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<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} </math> | ||
| − | c) <math> \sum_{m=-\infty}^{\infty} | + | c) |
| + | <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 20:32, 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} $
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} $
