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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> | ||
<br> | <br> | ||
+ | |||
+ | c)<br> | ||
+ | minimum sampling frequency <math>\dfrac{1}{T}\ge\dfrac{2}{a}</math> <math>f\ge\dfrac{2}{a}</math> <math>T\le\dfrac{a}{2}</math><br> | ||
+ | <br> | ||
+ | |||
+ | d)<br> | ||
+ | <math>T=\dfrac{a}{2}</math><br> | ||
+ | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS5-3.PNG<br> | ||
+ | <br> | ||
+ | |||
+ | e)<br> | ||
+ | <math>T=a</math><br> | ||
+ | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS5-4.PNG<br> | ||
+ | <br> | ||
+ | |||
---- | ---- | ||
+ | ===Similar Problem=== | ||
+ | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_13/CS-5.pdf?dl=1 2013 QE CS5 Prob1]<br> | ||
+ | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_09/CS-5%20QE%2009.pdf?dl=1 2009 QE CS5 Prob1]<br> | ||
+ | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_08/CS-5%20QE%2008.pdf?dl=1 2008 QE CS5 Prob3]<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 11:00, 25 February 2019
Communication Signal (CS)
Question 5: Image Processing
August 2017 Problem 2
Solution
a)
$ sinc^2(\dfrac{t}{a}) \Rightarrow |a|\Lambda(af) $ (CTFT)
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}\ge\dfrac{2}{a} $ $ f\ge\dfrac{2}{a} $ $ T\le\dfrac{a}{2} $
d)
$ T=\dfrac{a}{2} $
e)
$ T=a $
Similar Problem
2013 QE CS5 Prob1
2009 QE CS5 Prob1
2008 QE CS5 Prob3