Discrete-Time Fourier Transform

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Discrete-Time Fourier Transform (DTFT)

A discrete-time signal can be represented in the frequency domain using discrete-time Fourier transform. Therefore, the Fourier transform of a discretetime sequence is called the discrete-time Fourier transform (DTFT).

Mathematically, x(n) is a discrete-time sequence, then its discrete-time Fourier transform is defined as −

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

The discrete-time Fourier transform $\mathrm{X(\omega)}$ of a discrete-time sequence $\mathrm{x(n)}$ represents the frequency content of the sequence $\mathrm{x(n)}$. Therefore, by taking the Fourier transform of the discrete-time sequence, the sequence is decomposed into its frequency components. For this reason, the DTFT $\mathrm{X(\omega)}$ is also called the signal spectrum.

Condition for Existence of Discrete-Time Fourier Transform

The Fourier transform of a discrete-time sequence $\mathrm{x(n)}$ exists if and only if the sequence $\mathrm{x(n)}$ is absolutely summable, i.e.,

$$\mathrm{\sum_{n=-\infty}^{\infty}\:|x(n)|\:\lt\:\infty}$$

The discrete-time Fourier transform (DTFT) of the exponentially growing sequences do not exist, because they are not absolutely summable.

Also, the DTFT method of analyzing systems can only be applied to asymptotically stable systems. It cannot be applied to unstable systems. In other words, the DTFT can only be used to analyze systems whose transfer function has poles inside the unit circle.

Numerical Examples

Example 1

Find the discrete-time Fourier transform of the sequence

$$\mathrm{x(n) \:=\: u(n)}$$

Solution

The given discrete-time sequence is,

$$\mathrm{x(n) \:=\: u(n) \:=\: \begin{cases} 1, & n \:\geq\: 0 \\\\ 0, & n \:\lt\: 0 \end{cases}}$$

Now, from the definition of DTFT, we have,

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

$$\mathrm{\therefore\:F[u(n)] \:=\:\sum_{n=-\infty}^{\infty}\:u(n)e^{-j\omega n}\:=\:\sum_{n=0}^{\infty}\:(1) e^{-j \omega n}}$$

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

Example 2

Find the DTFT of the sequence

$$\mathrm{x(n) \:=\: u(n\:-\:k)}$$

Solution

The given discrete-time sequence is,

$$\mathrm{x(n) \:=\: u(n\:-\:k) \:=\: \begin{cases} 1, & n \:\geq\: k \\\\ 0, & n\: \lt\: k \end{cases}}$$

Now, from the definition of DTFT, we have,

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

$$\mathrm{\therefore\:F[u(n\:-\:k)] \:=\: \sum_{n=-\infty}^{\infty}\: u(n\:-\:k) e^{-j \omega n}\:=\:\sum_{n=k}^{\infty}\:(1)e^{-j\omega n}}$$

$$\mathrm{\Rightarrow\: F[u(n\:-\:k)] \:=\: e^{-j\omega k} \:+\: e^{-j\omega (k+1)} \:+\: e^{-j\omega (k+2)} \:+\: \dots}$$

$$\mathrm{\Rightarrow\: F[u(n\:-\:k)] \:=\: e^{-j\omega k} \left( 1 \:+\: e^{-j\omega} \:+\: e^{-j2\omega} \:+\: e^{-j3\omega} \:+\: \dots \right)}$$

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

Example 3

Find the DTFT of the sequence

$$\mathrm{x(n) \:=\: \delta(n\:-\:k)}$$

Solution

The given discrete-time sequence is,

$$\mathrm{x(n) \:=\: \delta(n\:-\:k) \:=\: \begin{cases} 1, & n\: =\: k \\\\ 0, & n \:\neq\: k \end{cases}}$$

Hence, from the definition of the discrete-time Fourier transform, we have,

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

$$\mathrm{\therefore\:F[\delta(n\:-\:k)] \:=\: \sum_{n=-\infty}^{\infty}\: \delta(n\:-\:k) e^{-j \omega n}\:=\:[e^{-j\omega n}]_{n=k}}$$

$$\mathrm{\Rightarrow\:F[\delta(n\:-\:k)] \:=\: e^{-j \omega k}}$$

Example 4

Find the discrete-time Fourier transform

$$\mathrm{x(n) \:=\: \{1,\: 3,\: -2,\: 5,\: 2\}}$$

Solution

The given discrete-time sequence is,

$$\mathrm{x(n) \:=\: \{1,\: 3,\: -2,\: 5,\: 2\}}$$

The DTFT of a sequence is defined as:

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

$$\mathrm{\Rightarrow\:X(\omega) \:=\: x(0) \:+\: x(1) e^{-j \omega} \:+\: x(2) e^{-j 2 \omega} \:+\: x(3) e^{-j 3 \omega} \:+\: x(4) e^{-j 4 \omega}}$$

$$\mathrm{\therefore\:X(\omega) \:=\: 1 \:+\: 3 e^{-j \omega} \:-\: 2 e^{-j 2 \omega} \:+\: 5 e^{-j 3 \omega} \:+\: 2 e^{-j 4 \omega}}$$