- #1
FeDeX_LaTeX
Gold Member
- 437
- 13
Hello,
In the past couple of days I have been looking at how to transform a function f(t) into another function F(s) via the Laplace transform, and have practiced performing simple Laplace transformations such at f(t) = sin(at), sinat, cos(at), eatf(t) and so on.
I looked on Wikipedia at a formula for the Fourier transform and it's a bit confusing; I know there are several types, here is a formula I found;
[tex]f(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi ix\xi}dx[/tex] for every real number ξ.
So if I say that f(x) = 1, then I end up with [tex]f(\xi) = \int_{-\infty}^{\infty}e^{-2\pi ix \xi}dx[/tex].
Integrating, we get
[tex]f(\xi) = -\frac{1}{2\pi ix \xi}e^{-2\pi ix \xi}[/tex].
However I'm stuck on how to evaluate this now, as I have to find the evaluation of the function at negative infinity and subtract that from the function evaluated at infinity (assuming ξ > 0, but when I do that I end up with an [tex]e^{\infty}[/tex] term when evaluating the function at negative infinity. Does this simply mean that it does not converge? Can you show me an example of a function that does converge? I tried using the Fourier transform on sin(x) earlier today and I ended up with [tex]f(\xi) = 0[/tex]...
Thanks.
In the past couple of days I have been looking at how to transform a function f(t) into another function F(s) via the Laplace transform, and have practiced performing simple Laplace transformations such at f(t) = sin(at), sinat, cos(at), eatf(t) and so on.
I looked on Wikipedia at a formula for the Fourier transform and it's a bit confusing; I know there are several types, here is a formula I found;
[tex]f(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi ix\xi}dx[/tex] for every real number ξ.
So if I say that f(x) = 1, then I end up with [tex]f(\xi) = \int_{-\infty}^{\infty}e^{-2\pi ix \xi}dx[/tex].
Integrating, we get
[tex]f(\xi) = -\frac{1}{2\pi ix \xi}e^{-2\pi ix \xi}[/tex].
However I'm stuck on how to evaluate this now, as I have to find the evaluation of the function at negative infinity and subtract that from the function evaluated at infinity (assuming ξ > 0, but when I do that I end up with an [tex]e^{\infty}[/tex] term when evaluating the function at negative infinity. Does this simply mean that it does not converge? Can you show me an example of a function that does converge? I tried using the Fourier transform on sin(x) earlier today and I ended up with [tex]f(\xi) = 0[/tex]...
Thanks.