Differentiation in Frequency Domain Property of Discrete-Time Fourier Transform

Discrete-Time Fourier Transform

The Fourier transform of a discrete-time sequence is known as the discrete-time Fourier transform (DTFT).

Mathematically, the discrete-time Fourier transform (DTFT) of a discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$is defined as −

$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$

Differentiation in Frequency Domain Property of DTFT

Statement - The differentiation in frequency domain property of discrete-time Fourier transform states that the multiplication of a discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ by n is equivalent to the differentiation of its discrete-time Fourier transform in frequency domain. Therefore, if,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$

Then

$$\mathrm{\mathit{n}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{j}\frac{\mathit{d}}{\mathit{d\omega }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$

Proof

From the definition of DTFT, we have,

$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$

Differentiating both sides with respect to ω, we get,

$$\mathrm{\frac{\mathit{d}}{\mathit{d\omega }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\frac{\mathit{d}}{\mathit{d\omega}}\mathrm{\left[\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}} \right ]}}$$

$$\mathrm{\Rightarrow\frac{\mathit{d}}{\mathit{d\omega}}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\frac{\mathit{d}}{\mathit{d\omega}}\mathit{e^{-\mathit{j\omega n}}}\:\mathrm{=}\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathrm{\left ( -\mathit{jn}\right )}\mathit{e^{-\mathit{j\omega n}}}}$$

$$\mathrm{\Rightarrow\frac{\mathit{d}}{\mathit{d\omega}}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:-\mathit{j}\sum_{\mathit{n=-\infty}}^{\infty}\mathit{n}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}\:\mathrm{=}\:-\mathit{j}\mathit{F}\mathrm{\left[\mathit{n}\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}}$$

$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{n}\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{j}\frac{\mathit{d}}{\mathit{d\omega }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$

Numerical Example

Using the differentiation in frequency domain property of DTFT, find the DTFT of the following sequence,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{u}\right)}}$$

Solution

The given discrete-time sequence is,

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{u}\right)}}$$

Taking DTFT on both sides, we have,

$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{F}\mathrm{\left[\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{u}\right)} \right]}}$$

Now, using the differentiation in frequency domain property of DTFT $\mathrm{\left [ i.e,\:\mathit{n}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{j}\frac{\mathit{d}}{\mathit{d\omega }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)} \right ]}$ ,we get,

$$\mathrm{\mathit{F}\mathrm{\left[\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right]}\:\mathrm{=}\:\mathit{j}\frac{\mathit{d}}{\mathit{d\omega }}\mathrm{\left\{\mathit{F}\mathrm{\left[\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right]} \right\}}}$$

$$\mathrm{\Rightarrow\mathit{F}\mathrm{\left[\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right]}\:\mathrm{=}\:\mathit{j}\frac{\mathit{d}}{\mathit{d\omega }}\mathrm{\left[\frac{1}{1-\mathrm{\left ( \frac{1}{3} \right)}\mathit{e^{-j\omega}}}\right ]}}$$

$$\mathrm{\Rightarrow\mathit{F}\mathrm{\left[\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right]}\:\mathrm{=}\:\mathit{j}\mathrm{\left[ \frac{\mathrm{\left\{ 1-\mathrm{\left(\frac{1}{3}\right)}\mathit{e^{-\mathit{j\omega}}}\right\}}\mathrm{\left( 0 \right)}-\mathrm{\left( 1\right)}\mathrm{\left\{ \mathrm{\left[-\mathrm{\left ( \frac{1}{3}\right)}\mathit{e}^{-\mathit{j\omega}}\mathrm{\left ( -\mathit{j}\right)}\right ]}\right\}}}{\mathrm{\left\{1-\mathrm{\left ( \frac{1}{3} \right)}\mathit{e^{-j\omega }} \right\}}^{\mathrm{2}}} \right ]}}$$

$$\mathrm{\Rightarrow\mathit{F}\mathrm{\left[\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right]}\:\mathrm{=}\:\mathit{j}\mathrm{\left[\frac{-\mathit{j\mathrm{\left( \frac{1}{3}\right)}\mathit{e^{-j\omega }}}}{\mathrm{\left\{1-\mathrm{\left( \frac{1}{3}\right)} \mathit{e^{-j\omega}}\right\}}^{\mathrm{2}}}\right ]}}$$

$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{n}\mathrm{\left( \frac{1}{3}\right)}^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)} \right]}\:\mathrm{=}\:\frac{\mathrm{\left ( \frac{1}{3} \right )}\mathit{e^{-j\omega}}}{\mathrm{\left[1-\mathrm{\left ( \frac{1}{3}\right)} \mathit{e^{-j\omega }}\right ]}^{\mathrm{2}}}}$$