Fourier Cosine and Sine transforms

Consider the Fourier cosine integral representation of a function f(x)

0 0

2( ) ( )cos cos , 0f x f t st sxdtds x

0 0

2 2( )cos cosf t st dt sxds

0

0

2( ) ( )cos ( )....(1)

2( ) ( ) cos .......(2)

c c

c

Let F f x f t st dt F s

Then f x F f x sxds

The transform { ( )}cF f x defined by (1) is called the Fourier cosine transform of

f(x).The formula (2) is called the inverse Fourier cosine transform of

{ ( )} ( )c cF f x F s and is denoted by 1( ) ( ) .c cf x F F s

Similarly, using the

Fourier sine integral representation of f(x) given by

0 0

2( ) ( )sin sin ,f x f t st sxdt ds

we can define the Fourier sine transform of f(x)

denoted by ( )sF f x as 0

2( ) ( )sin ( )s sF f x f t st dt F s

Then the inverse Fourier sine transform of ( )sF s is defined as

1

0

2( ) { ( )} ( )sins s sf x F F s F s sxds

Definition: A function f(x) is said to be self reciprocal under Fourier cosine (sine)

transform if ( ) ( ) ( ) ( )cF f x f s Fs f x f s

Properties of Fourier cosine/ sine transforms

(1) Both Fourier cosine and sine transforms are linear.

c 1 2 1 c 2 c

s 1 2 1 s 2 s 1 2

F {c f (x) c g(x)} c F {f (x)} c F {g(x)}

F{c f (x) c g(x)} c F{f (x)} c F{g(x)} where c and c are cons tan ts

c c c

c s s

s s s

s c s

1(2). F {f (x)cosax} F (s a) F (s a)

2

1F {f (x)sin ax} F (s a) F (a s)

2

1(3). F{f (x)cosax} F (s a) F (s a)

2

1F{f (x)sin ax} F (s a) F (s a)

2

The proof of (2) and (3) follows directly from trigonometric identities.

c c

s s

c s

s c

1(4). F {f (ax)} F (s / a)

a

1F{f (ax)} F (s / a)

a

(5). If f (x) 0 as x , then

2F {f (x)} f (0) s F (s)

F{f (x)} s F (s)

c

0

2Proof : Consider F {f (x)} f (x)cossx dx

Integrating by parts, we get

00

2'( ) ( )cos | ( )sin ( )CF f x f x sx f x sx s dx

0

20 (0) ( )sin f s f x sx dx

2

( ) (0)SsF f x f

provided ( ) 0 as f x x

Also,

0

2'( ) '( )sin SF f x f x sx dx

Integrating by parts, we get

00

2'( ) ( )sin | ( ) cos ( )SF f x f x sx f x sx s dx

2

0 0 ( ) ( ).2

C CsF s sF s

(6) 22

''( ) '(0) ( )C CF f x f s F s

22

''( ) (0) ( )S SF f x sf s F s

provided ( ) and '( ) 0 f x f x as x

(7) ( ) and ( )S CC SdF dF

F xf x F xf xds ds

(8)

( ) (s), ( ) ( ),

( ) (s) and ( ) ( ) exist,

C C C C

S S S S

If F f x F F g x G s

F f x F F g x G s

then 0

0 0

( ) ( ) ( ) ( ) ( ) ( )C C S SF s G s ds F s G s ds f x g x dx

and 2 2 2

C C

0 0 0

F ( ) F ( ) ( )s ds s ds f x dx

which is called Parsivals identity.

Problems:

(1) Find C SF , Fax axe e and hence find C SF and Fax axxe xe

Solution: By definition,

C0

2F cos ax axe e sx dx

2 2 2 20

2 2cos sin , 0

axe aa sx s sx a

a s a s

S0

2F sin ax axe e sx dx

2 2 2 20

2 2sin cos , 0.

axe sa sx s sx a

a s a s

Therefore,

2 2

C 2 2 2 2 2

2 2F

( )

ax ax

S

d d s a sxe F e

ds ds a s a s

S 2 2 2 2 22 2 2

F( )

ax ax

C

d d a asxe F e

ds ds a s a s

2). Find 2 2 2 2

&a x a xc sF e F xe

.

Solution : We have from definition,

2 2 2 2 2 2 2 2

2

2

2 2

2 2 2 2 2 2

0 0

4

4 4

3

2 2cos

1

2

1

2 2 2

a x a x a x isx a x

c

s

a

s s

a x a x a as c

F e e sxdx e e dx F e

ea

d d sF xe F e e e

ds ds a a

Note that

2 2

2 2

2 2

2 2

2 2

2 2

1for ,

2

& .

x s

a x

c c

x s

a x

s s

a F e F e e

F xe F xe xe

2

2

2

2

is self reciprocal under Fourier cosine transform and

is self reciprocal under Fourier sine transform .

x

x

e

xe

3) Find 1 1& ,0 1.a ac sF x F x a

1

0

1 1 2

0 0

Consider ;

Let or . Then

1 ( ) ( ) ( )( )

( ) ( )cos sin

2 2

a isx

ai

a isx a y a

a a a a a

a a

x e dx

dyisx y dx

is

a a ax e dx y e dy i e

i s s sis

a a a ai

s s

1 1

0 0

1 1

0

1 1

0

Thus

( ) ( )cos cos cos sin

2 2

Equating real and imaginary parts, we get

2 2 ( )cos cos and

2

2 2 ( )sin sin

a a

a a

a a

c a

a a

s a

a a a ax sxdx i x sxdx i

s s

a aF x x sxdx

s

a aF x x sxdx

s

.2

Note:

11 1 1 1 1 For , 2 2

1 is self reciprocal under Fourier sine and cosine transforms.

a

c c sa F x F Fx s x

x

Note:

0

1

0

0

2Consider ( ) ( )cos ( ).

2 Then ( ) ( ) ( )cos (1)

Interchanging and in (1), we get

2( ) ( )cos ( )

Similarly, if ( ) ( ) then, ( ) ( ).

c c

c c c

c c c

s s s s

F f x f x sxdx F s

f x F F s F s sxds

s x

f s F x sxdx F F s

F f x F s F F s f s

4) Find

2 2

2 2

2 2

2 2

-

2

- -

2

1 and .

1 1

Solution: We have shown that

2( ).

2( ) .

Or .2

1For 1, .

1 2

1 2 2

c s

ax

c c

as

c c c

as

c

s

c

s

s

xF F

x x

aF e F s

a s

aF F x F e

a x

aF e

a x

a F es

x dF e e

x ds

.s

Exercises:

1) Find the Fourier cosine transform of cos , sinax axe ax e ax and hence find the

Fourier cosine transforms of

2

4 4 4 4

1 and .

x

x k x k

2) Find the Fourier sine transform of 1

and , 0.axe

ax x