/

0

cos)(4

T

n tdtntfT

a 0=nb

forodd periodic function;=

2/

0

sin)(4

T

n tdtntfT

b 00

==naa

How?? [Even function]

2

T

2

T

==

2/

0

2/

2/

cos)(4

cos)(2

TT

T

n tdtntfT

tdtntfT

a

(even) (even)

| |

(even)

0sin)(2

2/

2/

==

T

T

n tdtntfT

b

(even) (odd)

| |

(odd)

)(tf

t

How?? [Odd function]

2

T

2

T

==

2/

0

2/

2/

sin)(4

sin)(2

TT

T

n tdtntfT

tdtntfT

b

(odd)(odd)

| |

(even)

0cos)(2

2/

2/

==

T

T

n tdtntfT

a

(odd) (even)

| |

(odd)

)(tf

t

0)(2

2/

2/

0 ==

T

T

dttfT

a

(odd)

Example 4

Given


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8

Then we compute the coefficients. Sincef(t) is

an odd function, then

0)(2

2

2

0 ==

dttfT

a

0cos)(2

2

2

==

tdtntfT

an

and

n

n

n

n

n

n

n

nn

n

tn

n

n

n

tndt

n

tn

n

tnt

tdtntdtnt

tdtntfT

tdtntfT

bn

cos2sin2cos

cos2cossincos

coscoscos

sin1sin4

4

sin)(4

sin)(2

22

1

0

22

2

1

1

0

1

0

2

1

1

0

2

0

2

2

=+=

+=

++

=

+=

==

since 0sin2sin == nn

=

+

=

=

=

=

++=

1

1

1

1

0

2sin

)1(2

2sin

cos2

)sincos(2

)(

n

n

n

n

nn

tn

n

tn

n

n

tnbtnaa

tf

Finally,

Example 5

Compute the Fourier series expansion off(t).

SolutionThe function is described by

T= 3

3

22 ==

T

and


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9

3

2sin

2

3

2sinsin2

2

sin2

3sin2

3

4

sin2

3sin2sin

3

4

sin2

3

4sin

3

4

cos2cos13

4

cos)(4

cos)(2

2/3

1

1

0

2/3

1

1

0

2/3

0

3

0

n

n

nn

n

nn

n

nn

nn

n

tn

n

tn

tdtntdtn

tdtntfT

tdtntfT

an

=

=

=

+=

+

=

+=

==

;3

2 =

=

=

=

=

+=

++=

1

1

1

0

3

2cos

3

2sin

12

3

4

3

2cos

3

2sin

2

3

4

)sincos(2

)(

n

n

n

nn

tnn

n

tnn

n

tnbtnaa

tf

Finally,

and 0=nb sincef(t) is an even function.

Function defines over a finite interval

Fourier series only support periodic functions In real application, many functions are non-periodic The non-periodic functions are often can be defined

over finite intervals, e.g.

y=

ty = 1 y = 1

y = 2

y=

t2

Therefore, any non-periodic function must beextended to a periodic function first, before

computing its Fourier series representation

Normally, we prefer symmetry (even or odd) periodicextension instead of normal periodic extension, sincesymmetry function will provide zero coefficient of

eitheran orbn

This can provide a simpler Fourier series expansion

)(ty

t

)(tf

t

)(even tf

)(odd tf

t

t

lttytf


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10

If the function is extended as an even function,then the coefficient bn= 0, hence

=

+=

1

0 cos2

)(n

n tnaa

tf

which only contains the cosine harmonics.

Therefore, this approach is called as the half-range Fourier cosine series

If the function is extended as an odd function,then the coefficient an= 0, hence

=

=

1

sin)(n

n tnbtf

which only contains the sine harmonics.

Therefore, this approach is called as the half-range Fourier sine series

Example 6


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11

Solution

Since we want to seek the half-range cosine series,the function to is extended to be an even function:

T= 2

==

T

2

t

f(t)

t

f(t)

1

22

33 1

1

11

1

1

2

1

Hence, the coefficients are

[ ] 02)12(24

)(

4 10

2

1

0

2/

00

=

=

==

ttdttdttfTa

T

1

0

22

1

0

1

0

1

0

2/

0

cos4

sin2

2sin

2sin)12(

2

cos)12(2

4cos)(

4

+=

=

==

n

tn

n

n

dtn

tn

n

tnt

tdtnttdtntfT

a

T

n

=

=

even,0

odd,/8)1(cos422

22n

nn

n

n

0=nb

Therefore,

=

=

=

=

+=

+=

odd1

22

odd1

22

1

0

cos18

cos8

0

cos)(

nnnn

n

n

tn

n

tn

n

tnaatf

Parsevals Theorem

Parservals theorem states that the averagepower in a periodic signal is equal to the sum

