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Circular Convolution Property of Fourier Series

Module by: Justin Romberg

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

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=ft⊛gt v t ⊛ f t g t We find that the Fourier Series representation of yt y t , a n a n , is such that a n = c n d n a n c n d n . ft⊛gt ⊛ f t g t is the circular convolution of two periodic signals and is equivalent to the convolution over one interval, i.e. ft⊛gt=∫0T∫0Tfτgt-τdτdt ⊛ 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 =1T∫0Tvtⅇ-ⅈ ω 0 ntdt=1T2∫0T∫0Tfτgt-τdτⅇ-ⅈ ω 0 ntdt=1T∫0Tfτ1T∫0Tgt-τⅇ-ⅈ ω 0 ntdtdτ=∀ν,ν=t-τ:1T∫0Tfτ1T∫-τT-τgνⅇ-ⅈ ω 0 ν+τdνdτ=1T∫0Tfτ1T∫-τT-τgνⅇ-ⅈ ω 0 nνdνⅇ-ⅈ ω 0 nτdτ=1T∫0Tfτdnⅇ-ⅈ ω 0 nτdτ= d n 1T∫0Tfτⅇ-ⅈ ω 0 nτdτ= c n d n a n 1 T t 0 T v t ω 0 n t 1 T 2 t 0 T τ 0 T f τ g t τ ω 0 n t 1 T τ 0 T f τ 1 T t 0 T g t τ ω 0 n t ν ν t τ 1 T τ 0 T f τ 1 T ν τ T τ g ν ω 0 ν τ 1 T τ 0 T f τ 1 T ν τ T τ g ν ω 0 n ν ω 0 n τ 1 T τ 0 T f τ d n ω 0 n τ d n 1 T τ 0 T f τ ω 0 n τ c n d n (1)
Example 1 
Take a look at a square pulse with a period, T 1 =T4 T 1 T 4 :
sqpulse.png
Figure 1
For this signal c n =1Tifn=012sinπ2nπ2notherwise c n 1 T n 0 1 2 2 n 2 n
Problem 1
What signal has Fourier coefficients a n = c n 2=14sin2π2nπ2n2 a n c n 2 1 4 2 n 2 2 n 2 ?
[ Click for Solution 1 ]
Solution 1
exfig.png
Figure 2: A triangle pulse train with a period of T4 T 4 .
[ Hide Solution 1 ]

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