This post gives an extremely terse and brief overview of Fourier analysis (Fourier series and transform). Full derivations are given but other supplementary material should be used in addition to this post.

Fourier Series

Fourier’s theorem

Statement

A function is periodic with period \(2\pi\) if

\[f(x) = f(x + 2\pi).\]

\(f(x)\) need only be given on the half-open interval \(x \in [0, 2\pi)\) for it to be specified everywhere.

Fourier’s theorem states that any such \(f(x)\) which is well behaved can be written as one of the three following forms:

\[\fcolorbox{black}{lavender}{ $ \begin{aligned} f(x) &= \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left[ a_n \mathrm{cos}(nx) + b_n \mathrm{sin}(nx) \right]\\ f(x) &= \frac{a_0}{2} + \sum_{n = 1}^{\infty} A_n \mathrm{cos}(nx + \phi_n)\\ f(x) &= \sum_{n = -\infty }^{\infty} c_n e^{inx} \end{aligned} $ }\]

The different sets of coefficients \((a_n, b_n), \ (A_n, \phi_n), \ c_n\) can be converted between easily via trigonometric relations.

Formulae for the coefficients

We are going to make use of the integrals

\[\begin{aligned} I_{n,m} &= \int_0^{2\pi } \mathrm{cos} (nx) \ \mathrm{cos} (mx) \ dx = \delta_{n,m}\pi \\ J_{n,m} &= \int_0^{2\pi } \mathrm{cos} (nx) \ \mathrm{sin} (mx) \ dx = 0 \\ K_{n,m} &= \int_0^{2\pi } \mathrm{sin} (nx) \ \mathrm{sin} (mx) \ dx = \delta_{n,m}\pi. \end{aligned}\]

If we integrate \(f(x)\) over its range, \(\int_0^{2\pi} f(x) \ dx\), we see that since the average value of \(\mathrm{sin}(nx)\) and \(\mathrm{cos}(nx)\) is zero. The integral reduces to \(\int_0^{2\pi} \frac{a_0}{2} \ dx\), so we find

\[a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \ dx.\]

Note: we can integrate over \(2\pi\) at any point in the function’s range, so this is perfectly equivalent to

\[\frac{a_0}{2} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \ dx.\]

In other words, \(\frac{a_0}{2}\) is the average value of \(f(x)\).

For the \(a_n\) coefficients, multiply \(f(x)\) by \(\cos (mx)\) and integrate over the range:

\[\begin{aligned} \int_0^{2\pi} \cos(mx) f(x) \ dx &= \int_0^{2\pi} \sum_{n=1}^{\infty} a_n \cos(nx) \cos(mx) \ dx \\ &= \sum_{n=1}^{\infty}\int_0^{2\pi} a_n \cos(nx) \cos(mx) \ dx \\ &= a_m \int_0^{2\pi} \cos^2(mx) \ dx = \pi a_m, \end{aligned}\]

finally, relabelling \(m\) to \(n\) gives

\[\fcolorbox{black}{lavender}{$ \begin{aligned} a_n = \frac{1}{\pi} \int_0^{2\pi} \cos(nx) f(x) \ dx. \end{aligned} $}\]

Exactly the same argument holds for the \(b_n\), so similarly

\[\fcolorbox{black}{lavender}{$ \begin{aligned} b_n = \frac{1}{\pi} \int_0^{2\pi} \sin(nx) f(x) \ dx. \end{aligned} $ }\]

Examples

Square wave example

Given the function

\[f(x)=\begin{cases} -1 & -\pi \leq x < 0\\ \phantom{-} 1 & \phantom{-} 0 \leq x < \pi \end{cases}\]

We see that since it is odd, \(a_n\) must be zero, and we find the \(b_n\) to be \(b_n = \frac{2}{n\pi} \left( 1 - (-1)^n \right)\)

So the Fourier series representation of \(f(x)\) is

\[f(x) = \sum_{n=1}^{\infty} \frac{2}{n\pi} \left( 1 - (-1)^n \right) \sin(nx).\]

Triangle wave example

Given the function

\[f(x)=\begin{cases} -x & -\pi \leq x < 0\\ \phantom{-} x & \phantom{-} 0 \leq x < \pi \end{cases}\]

We see that since it is even, \(b_n\) must be zero, and we find the \(a_n\) to be \(a_n = \frac{2}{n^2 \pi} \left( (-1)^n - 1 \right)\)

So the Fourier series representation of \(f(x)\) is

\[f(x) = \sum_{n=1}^{\infty} \frac{2}{n^2 \pi} \left( (-1)^n - 1 \right) \cos(nx) .\]

Extension to other periods

Suppose \(f(x) = f(x + T),\) that is \(f(x)\) is periodic with some arbitrary period. Then we would write

\[f(x) = \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left[ a_n \mathrm{cos}\left(\frac{2\pi nx}{T} \right) + b_n \mathrm{sin}\left(\frac{2\pi nx}{T} \right) \right]\]

And the formulae for the coefficients would become

\[\begin{aligned} a_0 &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \ dx \\ a_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos \left(\frac{2\pi nx}{T} \right) f(x) \ dx \\ b_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \sin \left(\frac{2\pi nx}{T} \right) f(x) \ dx. \\ \end{aligned}\]

