Properties_of_the_Classical_Fourier_Transform.pdf

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De nition of the Classical Fourier Transform Let f ( x ) be a (possibly
complex-valued) function de ned for 1 x 1 and square integrable,
There are certain regularity requirements inherent in this de nition whichwe
on proof of Fourier transform properties . The transform produces from
^
f ( ), by the formula
f ( x ), another function,
i x
2
2
2
2
classical, or continuous Fourier transform as the factor
of the discrete Fourier transform . With the transform as wehave de ned
^
f ( ) is square integrable just in case f ( x ) is square integrable
and we have the Plancherel identity
j f ( x )j
dx =
^
f ( )
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It is also true that if f ( x ) and g ( x ) are both square integrable functions and
^
f ( ) and g ( ) are their respective transforms, then
^
f ( ) g ( ) d =
^
f; g
f(x) g ( x ) dx =
The second property reduces to the rst when we set g ( x ) = f ( x ), of course.
unitary .
Fourier Transform Properties The properties of the Fourier transform
^
f ( ) or
calF ( f )( ), whichever is more convenient under the circumstances.
f ( x ) . We have
i x
f ( x )
( ) =
f ( x ) dx
i ( a ) x
f ( x ) dx = calF ( f )( a ) :
Property II: Fourier Transform of ( ix )
f ( x )
i x
calF ((( ix )
f ( x ))) ( ) =
( ix )
f ( x ) dx
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i x
f ( x ) dx
calF ( f )( ) :
2
F ( x ), as indicated, lie in L
(1 ; 1) then
lies in L
(1 ; 1) and,
i x
2
i x
f ( x ) dx =
calF ( f )( ) :
2
i x
( ) =
( x ) dx
2
i x
f ( x ) dx = i calF ( f )( ) :
2
This process can be repeated to see that if f ( x ) ; f
( x ) ; :::; f
( x ) all lie in
( ) = ( i )
calF ( f )( ) ; j = 1 ; 2 ; :::; k:
PropertyV:Fourier Transform of f ( x a ) The behavior of the Fourier
real number a
i x
calF ( f ( x a )) ( ) =
f ( x a ) dx
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ia
i ( x a )
ia
= e
f ( x a ) d ( x a ) = e
calF ( f )( ) :
2
Property VI: Fourier Transform of the Convolution ( f g )( x ) The
convolution product of two functions f ( x ) ; g ( x )in L
(1 ; 1) is de ned by
The integral is de ned for all real x because f ( y ) and g ( x y ) lie in L
(1 ; 1)
if f ( x ) and g ( x ) lie in that space and the Schwarz inequality then shows
i x
calF (( f g )( x )) ( ) =
f ( y ) g ( x y ) dy dx
i (r+y)
i y
i r
2
2
2 calF ( f )( ) calF ( g )( ) :
f ( x ) = x
; f ( x ) = e
; f ( x ) = sin ax , etc. (it is possible to interpret the
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transforms of the rst and second of these as distributions ,however). Even
the transforms of such functions such as f ( x ) =
1+ x
ax
i x
ax
2
(ax
2
^
f ( ) =
dx =
a
x
a
4 a
4 a
Using the methods of contour integration in the complex plane one can show
a
2 a
ax
dx =
for any real . To compute the last integral we note that
ax
ax
ay
2
a
ar
dx dy =
ar
ar
rdrd =
2 ar dr d
2 a
ar
ar
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