Contents
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)
- Is one of them that the second omega should be positive not negative ~AS
- Actually, it does not make any difference since cosine is an even function. -pm
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)
- Shouldn't the 1/2pi not come in until the third line, like this?
$ x(t) = \mathcal{F}^{-1} (\mathcal{X}(\omega)) \, $
$ = \pi \mathcal{F}^{-1} (\delta(\omega + \omega _o) + \delta(\omega - \omega _o)) $
$ = \frac{1}{2\pi} \pi \int^{\infty}_{-\infty} (\delta(\omega + \omega _o) + \delta(\omega - \omega _o)) e^{j\omega t} d\omega $
- Yes, you are right. -pm
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.
- The conclusion is incorrect. Can somebody explain what the mistake is? --Mboutin 09:29, 22 October 2008 (UTC)
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.
- Although the conclusion is correct, the justification is wrong. Hint: notice that the signal x[n] is a DT signal. --Mboutin 09:30, 22 October 2008 (UTC)
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) $
- Looks pretty good to me, though I would have prefered the sentence "Taking the I.F.T. of both sides of the equation, we get:" instead of "After going through a inverse fourier transformation:", which is bit imprecise. Also, some "if and only if" symbols ($ \Leftrightarrow $) in the right places would improve the clarity of the explanation. Still I would give full credit. --Mboutin 09:34, 22 October 2008 (UTC)
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}) } $
- One could simply use Formula 42 in the table to obtain the Fourier transform of $ x[n] $, instead of computing it. The final answer is correct and all the important steps are there. Repeating the $ \mathcal{Y}(\omega) $ in the last three lines makes the answer a bit less clear (and takes more time to write). Still I would give full credit for that answer. --Mboutin 09:39, 22 October 2008 (UTC)
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 } $
- The answer is wrong. The justification seems unrelated to the actual question, though it's hard to say what the person is actually doing. More important, the answer is supposed to be a function of "n", but what is given is a function of "$ \omega $" as well, so something is really wrong . Can somebody suggest a better answer? --Mboutin 09:55, 22 October 2008 (UTC)
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 \frac{u}{a}} du , a<0 $
$ = \frac{e^{j\omega \frac{b}{a}}}{-a} \mathcal{X}(\frac{\omega}{a}), a<0 $
- Good job! This gets full credit. However, you could improve the logic/flow of your answer by adding the word "since" before the first $ a<0 $, and removing the $ a<0 $ from the last line. --Mboutin 10:00, 22 October 2008 (UTC)
Corrections
I typed it yesterday night, and was planning to makes some corrections tonight. I never thought it would be graded that fast though......
BTW, corrections:
Question 1
The answer should be $ 8\cos(2\omega) + 2\cos(\omega)\, $
Corrections again, i never thought i made another mistake. Thanks for those who pointed it out for me.
$ =\sum^{\infty}_{n = -\infty}(n^2(u[n-2] - u[n+2])e^{ -j\omega n}\, $
$ = \sum^{1}_{n = -2}n^2 e^{ -j\omega n}(-1)\, $
$ = -(4e^{2 j\omega } + e^{ j\omega } + e^{- j\omega })\, $
$ = -4e^{2 j\omega }- 2\frac{e^{ j\omega } + e^{- j\omega }}{2} \, $
$ = -4e^{2 j\omega } - 2\cos(\omega)\, $
$ =-2cos(\omega) - 4(cos(2\omega) + j \sin(2\omega))\, $
I think this is the right answer. The sum only goes to 1.
$ =\sum^{\infty}_{n = -\infty}(n^2(u[n-2] - u[n+2])e^{ j\omega n}\, $
$ = \sum^{1}_{n = -2}n^2 e^{ j\omega n}\, $
$ = 4e^{-2 j\omega } + e^{ j\omega } + e^{- j\omega }\, $
$ = 8\frac{e^{-2 j\omega }}{2}+ 2\frac{e^{ j\omega } + e^{- j\omega }}{2} \, $
$ = 4e^{-2 j\omega } + 2\cos(\omega)\, $
Question 3
I still can't really understand question relating real, imaginary , even and odd signal. Can anyone explain this?
The Fourier Transform of a DT signal has to be periodic and since X(w) is not, the answer can not be right.
Question 4
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{2}{(1-\frac{1}{4}e^{-j\omega} ) } + \frac{4}{(1-\frac{1}{2}e^{-j\omega})} $
Fourier transfomation of $ (x[n])\, $ where $ x[n] = (a)^n u[n] \, $ is $ \frac{1}{1-ae^{-j\omega}}, a<1 \, $, thus the unit impulse response for $ \mathcal{X}(\omega)\, $ is
$ x[n] = 4(\frac{1}{2})^nu[n] - 2(\frac{1}{4})^nu[n] $
Other Thoughts/Opinions
Question 1
$ =\sum^{\infty}_{n = -\infty}(n^2(u[n-2] - u[n+2])e^{ j\omega n}\, $
Note that u[n-2] is 0 up until n = 2, and u[n+2] is 0 only up until n = -2. So, there's a minus sign missing in the next step:
$ = \sum^{1}_{n = -2}-n^2 e^{ j\omega n}\, $
$ = -4e^{-2 j\omega } - e^{ j\omega } - e^{- j\omega }\, $
$ = -2 \cos(\omega) - 4(\cos(2\omega) + j \sin(2\omega)) $ (Euler's formula)
--Thomas34 19:27, 26 October 2008 (UTC)
Question 3
Since g[n] is odd, $ x[-n] = \frac{1}{(g[-n])^2} = \frac{1}{(-g[n])^2} = \frac{1}{(g[n])^2} = x[n] $, so x[n] is even.
Since g[n] is imaginary, g[n] = jh[n], where h[n] is real. $ x[n] = \frac{1}{(jh[n])^2} = \frac{1}{j^2 (h[n])^2} = \frac{-1}{(h[n])^2} $, so x[n] is real.
Thus, the FT of x[n], namely X(w), is also both real and even.
Bob's answer in part a is obviously imaginary, and therefore cannot satisfy our condition that X(w) must be real.
Alice's answer in part b is odd ($ X_A (-\omega) = \frac{1}{sin(-\omega)} = \frac{-1}{sin(\omega)} = -X_A (\omega) $), and therefore cannot satisfy our condition that X(w) must be even.
Devin's answer, however, is both real and even ($ X(-\omega) = \frac{1}{(-\omega)^2} = \frac{1}{\omega^2} = X(\omega) $), so it is potentially correct.
--Thomas34 19:27, 26 October 2008 (UTC)
About Devin's answer, she went over this in class and $ \frac{1}{\omega^2} $ can't be correct because it's not periodic, and all DT fourier transforms are periodic.
--tsafford 10:15, 29 October 2008 (UTC)
Question 4
I agree with your corrected answer; your solution appears more or less correct to me, except for a missed absolute value sign: The Fourier transfomation of $ (x[n])\, $ where $ x[n] = (a)^n u[n] \, $ is $ \frac{1}{1-ae^{-j\omega}}, |a|<1 \, $. (Else, we could have a=-2, e.g., which would diverge.)
--Thomas34 19:31, 26 October 2008 (UTC)
