FOURIER TRANSFORMATION Fourier Transform:

We know the complex form of Fourier integral i.e.

f (x) =

1 2π

−∞

−∞

∫ ∫

f (t) ei λ (t − x) d t d λ

On rewriting this (i.e. by replacing λ by s ), we get

1 ∞ ∞ 1 = f (x) = f (t) ei s (t − x) d t d s ∫ ∫ 2 π −∞ −∞ 2π Now if

−∞

f (t) ei st d t = F (s) , then from (1) we get f (x) =

1 2π

−∞

F (s) e−i s x d s

−∞

e

−i s x

d s ∫ f (t) ei st d t −∞

(1)

(2)

The function F (s) defined above is known as Fourier Transform of f (x) . Also the function f (x) given by (2) is called inverse Fourier transform of F (s) . The expression given by (2) is also known as inversion formula of Fourier transform.

Fourier Sine Transform:

Again Fourier sine transform can be deducted from Fourier sine integral i.e. ∞

∞ 2 f (x) = ∫  ∫ f (t)sin λ t d t  sin λ x d λ  π 0  0

On rewriting this (i.e. by replacing λ by s ), we get ∞

∞ 2 f (x) = ∫  ∫ f (t)sin s t d t  sin s x d λ  π 0  0

(3)

Dr. Jogendra Kumar

Page 1

FOURIER TRANSFORMATION Now let

Fs (s) = ∫ f (t) sin s t d t , then from (1) 0

f (x) =

The function

2 ∞ Fs (s)sin s x d s π ∫0

(4)

Fs (s) is known as Fourier sine transforms of f (x) in 0 < x < ∞. . Also the function

f (x) given by (4) is called inverse Fourier sine transforms of Fs (s) .

Fourier Cosine Transform:

Again Fourier cosine transform can be deducted from Fourier cosine integral i.e. ∞

∞ 2 f (x) = ∫  ∫ f (t) cos λ t d t  cos λ x d λ  π 0  0

On rewriting this (i.e. by replacing λ by s ), we get ∞

∞ 2 f (x) = ∫  ∫ f (t) cos s t d t  cos s x d s  π 0  0

Now let

(5)

Fc (s) = ∫ f (t) cos s t d t , then from (1)

f (x) =

The function

0

2 ∞ Fc (s) cos s x ds π ∫0

(6)

Fc (s) is known as Fourier cosine transforms of f (x) in 0 < x < ∞. . Also the function

f (x) given by (6) is called inverse Fourier cosine transforms of Fc (s) .

Dr. Jogendra Kumar

Page 2

FOURIER TRANSFORMATION Properties of Fourier Transform:

Linearity Property: If F (s) and G(s) are complex Fourier transforms of f (x) and

g(x) respectively, then

F [ a f (x) + b g (x) ] =a F (s) + b G(s)

Where a and b are constants.

Change of scale property:

If F (s) is the complex Fourier transforms of

= F {f (a x)}

1 s F  , a a

a≠0

f (x) , then

Proof: We know that

F{f (x)} = F(s) = ∴

−∞

f (x) ei s x d x

(1)

dt Pu ta x =t ⇒ d x = a

F{f (a x)} =∫ f (a x) ei s x d x −∞

=

i st 1 ∞ a f e dt (t) a ∫−∞ s

i  t 1 ∞ = ∫ f (t) e  a  d t a −∞ 1 s = F  a a

Shifting Property:

If F (s) is the complex Fourier transforms of

F{f ( x − a)} = ei s a F ( s )

f (x) , then

Dr. Jogendra Kumar

Page 3

FOURIER TRANSFORMATION Proof: We know that

∫ ∴ F{f (x − a)} = ∫ =∫ = F(s) = F{f (x)}

−∞

−∞ ∞

−∞

f (x) ei s x d x f (x − a) ei s x d x

Pu t x − a = t ⇒ d x = d t

f (t) ei s (t + a ) d x ∞

= ei a s ∫ f (t) ei s t d x −∞

=e

ias

F (s)

Modulation Theorem:

If F (s) is the complex Fourier transforms of

1 [ F (s+ a) + F(s− a)] 2

F{f ( x) cos a = x}

f (x) , then

Proof: If F (s) is the complex Fourier transforms of

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

−∞

f (x) ei s x d x

f (x) , then

isx ∴ F{f ( x) cos s x} = ∫ f (x) cos a x .e d x −∞

=∫

−∞

 ei a x + e − i a x f (x) ei s x  2 

(

 d x 

)

1 ∞ f (x) ei ( s + a ) x + ei ( s − a ) x d x ∫ −∞ 2 ∞ 1 ∞ i(s+ a) x f e d x f (x) ei ( s − a ) x d x  (x) = + ∫ ∫   −∞ −∞ 2 1 = [ F(s+ a) + F(s− a)] 2 =

Dr. Jogendra Kumar

Page 4

FOURIER TRANSFORMATION Note: If Fs (s) and Fc (s) are the Fourier sine transforms and Fourier cosine transforms of

f (x)

respectively, then the following results hold true. (i)

(ii)

(iii)

x} Fs {f ( x) cos a=

1 [ Fs (s+ a) + Fs (s− a)] 2

x} Fc {f ( x) sin a=

1 [ Fs (s+ a) − Fs (s− a)] 2

x} Fs {f ( x) sin a=

1 [ Fc (s− a) − Fc (s+ a)] 2

All these result can be proved similarly as above result.