of the average power in its DC component and

the average powers in its harmonics

=

+ +

+ + +

2

0a

ta cos1

ta 2cos2

tb sin1

tb 2sin2

f(t)

t

PavgP

dc

Pa1 Pb1

Pa2 Pb2

For sinusoidal (cosine or sine) signal,

R

V

R

V

R

VP

2

peak

2

peak2

rms

2

12=

==

For simplicity, we often assumeR = 1,which yields

2

peak2

1VP=


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12

For sinusoidal (cosine or sine) signal,

+++++

=

+++++=

2

2

2

2

2

1

2

1

2

0

dcavg

2

1

2

1

2

1

2

1

2

2211

babaa

PPPPPPbaba

=

++=1

222

0avg )(2

1

4

1

n

nnbaaP

Exponential Fourier series

Recall that, from the Eulers identity,xjxe

jxsincos =

yields

2cos

jxjxee

x

+

=

2sin

j

eex

jxjx

=and

Then the Fourier series representation becomes

=

=

=

=

=

=

++

+=

++

+=

++=

+

++=

++=

11

0

1

0

1

0

1

0

1

0

222

222

222

222

)sincos(2

)(

n

tjnnn

n

tjnnn

n

tjnnntjnnn

n

tjntjn

n

tjntjn

n

n

tjntjn

n

tjntjn

n

n

nn

ejba

ejbaa

ejba

ejbaa

eejb

eea

a

j

eeb

eea

a

tnbtnaa

tf

Here, let we name

=

=

++

+=

11

0

222)(

n

tjnnn

n

tjnnn ejba

ejbaa

tf

2

nnn

jbac

= ,2

nnn

jbac

+

=

Hence,

=

=

=

=

=

=

=

=++=

++=

++=

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

nn

tjn

n

ececcec

ececc

ececc

1

0

1

11

0

110

and .2

0

0

a

c =

c0 cncn

Then, the coefficient cn can be derived from

=

=

=

=

=

T

tjn

T

TT

TT

nnn

dtetfT

dttnjtntfT

tdtntfjtdtntfT

tdtntfT

jtdtntf

T

jbac

0

0

00

00

)(1

]sin)[cos(1

sin)(cos)(1

sin)(2

2cos)(

2

2

1

2

In fact, in many cases, the complex Fourierseries is easier to obtain rather than the

trigonometrical Fourier series

In summary, the relationship between thecomplex and trigonometrical Fourier series

are:

2

nnn

jbac

=

2

nnn

jbac

+

=

==T

dttfT

ac

0

00 )(

1

2

=

T

tjn

n dtetfT

c0

)(1

nncc =

or


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13

Example 8

Obtain the complex Fourier series of the followingfunction

2 44 2 0

f(t)

=e

t

2e

1

)(tf

t

Since , . Hence

Solution

[ ]

2

1

2

1

2

1

)(1

22

0

2

0

0

0

==

=

=

ee

dte

dttfT

c

t

t

T

1=2=T

)1(2

1

)1(2

1

)1(2

112

1

2

1

2

1

)(1

222)1(2

2

0

)1(

2

0

)1(

2

0

0

jn

e

jn

ee

jn

ejn

e

dtedtee

dtetfT

c

njjn

tjn

tjnjntt

T

tjn

n

=

=

=

=

==

=

since 1012sin2cos2

===

njnenj

jnt

nn

tjn

n ejn

eectf

=

=

==

)1(2

1)(

2

Therefore, the complex Fourier series off(t) is

0

2

0

2

0 2

1

)1(2

1c

e

jn

ec

n

nn=

=

=

=

=

*Notes: Even though c0 can be found by substituting

cn with n = 0, sometimes it doesnt works (as shown

in the next example). Therefore, it is always better tocalculate c0 alone.

2

2

12

1

n

e

cn

+

=

cn is a complex term, and it depends on n.Therefore, we may plot a graph of|cn| vs n.

In other words, we have transformed the function

f(t)in the time domain (t), to the function cn in the

frequency domain (n).

Example 9

Obtain the complex Fourier series of the function inExample 1.


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14

Solution

2

11

2

1)(

11

00

0 === dtdttfTc

T

=

)1(22

1

012

1)(

1

1

0

2

1

1

00

=

=

+==

jntjn

tjn

T

tjn

n

en

j

jn

e

dtedtetfT

c

)1(2

=

jn

ne

n

jc

Butnjn

nnjne )1(cossincos ===

Thus,

==even,0

odd,/]1)1[(

2 n

nnj

n

j n

Therefore,

=

=

==

odd0

2

1)(

nn

n

tjn

n

tjn

n en

jectf

*Here notice that . 00 cc nn =

=

even,0

odd,1

n

n

ncn

The plot of|cn| vs n is shown below

2

10=c

0.5

The Fourier Transform

BET2533 Eng. Math. III

R. M. Taufika R. Ismail

FKEE, UMP

Introduction Fourier transform is another method to transform a

signal from time domain to frequency domain

The basic idea of Fourier transform comes from thecomplex Fourier series

Practically, many signals are non-periodic Fourier transform use the principal of the Fourier

series, with assumption that the period of the non-

periodic signal is infinity (T).

