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Frequency Convolution Theorem
Convolution
The convolution of two signals $\mathit{x\left ( t \right )}$ and $\mathit{h\left ( t \right )}$ is defined as,
$$\mathrm{\mathit{y\left(t\right)\mathrm{=}x\left(t\right)*h\left(t\right)\mathrm{=}\int_{-\infty }^{\infty}x\left(\tau\right)\:h\left(t-\tau\right)\:d\tau}}$$
This integral is also called the convolution integral.
Frequency Convolution Theorem
Statement - The frequency convolution theorem states that the multiplication of two signals in time domain is equivalent to the convolution of their spectra in the frequency domain.
Therefore, if the Fourier transform of two signals $\mathit{x_{\mathrm{1}}\left ( t \right )}$ and $\mathit{x_{\mathrm{2}}\left ( t \right )}$ is defined as
$$\mathrm{\mathit{x_{\mathrm{1}}\left(t\right)\overset{FT}{\leftrightarrow} X_{\mathrm{1}}\left(\omega\right)} }$$
And
$$\mathrm{\mathit{x_{\mathrm{2}}\left(t\right)\overset{FT}{\leftrightarrow} X_{\mathrm{2}}\left(\omega\right)}}$$
Then, according to the frequency convolution theorem,
$$\mathrm{\mathit{x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right)\overset{FT}{\leftrightarrow}\frac{\mathrm{1}}{\mathrm{2}\pi }\left [ X_{\mathrm{1}}\left(\omega\right)* X_{\mathrm{2}}\left(\omega\right)\right ]}}$$
Proof
From the definition of Fourier transform, we have,
$$\mathrm{\mathit{F\left [ x\left(t\right) \right ]\mathrm{=}X\left(\omega \right)=\int_{-\infty }^{\infty }x(t)e^{-j\omega t}\:dt}}$$
Therefore,
$$\mathrm{\mathit{F\left [ x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]=\int_{-\infty }^{\infty }\left [ x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]e^{-j\omega t}\:dt}}$$
Now, from the definition of inverse Fourier transform, we get,
$$\mathrm{\mathit{x\left(t\right)=F^{-\mathrm{1}}\left [X\left(\omega\right) \right ]=\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X\left(\omega\right)e^{j\omega t}\:d\omega} }$$
$$\mathrm{\mathit{\therefore F\left [ x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]=\int_{-\infty }^{\infty }\left [ \frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X_{\mathrm{1}}(p) e^{jpt}\:dp\right ]x_{\mathrm{2}}\left(t\right)e^{-j\omega t}\:dt}}$$
By rearranging the order of integration, we get,
$$\mathrm{\mathit{F\left [ x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]= \frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X_{\mathrm{1}}\left(p\right)\left [ \int_{-\infty }^{\infty }x_{\mathrm{2}}\left(t\right)\:e^{-j\omega t}\:e^{jpt}\:dt\right ]dp}}$$
$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]= \frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X_{\mathrm{1}}\left(p\right)\left [ \int_{-\infty }^{\infty }x_{\mathrm{2}}\left(t\right)\:e^{-j\left(\omega -p\right)t}\:dt\right ]dp }}$$
$$\mathrm{\mathit{\Rightarrow F\left [ x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]= \frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\infty }^{\infty }X_{\mathrm{1}}\left(p\right)X_{\mathrm{2}}\left(\omega -p\right)\:dp}}$$
Hence, by the definition of convolution, we can write the above expression as,
$$\mathrm{ \mathit{F\left [ x_{\mathrm{\mathrm{1}}}\left(t\right).x_{\mathrm{2}}\left(t\right) \right ]= \frac{\mathrm{1}}{\mathrm{2}\pi }\left [ X_{\mathrm{1}}\left(\omega\right) *X_{\mathrm{2}}\left(\omega\right)\right ]}}$$
Or, it can also be represented as,
$$\mathrm{\mathit{x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right)\overset{FT}{\leftrightarrow} \frac{\mathrm{1}}{\mathrm{2}\pi }\left [ X_{\mathrm{1}}\left(\omega\right) *X_{\mathrm{2}}\left(\omega\right)\right ]}}$$
This is the frequency convolution theorem in radian frequency. In terms of frequency, we can write it as,
$$\mathrm{\mathit{x_{\mathrm{1}}\left(t\right).x_{\mathrm{2}}\left(t\right)\overset{FT}{\leftrightarrow}X_{\mathrm{1}}\left(f\right)*X_{\mathrm{2}}\left(f\right);\:\:\left ( \because f=\frac{\omega }{\mathrm{2}\pi } \right )}}$$
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