Fourier series

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 $n \in \mathbb{Z}$.