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

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

Dr. Paul Garrett