fourier series notes corrected.pdf

8/20/2019 Fourier Series notes corrected.pdf

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Fourier series

Preliminaries

Periodic functions:

A function ( ) f t is said to be periodic with period T if there exists a positive

constant T such that ( ) ( ) f t T f t for all .t

Note: If ( ) f t is periodic with period T , then ( ) ( ) f t nT f t for all integers.n

The smallest positive number T satisfying this property is called the primitive

period or simply the period of the function ( ). f t The graph of a period function ( ) f t with period T periodically repeats in an

interval of width .T Hence it is sufficient to study the properties (nature) of thefunction in an interval of length T , in particular in the interval [0, ],T which is

called one period of the function.

Example (1): The trigonometric functions ( ) sin f x x and ( ) cos f x x are periodic with period 2 .T

Graph of


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Graph of ( ) cos f x x

Example (2): The functions sin kt and coskt have period2

,T k

since

2 2sin ( sin( 2 ) sin . f t k t kt kt

k k

Let ( ) f t be a periodic function of t with the period .T Define a new variable x

as follows.

2

x t

T or

2 x t

T

i.e.,

2

T t x

Then ( ) ( )2

T f t f x g x

is a function of . x

Consider ( 2 ) ( 2 ) ) ( ) ( )2 2

T T g x f x f x T f t T f t

( )2

T f x g x

Therefore ( ) g x is a periodic function of x with period 2 .


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Note: If ( ) f x and ( ) g x are periodic functions of x with period1T and 2T

respectively, then 1 2( ) ( )c f x c g x is also periodic with period 1 2( , ).T lcm T T

Even an Odd functions:

A function ( ) f t is said to be even if ( ) ( ) f t f t and odd if ( ) ( ). f t f t

Note that the graph of an even function is symmetric about y axis, whereas thegraph of an odd function is symmetric about the origin.

Exmple (1):2( ) , f x x ( ) f x x and ( ) cos f x x are all even functions of

. x

Graph of2( ) f x x Graph of ( ) f x x Graph of ( ) cos f x x

Example (2): ( ) , sin , f x x x 3( ) , f x x ( ) sin , f x x ( ) tan f x x are allodd functions of . x

Graph of ( ) f x x Graph of ( ) sin f x x

x


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Products of two even and two odd functions are even and the product of an even

function with an odd function is odd.

Also, 0

2 ( ) , if ( ) is even

( )0, if ( ) is odd

a

a

a

f t dt f t

f t dt f t

.

Generalized formula for integration by parts

If u and v are functions of , x then

1 2 3 4' '' ''' ...uvdx uv u v u v u v

where ' ''' , '' , ''' ,...du du duu u udx dx dx

and

1 2 1 3 2, , ,...v vdx v v dx v v dx

Example:2 2 cos 2 sin 2 cos 2sin 2 2 2

2 4 8

x x x x x dx x x

Orthogonality of trigonometric functions:

0, 02cos

, 0

c T

c

nnt dt

T nT

2sin 0 for all

c T

c

nt dt n

T

2 2cos sin 0 for all and

c T

c

m nt t dt m n

T T


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0, m2 2

cos cos, m

2

c T

c

nm n

t t dt T T T n

0, m2 2

sin sin, m

2

c T

c

nm n

t t dt T T T n

Also,

cos 1 ,n

n for all n , sin 0,n for all n

2

0, oddcos

2 1 , evenn

nn

n

,

1

2

0, evensin

2 1 , oddn

nn

n

Definition: A function ( ) f t is said to be piecewise continuous in an interval

[ , ]a b if it is discontinuous at finite number of points in the interval and wherever

it is discontinuous, it has finite left and right hand limits.

Fourier series:

In solving many boundary valued problems involving ordinary and partialdifferential equations it is required to represent some functions as a sum of

trigonometric functions cosine and sine. Such a series representation of a function

( ) f t (which may be discontinuous); if exists, is called the trigonometric series

expansion of ( ) f t .

