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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 | ||
| − | + | <math> x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} </math> | |
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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> | = 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> | ||
| − | = int-<sub>pi</sub><sup>pi</sup> (u(w-W+pi) - u(w-W-pi)) dW | + | = int<sub>-</sub><sub>pi</sub><sup>pi</sup> (u(w-W+pi) - u(w-W-pi)) dW |
since W-pi <= w < pi-W, and -pi <= W < pi | since W-pi <= w < pi-W, and -pi <= W < pi | ||
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.
