Multiplication Property of Continuous Time Fourier Transform

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,

$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty }X_{\mathrm{1}}\left ( p \right )\left [\int_{-\infty }^{\infty} x_{\mathrm{2}}\left ( t \right )e^{jpt}e^{-j\omega t}dt \right ]dp}}$$

$$\mathrm{\mathit{\therefore F\left [ x_{\mathrm{1}}\left ( t \right )\cdot x_{\mathrm{2}}\left ( t \right ) \right ]\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi}\left [ X_{\mathrm{1}}\left ( \omega \right )\ast X_{\mathrm{2}}\left ( \omega \right ) \right ]}}$$

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,

$$\mathrm{\mathit{X\left ( \omega \right )\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty}\frac{\mathrm{4}\sin \mathrm{2}p}{\mathrm{2}p}\left [ \pi \delta \left ( \omega-\mathrm{2}\pi -p \right ) \mathrm{\mathrm{\mathrm{+}}} \pi \delta \left ( \omega\mathrm{\mathrm{\mathrm{+}}}\mathrm{2}\pi -p \right )\right ]dp}}$$

$$\mathrm{\Rightarrow \mathit{X\left ( \omega \right )\mathrm{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty}\frac{\mathrm{4}\sin \mathrm{2}p}{\mathrm{2}p}\: \pi \delta \left ( \omega-\mathrm{2}\pi -p \right )dp\mathrm{\mathrm{\mathrm{+}}}\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{-\infty }^{\infty}\frac{\mathrm{4}\sin \mathrm{2}p}{\mathrm{2}p}\: \pi \delta \left ( \omega\mathrm{\mathrm{+}}\mathrm{2}\pi -p \right )dp }}$$

$$\mathrm{\Rightarrow \mathit{X\left ( \omega \right )\mathrm{\mathrm{=}} \frac{\mathrm{2}\sin \mathrm{2}\left ( \omega -\mathrm{2}\pi \right )}{\mathrm{2}\left ( \omega -\mathrm{2}\pi \right )}\mathrm{\mathrm{\mathrm{+}}}\frac{\mathrm{2}\sin \mathrm{2}\left ( \omega \mathrm{\mathrm{+}}\mathrm{2}\pi \right )}{\mathrm{2}\left ( \omega \mathrm{\mathrm{+}}\mathrm{2}\pi \right)}}}$$

$$\mathrm{\therefore \mathit{F\left [ x\left ( t \right ) \right ]\mathrm{\mathrm{=}}X\left ( \omega \right )\mathrm{\mathrm{=}}\mathrm{2}\sin c\left [ \mathrm{2}\left ( \omega -\mathrm{2}\pi \right ) \right ]\mathrm{\mathrm{+}}\mathrm{2}\sin c\left [ \mathrm{2}\left ( \omega \mathrm{\mathrm{+}}\mathrm{2}\pi \right ) \right ] }}$$

Updated on 17-Dec-2021 07:20:33

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