Discrete Fourier series

In digital signal processing, the term Discrete Fourier series (DFS) is any periodic discrete-time signal comprising harmonically-related (i.e. Fourier) discrete real sinusoids or discrete complex exponentials, combined by a weighted summation. A specific example is the inverse discrete Fourier transform (inverse DFT).

Definition

The general form of a DFS is:

Discrete Fourier series

$\begin{array}{r}s[n]=\sum _{k}S[k]\cdot e^{i2\pi \frac{k}{N}n},n\in \mathbb{\mathbb{Z}},\end{array}$

(Eq.1)

which are harmonics of a fundamental frequency $1/N,$ for some positive integer $N.$ The practical range of $k,$ is $[0,N-1],$ because periodicity causes larger values to be redundant. When the $S[k]$ coefficients are derived from an $N$-length DFT, and a factor of $1/N$ is inserted, this becomes an inverse DFT.  And in that case, just the coefficients themselves are sometimes referred to as a discrete Fourier series.

Example

A common practice is to create a sequence of length $N$ from a longer $s[n]$ sequence by partitioning it into $N$-length segments and adding them together, pointwise.(see DTFT § L=N×I) That produces one cycle of the periodic summation:

$s_{{}_N}[n]\triangleq \sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}s[n-mN],n\in \mathbb{\mathbb{Z}}.$

Because of periodicity, $s_{{}_N}$can be represented as a DFS with $N$ unique coefficients that can be extracted by an $N$-length DFT.   

$\begin{array}{rlrl}& s_{{}_N}[n]=\frac{1}{N}\sum _{k=0}^{N-1}{S}_N[k]\cdot e^{i2\pi \frac{k}{N}n};n\in \mathbb{\mathbb{Z}}& & \text{DFS}\\ & S_N[k]\triangleq \sum _{n=0}^{N-1}{s}_{{}_N}[n]\cdot e^{-i2\pi \frac{k}{N}n};k\in [0,N-1]& & \text{DFT}\end{array}$

The coefficients are useful because they are also samples of the discrete-time Fourier transform (DTFT) of the $s[n]$ sequence:

$S_{1/T}(f)\triangleq \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-i2\pi fnT}Ts(nT)=\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}S(f-\frac{m}{T}),f\in \mathbb{ℝ}$

Here, $s(nT)$ represents a sample of a continuous function $s(t),$ with a sampling interval of $T,$ and $S(f)$ is the Fourier transform of $s(t).$ The equality is a result of the Poisson summation formula. With definitions $s[n]\triangleq Ts(nT)$ and $S[k]\triangleq S_{1/T}\left(\frac{k}{NT}\right)$:

$S[k]=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-i2\pi \frac{k}{N}n}s[n];k\in \mathbb{\mathbb{Z}}.$

Due to the $N$-periodicity of the $e^{-i2\pi \frac{k}{N}n}$ kernel, the summation can be "folded" as follows:

$\begin{array}{rl}S[k]& =\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}(\sum _{n=0}^{N-1}{e}^{-i2\pi \frac{k}{N}n}s[n-mN])\\ & =\sum _{n=0}^{N-1}{e}^{-i2\pi \frac{k}{N}n}\underset{s_{{}_N}[n]}{\underbrace{(\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}s[n-mN])}}\\ & =S_N[k].\end{array}$