Fourier introduced such an expansion of periodic functions in terms of sine and

cosine functions and hence it is called a Fourier series expansion. Many functions

including some discontinuous periodic functions can be expanded in a Fourier

series and hence are, in certain sense more universal than Taylor series expansions,which cannot be established for discontinuous functions. Fourier series solution

method is a powerful tool in solving some ordinary and partial differential

equations given with the initial or boundary conditions.


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Definition:

Let ( ) f t be a periodic function of t with period T , and is defined in an interval

[ , ]c c T . Then the expansion of the form

0

1

2 2( ) cos sinn n

n

n n f t a a t b t

T T

, if exists called the Fourier series or

Fourier expansion of ( ) f t . Here 0, andn na a b are called Fourier coefficients.

Euler’s formulae:

Given a periodic function ( ) f t with period T , represented in [ , ]c c T by a Fourier

series,

0

1

2 2( ) cos sin (1)n n

n

n n f t a a t b t

T T

, to determine

the coefficients0, andn na a b , we proceed as follows,

Integrating (1) we get,

0

1

0 1

0

2 2( ) cos sin

2 2 cos sin

0

c T c T

n n

nc c

c T c T c T

n nnc c c

n n f t dt a a t b t dt

T T

n na dt a t dt b t dt

T T

a T

0

1( ) (2)

c T

c

a f t dt T

Multiplying (1) by2

cos m

t T

integrating, we get

2( )cos

c T

c

m f t t dt

T


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0

1

2cos

2 2 2 2 cos cos sin cos

0 02

c T

c

c T c T

n n

n c c

m

ma t dt

T

n m n ma t t dt b t t dt

T T T T T

a

2 2( )cos (3)

c T

n

c

na f t t dt

T T

Multiplying (1) by2

sin m

t T

integrating, we get

0

1

2 2( )sin sin

2 2 2 2 cos sin sin sin

0 0

c T c T

c c

c T c T

n n

n c c

m m f t t dt a t dt T T

n m n ma t t dt b t t dt

T T T T

2m

T b

2 2( )sin (4)

c T

n

c

nb f t t dt

T T

The Fourier coefficients 0, andn na a b are given by the formulae (2), (3) and (4).

These are called the Euler’s formulae for Fourier coefficients.

Note:

(1)

For2

T c

, the formulae for the Fourier coefficients

0, andn na a b becomes

2

20

20

0, if ( ) is odd

1( )

2( ) , if ( ) is even

T

T

T

f t

a f t dt T f t dt f t

T


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01

2

0

2

2

( ) cos sin

1 where ( ) ,

2

1 ( )cos

1 and ( )sin .

n n

n

c

cc

n

c

c

n

c

f t a a nt b nt

a f t dt

a f t ntdt

b f t ntdt

For2

c

if f(t) is even function of t , then Fourier series expansion of f(t) is

0

1

( ) cosnn

f t a a nt

where 0 0 0

1 2( ) , ( )cos .na f t dt a f t nt dt

If f (t) is odd function of t , then Fourier series expansion of f(t) is

1

( ) sinnn

f t b nt

where0

2( )sin .nb f t nt dt

3) Let f(t) has Fourier series expansion 01

2 2( ) cos sinn n

n

n t n t f t a a b

T T

Consider the terms2 2

cos sinn nn t n t

a bT T

which is called the nth harmonic in

the Fourier series expansion of f(t) for

2 2 1cos , sin tan nn n n n n n n n n n

n

ba r b r or r a b and

a

the nth harmonic

becomes2 2 2

cos cos sin sin cosn n n n n nn t n t n t

r r r T T T

. Then

2 2

n n nr a b is called the amplitude of the nth harmonic and 1tan nn

n

ba

is

called phase angle of the nth harmonic.


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Dirichlet’s condition for the convergence of the Fourier series

If a periodic function f (t) of period T, is piecewise continuous in the interval

[c, c + T ] and has left and right hand derivative at each point of that interval then

The Fourier series expansion 01

2 2cos sin

n n

n

n t n t a a b

T T

is convergent.

