The document provides information about Fourier transforms. It defines the Fourier integral transform using a kernel function k(s,x). It presents the Fourier integral theorem relating a function f(x) to its Fourier integral. It gives examples of using the Fourier integral formula to express functions as Fourier integrals and evaluates related integrals. It also defines the complex form of Fourier integrals and Fourier transforms, and presents the inversion formulas. It discusses Fourier sine and cosine transforms and their inversion formulas. It provides problems demonstrating the use of Fourier integral and transform formulas to represent functions and prove identities.
3 examples of PDE, for Laplace, Diffusion of Heat and Wave function. A brief definition of Fouriers Series. Slides created and compiled using LaTeX, beamer package.
Presentation on Fourier Series
contents are:-
Euler’s Formula
Functions having point of discontinuity
Change of interval
Even and Odd functions
Half Range series
Harmonic analysis
3 examples of PDE, for Laplace, Diffusion of Heat and Wave function. A brief definition of Fouriers Series. Slides created and compiled using LaTeX, beamer package.
Presentation on Fourier Series
contents are:-
Euler’s Formula
Functions having point of discontinuity
Change of interval
Even and Odd functions
Half Range series
Harmonic analysis
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1. CHAPTER 4
FOURIER TRANSFORMS
INTEGRAL TRANSFORM
b
The integral transform of a function f (x ) is defined by ∫ f ( x).k (s , x)dx where
a
k(s , x) is a known function of s and x and it is called the kernel of the transform.
When k(s , x) is a sine or cosine function, we get transforms called Fourier sine or
cosine transforms.
FOURIER INTEGRAL THEOREM
If f (x) is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then
∞ ∞
1
π ∫ −∫
f ( x) = f (t ) cos λ (t − x) dt dλ
0 ∞
At a point of discontinuity the value of the integral on the left of above equation is
1
{ f ( x + 0) − f ( x − 0)}.
2
EXAMPLES
1 for x ≤ 1
1. Express the function f ( x) = as a Fourier Integral. Hence evaluate
0 for x > 1
∞ ∞
sin λ cos λx sin λ
∫ λ
0
dλ and find the value of ∫
0
λ
dλ .
Solution:
We know that the Fourier Integral formula for f (x) is
∞ ∞
1
f ( x) =
π ∫ ∫ f (t ) cos λ (t − x) dt dλ
0 −∞
……………….(1)
Here f (t ) = 1 for t ≤ 1 i.e., f(t) = 1 in -1 < t < 1
f (t ) = 0 for t > 1
f (t ) = 0 in − ∞ < t < −1 and 1 < t < ∞
∞ 1
1
∴ Equation (1) ⇒ f ( x) =
π ∫ ∫ cos λ (t − x) dt dλ
0 −1
∞ 1
1 sin λ (t − x)
π ∫
= dλ
0
λ −1
∞
1 sin λ (1 − x) − sin λ (−1 − x)
π∫
= dλ
0
λ
∞
1 sin λ (1 − x) + sin λ (1 + x )
= ∫ dλ
π 0 λ
1
2. ∞
2 sin λ cos λx
π∫
∴ f ( x) = dλ .………………(2)
0
λ
[Using sin (A+B) + sin (A-B) = 2 sin A cos B]
This is Fourier Integral of the given function. From (2) we get
∞
sin λ cos λx π
∫
0
λ
dλ =
2
f ( x) ……………….(3
1 for x ≤ 1
But f ( x) = ………………..(4)
0 for x > 1
Substituting (4) in (3) we get
π
∞
sin λ cos λx for x ≤ 1
∫ λ dλ = 2
0 0 for x > 1
∞
sin λ π
Putting x = 0 we get ∫0
λ
dλ =
2
2. Find the Fourier Integral of the function
0 x<0
1
f ( x) = x=0
2
e − x x > 0
Verify the representation directly at the point x = 0.
Solution:
The Fourier integral of f (x) is
∞ ∞
1
f ( x) =
π ∫ ∫ f (t ) cos λ (t − x) dt dλ
0 −∞
……………….(1)
1
∞ 0 ∞
= ∫ −∫∞
π 0
f (t ) cos λ (t − x )dt + ∫ f (t ) cos λ (t − x)dt d λ
0
1
∞ 0 ∞
= ∫ ∫ 0. cos λ (t − x)dt + ∫ e −t cos λ (t − x)dt dλ
π 0 − ∞ 0
∞ ∞
1 e −t
= ∫ 2 [ − cos( λt − λx ) + λ sin(λt − λx)] dλ
π 0 λ +1 0
∞
1 cos λx + λ sin λx
π∫
f (x) = dλ ……….………(2)
0 λ2 + 1
2
3. Putting x = 0 in (2), we get
∞
1
f (0) = ∫ 2
1
π 0 λ +1
1
d λ = tan −1 ( λ ) 0
π
∞
[ ]
1
[
= tan −1 ( ∞ ) − tan −1 (0)
π
]
1 π 1
= =
π 2 2
1
The value of the given function at x = 0 is . Hence verified.
2
FOURIER SINE AND COSINE INTEGRALS
The integral of the form
2∞ ∞
f ( x) = ∫ sin λx ∫ f (t ) sin λt dt dλ
π0 0
is known as Fourier sine integral.
