# 6.5: Continuous Time Circular Convolution and the CTFS

## Introduction

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

## Signal Circular Convolution

Given a signal $$f(t)$$ with Fourier coefficients $$c_n$$ and a signal $$g(t)$$ with Fourier coefficients $$d_n$$, we can define a new signal, $$v(t)$$, where $$v(t)=f(t) \circledast g(t)$$. We find that the Fourier Series representation of $$v(t)$$, $$a_n$$, is such that $$a_n=c_nd_n$$. $$f(t) \circledast g(t)$$ is the circular convolution (Section 7.5) of two periodic signals and is equivalent to the convolution over one interval, i.e. $$f(t) \circledast g(t)=\int_{0}^{T} \int_{0}^{T} f(\tau) g(t-\tau) d \tau d t$$.

Note

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

This is proved as follows

\begin{align} a_{n} &=\frac{1}{T} \int_{0}^{T} v(t) e^{-\left(j \omega_{0} n t\right)} \mathrm{d} t \nonumber \\ &=\frac{1}{T^{2}} \int_{0}^{T} \int_{0}^{T} f(\tau) g(t-\tau) \mathrm{d} \tau e^{-\left(\omega j_{0} n t\right)} \mathrm{d} t \nonumber \\ &=\frac{1}{T} \int_{0}^{T} f(\tau)\left(\frac{1}{T} \int_{0}^{T} g(t-\tau) e^{-\left(j \omega_{0} n t\right)} \mathrm{d} t\right) \mathrm{d} \tau \nonumber \\ &=\forall \nu, \nu=t-\tau:\left(\frac{1}{T} \int_{0}^{T} f(\tau)\left(\frac{1}{T} \int_{-\tau}^{T-\tau} g(\nu) e^{-\left(j \omega_{0}(\nu+\tau)\right)} \mathrm{d} \nu\right) \mathrm{d} \tau\right) \nonumber \\ &=\frac{1}{T} \int_{0}^{T} f(\tau)\left(\frac{1}{T} \int_{-\tau}^{T-\tau} g(\nu) e^{-\left(j \omega_{0} n \nu\right)} \mathrm{d} \nu\right) e^{-\left(j \omega_{0} n \tau\right)} \mathrm{d} \tau \nonumber \\ &=\frac{1}{T} \int_{0}^{T} f(\tau) d_{n} e^{-\left(j \omega_{0} n \tau\right)} \mathrm{d} \tau \nonumber \\ &=d_{n}\left(\frac{1}{T} \int_{0}^{T} f(\tau) e^{-\left(j \omega_{0} n \tau\right)} \mathrm{d} \tau\right) \nonumber \\ &=c_{n} d_{n} \end{align}

## Exercise

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

For this signal

$c_{n}=\left\{\begin{array}{l} \frac{1}{T} \text { if } n=0 \\ \frac{1}{2} \frac{\sin \left(\frac{\pi}{2} n\right)}{\frac{\pi}{2} n} \text { otherwise } \end{array}\right.$

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 $$a_{n}=c_{n}^{2}=\frac{1}{4} \frac{\sin ^{2}}{\left(\frac{\pi}{2} n\right)}$$.

Exercise $$\PageIndex{1}$$

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

$$a_{n}=\left\{\begin{array}{ll} \text { undefined } & n=0 \\ \frac{1}{8} \frac{\sin ^{3}\left(\frac{\pi}{2} n\right)}{\left(\frac{\pi}{2} n\right)^{3}} & \text { otherwise } \end{array}\right.$$