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e)<br>
 
e)<br>
 
This is a sharpen filter. The image will become more sharpen as <math>\lambda</math> increases.
 
This is a sharpen filter. The image will become more sharpen as <math>\lambda</math> increases.
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===Similar Problem===
 +
[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2014/CS-5.pdf?dl=1 2014 QE CS5 Prob2]<br>
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[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_12/CS-5%20QE%2012.pdf?dl=1 2012 QE CS5 Prob2]<br>
 +
[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_11/CS-5%20QE%2011.pdf?dl=1 2011 QE CS5 Prob1]<br>
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 +
 +
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[[QE_2017_CS-5|Back to QE CS question 5, August 2017]]
 
[[QE_2017_CS-5|Back to QE CS question 5, August 2017]]
  
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Latest revision as of 11:58, 25 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.


Similar Problem

2014 QE CS5 Prob2
2012 QE CS5 Prob2
2011 QE CS5 Prob1



Back to QE CS question 5, August 2017

Back to ECE Qualifying Exams (QE) page

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