| Line 31: | Line 31: | ||
<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>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> | <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> | ||
---- | ---- | ||
[[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]] | ||
Revision as of 11:26, 19 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) $
