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
Signals and Systems – Multiplication Property of Fourier Transform
For a continuous-time function $\mathit{x(t)}$, the Fourier transform of $\mathit{x(t)}$ can be defined as
$$\mathrm{\mathit{X\left ( \omega \right )\mathrm{\mathrm{=}}\int_{-\infty }^{\infty }x\left ( t \right )e^{-j\omega t}dt}}$$
And the inverse Fourier transform is defined as,
$$\mathrm{\mathit{F^{\mathrm{-1}}\left [ X\left ( \omega \right ) \right ]\mathrm{\mathrm{=}}x\left ( t \right )\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X\left ( \omega \right )e^{j\omega t}d\omega }}$$
Multiplication Property of Fourier Transform
Statement – The multiplication property of continuous-time Fourier transform (CTFT) states that the multiplication of two functions in time domain is equivalent to the convolution of their spectra in the frequency domain. The multiplication property is also called frequency convolution theorem of Fourier transform. Therefore, if
$$\mathrm{\mathit{x_{\mathrm{1}}(t)\overset{FT}{\leftrightarrow}X_{\mathrm{1}}\left ( \omega \right )\: \mathrm{and} \: x_{\mathrm{2}}(t)\overset{FT}{\leftrightarrow}X_{\mathrm{2}}\left ( \omega \right )} }$$
Then, according to the multiplication property,
$$\mathrm{\mathit{x_{\mathrm{1}}(t)\cdot x_{\mathrm{1}}(t)\overset{FT}{\leftrightarrow}\frac{\mathrm{1}}{\mathrm{2}\pi }\left [ X_{\mathrm{1}}\left ( \omega \right )\ast X_{\mathrm{2}}\left ( \omega \right ) \right ]}}$$
Proof
From the definition of Fourier transform, we have,
$$\mathrm{\mathit{F\left [ x\left ( t \right ) \right ]\mathrm{\mathrm{=}}X\left ( \omega \right )\mathrm{\mathrm{=}}\int_{-\infty }^{\infty}x\left ( t \right )e^{-j\omega t}dt}}$$
$$\mathrm{\mathit{\therefore F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\int_{-\infty }^{\infty}\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]e^{-j\omega t}dt}}$$
Now, from the definition of inverse Fourier transform, we have,
$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\int_{-\infty }^{\infty}\left [\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty }X_{\mathrm{1}}\left ( p \right )e^{jpt} dp \right ]x_{\mathrm{2}}\left ( t \right )e^{-j\omega t}dt}}$$
By interchanging the order of integration in RHS of the above expression, we get,
$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty }X_{\mathrm{1}}\left ( p \right )\left [\int_{-\infty }^{\infty} x_{\mathrm{2}}\left ( t \right )e^{jpt}e^{-j\omega t}dt \right ]dp}}$$
$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty }X_{\mathrm{1}}\left ( p \right )\left [\int_{-\infty }^{\infty} x_{\mathrm{2}}\left ( t \right )e^{-j\left ( \omega -p \right )t}dt \right ]dp}}$$
$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty }X_{\mathrm{1}}\left ( p \right )X_{\mathrm{2}}\left ( \omega -p \right )dp}}$$
$$\mathrm{\mathit{\therefore F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi}\left [ X_{\mathrm{1}}\left ( \omega \right )\ast X_{\mathrm{2}}\left ( \omega \right ) \right ]}}$$
Or, it can also be represented as,
$$\mathrm{\mathit{x_{1}\left ( t \right )\cdot x_{\mathrm{1}}\left ( t \right )\overset{FT}{\leftrightarrow} \frac{\mathrm{1}}{\mathrm{2}\pi}\left [ X_{\mathrm{1}}\left ( \omega \right )\ast X_{\mathrm{2}}\left ( \omega \right ) \right ]}}$$
Also,
$$\mathrm{\mathit{\mathrm{2}\pi\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{1}}\left ( t \right ) \right ]\overset{FT}{\leftrightarrow} \left [ X_{\mathrm{1}}\left ( \omega \right )\ast X_{\mathrm{2}}\left ( \omega \right ) \right ]}}$$
$$\mathrm{\mathit{\Rightarrow \left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{1}}\left ( t \right ) \right ]\overset{FT}{\leftrightarrow} \left [ X_{\mathrm{1}}\left ( f \right )\ast X_{\mathrm{2}}\left ( f \right ) \right ];\; \; \left ( \because f\mathrm{\mathrm{=}}\frac{\omega }{\mathrm{2}\pi } \right )}}$$
