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==Question ==
 
==Question ==
  
 +
a)
  
 +
<math>
 +
\begin{align}
 +
y[m,n] =& -\frac{1}{8}x[m+1,n-1] + \frac{1}{2}x[m,n-1] - \frac{1}{8}x[m-1,n-1] \\
 +
& -\frac{1}{4}x[m+1,n] + x[m,n] -\frac{1}{4}x[m,n-1] \\
 +
& -\frac{1}{8}x[m+1,n+1] + \frac{1}{2}x[m,n+1] -\frac{1}{8}x[m-1,n+1]
 +
\end{align}</math>
  
 +
b) Yes. The coefficient matrix of h[m,n] can be written as product of two vectors.
  
 +
<math>
 +
\begin{pmatrix}
 +
-\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} \\
 +
-\frac{1}{4} & 1 & -\frac{1}{4} \\
 +
-\frac{1}{8} & \frac{1}{2} & -\frac{1}{8}
 +
\end{pmatrix} =
 +
\begin{pmatrix}
 +
\frac{1}{2}  \\
 +
1 \\
 +
\frac{1}{2}
 +
\end{pmatrix} \cdot
 +
\begin{pmatrix}
 +
-\frac{1}{4} & 1 & -\frac{1}{4}
 +
\end{pmatrix}
 +
</math>
  
 +
Therefore the filter can be separate into two 1-D filters.
 +
 +
<math>h_1[m] = -\frac{1}{4}\delta[m+1] + \delta[m] -\frac{1}{4}\delta[m-1]</math>
 +
 +
<math>h_2[n] = \frac{1}{2}\delta[n+1] + \delta[n] +\frac{1}{2}\delta[n-1]</math>
 
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Revision as of 12:50, 4 December 2011

Homework 8, ECE438, Fall 2011, Prof. Boutin


Question

a)

$ \begin{align} y[m,n] =& -\frac{1}{8}x[m+1,n-1] + \frac{1}{2}x[m,n-1] - \frac{1}{8}x[m-1,n-1] \\ & -\frac{1}{4}x[m+1,n] + x[m,n] -\frac{1}{4}x[m,n-1] \\ & -\frac{1}{8}x[m+1,n+1] + \frac{1}{2}x[m,n+1] -\frac{1}{8}x[m-1,n+1] \end{align} $

b) Yes. The coefficient matrix of h[m,n] can be written as product of two vectors.

$ \begin{pmatrix} -\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} \\ -\frac{1}{4} & 1 & -\frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} & -\frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ 1 \\ \frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} -\frac{1}{4} & 1 & -\frac{1}{4} \end{pmatrix} $

Therefore the filter can be separate into two 1-D filters.

$ h_1[m] = -\frac{1}{4}\delta[m+1] + \delta[m] -\frac{1}{4}\delta[m-1] $

$ h_2[n] = \frac{1}{2}\delta[n+1] + \delta[n] +\frac{1}{2}\delta[n-1] $


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