Line 15: Line 15:
 
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>
\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>
+
 
+
  
 
c) <br>
 
c) <br>
 
<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>
 
<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>
 +
which is the DC point of the image.
 +
 +
d) No, it can't provide sufficient information.
 +
From the expression in a) and b), we see that <math> p_0(e^{jw}) </math> and  <math> p_1(e^{jw}) </math> are only slices of the DSFT. It lost the information when <math> \mu </math> and <math> \nu <math> are not zero.
 +
A simple example would be:
 +
Let <math>

Revision as of 20:38, 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} $ which is the DC point of the image.

d) No, it can't provide sufficient information. From the expression in a) and b), we see that $ p_0(e^{jw}) $ and $ p_1(e^{jw}) $ are only slices of the DSFT. It lost the information when $ \mu $ and $ \nu <math> are not zero. A simple example would be: Let <math> $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett