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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 11:50, 20 April 2011

Practice Question on sampling and reconstruction (related to Nyquist rate)

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.


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