(New page: Category:2010 Fall ECE 438 Boutin ---- == Solution to Q2 of Week 14 Quiz Pool == ---- <math>y(m,n) = x(m,n) + \lambda \left( x(m,n) - \frac{1}{9} \sum_{k=-1}^{1}\sum_{l=-1}^{1}x(m-k,...) |
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\begin{cases} | \begin{cases} | ||
1+\lambda-\frac{\lambda}{9} & m=0,n=0 \\ | 1+\lambda-\frac{\lambda}{9} & m=0,n=0 \\ | ||
| − | -\frac{\lambda}{9} & -1 | + | -\frac{\lambda}{9} & -1 \le m \le 1, -1 \le n \le 1, m \ne 0, n \ne 0\\ |
0 & \mbox{otherwise} | 0 & \mbox{otherwise} | ||
\end{cases} | \end{cases} | ||
Latest revision as of 13:34, 28 November 2010
Solution to Q2 of Week 14 Quiz Pool
$ y(m,n) = x(m,n) + \lambda \left( x(m,n) - \frac{1}{9} \sum_{k=-1}^{1}\sum_{l=-1}^{1}x(m-k,n-l) \right) $
a. Yes, the system is linear and space invariant.
b.
$ \begin{align} h(m,n) &= \delta(m,n) + \lambda \delta(m,n) - \frac{\lambda}{9} \sum_{k=-1}^{1}\sum_{l=-1}^{1}\delta(m-k,n-l) \\ &= (1 + \lambda) \delta(m,n) - \frac{\lambda}{9} \sum_{k=-1}^{1}\sum_{l=-1}^{1}\delta(m-k,n-l) \\ \\ h(m,n) &= \begin{cases} 1+\lambda-\frac{\lambda}{9} & m=0,n=0 \\ -\frac{\lambda}{9} & -1 \le m \le 1, -1 \le n \le 1, m \ne 0, n \ne 0\\ 0 & \mbox{otherwise} \end{cases} \end{align} $
c. Taking CSFT on both sides,
$ \begin{align} H(u,v) &= (1 + \lambda) - \frac{\lambda}{9} \sum_{k=-1}^{1}\sum_{l=-1}^{1}e^{-j2\pi u k}e^{-j2\pi v l} \\ &= (1 + \lambda) - \frac{\lambda}{9} \left( \sum_{k=-1}^{1}e^{-j2\pi u k} \right) \left( \sum_{l=-1}^{1}e^{-j2\pi v l} \right) \\ &= (1 + \lambda) - \frac{\lambda}{9} \left( e^{j2\pi u} + 1 + e^{-j2\pi u} \right) \left( e^{j2\pi v} + 1 + e^{-j2\pi v} \right) \\ &= (1 + \lambda) - \frac{\lambda}{9} \left( 1 + 2cos(2\pi u) \right) \left( 1 + 2cos(2\pi v) \right) \\ \end{align} $
d. For large values of $ \lambda $, the filter performs sharpening.
e. For -1 < $ \lambda $ < 0, the filter performs blurring.
Credit: Prof. Bouman
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