Recall the complex form of Fourier series:

=

=

n

tjn

nectf0)(

=

2/

2/

0)(1

T

T

tjn

n dtetfT

c

where

(2.1)

(2.2)

(2.3)

Substituting (2.2) into (2.1) gives

=

=

n

tjn

T

T

tjnedtetf

Ttf 00

2/

2/

)(1

)(


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15

The spacing between adjacent harmonics is

Tnn

2)1( 000 ==+=

or

2

=T

Then (2.3) becomes:

=

=

=

=

=

=

n

tjn

T

T

tjn

n

tjnT

T

tjn

n

tjnT

T

tjn

edtetf

edtetf

edtetfT

tf

00

00

00

2/

2/

2/

2/

2/

2/

)(2

1

)(2

)(1

)(

Now if we let T, the summation becomesintegration, becomes , and the discreteharmonic frequency n0becomes a

continuous frequency , i.e.:

=

0n

d

n

Thus, as T,

=

=

dedtetf

edtetftf

tjtj

n

tjn

T

T

tjn

)(2

1

)(2

1)( 00

2/

2/

)(F

Therefore

=

deFtf tj)(2

1)(

where

=

detfF tj)()(

We say thatF() is the Fourier transform off(t), andf(t) is the inverse Fourier transform of

F()

Or in mathematical expression,

==

deFFtf tj)(

2

1)}({)( 1F

and

==

detftfF tj)()}({)( F

where Fand F1 is the Fourier transform and

inverse Fourier transform operator

respectively:F

1F

)(F)(tf


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16

Generally, the Fourier transformF() existswhen the Fourier integral converges

A condition for a functionf(t) to have a Fouriertransform is,f(t) can be completely integrable,i.e.

This condition is sufficient but not necessary 0.

je

??sincos ==

jej

+ )( jae

00)(

===+ jjja

eeee


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17

Example 1

Using the definition, find the Fourier transformof(t). Then deduce the Fourier transform of 1.

Solution

1)()}({)( 00 =====

eedtettFjtj

F

The Fourier transform off(t) = (t) is

*Recall the sifting property: )()()( afdttfat =

Then, using the duality principle, the Fouriertransform of1 is

)(2)(2)(2}1{ === fF

since . )()( =

Example 2

Find the Fourier transform of the followingfunction:

(a) (b) (c) tcos)( at tje

Solution

ajtj edteatat

== )()}({F

(a) Using the definition,

(b) From the result in (a), by replacing = a,

and using the duality principle,

)(2))((2)(2}{ === fe tjF

(c) From the result in (b),

[ ]

[ ]

[ ])()(

)(2)(22

1

}{}{2

1

2}{cos

++=

++=

+=

+

=

tjtj

tjtj

ee

eet

FF

FF

Example 3

Using the definition, find the Fourier transform

of the following:

(a) (b)

where Rand > 0.

)(tue t ||te


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18

Solution

jjj

e

dtedteetue

tj

tjtjtt

+

=

+

=

+=

==

+

+

1

)(

10

)(

)}({

0

)(

0

)(

0

F

(a)

(b)

22

0

)(0

)(

0

)(

0

)(

0

0

||

211

)(

}{

+

=

+

+

=

++

=

+=

+=

+

+

jj

j

e

j

e

dtedte

dteedteee

tjtj

tjtj

tjttjttF

Properties of the Fourier transform

Linearity Time scaling Time shifting Frequency shifting Time differentiation Frequency differentiation Duality Convolution

)()()}()({ bGaFtbgtaf =F

1. Linearity

IfF() = F{f(t)}, G() = F{g(t)}, a and b are

constants, then:

=

aF

aatf

||

1)}({F

2. Time scaling

)()}({ 00

Fettftj

=F

that is, a delay in the time domain corresponds to

a phase shift in the frequency domain.

)(||)|(|)1()( == FFFe j

= ||)( FF*Phase shift: Say . Then

.

