Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Fourier Transform of Unit Step Function
Fourier Transform
For a continuous-time function $x(t)$, the Fourier transform is defined as,
$$\mathrm{X(\omega)=\int_{−\infty }^{\infty}x(t)e^{−j\omega t}\:dt}$$
Fourier Transform of Unit Step Function
The unit step function is defined as,
$$\mathrm{u(t)=\begin{cases}1 & for\:t≥ 0\0 & for\:t< 0\end{cases}}$$
Because the unit step function is not absolutely integrable, thus its Fourier transform cannot be found directly.
In order to find the Fourier transform of the unit step function, express the unit step function in terms of signum function as
$$\mathrm{u(t)=\frac{1}{2}+\frac{1}{2}sgn(t)=\frac{1}{2}[1+sgn(t)]}$$
Given that
$$\mathrm{x(t)=u(t)=\frac{1}{2}[1+sgn(t)]}$$
Now, from the definition of the Fourier transform, we have,
$$\mathrm{F[u(t)]=X(\omega)=\int_{−\infty }^{\infty}x(t)e^{-j\omega t} dt=\int_{−\infty }^{\infty} u(t)e^{-j\omega t} dt}$$
$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty }^{\infty} \frac{1}{2}[1+sgn(t)]e^{-j\omega t}dt}$$
$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2}\left [ \int_{−\infty }^{\infty} 1 \cdot e^{-j\omega t} dt + \int_{−\infty }^{\infty} sgn(t) \cdot e^{-j\omega t} dt\right ]}$$
$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2}\{ F[1]+ F[sgn(t)]\}}$$
The Fourier transform of the constant amplitude and the signum function is given by,
$$\mathrm{F[1]=2\pi\delta(\omega)\:\:and\:\:F[sgn(t)]=\frac{2}{j\omega}}$$
$$\mathrm{\therefore\:F[u(t)]=X(\omega)=\frac{1}{2}\left [2\pi\delta(\omega) + \frac{2}{j\omega}\right ] }$$
Therefore, the Fourier transform of the unit step function is,
$$\mathrm{F[u(t)]=\left (\pi\delta(\omega) + \frac{1}{j\omega}\right )}$$
Or, it can also be represented as,
$$\mathrm{u(t)\overset{FT}{\leftrightarrow}\left ( \pi\delta(\omega) +\frac{1}{j\omega}\right )}$$
Magnitude and phase representation of Fourier transform of the unit step function −
$$\mathrm{Magnitude,|X(\omega)|=\begin{cases}∞ & at \:\omega = 0\0 & at\:\omega= −\infty & \omega= \infty\end{cases}}$$
The graphical representation of the unit step function with its frequency spectrum is shown in the figure.
39K+ Views
