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)

Alumni Liaison

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

Francisco Blanco-Silva