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# Continuous Time Circular Convolution and the CTFS

Module by: Justin Romberg. E-mail the author

Summary: This module looks at the basic circular convolution relationship between two sets of Fourier coefficients.

Note: You are viewing an old version of this document. The latest version is available here.

## 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 ft f t with Fourier coefficients c n c n and a signal gt g t with Fourier coefficients d n d n , we can define a new signal, vt v t , where vt=ftgt v t f t g t We find that the Fourier Series representation of vt v t , a n a n , is such that a n = c n d n a n c n d n . ftgt f t g t is the circular convolution of two periodic signals and is equivalent to the convolution over one interval, i.e. ftgt=0T0Tfτgtτd τ d t f t g t t 0 T τ 0 T f τ g t τ .

### Note:

Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients.
This is proved as follows
a n =1T0Tvte(j ω 0 nt)d t =1T20T0Tfτgtτd τ e(ω j 0 nt)d t =1T0Tfτ(1T0Tgtτe(j ω 0 nt)d t )d τ = ν ,ν=tτ:1T0Tfτ(1TτTτgνe(j ω 0 (ν+τ))d ν )d τ =1T0Tfτ(1TτTτgνe(j ω 0 nν)d ν )e(j ω 0 nτ)d τ =1T0Tfτdne(j ω 0 nτ)d τ = d n (1T0Tfτe(j ω 0 nτ)d τ )= c n d n a n 1 T t 0 T v t j ω 0 n t 1 T 2 t 0 T τ 0 T f τ g t τ ω j 0 n t 1 T τ 0 T f τ 1 T t 0 T g t τ j ω 0 n t ν ν t τ 1 T τ 0 T f τ 1 T ν τ T τ g ν j ω 0 ν τ 1 T τ 0 T f τ 1 T ν τ T τ g ν j ω 0 n ν j ω 0 n τ 1 T τ 0 T f τ d n j ω 0 n τ d n 1 T τ 0 T f τ j ω 0 n τ c n d n
(1)

## Exercise

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

For this signal c n ={1T  if  n=012sinπ2nπ2n  otherwise   c n 1 T n 0 1 2 2 n 2 n

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=14sin2π2nπ2n2 a n c n 2 1 4 2 n 2 2 n 2

### 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 = undefined n = 0 1 8 s i n 3 ( π 2 n ) ( π 2 n ) 3 otherwise a n = undefined n = 0 1 8 s i n 3 ( π 2 n ) ( π 2 n ) 3 otherwise

## Conclusion

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

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