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

The signal

$ x(t)= \cos ( \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?

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Answer 1

X(w) = F(cos(pi t)) * F(sin(3pi t) / (pi t))

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

Let Z(w) = [u(w+3pi) - u(w-3pi)]

X(w) = pi[delta(w-pi) + delta(w+pi)] * Z(w)

       = pi[Z(w-pi) + Z(w+pi)]

       = pi[u(w+2pi) - u(w-4pi) + u(w+4pi) - u(w-2pi)]

Thus the signal is band limited abs(w_m) < 4pi

Nyquist rate = 2w_m = 8pi

w_s must be greater than 8pi

T = 2pi/w_s < 2pi/8pi = 1/4

T < 1/4

--Ssanthak 12:35, 20 April 2011 (UTC)

TA's comment: The condition on the sampling frequency is correct. However you have a mistake when using the multiplication property of the FT, it is missing the 1 / 2π factor.

Answer 2

I got the same answer, but with my FT multiplied by 1/4π.  (I'm pretty sure you just flipped the 1/2π on accident.)

--Kellsper 18:34, 21 April 2011 (UTC)

Write it here

Answer 3

Write it here.

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