# Calculation of Fourier Transform Derivative d/dw (F{x(t)})=d/dw(X(w))

#### Alexei Nomazov

Hello to my Math Fellows,

Problem:
I am looking for a way to calculate w-derivative of Fourier transform, d/dw (F{x(t)}), in terms of regular Fourier transform, X(w)=F{x(t)}.

Definition Based Solution (not good enough):
from

I can find that w-derivative of Fourier transform for x(t) is Fourier transform of t*x(t) multiplied by -j:
d/dw (F{x(t)})=d/dw(X(w))=-j*F{t*x(t)}

Question:
But, taking into account the differentiation and duality properties of Fourier transform:

is it possible to express the derivative, d/dw (F{x(t)}), in frequency domain using terms of X(w) ???

Many Thanks,
Desperate Engineer.

Last edited by a moderator:

#### romsek

Math Team
$X(\omega) = \displaystyle \int_{-\infty}^\infty~x(t) e^{-j\omega t}~dt$

$\dfrac{d}{d\omega} X(\omega) = \displaystyle \int_{-\infty}^\infty~x(t) \dfrac {d}{d\omega}e^{-j\omega t}~dt =$

$\displaystyle \int_{-\infty}^\infty~(-jt) x(t) e^{-j\omega t}~dt= \mathscr{F}\{(-jt)x(t)\}$

repeated derivatives add another power of $(-jt)$ obtaining the full property

$\dfrac{d^n}{d\omega^n} X(\omega) = \mathscr{F}\{(-jt)^n x(t)\}$

This is property 11 on your sheet.