# Parseval's Relation w/ Fourier Transform

1. Oct 22, 2011

### xago

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 $\tilde{f} (w) = \frac{2a}{w^{2}+a^{2}}$

Last edited by a moderator: May 5, 2017
2. Oct 22, 2011

### 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.

Last edited: Oct 22, 2011
3. Oct 22, 2011

### 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

Last edited: Oct 22, 2011
4. Oct 22, 2011

### xago

Can anyone shed some light on this?

5. Oct 22, 2011

### vela

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

6. Oct 22, 2011

### xago

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

7. Oct 22, 2011

### Dickfore

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

8. Oct 22, 2011

### xago

oops

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

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

9. Oct 22, 2011

### vela

Staff Emeritus
Yes.

10. Oct 22, 2011

### 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.

Last edited by a moderator: Apr 26, 2017
11. Oct 22, 2011

### 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!