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Fourier transform

So I'm having some trouble getting on top of fourier transforms.

f(x)=e−x22γ22πγf(x) = \dfrac{e^{\frac{-x^2}{2\gamma^2}}}{\sqrt{2\pi} \gamma}

and ∫−∞∞e−t2dt=π\displaystyle\int^{\infty}_{-\infty} e^{-t^2} dt = \sqrt{\pi}

f^(s)=∫−∞∞f(x)cos(sx)dx\hat{f}(s) = \displaystyle\int^{\infty}_{-\infty} f(x)cos(sx) dx since we have an even function.

But... I feel like I'm missing something major... the integral I'm required to do seems massive... is there an easier way that I'm missing?
Reply 1
Avatar for DFranklin
DFranklin
18
It's not clear from what you've written what you know, what it is you are supposed to show, etc.

For this particular case, there's a trick (which it would be very hard to come up with yourself);

Differentiate f(x); you'll get something like f'(x) = Ax f(x) for some constant A (based on gamma).
Now, take the FT of both sides, recalling that there are rules for taking the FT of a derivative, and for finding the FT of x f(x) in terms of the derivative of the FT of f.

This gives you a differential equation that your Fourier transform must follow; solve it and you're done. (For the arbit constant, note that you know the value of the FT when s = 0).
Got it, thanks! :h:
Ah, yes - well, I used a different method (it seems it was in my notes; it's the Gaussian function) and I get to the last stage (where I have to use the integral that I'm given) and I can't seem to understand the last step (ie, substitution... lol)

f(x)=e−x22γ22πγf(x) = \dfrac{e^{\frac{-x^2}{2\gamma^2}}}{\sqrt{2\pi} \gamma}

f^(s)=12π∫−∞∞e−x22γ22πγeisx dx\hat{f}(s) = \dfrac{1}{\sqrt {2\pi}} \displaystyle \int_{-\infty}^{\infty} {\dfrac{e^{\frac{-x^2}{2\gamma^2}}}{\sqrt{2\pi} \gamma} e^{isx}} \ dx

Unparseable latex formula:

= \dfrac{1}{\sqrt {2\pi }. {\sqrt{2\pi} \gamma}} \displaystyle \int_{-\infty}^{\infty} {e^{\frac{-x^2}{2\gamma^2}}}. e^{isx}} \ dx



Unparseable latex formula:

= \dfrac{1}{2\pi \gamma}} \displaystyle \int_{-\infty}^{\infty} {e^{\frac{-x^2 + isx}{2\gamma^2}}}} \ dx



Unparseable latex formula:

= \dfrac{1}{2\pi \gamma}} \displaystyle \int_{-\infty}^{\infty} {e^{-\frac{1}{2}(\frac{x^2}{\gamma^2} - 2isx)}}} \ dx



Unparseable latex formula:

= \dfrac{1}{2\pi \gamma}} \displaystyle \int_{-\infty}^{\infty} {e^{-\frac{1}{2\gamma^2}(x^2 -2isx\gamma^2)}}} \ dx = \dfrac{1}{2\pi \gamma}} \displaystyle \int_{-\infty}^{\infty} {e^{-\frac{1}{2\gamma^2}((x-is\gamma^2)^2 - (is\gamma^2)^2)}}} \ dx


(Here I completed the square and took out the exponential that's only in terms of ssand γ\gamma).

