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b)<br> | b)<br> | ||
| − | <math>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)<math><br> | + | <math>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)</math><br> |
<math>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)))</math><br> | <math>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)))</math><br> | ||
https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS5-1.PNG<br> | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS5-1.PNG<br> | ||
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d)<br> | 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^ | + | <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> |
<br> | <br> | ||
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. | ||
| + | |||
---- | ---- | ||
| + | ===Similar Problem=== | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2014/CS-5.pdf?dl=1 2014 QE CS5 Prob2]<br> | ||
| + | [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> | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
[[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 10:58, 25 February 2019
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))) $
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
