This module relates circular convolution of periodic signals in the time domain to multiplication in the frequency domain.

Given a signal $f\left(t\right)$ with Fourier coefficients ${c}_{n}$ and a signal $g\left(t\right)$ with Fourier coefficients ${d}_{n}$, we can define a new signal, $v\left(t\right)$, where $v\left(t\right)=f\left(t\right)\u229bg\left(t\right)$ We find that the Fourier Series representation of $v\left(t\right)$, ${a}_{n}$, is such that ${a}_{n}={c}_{n}{d}_{n}$. $f\left(t\right)\u229bg\left(t\right)$ is the circular convolution of two periodic signals and is equivalent to the convolution over one interval, i.e. $f\left(t\right)\u229bg\left(t\right)={\int}_{0}^{T}{\int}_{0}^{T}f\left(\tau \right)g\left(t-\tau \right)d\tau dt$.

Note:

Circular convolution in the time domain is equivalent to
multiplication of the Fourier coefficients.
$$\begin{array}{rcl}\hfill {a}_{n}& \hfill =\hfill & \frac{1}{T}{\int}_{0}^{T}v\left(t\right){e}^{-(j{\omega}_{0}nt)}dt\hfill \\ \hfill & \hfill =\hfill & \frac{1}{{T}^{2}}{\int}_{0}^{T}{\int}_{0}^{T}f\left(\tau \right)g\left(t-\tau \right)d\tau {e}^{-(\omega {j}_{0}nt)}dt\hfill \\ \hfill & \hfill =\hfill & \frac{1}{T}{\int}_{0}^{T}f\left(\tau \right)(\frac{1}{T}{\int}_{0}^{T}g\left(t-\tau \right){e}^{-(j{\omega}_{0}nt)}dt)d\tau \hfill \\ \hfill & \hfill =\hfill & \forall \nu ,\nu =t-\tau :\left(\frac{1}{T}{\int}_{0}^{T}f\left(\tau \right)(\frac{1}{T}{\int}_{-\tau}^{T-\tau}g\left(\nu \right){e}^{-(j{\omega}_{0}(\nu +\tau ))}d\nu )d\tau \right)\hfill \\ \hfill & \hfill =\hfill & \frac{1}{T}{\int}_{0}^{T}f\left(\tau \right)(\frac{1}{T}{\int}_{-\tau}^{T-\tau}g\left(\nu \right){e}^{-(j{\omega}_{0}n\nu )}d\nu ){e}^{-(j{\omega}_{0}n\tau )}d\tau \hfill \\ \hfill & \hfill =\hfill & \frac{1}{T}{\int}_{0}^{T}f\left(\tau \right){d}_{n}{e}^{-(j{\omega}_{0}n\tau )}d\tau \hfill \\ \hfill & \hfill =\hfill & {d}_{n}(\frac{1}{T}{\int}_{0}^{T}f\left(\tau \right){e}^{-(j{\omega}_{0}n\tau )}d\tau )\hfill \\ \hfill & \hfill =\hfill & {c}_{n}{d}_{n}\hfill \end{array}$$

1

Take a look at a square pulse with a period of T.

For this signal $${c}_{n}=\{\begin{array}{l}\frac{1}{T}\text{if}n=0\\ \frac{1}{2}\frac{\mathrm{sin}\left(\frac{\pi}{2}n\right)}{\frac{\pi}{2}n}\text{otherwise}\end{array}$$

Take a look at a triangle pulse train with a period of T.

This signal is created by circularly convolving the square pulse with itself. The Fourier coefficients for this signal are ${a}_{n}={{c}_{n}}^{2}=\frac{1}{4}\frac{{\mathrm{sin}}^{2}}{\left(\frac{\pi}{2}n\right)}$

Exercise 1

Find the Fourier coefficients of the signal that is created when the square pulse and the triangle pulse are convolved.

Solution

${a}_{n}=\left\{\begin{array}{cc}\text{undefined}\hfill & n=0\hfill \\ \frac{1}{8}\frac{si{n}^{3}\left(\frac{\pi}{2}n\right)}{{\left(\frac{\pi}{2}n\right)}^{3}}\hfill & \text{otherwise}\hfill \end{array}\right.$

Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients in the frequency domain.