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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]=X[l]\ast Y^{\ast}[-l]=X[-l]\ast Y^{\ast}[l]=Y[l]\ast X^{\ast}[-l]=r_{yx}[l]</math><br>
 
<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>
 
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[[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 11:53, 19 February 2019


ECE Ph.D. Qualifying Exam

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] $


Back to QE CS question 2, August 2011

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