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===Solution=== | ===Solution=== | ||
a)<br> | a)<br> | ||
| − | <math>8\dfrac{sin(\dfrac{3\pi}{8}n)sin(\\dfrac{\pi}{8}n)}{\pi n} \ | + | <math>8\dfrac{sin(\dfrac{3\pi}{8}n)sin(\dfrac{\pi}{8}n)}{\pi n} </math> <br> |
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-1.PNG<br> | ||
| + | <math>\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})</math><br> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-2.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | b)<br> | ||
| + | <math>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})</math><br> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-3.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | c)<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> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-3.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | d)<br> | ||
| + | <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> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-4.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | e)<br> | ||
| + | <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> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-5.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | f)<br> | ||
| + | <math>Y(\omega)=\dfrac{1}{2}(H_0^2(\omega)-H_0^2(\pi-\omega))X(\omega)</math><br> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-6.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | ---- | ||
| + | ===Similar Problem=== | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2017/CS-2?dl=1 2017 QE CS2 Prob1]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2014/CS-2.pdf?dl=1 2014 QE CS2 Prob1]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_13/CS-2.pdf?dl=1 2013 QE CS2 Prob1]<br> | ||
---- | ---- | ||
[[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
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} $
$ \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}) $
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}) $
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}) $
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)) $
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)) $
f)
$ Y(\omega)=\dfrac{1}{2}(H_0^2(\omega)-H_0^2(\pi-\omega))X(\omega) $
Similar Problem
2017 QE CS2 Prob1
2014 QE CS2 Prob1
2013 QE CS2 Prob1
