Fourier seriesΒΆ
We consider a one-dimensional periodic signal \(f \left( x \right) \in \mathbb{C}\) with \(x \in \left[ 0, L \right)\). The Fourier series expansion of this signal is given by
\[f \left( x \right)
=
\sum_{k = - \infty}^{\infty}
F_k
\twiddle{2 \pi}{k x}{L},\]
where \(k \in \mathbb{Z}\), \(F_k \in \mathbb{C}\), and \(I\) is the imaginary unit \(\sqrt{-1}\). A weighted average of this relation in the given range:
\[\frac{1}{L}
\int_{0}^{L}
f \left( x \right)
\twiddle{- 2 \pi}{k x}{L}
dx\]
yields
\[ \begin{align}\begin{aligned}&
\frac{1}{L}
\int_{0}^{L}
\sum_{l = - \infty}^{\infty}
F_l
\twiddle{2 \pi}{l x}{L}
\twiddle{- 2 \pi}{k x}{L}
dx\\=
&
\frac{1}{L}
\sum_{l = - \infty}^{\infty}
F_l
\int_{0}^{L}
\twiddle{2 \pi}{\left( l - k \right) x}{L}
dx
\,\,
\left(
\text{N.B. integral and summation are interchanged}
\right)\\=
&
\sum_{l = - \infty}^{\infty}
F_l
\delta_{lk}\\=
&
F_k,\end{aligned}\end{align} \]
which is used to find \(F_k\) from \(f \left( x \right)\). Here we utilize the periodicity:
\[\frac{1}{L}
\int_{0}^{L}
\twiddle{2 \pi}{n x}{L}
dx
=
\delta_{n0}\]
with \(n \in \mathbb{Z}\).