DSPRelated.com
Free Books
   Free Books    Spectral Audio Signal Processing  

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:

\begin{eqnarray*} \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), \end{eqnarray*}

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

Next Section:
Differentiation Theorem Dual
Previous Section:
Radians versus Cycles