| Line 11: | Line 11: | ||
we have: | we have: | ||
| − | < | + | <math> p_0(e^{jw}) = X(e^{j\mu},e^{jw}) |_{\mu=0} |
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> p_1(e^{jw}) = X(e^{jw},e^{j\nu}) |_{\nu=0} |
Revision as of 20:29, 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: <math> p_1(e^{jw}) = X(e^{jw},e^{j\nu}) |_{\nu=0} $