Its sum is f(t) except at a point t0 at which f(t), is discontinuous and at t0, it

converges to the average of left and right hand limit of f(t) at t0. i.e., to

0 01

f(t + 0) +f(t - 0) .2

Parsevals identity:

If a periodic function f (t) has Fourier series expansion

0

1

2 2( ) cos sinn n

n

n t n t f t a a bT T

which is uniformly convergent in

[c, c + T ] then 2 2 201

2( ) 2

c T

n n

nc

f t dt a a bT

Proof : Let f(t) has a Fourier series expansion

0

1

2 2( ) cos sinn n

n

n t n t f t a a b

T T

which is uniformly convergent.

Consider 2

0

1

2 2( ) ( ) cos sin

c T c T

n n

nc c

n t n t f t dt f t a a b dt

T T

f(t)

f (t0 +0)

f(t0 –0)

tt0c c + T


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0

1

2 2( ) ( ) cos ( ) sin

c T c T c T

n n

nc c c

n t n t a f t dt a f t dt b f t dt

T T

since Fourier

series is uniformly convergent

0 0

1 2 2n n n n

n

T T a T a a a b b

2 2 201

22

n n

n

T a a b

Therefore, 2 2 201

2( ) 2

c T

n n

nc

f t dt a a bT

which is called Parseval’s identity.

Problems:

1.

Expand 2( ) , f x x x x f(x+2π)=f(x) , as a Fourier series.

Solution: Let0

1

( ) ( cos sin )n nn

f x a a nx b nx

. Then

2 2

0

0

1 2( )

2 2

a x x dx x dx

since, x is odd and x

2 is even function.

3 3 2

0

1

3 3 3

x

2 20

1 2cos cosna x x nx dx x nx dx

2

2 3

0

2 sin cos sin

2 2

nx nx nx

x xn n n

= 2 2 2 32 sin sin 0 cos cos0 2

0 2 0 sin sin 0n n n

n nn n n n n


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2 3

12 20 0 2 0 .0

n

n n

1

2 2

4 1 41

n

n

n n

.

20

1 2sin sin

nb x x nxdx x nxdx

=

0

2

2

cos sin1

nx

n

nx x

n

= 2

2 cos 10 sin sin

nn no

n n

= 12 2 2

cos 1 1n n

nn n n

.

2

1 1

21

4 21 cos 1 sin

3

n n

n

f x nx nxn n

.

2. Obtain the Fourier series expansion of

21, 0 2

4 f x x x

f(x+2π)=f(x) and hence obtain 21i n 1

21

n

iin

21

2 1iii

n

Solution:

Here f x is an even function.


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01

cosnn

f x a a nx

, where

2

2 22 3 3

0

0

0

3

1 1 1 1

2 4 8 24 123

a x dx x

2

2

0

1 1cos

4n

a x nxdx

= 2

2

2 3

0

1

4

sin cos sin2 2 1

nx nx nx x x

n n n

= 21 2

0 cos 2 cos0 04 nn

= 2 21 2 1

24 n n

2

21

1cos .

12 n f x nx

n

At x = 0, 2 2

21

10

4 12 n f

n

2 2 2

21

1................. 1

4 12 6n n

At , 0 x f x

2

21

10 cos

12 nn

n

22

1

1

12

n

n n

12

21

1..................... 2

12

n

n n

(1) + (2) gives


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2 2 2

21

12

6 12 42 1n n

2

2

1

1

82 1n n

.

3. Expand

, 0 1

0, 1

2 , 1 2

x x

f x x

x x

2 f x f x , as a Fourier series and

hence deduce that1 1 1

1 ..............4 3 5 7

Solution:

f x is odd,

1

sinnn

f x b n x

Where 2

0

2sin

2n

b f x n xdx

=

1 2

0 1sin 2 sin x xdx x xdx

= 21

2 2 2 2

0 1

cos sin cos sin1 2 1

n x n x n x n x x x

n n n n

2-2 -1 1


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= 1 cos1

cos 0 0 0 0n

nn n

= 12cos 2 2

1 1n nn

n n n

1

1

12 sin

n

n

f x n xn

1

2 2 f

1

1

12 sin

2

n

n

n

n

1

2

0 is evensin

2 1 is oddn

nn

n

1

1

1sin

4 2

n

n

n

n

1

1

1

2 1

n

n n

1 1 11

3 5 7 .

