6.5: Continuous Time Circular Convolution and the CTFS

• • Richard Baraniuk et al.
• Victor E. Cameron Professor (Electrical and Computer Engineering) at Rice University

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.

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