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F
(ω)
−2π
W
2π
W
ω
F I GU R E 12 . 8
A function
F
(
ω
).
F
P
(
ω
)
−2π
W
2π
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
)
.