## Contents

The signal

$ x(t)= e^{j \frac{\pi}{2} t }\frac{\sin (6 \pi t)}{ 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

Using the table and a time shift

X(w) = pi[u(w+6pi-pi/2) - u(w-6pi-pi/2)]

= pi[u(w+11pi/2) - u(w-13pi/2)]

Thus the signal is band limited wm = 13pi/2, so ws > Nyquist Rate = 2wm = 13pi

Since T = 2pi/ws

T < 2pi/13pi = 2/13

T < 2/13 in order to recover the orginal signal

--Ssanthak 12:38, 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

Write it here

### Answer 3

Write it here.