# Fourier Transform of Signum Function

Signals and SystemsElectronics & ElectricalDigital Electronics

## Fourier Transform

The Fourier transform of a continuous-time function $x(t)$ can be defined as,

$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$

## Fourier Transform of Signum Function

The signum function is represented by $sgn(t)$ and is defined as

$$\mathrm{sgn(t)=\begin{cases}1 & for\:t>0\-1 & for\:t<0 \end{cases}}$$

As the signum function is not absolutely integrable. Hence, its Fourier transform cannot be found directly. Therefore, to find the Fourier transform of the signum function, consider the function as given below.

$$\mathrm{x(t)=e^{-a|t|}sgn(t);\:\:a\rightarrow 0}$$

Therefore, the signum function can be obtained as,

$$\mathrm{x(t)=sgn(t)=\lim_{a\rightarrow \:0}e^{-a|t|}sgn(t)}$$

$$\mathrm{\Rightarrow\:x(t)=\lim_{a\rightarrow \:0}[e^{-at}u(t)-e^{at}u(-t)]}$$

From the definition of the Fourier transform, we have,

$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}\:dt=\int_{−\infty}^{\infty}sgn(t)e^{-j\omega t}\:dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}\left(\lim_{a\rightarrow \:0}[e^{-at}u(t)-e^{at}u(-t)] \right)e^{-j\omega t}\:dt}$$

$$\mathrm{\Rightarrow\:X(\omega)=\lim_{a\rightarrow \:0}\left[\int_{−\infty}^{\infty}e^{-at}e^{-j\omega t}u(t)dt- \int_{−\infty}^{\infty}e^{at}e^{-j\omega t}u(-t)dt\right]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\lim_{a\rightarrow \:0}\left[\int_{0}^{\infty}e^{-(a+j\omega)t}dt- \int_{−\infty}^{0}e^{(a-j\omega)t}dt\right]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\lim_{a\rightarrow \:0}\left[\int_{0}^{\infty}e^{-(a+j\omega)t}dt- \int_{0}^{\infty}e^{-(a-j\omega)t}dt\right]}$$

$$\mathrm{\Rightarrow\:X(\omega)=\lim_{a\rightarrow \:0}\left\{\left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} -\left[\frac{e^{-(a-j\omega)t}}{-(a-j\omega)}\right]_{0}^{\infty}\right \}}$$

$$\mathrm{\Rightarrow\:X(\omega)=\lim_{a\rightarrow \:0}\left\{\left[\frac{e^{-\infty}-e^{0}}{-(a+j\omega)} \right]- \left[\frac{e^{-\infty}-e^{0}}{-(a-j\omega)} \right] \right\}}$$

$$\mathrm{=\lim_{a\rightarrow \:0}\left[ \frac{1}{(a+j\omega)}-\frac{1}{(a-j\omega)}\right]}$$

On solving the limits, we get,

$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{j\omega}-\frac{1}{(-j\omega)}=\frac{2}{j\omega}}$$

Therefore, the Fourier transform of the signum function is,

$$\mathrm{X(\omega)=F[sgn(t)]=\frac{2}{j\omega}}$$

Or, it can also be represented as,

$$\mathrm{sgn(t)\overset{FT}{\leftrightarrow}\frac{2}{j\omega}}$$

The magnitude and phase representation of Fourier transform of the Signum function −

$$\mathrm{Magnitude, |X(\omega)| =\sqrt{0+\left(\frac{2}{\omega}\right)^{2}}=\frac{2}{\omega};\:\:for\:all\:\omega}$$

$$\mathrm{Phase,\angle\:X(\omega) =\begin{cases}\frac{\pi}{2}; & for\:\omega<0 \ -\frac{\pi}{2}; & for\:\omega>0 \end{cases}}$$

The graphical representation of the signum function with its magnitude and phase spectra is shown in the figure below.

Updated on 09-Dec-2021 06:50:10