| Line 25: | Line 25: | ||
</math> | </math> | ||
| + | b. Place the center of filter (i.e. where m=0,n=0) upon the pixel of image. Multiply h[m,n] with x[m,n] of the correspondent position and sum the value. We can get | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | y[0,0]=&0*h[-1,1]+0*h[0,1]+0*h[1,1]+ \\ | ||
| + | &0*h[-1,0]+1*h[0,0]+0*h[1,0] \\ | ||
| + | &1*h[-1,-1]+1*h[0,-1]+1*h[1,-1] \\ | ||
| + | =&1-\frac{1}{8}+\frac{1}{2}-\frac{1}{8} \\ | ||
| + | =&\frac{5}{4} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | c. | ||
---- | ---- | ||
Back to [[ECE438_Week14_Quiz|Lab Week 14 Quiz Pool]] | Back to [[ECE438_Week14_Quiz|Lab Week 14 Quiz Pool]] | ||
Revision as of 10:31, 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} $
b. Place the center of filter (i.e. where m=0,n=0) upon the pixel of image. Multiply h[m,n] with x[m,n] of the correspondent position and sum the value. We can get
$ \begin{align} y[0,0]=&0*h[-1,1]+0*h[0,1]+0*h[1,1]+ \\ &0*h[-1,0]+1*h[0,0]+0*h[1,0] \\ &1*h[-1,-1]+1*h[0,-1]+1*h[1,-1] \\ =&1-\frac{1}{8}+\frac{1}{2}-\frac{1}{8} \\ =&\frac{5}{4} \end{align} $
c.
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