Differentiation Theorem Dual


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

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

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

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

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


Next Section:
Scaling Theorem
Previous Section:
Differentiation Theorem