Contents
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?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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 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
Therefore, the Nyquist rate is still 2w_m = 2(3pi) = 6pi
Answer 3
Write it here.
