Unit III - KHIT

KALLAM HARANADHAREDDY INSTITUTE OF TECHNOLOGY (Affiliated to JNTUK, Kakinada; Approved by AICTE, New Delhi)

NH-5,Chowdavaram, Guntur-522019

MATHEMATICS – III Common to all Branches

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Prepared by K.V.B. Rajakumar

I YEAR - II SEMESTER

Unit – III

FOURIER TRANSFORMS AND FOURIER SERIES

Objectives:

To introduce

Fourier transform of a given function and the corresponding inverse.

Fourier sine and cosine transform of a given function and their corresponding inverses.

Finite Fourier transforms of a given function and their corresponding inverses.

Syllabus:

Fourier Transforms: Fourier integral theorem (without proof) – Fourier sine and cosine

integrals – Sine and cosine transforms – Properties – inverse transforms – Finite Fourier

transforms.

Outcomes:

Students will be able to

Find the Fourier transform of the given function in infinite cases.

Find the Fourier sine and cosine transforms of the given function in infinite cases.

Learning Material

Fourier Transforms are widely used to solve Partial differential equations and in various

boundary value problems of Engineering such as vibration of strings, Conduction of heat,

Oscillation of an elastic beam, transmission lines so on..

1 FOURIER TRANSFORMS 2-31

2 FOURIER SERIES 32-44




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Fourier Integral Theorem:

If f(x) satisfies Dirichlet’s conditions for expansion of Fourier series in (-C, C) an ( )

f x dx

converges, then

1( ) ( )

2

f x f t cosα(t - x) dt dα is known as Fourier Integral of f(x).

Fourier Sine Integral: If f(x) satisfies Dirichlet’s conditions for expansion of Fourier series

in (-C, C) and ( )

f x dx converges, if f(t) is odd function then

0 0

2( ) ( )

f x f t sinαt sinα x dt dα .

Fourier Cosine Integral: If f(x) satisfies Dirichlet’s conditions for expansion of Fourier

series in (-C, C) and ( )

f x dx converges, if f (t) is even function then

0 0

2( ) ( )

f x f t Cosαt Cosα x dt dα .

Complex form or exponential form of Fourier Integral: The complex form of Fourier

integral is known as

i ( t x)1( ) ( )

2

f x f t e dt dα

Note:

2 20

2 20

a-axe Cosbxdx =a +b

b-axe Sin xdx =a +b

Examples:

1. Using Fourier integral, prove that 2 2

0

2

-ax a Cosαxe dα

a

The Fourier Cosine integral of f(x) is 0 0

2( ) ( )

f x f t Cosαt Cosα x dt dα

Let us take ( )-a t

f t = e , a > 0

0 0

2

-ax -a te e Cosαt dt Cosα x dt dα

0 0

2

-ax -a te e Cosαt Cosα x dt dα




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2 20

2

aCosα x dα

a

2 20

2 20

2a

2

-ax

Cosα xdα

a

aCosα xdα=e

a

2. Using Fourier integral, prove that 2 2

02

-axSin αx

dα= ea

The Fourier Sine integral of f(x) is 0 0

2( ) ( )

f x f t sinαt sinα x dt dα

Let us take ( )-a t

f t = e , a > 0

0 0

2

-ax -a te e Sinαt Sinα x dt dα

0 0

2

-ax -a te e Sinαt dt Sinα x dt dα

2 20

2

Sinα x dα

a

2 20

2 20

2

2

-ax

Sinα xdα

a

Sinα xdα=e

a

3. Using Fourier integral, prove that 2

4

0

2

2 4

-ax

Cosαxe Cosx dα

The Fourier Cosine integral of f(x) is 0 0

2( ) ( )

f x f t Cosαt Cosα x dt dα

Let us take ( )- t

f t = e Cost then

0 0

2

- x - t

e Cos x e Cost Cosαt Cosα x dt dα

0 0

2

- x - te Cos x e Cost Cosαt dt Cosα x dα

We have 2 Cos C Cos D = Cos (C+D) + Cos(C-D)

