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F (ω)
−2π W
W
ω
F I GU R E 12 . 8
A function
F
(
ω
).
F P (
ω
)
−2π W
W
ω
F I GU R E 12 . 9
The periodic extension F P (
).
ω
c n } n =−∞
The coefficients of the expansion
{
are then given by
2 π W
1
e jn 2 W ω d
c n =
F P (ω)
ω
(39)
4
π
W
2
π
W
However, in the interval
(
2
π
W
,
2
π
W
),
F
(ω)
is identical to F P (ω)
; therefore,
2 π W
1
e jn 2 W ω d
c n =
F
(ω)
ω
(40)
4
π
W
2
π
W
The function F
(ω)
is zero outside the interval
(
2
π
W
,
2
π
W
)
, so we can extend the limits to
infinity without changing the result:
1
2
1
2 W
e jn 2 W ω d
c n =
F
(ω)
ω
(41)
π
−∞
n
The expression in brackets is simply the inverse Fourier transform evaluated at t
=
2 W ;
therefore,
2 W f n
1
c n =
(42)
2 W
c n } n =−∞
Knowing
{
and the value of W , we can reconstruct F P (ω)
. Because F P (ω)
and
F
(ω)
are identical in the interval
(
2
π
W
,
2
π
W
)
, we can also reconstruct F
(ω)
in this interval.
c n } n =−∞
1
But
{
are simply the samples of f
(
t
)
every
2 W seconds, and F
(ω)
is zero outside this
interval. Therefore, given the samples of a function f
obtained at a rate of 2 W samples per
second, we should be able to exactly reconstruct the function f
(
t
)
(
t
)
.
 
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