Revision as of 08:15, 14 April 2011

Practice Question on the Nyquist rate of a signal

Is the following signal band-limited? (Answer yes/no and justify your answer.)

$ x(t)= e^{-j \frac{\pi}{2}t} \frac{\sin (3 \pi t)}{\pi t} \ $>

If you answered "yes", what is the Nyquist rate for this signal?

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Answer 1

$ \mathcal X (\omega) = \mathfrak {F} \Big(e^{-j \frac{\pi}{2}t}\Big) * \mathfrak F \Big(\frac{\sin (3 \pi t)}{\pi t}\Big)=2\pi\delta(\omega-\frac{\pi}{2}) * \Big(u(\omega+3\pi)-u(\omega-3\pi)\Big) = 2\pi \Big(u(\frac{\pi}{2}+3\pi)-u(\frac{\pi}{2}-3\pi)\Big) $

$ \mathcal X (\omega) = 2\pi $

So this signal is not band limited.

As such, there can be no Nyquist rate for this signal.

--Cmcmican 23:30, 30 March 2011 (UTC)

INstructor's comment: Reality check: the only signal whose Fourier transform is a constant is an impulse. So your answer cannot possibly be correct. -pm

Answer 2

I cannot figure out how to use equation editor, so sorry this answer is not as pretty as the one above...

X(w) = F(exp(-j(pi/2)t) * F(sin((3pi)t)/(tpi))

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

        = 2pi (u((w + pi/2) + 3pi) - u((w + pi/2) - 3pi))

In other words, it is the FT of a sinc function with w_m = 3pi, but shifted to the left by pi/2.  Graphed, it would look like a box with

X(w) = {       1,     -7pi/2 < w < 5pi/2

                  0,     else

So it is band limited, and the Nyquist rate is still 2w_m = 2(3pi) = 6pi

TA's comment: That's correct. Good job!

Answer 3

Write it here.

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