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

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 August 11th, 2019, 03:14 AM #1 Newbie   Joined: Jun 2017 From: Israel Posts: 2 Thanks: 0 Calculation of Fourier Transform Derivative d/dw (F{x(t)})=d/dw(X(w)) 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 skipjack; August 11th, 2019 at 11:00 AM.
 August 11th, 2019, 10:52 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,645 Thanks: 1476 $X(\omega) = \displaystyle \int_{-\infty}^\infty~x(t) e^{-j\omega t}~dt$ $\dfrac{d}{d\omega} X(\omega) = \displaystyle \int_{-\infty}^\infty~x(t) \dfrac {d}{d\omega}e^{-j\omega t}~dt =$ $\displaystyle \int_{-\infty}^\infty~(-jt) x(t) e^{-j\omega t}~dt= \mathscr{F}\{(-jt)x(t)\}$ 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.

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