Digital Signal Processing Reference
In-Depth Information
Since n is an integer, this reduces to:
4 cos(283n=103):
We do not stop here. Since 83n=103 is equivalent to (10320)n=103, we can elim-
inate another 2n term, which leaves us with a frequency component that appears
to be20 Hz. We do this because it will show up as20 Hz anyway; the signal's
spectrum contains information fromf
s
=2 to f
s
=2, with the same information on
both sides, reected around 0. This frequency component 83=103 is greater than
103=2 (i.e., f
s
=2), so it appears in the spectrum as20=103. Since the left half of
the spectrum for a real signal is a mirror of the right half, this shows up as +20
Hz, too. In other words, cos() = cos(), for any real angle . Likewise, when we
remove multiples of 103 from the other components, we get:
x[n] = 4 cos(220n=103) + 6 cos(210n=103) + 8 cos(0) + 10 cos(210n=103):
The information in both the second and the fourth components appears at 10 Hz,
meaning that we have lost information. Figure 5.11 shows what the spectrum would
look like if we did sample at 103 Hz. An interesting point is that the component at
0 Hz appears to be twice as large, as will any 0 Hz component. The amplitudes are
relative to each other, and since we expect, say, 10 Hz to have an equal amplitude at
10 Hz, we should expect this for 0 Hz and0 Hz, too. Using 2081 samples/second
as we did in Figure 5.9, we avoided doubling the amplitude shown at 1040 Hz.
Sampled signal (after DFT) fs=103, N=1030
1
0.8
0.6
0.4
0.2
0
0
10
20
30
40
50
60
Frequency (Hz)
Figure 5.11: Example signal sampled at 103 Hz.
All real signal information will show up in the spectrum between 0 and f
s
=2,
even if it has a frequency component above f
s
=2. This is the idea behind the Nyquist
sampling rate; f
s
=2 must be at least as large as the bandwidth. Since we are likely
to know the bandwidth and need to solve for f
s
, this becomes the relation f
s
2B.
We can think of the signal information mapping to the spectrum, from 0 to f
s
=2, in
