Fourier Series

Definition

Let $f: \mathbb{R} \rightarrow \mathbb{C}$ be a reasonably well behaved $L$ -periodic function (i.e. $f(x) = f(x + L)$ for all $x$ ). Then $f$ can be expressed as an infinite series

$f(x) = \sum_{n \in \mathbb{Z}} A_n e^{2 \pi i n x / L}$

The values of the coefficients $A_n$ can be found using the orthogonality of basic exponential sinusoids. For any $m \in \mathbb{Z}$ ,

$\begin{align*} \int_{-L/2}^{L/2} f(x) e^{-2 \pi i m x / L} dx &= \int_{-L/2}^{L/2} \sum_{n \in \mathbb{Z}} A_n e^{2 \pi i n x / L} e^{-2 \pi i m x / L} dx \\ &= \sum_{n \in \mathbb{Z}} A_n \int_{-L/2}^{L/2} e^{2 \pi i (n - m) x / L} dx \\ &= \sum_{n \in \mathbb{Z}} A_n L \delta_{nm} \\ &= L A_m \end{align*}$

where the second to last line uses Kronecker delta notation. Thus by the periodicity of $f$ and the exponential function

$\begin{align*} A_n &= \frac{1}{L} \int_{-L/2}^{L/2} f(x) e^{-2 \pi i n x / L} dx \\ &= \frac{1}{L} \int_{0}^{L} f(x) e^{-2 \pi i n x / L} dx \end{align*}$

Relation to the Fourier Transform

Now suppose we have a function $f: \mathbb{R} \rightarrow \mathbb{C}$ , and we construct an $L$ -periodic version

$f_L(x) = \sum_{n \in \mathbb{Z}} f(x + nL)$

The coefficients of the Fourier Series of $f_L$ are

$\begin{align*} L A_n &= \int_0^L \sum_{m \in \mathbb{Z}} f(x + mL) e^{-2 \pi i n x / L} dx \\ &= \sum_{m \in \mathbb{Z}} \int_0^L f(x + mL) e^{-2 \pi i n x / L} dx \\ &= \sum_{m \in \mathbb{Z}} \int_{mL}^{(m + 1)L} f(x) e^{-2 \pi i n (x - mL) / L} dx \\ &= e^{-2 \pi i n m} \sum_{m \in \mathbb{Z}} \int_{mL}^{(m + 1)L} f(x) e^{-2 \pi i x n / L} dx \\ &= \int_\mathbb{R} f(x) e^{-2 \pi i x n / L} dx \\ &= \hat{f}(n/L) \end{align*}$

$f_L(x) = \frac{1}{L} \sum_{n \in \mathbb{Z}} \hat{f}(n / L) e^{2 \pi i n x / L}$

Thus the periodic summation of $f$ is completely determined by discrete samples of $\hat{f}$ . This is remarkable in that an uncountable set of numbers (all the values taken by $f_L$ over one period) can be determined by a countable one (the samples of $\hat{f}$ ). Even more incredible, if $f$ has finite bandwidth then only a finite number of the samples will be nonzero. So the uncountable set of numbers is determined by a finite one.

As an aside, by taking $x = 0$ we can derive the Poisson summation formula:

$\sum_{n \in \mathbb{Z}} f(nL) = \frac{1}{L} \sum_{n \in \mathbb{Z}} \hat{f}(n / L)$

Derivation of the Discrete Time Fourier Transform

We can apply the above results to $\hat{f}$ as well. Recall that the Fourier transform of $\hat{f}$ is $f(-x)$ . Thus the periodic summation of $\hat{f}$ is

$\begin{align*} \hat{f}_L(x) &= \frac{1}{L} \sum_{n \in \mathbb{Z}} f(-n / L) e^{2 \pi i n x / L} \\ &= \frac{1}{L} \sum_{n \in \mathbb{Z}} f(n / L) e^{-2 \pi i n x / L} \end{align*}$

This is precisely the definition of the discrete time Fourier transform.

If $f$ is time-limited, then we’ll have only a finite number of nonzero samples. But then $\hat{f}$ is necessarily not bandwidth limited, so the tails of $\hat{f}$ will overlap in the periodic summation. On the other hand, if $f$ is bandwidth limited, for sufficiently large $L$ we can recover $\hat{f}$ . To do so perfectly requires an infinite number of samples, but in practice reasonably bandwidth limited signals can still be recovered quite well from a finite number of samples.

Interpretation of the Discrete Fourier Transform

Forward DFT

Suppose we take samples of a function $f$ at integer multiples of a time $T$ (or distance, etc.). As we saw above,

$\hat{f}_{1/T}(x) = T \sum_{n \in \mathbb{Z}} f(n T) e^{-2 \pi i n x T}$

So when $f$ is time limited such that only the $N$ samples $f(0)$ , $\ldots$ , $f((N - 1) T)$ are nonzero,

$\hat{f}_{1/T}(x) = T \sum_{n = 0}^{N - 1} f(n T) e^{-2 \pi i n x T}$

In this case the discrete Fourier transform gives us $N$ evenly spaced samples from one period of $\hat{f}_{1/T}$ . Namely, for $0 \leq k < N$ ,

$\hat{f}_{1/T}(k / NT) = T \sum_{n = 0}^{N - 1} f(n T) e^{-2 \pi i n k / N}$

Backward DFT

Similarly,

$f_{NT}(x) = \frac{1}{NT} \sum_{n \in \mathbb{Z}} \hat{f}(n / NT) e^{2 \pi i n x / (NT)}$

So when $f$ is bandwidth limited such that only the $N$ samples $\hat{f}(0)$ , $\ldots$ , $\hat{f}((N - 1) / NT)$ are nonzero,

$f_{NT}(x) = \frac{1}{NT} \sum_{n = 0}^{N - 1} \hat{f}(n / NT) e^{2 \pi i n x / (NT)}$

And in this case the inverse discrete Fourier transform gives us $N$ evenly spaced samples from one period of $f_{NT}$ . Namely, for $0 \leq k < N$ ,

$f_{NT}(kT) = \frac{1}{NT} \sum_{n = 0}^{N - 1} \hat{f}(n / NT) e^{2 \pi i n k / N}$

Commentary

If $f$ were both time and bandwidth limited as discussed above, then $f_{NT} = f$ and $\hat{f}_{1/T} = \hat{f}$ . So the samples of $\hat{f}$ we get from the forward DFT could be fed into the backward DFT to exactly recover our original samples of $f$ . Unfortunately no such functions exist, but in practice it works pretty well for signals that strike a balance between time and bandwidth limits. The penalty for the imprecision is some aliasing caused by overlapping tails in the periodic summations. So make sure $1/T$ is large enough to avoid significant aliasing in the frequency domain, and $NT$ is large enough to avoid significant aliasing in the time (or space, etc. domain).

We also see why the frequency spectra obtained from the DFT is periodic: we’re not getting samples of $\hat{f}$ , but of its periodic summation. If $\hat{f}$ is localized near the origin, it can be more instructive to view half the samples of $\hat{f}_{1/T}$ provided by the DFT as representing positive frequencies, and the other half as negative frequencies.