| Line 11: | Line 11: | ||
we have: | we have: | ||
| − | <math> p_0(e^{jw}) = X(e^{j\mu},e^{jw}) |_{\mu=0} | + | <math> p_0(e^{jw}) = X(e^{j\mu},e^{jw}) |_{\mu=0} </math> |
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} </math> |
| + | |||
| + | c) <math> \sum_{m=-\infty}^{\infty} | ||
Revision as of 21:30, 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_{m=-\infty}^{\infty} $
