Question

(Fourier Transform) I'm trying to prove a property of Fourier Transforms and am having a little trouble doing so. I got the idea of making a substitution to try to solve it, but am otherwise blundering around in the dark a little. More information below.

Answer 1

Original prompt:

\[\text{Show that}\]\[E_\alpha \left\{f(ct) \right\}=\frac{1}{|c|}F_e \bigg(\frac{\alpha}{c}\bigg), \ \ \ c \neq 0\]\[\text{This is a} \ \textit{scaling} \ \text{formula}.\]

Answer 2

What I did was let z = ct, and thus dz = c, t = z/c, and (I'm not sure if I can say this last one, but it's really tempting given the form of the solution) dt = 1/c.

Answer 3

\[E_\alpha \left\{f(ct)\right\}=\int\limits_{-\infty}^{\infty}f(ct)e^{-i \alpha t}dt=\frac{1}{c}\int\limits_{-\infty}^{\infty}f(z)e^{-\frac{i \alpha z}{c}}dz\]

I'm both not totally sure if that last part is totally right, and I'm not sure if I'm on the right track or what exactly I should do to prove this property.

Answer 4

@SithsAndGiggles

Answer 5

ya, now simplify the integral what u are doing is rght

Answer 6

alpha = 2pif?