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d) No, it can't provide sufficient information.  
 
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.  
+
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:  
 
A simple example would be:  
 
Let  
 
Let  
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x(m,n) =  
 
x(m,n) =  
 
\left[ {\begin{array}{*{20}{c}}
 
\left[ {\begin{array}{*{20}{c}}
1 2 \\
+
1 ~ 2 \\
3 4\\
+
3 ~ 4\\
 
\end{array}} \right] </math>
 
\end{array}} \right] </math>

Revision as of 21:39, 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 $ are not zero. A simple example would be: Let $ x(m,n) = \left[ {\begin{array}{*{20}{c}} 1 ~ 2 \\ 3 ~ 4\\ \end{array}} \right] $

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

Dr. Paul Garrett