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===Similar Problem===
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[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2017/CS-2?dl=1 2017 QE CS2 Prob1]<br>
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[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2014/CS-2.pdf?dl=1 2014 QE CS2 Prob1]<br>
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[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_13/CS-2.pdf?dl=1 2013 QE CS2 Prob1]<br>
 
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[[QE2011_CS-2_ECE538|Back to QE CS question 2, August 2011]]
 
[[QE2011_CS-2_ECE538|Back to QE CS question 2, August 2011]]
  
 
[[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 10:53, 25 February 2019


ECE Ph.D. Qualifying Exam

Communication Signal (CS)

Question 2: Signal Processing

August 2011 Problem 1


Solution

a)
$ 8\dfrac{sin(\dfrac{3\pi}{8}n)sin(\dfrac{\pi}{8}n)}{\pi n} $
Wan82_CS2-1.PNG
$ \Rightarrow x[n]=16\dfrac{sin(\dfrac{3\pi}{8}n)}{\pi n}\dfrac{sin(\dfrac{\pi}{8}n)}{\pi n}cos(\dfrac{\pi n}{2}) $
Wan82_CS2-2.PNG

b)
$ X_0(\omega)=\dfrac{1}{2}H_0(\dfrac{\omega}{2})X(\dfrac{\omega}{2})+\dfrac{1}{2}H_0(\dfrac{\omega-2\pi}{2})X(\dfrac{\omega-2\pi}{2}) $
Wan82_CS2-3.PNG

c)
$ X_1(\omega)=\dfrac{1}{2}H_0(\dfrac{\omega}{2})X(\dfrac{\omega}{2})+\dfrac{1}{2}H_0(\dfrac{\omega-2\pi}{2})X(\dfrac{\omega-2\pi}{2}) $
Wan82_CS2-3.PNG

d)
$ Y_0(\omega)=H_0(\omega)(\dfrac{1}{2}H_0(\omega)X(\omega)+\dfrac{1}{2}H_0(\omega-\pi)X(\omega-\pi)) $
Wan82_CS2-4.PNG

e)
$ Y_1(\omega)=-H_1(\omega_0)(\dfrac{1}{2}H_1(\omega)X(\omega)+\dfrac{1}{2}H_1(\omega-\pi)X_1(\omega-\pi)) $
Wan82_CS2-5.PNG

f)
$ Y(\omega)=\dfrac{1}{2}(H_0^2(\omega)-H_0^2(\pi-\omega))X(\omega) $
Wan82_CS2-6.PNG


Similar Problem

2017 QE CS2 Prob1
2014 QE CS2 Prob1
2013 QE CS2 Prob1


Back to QE CS question 2, August 2011

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva