Differentiation Theorem Dual

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

$$X^\prime(\omega) \isdefs \frac{d}{d\omega} X(\omega)$$ (B.6)

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

$$\zbox {-jt x(t) \;\longleftrightarrow\;\frac{d}{d\omega}X(\omega)}$$ (B.7)

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:

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

An alternate method of proof is given in §2.3.13.

The transform-pair may be alternately stated as follows:

$$\zbox {-t x(t) \;\longleftrightarrow\;\frac{d}{d(j\omega)}X(\omega)}$$ (B.8)