Unparseable latex formula:

= \dfrac{e^{\frac{-s^2 \gamma^2}{2}}}{2\pi \gamma}} \displaystyle \int_{-\infty}^{\infty} {e^{-\frac{1}{2\gamma^2} (x-is\gamma^2)^2}}} \ dx



Now, if I let t=γ(x−isγ2)t={\gamma(x-is\gamma^2)} and use ∫−∞∞e−t2dt=π\displaystyle\int^{\infty}_{-\infty} e^{-t^2} dt = \sqrt{\pi}, then I should get the correct answer, which is

f^γ(s)=12πe−s2γ22\hat{f}_{\gamma}(s) = \dfrac{1}{\sqrt{2\pi}}e^{\frac{-s^2 \gamma^2}{2}}

However, the problem I've got is that my γ\gamma doesn't cancel and I need a 2\sqrt{2} on the top of the fraction, too. Have I gone wrong somewhere in my calculation? I can't spot it. :s-smilie:
(edited 13 years ago)
Reply 4
Avatar for DFranklin
DFranklin
18
When you complete the square, you are left with (x−isγ)2(x-is\gamma)^2, not (x−isγ2)2(x-is\gamma^2)^2.

As far as the sqrt(2), the 1/2 in e121γ2...\displaystyle e^{\frac{1}{2}\frac{1}{\gamma^2}} ... should provide it.

The reason I wasn't too keen to advocate the solution: you are tacitly assuming that ∫−∞∞e−t2 dt=∫−∞−isγ∞−isγe−t2 dt\int_{-\infty}^\infty e^{-t^2} \, dt = \int_{-\infty-is\gamma}^{\infty-is\gamma} e^{-t^2} \, dt, which I think will be hard without contour integration.
Original post by DFranklin
When you complete the square, you are left with (x−isγ)2(x-is\gamma)^2, not (x−isγ2)2(x-is\gamma^2)^2.


But isn't
Unparseable latex formula:

(x-is\gamma)^2 = x^2 -2is\gammax + i^2 s^2\gamma^2

? The x coefficient needs to be 2isγ22is\gamma^2, from what I've written above, anyway. :s-smilie:
Reply 6
Avatar for DFranklin
DFranklin
18
Yes, sorry. (Doesn't seem to be my best day for silly mistakes).

You need t2=12(x−isγ2γ2)2t^2 = \frac{1}{2}\left(\frac{x-is\gamma^2}{\gamma^2}\right)^2, and so

t=1γ2(x−isγ2)t=\frac{1}{\gamma\sqrt{2}}(x-is\gamma^2), not γ(x−isγ2)\gamma(x-is\gamma^2)
Reply 7
Avatar for DFranklin
DFranklin
18
Yes, sorry. (Doesn't seem to be my best day for silly mistakes).

You need t2=12(x−isγ2γ)2t^2 = \frac{1}{2}\left(\frac{x-is\gamma^2}{\gamma}\right)^2, and so

t=1γ2(x−isγ2)t=\frac{1}{\gamma\sqrt{2}}(x-is\gamma^2), not γ(x−isγ2)\gamma(x-is\gamma^2)
Original post by DFranklin
Yes, sorry. (Doesn't seem to be my best day for silly mistakes).

You need t2=12(x−isγ2γ2)2t^2 = \frac{1}{2}\left(\frac{x-is\gamma^2}{\gamma^2}\right)^2, and so

t=1γ2(x−isγ2)t=\frac{1}{\gamma\sqrt{2}}(x-is\gamma^2), not γ(x−isγ2)\gamma(x-is\gamma^2)


But we've got
Unparseable latex formula:

\displaystyle \int_{-\infty}^{\infty} {e^{-\frac{1}{2\gamma^2} (x-is\gamma^2)^2}}} \ dx



Unparseable latex formula:

\implies \ t^2 = \frac{1}{2\gamma^2} (x-is\gamma^2)^2}}


Unparseable latex formula:

\iff t^2 = \dfrac{1}{2} \left(\dfrac{x-is\gamma^2}{\gamma}\right)^2}}



Right? :K:
Reply 9
Avatar for DFranklin
DFranklin
18
I'm not sure what your question is. In your original post, you said you wanted to make the substitution t=γ(x−isγ2)t=\gamma(x-is\gamma^2). I was pointing out why this was going to be wrong.
Nevermind, finally realised what I was doing. I'm not a late night person. Thanks very much, helped a lot!

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