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August 11th, 2019, 03:14 AM   #1
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Calculation of Fourier Transform Derivative d/dw (F{x(t)})=d/dw(X(w))

Hello to my Math Fellows,

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):

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)}

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.
Alexei Nomazov is offline  
August 11th, 2019, 10:52 AM   #2
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$X(\omega) = \displaystyle \int_{-\infty}^\infty~x(t) e^{-j\omega t}~dt$

$\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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