Differentiation Theorem Dual

Theorem: Let $x(n)$ denote a signal with DTFT $X(e^{j\omega})$, and let

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

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

$$\zbox {-jn x(n) \;\longleftrightarrow\;\frac{d}{d\omega}X(e^{j\omega})}$$

where $X(e^{j\omega})$ denotes the DTFT of $x(n)$.

Proof: Using integration by parts, we obtain

$$\begin{eqnarray*} \hbox{\sc IDTFT}_{n}(X^\prime) &\isdef & \frac{1}{2\pi}\int_{... ...i}^\pi X(e^{j\omega}) (jn)e^{j\omega n} d\omega\\ &=& -jn x(n). \end{eqnarray*}$$

An alternate method of proof is given in §2.4.3.

Corollary: Perhaps a cleaner statement is as follows:

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