Difference between revisions of "Midterm Exam Div 2 Fall 2006 Solution ECE301Fall2008mboutin" - Rhea

Revision as of 21:36, 21 October 2008

Question 1

$ \mathcal{F} (n^2u[n-2] - n^2 u[n+2]) = \sum^{\infty}_{n = -\infty}(n^2u[n-2] - n^2 u[n+2])e^{ j\omega n}\, $

$ =\sum^{\infty}_{n = -\infty}(n^2(u[n-2] - u[n+2])e^{ j\omega n}\, $

$ = \sum^{2}_{n = -2}(n^2 e^{ j\omega n}\, $

$ = 4e^{2 j\omega } + 4e^{-2 j\omega } + e^{ j\omega } + e^{- j\omega }\, $

$ = 8\frac{e^{2 j\omega } + e^{-2 j\omega }}{2}+ 2\frac{e^{ j\omega } + e^{- j\omega }}{2} \, $

$ = 8\cos(2\omega) + 2\cos(-\omega)\, $

Question 2

$ x(t) = \mathcal{F}^{-1} (\mathcal{X}(\omega)) \, $

$ = \frac{1}{2\pi} \pi \mathcal{F}^{-1} (\delta(\omega + \omega _o) + \delta(\omega - \omega _o)) $

$ = \frac{1}{2} \int^{\infty}_{-\infty} (\delta(\omega + \omega _o) + \delta(\omega - \omega _o)) e^{j\omega t} d\omega $

$ = \frac{1}{2} e^{j\omega _o t} \int^{\infty}_{-\infty} \delta(\omega + \omega _o) d\omega + \frac{1}{2} e^{-j\omega _o t} \int^{\infty}_{-\infty}\delta(\omega - \omega _o)) d\omega $

$ = \frac{1}{2} e^{j\omega _o t} + \frac{1}{2} e^{-j\omega _o t} d\omega $

$ = \frac{1}{2} ( e^{j\omega _o t} + e^{-j\omega _o t} ) $

$ = \cos(\omega _o t) \, $

Question 3

a

Only a Real and odd signal can be Fourier transformed into an imaginary and odd function. The Transformed output is a imaginary and even signal,thus the Fourier transform is wrong.

b

A pure imaginary signal and odd function will Fourier transform into a real and odd signal. As the function $ \frac{1}{\sin(\omega)} \, $ is a real and odd signal, Alice answer could be right.

c

As stated above, A pure imaginary signal and odd function will Fourier transform into a real and odd signal. Ths transformed signal is a real and even signal, which is transformed from a even and real signal. Thus Devin answer is wrong.

Question 4

a

$ \mathcal{Y}(\omega) = \mathcal{H}(e^{j\omega}) \mathcal{X}(\omega) $

$ \mathcal{Y}(\omega) = \frac{2}{1-\frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega} } \mathcal{X}(\omega) $

$ \mathcal{Y}(\omega) -\frac{3}{4}e^{-j\omega} \mathcal{Y}(\omega)+ \frac{1}{8}e^{-2j\omega} \mathcal{Y}(\omega)= 2 \mathcal{X}(\omega) $

After going through a inverse fourier transformation:

$ Y[n] -\frac{3}{4} Y[n-1]+ \frac{1}{8}Y[n-2] = 2 \mathcal{X}(\omega) $

b

$ x[n] = (\frac{1}{4})^n u[n] $

$ \mathcal{F} (x[n]) = \sum^{\infty}_{n=-\infty}(\frac{1}{4})^n u[n] e^{-j\omega n} $

$ = \sum^{\infty}_{n=0}(\frac{1}{4} e^{-j\omega })^n $

$ \mathcal{X}(\omega)= \frac{1}{1 - \frac{1}{4} e^{-j\omega }} $

$ \mathcal{Y}(\omega) = \mathcal{H}(e^{j\omega}) \mathcal{X}(\omega) $

$ \mathcal{Y}(\omega) = (\frac{2}{1-\frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega} } )(\frac{1}{1 - \frac{1}{4} e^{-j\omega }}) $

$ \mathcal{Y}(\omega) = (\frac{2}{(1-\frac{1}{4}e^{-j\omega} ) (1-\frac{1}{2}e^{-j\omega}) } )(\frac{1}{1 - \frac{1}{4} e^{-j\omega }}) $

$ \mathcal{Y}(\omega) = \frac{2}{(1-\frac{1}{4}e^{-j\omega} )^2 (1-\frac{1}{2}e^{-j\omega}) } $

c

$ \mathcal{X}(\omega) = \frac{2}{1-\frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega} } $

$ = \frac{2}{(1-\frac{1}{4}e^{-j\omega} ) (1-\frac{1}{2}e^{-j\omega})} $

$ = -\frac{4}{(1-\frac{1}{2}e^{-j\omega} ) } + \frac{4}{(1-\frac{1}{2}e^{-j\omega})} $

As shown above, Fourier transfomation of $ \mathcal{F}(x[n]) = \mathcal{F}((a)^n u[n]) = \frac{1}{1-a}, a<0 \, $, thus the unit impulse response for $ \mathcal{X}(\omega)\, $ is

$ x[n] = 4(\frac{1}{2})^ne^{-j\omega n } - 2(\frac{1}{4})^ne^{-j\omega n } $