The integral of the form
∞ ∞
2
f ( x) = ∫ cos λx ∫ f (t ) cos λt dt dλ
π 0 0
is known as Fourier cosine integral.
PROBLEMS
1. Using Fourier integral formula, prove that
2(b 2 − a 2 ) ∞ u sin xu
∫ (u 2 + a 2 )(u 2 + b 2 ) du (a, b > 0)
− ax − bx
e −e =
π 0
Solution:
The presence of sin xu in the integral suggests that the Fourier sine integral formula
has been used.
Fourier sine integral representation is given by
∞ ∞
2
f ( x) = ∫ sin ux ∫ f (t ) sin ut dt du
π 0 0
∞
∞
sin ux du ∫ ( e − at − e −bt ) sin ut dt
2
e − ax − e −bx = ∫
π 0 0
3
4. ∞ ∞
2 e − at − bt
= ∫ sin ux du 2 { − a sin ut − u cos ut} − 2e 2 { − b sin ut − u cos ut}
π 0 a + u
2
b +u 0
∞
2 u u
= ∫ sin ux du a 2 + u 2 − b 2 + u 2
π 0
∞
2(b 2 − a 2 ) u sin ux
=
π ∫ (u 2 + a 2 )(u 2 + b 2 ) du
0
2. Using Fourier integral formula, prove that
2
∞
(λ 2
+ 2 ) cos xλ
∫ dλ
−x
e cos x =
π 0 λ2 + 4
Solution:
The presence of cos xλ in the integral suggests that the Fourier cosine integral
formula for e − x cos x has been used.
Fourier cosine integral representation is given by
∞ ∞
2
π∫
f ( x) = cos λx ∫ f (t ) cos λt dt dλ
0 0
2∞ ∞ − t
∴e −x
cos x = ∫ cos xλ dλ ∫ e cos t cos λt dt
π 0 0
2
∞
1 ∞
= ∫ cos xλ dλ ∫ e −t { cos(λ + 1)t + cos(λ − 1)t } dt
π 0 2 0
∞
=
2
π 0
1 −t
[
∫ cos xλ dλ (λ + 1) 2 + 1 e { − cos(λ + 1)t + (λ + 1) sin(λ + 1)t} ] ∞
0
+
1
(λ − 1) + 1
2
[
e −t { − cos(λ − 1)t + (λ − 1) sin(λ − 1)t } 0
∞
] )
∞
1 1 1
= ∫ + cos xλ dλ
π 0 (λ + 1) + 1 (λ − 1) + 1
2 2
∞
2 (λ2 + 2) cos xλ
= ∫ dλ.
π 0 λ2 + 4
COMPLEX FORM OF FOURIER INTEGRALS
4
5. The integral of the form
∞ ∞
1
∫ e − iλx ∫ f (t ) e
iλt
f ( x) = dt d λ
2π −∞ −∞
is known as Complex form of Fourier Integral.
FOURIER TRANSFORMS
COMPLEX FOURIER TRANSFORMS
∞
1
The function F [ f ( x)] = ∫∞ f (t ).e dt is called the Complex Fourier transform
ist
2π −
of f (x ) .
INVERSION FORMULA FOR THE COMPLEX FOURIER TRANSFORM
∞
1
∫∞F [ f ( x)].e ds is called the inversion formula for the
−isx
The function f ( x) =
2π −
Complex Fourier transform of F [ f ( x)] and it is denoted by F −1 [ F ( f ( x))].
FOURIER SINE TRANSFORMS
∞
2
The function FS [ f ( x )] = ∫ f (t ).sin st dt is called the Fourier Sine Transform of
π 0
the function f (x ) .
∞
2
The function f ( x) =
π ∫ F [ f ( x)]. sin sx ds is called the inversion formula for the
0
S
Fourier sine transform and it is denoted by FS
−1
[ FS ( f ( x))].
FOURIER COSINE TRANSFORMS
∞
2
The function FC [ f ( x)] =
π ∫
f (t ). cos st dt is called the Fourier Cosine
0
Transform of f (x) .
∞
2
The function f ( x) =
π ∫ F [ f ( x)]. cos sx ds
0
C is called the inversion formula for the
Fourier Cosine Transform and it is denoted by FC
−1
[ FC ( f ( x))].
PROBLEMS
1. Find the Fourier Transform of
1 − x 2 in x ≤ 1
f ( x) =
0
in x > 1
5
6. ∞
sin s − s cos s s 3π
Hence prove that ∫
0 s 3
cos ds =
2 16
.
Solution:
We know that the Fourier transform of f (x) is given by
∞
1
F [ f ( x )] = ∫ f ( x).e
isx
dx
2π −∞
−1 1 ∞
1 1 1
= ∫ f ( x).e dx + ∫ f ( x).e dx + ∫ f ( x).e
isx isx isx
dx
2π −∞ 2π −1 2π 1
−1 1 ∞
1 1 1
= ∫ 0.e dx + ∫ (1 − x ).e dx + ∫ 0.e
isx 2 isx isx
dx
2π −∞ 2π −1 2π 1
1
1
= ∫ (1 − x
2
).e isx dx
2π −1
1
1 e isx e isx e isx
= (1 − x 2 ) − ( −2 x ) 2 2 − 2 3 3
2π is i s i s −1
1 − 2 is 2 is − 2 −is 2 e −is
= 2 e + 3e + 2 e −
2π s is s i s3
1 − 2 is 2
= s (e + e −is ) + 3 (e is − e −is )
2π
2
is
1 − 4 4 1 4
= s 2 cos s + s 3 sin s = s 3 (sin s − s cos s )
2π 2π
By using inverse Fourier Transform we get
∞
1 1 4
f ( x) =
2π
.
