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:Instructor's comment: Why do you say that we "must" sample above Nyquist? Is it possible that one could still be able to reconstruct when sampling below Nyquist? -pm | :Instructor's comment: Why do you say that we "must" sample above Nyquist? Is it possible that one could still be able to reconstruct when sampling below Nyquist? -pm | ||
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+ | I guess must was a bad choice of words, we should sample above Nyquist to guarantee that we can reconstruct the signal. In this case I do not believe we can sample below the Nyquist rate because the signal is present in all frequencies from -3pi to 3pi. If the signal was asymmetric then we could sample below Nyquist provided the copies never overlap. | ||
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+ | --[[User:Ssanthak|Ssanthak]] 15:50, 20 April 2011 (UTC) | ||
=== Answer 2 === | === Answer 2 === | ||
Revision as of 10:50, 20 April 2011
Contents
The signal
$ x(t)= \frac{\sin (3 \pi t)}{\pi t} $
is sampled with a sampling period T. For what values of T is it possible to reconstruct the signal from its sampling?
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Answer 1
from the Table
x(w) = u(w+3pi)-u(w-3pi)
Thus the signal is bandlimited with a wm = 3pi
We must sample above the Nyquist Rate which is equal to 2wm or 6pi
ws > 6pi
T = 2pi/ws < 2/6 = 1/3
The signal can be reconstructed for all T < 1/3.
--Ssanthak 12:09, 20 April 2011 (UTC)
- Instructor's comment: Why do you say that we "must" sample above Nyquist? Is it possible that one could still be able to reconstruct when sampling below Nyquist? -pm
I guess must was a bad choice of words, we should sample above Nyquist to guarantee that we can reconstruct the signal. In this case I do not believe we can sample below the Nyquist rate because the signal is present in all frequencies from -3pi to 3pi. If the signal was asymmetric then we could sample below Nyquist provided the copies never overlap.
--Ssanthak 15:50, 20 April 2011 (UTC)
Answer 2
Write it here
Answer 3
Write it here.