Practice Question on Nyquist rate

What is the Nyquist rate of the signal

$ x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} ? $

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Answer 1

Use CTFT to find the frequency response

Using table, we know FT(sin(pi t)/(pi t)) --> u(w+W) - u(w-W)

X(w) = [u(w+pi) - u(w-pi)] * [u(w+pi) - u(w-pi)]

       = int_-infinity^infinity ( [u(W+pi) - u(W-pi)] [u(w-W+pi) - u(w-W-pi)] )dW

       =  int-_pi^pi (u(w-W+pi) - u(w-W-pi)) dW

       since      W-pi <= w < pi-W,     and     -pi <= W < pi

            -2pi <= w < 2pi

I'm not sure if I did the convolution right... help please (if you can read it)

                int_pi^-w-pidW     if -2pi <= w < 0

X(w) = {     int_w-pi^pidW      if  0 <= w < 2pi

                 0                   else

                 -w-2pi            if -2pi <= w < 0

        = {     2pi-w             if 0 <= w < 2pi

                 0                   else

Regardless, wm = 2pi so NR = 4pi

--Kellsper 22:36, 20 April 2011 (UTC)

Answer 2

Write it here

Answer 3

Write it here.

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