Time Convolution Theorem

Convolution

The convolution of two signals 𝑥(𝑡) and ℎ(𝑡) is defined as,

$$\mathrm{y\left ( t \right )=x\left( t \right )\ast h\left ( t \right )=\int_{-\infty }^{\infty}x\left ( \tau \right )h\left ( t-\tau \right )d\tau}$$

This integral is also called the convolution integral.

Time Convolution Theorem

Statement – The time convolution theorem states that the convolution in time domain is equivalent to the multiplication of their spectrum in frequency domain. Therefore, if the Fourier transform of two time signals is given as,

$$\mathrm{x_{1}\left ( t \right )\overset{FT}{\leftrightarrow}X_{1} \left ( \omega \right )}$$

And

$$\mathrm{x_{2}\left ( t \right )\overset{FT}{\leftrightarrow}X_{2} \left ( \omega \right )}$$

Then, according to the time convolution theorem,

$$\mathrm{x_{1}\left ( t \right )\ast x_{2}\left ( t \right )\overset{FT}{\leftrightarrow}X_{1} \left ( \omega \right )\cdot X_{2} \left ( \omega \right )}$$

Proof

From the definition of Fourier transform, we have,

$$\mathrm{F\left [ x\left ( t \right ) \right ]=\int_{-\infty }^{\infty }x\left ( t \right )e^{-j\omega t}dt}$$

Therefore,

$$\mathrm{F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]e^{-j\omega t}dt}$$

Also, by the definition of convolution, we have,

$$\mathrm{x_{1}\left ( t \right )\ast x_{2}\left ( t \right )=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right )x_{2}\left ( t-\tau \right )d\tau }$$

$$\mathrm{\therefore F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }\left [ \int_{-\infty }^{\infty }x_{1}\left ( \tau \right )x_{2}\left ( t-\tau \right )d\tau \right ]e^{-j\omega t}dt}$$

By rearranging the order of integrations, we get,

$$\mathrm{F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right )\left [ \int_{-\infty }^{\infty }x_{2}\left ( t-\tau \right )e^{-j\omega t}dt \right ]d\tau}$$

On substituting (𝑡 − 𝜏) = 𝑢, in the second integration, we get, 𝑡 = 𝑢 + 𝜏 and 𝑑𝑡 = 𝑑𝑢

$$\mathrm{\therefore F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right )\left [ \int_{-\infty }^{\infty }x_{2}\left ( u \right )e^{-j\omega \left ( u+\tau \right )}du \right ]d\tau}$$

$$\mathrm{\Rightarrow F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right )\left [ \int_{-\infty }^{\infty }x_{2}\left ( u \right )e^{-j\omega u}du \right ]e^{-j\omega \tau }d\tau}$$

$$\mathrm{\Rightarrow F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right ) X_{2}\left ( \omega \right )e^{-j\omega \tau }d\tau =X_{2}\left ( \omega \right )\int_{-\infty}^{\infty}x_{1}\left ( \tau \right )e^{-j\omega \tau }d\tau }$$

$$\mathrm{\Rightarrow F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=\int_{-\infty }^{\infty }x_{1}\left ( \tau \right ) X_{2}\left ( \omega \right )e^{-j\omega \tau }d\tau =X_{2}\left ( \omega \right )\int_{-\infty}^{\infty}x_{1}\left ( \tau \right )e^{-j\omega \tau }d\tau }$$

$$\mathrm{\Rightarrow F\left [ x_{1}\left ( t \right )\ast x_{2}\left ( t \right ) \right ]=X_{2}\left ( \omega \right )X_{1}\left ( \omega \right )}$$

Therefore, it proves that,

$$\mathrm{x_{1}\left ( t \right )\ast x_{2}\left ( t \right )\overset{FT}{\leftrightarrow}X_{1}\left ( \omega \right )\cdot X_{2}\left ( \omega \right )}$$

The above expression is known as Time Convolution Theorem.