Free Books
   Free Books    Mathematics of the DFT  

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 $ x^\prime(t)$ exist, where $ x^\prime(t)$ denotes the time derivative of $ x(t)$. Then we have

$\displaystyle \zbox {x^\prime(t) \;\longleftrightarrow\;j\omega X(\omega)} $

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

$\displaystyle \zbox {\hbox{\sc FT}_{\omega}(x^\prime) = j\omega X(\omega)} $


Proof: This follows immediately from integration by parts:
\begin{eqnarray*} \hbox{\sc FT}_{\omega}(x^\prime) &\isdef & \int_{-\infty}^\in... ...\infty x(t) (-j\omega)e^{-j\omega t} dt\\ &=& j\omega X(\omega) \end{eqnarray*}
since $ x(\pm\infty)=0$. The differentiation theorem is implicitly used in §E.6 to show that audio signals are perceptually equivalent to bandlimited signals which are infinitely differentiable for all time.
Next Section:
Scaling Theorem
Previous Section:
Fourier Series (FS) and Relation to DFT