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

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

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

$$\zbox {\hbox{\sc FT}_{\omega}({\dot x}) = j\omega X(\omega)}$$ (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 $x(\pm\infty)=0$ .