2π
∫
−∞ s 3
(sin s − s cos s ).e −isx ds
∞
1 4
=
2π ∫s
−∞
3
(sin s − s cos s ).(cos sx − i sin sx ) ds
∞
1 4
=
2π ∫
−∞ s3
(sin s − s cos s ) cos sx ds
∞
1 4
−
2π ∫
−∞ s3
(sin s − s cos s ) i sin sx ds
6
7. The second integral is odd and hence its values is zero.
∞
2 sin s − s cos s
π −∫
∴ f ( x) = cos sx ds
∞ s3
∞
4 sin s − s cos s
= ∫ cos sx ds
π 0 s3
∞
sin s − s cos s π
i.e., ∫
0 s 3
cos sx ds = f ( x)
4
1
Putting x = , we get
2
∞
sin s − s cos s s π 1 π 1 3π
∫0 s 3
cos ds =
2
f = 1 − =
4 2 4 4 16
.
∞
sin s − s cos s s 3π
∫
0 s 3
cos ds =
2 16
.
2. Find the Fourier sine transform of e − x , x ≥ 0 (or) e − x , x > 0. Hence evaluate
∞
x sin mx
∫ 1 + x 2 dx.
0
Solution:
The Fourier sine transform of f(x) is given by
∞
2
FS [ f ( x )] = ∫ f ( x).sin sx dx
π 0
−x −x
Here e =e for x > 0
∞
[ ]
FS e − x =
2
π ∫e
−x
. sin sx dx
0
2 s ∞ − ax b
= ∫ e sin bx dx = 2
π s2 +1 0 a + b2
Using inverse Fourier sine transform we get
∞
∫ F [e ]. sin sx ds
2 −x
f ( x) = s
π 0
∞
2 2 s
=
π ∫
0
. 2
π s +1
. sin sx ds
∞
2 s
= ∫ s 2 + 1 sin sx ds
π 0
7
8. ∞
π s
i.e., f ( x) = ∫ 2 . sin sx ds
2 0 s +1
∞
s. sin sx π
i.e., ∫0 s +1
2
ds = e − x
2
Replacing x by m we get
∞
s. sin ms π
i.e., ∫
0 s +1
2
ds = e − m
2
∞
x. sin mx π
i.e., ∫
0 x +1
2
dx = e −m
2
[since s is dummy variable, we can replace it by x]
e − ax
3. Find the Fourier cosine transform of .
x
Solution:
∞
2
We know that FC [ f ( x )] = ∫ f ( x). cos sx dx
π 0
− ax
e
Here f ( x) = .
x
∞
2 e − ax
∴ FC [ f ( x )] = ∫ x . cos sx dx
π 0
Let FC [ f ( x)] = F ( s )
∞
2 e − ax
Then F (s) =
π ∫ x . cos sx dx
0
………………(1)
Differentiating on both sides w.r.t. ‘s’ we get,
∞
dF ( s ) d 2 e − ax
ds
=
ds π ∫ x . cos sx dx
0
∞
2 ∂ e −ax
= ∫ ∂s x . cos sxdx
π0
∞ ∞
2 e −ax 2 − ax
= ∫ x (− sin sx).x dx = − π ∫ e sin sx dx
π0 0
dF ( s ) 2 s ∞
b
∫e
− ax
=− . 2 sin bx dx =
ds π a + s2 0 a + b2
2
Integrating w.r.t. ‘s’ we get
8
9. 2 s
F (s) = − .∫ 2 ds
π s + a2
2 1 1
=− . . log ( s 2 + a 2 ) = − . log ( s 2 + a 2 )
π 2 2π
e − ax − e −bx
4. Find the Fourier cosine transform of .
x
Solution:
We know that the Fourier cosine transform of f(x) is
∞
2
FC [ f ( x )] = ∫ f ( x). cos sx dx
π 0
− ax
e − e −bx
Here f ( x) =
x
∞
e − ax − e −bx 2 e − ax − e − bx
∴ FC
x
=
π ∫ x . cos sx dx
0
∞ ∞
2 e − ax 2 e −bx
=
π ∫ x . cos sx dx − π ∫ x cos sx dx
0 0
e −ax e −bx
= Fc − Fc
x x
1 1
=− log ( s 2 + a 2 ) + log ( s 2 + b 2 )
2π 2π
1 s2 + b2
= log 2
s + a2
2π
e − as
5. Find f (x) , if its sine transform is . Hence deduce that the inverse sine
s
1
transform of .