4. Expand

0, 02

( ) , ( ),

sin , 0

t

f t f t f t t

E t t

as a Fourier

series.

Solution:

( ) f t is neither even nor odd.

0


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01

( ) cos sinn nn

f t a a n t b n t

where

2

0

0

2

0

1 1

( ) sin2

cos .

2

T

T a f t dt E tdt T

E t E

2

0

2

2( )cos sin(1 ) sin(1 )

2

T

n

T

E a f t n tdt n t n t dt

T

0

0

0

cos(1 ) cos(1 ), 1

2 (1 ) (1 )

sin 2 , 12

cos(1 ) cos(1 ) 1 1, 1

2 (1 ) (1 ) (1 ) 1

cos2, 1

2 2

E n t n t n

n n

E tdt n

E n nn

n n n n

E t n

E

22 2 2

, is even2 (1 ) 1 1 .

0, is odd

E n

n n n

n


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2

0

2

0

0

2( )sin sin sin

cos( 1) cos( 1)2

sin( 1) sin( 1)1

2 ( 1) ( 1)

0, 1

W

n

W

w

Ewb f t wt dt wt nwt dt

T

Ewn wt n wt dt

Ew n wt n wt n

n w n w

n

2

1

0

0

0

2 2

sin

(1 cos2 )2

sin2

2 2

( 0)2 2

2 cos2 cos4( ) sin .....

2 2 1 4 1

w

W

w

Ewb wt dt

Ewwt dt

Ew wt t

w

Ew E

w

E E E wt wt f t wt

5.

Expand 2f(x) = x , 0 x 2, f(x+2) = f(x) as a Fourier series and hence

evaluate2

1

1

n n

.

f(x)

x42-2


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Solution: Note that f(x) is neither even nor odd.

01

( ) ( cos sin )n nn

f x a a nx b nx

where

22 32

0

0 0

1 1 4

2 2 3 3

xa x dx

22 2

2

2 2 3 3 2 2

0 0

2 sin cos sin 4cos 2 2

2n

x n x n x n xa x n x dx x

n n n n

22 2

2

2 2 3 30 0

3 3

2 cos sin s

sin 2 22

4 2cos2 0 0 (cos2 cos0)

4.

n

x n x n x co n x

b x n x dx xn n n

n nn n

n

2 21

4 4 4( ) cos sin

3 n f x n x n x

n n

A x=0, f(x) is discontinuous

2 21

2 21

2 2

21

(0 ) (0 ) 4 4 4cos0 sin0

2 3

4 0 4 4 1

2 3

1 42

3 4 6

n

n

n

f f

n n

n

n


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2lO

y

x

l-l

Half range Expansions

While solving various physical and engineering problems, there is a practical need

for expanding functions defined over a finite range. Such an expansion is possible

if functions under consideration can be extended to a periodic function which is

either even or odd.

Consider a piecewise continuous function ( ), f x defined in a finite interval

(0, ).l Then it is possible to extend ( ) f x to a periodic function, which is even or

odd.

Consider the function ( ) g x defined as follows:

( ), 0

( ) ; ( 2 ) ( ).( ), 0

f x x l

g x g x l g x f x l x

Then ( ) g x is called an even periodic extension of ( ) f x .

Graph of ( ) f x

Graph of even periodic extension of f(x).

