Databases Reference
In-Depth Information
What happens if we do sample at a rate less than 2
W
samples per second (that is,
σ
0
is less
than 4
π
W
)? Again we can see the results most easily in a pictorial fashion. The result for
σ
0
equal to 3
W
is shown in Figure
12.11
. Filtering this signal through an ideal low-pass filter,
we get the distorted signal shown in Figure
12.12
. Therefore, if
π
σ
0
is less than 4
π
W
, we can
(
)
no longer recover the signal
f
from its samples. This distortion is known as
aliasing
.In
order to prevent aliasing, it is useful to filter the signal prior to sampling using a low-pass filter
with a bandwidth less than half the sampling frequency.
Once we have the samples of a signal, sometimes the actual times they were sampled at
are not important. In these situations we can normalize the sampling frequency to unity. This
means that the highest frequency component in the signal is at 0.5Hz, or
t
radians. Thus,
when dealing with sampled signals, we will often talk about frequency ranges of
π
−
π
to
π
.
12.8 Discrete Fourier Transform
The procedures that we gave for obtaining the Fourier series and transform were based on the
assumption that the signal we were examining could be represented as a continuous function of
time. However, for the applications that we will be interested in, we will primarily be dealing
with samples of a signal. To obtain the Fourier transform of nonperiodic signals, we started
from the Fourier series and modified it to take into account the nonperiodic nature of the signal.
To obtain the discrete Fourier transform (DFT), we again start from the Fourier series. We
begin with the Fourier series representation of a sampled function, the discrete Fourier series.
Recall that the Fourier series coefficients of a periodic function
f
(
)
t
with period
T
are
given by
T
1
T
e
−
jk
w
0
t
dt
c
k
=
f
(
t
)
(56)
0
Suppose instead of a continuous function, we have a function sampled
N
times during each
period
T
. We can obtain the coefficients of the Fourier series representation of this sampled
function as
T
N
−
1
t
N
T
e
−
jk
w
0
t
dt
1
T
n
F
k
=
f
(
t
)
0
δ
−
(57)
0
n
=
f
n
N
T
e
−
j
2
π
kn
N
−
1
1
T
=
(58)
N
n
=
0
2
T
where we have used the fact that
w
0
=
, and we have replaced
c
k
by
F
k
. Taking
T
=
1for
convenience and defining
f
n
N
f
n
=
we get the coefficients for the discrete Fourier series (DFS) representation:
N
−
1
f
n
e
−
j
2
π
kn
F
k
=
(59)
N
n
=
0
