# Fourier analysis of the SQUARE WAVE

## The Problem

Let f(x) = {1 if -p<x<0, 0 if 0<x<p, and periodic of period 2p}. This
function is a square wave; a plot shows the value 1 from x=p to
x = 0 followed by the value 0 from x = 0 to x=p and the shape
repeats in each interval of length 2p. The problem is to find
the coefficients for the Fourier expansion of the square wave.

## The Solution

The Fourier coefficients are found by integration:

a_{0} = 1,

a_{n}>0 = 0, cosine terms are absent,

b_{n} = {0 if n is even and -2/np if n is odd} (only
odd sine terms are present).

The Fourier expansion, through n=2m+1 is

F(m,x) = 1/2 - 2/p[sin(x) + 1/3 sin(3x) + 1/5 sin(5x) + ... + 1/(2m+1) sin((2m+1)x)].

The following plot illustrates the expansion.

Fourier expansion of the square wave. The function f(x) is defined to be f(-p<x<0) = 0 and f(0<x<p)=0 and
it is periodic f(x+2p) = f(x). Two Fourier expansions are
shown. The two-term expansion, F(1,x) = 1/2 - 0.637 sin(x) is a pure fundamental of
frequency 2p shifted upward by the constant 1/2. A much closer
approximation is the 7-term series, 1/2 -
0.637sin(x)-0.212sin(3x)-0.127sin(5x)-0.091sin(7x)

-0.071sin(9x)-0.058sin(11x), containing 5 odd harmonics with diminishing amplitudes.