O

y

x

l


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2lO

y

x

l-l

The function ( ) g x can be expanded as Fourier cosine series

0

1

( ) cosnn

n g x a a x

l

where0

0

1 ( )

l

a g x dxl

and0

2 ( ) cos

l

nna g x x dx

l l

But for 0 , x l ( ) ( ). g x f x We have

0

0

1( )

l

a f x dxl

and0

2( ) cos

l

n

na f x x dx

l l

( ) ( ) f x g x 0

1

cos ,nn

na a x

l

for 0 . x l

Such an expansion of ( ) f x is called the half range Fourier Cosine series

expansion of ( ). f x

Also, if( ), 0

( ) ; ( 2 ) ( ).( ), 0

f x x l g x g x l g x

f x l x

Then ( ) g x is called an odd periodic extension of ( ) f x .

Graph of ( ) f x

Graph of odd periodic extension of f(x)

O

y

xl


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The function ( ) g x can be expanded as Fourier Sine series

1

( ) sinnn

n g x b x

l

where

0

2( ) sin

l

n

nb g x x dx

l l

But for 0 , x l ( ) ( ). g x f x We have

0

2( ) sin

l

n

nb f x x dx

l l

( ) ( ) f x g x 1

sin ,nn

nb x

l

for 0 . x l

Which is called the half range Fourier Sine series expansion of ( ). f x

Problems:

1. Expand ( ) ,0 f x x x as half range Fourier cosine and sine series. Also

draw the graph of the corresponding periodic extensions of ( ). f x

Solution:

Graph of even periodic extension of ( ) f x .

0

1

( ) cosnn

f x a a nx

, where


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0

0 0

1 1( ) .

2a f x dx xdx

2

00 0

2

2

2 2 2 sin ( cos )( )cos cos (1)

0,2 ( 1) 1

.4, odd

n

n

x nx nxa f x nxdx x nxdx

n nn even

n nn

2 2

4 cos cos3( ) .

2 1 3

x x f x x

Graph of odd periodic extension of ( ) f x .

1

( ) sinnn

f x b nx

, where

2

00 0

1

2 2 2 ( cos ) ( sin )( )sin sin (1)

2 ( 1) 0 2( 1) .

n

nn

x nx nxa f x nxdx x nxdx

n n

n n

1

1

2( ) ( 1) sin .n

n

f x x nxn


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2. Expand

, 02

( )

1 ,2

x l x

l f x

x l x l

l

as half range Fourier cosine and sine

series. Also draw the graph of the corresponding periodic extensions of( ). f x

Solution:

Graph of even periodic extension of ( ) f x .

( ) f x 01

cosnn

na a x

l

Where

2

0

0 0

2

1 1( ) 1

l

l l

l

x xa f x dx dx dxl l l l

22 2

02

1 11

2 2 4

l l

l

x x

l l l

2

0 0

2

2 2( )cos cos 1 cos

l

l l

n

l

n x n x na f x x dx x dx x dxl l l l l l l

2 2

22cos cos 1

2

nn

n

1


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2 2

0, 2, 6, 10, 14, ...

8, 2, 6, 10, 14, ...

n

nn

2 2 2

1 8 1 2 1 6( ) cos cos

4 2 6 f x x x

l l

.

Graph of odd periodic extension of ( ) f x

( ) f x 1 sinn

n

n

b xl

Where

0

2( )sin

l

n

na f x x dx

l l

2

0

2

2 sin 1 sin

l

l

l

x n x n x dx x dx

l l l l l

1

2 22

2 2

0, even4

sin 42 1 , odd

n

nn

xn n

n

O

y

x


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2 2 2 2

4 1 1 3 1 5 ( ) sin sin sin

1 3 5 f x x x x

l l l

Expand ( ) 1 ,0 x

f x x l l

as a Fourier cosine and sine series. Also draw the

graph of the corresponding periodic extensions of ( ). f x

Solution:

Graph of even periodic extension of ( ). f x

Let 01

( ) cosnn

n f x a a xl

.

Then0

0

2 1( ) .

2

l

a f x dxl

0 0

2 2

2 2

2 2( )cos 1 cos

0,

2 1 ( 1) .4,

l l

n

n

n x na f x xdx xdx

l l l l l

n even

n n odd n

2 21

1 4 (2 1)( ) cos

2 (2 1)n

n x f x

n l

.

