Differentiation in Frequency Domain Property of Discrete-Time Fourier Transform

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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 x(n) is defined as −

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

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 x(n) by n is equivalent to the differentiation of its discrete-time Fourier transform in frequency domain. Therefore, if,

$$\mathrm{x(n)\:\overset{FT}\longleftrightarrow\:X(\omega)}$$

Then

$$\mathrm{nx(n)\:\overset{FT}\longleftrightarrow\:j\:\frac{d}{d\omega}\:X(\omega)}$$

Proof

From the definition of DTFT, we have,

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

Differentiating both sides with respect to ω, we get,

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

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

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

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

Numerical Example

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

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

Solution

The given discrete-time sequence is,

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

Taking DTFT on both sides, we have,

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

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

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

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

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

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

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