Hello guys,
just a question:
Is it ok if the fourier transform of the real-valued function of real variable has only complex values?
Can it be right?
Thank you
Aside from tables the only way I know is to do an inverse FT. You would need to have closed form expressions for your coefficients, though, so if you can't find them (or you're working numerically) I don't know what method you could use.
Do you have a specific problem you are working with?
-Dan
I was doing homework and I wanted to be sure that I am doing it right, we had only 2 lectures about it so far, so I guess we use the most basic stuff ..
It was only definition, successive conditions for existence of FT, and some basic relations about linearity and derivatives .. so we're just working with nice smooth functions,
analytical approach ..
I found something now about that inverse transform, so when I get $\displaystyle \mathcal{F}[f](\omega)$ (FT of f), I just need to find
$\displaystyle P.V.\int_{-\infty}^{\infty}\mathcal{F}[f](\omega)\mathrm{e}^{i\omega x}\,\mathrm{d}\omega$
and see if I get the original function, right?
yeah, it took a while but at least there's a way how to check it ..
if I may ask in this thread, one of my exercises says:
Compute the integral
$\displaystyle \int_0^{\infty}\frac{1}{x}(\cos{(ax)}-1)\sin{(bx)}\,\mathrm{d}x$
using the FT of the function
$\displaystyle f(x)=\left\{\begin{array}{cll} 1 &\,,\,& x\in [0,a] \\ -1 &\,,\,& x\in [-a,0) \\ 0 &\,,\,& x\in\mathbb{R}-[-a,a] \end{array}\right.$
where $\displaystyle a,b > 0$
I found $\displaystyle \mathcal{F}[f](\omega)=\frac{i}{\pi\omega}(\cos{(a\omega)}-1)\,,\,\omega\neq 0\,,$ and $\displaystyle \mathcal{F}[f](0)=0$,
and it looks that it could go through the inverse FT theorem somehow, but I didn't find the way, I don't see it there