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FOURIER TRANSFORM METHOD

Josph Fourier, a French Mathematicians, had invented a
method called Fourier Transform in 1801, to explain the flow of
heat around an anchor ring
...
It can provide a
means of solving unwieldy equation that describe dynamic responses
to electricity, heat or light
...

Fourier Transform has become indispensable in the numerical
calculations needed to design electrical circuits, to analyze the
mechanical vibrations, and to study wave propagation
...
In order to deal with such
problems, it is necessary to generalize Fourier series to include
infinite intervals and to introduce the concept of Fourier Integral
...


Definition
...
f (α ) is the Fourier
transform of f (x ) and f ( x) = F −1[ f (α )] , that is, inverse Fourier
f (α )
...


Example-1
...
f (α ) = F [ f ( x)] =

=

1
2π

∞

∫e

−( x −iα )
2

2

e

−α
2

1
2π

2

dx =

−∞

e

∫

∞

−∞

−α 2
2

2π

f ( x )e iαx dx =

∞

∫e

−( x −iα ) 2
2

1
2π

−

∞

x2
2

∫e

−

−∞

dx

−∞

dx
( x − iα )
= dt
...


π
1

= e

−α
2

2


...
Find the Fourier transform of

f ( x) = e

−a x

, − ∞ < x < ∞
...


1
2π

f (α ) =
=
=

1
2π
2

0

∫e

∫

∞

f ( x )e dx =

−∞

( a + iα ) x

dx +

−∞

1
2π

∞

0

iα x

1
[ ∫ e ax e iαx dx + ∫ e −ax e iαx dx ]
2π −∞
0

∞

1
1
1
−
[
]
2π a + iα − a + iα

( − a + iα ) x
dx =
∫e
0

a
]
...
Find the Fourier transform of
⎪0, x > a ⎪
⎩
⎭
∞

hence evaluate,

∫

sin αa cosαx

−∞

Solution
...


f (α ) =

f ( x) =

is,

1

π

∞

∫

1
2π

∫e

α

sin αa

α

− i αx

∫

dα

α

dα =

2π

a

∫ (1)e

iα x

dx =

−a

2a
, α = 0
...
That
∫∞ α 2π
⎪0, x > a ⎪
−
⎩
⎭
∞

⎧1, x ≤ a ⎫
⎪
⎪
=⎨
⎬
...
Also by setting
⎪0, x > a ⎪
⎩
⎭

sin αa

−∞

1

1 ⎡ e iαa − e −iαa ⎤ 2 sin αa
, α > 0
...
Since the integrand is even, so we have,


...
These transformations of interest mainly as
tools in solving ordinary differential equations, partial differential

equations and integral equations, and they often also helpful in
handling and applying special functions
...

FOURIER COSINE TRANSFORM
For an even function

f (x ),

the Fourier integral is the Fourier

Cosine integral,
∞

f ( x) = ∫ A( w) cos wxdw

(1)

0

where
A( w)) =

2

π

∞

∫ f ( x) cos wxdx

(2)

0

Setting A( w) =

2

π

Fc ( w), where c stands for cosine
...


Formula (3) gives from

Fourier Cosine transform of
from

Fc (w),

f (x )
...


Equations (3) and (4) constitute the Fourier

Cosine transform pair
...
Then from (5),

∞

∫ f ( x) sin wxdx
0

(7)
2

∞

and from (5), f ( x) = π ∫ Fs ( w) sin wxdw
0

(8)

Equation (7) is called Fourier Sine transform of
(8) is called the inverse Fourier Sine transform of

f (x )
...


(7) and (8) constitute the Fourier Sine transform pair
...
Find the Fourier Cosine and Sine transforms of the
⎧k , 0 < x < a ⎫
f ( x) = ⎨
⎬
...
Fourier Cosine transform
2

Fc ( w)) =

π

∞

∫
0

2 ⎛ sin wx ⎞
f ( x) cos wxdx =
k
...

π ⎝ w ⎠

Fourier Sine transform
Fs ( w)) =

2

∞

f ( x) sin wxdx =
π∫
0

2

a

k
...

π ⎝
w
⎠

EXISTENCE OF FOURIER COSINE AND SINE
TRANSFORM
∞

If

f (x )

is absolutely integrable [ ∫ f ( x) dx exist ] on the positive
−∞

x-axis and that

f (x ) and f ′(x )

be piecewise continuous on every finite

interval, then the Fourier Cosine and Sine transforms of

f (x )

exist
...

c

0

c

Fs (af ( x) + bg ( x) ) = aFs f ( x) + bFs g ( x)
...


COSINE AND SINE TRANSFORMS OF DERIVATIVES

Theorem-1
...


Then

a
...


F s [ f ′ ( x ) ] = − wF c [ f ( x ) ]
...
By definition,

Fc [ f ′ ( x ) ] =
gives)

2

π

∞

∫
0

f ′ ( x ) cos wxdx

(Integration by parts

2⎡
⎢ f ( x ) cos wx
π⎣

Fc [ f ′( x ) ] =

∞
0

⎤
+ w ∫ f ( x ) sin wxdx ⎥ =
0
⎦
∞

∞
⎤
2⎡
2
0 − f (0)
...


Similarly,

F s [ f ′( x ) ] =

2

π

∞

∫ f ′( x ) sin wxdx

(Integration by parts gives)

0

Fs [ f ′( x ) ] =

2⎡
⎢ f ( x ) sin wx
π⎣

∞
0

⎤
− w ∫ f ( x ) cos wxdx ⎥ =
0
⎦
∞

∞
⎤
2 ⎡
0 − w ∫ f ( x ) cos wxdx ⎥ = − wF c [ f ( x ) ]
...


From formula (a),

Fc [ f ′′(x)] = wFs [ f ′( x)] −

2

π

Fc [ f ′′( x)] = − w2 Fc [ f ( x)] −

f ′(0) = w(− wFc [ f ( x)]) −

2

π

2

π

f ′(0) , so

f ′(0)
...


Find

the

Fourier

Cosine

transform

of

f ( x) = e − ax , a > 0
...
By definition,
∞

∞

⎛ eiwx +e−iwx ⎞
⎟dx=
Fc [ f (x)] = ∫ f (x)coswxdx
= ∫e ⎜
⎜
π0
π0 ⎝ 2 ⎟
⎠
∞
∞
2 1 ⎡ −(a−iw)x
2 1⎤⎡ e−(a−iw)x e−(a+iw)x ⎤ ∞

...
⎥⎢
+
⎥ =
π 2 ⎣0
π 2⎦⎣−(a −iw −(a +iw ⎦0
)
)
0
2

2

−ax

1 2⎡ 1
1 ⎤ 2 ⎡ a ⎤ −∞

...
(e = 0)
2 π ⎣a −iw a +iw⎦
⎦
⎣
Secondly, it can also be solved as,

f ( x ) = e − ax

Since,

f ′′( x ) = a 2e − ax ,

so

[

]

This

[ ]

implies,

F [ f ′(x)] = F a2e−ax = a2F e−ax = a2F [ f (x)]
c
c
c
c

Thus,

Fc
a

2

(a
Fc

[f

′′ ( x ) ] = − w

Fc

[f

2

+ w

[f

2

( x )]= − w
2

)F [ f

( x )]=

c

Fc
2

[f

Fc

( x ) ]−

[f

( x )]= −

2 ⎡
π ⎢a
⎣

2

a
+ w

2

π

( x ) ]−
2

π
2

(− a )

⎤
⎥

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