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. Notice that
$ h[m,n]= \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} \begin{pmatrix} -\frac{1}{4} & 1 & -\frac{1}{4} \end{pmatrix} $
Therefore h[m,n] can be separated as outer product of two column vector given by
h[m,n]=h_1[m]h_2[n]
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