The Fourier series of a periodic function f(x) with period x = 2π is of the form
f(x) = a0 /2 + Σ (ancos(nx) + bnsin(nx)); n=1,2,3....∞
To calculate the coefficients of the series, one starts with the following assumed identities:
∫f(x)cos(mx)dx = ∫cos(mx) Σ(a0/2 + Σ ancos(nx) + bnsin(nx))dx
∫f(x)sin(mx)dx = ∫sin(mx) Σ(a0/2 + Σ ancos(nx) + bnsin(nx))dx
where one integrates over one base period (m = 1).
Suppressing constants, the following types of integral are to be evaluated, summed over index n:
With m = 1,2,3...∞ and n = 1,2,3...∞: order of the harmonic (fundamental m, n = 1)
cos (mx)
sin (mx)
cos (mx) * (a*cos (nx) + b*sin (nx))
sin (mx) * (a*cos (nx) + b*sin(nx))
All integrals are zero except of those few where the indices are identical: m = n and the function types are the same (sine or cosine). Therefore every sum for a specific index n has only one member and the coefficients can easily be derived from the reduced equations as:
a0 = 2/T∫f(t) dt
an= 2/T∫cos(nx) f(t) dt
bn = 2/T∫sin(nx) f(t) dt
This simulation demonstrates the different types of functions and their integral.