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

The signal

$ x(t)= e^{j \pi 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?


Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!


Answer 1

x(w) = (1/2pi) F(e^jtpi)*F(sin(3tpi)/tpi)

        = (1/2pi) [2pi delta(w-pi)] * [u(w+3pi)-u(w-3pi)]

        = u(w-pi+3pi) - u(w-pi-3pi)

        = u(w+2pi) - u(w-4pi)

wm=4pi

Nyquist Rate = 2wm = 8pi

Since we should sample ws > 8pi

ws = 2pi/T > 8pi

T < 1/4 in order to be able to reconstruct the signal using Nyquist.
--Ssanthak 13:01, 21 April 2011 (UTC)

Instructor's comment: But would it be possible to sample below the Nyquist rate and still be able to reconstruct the signal from its samples? -pm

Answer 2

The signal could still be reconstructed as long as T < 1/3, since the unshifted signal would have wm = 3pi, and therefore T < (1/2)(2pi/wm) = 1/3.  As long as ws is slightly bigger than 3pi, there will not actually be overlap in the frequency response, so it can be filtered later.

--Kellsper 18:05, 21 April 2011 (UTC)

                                                   

Answer 3

Write it here.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn