Parseval's Relation w/ Fourier Transform

[xago]

Homework Statement

[PLAIN]http://img600.imageshack.us/img600/161/parcq.png

Homework Equations

Parseval's Theorem using FT's for this is $∫^{\infty}_{-\infty}$ $|f(t)|^{2}$dx = $∫^{\infty}_{-\infty}$ $|\tilde{f}(w)|^{2}$dw

The Attempt at a Solution

From what I know, the Fourier transform of f(t) = $e^{-a|t|}$ is $\tilde{f} (w) = \frac{2a}{w^{2}+a^{2}}$

So for my answer I would simply evaluate $∫^{\infty}_{-\infty}$ $|f(t)|^{2}$dx for f(t) = $e^{-a|t|}$

However in the question there is no "2a" term on the top so I'm confused as how to proceed
$|\tilde{f}(w)|^{2}$ does not equal $\frac{dw}{w^{2}+a^{2}}$ as given in the question where $\tilde{f} (w) = \frac{2a}{w^{2}+a^{2}}$

 

[Dickfore]

So, you need to integrate:
$$\int_{-\infty}^{\infty}{e^{-2 a |t|} \, dt}$$
Parseval's Theorem guarantees that this integral (provided that $f(t) = e^{-a |t|}$ is truly the inverse Fourier transform of $(\omega^2 + a^2)^{-1}$) is the same as the integral you are supposed to calculate.

Then, split the limits of integration so that you get rid of the absolute value sign and do the integrals. They are table.

 

[xago]

But the Fourier Transform of f(t) = $e^{-a|t|}$ is $\tilde{f} (w) = \frac{2a}{w^{2}+2a^{2}}$ and not $\frac{1}{w^{2}+a^{2}}$ so i don't understand how you can say $∫^{\infty}_{-\infty}$ $|f(t)|^{2}$dx = $∫^{\infty}_{-\infty}$ $|\tilde{f}(w)|^{2}$dw when $|f(t)|^{2}$ = $e^{-2a|t|}$ and $|f(w)|^{2}$ = $(\frac{2a}{w^{2}+a^{2}})^{2}$

Basically it says evaluate $∫^{\infty}_{-\infty}$ $\frac{dw}{w^{2}+a^{2}}$ and not $∫^{\infty}_{-\infty}$ $\frac{2a}{w^{2}+a^{2}}$dw

 

[xago]

Can anyone shed some light on this?

 

[vela]

Hint:
$$\frac{1}{\omega^2+a^2} = \frac{1}{2a}\left(\frac{2a}{\omega^2+a^2}\right)$$

 

[xago]

Hmm :S , so is it just as simple as evaluating $\frac{1}{2a}∫^{\infty}_{-\infty}$ $e^{-2a|t|}$dt ?

 

[Dickfore]

no. Because you need $|f(t)|^2$ in the integral.

 

[xago]

oops

$∫^{\infty}_{-\infty}$ $(\frac{e^{-a|t|}}{2a})^{2}$dt

=$\frac{1}{4a^{2}}∫^{\infty}_{-\infty}$ $e^{-2a|t|}$dt
?

 

[vela]

Yes.

 

[Dickfore]

BTW, methods of contour integration should give the result:
$$2 \pi i \lim_{z \rightarrow i a}{\frac{d}{dz} \left( \frac{1}{(z + i a)^2}\right)}$$

and Wolfram Alpha gives the following value for the integral:
http://www.wolframalpha.com/input/?i=Integrate[1/(x^2+a^2)^2,{x,-Infinity,Infinity}]"

EDIT:
Also, your Parseval's Theorem is wrong. It should be:
$$\int_{-\infty}^{\infty}{\vert f(t) \vert^2 \, dt} = \int_{-\infty}^{\infty}{\vert \tilde{F}(\omega) \vert^2 \, \frac{d \omega}{2 \pi}}$$
(there's a factor of 2 pi missing in your definition), PROVIDED THAT, your Fourier transform is:
$$\tilde{F}(\omega) = \int_{-\infty}^{\infty}{f(t) \, e^{i \omega t} \, dt}$$
and the inverse Fourier transform is:
$$f(t) = \int_{-\infty}^{\infty}{\tilde{F}(\omega) e^{-i \omega t} \, \frac{d \omega}{2 \pi}}$$

This convention is standard in Physics, but not in EE. See if your function and Fourier transform satisfy these.

 

[xago]

Well I ended up getting $\frac{1}{4a^{3}}$ which is the same as the Wolfram Alpha solution minus the factor of 2$\pi$, so Ill include that in my FT's this time. Thanks for your help!