## Contents

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?

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### 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)

w_{m}=4pi

Nyquist Rate = 2w_{m} = 8pi

Since we should sample w_{s} > 8pi

w_{s} = 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 w_{m} = 3pi, and therefore T < (1/2)(2pi/w_{m}) = 1/3. As long as w_{s} 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.