(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  =
 
= 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>  
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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  ===
Write it here.
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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)) --&gt; u(w+W) - u(w-W)
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X(w) =&nbsp;[u(w+pi) - u(w-pi)] * [u(w+pi) - u(w-pi)]
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&nbsp;&nbsp; &nbsp; &nbsp; = int<sub>-infinity</sub><sup>infinity</sup>&nbsp;( [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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&nbsp;&nbsp; &nbsp; &nbsp; =&nbsp; int-<sub>pi</sub><sup>pi</sup>&nbsp;(u(w-W+pi) - u(w-W-pi)) dW
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&nbsp;&nbsp; &nbsp; &nbsp; since &nbsp; &nbsp; &nbsp;W-pi &lt;= w &lt; pi-W, &nbsp; &nbsp; and &nbsp; &nbsp; -pi &lt;= W &lt; pi
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&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;-2pi &lt;= w &lt; 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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&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;int<sub>pi</sub><sup>-w-pi</sup>dW &nbsp; &nbsp; if -2pi &lt;= w &lt; 0
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X(w) = { &nbsp; &nbsp; int<sub>w-pi</sub><sup>pi</sup>dW &nbsp; &nbsp; &nbsp;if &nbsp;0 &lt;= w &lt; 2pi
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&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; else
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&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; -w-2pi &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;if -2pi &lt;= w &lt; 0
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&nbsp;&nbsp; &nbsp; &nbsp; &nbsp;= { &nbsp; &nbsp; 2pi-w&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if 0 &lt;= w &lt; 2pi
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&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; else
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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.
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Write it here.  
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Revision as of 17:36, 20 April 2011

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} ? $


Share your answers below

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.


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