A Square wave is shown opposite. Because of the discontinuities, f(t)
cannot be expressed as a single function and 3 pieces must be used. These are: f(t) = 1 for T/2 < t < T/4 f(t) = + 1 for T/4 £ t £ T/4 f(t) =  1 for T/4 < t < T/2 

Since the square wave is an even function, then b_{n} = 0 for all n and only a_{n} as given in the equation opposite needs to be calculated. 

Because of the symmetries in the square, it is only necessary to integrate from 0 to T/4. However here the long approach is used. Substituting the expression for f(t) into that for a_{n} produces the equation opposite. 

Carrying out the integrations { òcos nwt = (1/nw) sin nwt } give the expression shown opposite. 

Inserting the limits into the result of the integration produces: 

Using the odd property of the sin (sin(x) =  sin(x)) and wT = 2p then the resulting expression for a_{n} is: 

The Equation for the Square wave when expressed as a Fourier Series is given opposite: 
