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:TA's comment: Well, in this problem the given signal is a complex-valued signal which basically what makes Nyquist theorem a little tricky to apply here. One way to look at this is by trying to avoid aliasing in frequency domain, which implies the condition that <math>\omega_s-7\pi/2>5\pi/2</math> or equivalently <math>\omega_s>6\pi</math>. | :TA's comment: Well, in this problem the given signal is a complex-valued signal which basically what makes Nyquist theorem a little tricky to apply here. One way to look at this is by trying to avoid aliasing in frequency domain, which implies the condition that <math>\omega_s-7\pi/2>5\pi/2</math> or equivalently <math>\omega_s>6\pi</math>. | ||
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| + | I thought that the Nyquist rate would have to be 7pi in this case, even though aliasing wouldn't occur until 6pi? In homework 9 question 2, we were able to violate Nyquist and still recover a signal. -sr | ||
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Revision as of 08:27, 21 April 2011
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 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
X(w) = F(e^(-jtpi/2))*F(sin(3tpi/tpi))
= 2pi(delta(w-pi/2)*[u(w+3pi)-u(w-3pi)])
= 2pi[u(w+7pi/2)-u(w-5pi/2)]
This will appear to be a box starting at w = -7pi/2 and ending at w = 5pi/2
wm = max(abs(w)) = 7pi/2
Nyquist = 2wm = 7pi
I see how you can shift the box to make the max w be 3pi and Nyquist 6pi, however I simply thought this was a case where the signal was not symetric and thus Nyquist could be violated a little bit. How do we know when to shift it and when to just leave it
--Ssanthak 11:27, 19 April 2011 (UTC)
- TA's comment: Well, in this problem the given signal is a complex-valued signal which basically what makes Nyquist theorem a little tricky to apply here. One way to look at this is by trying to avoid aliasing in frequency domain, which implies the condition that $ \omega_s-7\pi/2>5\pi/2 $ or equivalently $ \omega_s>6\pi $.
I thought that the Nyquist rate would have to be 7pi in this case, even though aliasing wouldn't occur until 6pi? In homework 9 question 2, we were able to violate Nyquist and still recover a signal. -sr
