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Question about inverse Fourier transform


Johnny5

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I want to derive the formula for the inverse Fourier transform.

 

Does anyone know how to do this?

 

Thank you

 

PS: I was looking at this site here , and they talk about convolutions. I don't know if convolutions would help or not. I remember using the convolution theorem, but it's been awhile.

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I'll write up something soon (in next 1-2 days[/i']), as I'm bit busy at the moment.

 

 

Take your time, in the meantime I am going to refresh my memory on Fourier series. 2 days is about right for me to have it again, so it's just as well.

 

Thank you Dapthar.

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Hmmm, from all the mathematics I've taken, I've always taken the Fourier transform as a definition. I didn't even know that there was an explanation to the equation that defines the concept. Usually what I would be asked to do is show that the fourier transform or inverse fourier transform of a pair (say rect(x) and sinc(x)) is so using the "first principles" which is the given definition. However, the proving of the definition itself will be interesting, so I'll be looking forward to seeing stuff here ^^

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Well, he's not so much proving the definition of a Fourier transform as deriving the definition of the inverse Fourier transform from the definition of a Fourier transform.

 

That's correct Dave. You cannot prove a definition, definitions are stipulated to be true, so there would be no point to that.

 

Dave said it correctly. I want to derive the formula for the inverse Fourier transform, not derive the definition of the Fourier transform.

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I want to derive the formula for the inverse Fourier transform.

 

Does anyone know how to do this?

 

Thank you

 

PS: I was looking at this site here ' date=' and they talk about convolutions. I don't know if convolutions would help or not. I remember using the convolution theorem, but it's been awhile.[/quote']

 

Yeah, it comes out of Fourier's inversion theorem.

 

[math]f(t)=\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{i\omega t}d\omega}\int^{\infty}_{-\infty}f(t)e^{-i\omega t}dt

[/math]

 

 

The integral with respect to t is just the fourier transform and so when it is multiplied by [math]e^{i\omega t}[/math] and integrated with respect to omega we end up with f(t) again. The inversion theorem comes from an extra bit on the derivation of the transform itself. That's the derivation I know of at any rate. The transform bit goes in the omega integral, it didn't look right when I used brackets, so I used the notation used when I learnt it. And so the inverse is

 

[math]f(t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty} \tilde{f} (\omega)e^{i\omega t}d\omega}

[/math]

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