(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q3 of Week 14 Quiz Pool == ---- ---- Back to Lab Week 14 Quiz Pool Back to [[ECE438_Lab_Fall_2010|ECE ...) |
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| + | a. According to the table, we have | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | h[m,n]=&-\frac{1}{8}\delta [m+1,n-1]+\frac{1}{2}\delta [m,n-1]-\frac{1}{8}\delta [m-1,n-1] \\ | ||
| + | &-\frac{1}{4}\delta [m+1,n]+\delta [m,n]-\frac{1}{4}\delta [m,n-1] \\ | ||
| + | &-\frac{1}{8}\delta [m+1,n+1]+\frac{1}{2}\delta [m,n+1]-\frac{1}{8}\delta [m-1,n+1] | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Replace <math>\delta [m,n]</math> with general input signal <math>x[m,n]</math> we get the difference equation of the filter. | ||
| + | |||
| + | <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> | ||
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Revision as of 11:09, 29 November 2010
Solution to Q3 of Week 14 Quiz Pool
a. According to the table, we have
$ \begin{align} h[m,n]=&-\frac{1}{8}\delta [m+1,n-1]+\frac{1}{2}\delta [m,n-1]-\frac{1}{8}\delta [m-1,n-1] \\ &-\frac{1}{4}\delta [m+1,n]+\delta [m,n]-\frac{1}{4}\delta [m,n-1] \\ &-\frac{1}{8}\delta [m+1,n+1]+\frac{1}{2}\delta [m,n+1]-\frac{1}{8}\delta [m-1,n+1] \end{align} $
Replace $ \delta [m,n] $ with general input signal $ x[m,n] $ we get the difference equation of the filter.
$ \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} $
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