0 0

21 1-

- x - te Cos x Cos + t +Cos t e dt Cosα x dα

0 0 0

21 1-

- x - t - te Cos x e Cos + t dt + e Cos tdt Cosα x dα




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2 2

0

2 1 1

1 1 1 1

- xe Cos x Cosα x dα

2

4

0

2

2 4

-ax

Cosαxe Cosx dα

4. Using Fourier integral, prove that 3

4

04 2

-axSin αx

dα= e Cos x

0 0

2( ) ( )

f x f t sinαt sinα x dt dα

Let us take ( )- t

f t = e Cost

0 0

2( )

-axe Cos x f t sinαt sinα x dt dα

0 0

2

- t-axe Cos x e Cost sinαt sinα x dt dα

0 0

2

- t-axe Cos x Sinα x e Cost sinα t dt dα

0 0

2

2 S S 1 S 1

2 S S 1 S 1

- t-axe Cos x Sinα x e sinα t Cosα t dt dα

inα t Cosα t inα t inα t

inα t Cosα t inα t inα t

0 0

2S 1 S 1

- t-axe Cos x Sinα x e inα t inα t dt dα

0 0

2S 1 S 1

- t-axe Cos x Sinα x e inα t inα t dt dα

0 0 0

1S 1 S 1

- t - t-axe Cos x = Sinα xdα e inα t e inα t

2 2

0

1 1 1

1 1 1 1

-axe Cos x Sinα x dα

3

4

0

1 2

4

-axe Cos x Sinα x dα

3

4

04 2

-axSin αx

dα= e Cos x

5. Using Fourier integral, prove that

0,,

sin2

0

2222

22

badba

xabee bxax

Since the integrand on R.H.S contains sine term, we use Fourier sine integral formula.

We know that fouries sine integral for f(x) is given by

00

sin)(sin2

)( ptdtdptfpxxf




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Replacing p with λ we get

00

sin)(sin2

)(

tdtdtfxxf

Here f(x) = e-ax

- e- bx

→ f(t) = e-at

- e-bt

Substituting (2) in (1) , we get

dtdteexxf btat

00

sinsin2

)(

dttbb

etta

a

exxf

btat

0

22

0

22

0

cossincossinsin2

)(

d

baxxf

2222

0

sin2

)(

d

ba

abxxf

2222

22

0

sin2

)(

d

ba

xabee bxax

0

2222

22 sin2




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Problem: Find the Fourier transform of f(x) defined by 1, | |

( )0, | |

if x af x

if x a

And hence evaluate

0

sindp

p

p and

dpp

pxapcossin

Sol: We have F[f(x)]=

dxxfeipx )( =

a

ipx dxxfe )( +

a

a

ipx dxxfe )( +

a

ipx dxxfe )(

=

a

a

ipx dxe =p

apsin2

By the inversion formula, we know that f(x)= 2

1

dppFe ipx )(

2

1

dpp

ape ipx sin2

= 1, | |

0, | |

if x a

if x a

2

1

dpp

appx

sin2cos -

2

1

dpp

appx

sin2sin =

1, | |

0, | |

if x a

if x a

Since the second integral is an odd function,

dpp

pxapcossin= ,

1, | |

0, | |

if x a

if x a

Put x=0, we get, 2

1

dpp

apsin2=

1, 0

0, 0

if a

if a

0

sindp

p

p =

2

, a>0

= 0, a<0




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And put x=0 and a=1 then we get

0

sindp

p

p=

2

Problem: Find the fourier sine transform of e-ax

, a>0 and hence deduce that

0

22

sindp

pa

pxp

Sol: xfFs =

0

sin)( dxpxxf =

0

sin dxpxe ax=

22 pa

p

By the inversion formula, we know that f(x)=

2

0

sin dppxxfFs

=

2

0

22sin dppx

pa

p

0

22 2

sin axedppa

pxp




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FOURIER SERIES




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Documents

Unit III - KHIT