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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.

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 =1T0Tfτ(1TτTτgνe(j ω 0 (ν+τ))d ν )d τ   ,   ν=tτ   =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.

Figure 1
Figure 1 (sqpulse.png)

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.

Figure 2
Figure 2 (exfig.png)

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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