Numerical Example
Using Multiplication property of Fourier transform, find the Fourier transform of the function given as,
$$\mathrm{\mathit{x\left ( t \right )\mathrm{\mathrm{=}}\left [ u(t\mathrm{\mathrm{\mathrm{\mathrm{+}}}}\mathrm{2})-u\left ( t-\mathrm{2} \right ) \right ]\cos \mathrm{2}\pi t}} $$
Solution
From the definition of Fourier transform of cosine function, we get,
$$\mathrm{\mathit{F\left[ \cos \mathrm{2}\pi t \right ]\mathrm{\mathrm{=}}\pi \delta \left ( \omega -\mathrm{2}\pi \right )\mathrm{\mathrm{\mathrm{+}}}\pi \delta \left ( \omega \mathrm{\mathrm{\mathrm{+}}}\mathrm{2}\pi \right )}}$$
And by the definition of Fourier transform of unit step function, we have,
$$\mathrm{\mathit{F\left [ u\left ( t\mathrm{\mathrm{\mathrm{+}}}\mathrm{2} \right )-u\left ( t-\mathrm{2} \right ) \right ]\mathrm{\mathrm{=}}\int_{-\mathrm{2}}^{\mathrm{2}}\mathrm{1}\cdot e^{-j\omega t\: }dt\mathrm{\mathrm{=}}\left [ \frac{e^{-j\omega t}}{-j\omega} \right ]_{-\mathrm{2}}^{\mathrm{2}}}} $$
$$\mathrm{\Rightarrow \mathit{F\left [ u\left ( t\mathrm{\mathrm{\mathrm{+}}}\mathrm{2} \right )-u\left ( t-\mathrm{2} \right ) \right ]\mathrm{\mathrm{=}}\left [ \frac{e^{-j\mathrm{2}\omega }-e^{j\mathrm{2}\omega }}{-j\omega } \right ]\mathrm{\mathrm{=}}\left [ \frac{e^{j\mathrm{2}\omega }-e^{-j\mathrm{2}\omega }}{j\omega } \right ]}}$$
$$\mathrm{\Rightarrow \mathit{F\left [ u\left ( t\mathrm{\mathrm{\mathrm{+}}}\mathrm{2} \right )-u\left ( t-\mathrm{2} \right ) \right ]\mathrm{\mathrm{=}}\left [ \frac{\mathrm{2}\left ( e^{j\mathrm{2}\omega }-e^{-j\mathrm{2}\omega } \right )}{\mathrm{2}j\omega } \right ]\mathrm{\mathrm{=}}\frac{\mathrm{4}\: \sin \mathrm{2}\omega }{\mathrm{2}\omega }\mathrm{\mathrm{=}}\mathrm{4}\: \sin c(\mathrm{2}\omega )}}$$
Now, the Fourier transform of the given function is,
$$\mathrm{\mathit{F\left [ x\left ( t \right ) \right ]\mathrm{\mathrm{=}}F\left [ \left [ u\left ( t\mathrm{\mathrm{\mathrm{+}}}\mathrm{2} \right )-u\left ( t-\mathrm{2} \right ) \right ]\cos \mathrm{2}\pi t \right ]}}$$
By using the multiplication property $\mathrm{\mathit{\left [ i.e.,\: \: x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{1}}\left ( t \right )\overset{FT}{\leftrightarrow}\frac{\mathrm{1}}{\mathrm{2}\pi }\left [ X_{\mathrm{1}}\left ( \omega \right )\ast X_{\mathrm{2}}\left ( \omega \right ) \right ] \right ]}}$ of Fourier transform, we have,
$$\mathrm{\mathit{X\left ( \omega \right )\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty}\frac{\mathrm{4}\sin \mathrm{2}p}{\mathrm{2}p}\left [ \pi \delta \left ( \omega-\mathrm{2}\pi -p \right ) \mathrm{\mathrm{\mathrm{+}}} \pi \delta \left ( \omega\mathrm{\mathrm{\mathrm{+}}}\mathrm{2}\pi -p \right )\right ]dp}}$$
$$\mathrm{\Rightarrow \mathit{X\left ( \omega \right )\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty}\frac{\mathrm{4}\sin \mathrm{2}p}{\mathrm{2}p}\: \pi \delta \left ( \omega-\mathrm{2}\pi -p \right )dp\mathrm{\mathrm{\mathrm{+}}}\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty}\frac{\mathrm{4}\sin \mathrm{2}p}{\mathrm{2}p}\: \pi \delta \left ( \omega\mathrm{\mathrm{+}}\mathrm{2}\pi -p \right )dp }}$$
$$\mathrm{\Rightarrow \mathit{X\left ( \omega \right )\mathrm{\mathrm{=}} \frac{\mathrm{2}\sin \mathrm{2}\left ( \omega -\mathrm{2}\pi \right )}{\mathrm{2}\left ( \omega -\mathrm{2}\pi \right )}\mathrm{\mathrm{\mathrm{+}}}\frac{\mathrm{2}\sin \mathrm{2}\left ( \omega \mathrm{\mathrm{+}}\mathrm{2}\pi \right )}{\mathrm{2}\left ( \omega \mathrm{\mathrm{+}}\mathrm{2}\pi \right)}}}$$
$$\mathrm{\therefore \mathit{F\left [ x\left ( t \right ) \right ]\mathrm{\mathrm{=}}X\left ( \omega \right )\mathrm{\mathrm{=}}\mathrm{2}\sin c\left [ \mathrm{2}\left ( \omega -\mathrm{2}\pi \right ) \right ]\mathrm{\mathrm{+}}\mathrm{2}\sin c\left [ \mathrm{2}\left ( \omega \mathrm{\mathrm{+}}\mathrm{2}\pi \right ) \right ] }}$$
6K+ Views