s
Solution:
We know that the inverse Fourier sine transform of FS [ f (x )] is given by
∞
2
f ( x) =
π ∫ F [ f ( x)]. sin sx ds
0
S
e − as
Here FS [ f ( x )] =
s
9
10. ∞
2 e − as
∴ f ( x) =
π ∫ s . sin sx ds
0
Differentiating w.r.t. ‘x’ on both sides, we get,
d [ f ( x )]
∞
2 e − as ∂
dx
=
π ∫ s . ∂x (sin sx) ds
0
∞ ∞
2 e − as 2 − as
=
π ∫ s . cos sx s ds = π ∫ e . cos sx ds
0 0
2 a ∞ −ax a
= ∫ e cos bx dx = 2
π a + x2
2
0 a + b2
d [ f ( x )] 2 a
=
dx π x + a2
2
2 1 2 1 −1 x
∴ f ( x) = a
π ∫ x 2 + a 2 dx = a π a tan a
2 x
∴ f ( x) = tan −1
π a
1
To find the inverse Fourier sine transform of :
s
Put a = 0, in (1), we get
2 2 π π
f ( x) = tan −1 (∞) = . =
π π 2 2
PROPERTIES
1. Linearity Property
If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then
F [ a f ( x ) + b g ( x)] = a F ( s) + b G ( s )
Proof:
∞
1
F [ a f ( x) + b g ( x)] = ∫∞[ a f ( x) + b g ( x)] e dx
isx
2π −
∞ ∞
1 1
= ∫ a f ( x).e dx + ∫ b g ( x).e
isx isx
dx
2π −∞ 2π −∞
∞ ∞
a b
= ∫ f ( x).e isx dx + ∫ g ( x).e
isx
dx
2π −∞ 2π −∞
= a F ( s ) + b G ( s)
10
11. 2. Change of Scale Property
1 s
If F(s) is the Fourier transform of f (x ) then F [ f (ax)] = F ,a>0
a a
Proof:
∞
1
F [ f (ax)] = ∫ f (ax).e
isx
dx
2π −∞
Put ax = y
dy
a dx = dy i.e., dx =
a
When x = −∞, y = −∞ and x = ∞, y=∞
∞ y ∞ s
1 dy 1 1 i y
F [ f (ax)] =
is
2π
∫ f ( y).e
−∞
a
. =
a a 2π
∫ f ( y).e
−∞
a
.dy
1
= F ( s a)
a
3. Shifting Property ( Shifting in x )
If F(s) is the Fourier transform of f (x ) then F [ f ( x − a )] = e ias F ( s )
Proof:
∞
1
F [ f ( x − a)] = ∫ f ( x − a).e
isx
dx
2π −∞
Put x-a = y
dx = dy
When x = −∞, y = −∞ and x = ∞, y=∞
∞ ∞
1 e ias
F [ f ( x − a)] = ∫ f ( y ).e is ( y + a ) . dy = ∫ f ( y).e
isy
.dy
2π −∞ 2π −∞
∞
e ias
= ∫ f ( x).e .dx = e isa F ( s )
isx
2π −∞
4. Shifting in respect of s
If F(s) is the Fourier transform of f (x ) then F e iax f ( x) = F ( s + a ) [ ]
Proof:
∞
Fe[ iax
]
f ( x) =
1
∫e
iax
f ( x) e isx dx
2π −∞
11
12. ∞
1
∫ f ( x).e
i( s+a) x
= dx = F ( s + a)
2π −∞
5. Modulation Theorem
1
If F(s) is the Fourier transform of f (x ) then F [ f ( x) cos ax ] = [ F ( s + a ) + F ( s − a )]
2
Proof:
∞
1
F [ f ( x) cos ax ] = ∫ f ( x). cos ax.e
isx
dx
2π −∞
∞
1 e iax + e −iax
= ∫
2π −∞
f ( x).e isx
2
dx
∞ ∞
1 1 1 1
∫∞ f ( x).e dx + 2 . 2π ∫ f ( x).e
i( s+a ) x i ( s −a ) x
= . dx
2 2π − −∞
1 1 1
= f ( s + a ) + f ( s − a ) = [ f ( s + a ) + f ( s − a )]
2 2 2
1
F [ f ( x) cos ax ] = [ F ( s + a ) + F ( s − a)]
2
COROLLARIES
1
(i ) FC [ f ( x) cos ax ] = [ FC ( s + a) + FC ( s − a)]
2
1
(ii ) FC [ f ( x) sin ax ] = [ FS (a + s) + FS (a − s )]
2
1
(iii ) FS [ f ( x) cos ax ] = [ FS ( s + a ) + FS ( s − a )]
2
1
(iv ) FS [ f ( x) sin ax ] = [ FC ( s − a ) − FC ( s + a)]
2
6. Conjugate Symmetry Property
If F(s) is the Fourier transform of f (x ) then F f ( − x) = F ( s) [ ]
Proof:
∞
1
We know that F ( s ) = ∫ f ( x). e
isx
dx
2π −∞
Taking complex conjugate on both sides we get
∞
1
∫∞ f ( x). e dx
−isx
F (s) =
2π −
12
13. Put x = -y
dx = -dy
When x = −∞, y = ∞ and x = ∞, y = −∞
−∞
1
∴ F (s) = ∫ f (− y) .e (− dy )
isy
2π ∞
−∞
1
=− ∫ f (− y). e
isy
dy
2π ∞
[ ]
∞
1
= ∫ f (− x). e dx = F f (− x)
isx
2π −∞
7. Transform of Derivatives
If F(s) is the Fourier transform of f (x ) and if f (x) is continuous, f ′(x) is piecewise
continuously differentiable, f (x ) and f ′(x) are absolutely integrable in (−∞ , ∞) and
lim [ f ( x)] = 0 , then
x → ±∞
F ( f ′( x ) ) = −is F ( s )
Proof:
By the first three conditions given, F { f (x)} and F { f ′(x)} exist.
