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 June 16th, 2019, 07:45 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 How do I deal with a Fourier transform that contains derivatives? I want to calculate this: $\displaystyle \frac{1}{2 \pi } \int_{- \infty}^{\infty} |X(ω)|^2 dω$ Should I just normally calculate the derivative inside of X(ω) and then use that final form of Χ(ω) to calculate the integral? Or can I use somehow the property which says that the Fourier transformation of the nth derivative of x(t) is just the Fourier transformation of x(t) times $\displaystyle (jω)^n$??? $\displaystyle F[\frac{d^n}{dx^n}x(t)] = (jω)^n X(ω)$ I can't find a way to use this property in this situation since the derivative is already inside the Fourier itself. By the way the text says "The Fourier transform of x(t) is given by X(ω). Find the energy of that signal."
 June 16th, 2019, 11:08 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,496 Thanks: 1371 $i t f(t) \Leftrightarrow \dfrac{d}{d\omega}F(\omega)$ Taking the inverse Fourier transform of what's inside the brackets we have $\mathscr{F}^{-1}\left\{ \left(\pi \delta (\omega )+\dfrac{1}{i \omega }\right) \left(1-\exp (-i 4 \omega ) \right)\right \}= \sqrt{2\pi }\left(u(t+4)-u(t) \right)$ and thus $f(t) = \sqrt{2\pi }\left(u(t+4)-u(t) \right) t^4$ The energy in $f(t)$ is given by $E = \displaystyle \int_{-\infty}^\infty~\left|x(t)\right|^2~dt = 2\pi \int_{-4}^0~t^8~dt$ I leave you to do that simple integral. Thanks from topsquark
 June 16th, 2019, 11:31 AM #3 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 So I need to go back to the time domain. I thought that I shouldn't do that

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