Fourier Transforms of Special Functions

Fourier Transforms of Special Functions

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Content Introduction More on Impulse Function Fourier Transform Related to Impulse Function Fourier Transform of Some Special Functions Fourier Transform vs. Fourier Series

Introduction Sufficient condition for the existence of a

Fourier transform

dttf |)(|

That is, f(t) is absolutely integrable. However, the above condition is not the

necessary one.

Some Unabsolutely Integrable Functions

Sinusoidal Functions: cos t, sin t,…Unit Step Function: u(t).

Generalized Functions:– Impulse Function (t); and– Impulse Train.


More onImpulse Function

Dirac Delta Function

000

)(tt

t and 1)(

dtt

0 t

Also called unit impulse function.

Generalized Function The value of delta function can also be defined

in the sense of generalized function:

)0()()(

dttt (t): Test Function

We shall never talk about the value of (t). Instead, we talk about the values of integrals

involving (t).

Properties of Unit Impulse Function

)()()( 00 tdtttt

Pf)

dtttt )()( 0

Write t as t + t0

dtttt )()( 0

)( 0t


)0(||

1)()(

adttat

Pf)

dttat )()(

Write t as t/aConsider a>0

dt

att

a)(1

)0(||

1

a

dttat )()(

Consider a<0

dt

att

a)(1

)0(||

1

a


)()0()()( tfttf

Pf)

dttttf )()]()([

dtttft )]()()[(

)0()0( f

dtttf )()()0(

dtttf )()]()0([


)()0()()( tfttf

Pf)

dttat )()(

)(||

1)( ta

at

)0(||

1

a

dttt

a)()(

||1

dttt

a)()(

||1


)()0()()( tfttf )(

||1)( ta

at

0)( tt )()( tt

Generalized Derivatives

The derivative f’(t) of an arbitrary generalized function f(t) is defined by:

dtttfdtttf )(')()()('

Show that this definition is consistent to the ordinary definition for the first derivative of a continuous function.

dtttf )()(' dtttfttf

)(')()()(

=0

Derivatives of the -Function

)0(')(')()()('

dtttdttt

0

)()0(' ,)()('

tdttd

dttdt

)0()1()()( )()( nnn dttt

0

)()( )()0( ,)()(

tn

nn

n

nn

dttd

dttdt

Product Rule)(')()()(')]'()([ ttfttfttf

dttttf )(')]()([

Pf)dttttf )(')]()([

dtttft )](')()[(

dtttfttft )}()(')]'()(){[(

dtttftdtttft )]()'()[()]'()()[(

dtttftdtttft )]()'()[()]()()[('

dtttfttft )()](')()()('[

Product Rule)()0(')(')0()(')( tftfttf

)()'()]'()([)(')( ttfttfttf

Pf)

)]'()0([ tf )(')0( tf

)()0(' tf

Unit Step Function u(t)

Define

0)()()( dttdtttu

0 t

u(t)

0001

)(tt

tu

Derivative of the Unit Step Function

Show that )()(' ttu

dtttu )()('

0)(' dtt

)]0()([ )0(

dtttu )(')(

dttt )()(

Derivative of the Unit Step Function

0 t

u(t)Derivative

0 t

(t)


Fourier Transform Related toImpulse Function

Fourier Transform for (t)

1)( Ft

dtett tj)()]([F 10

t

tje

0 t

(t)

0

1

F(j)

F


Show that

det tj

21)(

]1[)( 1 Ft

de tj121

de tj

21

de tj

21

The integration converges to

in the sense of generalized function.

)(t


Show that

0

cos1)( tdt

det tj

21)(

dtjt )sin(cos

21

tdjtd sin

2cos

21

0

cos1 td Converges to (t) in the sense of generalized function.

Two Identities for (t)

dxey jxy

21)(

0cos1)( xydxy

These two ordinary integrations themselves are meaningless.

They converge to (t) in the sense of generalized function.