-l l


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Graph of odd periodic extension of ( ) f x

Let1

( ) sin .nn

n f x b x

l

Then0 0

2 2 2( )sin (1 )sin

l l

n

n x nb f x xdx xdx

l l l l l n

Thus1

2( ) sin .

n

n f x x

n l

Problems:

Obtain the half range Fourier Cosine and sine series expansions of the

functions. Also draw the graph of corresponding periodic extensions.1)

( ) sin ,0 f x x x x .

2) ( ) ( ),0 f x x x x .

3)

, 02

( )

,2

x x

f x

x x

.

4)

( ) 2 ,0 2 f x x x .

5) ( ) cos ,0 f x x x l

l .

6) ( ) sin ,0 f x x x l l

.

7)

If ( ) f x is piecewise continuous for 0 , x l having half range expansions

l-l

0

2l


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Page 28 of 39

01 1

( ) cos & sinn nn n

n n f x a a x f x b x

l l

then show that

2 22 2 2

0

1 10 0

2 2( ) 2 & ( )

l l

n n

n n

f x dx a a f x dx bl l

.

Fourier integral representation:

Let f(x) be a piecewise continuous and absolutely integrable function of x in

( , ). , ., ( ) .i e f x dx exists

Then f(x) can be represented by an integral as

0

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

( ) ( )cos , ( ) ( )sin .

f x A S sx B s nx ds

where A s f t st dt B s f t st dt

Such an integral representation is called the Fourier integral representation of f(x).

The integral on RHS of (1) converges to f(x0) if f(x) is continuous at x0 and to

average of left and right hand limits if f(x) is discontinuous at x0.

Proof: Consider a periodic function f l(x) defined in (-l, l) such that f l(x) =f(x) for

l x l

Then, f l(x) can be represented by a Fourier series as

0

1

( ) cos sin (2)l n nn

n x n x f x a a b

l l

Where


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0

1 1 1( ) , ( )cos , ( )sin

2

l l l

l n l n l

l l l

n t n t a f t dt a f t dt b f t dt

l l l l l

Substituting in (2) we get,

1

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

2

l l l

l l l l

nl l l

n t n x n t n x f x f t dt f t dt f t dt

l l l l l l l

1

1 1( ) ( ) ( )(cos cos sin sin )2

l l

l l l

nl l

n t n x n t n x f x f t dt f t dt l l l l l l

n

n Let s

l

Then1

1 nn n n

s s s s or

l l

On substitution we get

1

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

2

l l

nl l l n n n n n

nl l

s f x f t dt f t s t s x s t s x dt s

1

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

2

l l l

l l n n l n n l n n

nl l l

f x f t dt s s x f t s t dt s x f t s t s

Let , then ( ) ( )l l f x f x .

( ) 0 0.

l

l n n

l

f t dt s s

Now taking the limit as , we getl


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Which is called the complex form of Fourier integral representation of ( ). f x

Note (2): If ( ) f x is an even function of , x then

0

( ) ( )cos 2 ( )cos A s f t st dt f t st dt

( ) ( )sin 0 B s f t st dt

Therefore Fourier integral becomes

0

2( ) ( )cos f x A s sx ds

0 0

2( ) cos cos f t st sx dt ds

Which is called the Fourier Cosine integral representation of ( ). f x

If ( ) f x is an odd function of , x then

( ) ( )cos 0 A s f t st dt

0

( ) ( )sin 2 ( )sin A s f t st dt f t st dt

0

2( ) ( )sin f x B s sx ds

0 0

2( ) sin sin f t st sx dt ds

Which is called the Fourier Sine integral representation of ( ). f x


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Fourier transforms

Consider the Fourier integral representation of the function ( ) f x given by

( )1

( ) ( )2

i s x t

f x f t e dt ds

)1 1( )

2 2

isx ist e f t e dt ds

Let1

( ) ( ) (1)2

ist F s f t e dt

Then1

( ) ( ) (2)2

isx f x F s e ds

The integral defined by (1) is called the Fourier transform of the function ( ) f x

and is denoted by ( ) . F f x Given ( ) ( ) , F s F f x the formula (2) defined( ), f x which is called the inverse Fourier transform of ( ) F s and is denoted by

1 ( ( ) F F s .