3. Time shifting

)(})({ 00

= Fetftj

F

4. Frequency shifting

)()()(

Fjdt

tfd nn

n

=

F

5. Time differentiation

n

nnn

d

Fdjtft

)()}({ =F

6. Frequency differentiation

)(2)}({ = ftFF

7. Duality


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19

)()()}()({ GFtgtf =F

8. Convolution

)()(2

1)}()({

GFtgtf =F

Example 4

Derive the Fourier transform of a singlerectangular pulse of width and heightA, as

shown below:

Solution

2sinc

2/

)2/sin()2/sin(2

)(

)()}({)(

2/2/

2/

2/

2/

2/

AAA

j

eeA

j

eAdtAe

dtetftfF

jj

tjtj

tj

===

=

==

==

F

Using the definition of the Fourier transform,

Example 5

From the result in Example 4, using the time

shifting theorem, obtain the Fourier transform

of the functiong(t):

Then write as shown below:

Solution

)()()( 21 tgtgtg +=

SetA = 1 and = 1 of the functionf(t) to get thefollowing waveform:

= +

2sinc)(

=F

where and . Hence,

Therefore, the Fourier transform ofg(t) is

)()(21

1 += tftg )()( 21

2 = tftg

)()(

)()()(

21

21

21

+=

+=

tftf

tgtgtg

j

jj

Fee

FeFe

tftf

tgG

jj

jj

)1(cos2)2/(sin4

2sinc

2sin2

)()(

)()(

)}({)}({

)}({)(

2

2/2/

2/2/

21

21

===

=

=

+=

=

FF

F

[time shifting theorem]


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20

Method ofDifferentiation in time

domain

The simplest Fourier transform is on the deltafunction, where F{(t)} = 1

Using this idea, before we transformed afunction, we differentiate it until its derivative isexpressed in delta functions form

The important properties in implementing thismethod are:

)()()(

Fjdt

tfd nn

n

=

F )()}({ 00 Fettf tj=Fand

Example 6

Using time differentiation, find the Fourier

transform of the following function

Solution

The idea is we differentiatef(t) until its

derivative is expressed in the terms of delta

functions. This required twice differentiation

as shown below:

dt

df

2

2

dt

fd

f= t+ 1 f= 1 t

Hence, )1()(2)1()(

2

2

+

+= tttdt

tfd

Taking the Fourier transform for both sides of eqn,

2)()(

2)()(

)}1()(2)1({)(

2

2

2

2

+=

+=

++=

jj

jj

eeF

eeFj

tttdt

tfdFF

2

)cos1(2)(

=FThus,

Example 7Given the Fourier transform off(t) is

jF

+

=

5

3)(

Determine the Fourier transform of the

following:

(a) (b)

(c) (d)

2

tf )12( tf

)()(cos tft )(tfe t


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21

Solution

(a) Using time scaling property,

25

6

)2(5

32)2(2

1

221

21

jjF

Ft

f

+

=

+

==

=

F

(b) This involve the time scaling and time

shifting properties.))

j

ee

j

eF

tftf

jj

j

+

=

+

=

=

=

10

3

)2/(5

3

2

1

22

1

2)12(

2/2/

2/

21FF

* Note that the time shift is , not 1. The timeshift must be determined when the

coefficient oftis 1.

(c) From the Eulers identity,

[ ])()(2

1)(

2)()(cos tfetfetf

eetft

jtjtjtjt

+=

+=

Then, using the frequency shifting property,

{ } { } { }[ ]

[ ]

++

+

+=

++=

+=

)1(5

3

)1(5

3

2

1

)1()1(2

1

)()(

2

1)()(cos

jj

FF

tfetfetft jtjt FFF

(d) The frequency shifting property cannot

be apply directly, since there are noterm e jat. However, note that j 2 = 1,

that is

Then, using the frequency shifting property,

jjj

jF

tfetfe jjtt

+

=

+

=

=

=

6

3

)(5

3

)(

)()( FF

)()()(2

tfetfetfe jjttjt ==

The convolution integralA linear system:

h(t) y(t) =x(t)*h(t)x(t)

In time domainH() Y() =X()H()X()

In frequency domain

Convolution :

== dtfgdtgftgtf )()()()()()(

Which means the integral is evaluated indomain


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22

The steps of convolution:1. Create a dummy variable (i.e. ) and makeeach waveform a function of.

2. Time-invert one of the waveforms (doesnt

matter which).

3. Shift the inverted waveform by t. this allow

the function to travel along -axis.

)()( ftf )()( gtg

)()( gg

)())(()( = tgtgg

4. Find the product of the intersection

5. For a given time t, integrate the product for

the intersection range a < < b.

)()(

tgf

b

a

dtgf )()(

6. Combine all the integrals after the travelling

waveform moved for < t< .

Example 8

Find the convolution of the two signals:

Solution

1. Writefandgas functions of.

g()

=

f() g()

2. Choose one of the functions as the

travelling wave. Suppose that we choosef.Reflectf() about the vertical axis to getf().

f() g()

g()

=

3. Writef() asf(t) . Since tvaries from

to +, it means we bring f to and shift it

to + direction.

