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<math>y(n)=sinc^2(\dfrac{nT}{a}) \Rightarrow X_s(f)=\dfrac{1}{T}\sum_{k=-\infty}^{\infty} X(f-kF)=\dfrac{|a|}{T}\sum_{k=-\infty}^{\infty}\Lambda(a(f-\dfrac{k}{T}))</math><br>
 
<math>y(n)=sinc^2(\dfrac{nT}{a}) \Rightarrow X_s(f)=\dfrac{1}{T}\sum_{k=-\infty}^{\infty} X(f-kF)=\dfrac{|a|}{T}\sum_{k=-\infty}^{\infty}\Lambda(a(f-\dfrac{k}{T}))</math><br>
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minimum sampling frequency <math>\dfrac{1}{T}>=\dfrac{2}{a}</math>  <math>f>=\dfrac{2}{a}</math>  <math>T<=\dfrac{a}{2}</math><br>
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<math>T=\dfrac{a}{2}</math><br>
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https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS5-3.PNG<br>
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e)<br>
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<math>T=a</math><br>
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https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS5-4.PNG<br>
 
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Revision as of 15:24, 19 February 2019


ECE Ph.D. Qualifying Exam

Communication Signal (CS)

Question 5: Image Processing

August 2017 Problem 2


Solution

a)
$ sinc^2(\dfrac{t}{a}) \Rightarrow |a|\Lambda(af) $ (CTFT)
Wan82_CS5-2.PNG

b)
$ y(n)=sinc^2(\dfrac{nT}{a}) \Rightarrow X_s(f)=\dfrac{1}{T}\sum_{k=-\infty}^{\infty} X(f-kF)=\dfrac{|a|}{T}\sum_{k=-\infty}^{\infty}\Lambda(a(f-\dfrac{k}{T})) $

c)
minimum sampling frequency $ \dfrac{1}{T}>=\dfrac{2}{a} $ $ f>=\dfrac{2}{a} $ $ T<=\dfrac{a}{2} $

d)
$ T=\dfrac{a}{2} $
Wan82_CS5-3.PNG

e)
$ T=a $
Wan82_CS5-4.PNG


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Alumni Liaison

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

Buyue Zhang