 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. Tags calculation, d or dw, derivative, fourier, fourier transform, fxtd or dwxw, transform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Applied Math 0 December 16th, 2015 02:03 PM bonildo Calculus 5 September 27th, 2014 11:27 AM progrocklover Real Analysis 1 March 24th, 2011 09:29 PM beckie Real Analysis 3 June 20th, 2010 01:58 PM eliotsbowe Real Analysis 0 June 19th, 2010 07:13 AM

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