Digital Signal Processing Reference
In-Depth Information
To visualize these concepts, let us consider a real sine wave of 70 Hz, sampled at
65 Hz (Figure 2.5(a)). The sampled signal is thus a sine wave of 5 Hz frequency. In
fact, the spectrum of amplitude of the continuous time sine wave is even, and
consists of two pulses at 70 Hz and -70 Hz (Figure 2.5(b)). To sample correctly, at
least two sampling points per period of the continuous time sine wave are necessary.
After sampling, we obtain the spectrum as in Figure 2.5(c), which has a peak at
5 Hz.
The spectral aliasing phenomenon also explains why the wheel of a vehicle
being filmed seems to rotate slowly in the other direction or even seems stationary
when the images are sampled at 24 Hz (25 Hz on television). The spectral aliasing
can also be found for example in the tuning of the ignition point of a spark ignition
engine motor through a stroboscope.
Generally, one must be careful when a continuous time signal is sampled, so that
the Shannon condition is fulfilled, at the risk of seeing the low frequencies generated
by the aliasing of high frequencies of the continuous time signal in the sampled
signal. In practice, before any sampling, the signal observed must be filtered by a
low-pass analog filter, known as
anti-aliasing,
whose cut-off frequency is 2.5 to 5
time less than the sampling frequency.
2.2.9.
Practical calculation, FFT
(
)
kZ
()
In practice, we record a signal
, which has a
N
finite number of
xxk
∈
=
(
)
0
points, that is
()
xk
≤≤ −
. It is thus tempting to approach the Fourier transform
kN
1
x
, that is:
+∞
=
∑
()
()
−
j
2
π
k
xv
ˆ
xke
k
=−∞
by the finite sum:
N
−
1
=
∑
−
j
2
π
vk
()
()
Xv
xke
k
=
0
This amounts to approaching
x
by the Fourier transform of the signal
x
multiplied by the rectangular window 1
0,
N
-1
i.e.
0,
−
. We will see the effects and
x
1
N
1
inconveniences of this truncation in section 2.3.
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