4&5. Along the journey off(t), observe the

intersections betweenfandg.

Then integrate along the intersection ranges:

f(t) g()

tt 1 0 1

f(t) g()

tt 1

(i) < t< 0:No overlap. f*g= 0

(ii) 0 < t< 1

0 1

22))(1(

2

0

2

0

tdgf

tt

=

==


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23

f(t

)g()

tt 1

(iii) 1 < t< 2

0 1

)2(22

))(1(

1

1

21

1

tt

dgf

tt

=

==

(iv) 2 < t< :No overlap. f*g= 0

6. Combine all the integral for < t< to

obtainf * g:

f*g(t)

t0 2

>


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24

h(t)

t t+ 1t 1

(iii) 1 < t< 0

ttddhx

t

=++=+= +

)01()10()1)(1()1)(1(

1

0

0

1

x(

)

0

11

1

1

h(t)

t t+ 1t 1

(iv) 0 < t< 1

ttddhx

t

=++=+=

)01()10()1)(1()1)(1(

1

0

0

1

x()

0

11

1

1

h(t)

t t+ 1t 1

(v) 1 < t< 2

2)1(1)1)(1(

1

1

+===

ttdhx

t

x()

0

11

1

1

(vi) 2 < t< :No overlap. x *h = 0

Combine all the integral for < t< to

obtainx * h :

x*h (t)

t

0

2


25/46

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25

Introduction

Jean BaptisteJoseph Fourier(*1768-1830)

French MathematicianLa Thorie Analitique

de la Chaleur (1822)

145

Fourier Series

Any periodic function can be expressed as asum of sines and/or cosines

Fourier Series

146

Fourier Transform

Even functions thatare not periodichave a finite area under curvecan be expressed as an integral of sines and

cosines multiplied by a weighing function

Both the Fourier Series and the FourierTransform have an inverse operation:

Original Domain Fourier Domain147

Contents

Complex numbers etc. ImpulsesFourier Transform (+examples)Convolution theoremFourier Transform of sampled functionsSampling theoremAliasingDiscrete Fourier TransformApplication Examples 148

Complex numbersComplex number

Its complex conjugate

149

Complex numbers polarComplex number in polar coordinates

150


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26

Eulers formula

151

Sin ()

Cos ()

?

?

152

Re

Im

Complex math

Complex (vector) addition

Multiplication with i

is rotation by 90 degrees

153

Contents

Complex number etc. ImpulsesFourier Transform (+examples)Convolution theoremFourier Transform of sampled functionsSampling theoremAliasingDiscrete Fourier TransformApplication Examples 154

Unit impulse (Dirac delta function)Definition

ConstraintSifting propertySpecifically for t=0

155

Discrete unit impulseDefinition

ConstraintSifting propertySpecifically for x=0

156


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27

What does this look like?

Impulse train

157

T = 1

Contents


Fourier Series

with

159

Series of

sines and

cosines,

see Eulers

formula

Periodic

with

period T

Fourier Transform

Continuous Fourier Transform (1D)

Inverse Continuous Fourier Transform (1D)

160

161

Symmetry: The only difference between the FourierTransform and its inverse is the sign of the exponential.

Fourier Euler

Fourier and Euler


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28

If f(t) is real, then F() is complexF() is expansion of f(t) multiplied by sinusoidal

terms

t is integrated over, disappearsF() is a function of only , which determines

the frequencyof sinusoidals

Fourier transform frequency domain163

Examples Block 1

164

-W/2 W/2

A

Examples Block 2

165

Examples Block 3

166

?

Examples Impulse

167

constant

Examples Shifted impulse

168

Euler


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29

Examples Shifted impulse 2

169

Real part Imaginary part

impulse constant

Examples - Impulse train

170

Examples - Impulse train 2

171

Intermezzo: Symmetry in the FT

172

173

ContentsComplex number etc. ImpulsesFourier Transform (+examples)Convolution theoremFourier Transform of sampled functionsSampling theoremAliasingDiscrete Fourier TransformApplication Examples 174


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30

Fourier + Convolution

What is the Fourier domain equivalent ofconvolution?

175

What is

176

Intermezzo 1

Whatis ?

Let , so

177

Intermezzo 2

Property of Fourier Transform

178

Fourier + Convolution cont

d

179

Recapitulation 1Convolution in one domain is multiplication in

the other domain

And (see book)

180


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31

Recapitulation 2

Shift in one domain is multiplication withcomplex exponential in the other domain

And (see book)

181

Contents


Sampling

Sampled function can be written as

Obtain value of arbitrary sample k as

183

Sampling - 2

184

Sampling - 3

185

Sampling - 4

186


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32

FT of sampled functions

Fourier transform of sampled function

Convolution theoremFrom FT of impulse train

187

(who?)

