(New page: = Practice Question on Nyquist rate = What is the Nyquist rate of the signal <math>x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} ?</math> ---- == Share your answer...) |
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| − | = Practice Question on Nyquist rate = | + | = [[:Category:Problem_solving|Practice Question]] on Nyquist rate = |
| − | What is the Nyquist rate of the signal | + | |
| + | What is the Nyquist rate of the signal | ||
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| + | <math> x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} </math> | ||
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=== Answer 1 === | === Answer 1 === | ||
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| + | Use CTFT to find the frequency response | ||
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| + | Using table, we know FT(sin(pi t)/(pi t)) --> u(w+W) - u(w-W) | ||
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| + | X(w) = [u(w+pi) - u(w-pi)] * [u(w+pi) - u(w-pi)] | ||
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| + | = int<sub>-infinity</sub><sup>infinity</sup> ( [u(W+pi) - u(W-pi)] [u(w-W+pi) - u(w-W-pi)] )dW<sup></sup><sub></sub><sub></sub> | ||
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| + | = int<sub>-</sub><sub>pi</sub><sup>pi</sup> (u(w-W+pi) - u(w-W-pi)) dW | ||
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| + | since W-pi <= w < pi-W, and -pi <= W < pi | ||
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| + | -2pi <= w < 2pi | ||
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| + | I'm not sure if I did the convolution right... help please (if you can read it) | ||
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| + | int<sub>pi</sub><sup>-w-pi</sup>dW if -2pi <= w < 0 | ||
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| + | X(w) = { int<sub>w-pi</sub><sup>pi</sup>dW if 0 <= w < 2pi | ||
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| + | 0 else | ||
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| + | <br> | ||
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| + | -w-2pi if -2pi <= w < 0 | ||
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| + | = { 2pi-w if 0 <= w < 2pi | ||
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| + | 0 else | ||
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| + | <br> | ||
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| + | Regardless, wm = 2pi so NR = 4pi | ||
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| + | --[[User:Kellsper|Kellsper]] 22:36, 20 April 2011 (UTC)<br> | ||
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=== Answer 2 === | === Answer 2 === | ||
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Write it here | Write it here | ||
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=== Answer 3 === | === Answer 3 === | ||
| − | Write it here. | + | |
| + | Write it here. | ||
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Latest revision as of 09:30, 11 November 2011
Contents
Practice Question on Nyquist rate
What is the Nyquist rate of the signal
$ x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} $
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
Use CTFT to find the frequency response
Using table, we know FT(sin(pi t)/(pi t)) --> u(w+W) - u(w-W)
X(w) = [u(w+pi) - u(w-pi)] * [u(w+pi) - u(w-pi)]
= int-infinityinfinity ( [u(W+pi) - u(W-pi)] [u(w-W+pi) - u(w-W-pi)] )dW
= int-pipi (u(w-W+pi) - u(w-W-pi)) dW
since W-pi <= w < pi-W, and -pi <= W < pi
-2pi <= w < 2pi
I'm not sure if I did the convolution right... help please (if you can read it)
intpi-w-pidW if -2pi <= w < 0
X(w) = { intw-pipidW if 0 <= w < 2pi
0 else
-w-2pi if -2pi <= w < 0
= { 2pi-w if 0 <= w < 2pi
0 else
Regardless, wm = 2pi so NR = 4pi
--Kellsper 22:36, 20 April 2011 (UTC)
Answer 2
Write it here
Answer 3
Write it here.
