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)] si
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)
∫
Question-Answer Section Hence evaluate
∞
−∞
∫
∞
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
