Derivation of Fourier Series

1.The Basic Principle

Fourier series can be used to model different forms of waves. The basic principle can be made clear in the following video:

Here, we only talk about the derivation of the coefficients of Fourier Series.

2.The form of the Fourier Series

The most basic periodic function is the Trigonometric series:
$$
f(t) = A sin(wt+\phi)
$$
If we take a linear superposition of trigonometric series of different magnitudes, frequencies, and initial phases in a certain way, we can obtain a function of any waveform:
$$
f(t) = A_0 + \sum A_n sin(W_nt+\phi_n)
$$
For the convenience of the following derivation, Fourier suggests a given initial frequency \(w_0\), and the another different frequencies can be expressed as integer multiples of \(w_0\):
$$
f(t) = A_0 + \sum A_n sin(nW_0t+\phi_n)
$$
Then we use trigonometric formula to simplify the equation:

$$
f(t) = A_0 + \sum [\overbrace{A_ncos\phi_n}^{Constant} sin(nW_0t) + \overbrace{A_nsin\phi_n}^{Constant} cos(nW_0t)]
$$

Merging the constant terms and rewrite the equation and then we have the standard Fourier series expression as below:
$$
f(t) = A_0 + \sum [a_n sin(nW_0t)+b_n cos(nW_0t)]
$$

3.The Derivation of The Coefficients

3.1The Derivation of \(A_0\)

As is known to all, The integral of a trig function over a period is zero. So obtaining the expression of \(A_0\) becomes simple, we just need to integrate both sides of the equal sign:
\[
\int_0^{T_0}f(t)dt = \int_0^{T_0} A_0 dt + \int_0^{T_0} a_n sin(nW_0t)dt + \int_0^{T_0} b_n cos(nW_0t)dt
\\ =2\pi A_0+0+0
\]
$$
A_0 = \frac{1}{2\pi}\int_0^{T_0}f(t)dt
$$

3.2The Derivation of \(a_n, b_n\)

3.2.1The Orthogonality of Trig Functions

Def: If the integral of the product of any two different functions belonging to the trig function system \( [1. cosx, sinx, cos2x, sin2x, …, cosnx, sinnx] \)over a period is equal to zero, then the trig functions are orthogonal.

How to prove it? Firstly, we can write all 5 situations as below:
\[
\int_0^{T_0} 1\times cosnxdx = 0……n=1,2,3,…
\\ \int_0^{T_0} 1\times sinnxdx = 0……n=1,2,3,…
\\ \int_0^{T_0} sinkx\times cosnxdx = 0……k,n=1,2,3,…
\\ \int_0^{T_0} coskx\times cosnxdx = 0……k,n=1,2,3,…k\neq n
\\ \int_0^{T_0} sinkx\times sinnxdx = 0……k,n=1,2,3,…k\neq n
\]

Take the prove of the 4th equation as an example.

Prove:
$$
\int_0^{T_0} coskx\times cosnxdx = 0……k,n=1,2,3,…k\neq n
$$
①Use Trig formula:
$$
coskx\cdot cosnx = \frac{cos(kx-nx)+cos(kx+nx)}{2}
$$
②Integrate both sides of the equal sign:
\[
\int_0^{T_0}coskx\cdot cosnx dx = \frac{1}{2}(\int_0^{T_0}cos(kx-nx)dx+\int_0^{T_0}cos(kx+nx)dx)
\\ =0+0=0
\]

Q.E.D.

3.2.2Solve for the Coefficients

①Multiply both sides of equal sign by \(coskw_0t\)
$$
f(t) cos(kw_0t) = A_0cos(kw_0t) + \sum [a_n sin(nW_0t)+b_n cos(nW_0t)]cos(kw_0t)
$$
②Integrate both sides of the equal sign:
\[
\int_0^{T_0}f(t) cos(kw_0t) dt = \int_0^{T_0}A_0cos(kw_0t)dt
\\ + \sum [a_n \overbrace{\int_0^{T_0}sin(nW_0tcos(kw_0t)dt}^{=0}
\\ +b_n\overbrace{\int_0^{T_0}cos(nW_0t)cos(kw_0t)dt}^{=0}]
\]
③Use the orthogonality of Trig functions:
$$
\int_0^{T_0}f(t) cos(kw_0t) dt = a_n\int_0^{T_0} cos^2(nw_0t) dt
$$
④Use trigonometric formula to simplify the equation:
$$
\int_0^{T_0}f(t) cos(kw_0t) dt = a_n\int_0^{T_0} \frac{1 + cos^2(2nw_0t)}{2} dt=a_n\frac{T}{2}
$$
So, we can give \(a_n\), and \(b_n\) is the same way.

\[
a_n = \frac{2}{T_0} \int_{0}^{T_0} p(t) cos nw_0t dt …… n=1,2,3,…
\\ b_n = \frac{2}{T_0} \int_{0}^{T_0} p(t) sin nw_0t dt …… n=1,2,3,…
\]