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<math>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})</math><br>
 
<math>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})</math><br>
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https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-3.PNG<br>
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d)<br>
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<math>Y_0(\omega)=H_0(\omega)(\dfrac{1}{2}H_0(\omega)X(\omega)+\dfrac{1}{2}H_0(\omega-\pi)X(\omega-\pi))</math><br>
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e)<br>
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<math>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})</math><br>
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f)<br>
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<math>Y(\omega)=\dfrac{1}{2}(H_0^2(\omega)-H_0^2(\pi-\omega))X(\omega)</math><br>
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https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-6.PNG<br>
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Revision as of 11:20, 19 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


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