Note (1): A function ( ) f x is said to be self-reciprocal under Fourier transforms if

( ) ( ). F f x F s

Note (2): If ( ) ( ), F f x F s then 1( ) ( ) f x F F s is called spectralrepresentation of ( ) F s and ( ) F s is called spectral density of ( ). f x Here s is

called the frequency of the transform. The graph of ( ) F s is called amplitude

spectrum of ( ) f x and2

( ) F s is called energy of the spectrum.

Properties of the Fourier transforms:

(1) Fourier transform is linear

i.e., if 1 2andc c are constants then

1 2 1 2( ) ( ) ( ) ( ) F c f x c g x c F f x c F g x


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Proof: Follows from definition and linearity of the integral.

(2) If ( ) ( ) F f x F s then ( ) ( )iax F e f x F s a

Proof: Consider( )1( ) ( )

2

i s a t F s a f t e dt

1( )

2

ist iat f t e e dt

1

( )2

iat ist f t e e dt

( )iax F e f x

(3)

If ( ) ( ) F f x F s then ( ) ( )isa F f x a e F s Proof: Exercise

(4) If ( ) ( ) F f x F s then

1( )

s F f ax F

a a

Proof: Exercise

(5)

If ( ) ( ) F f x F s then

(i) 1

( )cos ( ) ( )

2

F f x ax F s a F s a

(ii) ( )sin ( ) ( )

2

i F f x ax F s a F s a

(iii) ( ) ( ) F f x F s

(iv) ( ) ( ) F f x F s

(v) ( ) ( ) F f x F s

(vi) ( ) ( )n

n n

nd F x f x i F sds

(vii) ( ) ( ) ( ) ( )n n F f x is F f x


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Convolution:

For functions ( ) & ( ), f x g x we define the convolution of ( ) & ( ), f x g x denoted by

( ) f g x as 1

( ) ( ) ( ) ;

2

f g x f t g x t dt

provided the integral exists.

Note that . f g g f

Convolution Theorem:

( ) ( ) ( ) . F f g x F f x F g x

Proof:

Consider

( )

1( ) ( )( )

2

1 1( ) ( )

2 2

1 1( ) ( )

2 2

isx

isx

ist is x t

F f g x f g x e dx

f t g x t dt e dx

f t e g x t e dx dt

Let x t u . Then

( )

( )

1 1( ) ( ) ( )

2 2

1 1( ) ( )

2 2

( ) ( ) .

ist is u

ist is u

F f g x f t e g u e du dt

f t e dt g u e du

F f x F g x

Parseval’s Identity:

If ( ) ( ) F f x F s then 2 2| ( ) | | ( ) | f x dx F s ds

.

Proof:


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1

( ) ( ), { ( )} ( ),

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 1. ., ( ) ( ) ( ) ( )2 2

isx

If F f x F s F g x G s then

F s G s F f x g x or

f x g x F F s G s

i e f t g x t dt F s G s e ds

For 0, x we get

( ) ( ) ( ) ( ) f t g t dt F s G s ds

( ) ( ) ( ) ( ) Let g x f x or g x f x .

2 2

( ) ( ) ( ) ( ) ( ).

( ) ( ) ( ) ( ) ( ) ( )

| ( ) | | ( ) |

ThenG s F g x F f x F f x F s

f t g t dt f t f t dt F s F s ds

f t dt F s ds

Problems:

1) Find the Fourier transform of1, | |

( )0, | |

x a f x

x a

. Hence deduce that

2sin sin

2 2

t t dt and dt

t t

.

Solution:

1

( ) ( )

2

1 1 2 sin1 ( )

2 2

isx

aa isxisx

a a

F f x f x e dx

e ase dx F s

is s

.


