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

Revision as of 05:28, 22 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)\, $

• Overall, this is pretty good: well explained, all needed steps are there, and no extraneous details. Notice how the answer flows like a text, and all the steps are logical. However, there are two small computational mistakes which would cost a total of about three points: can anybody see them? --Mboutin 09:11, 22 October 2008 (UTC)

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) \, $

• This problem is attacked the right way: taking the inverse Fourier transform, instead of the Fourier transform. The explanation is crystal clear and logical. No important step is missing. This deserves a perfect score! --Mboutin 09:14, 22 October 2008 (UTC)

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.

• This could be phrased a bit more formally but the text is perfectly clear. The first statement is correct. The second statement is also correct. However, the conclusion does not follow from these two statements. Can somebody propose a better solution? --Mboutin 09:28, 22 October 2008 (UTC)

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 } $

Question 5

$ \mathcal{F}(x(at+b) = \int^{\infty}_{-\infty} x(at+b) e^{-j\omega t} dt $

Let $ u = at+b \, $

$ t = \frac{u+b}{a} \, $

$ du = a dt \, $

$ \mathcal{F}(x(at+b) = \int^{\infty}_{-\infty} x(u) e^{-j\omega \frac{u+b}{a}} \frac{du}{a} $

$ = \frac{e^{j\omega \frac{b}{a}}}{-a}\int^{\infty}_{-\infty} x(u) e^{-j\omega u} du , a<0 $

$ = \frac{e^{j\omega \frac{b}{a}}}{-a} \mathcal{X}(\frac{\omega}{a}), a<0 $