Line 23: Line 23:
  
 
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>

Revision as of 16:50, 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)<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))) $
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.


Back to QE CS question 5, August 2017

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett