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In-Depth Information
Let us see how we can do this:
∞
1
2
e
−
j
ω
t
d
f
(
t
)
=
F
(ω)
ω
(43)
π
−∞
2
π
W
1
2
e
−
j
ω
t
d
=
F
(ω)
ω
(44)
π
−
2
π
W
2
π
W
1
2
e
−
j
ω
t
d
=
F
P
(ω)
ω
(45)
π
−
2
π
W
2
π
W
∞
1
2
c
n
e
jn
2
W
ω
e
−
j
ω
t
d
=
ω
(46)
π
−
2
π
W
n
=−∞
c
n
2
π
W
−
∞
1
2
n
2
W
)
d
e
j
w(
t
−
=
ω
(47)
π
2
π
W
n
=−∞
Evaluating the integral and substituting for
c
n
from Equation (
42
), we obtain
f
n
2
W
Sinc
2
W
t
∞
n
2
W
f
(
t
)
=
−
(48)
n
=−∞
where
sin
(π
x
)
Sinc
[
x
]=
(49)
π
x
1
Thus, given samples of
f
(
t
)
taken every
2
W
seconds, or, in other words, samples of
f
(
t
)
obtained at a rate of 2
W
samples per second, we can reconstruct
f
(
t
)
by interpolating between
the samples using the Sinc function.
12.7.2 Ideal Sampling—Time Domain View
Let us look at this process from a slightly different point of view, starting with the sampling op-
eration. Mathematically, we can represent the sampling operation by multiplying the function
f
(
t
)
with a train of impulses to obtain the sampled function
f
S
(
t
)
:
∞
1
2
W
f
S
(
)
=
(
)
δ(
−
),
<
(50)
t
f
t
t
nT
T
n
=−∞
To obtain the Fourier transform of the sampled function, we use the convolution theorem:
f
∞
∞
F
(
t
)
δ(
t
−
nT
)
=
F
[
f
(
t
)
]
⊗
F
δ(
t
−
nT
)
(51)
n
=−∞
n
=−∞
Let us denote the Fourier transform of
f
. The Fourier transform of a train of
impulses in the time domain is a train of impulses in the frequency domain (see Problem 5 at
the end of this chapter):
(
t
)
by
F
(ω)
∞
∞
2
T
F
δ(
t
−
nT
)
=
σ
0
δ(w
−
n
σ
0
)
0
=
(52)
n
=−∞
n
=−∞
