(New page: a) Since <math>X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-j(m\mu+n\nu)}</math> and <span class="texhtml"><math>p_0(e^{jw}) = \sum_{m=-\infty...)
 
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a) Since <math>X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
+
a) Since
x(m,n)e^{-j(m\mu+n\nu)}</math> and&nbsp;<span class="texhtml"><math>p_0(e^{jw}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
+
 
x(m,n)e^{-jnw}</math> </span>
+
<math>X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
 +
x(m,n)e^{-j(m\mu+n\nu)}</math>  
 +
 
 +
and
 +
 
 +
<span class="texhtml"><math>p_0(e^{jw}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
 +
x(m,n)e^{-jnw}</math>,&nbsp;</span>
 +
 
 +
we have:
 +
 
 +
<math>P_0(e^{jw}) = X(e^{j\mu}, e^{jw})|_\mu=0</math>

Revision as of 20:24, 10 November 2014

a) Since

$ X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-j(m\mu+n\nu)} $

and

$ p_0(e^{jw}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-jnw} $, 

we have:

$ P_0(e^{jw}) = X(e^{j\mu}, e^{jw})|_\mu=0 $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood