Differentiation Theorem

Let $ x(t)$ denote a function differentiable for all $ t$ such that $ x(\pm\infty)=0$ and the Fourier transforms (FT) of both $ x(t)$ and $ {\dot x}(t)$ exist, where $ {\dot x}(t)$ denotes the time derivative of $ x(t)$ . Then we have

$\displaystyle \zbox {{\dot x}(t) \;\longleftrightarrow\;j\omega X(\omega)}$ (B.4)

where $ X(\omega)$ denotes the Fourier transform of $ x(t)$ . In operator notation:

$\displaystyle \zbox {\hbox{\sc FT}_{\omega}({\dot x}) = j\omega X(\omega)}$ (B.5)

Proof: This follows immediately from integration by parts:

\hbox{\sc FT}_{\omega}({\dot x})
&\isdef & \int_{-\infty}^\infty {\dot x}(t) e^{-j\omega t} dt\\
&=& \left. x(t)e^{-j\omega t}\right\vert _{-\infty}^{\infty} -
\int_{-\infty}^\infty x(t) (-j\omega)e^{-j\omega t} dt\\
&=& j\omega X(\omega),

since $ x(\pm\infty)=0$ .

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