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Convolution Property of Fourier Transform – Statement, Proof & Examples
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,
$$\mathrm{x(t)=-2e^{-t}u(t)+te^{-t}u(t)+2e^{-2t}u(t)+te^{-2t}u(t)}$$
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