Que. Find the Fourier transform of

1 for x < 1 , f (x) =   0 for x > 1

Solution: By definition, we have

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

−∞

0

sin x d x. x

f (x) ei s x d x

i.e. − 1 < x < 1  1 for x < 1, 0 for x > 1, i.e. x < −1 and x > 1

Now since f (x) = 

∴ F{f ( = x)}

∫ ∫

−1

−∞

=

−1

−∞

1

f (x).ei s x d x + ∫ f (x).ei s x d x + ∫ f (x).ei s x d x −1

0.e

isx

1

d x + ∫ 1.e −1

isx

1

d x + ∫ o .e 1

isx

dx

1

ei s x ei s − e − i s 2  ei s − e − i s  = ∫ e= dx = =   −1 i s −1 is s  2i  2sin s = = F (s) s 1

isx

Dr. Jogendra Kumar

Page 5

FOURIER TRANSFORMATION Now by inverse Fourier transforms, we have

f (x) =

1 2π

−∞

F (s) e−i s x d s

∞ sin s 1 ∞ 2sin s −i s x ⇒ = e d s e −i s x d s π f (x) ∫ ∫ −∞ −∞ 2π s s Now putting x = 0 both sides, we get ∞ sin s [ f(0) 1by definition of f(x)] = d s π= f (0) π ∫−∞ s= ∞ sin s ∞ sin s π  sin s  ⇒ 2∫ d s= d s= is even function  π ⇒ ∫  0 0 2  s s s  ∞ sin x π d x= ⇒ ∫ ( In definite integral variable can be replaced by other variable ) 0 x 2

= ∴ f (x)

(2) Find the Fourier transform of 2 1 − x for x ≤ 1 , f (x) =   0 for x > 1

Hence evaluate

Solution: By definition, we have

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

∴ F{f ( = x)} =

∫ ∫

−1

−∞ −1

−∞

−∞

0

x  x cos x − sin x    cos d x. 3 x 2  

f (x) ei s x d x

1

f (x).ei s x d x + ∫ f (x).ei s x d x + ∫ f (x).ei s x d x −1

1

1

0.ei s x d x + ∫ (1 − x 2 ).ei s x d x + ∫ o .ei s x d x −1

1

Dr. Jogendra Kumar

Page 6

FOURIER TRANSFORMATION 1

F {f (x)} =∫ (1 − x ) e 2

isx

−1

ei s x (1 − x 2 ) = is

ei s x d x =(1 − x ) is +2

 ei s + e − i s 0 − 2 = 2  s

1

−1

− ∫ (−2 x)

1

1

−1

1

2

x ei s x

(i s )

2 −

ei s x −2 (i s )3

  ei s − e − i s  + 2 3   is

1 1 = 4  3 sin s − 2 cos s s

−1

ei s x dx is

1

−1

 1  ei s − e − i s  4 =  3    s  2i

 1  ei s + e − i s − 2   s  2

  

4  − 3 (s cos s − sin s) s = s 

Now by inverse Fourier transform, we have

f (x) =

1 2π

−∞

F (s) e − i s x d s

1 ∞ 4 2 ∞  s cos s − sin  ∴ f (x) = − 3 ( s cos s − sin s )  e − i s x d s = − ∫  ∫  2 π −∞  s π −∞  s3  Now putting x =

( )

s   −i s x  e d s 

1 both sides, we get 2

 s cos s − sin s   − i 2s = − ∫   e d s 2 s3 π −∞   ∞  s cos s − sin s   s s π  1 3π ⇒ ∫  − 1 −  = −   cos − i sin  d s = 3 −∞ s 2 2 2  4 8   f 1

−∞

2

s 3π  s cos s − sin s  (On comparing real and imaginary parts both sides) −   cos d s = 3 s 2 8  

Dr. Jogendra Kumar

Page 7

FOURIER TRANSFORMATION Que. Find the Fourier transform of e

− a2 x2

, a > 0 . Hence deduce that e

x2 2

is self reciprocal in respect of

Fourier transform.

