Convolution Property of Fourier Transform – Statement, Proof & Examples

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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,

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

Proof

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

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

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

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

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

By interchanging the order of integration, we get,

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

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

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

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

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

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

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

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

Or, it can also be represented as,

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

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,

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

Therefore,

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

By taking partial fractions, we get,

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

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

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

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

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

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

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

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