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 Impulse Function, Constant Amplitude and Complex Exponential Function
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 Unit Impulse Function
The unit impulse function is defined as,
$$\mathrm{\delta(t)=\begin{cases}1 & for\:t=0 \0 & for\:t ≠ 0 \end{cases}}$$
If it is given that
$$\mathrm{x(t)=\delta(t)}$$
Then, from the definition of Fourier transform, we have,
$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}\delta(t)e^{-j\omega t}dt}$$
As the impulse function exists only at t= 0. Thus,
$$\mathrm{X(\omega)=\int_{−\infty}^{\infty}\delta(t) e^{-j\omega t}dt=\int_{−\infty}^{\infty}1\cdot e^{-j\omega t}dt=e^{-j\omega t}|_{t=0}=1}$$
$$\mathrm{\therefore\:F[\delta(t)]=1\:\:or\:\:\delta(t) \overset{FT}{\leftrightarrow}1}$$
That is, the Fourier transform of a unit impulse function is unity.
The magnitude and phase representation of the Fourier transform of unit impulse function are as follows −
$$\mathrm{Magnitude,|X(\omega)|=1;\:\:for\:all\:\omega}$$
$$\mathrm{Phase,\angle X(\omega)=0;\:\:for\:all\:\omega}$$
The graphical representation of the impulse function with its magnitude and phase spectra are shown in the figure.
Fourier Transform of Constant Amplitude
If the function is given as
$$\mathrm{x(t)=1}$$
Then, the function $X(t)$ is a constant function and it is not absolutely integrable, hence its Fourier transform cannot be found directly. Therefore, the Fourier transform of $X(t)=1$ is determined through inverse Fourier transform of impulse function $[\delta(\omega)]$.
From the definition of inverse Fourier transform, we have,
$$\mathrm{x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)e^{j\omega t}d\omega}$$
Let
$$\mathrm{X(\omega)=\delta(\omega)}$$
Where,
$$\mathrm{\delta(\omega)=\begin{cases}1 & for\:\omega=0 \0 & for\:\omega ≠ 0\end{cases}}$$
$$\mathrm{\therefore\:x(t)=F^{-1}[X(\omega)]=F^{-1}[\delta(\omega)]}$$
$$\mathrm{\Rightarrow\:x(t)=x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}\delta(\omega)e^{j\omega t}d\omega=\frac{1}{2\pi}\cdot(1)=\frac{1}{2\pi}}$$
$$\mathrm{\therefore\:F^{-1}[\delta(\omega)]=\frac{1}{2\pi}}$$
$$\mathrm{\Rightarrow\:F^{-1}[2\pi\delta(\omega)]=1}$$
Hence, the Fourier transform of a constant function is,
$$\mathrm{F[1]=2\pi\delta(\omega)\:or\:\:1\overset{FT}{\leftrightarrow}2\pi\delta(\omega)}$$
When the amplitude of the constant function is A, then the Fourier transform of the function becomes
$$\mathrm{A\overset{FT}{\leftrightarrow}2\pi A\delta(\omega)}$$
Fourier Transform of Complex Exponential Function
Consider the complex exponential function as,
$$\mathrm{x(t)=e^{j\omega_{0}t}}$$
The Fourier transform of a complex exponential function cannot be found directly. In order to find the Fourier transform of complex exponential function $x(t)$, consider finding the inverse Fourier transform of shifted impulse function in frequency domain $[\delta(\omega-\omega_{0})]$.
Let
$$\mathrm{X(\omega)=\delta(\omega-\omega_{0})}$$
Then, from the definition of inverse Fourier transform, we have,
$$\mathrm{x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)e^{j\omega t}d{\omega}}$$
$$\mathrm{\Rightarrow\:x(t)=F^{-1}[X(\omega)]=F^{-1}[\delta(\omega-\omega_{0})]}$$
$$\mathrm{\Rightarrow\:x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}\delta(\omega-\omega_{0})e^{j\omega t}d{\omega}=\frac{1}{2\pi}e^{j\omega_{0} t}}$$
Therefore, the inverse Fourier transform of $\delta(\omega-\omega_{0})$ is,
$$\mathrm{F^{-1}[\delta(\omega-\omega_{0})]=\frac{1}{2\pi}e^{j\omega_{0} t}}$$
$$\mathrm{\Rightarrow\:F^{-1}[2\pi\delta(\omega-\omega_{0})]=e^{j\omega_{0} t}}$$
Hence, the Fourier transform of the complex exponential function is given by,
$$\mathrm{[e^{j\omega_{0} t}]=2\pi\delta(\omega-\omega_{0})}$$
Or, it can also be represented as,
$$\mathrm{e^{j\omega_{0} t}\overset{FT}{\leftrightarrow}2\pi\delta(\omega-\omega_{0})}$$
16K+ Views
