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Commit 3e38b018 authored by Erik Strand's avatar Erik Strand
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Add Fourier series notes

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---
title: 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](https://en.wikipedia.org/wiki/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
$$
\begin{align*}
L A_n &= \int_0^L \sum_{m \in \mathbb{Z}} f(x + mL) e^{-2 \pi i n x / L} dx \\
&= \sum_{m \in \mathbb{Z}} \int_0^L f(x + mL) e^{-2 \pi i n x / L} dx \\
&= \sum_{m \in \mathbb{Z}} \int_{mL}^{(m + 1)L} f(x) e^{-2 \pi i n (x - mL) / L} dx \\
&= e^{-2 \pi i n m} \sum_{m \in \mathbb{Z}} \int_{mL}^{(m + 1)L} f(x) e^{-2 \pi i x n / L} dx \\
&= \int_\mathbb{R} f(x) e^{-2 \pi i x n / L} dx \\
&= \hat{f}(n/L)
\end{align*}
$$
so
$$
f_L(x) = \frac{1}{L} \sum_{n \in \mathbb{Z}} \hat{f}(n / L) e^{2 \pi i n x / L}
$$
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.
## Derivation of the Discrete Time Fourier Transform
We can apply the result above to $$\hat{f}$$ as well. Recall that the Fourier transform of
$$\hat{f}$$ is $$f(-x)$$. Thus the periodic summation of the Fourier transform of $$f$$ 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.
## Poisson Resummation
By taking $$x = 0$$ we see that
$$
\sum_{n \in \mathbb{Z}} f(nL) = \frac{1}{L} \sum_{n \in \mathbb{Z}} \hat{f}(n / L)
$$
This is known as the
[Poisson summation formula](https://en.wikipedia.org/wiki/Poisson_summation_formula).
......@@ -2,7 +2,7 @@
title: Fourier Transforms
---
To really make sense of chapter 2 I needed to review the properties of Fourier Transforms. These
To really make sense of chapter 2 I needed to review the properties of Fourier transforms. These
notes are based on my prior knowledge and some helpful websites:
- [Properties of Fourier Transform](http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html)
......@@ -177,7 +177,7 @@ for spectral power analysis).
## Transforms of Gaussians
A Fourier Transform that comes up frequently is that of a Gaussian. It can be calculated by
A Fourier transform that comes up frequently is that of a Gaussian. It can be calculated by
completing a square.
$$
......@@ -214,5 +214,5 @@ $$
\end{align*}
$$
This depends on the variance, which is inverted by the Fourier Transform. So since the power is
This depends on the variance, which is inverted by the Fourier transform. So since the power is
invariant, the normalization cannot in general be conserved.
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