FT of sampled functions

188

Sifting property

189

Contents


Sampling theoremBand-limited function

Sampled function lower value of 1/T

would cause triangles

to merge

191

Sampling theorem 2Sampling theorem:If copy of can be isolated from the periodic

sequence of copies contained in

, can be completely recoveredfrom the sampled version.

Since is a continuous, periodic functionwith period 1/T, one complete period from

is enough to reconstruct the signal.

This can be done via the Inverse FT.192


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33

Extracting a single period from that isequal to is possible ifSampling theorem 3

193

Nyquist frequency

Contents


Aliasing

If , aliasing can occur

195

Contents


Discrete Fourier TransformFourier Transform of a sampled function is an

infinite periodic sequence of copies of the

transform of the original continuous function

197

Discrete Fourier TransformContinuous transform of sampled function

198


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34

is discrete function is continuous and infinitely periodic with

period

1/T199

We need only one period to characterise If we want to take M equally spaced samples

from

in the period = 0 to = 1/, this can

be done thus

200

Substituting

Into

yields201

Contents


ExamplesCardiac tagging analysisDither removal

203

Formulation in 2D spatial

coordinatesContinuous Fourier Transform (2D)


204


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35

The Fourier Transform II

Contents

Fourier Transform of sine and cosine 2D Fourier Transform Properties of the Discrete Fourier Transform

206

207

Eulers formula

208

Recall

209

Recall

210

Cos(t)

Sin(t)

1/2 1/2i


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36

Contents


211 212

Formulation in 2D spatial coordinates

Continuous Fourier Transform (2D)


with angular frequencies

Discrete Fourier Transform

213

Forward

Inverse

214

Formulation in 2D spatial coordinates

Discrete Fourier Transform (2D)

Inverse Discrete Transform (2D)

Spatial and Frequency intervals Inverse proportionality (Smallest) Frequency step depends on largest

distance covered in spatial domain

Suppose function is sampled M times in x,with step , distance is covered,

which is related to the lowest frequency that

can be measured

215

Examples

216


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37

Fourier Series

with

217

Series of

sines and

cosines,

see Eulers

formula

Periodicwith

period T

Examples

218

Periodicity

2D Fourier Transform is periodic in bothdirections

219

Periodicity

2D Inverse Fourier Transform is periodic inboth directions

220

Fourier Domain

221

Inverse Fourier Domain Periodic?

222

Periodic!


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38

Contents


223

Properties of the 2D DFT

224

225

Real

ImaginarySin (x)Sin (x+/2)

Real

226

Real

ImaginaryF(Cos(x))F(Cos(x)+k)

Even

227

Real

Odd

Sin (x)Sin(y)Sin (x)Imaginary

228

Real

Imaginary(Sin (x)+1)(Sin(y)+1)


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39

Symmetry: even and odd

Any real or complex function w(x,y) can beexpressed as the sum of an even and an odd

part (either real or complex)

229

Properties

Even function

Odd function

230

Properties - 2

231

Consequences for the Fourier Transform

FT of real function is conjugate symmetric

FT of imaginary function is conjugateantisymmetric

232

Im Re


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40

Re Im

FT of even and odd functions

FT of even function is real

FT of odd function is imaginary

237 238

Real

ImaginaryCos (x)

Even

239

Real

ImaginarySin (x)

Odd

Fourier Series & The Fourier Transform

What is the Fourier Transform?

Fourier Cosine Series for even functions

Fourier Sine Series for odd functions

The continuous limit: the Fourier

transform (and its inverse)

Some transform examples and the Dirac

delta function

( ) ( ) exp( )F f t i t dt

= 1

( ) ( ) exp ( )2

f t F i t d

= Prof. Rick Trebino, Georgia Tech


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41

What do we hope to achieve with the Fourier

Transform?

We desire a measure of the frequencies present in a wave. This willlead to a definition of the term, the spectrum.

Plane waves have only

one frequency, .

This light wave has manyfrequencies. And the

frequency increases in

time (from red to blue).

It will be nice if our measure also tells us when each frequency occurs.

Lightelectric

field

Time

Lord Kelvin on Fouriers theorem

Fouriers theorem is notonly one of the most

beautiful results of modern

analysis, but it may be said

to furnish an indispensable

instrument in the treatmentof nearly every recondite

question in modern physics.

Lord Kelvin

Joseph Fourier

Fourier was obsessed

with the physics of heat

and developed the

Fourier series andtransform to model

heat-flow problems.

Joseph Fourier 1768 - 1830

Anharmonic waves are sums of sinusoids.

Consider the sum of two sine waves (i.e., harmonic waves) of

different frequencies:

The resulting wave is periodic, but not harmonic.

