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).

The general form of a DFS is:

which are harmonics of a fundamental frequency ${\displaystyle 1/N,}$ for some positive integer ${\displaystyle N.}$ The practical range of ${\displaystyle k,}$ is ${\displaystyle [0,N-1],}$ because periodicity causes larger values to be redundant. When the ${\displaystyle S[k]}$ coefficients are derived from an ${\displaystyle N}$-length DFT, and a factor of ${\displaystyle 1/N}$ is inserted, this becomes an inverse DFT.^{[1]: p.542 (eq 8.4)  [2]: p.77 (eq 4.24)} And in that case, just the coefficients themselves are sometimes referred to as a discrete Fourier series.^{[3]: p.85 (eq 15a)}

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

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

Because of periodicity, ${\displaystyle s_{{}_N}}$can be represented as a DFS with ${\displaystyle N}$ unique coefficients that can be extracted by an ${\displaystyle N}$-length DFT.^{[1]: p 543 (eq 8.9) : pp 557-558    [2]: p 72 (eq 4.11)}

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

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

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

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

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

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

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

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