Solution: By definition

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

2 2

F{e − a x= }

−∞

−∞

f (x) ei s x d x

e − a x ei s x d= x 2 2

−∞

 i s x i2 s2  − a 2  x 2 − 2 + 4  a 4a  

e

isx   − a2  x2 − 2  a  

i2 s2 4 a2

e e dx ∫ ∫=

=

−∞

=e

−∞

s2 4 a2

−∞

e

 is  − a 2  x −  a 2  2 

−∞

 is  − a 2  x − 2   2a 

e

2

e

s2 4 a2

dx

dx

 is  1 put a  x − 2  =⇒ t d x= dt 2a  a 

dx

s2

π

 i s x i2 s2 i2 s2  − a 2  x 2 − 2 + 4 −  a 4a 4 a 4  

2

π

1 − 4 a2 ∞ − t 2 e = = ∫−∞ e d t a

2 2

e

d= x

a

e

s2 4 a2

{

 ∫ e− t d t =

π

2

−∞

}

s2 4 a2

F{e − a x } = e a

Now taking a 2 = −

x2 2

1 in above transform, we get 2

= F{e }

π

s2 4.1

= e 1 2

So, we can conclude that e

x2 2

2

2π e

s2 2

.Since the functions e

is self reciprocal of its Fourier transform e

x2 2

s2 2

and e

s2 2

are the function.

.

Dr. Jogendra Kumar

Page 8

FOURIER TRANSFORMATION Que. Find the Fourier transform of (i) e − 2 (x −3) and (ii) e − x cos 3 x . 2

2

Solution: (i) From above problem, we have −

x2 2

s2 2

= 2π e F (s) , say

= F{e }

Now consider e − 2 x which can be rewrite as 2

(2 x)2 2

1 s e= e= f (2 x) . Thus F{ f (2 x)} = F   by change of scale property i.e. 2 2 −

−2 x 2

( If F (s) is the complex Fourier transforms of −

(2 x )2 2

f (x) , then = F {f (a x)}

( s 2)

a ≠ 0)

2

1 πe 2 {e } 2= F{e } F= Therefore, = 2 −2 x

2

1 s F  , a a

π 2

e

s2 8

Again F {e −2( x −3) = = ei a s F (s) , by shifting property i.e. } F{f (x − 3)} 2

(If F (s) is the complex Fourier transforms of

π

s2 8

ias −2( x −3) F{e= F (s) e3 i s = e } e= 2 2

(ii)

π

2

e

f (x) , then F{f ( x − a)} = ei s a F ( s ) )

 s2   3 i s − 8   

We know that −

x2 2

= F{e }

s2 2

= F (s) , say. Now by modulation theorem, we have 2π e

If F (s) is the complex Fourier transforms of

f (x) , then F{f ( x) cos a = x}

1 [ F (s+ a) + F(s− a)] 2

Dr. Jogendra Kumar

Page 9

FOURIER TRANSFORMATION ∴

x2 2

1 [ F (s+ 3) + F(s− 3)] 2 (s + 3)2 (s −3)2  − − 1 2 = + 2π e 2   2π e 2   2 2 2 (s −3)  (s −3)2   − (s +3)  − (s +3) − − 1 π 2 π e 2 + e 2 = =  e 2 + e 2  2 2    

F{e

cos 3 x} =

Que. Find the Fourier cosine transform of e − x . 2

Solution. By the definition of Fourier cosine transform, we have ∞

Fc {f (x)} = ∫ f (x) cos s x d x 0

Now, let I

e cos s x d x (1) ( f (x) ∫= ∞

− x2

0

e− x

2

)

Therefore,

(

)

∞ 2 2 1 ∞ dI e − x ( x sin s x) d x = =− −2 x e − x sin s x d x ∫ ∫ 0 ds 2 0 ∞ 1 − x2 1 ∞ 2 e sin s x − ∫ e − x ( s cos s x ) 0 2 2 0 s ∞ 2 s 0 − ∫ e − x cos s x d x = = − I 2 0 2 dI s 0 ⇒ + I= ds 2

=

which is linear differential equation of first order and can be solve by separable method.

Dr. Jogendra Kumar

Page 10

FOURIER TRANSFORMATION s2 I s2 − + log c ⇒ log = − log I = c 4 4

dI s dI s + I= = − ds ⇒ 0⇒ ds 2 I 2 −

ce ⇒ I=

s2 4

(2)

Now by putting s = 0 in (1) and (2), we get

I =

π

e dx ∫= − x2

2

0

and I = c respectively, which implies that c =

π 2

.

Thus from (2), we get

I=

π 2

e

s2 4

, which is Fourier cosine transform of given function.