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0

0

1 1 2 sin( ) ( ) cos sin

2 2

2 sin sincos sin is an odd function of

0, | |

sincos ( ) , | |

2 2

1

2 2

isx as f x F s e sx i sx ds s

as as sxds sx s

s s

x a

as sxds f x x a

s

0 , | |4

x a

.

0

For 0, ( ) 1.

sin.

2

x f x

asds

s

0

0

Let or . .

On substitution, we get

sin

/ 2

sintor .

t 2

t dt as t s ds

a a

t dt

t a a

dt

2 2

2 2

0

2

0

2

0

By Parseval's identity,

| ( ) | | ( ) |

2 sin 4 sini.e., 1

sin

2

sinFor 1, we get .

2

a

a

f x dx F s ds

as asdx ds ds

s s

as ads

s

t a ds

t


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2) Find the Fourier transform of , 0a x

e a and hence evaluate

2 2

0

cos xt dt

a t

and

{ }a x

F xe

0

2 2 0

2 2 2 2

2 2

2 2

0

2 2 2

0

1 2{ } cos

2

2cos sin

2 2(0 ( )) ( )

1( ) ( )

2

1(cos sin )

2

2 cos

cos cos

a x a x isx a x

ax

o

isx

F e e e dx e sxdx

ea sx s sx

a s

e aa F s

a s a s

f x F s e ds

a sx i sx ds

a s

a sxds

a s

sx xt ds

a s a

20

( )

2 2

a x f xdt e

t a a

Since

1

2 2 2 2 2

{ ( )} ( )

2 2 21, { } { }

( )

nnn

n

a x a x

d F x f x i F s

ds

d d a as for n F x e i F e i i

ds ds a s a s

3)

Find the Fourier transform of2 2

, 0a xe a and hence show that

2

2

x

e

is

self-reciprocal under Fourier transform.


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2 2 2 2 2 2

2 2

2

2

22

( )

2 4

42

1 1{ }

2 2

1

2

2

a x a x isx a x isx

is sax

a a

sis

a axa

F e e e dx e dx

e dx

ee dx

Let2

isax t

a then

dt dx

a

=

2 2

2 2

2 24 42

2 2

s s

a at t e dt e

e e dt a a

Put

2

1

21

2

t y or t y

dt y dy

2 2 2 2

2 2 2 214 4 4 42

0

12

2 22 2 2 2

s s s s

ya a a ae e e e e y dy

a a a

For1

2a or 2 1a

2 22 2

2 2{ } { } x s

a x F e F e e

2

2 s

e

is self-reciprocal under Fourier transform.

4)

Find the Fourier transform of1 1

( ) 0 1

x x f x x

and hence evaluate

2

0

sin xdx

x


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1

1

1

0

1

2 2

0

2

1 1{ ( )} ( ) (1 )(cos sin )

2 2

2(1 ) cos (1 )sin .

2 sin cos 2 (1 cos )(1 ) ( )

1( ) ( )

2

1 2 (1 cos )(cos

2

isx

isx

F f x f x e dx x sx i sx dx

x sx dx x sx is an odd function of x

sx sx s x F s

s s s

f x F s e ds

s sx

s

2

0

sin )

(1 ) 12 (1 cos )cos ( ) 2

20 1

i sx ds

x x s sx ds f x

s x

20

2

20

22

20 0

(1 cos )0, (1 0)

2 2

2sin ( / 2)

2

/ 2 2 2

2 sin sin2

4 2 2

sx For x ds

s

sds

s

Let s t or s t ds dt

t t dt or dt

t t

Exercises:

1.

Obtain the Fourier transform of2

3

0

1 1 cos sin( ) and hence evaluate cos( / 2)

0 1

x x x x x f x x dx

x x

2. Find the Fourier transform of ( )0 0

x x a f x

x a

3.

Find the Fourier transform of

2 2

( )0

a x x a f x

fourier series notes corrected.pdf

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