(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 | + | |
− | 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>, </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 $