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b) Similarly to a), we have:  
 
b) Similarly to a), we have:  
  
<math> p_1(e^{jw}) = X(e^{jw},e^{j\nu}) |_{\nu=0} </math>
+
<math> p_1(e^{jw}) = X(e^{jw},e^{j\nu}) |_{\nu=0}
 +
\sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} </math>
 +
 
  
 
c) <br>
 
c) <br>
<math> \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) | {\mu=0, \nu=0} </math>
+
<math> \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} </math>

Revision as of 20:33, 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} $

b) Similarly to a), we have:

$ p_1(e^{jw}) = X(e^{jw},e^{j\nu}) |_{\nu=0} \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} $


c)
$ \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang