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
(
)
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
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
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
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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