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|} | |} | ||
| − | c. | + | 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, | ||
| + | <math> | ||
| + | \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} | ||
---- | ---- | ||
Revision as of 16:41, 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
| 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} ---- Back to [[ECE438 Week13 Quiz|Lab Week 13 Quiz Pool]] Back to [[ECE438 Lab Fall 2010|ECE 438 Fall 2010 Lab Wiki Page]] Back to [[2010 Fall ECE 438 Boutin|ECE 438 Fall 2010]] [[Category:2010_Fall_ECE_438_Boutin]] $
