# Fourier Transform of a Gaussian With Non-Zero Mean

1. Nov 30, 2012

### malevolence19

1. The problem statement, all variables and given/known data
I am looking at finding the fourier transform of:
$$f(t)=\exp \left[ \frac{-(t-m)^2}{2 \sigma^2}\right]$$

2. Relevant equations

$$\hat{f}(t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt$$

3. The attempt at a solution
I did it a little differently that my professor suggested. He said use taylor series but when I do that I end up getting a triple sum (2 from the exponentials and 1 coming from an (x+m)^j; so I used binomial expansion) so I did it the traditional way you deal with gaussians: complete the square.

I combined the exponentials, completed the square and end up with an answer that looks like:
$$\hat{f}(t)=\frac{1}{2 \sigma}e^{-(\frac{\sigma^2 \omega^2}{2}+i \sigma \omega)}$$
But wolfram has this:
http://www.wolframalpha.com/input/?i=fourier+transform+of+e%5E%28-%28t-m%29%5E2%2F%282r%5E2%29%29

With $r= 2 \sigma^2$

I was just wondering if anyone could double check me against wolfram? I can type out steps if needed. This isn't a homework or anything, he just brought it up and I suggested completing the square instead. Apparently this means I have to present it to the class on Monday -.-

Last edited: Nov 30, 2012
2. Nov 30, 2012

### jbunniii

Comparing your expression for $f(t)$ with the input to the Fourier transform at the Wolfram link, I'm not sure why you put $r = 2\sigma^2$ instead of $r = \sigma$.

Also, check Wolfram's definition of the Fourier transform. It might not include your $1/\sqrt{2\pi}$ factor. There is no universal consensus regarding this. : You can see their definition by clicking on the "definition" link. Indeed, their definition omits the $1/\sqrt{2\pi}$

3. Nov 30, 2012

### malevolence19

That's what I meant $$r=\sigma$$. I typed it in two different ways and I had them mixed up :/

And omitting the 1/sqrt(2 pi) gives me:
$$\sqrt{\frac{\pi}{2 \sigma^2}}e^{-(\frac{\sigma^2 \omega^2}{2}+\sigma^2 i \omega)}$$
They have the sigma in the numerator some how.

4. Nov 30, 2012

### malevolence19

Ahhhh, nevermind I see it!

5. Nov 30, 2012

### jbunniii

By the way, you can also solve this problem by using the shift property of the Fourier transform: if $\mathcal{F}[f(t)] = \hat{f}(\omega)$, then $\mathcal{F}[f(t-m)] = \hat{f}(\omega)\exp(-im\omega)$.

6. Nov 30, 2012

### malevolence19

Thank you. I always forget about these nice shifting formulae.

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