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d)<br>
<math>H(e^{j\mu},e^{jv})=\dfrac{3}{2}-\dfrac{1}{18}\sum_{m=-1}^{1} e^{-j\mu}\sum_{n=-1}^{1} e^(-jv) =\dfrac{3}{2}-\dfrac{1}{18}(1+2cos\mu)(1+2cosv)</math><br>
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<math>H(e^{j\mu},e^{jv})=\dfrac{3}{2}-\dfrac{1}{18}\sum_{m=-1}^{1} e^{-j\mu}\sum_{n=-1}^{1} e^{-jv} =\dfrac{3}{2}-\dfrac{1}{18}(1+2cos\mu)(1+2cosv)</math><br>
 
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Revision as of 15:51, 19 February 2019


ECE Ph.D. Qualifying Exam

Communication Signal (CS)

Question 5: Image Processing

August 2017 Problem 1


Solution

a)
$ ay(m,n)=ax(m,n)+a\lambda(x(m,n)-\dfrac{1}{9}\sum_{k=-1}^{1} \sum_{l=-1}^{1} x(m-k,n-l)) $ linear

b)
$ y(m,n)=x(m,n)+\lambda(x(m,n)-\dfrac{1}{9}\sum_{k=-1}^{1} \sum_{l=-1}^{1} x(m-k,n-l))=1.5x(m,n)-\dfrac{1}{18}\sum_{k=-1}^{1} \sum_{l=-1}^{1} x(m-k,n-l) $
$ h(m,n)=1.5\delta(m,n)-\dfrac{1}{18}(\delta(m+1)+\delta(m)+\delta(m-1))(\delta(n-1)+\delta(n)+\delta(n+1))) $
Wan82_CS5-1.PNG

c)
Not a separable system.

d)
$ H(e^{j\mu},e^{jv})=\dfrac{3}{2}-\dfrac{1}{18}\sum_{m=-1}^{1} e^{-j\mu}\sum_{n=-1}^{1} e^{-jv} =\dfrac{3}{2}-\dfrac{1}{18}(1+2cos\mu)(1+2cosv) $

e)
This is a sharpen filter. The image will become more sharpen as $ \lambda $ increases.


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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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