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What is the Nyquist rate of the signal  
 
What is the Nyquist rate of the signal  
  
'''Failed to parse (lexing error): x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} ?'''
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<math> x(t) = \frac{ \sin ( \pi t )}{\pi t} \frac{ \sin ( \pi t )}{\pi t} </math>  
 
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^ Sorry, I think I broke the equation in the problem statement but I don't know how to fix it. -ke<br>  
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Revision as of 15:52, 21 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} $


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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.


Back to ECE301 Spring 2011 Prof. Boutin

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010