∞
1
F { f ′( x)} = ∫ f ′( x) e
isx
dx
2π −∞
∞
=
1
[e isx
] ∞
f ( x) −∞ −
is
∫e
isx
f ( x) dx, on int egrating by parts.
2π 2π −∞
= 0 − isF { f ( x)} , by the given condition.
= −is F ( s).
The theorem can be extended as follows.
If f , f ′, f ′′, , f ( n −1) are continuous, f (n ) is piecewise continuous, f , f ′, f ′′, , f ( n )
are absolutely integrable in (−∞ , ∞) and f , f ′, f ′′, , f ( n −1) → 0 as x → ±∞ , then
F ( f ( n ) ( x) ) = (−is ) n F ( s )
13
14. 8. Derivatives of the Transform
dF ( s )
If F(s) is the Fourier transform of f (x ) then F [ x. f ( x )] = (−i )
ds
Proof:
∞
1
F (s) = ∫ f ( x )e
isx
dx
2π −∞
∞
∫ ds [ f ( x)e ]dx
dF ( s ) 1 d
∴ = isx
ds 2π −∞
∞
i
= ∫ [ x. f ( x)]e dx = iF [ xf ( x)]
isx
2π −∞
dF ( s)
∴ ( −i ) = F [ x. f ( x)]
ds
[ ]
Extending, we get, F x n . f ( x) = (−i ) n
d n F (s)
ds n
DEFINITION
∞
1
∫ f ( x − u ) g (u )du is called the convolution product or simply the convolution
2π −∞
of the functions f (x) and g (x) and is denoted by f ( x) * g ( x ) .
9. Convolution Theorem
If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then the
Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier
transforms.
i.e., F [ f ( x ) * g ( x)] = F ( s ).G ( s)
Proof
∞
1
F [ f ( x ) * g ( x )] = ∫∞ f ( x) * g ( x)e dx
isx
2π −
1 1 ∞
∞
isx
= ∫∞ 2π −∫∞ f ( x − u) g (u )du e dx
2π −
1
∞
1 ∞
= ∫∞ 2π −∫∞ f ( x − u)e dxdu,
isx
g (u )
2π −
on changing the order of int egration.
14
15. ∞
∫ g (u )[e ]
1
= ius
F ( s) du , by the shifting property.
2π −∞
∞
1
= F ( s). ∫ g (u ).e
ius
du
2π −∞
= F ( s).G ( s )
Inverting, we get
F −1 [ F ( s ).G ( s)] = f ( x) * g ( x )
= F −1 { F ( s )} * F −1 { G ( s )}
10. Parseval’s Identity (or) Energy Theorem
If f (x) is a given function defined in (−∞ , ∞) then it satisfy the identity,
∞ ∞
∫ ∫ F ( s)
2 2
f ( x) dx = ds
−∞ −∞
where F(s) is the Fourier transform of f (x ) .
Proof:
We know that F −1 [ F ( s ).G ( s)] = f ( x) * g ( x )
∞ ∞
1 1
∫ F (s).G(s)e ∫ f (t ) g ( x − t )dt
−isx
ds =
2π −∞ 2π −∞
Putting x = 0, we get
∞ ∞
∫ F (s).G(s)ds = ∫ f (t ) g (−t )dt
−∞ −∞
………………..(1)
Let g ( − t ) = f (t ) .……………….(2)
i.e., g (t ) = f (−t ) ………………..(3)
∴ G ( s ) = F [ f (− x)] = F ( s ) by property (9)
i.e., ∴ G ( s) = F ( s) ………………..(4)
Substituting (2) and (4) in (1) we get
∞ ∞
∫ F (s).F (s) ds =
−∞
∫ f (t ). f (t ) dt
−∞
[ F (s).F (s) = F (s) ]
∞ ∞
∫ ∫
2 2 2
F ( s) ds = f ( x) dx
−∞ −∞
11. If f (x) and g (x) are given functions of x and FC [ f ( x)] and FC [ g ( x)] are their
Fourier cosine transforms and FS [ f ( x)] and FS [ g ( x)] are their Fourier sine transforms then
15
16. ∞ ∞ ∞
(i) ∫
0
f ( x ) g ( x)dx = ∫ FC [ f ( x)].FC [ g ( x)]ds = ∫ FS [ f ( x )].FS [ g ( x)]ds
0 0
∞ ∞ ∞
∫ f ( x) dx = ∫ FC [ f ( x)] ds = ∫ FS [ f ( x )] ds ,
2 2 2
(ii)
0 0 0
which is Parseval’s identity for Fourier cosine and sine transforms.
