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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 11:53, 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

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