Digital Signal Processing Reference
In-Depth Information
signal. After all, if we knew what the analog signal was, why would we bother to
sample it? However, any analog signal can be represented as a sum of sinusoids, so
it is reasonable to assume that any analog signal has a mathematical representation
in a form like this. The form for each sinusoid component is acos(2ft + ).
When we sample the signal, these components become acos(2fn=f
s
+ ). We
know that cos(2k + ) = cos(), as long as k is an integer. This is not surprising;
since the cosine function is periodic, it repeats itself over and over again. In fact, it
repeats itself innitely in both positive and negative directions (that is, k can be a
positive or negative integer). So a component of an analog signal will repeat itself,
given enough time. When f
s
= 100 samples/second, the rst component becomes:
4 cos(2fn=f
s
)
=
4 cos(21010n=100)
=
4 cos(2(1000 + 10)n=100)
=
4 cos(21000n=100 + 210n=100)
=
4 cos(210n + 2n=10):
Since n is an integer (i.e., let k = 10n), this reduces to:
4 cos(2n=10):
In other words, the 1010 Hz component appears as a 10 Hz component. It is left
to the reader to verify that the other components will appear as 20 Hz, 30 Hz, and
40 Hz when sampled at this rate. Also, the reader is encouraged to show that this
works when there is a nonzero phase angle, e.g., for 4 cos(21010n=f
s
+ =4). This
example works because there is no conict between frequencies. But what if there
already was a 10 Hz component? In that case, the information would be lost, since
both the 10 Hz and the 1010 Hz components would appear to be a single component
at 10 Hz. Avoiding overlap is one thing to look for when bandpass sampling.
As we saw above, bandpass sampling depends upon the sampling frequency. If
f
s
were, say, 103 samples per second, then the rst component becomes:
4 cos(2fn=f
s
)
=
4 cos(21010n=103)
=
4 cos(2(927 + 83)n=103)
=
4 cos(2927n=103 + 283n=103)
=
4 cos(29n + 283n=103):
