| (5 intermediate revisions by the same user not shown) | |||
| Line 20: | Line 20: | ||
a)<br> | a)<br> | ||
Because <math>x[-n]=x[n]</math> <math>y[-n]=y[n]</math><br> | Because <math>x[-n]=x[n]</math> <math>y[-n]=y[n]</math><br> | ||
| − | <math>r_{xy}[l]= | + | <math>r_{xy}[l]=X[l]\ast Y^{\ast}[-l]=X[-l]\ast Y^{\ast}[l]=Y[l]\ast X^{\ast}[-l]=r_{yx}[l]</math><br> |
| + | <br> | ||
| + | |||
| + | b)<br> | ||
| + | <math>z[n]=x[n]+jy[n]</math><br> | ||
| + | <math>r_{zz}[l]=(x[l]+jy[l])*(x[-l]+jy[-l])^*=x[l]*x^*[-l]+jy[l]*x^*[-l]-jx[l]*y^*[-l]+y[l]*y^*[-l]=r_{xx}[l]+r_{yy}[l]</math><br> | ||
| + | <br> | ||
| + | |||
| + | c)<br> | ||
| + | <math>x[n]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}(1+(-1)^n)=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}(1+e^{j\pi n})</math><br> | ||
| + | <math>\Rightarrow r_{xx}[l]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}(1+e^{j\pi n})=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}2cos^2\dfrac{\pi}{2}l</math> | ||
| + | <br> | ||
| + | |||
| + | d)<br> | ||
| + | <math>Y[n]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}cos(\dfrac{\pi}{2}n) \Rightarrow r_{yy}[l]=\dfrac{sin(\dfrac{\pi}{4})}{\pi l}cos(\dfrac{\pi}{2}l)</math><br> | ||
| + | <br> | ||
| + | |||
| + | e)<br> | ||
| + | <math>r_{zz}[l]=r_{xx}[l]+r_{yy}[l]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}2cos^2\dfrac{\pi}{2}l+\dfrac{sin(\dfrac{\pi}{4})}{\pi l}cos(\dfrac{\pi}{2}l)</math><br> | ||
| + | <br> | ||
| + | |||
| + | f)<br> | ||
| + | https://www.projectrhea.org/rhea/dropbox_/381ea5db244c12bb92e6de3206725a7a/Wan82_CS2-7.PNG<br> | ||
| + | <br> | ||
| + | |||
| + | ---- | ||
| + | ===Similar Problem=== | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2017/CS-2?dl=1 2017 QE CS2 Prob3]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2016/CS-2?dl=1 2016 QE CS2 Prob1]<br> | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/CS-2?dl=1 2015 QE CS2 Prob3]<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 11:54, 25 February 2019
Communication Signal (CS)
Question 2: Signal Processing
August 2011 Problem 2
Solution
a)
Because $ x[-n]=x[n] $ $ y[-n]=y[n] $
$ r_{xy}[l]=X[l]\ast Y^{\ast}[-l]=X[-l]\ast Y^{\ast}[l]=Y[l]\ast X^{\ast}[-l]=r_{yx}[l] $
b)
$ z[n]=x[n]+jy[n] $
$ r_{zz}[l]=(x[l]+jy[l])*(x[-l]+jy[-l])^*=x[l]*x^*[-l]+jy[l]*x^*[-l]-jx[l]*y^*[-l]+y[l]*y^*[-l]=r_{xx}[l]+r_{yy}[l] $
c)
$ x[n]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}(1+(-1)^n)=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}(1+e^{j\pi n}) $
$ \Rightarrow r_{xx}[l]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}(1+e^{j\pi n})=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}2cos^2\dfrac{\pi}{2}l $
d)
$ Y[n]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}cos(\dfrac{\pi}{2}n) \Rightarrow r_{yy}[l]=\dfrac{sin(\dfrac{\pi}{4})}{\pi l}cos(\dfrac{\pi}{2}l) $
e)
$ r_{zz}[l]=r_{xx}[l]+r_{yy}[l]=\dfrac{sin(\dfrac{\pi}{4}n)}{\pi n}2cos^2\dfrac{\pi}{2}l+\dfrac{sin(\dfrac{\pi}{4})}{\pi l}cos(\dfrac{\pi}{2}l) $
f)
Similar Problem
2017 QE CS2 Prob3
2016 QE CS2 Prob1
2015 QE CS2 Prob3