Proof:
∞ ∞
2∞
(i) ∫ FC [ f ( x )].FC [ g ( x )]ds = ∫ FC [ f ( x)] ∫ g ( x) cos sx dx ds
0 0 π 0
∞
2∞
= ∫ g ( x) ∫ FC [ f ( x)] cos sx ds dx,
0 π 0
Changing the order of integration
∞
= ∫ f ( x) g ( x)dx
0
Similarly we can prove the other part of the result.
(ii) Replacing g ( x) = f * ( x) in (i) and noting that FC [ f ( x)] = FC [ f ( x)] and
FS [ f ( x)] = FS [ f ( x)] , we get
∞ ∞ ∞
∫0
f ( x ). f ( x).dx = ∫ FC [ f ( x)].FC [ f ( x)] ds = ∫ FS [ f ( x )].FS [ f ( x )] ds
0 0
∞ ∞ ∞
∫ f ( x) .dx = ∫ FC [ f ( x)] ds = ∫ FS [ f ( x)] ds
2 2 2
i.e.,
0 0 0
12. If FC [ f ( x)] = FC ( s ) and FS [ f ( x)] = FS ( s ) , then
d
(i) { FC ( s)} = − FS { xf ( x)} and
ds
d
(ii) { FS ( s)} = − FC { xf ( x)}.
ds
Proof:
∞
2
π∫
FC ( s) = f ( x) cos sx dx
0
∞
d
{ FC ( s)} = ∫ f ( x)(− x sin sx)dx
ds 0
∞
= − ∫ {xf ( x)}sin sx dx
0
= − FS {xf ( x)}
Similarly the result (ii) follows.
16
17. PROBLEMS
a 2 − x 2
x <a
1. Show that the Fourier transform of f ( x ) = is
0
x >a>0
∞
2 sin as − as cos as sin t − t cos t π
2 . Hence deduce that ∫ dt = . Using Parseval’s
π s 3
0 t 3
4
∞ 2
sin t − t cos t π
identity show that ∫ 3 dt = .
0
t 15
Solution:
We know that
∞
1
F [ f ( x )] = ∫ f ( x).e
isx
dx
2π −∞
1
−a a ∞
= ∫ f ( x)e dx + ∫ f ( x)e dx + ∫ f ( x )e dx
isx isx isx
2π −∞ −a a
1
a a
1
= 0 + ∫ ( a 2 − x 2 )e isx dx + 0 = ∫ (a
2
− x 2 ).e isx dx
2π − a 2π −a
a
1 2 e isx e isx e isx
= ( a − x 2 )
is − (−2 x) 2 2 − 2 3 3
i s i s
2π − a
=
1 − 2a isa
2 e +e [−isa 2i
] [
+ 3 e −isa − e isa ]
2π s s
1 − 2a 4
= 2 [2 cos as ] + 3 sin as
2π s s
4 sin as − as cos as
=
2π
s3
2 sin as − as cos as
F (s) = 2
π s3
2 sin s − s cos s
When a = 1, F ( s ) = 2 ………………..(A)
π s3
Using inverse Fourier Transform, we get
17
18. ∞
1 2 1
f ( x) =
2π
.2.
π ∫s
−∞
3
[sin as − as cos as].e −isx ds
∞
1 2 1
{ sin as − as cos as}{ cos sx − i sin sx} ds
π −∫ s 3
= .2.
2π ∞
∞
2 sin as − as cos as
π −∫
f ( x) = cos sx ds
∞ s3
[The second integral is odd and hence its value is zero]
∞
4 sin as − as cos as
= ∫ cos sx ds
π 0 s3
[since the integrand is an even function of s]
Putting a = 1, we get
∞
4 sin s − s cos s
f ( x) = ∫ cos sx ds
π 0 s3
∞
4 sin t − t cos t
f ( x) = ∫ cos tx dt
π 0 t3
Putting x = 0, in the given function we get
∞
4 sin t − t cos t
π∫
dt = f (0) = 1
0 t3
∞
sin t − t cos t π
∴ ∫
0 t 3
dt =
4
∞ ∞
∫ ∫ F (s)
2 2
Using Parseval’s identity, f ( x) dx = ds [Using (A)]
−∞ −∞
2
2
∞ 2. (sin s − s cos s) 1
π
∫
− ∞ s3
ds = ∫ (1 − x 2 ) 2 dx
−1
∞ 2 1
8 sin s − s cos s
∫∞ ds = 2 ∫ (1 − x ) dx
2 2
π− s 3
0
∞ 2
16 sin s − s cos s 8
∫
π 0 s 3 ds = 2.
15
18
19. ∞ 2
sin s − s cos s π
i.e., ∫
0 s 3 ds =
15
∞ 2
sin t − t cos t π
i.e., ∫ t3
0
dt =
15
2. Find the Fourier Transform of f (x) if
1 − x , x < 1
f ( x) =
0,
x >1
∞ 4
sin t π
Hence deduce that ∫ dt =
0
t 3
Solution:
We know that
∞
1
F [ f ( x )] = ∫∞ f ( x).e dx
isx
2π −
1
∫ (1 − x ).e
1
= isx
dx
2π −1
Since x > 1, f ( x) = 0, i.e., in − ∞ < x < −1, and 1 < x < ∞, f ( x ) = 0.
