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# Convergence of Fourier Series

Module by: Michael Haag, Justin Romberg. E-mail the authors

Summary: This module discusses the covergence of the Fourier Series to show that it can be a very good approximation for all signals.

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## Introduction

Before looking at this module, hopefully you have become fully convinced of the fact that any periodic function, ft f t , can be represented as a sum of complex sinusoids. If you are not, then try looking back at eigen-stuff in a nutshell or eigenfunctions of LTI systems. We have shown that we can represent a signal as the sum of exponentials through the Fourier Series equations below:

ft=n c n ei ω 0 nt f t n c n ω 0 n t
(1)
c n =1T0Tfte(i ω 0 nt)d t c n 1 T t T 0 f t ω 0 n t
(2)
Joseph Fourier insisted that these equations were true, but could not prove it. Lagrange publicly ridiculed Fourier, and said that only continuous functions can be represented by Equation 1 (indeed he proved that Equation 1 holds for continuous-time functions). However, we know now that the real truth lies in between Fourier and Lagrange's positions.

## Understanding the Truth

Formulating our question mathematically, let f N t= n =NN c n ei ω 0 nt f N t n N N c n ω 0 n t where c n c n equals the Fourier coefficients of ft f t (see Equation 2).

f N t f N t is a "partial reconstruction" of ft f t using the first 2N+1 2 N 1 Fourier coefficients. f N t f N t approximates ft f t , with the approximation getting better and better as NN gets large. Therefore, we can think of the set N ,N=01:d f N td N N 0 1 f N t as a sequence of functions, each one approximating ft f t better than the one before.

The question is, does this sequence converge to ft f t ? Does f N tft f N t f t as N N ? We will try to answer this question by thinking about convergence in two different ways:

1. Looking at the energy of the error signal: e N t=ft f N t e N t f t f N t
2. Looking at limit   N d f N td N f N t at each point and comparing to ft f t .

### Approach #1

Let e N t e N t be the difference (i.e. error) between the signal ft f t and its partial reconstruction f N t f N t

e N t=ft f N t e N t f t f N t
(3)
If ft L 2 0 T f t L 2 0 T (finite energy), then the energy of e N t0 e N t 0 as N N is
0T| e N t|2d t =0Tft f N t2d t 0 t T 0 e N t 2 t T 0 f t f N t 2 0
(4)
We can prove this equation using Parseval's relation: limit   N 0T|ft f N t|2d t =limit   N N =| n ft n d f N td|2=limit   N |n|>N| c n |2=0 N t T 0 f t f N t 2 N N n f t n f N t 2 N n n N c n 2 0 where the last equation before zero is the tail sum of the Fourier Series, which approaches zero because ft L 2 0 T f t L 2 0 T . Since physical systems respond to energy, the Fourier Series provides an adequate representation for all ft L 2 0 T f t L 2 0 T equaling finite energy over one period.

### Approach #2

The fact that e N 0 e N 0 says nothing about ft f t and limit   N d f N td N f N t being equal at a given point. Take the two functions graphed below for example:

Given these two functions, ft f t and gt g t , then we can see that for all t t, ftgt f t g t , but 0T|ftgt|2d t =0 t T 0 f t g t 2 0 From this we can see the following relationships: energy convergencepointwise convergence energy convergence pointwise convergence pointwise convergence convergence in L 2 0 T pointwise convergence convergence in L 2 0 T However, the reverse of the above statement does not hold true.

It turns out that if ft f t has a discontinuity (as can be seen in figure of gt g t above) at t 0 t 0 , then f t 0 limit   N d f N t 0 d f t 0 N f N t 0 But as long as ft f t meets some other fairly mild conditions, then f t =limit   N d f N t d f t N f N t if ft f t is continuous at t= t t t .

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