# Convolution Property of Fourier Transform – Statement, Proof & Examples

Signals and SystemsElectronics & ElectricalDigital Electronics

## Fourier Transform

The Fourier transform of a continuous-time function 𝑥(𝑡) can be defined as,

$$\mathrm{X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt}$$

## Convolution Property of Fourier Transform

Statement – The convolution of two signals in time domain is equivalent to the multiplication of their spectra in frequency domain. Therefore, if

$$\mathrm{x_1(t)\overset{FT}{\leftrightarrow}X_1(\omega)\:and\:x_2(t)\overset{FT}{\leftrightarrow}X_2(\omega)}$$

Then, according to time convolution property of Fourier transform,

$$\mathrm{x_1(t)*x_2(t)\overset{FT}{\leftrightarrow}X_1(\omega)*X_2(\omega)}$$

### Proof

The convolution of two continuous time signals 𝑥1(𝑡) and 𝑥2(𝑡) is defined as,

$$\mathrm{x_1(t)*x_2(t)=\int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau}$$

Now, from the definition of Fourier transform, we have,

$$\mathrm{X(\omega)=F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}[x_1(t)*x_2(t)]e^{-j \omega t}dt}$$

$$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}[\int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau]e^{-j \omega t}dt }$$

By interchanging the order of integration, we get,

$$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}x_1(\tau)[\int_{-\infty}^{\infty}x_{2}(t-\tau)e^{-j \omega t}dt]d\tau }$$

By replacing (𝑡 − 𝜏) = 𝑢 in the second integration, we get,

$$\mathrm{t = (u + \tau)\: and\: dt = du}$$

$$\mathrm{\therefore F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}x_1(\tau)[\int_{-\infty}^{\infty}x_{2}(u)e^{-j \omega (u+\tau)}du]d\tau}$$

$$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}x_1(\tau)[\int_{-\infty}^{\infty}x_{2}(u)e^{-j \omega u}du]e^{-j\omega \tau}d\tau}$$

$$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=\int_{-\infty}^{\infty}x_1(\tau)X_2(\omega)e^{-j\omega\tau}d\tau.}$$

$$\mathrm{\Rightarrow F[x_1(t)*x_2(t)]=[\int_{-\infty}^{\infty}x_{1}(\tau)e^{-j\omega \tau}d\tau]X_{2}(\omega)-X_{1}(\omega).X_{2}(\omega)}$$

$$\mathrm{\therefore F[x_1(t)*x_2(t)]=X_1(\omega).X_2(\omega)}$$

Or, it can also be represented as,

$$\mathrm{x_1(t)*x_2(t)\overset{FT}{\leftrightarrow}X_1(\omega).X_2(\omega)}$$

## Numerical Example

Using Fourier transform, find the convolution of the signals given by,

$$\mathrm{x_1(t)=te^{-t}u(t)\:and\:x_2(t)=te^{-2t}u(t)}$$

### Solution

Given

$$\mathrm{x_1(t)=te^{-t}u(t)}$$

The Fourier transform of 𝑥1(𝑡) is,

$$\mathrm{X_1(\omega)=\frac{1}{(1+j\omega)^2}}$$

And

$$\mathrm{x_2(t)=te^{-2t}u(t)}$$

The Fourier transform of 𝑥2(𝑡) is,

$$\mathrm{X_2(\omega)=\frac{1}{(2+j\omega)^2}}$$

Now, according to the convolution property of Fourier transform, we have,

$$\mathrm{x_1(t)*x_2(t)\overset{FT}{\leftrightarrow}X_1(\omega).X_2(\omega)}$$

Therefore,

$$\mathrm{x_1(t)*x_2(t)=F^{-1}[X_1(\omega).X_2(\omega)]=F^{-1}[\frac{1}{(1+j\omega)^2.(2+j\omega)^2}]}$$

By taking partial fractions, we get,

$$\mathrm{X(\omega)=\frac{1}{(1+j\omega)^2.(2+j\omega)^2}}$$

$$\mathrm{=\frac{A}{(1+j\omega)}+\frac{B}{(1+j\omega)^2}+\frac{C}{(2+j\omega)}+\frac{D}{(2+j\omega)^2}}$$

On solving this, we get the values of A, B, C and D as

$$\mathrm{𝐴 = −2;\: 𝐵 = 1;\: 𝐶 = 2; \:𝐷 = 1}$$

$$\mathrm{\therefore X(\omega)=\frac{1}{(1+j\omega)^2.(2+j\omega)^2}}$$

$$\mathrm{=\frac{-2}{(1+j\omega)}+\frac{1}{(1+j\omega)^2}+\frac{2}{(2+j\omega)}+\frac{1}{(2+j\omega)^2}}$$

By taking inverse Fourier transform, we get the convolution of signals 𝑥1(𝑡) and 𝑥2(𝑡) as,

$$\mathrm{x(t)=-2e^{-t}u(t)+te^{-t}u(t)+2e^{-2t}u(t)+te^{-2t}u(t)}$$

Published on 06-Dec-2021 12:19:46