1
∫ [1 − x ] [ cos sx + i sin sx] dx
1
=
2π −1
1
∫ [1 − x ] cos sx dx
1
=
2π −1
The second integral becomes zero since it is an odd function.
1
2
=
2π
∫ (1 − x) cos sx dx
0
[ [1 − x ] cos x is an even function]
1
2 sin sx − cos sx
= (1 − x ) s − (−1) s 2
π 0
2 − cos s 1
= + 2
π s2 s
2 1
i.e., F ( s) = . (1 − cos s )
π s2
19
20. Using Parseval’s identity
∞ ∞
∫ ∫ F ( s)
2 2
f ( x) dx = ds
−∞ −∞
1 ∞
[1 − x ] dx = 2 ∫ (1 − cos s) ds
2
∫
2
−1
π −∞ s4
1 ∞
2 (1 − cos s) 2
2 ∫ [1 − x ] dx =
π −∫
2
ds
0 ∞ s4
∞
2 2 (1 − cos s ) 2
3 π −∫
= ds
∞ s4
put s = 2t when s = ∞, t = ∞
ds = 2dt when s = −∞, t = −∞
π ∞ (1 − cos 2t ) 2
3 −∫
= .2 dt
∞ 16t 4
π ∞ (1 − cos 2t ) 2
3 −∫
= dt
∞ 8t 4
∞
π (1 − cos 2t ) 2
= 2∫ dt
3 0 8t 4
π ∞ sin 4 t
3 ∫ t4
= dt
0
∞ 4
sin t π
i.e., ∫ t dt = 3
0
∞
dx
3. Evaluate ∫ (x
0
2
+ a )( x 2 + b 2 )
2 using transforms.
Solution:
2 a
We know that the Fourier cosine transform of f ( x) = e − ax is . 2 .
π s + a2
2 b
Similarly the Fourier cosine transform of f ( x) = e − ax is . 2 .
π s + b2
20
21. ∞ ∞
We know that ∫ FC [ f ( x)].FC [ g ( x)] ds = ∫ f ( x).g ( x) dx
0 0
∞ ∞
2 a 2 b
i.e., ∫
0
. 2 . . 2
π s +a π s +b
2 2
. ds = ∫ e − ax .e −bx dx
0
∞ ∞
2 ab
∫ (s 2 + a 2 )(s 2 + b 2 ) ds = ∫ e
−( a +b ) x
i.e., ds
π 0 0
∞
e −( a + b ) x 1 1
= = 0− =
− ( a + b) 0 − ( a + b) a + b
∞
dx π
i.e., ∫ (x
0
2 2 2 2
=
+ a )( x + b ) 2ab( a + b)
4. Find the Fourier transform of e − a x and hence deduce that
∞
cos xt π −a x
(i) ∫ 2 2 dt = e
0 a +t
2a
[
(ii) F xe
−a x
]=i 2 2as
π (s + a 2 ) 2
2
Solution:
∞
1
F [ f ( x )] = ∫ f ( x).e
isx
dx
2π −∞
1
0 ∞
= ∫ f ( x)e isx dx + ∫ f ( x)e isx dx
2π −∞ 0
e − ax
if 0 ≤ x < ∞
Here f ( x ) = ax
e
if − ∞ < x < 0
1 ax isx
0 ∞
= ∫ e. e dx + ∫ e −ax .e isx dx
2π −∞ 0
1 ( a +is ) x
0 ∞
= ∫ e. dx + ∫ e −( a −is ) x dx
2π −∞ 0
21
22. 1 e ( a +is ) x e −( a −is ) x
0 ∞
= +
2π (a + is) −∞ − (a − is ) 0
1 1 1
= a + is + a − is
2π
Fe[ −a x
]= 2 a
π s + a2
2
Using inversion formula, we get
∞
1 2 a
f ( x) =
2π
∫
−∞
. 2
π s +a 2
e −isx ds
∞
a cos sx − i sin sx
π −∫
= ds
∞ s2 + a2
∞
a cos x
=
π−∫∞ s 2 + a 2 ds
∞
cos sx π π −a x
∫s
0
2
+a 2
dx =
2a
f ( x) =
2a
.e (or )
∞
cos tx π −a x
∫s
0
2
+a 2
dt =
2a
.e
Putting a = 1, we get,
Fe[ ]=−x 2 1
.
π s2 +1
∞ ∞
cos sx π −x cos tx π −x
and ∫ s 2 + 1 ds = 2 e
0
(or ) ∫t
0
2
+1
dt = e
2
FINITE FOURIER TRANSFORMS
If f (x) is a function defined in the interval (0 , l) then the finite Fourier sine
transform of f (x) in 0 <x < l is defined as
22
23. nπx
l
FS [ f ( x)] = ∫ f ( x). sin dx where ‘n’ is an integer
0
l
The inverse finite Fourier sine transform of FS [ f (x )] is f (x) and is given by
2 ∞ nπx
f ( x) = ∑ FS [ f ( x )] sin
l n =1 l
The finite Fourier cosine transform of f (x ) in 0 < x < l is defined as
nπx
l
FC [ f ( x)] = ∫ f ( x). cos dx where ‘n’ is an integer
0
l
The inverse finite Fourier cosine transform of FC [ f (x)] is f (x) and is given by
