(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q4 of Week 13 Quiz Pool == ---- a. y[m,n] = h[m,n] ** x[m,n] Using definition of convolution, <br/> <math> \begin{align} y[m,n] ...)
 
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[[Category:2010 Fall ECE 438 Boutin]]
 
 
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== Solution to Q4 of Week 13 Quiz Pool ==
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== Solution to Q4 of Week 13 Quiz Pool ==
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----
  
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a. y[m,n] = h[m,n] ** x[m,n]
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a. y[m,n] = h[m,n] ** x[m,n]  
  
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Using definition of convolution, <br/>
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Using definition of convolution, <br> <math>\begin{align}
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<math>
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−
\begin{align}
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y[m,n] &= \sum_{k=-1}^{1} \sum_{l=-1}^{1} h[k,l] x[m-k,n-l] \\
 
y[m,n] &= \sum_{k=-1}^{1} \sum_{l=-1}^{1} h[k,l] x[m-k,n-l] \\
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\end{align}
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\end{align}</math>  
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</math>
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Expanding, <br/>
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Expanding, <br> y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,0] x[m+1,n] + h[-1,1] x[m+1,n-1] + h[0,-1] x[m,n+1] + h[0,0] x[m,n] + h[0,1] x[m,n-1] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]  
−
y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,0] x[m+1,n] + h[-1,1] x[m+1,n-1] + h[0,-1] x[m,n+1] + h[0,0] x[m,n] + h[0,1] x[m,n-1] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]  
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Sub values of h[m,n] from table, zero terms go away,<br/>
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Sub values of h[m,n] from table, zero terms go away,<br>  
  
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y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,1] x[m+1,n-1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,1] x[m-1,n-1] y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
 +
 +
b. We can rewrite h[m,n] as
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 +
{| width="20%" cellspacing="2" cellpadding="2" border="1" class="wikitable" style="text-align: center;"
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|+ m
 +
|-
 +
! n
 +
! -1
 +
! 0
 +
! 1
 +
|-
 +
! -1
 +
| 0.5
 +
| 0
 +
| -0.5
 +
|-
 +
! 0
 +
| 0
 +
| 1
 +
| 0
 +
|-
 +
! 1
 +
| -0.5
 +
| 0
 +
| 0.5
 +
|}
 +
 +
We compute the output: <br> y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1] <br> by considering 3X3 portions of x[m,n], where the element at m,n corresponds to 0,0 in h[m,n], so we would look at neighboring elements (if they exist) and multiply with corresponding neighbors in h[m,n] and then sum them to form y[m,n].
 +
 +
Example - (Indexed starting from 0) y[3,3] = 0.5 x[4,4] - 0.5 x[4,2] + x[3,3] - 0.5 x[2,4] + 0.5 x[2,2] x[2,2] = 0 x[2,4] = 1 x[3,3] = 1 x[4,2] = 1 x[4,4] = 1 so y[3,3] = 0.5 - 0.5 + 1 - 0.5 + 0 = 0.5
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 +
Similarly calculating values sequentially, results in y[m,n] - <br>
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 +
{| width="50%" cellspacing="3" cellpadding="2" border="1"
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|-
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0
 +
| -0.5
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
|-
 +
| 0
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| -0.5
 +
| -0.5
 +
| 0
 +
| 0
 +
| 0
 +
|-
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| 1.5
 +
| -0.5
 +
| -0.5
 +
| 0
 +
| 0
 +
|-
 +
| 0
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| 1.5
 +
| 1.5
 +
| -0.5
 +
| -0.5
 +
| 0
 +
|-
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| 1
 +
| 1
 +
| 1.5
 +
| 1.5
 +
| -0.5
 +
| -0.5
 +
|-
 +
| 0.5
 +
| 1
 +
| 0.5
 +
| 1
 +
| 1
 +
| 1
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| 1
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| 1
 +
| 1.5
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| 1
 +
| -0.5
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|-
 +
| 0
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 0
 +
|-
 +
| 0
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 0
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|-
 +
| 0
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 0
 +
|-
 +
| -0.5
 +
| 0.5
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1.5
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| 0.5
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|-
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| -0.5
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| -0.5
 +
| 0
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| 0
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| 0
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| 0
 +
| 0
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| 0
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| 0
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| 0.5
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| 0.5
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|}
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c. From the difference equation -
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y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
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 +
Taking Fourier Transform on both sides, <br/>
 