Essentially all waves are anharmonic.

Fourier

decomposingfunctions

Here, we write a

square wave as

a sum of sine waves.

sin(t)

sin(3t)

sin(5t)

Any function can be written as the

sum of an even and an odd function.

E(-x) = E(x)

O(-x) = -O(x)

E(x)O(x)

f(x)


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42

Fourier Cosine Series

Because cos(mt) is an even function (for all m), we can write an even function,

f(t), as:

where the set {Fm; m = 0, 1, } is a set of coefficients that define the series.

And where well only worry about the functionf(t) over the interval

(,).

f(t) =1

!

Fm cos(mt)

m=0

"

#

The Kronecker delta function

Finding the coefficients, Fm, in a Fourier Cosine Series

Fourier Cosine Series:

To findFm, multiply each side by cos(mt), where m is another integer, and integrate:

But:

So: only the m = m term contributes

Dropping the from the m:

yields the

coefficients for

anyf(t)!

0

1( ) cos( )m

m

f t F mt

=

=

0

1( ) cos( ' ) cos( ) cos( ' )

m

m

f t m t dt F mt m t dt

=

=

, '

'co s( ) cos( ' )

0 'm m

if m mmt m t dt

if m m

==

, '

0

1( ) cos( ' ) m m m

m

f t m t dt F

=

=

( ) cos( )mF f t mt dt

=

Fourier Sine Series

Because sin(mt) is an odd function (for all m), we can write

any odd function,f(t), as:

where the set {Fm; m = 0, 1, } is a set of coefficients that define theseries.

where well only worry about the functionf(t) over the interval (,).

f(t) =1

!

"Fm sin(mt)

m=0

#

$

Finding the coefficients, Fm, in a Fourier Sine Series

Fourier Sine Series:

To findFm, multiply each side by sin(mt), where m is another integer, and integrate:

But:

So: only the m = m term contributes

Dropping the from the m: yields the coefficients

for anyf(t)!

0

1( ) sin( )m

m

f t F mt

=

=

0

1( ) sin( ' ) sin( ) sin( ' )m

m

f t m t dt F mt m t dt

=

=

, '

'sin( ) sin( ' )

0 'm m

if m mmt m t dt

if m m

==

, '

0

1( ) sin( ' ) m m m

m

f t m t dt F

=

=

( ) sin( )mF f t mt dt

=

Fourier Series

even component odd component

where

and

0 0

1 1( ) cos( ) sin( )m m

m m

f t F mt F mt

= =

= +

Fm = f(t) cos(mt) dt! !Fm = f(t) sin(mt) dt"

So iff(t) is a general function, neither even nor odd, it can be written:


43/46

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43

We can plot the coefficients of a Fourier Series

We really need two such plots, one for the cosine series and another for the

sine series.

Fmvs. m

m5 25

201510

30

1.50

Discrete Fourier Series vs.

Continuous Fourier Transform

Fmvs. m

mAgain, we really need two such plots, one for the cosine series and another

for the sine series.

Let the integermbecome a real

number and let

the coefficients,Fm, become afunction F(m).

F(m)

The Fourier TransformConsider the Fourier coefficients. Lets define a function F(m) that

incorporates both cosine and sine series coefficients, with the sine seriesdistinguished by making it the imaginary component:

Lets now allow f(t) to range from to , so well have to integrate from

to , and lets redefine m to be the frequency, which well now call :

F() is called the Fourier Transform off(t). It contains equivalent

information to that in f(t). We say that f(t) lives in the time domain, and F

() lives in the frequency domain. F() is just another way of looking at a

function or wave.

( ) cos ( )f t mt dt ( ) s in( )i f t mt dt F(m) Fm iFm =

( ) ( ) exp ( )F f t i t dt

= The FourierTransform

The Inverse Fourier TransformThe Fourier Transform takes us from f(t) toF().How about going back?

Recall our formula for the Fourier Series off(t) :

Now transform the sums to integrals from to , and again replaceFm

withF

(

). Remembering the fact that we introduced a factor ofi

(andincluding a factor of 2 that just crops up), we have:

'

0 0

1 1( ) cos( ) sin( )m m

m m

f t F mt F mt

= =

= +

1( ) ( ) exp( )

2f t F i t d

= Inverse

Fourier

Transform

The Fourier Transform and its Inverse

The Fourier Transform and its Inverse:

So we can transform to the frequency domain and back. Interestingly,

these transformations are very similar.

There are different definitions of these transforms. The 2 can occur inseveral places, but the idea is generally the same.

Inverse Fourier Transform

FourierTransform( ) ( ) exp( )F f t i t dt

=

1( ) ( ) exp( )

2f t F i t d

=

There are several ways to denote the Fourier transform of a function.

