Engineering Analysis/Fourier Series

The L₂ space is an infinite function space, and therefore a linear combination of any infinite set of orthogonal functions can be used to represent any single member of the L₂ space. The decomposition of an L₂ function in terms of an infinite basis set is a technique known as the Fourier Decomposition of the function, and produces a result called the Fourier Series.

Fourier Basis

Let's consider a set of L₂ functions, $\phi$, as follows:

$\phi =\{1,\sin(\pi x),\cos(\pi x),\sin(2\pi x),\cos(2\pi x),\sin(3\pi x),\cos(3\pi x)...\}.\$

We can prove that over a range $[0,2\pi ]$, all of these functions are orthogonal:

$\int _{0}^{2\pi }1\cdot \cos(n\pi x)dx=0$

$\int _{0}^{2\pi }1\cdot \sin(n\pi x)dx=0$

$\int _{0}^{2\pi }\sin(n\pi x)\sin(m\pi x)dx=0,n\neq m$

$\int _{0}^{2\pi }\sin(n\pi x)\cos(m\pi x)dx=0$

$\int _{0}^{2\pi }\cos(n\pi x)\cos(m\pi x)dx=0,n\neq m$

Because $\phi$ is as an infinite orthogonal set in L₂, $\phi$ is also a valid basis set in the L₂ space. Therefore, we can decompose any function in L₂ as the following sum:

[Classical Fourier Series]

$\psi (x)=a_{0}(1)+\sum _{n=1}^{\infty }a_{n}\sin(n\pi x)+\sum _{m=1}^{\infty }b_{m}\cos(m\pi x)$

However, the difficulty occurs when we need to calculate the a and b coefficients. We will show the method to do this below:

a₀: The Constant Term

Calculation of a₀ is the easiest, and therefore we will show how to calculate it first. We use the value of a₀ which minimizes the error in approximating $f(x)$ by the Fourier series.

First, define an error function, E, that is equal to the squared norm of the difference between the function f(x) and the infinite sum above:

$E={\frac {1}{2}}\int _{0}^{2\pi }\|f(x)-a_{0}(1)-\sum _{n=1}^{\infty }a_{n}\sin(n\pi x)-\sum _{m=1}^{\infty }b_{m}\cos(m\pi x)\|^{2}dx$

For ease, we will write all the basis functions as the set φ, described above:

$\sum _{i=0}^{\infty }a_{i}\phi _{i}=a_{0}+\sum _{n=1}^{\infty }a_{n}\sin(n\pi x)+\sum _{m=1}^{\infty }b_{m}\cos(m\pi x)$

Combining the last two functions together, and writing the norm as an integral, we can say:

$E={\frac {1}{2}}\int _{0}^{2\pi }|\sum _{i=0}^{\infty }a_{i}\phi _{i}|^{2}dx$

We attempt to minimize this error function with respect to the constant term. To do this, we differentiate both sides with respect to a₀, and set the result to zero:

$0={\frac {\partial E}{\partial a_{0}}}=\int _{0}^{2\pi }(f(x)-\sum _{i=0}^{\infty }a_{i}\phi _{i}(x))(-\phi _{0}(x))dx$

The φ₀ term comes out of the sum because of the chain rule: it is the only term in the entire sum dependent on a₀. We can separate out the integral above as follows:

$\int _{0}^{2\pi }(f(x)-\sum _{i=0}^{\infty }a_{i}\phi _{i})(-\phi _{0})dx=-\int _{0}^{2\pi }f(x)\phi _{0}(x)dx+a_{0}\int _{0}^{2\pi }\phi _{0}(x)\phi _{0}(x)dx$

All the other terms drop out of the infinite sum because they are all orthogonal to φ₀. Again, we can rewrite the above equation in terms of the scalar product:

$0=-\langle f(x),\phi _{0}(x)\rangle +a_{0}\langle \phi _{0}(x),\phi _{0}(x)\rangle$

And solving for a₀, we get our final result:

$a_{0}={\frac {\langle f(x),\phi _{0}(x)\rangle }{\langle \phi _{0}(x),\phi _{0}(x)\rangle }}$

Sin Coefficients

Using the above method, we can solve for the a_n coefficients of the sin terms:

$a_{n}={\frac {\langle f(x),\sin(n\pi x)\rangle }{\langle \sin(n\pi x),\sin(n\pi x)\rangle }}$

Cos Coefficients

Also using the above method, we can solve for the b_n terms of the cos term.

$b_{n}={\frac {\langle f(x),\cos(n\pi x)\rangle }{\langle \cos(n\pi x),\cos(n\pi x)\rangle }}$