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] $

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)
Wan82_CS2-7.PNG


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva