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= Practice Question on the Nyquist rate of a signal = | = Practice Question on the Nyquist rate of a signal = | ||
Is the following signal band-limited? (Answer yes/no and justify your answer.) | Is the following signal band-limited? (Answer yes/no and justify your answer.) | ||
| − | <math | + | <math> x(t)= e^{-j \frac{\pi}{2}t} \frac{\sin (3 \pi t)}{\pi t} \ </math>> |
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| + | If you answered "yes", what is the Nyquist rate for this signal? | ||
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== Share your answers below == | == Share your answers below == | ||
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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! | 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! | ||
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=== Answer 1 === | === Answer 1 === | ||
| − | <math>\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)</math> | + | <math>\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)</math> |
| − | <math>\mathcal X (\omega) = 2\pi</math> | + | <math>\mathcal X (\omega) = 2\pi</math> |
| − | So this signal is not band limited. | + | So this signal is not band limited. |
| − | As such, there can be no Nyquist rate for this signal. | + | As such, there can be no Nyquist rate for this signal. |
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| + | --[[User:Cmcmican|Cmcmican]] 23:30, 30 March 2011 (UTC) | ||
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: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 | :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 | ||
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=== Answer 2 === | === Answer 2 === | ||
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| + | I cannot figure out how to use equation editor, so sorry this answer is not as pretty as the one above... | ||
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| + | X(w) = F(exp(-j(pi/2)t) * F(sin((3pi)t)/(tpi)) | ||
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| + | = 2pi (delta(w + pi/2) * (u(w + 3pi) - u(w - 3pi)) | ||
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| + | = 2pi (u((w + pi/2) + 3pi) - u((w + pi/2) - 3pi)) | ||
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| + | 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 from w = -7pi/2 to w = 5pi/2. | ||
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| + | Therefore, the Nyquist rate is still 2w_m = 2(3pi) = 6pi | ||
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=== Answer 3 === | === Answer 3 === | ||
| − | Write it here. | + | |
| + | Write it here. | ||
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| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
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| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 09:29, 9 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 shifted to the left by pi/2. Graphed, it would look like a box from w = -7pi/2 to w = 5pi/2.
Therefore, the Nyquist rate is still 2w_m = 2(3pi) = 6pi
Answer 3
Write it here.