This can be simplified by rewriting in terms of angular frequency \(T = \frac{2\pi}{\omega}\)

giving

\[\fcolorbox{black}{lavender}{ $ \begin{aligned} f(x) &= \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left[ a_n \mathrm{cos}(n \omega x) + b_n \mathrm{sin}(n \omega x) \right]\\ a_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos \left(n \omega x \right) f(x) \ dx \\ b_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \sin \left(n \omega x \right) f(x) \ dx. \\ \end{aligned}$ }\]

Fourier Transform

Derivation from Fourier Series

Earlier, we noted that the Fourier series can be written

\[f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx}.\]

Using the fact that \(\int_{-\pi}^{\pi} e^{-imx} e^{inx} \ dx = 2\pi \delta_{n,m}\) and integrating over the series representation, we extract the coefficients

\[c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-inx} f(x) \ dx.\]

For a period of \(2\pi L\), the series becomes

\[\begin{aligned} f(x) &= \sum_{n=-\infty}^{\infty} c_n e^{\frac{inx}{L}}\\ c_n &= \frac{1}{2\pi L} \int_{-\pi L}^{\pi L} e^{\frac{-inx}{L}} f(x) \ dx.\\ \end{aligned}\]

Now we define \(\hat{f}\left( \frac{n}{L}\right) = c_n,\) that is, a function that is only defined at the points \(\frac{n}{L}\). The series now looks like

\[\begin{aligned} f(x) &= \sum_{n=-\infty}^{\infty}\hat{f}\left( \frac{n}{L}\right) e^{\frac{inx}{L}}\\ \hat{f}\left( \frac{n}{L}\right) &= \frac{1}{2\pi L} \int_{-\pi L}^{\pi L} e^{\frac{-inx}{L}} f(x) \ dx.\\ \end{aligned}\]

Further, if we define \(\widetilde{f}\left( \frac{n}{L} \right) = \sqrt{2\pi} L \hat{f}\left( \frac{n}{L}\right)\) we get

\[\begin{aligned} f(x) &= \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty}\widetilde{f}\left( \frac{n}{L}\right) e^{\frac{inx}{L}} \frac{1}{L}\\ \widetilde{f}\left( \frac{n}{L} \right) &= \frac{1}{\sqrt{2\pi}} \int_{-\pi L}^{\pi L} e^{\frac{-inx}{L}} f(x) \ dx.\\ \end{aligned}\]

Rewriting \(\frac{n}{L}\) as \(k\) and taking \(L \rightarrow \infty\) finally gives us

\[\fcolorbox{black}{lavender}{ $ \begin{aligned} f(x) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ikx}\widetilde{f}(k) \ dx\\ \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} f(x) \ dx.\\ \end{aligned}$ }\]

Examples

Gaussian function

Suppose we have \(f(x) = e^{-\alpha x^2}.\) The Fourier transform of this is

\[\begin{aligned} \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} e^{-\alpha x^2} \ dx\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\alpha \left( x + \frac{ik}{2\alpha} \right)^2 - \frac{k^2}{4\alpha}} \ dx.\\ \end{aligned}\]

Shifting this by the substitution \(y = x + \frac{ik}{2\alpha}\) gives

\[\begin{aligned} \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} e^{-\frac{k^2}{4\alpha}} \int_{-\infty}^{\infty} e^{-\alpha y^2} \ dy\\ &= \frac{1}{\sqrt{2\pi}} e^{-\frac{k^2}{4\alpha}} \sqrt{\frac{\pi}{\alpha}}\\ &= \frac{1}{\sqrt{2\alpha}} e^{-\frac{k^2}{4\alpha}}. \end{aligned}\]

Another Gaussian, but stretched.

Square pulse

Given the function

\[f(x)= \begin{cases} \frac{1}{A} & -A \leq x \leq A\\ 0 & \phantom{--..} |x| > A\\ \end{cases}\]

its Fourier transform is

\[\begin{aligned} \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-A}^{A} \frac{1}{A} e^{-ikx} \ dx \\ &= \frac{1}{A\sqrt{2\pi}} \left[ \frac{e^{ikA} - e^{-ikA}}{ik} \right]\\ &= \sqrt{\frac{2}{\pi}} \frac{\sin(kA)}{kA}\\ &= \sqrt{\frac{2}{\pi}} \mathrm{sinc}(kA)\\ \end{aligned}\]

Convolutions

Definition

A convolution is an integral transform that takes two functions and composes them, forming a new function. The definition is

\[(f \star g)(x) = \int_{-\infty}^{\infty} f(x')g(x-x') \ dx'.\]

Fourier Transform of a Convolution

The Fourier transform of the convolution of two functions is \(\sqrt{2\pi}\) times the product of the Fourier transforms.

\[\begin{aligned} \mathcal{F}\{f \star g\} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} \left[ \int_{-\infty}^{\infty} f(x') g(x-x') \ dx' \right] dx\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x') \int_{-\infty}^{\infty} g(x-x') e^{-ikx} \ dx' dx\\ \end{aligned}\]

Setting \(y = x - x'\) gives

\[\begin{aligned} \mathcal{F}\{f \star g\} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x') \ dx' \int_{-\infty}^{\infty} e^{-ik(x'+y)} g(y) \ dy\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx'} f(x') \ dx' \int_{-\infty}^{\infty} e^{-iky} g(y) \ dy\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx'} f(x') \ dx' \int_{-\infty}^{\infty} e^{-iky} g(y) \ dy\\ &= \sqrt{2\pi} \widetilde{f}(k) \widetilde{g}(k). \end{aligned}\]

References

Riley, Hobson, Bence - Mathematical Methods for Physics and Engineering

Stroud - Engineering Mathematics