Differentiation Theorem Dual

Theorem: Let $x(n)$ denote a signal with Fourier transform $X(\omega)$, and let

$$X^\prime(\omega) \isdef \frac{d}{d\omega} X(\omega)$$

denote the derivative of $X$ with respect to $\omega$. Then we have

$$\zbox {-jt x(t) \;\longleftrightarrow\;\frac{d}{d\omega}X(\omega)}$$

where $X(\omega)$ denotes the Fourier transform of $x(t)$.

Proof: We can show this by direct differentiation of the definition of the Fourier transform:

$$\begin{eqnarray*} X^\prime(\omega) &\isdef & \frac{d}{d\omega} \int_{-\infty}^{\... ...(t)] e^{-j\omega t} dt\\ &=& \hbox{\sc FT}_\omega\{[-jtx(t)]\} \end{eqnarray*}$$

An alternate method of proof is given in §2.3.13.

The transform-pair may be alternately stated as follows:

$$\zbox {-t x(t) \;\longleftrightarrow\;\frac{d}{d(j\omega)}X(\omega)}$$