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| + | <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> | ||
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| + | <math>\mathcal X (\omega) = 2\pi</math> | ||
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| + | So this signal is not band limited. | ||
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| + | 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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=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 18:30, 30 March 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)
Answer 2
Write it here.
Answer 3
Write it here.