<math>
 
<math>
 
\begin{align}
 
\begin{align}
−
y[m,n] &= h[-1,-1] x[m+1,n+1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1] \\
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Y(\mu,\nu) &= \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{-j\nu} - \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{j\nu} + X(\mu,\nu) - \frac{1}{2}X(\mu,\nu) e^{j\mu}e^{-j\nu} + \frac{1}{2}X(\mu,\nu)e^{j\mu}e^{j\nu} \\
−
&= h[-1,-1] x[m+1,n+1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]
+
 
 +
\frac{Y(\mu,\nu)}{X(\mu,\nu)} &= \frac{1}{2}e^{-j\mu}e^{-j\nu} - \frac{1}{2}e^{-j\mu}e^{j\nu} + 1 - \frac{1}{2} e^{j\mu}e^{-j\nu} + \frac{1}{2}e^{j\mu}e^{j\nu} \\
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 +
H(\mu,\nu) &= 1 + \frac{1}{2} ( e^{-j(\mu + \nu)} + e^{j(\mu + \nu)} ) - \frac{1}{2} ( e^{-j(\mu - \nu)} + e^{j(\mu - \nu)} ) \\
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 +
H(\mu,\nu) &= 1 + cos(\mu + \nu) - cos(\mu - \nu)
 +
 
 
\end{align}
 
\end{align}
 
</math>
 
</math>
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to be continued...
 
 
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[[Category:2010_Fall_ECE_438_Boutin]]

Latest revision as of 19:54, 17 November 2010

Solution to Q4 of Week 13 Quiz Pool

a. y[m,n] = h[m,n] ** x[m,n]

Using definition of convolution,
$ \begin{align} y[m,n] &= \sum_{k=-1}^{1} \sum_{l=-1}^{1} h[k,l] x[m-k,n-l] \\ \end{align} $

Expanding,
y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,0] x[m+1,n] + h[-1,1] x[m+1,n-1] + h[0,-1] x[m,n+1] + h[0,0] x[m,n] + h[0,1] x[m,n-1] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]

Sub values of h[m,n] from table, zero terms go away,

y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,1] x[m+1,n-1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,1] x[m-1,n-1] y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]

b. We can rewrite h[m,n] as

m
n -1 0 1
-1 0.5 0 -0.5
0 0 1 0
1 -0.5 0 0.5

We compute the output:
y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
by considering 3X3 portions of x[m,n], where the element at m,n corresponds to 0,0 in h[m,n], so we would look at neighboring elements (if they exist) and multiply with corresponding neighbors in h[m,n] and then sum them to form y[m,n].

Example - (Indexed starting from 0) y[3,3] = 0.5 x[4,4] - 0.5 x[4,2] + x[3,3] - 0.5 x[2,4] + 0.5 x[2,2] x[2,2] = 0 x[2,4] = 1 x[3,3] = 1 x[4,2] = 1 x[4,4] = 1 so y[3,3] = 0.5 - 0.5 + 1 - 0.5 + 0 = 0.5

Similarly calculating values sequentially, results in y[m,n] -

0 0 0 0 0.5 0 -0.5 0 0 0 0
0 0 0 0.5 0.5 1 -0.5 -0.5 0 0 0
0 0 0.5 0.5 0.5 1 1.5 -0.5 -0.5 0 0
0 0.5 0.5 0.5 0.5 1 1.5 1.5 -0.5 -0.5 0
0.5 0.5 0.5 0.5 1 1 1 1.5 1.5 -0.5 -0.5
0.5 1 0.5 1 1 1 1 1 1.5 1 -0.5
0 1 1 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1 1 0
-0.5 0.5 1 1 1 1 1 1 1 1.5 0.5
-0.5 -0.5 0 0 0 0 0 0 0 0.5 0.5

c. From the difference equation - y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]

Taking Fourier Transform on both sides,
$ \begin{align} Y(\mu,\nu) &= \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{-j\nu} - \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{j\nu} + X(\mu,\nu) - \frac{1}{2}X(\mu,\nu) e^{j\mu}e^{-j\nu} + \frac{1}{2}X(\mu,\nu)e^{j\mu}e^{j\nu} \\ \frac{Y(\mu,\nu)}{X(\mu,\nu)} &= \frac{1}{2}e^{-j\mu}e^{-j\nu} - \frac{1}{2}e^{-j\mu}e^{j\nu} + 1 - \frac{1}{2} e^{j\mu}e^{-j\nu} + \frac{1}{2}e^{j\mu}e^{j\nu} \\ H(\mu,\nu) &= 1 + \frac{1}{2} ( e^{-j(\mu + \nu)} + e^{j(\mu + \nu)} ) - \frac{1}{2} ( e^{-j(\mu - \nu)} + e^{j(\mu - \nu)} ) \\ H(\mu,\nu) &= 1 + cos(\mu + \nu) - cos(\mu - \nu) \end{align} $

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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