Que. Find the Fourier sine transforms of e

− x

and hence show that

0

x sin m x π e− m d x , m > 0. = 1 + x2 2

Solution. We know that Fourier sine transform is defined in 0 < x < ∞ and given function can be redefined as

e − x

= Fs {e }

− x

e − ( − x) e x , for x < 0 = =  −x for x ≥ 0 e ,

e sin s x d x ∫ ∫= − x

0

0

e − x sin s x d x ∞

e− x s = (− sin s x − s cos s x) = = Fs (s) 2 1+ s 1 + s2 0 Using inversion formula, we have

2 ∞ 2 ∞ s f (x) = F (s)sin s x d s sin s x d s = s π ∫0 π ∫0 1 + s 2 ∞ s π π −x sin s x d s = f (x) = e ⇒ ∫ 2 0 1+ s 2 2 Dr. Jogendra Kumar

Page 11

FOURIER TRANSFORMATION Now putting x = m in above expression, we get

s π sin m s d s = e − m 2 0 1+ s 2 ∞ x π −m ⇒ ∫ sin m x d x = e (in definite integral we can replace variable by other variable) 2 0 1+ x 2

Que. Find the Fourier sine transforms of

e− a x . x

Solution. By definition Fourier sine transform, we have ∞

Fs {f (x)} = ∫ f (x) sin s x d x . Thus 0

 e− a x  Fs  =   x 

−a x ∞e = ∫0 x sin s x d x I , say

(1)

From (1), ∞ e− a x dI a x e− a= )d x⇒ cos s x d x 2 ∫0 x (x cos s x= ∫ 0 ds s + a2 dI a ds I a. 2 ⇒ = 2 ⇒ d= 2 d s s +a s + a2 s 1 I a. tan −1 + c ⇒ = a a s I tan −1 + c ⇒ = (2) a

dI = ds

Dr. Jogendra Kumar

Page 12

FOURIER TRANSFORMATION After putting s = 0 in (1), we get

I =0

Again put s = 0 in (2), we get

I = tan −1 0 + c = c ⇒ c = 0

( I = 0 for s = 0)

 e− a x  s Fs  tan −1 = a  x 

Que. Find the Fourier cosine transform of f (x) =

ϕ (x) =

1 .Hence derive Fourier sine transform of 1 + x2

x . 1 + x2

Solution: By definition Fourier cosine transforms, we have

 1  = Fc  2 1 + x 

1

cos s x d x ∫= 1+ x 0

2

∞ dI 1 (− x sin s x) d x = ∫ 0 1 + x2 ds ∞ dI x sin s x d x ⇒ = −∫ 0 (1 + x 2 ) ds

I , say

(1)

(2)

∞ dI x2 sin s x d x = −∫ 0 x (1 + x 2 ) ds

Dr. Jogendra Kumar

Page 13

FOURIER TRANSFORMATION 2 ∞ (1 + x − 1) dI = −∫ sin s x d x 0 x (1 + x 2 ) ds

2 ∞ (1 + x ) ∞ sin s x dI s x d x sin = −∫ + ∫0 x(1 + x 2 ) d x 0 x (1 + x 2 ) ds ∞ sin s x ∞ sin s x dI d x+∫ dx = −∫ 0 0 x (1 + x 2 ) ds x ∞ sin s x dI π dx = − +∫ ds 2 0 x(1 + x 2 )

⇒ ⇒ ⇒

d2 I ∴ = d s2

0

x sin s x dx = x(1 + x 2 )

0

(3)

sin s x dx I = (1 + x 2 )

d2 I 0 −I = d s2

Its solution is

= I c1 e s + c2 e − s

dI c1 e s − c2 e − s = ds

(4)

(5)

When s = 0 from (1), we have

I =

0

1 π −1 ∞ d x tan s = = 2 0 1+ x 2

(6)

When s = 0 from (4), we have

π

c1 + c2 = 2

(7)

Dr. Jogendra Kumar

Page 14

FOURIER TRANSFORMATION Now from (3) and (5) when s = 0 , we have

c1 − c2 = −

π

(8)

2

Hence from (7) and (8), we obtained = c1 0= and c2

π 2

∞ 1 π −s  1  = cos s x d x = Fc  I= e 2 2 ∫ 0 1+ x 2 1 + x 

 x  1 + x 

Now= Fs {ϕ (x)} F= s  2

0

x sin s x d x . Thus from (2), 1 + x2

∞ x dI  x  = −∫ − Fs  sin s x d x = 2 2 0 (1 + x ) ds 1 + x 

Therefore from (5), we get ∞ x π −s  x  = −∫ Fs  e sin s x d x = 2 2 0 (1 + x ) 2 1 + x 

Dr. Jogendra Kumar

Page 15

## fourier transformation

1. (x). (s). 2. 1. 2sin sin. (x). (x). 2. Now putting x 0 both sides, we get sin. (0). [ f(0) 1by definition of f(x)] sin sin sin. 2. 2 isx isx isx f. F. e d s s s f e ds e ds f s s s. d s.

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