Line 2: Line 2:
  
 
<math>X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
 
<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><br>
+
x(m,n)e^{-j(m\mu+n\nu)}</math><br>  
  
 
and  
 
and  
Line 11: Line 11:
 
we have:  
 
we have:  
  
<span class="texhtml">''P''<sub>0</sub>(''e''<sup>''j''''w'''</sup>''') = ''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j'''</sup>'''''w'') | <sub>μ</sub> = 0</span>
+
<span class="texhtml">''P''<sub>0</sub>(''e''<sup>''j''''w'''''</sup>''''') = '''''<b>X''(''e''<sup></sup>''j''μ,''e''<sup></sup>''j</b>'''''w'') | <sub>μ</sub> = 0'''</span>
 +
 
 +
b) Similarly to a), we have:
 +
 
 +
<math>p_1(e^{jw}) = X(e^{jw}, e^{j\nu})|\nu=0</math><br>

Revision as of 21:27, 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:

P0(ej'w) = X(ejμ,ejw) | μ = 0

b) Similarly to a), we have:

$ p_1(e^{jw}) = X(e^{jw}, e^{j\nu})|\nu=0 $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin