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<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 | ||
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we have: | we have: | ||
− | <span class="texhtml">''P''<sub>0</sub>(''e''<sup>''j''''w'''</sup>''') = ''X''(''e''<sup>''j''μ | + | <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 20:27, 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 $