1 2 ∞ nπx
f ( x) = FC (0) + ∑ FC [ f ( x )] cos
l l n =1 l
PROBLEMS
1. Find the finite Fourier sine and cosine transforms of f ( x) = x 2 in 0 < x < l.
Solution:
The finite Fourier sine transform is
nπx
l
FS [ f ( x)] = ∫ f ( x). sin dx
0
l
Here f ( x) = x 2
[ ] nπx
l
FS x 2 = ∫ x 2 . sin dx
0
l
l
nπx nπx nπx
− cos − sin cos
= x 2 l − 2 x l + 2 l
nπ n 2π 2 n 3π 3
l l2 l 3 0
− l3 2l 3 2l 3
= cos nπ + 3 3 cos nπ − 3 3
nπ nπ nπ
=
l3
nπ
2l 3
[
(−1) n +1 + 3 3 (−1) n − 1
nπ
]
The finite Fourier cosine transform is
23
24. nπx
l
FC [ f ( x)] = ∫ f ( x). cos dx
0
l
Here f ( x) = x 2
[ ] nπx
l
FC x 2 = ∫ x 2 . cos dx
0
l
l
nπx nπx nπx
sin − cos − sin
= x 2 l − 2 x l + 2 l
nπ n 2π 2 n 3π 3
l
l2 l3 0
2l 3
= cos nπ
n 2π 2
2l 3
= (−1) n
nπ
2 2
2. Find the finite Fourier sine and cosine transforms of f ( x ) = x in (0 , π ) .
Solution:
The finite Fourier sine transform of f ( x) = x in (0 , π ) is
π
FS [ f ( x)] = ∫ f ( x ). sin nx dx
0
Here f ( x) = x in (0 , π )
π π
− cos nx − sin nx
FS [ x ] = ∫ x. sin nx dx = x − 1 2
0 n n 0
π π
=− cos nπ = (−1) n +1 .
n n
The finite Fourier cosine transform of f ( x) = x in (0 , π ) is
π
FC [ f ( x)] = ∫ f ( x). cos nx dx
0
Here f ( x) = x in (0 , π )
π π
sin nx − cos nx
FC [ x ] = ∫ x. cos nx dx = x − 1 2
0 n n 0
=
1
n 2
1 1
[
cos nπ − 2 = 2 (−1) n − 1
n n
]
24
25. 2π (−1) p−1
3. Find f (x) if its finite sine transform is given by , where p is positive
p3
integer and 0 < x < π .
Solution:
We know that the inverse Fourier sine transform is given by
∞
2
f ( x) = ∑ FS [ f ( x )] sin px ………………..(1)
π p =1
2π (−1) p−1
Here FS [ f (x )] = ………………..(2)
p3
Substituting (2) in (1), we get
2 ∞ 2π (−1) p −1
f ( x) = ∑ sin px
π p =1 p3
∞
(−1) p −1
= 4∑ sin px
p =1 p3
2 pπ
cos
3 find FC [ f ( p )] if 0 < x <1.
−1
4. If
f ( p) =
(2 p + 1) 2
Solution:
∞
nπx
We know that FC
−1
[ f ( p)] = 1 FC (0) + 2 ∑ FC [ f ( x)] cos
l l n =1 l
2 pπ
cos
Here
f ( p) = 3
(2 p + 1) 2
Let FC [ f ( x)] = f ( p)
∞
nπx
∴ FC
−1
[ f ( p)] = 1 f C ( 0) +
2
∑ f ( p ) cos [ l = 1]
l l n =1 l
2 pπ
∞
cos
= 1 + 2∑ 3 . cos nπx
n =1 (2 p + 1) 2
25
26. UNIT-4
PART A
1. State the Fourier integral theorem.
Ans:
If f (x) is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then
∞ ∞
1
f ( x) =
π ∫ ∫ f (t ) cos λ (t − x) dt dλ
0 −∞
2. State the convolution theorem of the Fourier transform.
Ans:
If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then the
Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier
transforms.
i.e., F [ f ( x ) * g ( x)] = F ( s ).G ( s)
3. Write the Fourier transform pair.
Ans:
F [ f (x)] and F −1 [ F ( S )] are Fourier transform pairs.
4. Find the Fourier sine transform of f ( x) = e − ax (a > 0).
Ans:
∞
2
FS [ f ( x)] =
π ∫
f ( x ). sin sx dx
0
2
∞
2 s ∞ −ax b
∫ e . sin sx dx = ∫ e sin bx dx = 2
− ax
=
π 0
π s2 + a2
0 a + b2
2 s
=
π s2 + a2
5. If the Fourier transform of f (x ) is F(s) then prove that . F [ f ( x − a )] = e isa F ( s)
Ans:
∞
1
F [ f ( x − a)] = ∫∞ f ( x − a).e dx
isx
2π −
Put x-a = y
dx = dy
When x = −∞, y = −∞ and x = ∞, y = ∞
∞ ∞
1 e ias
F [ f ( x − a)] = ∫ f ( y).e
is ( y + a )
. dy = ∫ f ( y).e
isy
.dy
2π −∞ 2π −∞
∞
e ias
= ∫ f ( x).e .dx = e isa F ( s )
isx
2π −∞
6. State the Fourier transforms of the derivatives of a function.
26