(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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== Solution to Q3 of Week 14 Quiz Pool == | == Solution to Q3 of Week 14 Quiz Pool == | ||
---- | ---- | ||
| + | |||
| + | 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> | ||
| + | |||
| + | 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. Notice that | ||
| + | |||
| + | <math>\mathbf{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} | ||
| + | </math> | ||
| + | |||
| + | Therefore \mathbf{h}[m,n] can be separated as outer product of two column vector given by | ||
| + | |||
| + | <math>\mathbf{h}[m,n]=\mathbf{h}_1[m]\mathbf{h}_2[n]^T</math> | ||
| + | |||
| + | where | ||
| + | <math> | ||
| + | \mathbf{h}_1[m]= | ||
| + | \begin{pmatrix} | ||
| + | \frac{1}{2} \\ | ||
| + | 1 \\ | ||
| + | \frac{1}{2} | ||
| + | \end{pmatrix} | ||
| + | </math> | ||
| + | and | ||
| + | <math> | ||
| + | \mathbf{h}_2[n]= | ||
| + | \begin{pmatrix} | ||
| + | -\frac{1}{4} \\ | ||
| + | 1 \\ | ||
| + | -\frac{1}{4} | ||
| + | \end{pmatrix} | ||
| + | </math> | ||
| + | |||
| + | Then compute the CTFT of <math>h_1,h_2</math> we get | ||
| + | |||
| + | <math>H_1(u)=\frac{1}{2}e^{-ju(-1)}+e^{-ju(0)}+\frac{1}{2}e^{-ju(1)}</math> | ||
| + | |||
| + | <math>H_2(v)=-\frac{1}{4}e^{-jv(-1)}+e^{-jv(0)}-\frac{1}{4}e^{-jv(1)}</math> | ||
| + | |||
| + | According to the Separability property of CSFT, we get | ||
| + | |||
| + | <math>H(u,v)=H_1(u)H_2(v)=(1-\frac{1}{2}cosu)(1+cosv)</math> | ||
Latest revision as of 06:16, 30 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. Notice that
$ \mathbf{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 \mathbf{h}[m,n] can be separated as outer product of two column vector given by
$ \mathbf{h}[m,n]=\mathbf{h}_1[m]\mathbf{h}_2[n]^T $
where $ \mathbf{h}_1[m]= \begin{pmatrix} \frac{1}{2} \\ 1 \\ \frac{1}{2} \end{pmatrix} $ and $ \mathbf{h}_2[n]= \begin{pmatrix} -\frac{1}{4} \\ 1 \\ -\frac{1}{4} \end{pmatrix} $
Then compute the CTFT of $ h_1,h_2 $ we get
$ H_1(u)=\frac{1}{2}e^{-ju(-1)}+e^{-ju(0)}+\frac{1}{2}e^{-ju(1)} $
$ H_2(v)=-\frac{1}{4}e^{-jv(-1)}+e^{-jv(0)}-\frac{1}{4}e^{-jv(1)} $
According to the Separability property of CSFT, we get
$ H(u,v)=H_1(u)H_2(v)=(1-\frac{1}{2}cosu)(1+cosv) $
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