If the function is labeled by a lower-case letter, such asf,we can write:

f(t) F()

If the function is already labeled by an upper-case letter, such asE, wecan write:

or:

Fourier Transform Notation

( ) ( )E t E %( ) { ( )}E t E tF

Sometimes, this symbol isused instead of the arrow:


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44

The Spectrum

We define the spectrum, S(), of a waveE(t) to be:

2

( ) { ( )}S E t F

This is the measure of the frequencies present in a light wave.

Example: the Fourier Transform of a

rectangle function: rect(t)1/2

1/ 2

1/ 2

1/2

1( ) exp( ) [exp( )]

1[exp( / 2) exp(

e xp ( / 2 ) e xp (

2

sin(

F i t dt i ti

i ii

i i

i

= =

= /2)]

1 /2)=

( /2)

/2)=

( /2)

( sinc(F ) = /2)Imaginary

Component = 0

F()

Sinc(x) and why it's important

Sinc(x/2) is the Fouriertransform of a rectangle

function.

Sinc2(x/2) is the Fouriertransform of a t riangle

function.

Sinc2(ax) is the diffractionpattern from a slit.

It just crops up everywhere...

The Fourier Transform of the triangle

function, (t), is sinc2(/2)

0

2sinc ( / 2)

1

t0

( )t

1

1/2-1/2

The triangle function is just what it sounds like.

Well prove this when we learn about convolution.

Sometimes

people use

(t), too, for the

triangle

function.


decaying exponential: exp(-at) (t> 0)0

0 0

0

( exp( ) exp( )

exp( ) exp( [ )

1 1exp( [ ) [exp( ) exp(0)]

1 1[0 1]

F at i t dt

at i t dt a i t dt

a i ta i a i

a i a i

+

) =

= = + ]

= + ] =

+ +

= =

+ +

A complex Lorentzian!


Gaussian, exp(-at2

), is itself!

2 2

2

{exp( )} exp( ) exp( )

exp( / 4 )

at at i t dt

a

=

F

t0

2exp( )at

0

2exp( / 4 )a

The details are a HW problem!


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45

The Dirac delta function

Unlike the Kronecker delta-function, which is a function of two integers,

the Dirac delta function is a function of a real variable, t.

t

(t)

The Dirac delta function

Its best to think of the delta function as the limit of a series of peaked

continuous functions.

if 0( )

0 if 0

tt

t

=

t

f1(t)

f2(t)

fm(t) = m exp[-(mt)2]/

f3(t)

(t)

Dirac -function Properties( ) 1t dt

= t

(t)

( ) ( ) ( ) ( ) ( )t a f t dt t a f a dt f a

= =

exp( ) 2 (

exp[ ( ) ] 2 (

i t dt

i t dt

= )

= )

2()

The Fourier Transform of(t) is 1.

1 exp( ) 2 (i t dt

= )And the Fourier Transform of1 is 2():

( ) exp( ) exp( [0]) 1t i t dt i

= =

t

(t)

1

t

1

0

0

The Fourier transform ofexp(i0t)

{ }0 0exp( ) exp( ) exp( )i t i t i t dt

= F

0exp( [ ] )i t dt

=

The function exp(i0t) is the essential component of Fourier analysis. It is

a pure frequency.

F {exp(i0t)}

0 0

02 ( ) =

exp(i0t)

0t

tRe

Im

0

The Fourier transform ofcos(0t){ }0 0cos( ) cos( ) exp( )t t i t dt

= F

[ ]0 01

exp( ) exp( ) exp( )2

i t i t i t dt

= +

0 0

1 1exp( [ ] ) exp( [ ] )

2 2i t dt i t dt

= + +

0 0( ) ( ) = + +

+00 -

0

0{cos( )}tFcos(

0t)

t0


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Fourier Transform Symmetry Properties

Expanding the Fourier transform of a function,f(t):

( ) Re{ ( )} cos( ) Im{ ( )} sin( )F f t t d t f t t d t

= + Re{F()}

Im{F()}

= 0 ifRe{f(t)} is odd = 0 ifIm{f(t)} is even

Even functions of Odd functions of

( ) [Re{ ( )} Im{ ( )}] [cos( ) sin( )]F f t i f t t i t dt

= +

Im{ ( )} cos( ) Re{ ( )} sin( )i f t t dt i f t t dt

+

= 0 ifIm{f(t)} is odd = 0 ifRe{f(t)} is even

Expanding more, noting that: ( ) 0O t dt

= ifO(t) is an odd function

Some functions dont have Fouriertransforms.

The condition for the existence of a givenF() is:

Functions that do not asymptote to zero in both the + and

directions generally do not have Fourier transforms.

So well assume that all functions of interest go to zero at .

( )f t dt

<

the fourier series edited

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