Shifted Impulse Function

0)( 0tjett F

0)()]([ 0tjejFttf F

0

1

|F(j)|

F

Use the fact

0 t

(t t0)

t0


Fourier Transform of a Some Special Functions

Fourier Transform of a Constant

)(2)()( AjFAtf F

dAeAjF tj][)( F

dteA tj )(

212

)(2 A

Fourier Transform of a Constant

)(2)()( AjFAtf F

F

0 t

A A2()

0

F(j)

Fourier Transform of Exponential Wave

)(2)()( 00 jFetf tj F

)(2]1[ F

)]([])([ 00 jFetf tjF

)(2][ 00 tjeF

Fourier Transforms of Sinusoidal Functions

)()(cos 000 Ft

)()(sin 000 jjt F

F

(+0)

0

F(j)(0)

0 0

t

f(t)=cos0t

Fourier Transform of Unit Step Function

)()]([ jFtuFLet )()]([ jFtuF)0for (except 1)()( ttutu

]1[)]()([ FF tutu

)(2)]([)]([ tutu FF)(2)()( jFjF

F(j)=?

Can you guess it?


)(2)()( jFjF

Guess )()()( BkjF

)()()()()()( BBkkjFjF

)()()(2 BBk

k

0B() must be odd


Guess )()()( BkjF k

)()(' ttu

)()]([ jFtuF1)]([)]('[ ttu FF

)()]('[ jFjtuF)]()([ Bj

)()( Bjj

0

jB 1)(


Guess )()()( BkjF k

jB 1)(

jtu 1)()( F


jtu 1)()( F

F()

0

|F(j)|

0 t

1

f(t)


Fourier Transform vs. Fourier Series

Find the FT of a Periodic Function

Sufficient condition --- existence of FT

dttf |)(|

Any periodic function does not satisfy this condition.

How to find its FT (in the sense of general function)?


We can express a periodic function f(t) as:

Tectf

n

tjnn

2 ,)( 00

n

tjnnectfjF 0)]([)( FF

n

tjnn ec ][ 0F

n

n nc )(2 0

n

n nc )(2 0


We can express a periodic function f(t) as:

Tectf

n

tjnn

2 ,)( 00

n

n ncjF )(2)( 0

The FT of a periodic function consists of a sequence of equidistant impulses located at the harmonic frequencies of the function.

Example:Impulse Train

0 tT 2T 3TT2T3T

n

T nTtt )()( Find the FT of the impulse train.


0 tT 2T 3TT2T3T

n


n

tjnT e

Tt 0

1)(

c n


0 tT 2T 3TT2T3T

n


n

tjnT e

Tt 0

1)(

c n

n

T nT

t )(2)]([ 0F 0


0 tT 2T 3TT2T3T

n

T nT

t )(2)]([ 0F 0

0 0 20 3002030

2/T

F

Find Fourier Series Using Fourier Transform

n

tjnnectf 0)(

2/

2/0)(1 T

T

tjnn etf

Tc

T/2 T/2

f(t)t

T/2 T/2

fo(t)t

tjoo etfjF )()(

2/

2/)(

T

T

tjetf

)(10 jnF

Tc on

Find Fourier Series Using Fourier Transform

n

tjnnectf 0)(

2/

2/0)(1 T

T

tjnn etf

Tc

T/2 T/2

f(t)t

T/2 T/2

fo(t)t

tjoo etfjF )()(

2/

2/)(

T

T

tjetf

)(10 jnF

Tc on

Sampling the Fourier Transform of fo(t) with period 2/T, we can find the Fourier Series of f (t).

Example:The Fourier Series of a Rectangular Wave

0

f(t)

d1

t0

t

fo(t)1

dtejFd

d

tjo

2/

2/)(

2sin2 d

n

tjnnectf 0)(

)(10 jnF

Tc on

2sin2 0

0

dnTn

2sin1 0dn

n

Example:The Fourier Transform of a Rectangular Wave

0

f(t)

d1

t

n

tjnnectf 0)(

)(10 jnF

Tc on

2sin2 0

0

dnTn

2sin1 0dn

n

F [f(t)]=?

n

n ncjF )(2)( 0

)(2

sin2)( 00

ndnn

jFn

Fourier Transforms of Special Functions

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