Vectors and Fourier Series: Part 3

In Part 2, we adapted three tools developed for vectors to functions: a Basis in which to represent our function, Projection Operators to find the components of our function, and a Function Rebuilder which allows us to recreate our vector in the new basis. This is the third (and final!) post in a series of three:

• Part 1: Developing tools from vectors
• Part 2: Using these tools for Fourier series
• Part 3: A few examples using these tools

We can apply these tools to two problems that are common in Fourier Series analysis. First we'll look at the square wave and then the sawtooth wave. Since we've chosen a sine and cosine basis (a frequency basis), there are a few questions we can ask ourselves before we begin:

1. Will these two functions contain a finite or infinite number of components?
2. Will the amplitude of the components grow or shrink as a function of their frequency?

Let's try and get an intuitive answer to these questions first.

For 1., another way of asking this question is "could you come up with a way to combine a few sines and cosines to create the function?" The smoking guns here are the corners. Sines and cosines do not have sharp corners and so making a square wave or sawtooth wave with a finite number of them should be impossible.

For 2., one way of thinking about this is that the function we are decomposing are mostly smooth with a few corners. To get them to be smooth, we can't have more and more high frequency components, so the amplitude of the components should shrink.

Let's see if these intuitive answers are borne out.

Square Wave

We'll center the square wave vertically at zero and let it range from $[-L, L]$. In this case, the square wave function is

$f(x)=\begin{cases}1&-L\leq x\leq 0 \\-1&0\leq x\lt L\end{cases}.$

If we imagine this function being repeated periodically outside the range $[-L, L]$, it would be an odd (antisymmetric) function. Since sine functions are odd and cosine functions are even, an arbitrary odd function should only be built out of sums of other odd functions. So, we get to take one shortcut and only look at the projections onto the sine function (the cosine projections will be zero). You should work this out if this explanation wasn't clear.

Since the square wave is defined piecewise, our projection integral will also be piecewise:

$a_n = \text{Proj}_{s_n}(f(x)) \\= \int_{-L}^{0}dx\tfrac{1}{\sqrt{L}} \sin(\tfrac{n\pi x}{L}) (1)+\int_{0}^Ldx\tfrac{1}{\sqrt{L}} \sin(\tfrac{n\pi x}{L})(-1).$

Both of these integrals can be done exactly.

$a_n = \tfrac{1}{\sqrt{L}} \frac{-L}{n \pi}\cos(\tfrac{n\pi x}{L})|_{-L}^0 +\tfrac{1}{\sqrt{L}} \frac{L}{n \pi}\cos(\tfrac{n\pi x}{L})|_0^L \\= -\frac{\sqrt{L}}{n \pi}\cos(0)+\frac{\sqrt{L}}{n \pi}\cos(-n\pi)+\frac{\sqrt{L}}{n \pi}\cos(n\pi)-\frac{\sqrt{L}}{n \pi}\cos(0)\\= \frac{2\sqrt{L}}{n \pi}\cos(n\pi)-\frac{2\sqrt{L}}{n \pi}\cos(0).$

$\cos(n\pi)$ will be $(-1)^{n}$ and $\cos(0)$ is $1$. And so we have

$a_n=\frac{4\sqrt{L}}{n \pi}\begin{cases}1&n~\text{odd}\\ 0 &n~\text{even}\end{cases}.$

$f(x)=\sum_{1,3, 5,\ldots}\frac{4}{n \pi}\sin(\tfrac{n\pi x}{L})$

So we can see the answer to our questions. The square wave has an infinite number of components and those components shrink as $1/n$.

Sawtooth Wave

We'll have the sawtooth wave range from -1 to 1 vertically and span $[-L, L]$. In this case, the sawtooth wave function is

$g(x)=\frac{x}{L}.$

This function is also odd. The sawtooth wave coefficients will only have one contributing integral:

$a_n = \text{Proj}_{s_n}(f(x)) = \int_{-L}^Ldx\sqrt{\tfrac{1}{L}} \sin(\tfrac{n\pi x}{L})\frac{x}{L}.$

This integral can be done exactly with integration by parts.

$a_n = \sqrt{\tfrac{1}{L}}(-\frac{x}{n \pi}\cos(\tfrac{n\pi x}{L})+\frac{L}{n^2 \pi^2}\sin(\tfrac{n\pi x}{L}))|_{-L}^L\\=(-\frac{\sqrt{L}}{n \pi}\cos(n\pi)+\frac{\sqrt{L}}{n^2 \pi^2}\sin(n\pi))-(\frac{\sqrt{L}}{n \pi}\cos(n\pi)-\frac{\sqrt{L}}{n^2 \pi^2}\sin(n\pi))\\=\frac{2\sqrt{L}}{n^2 \pi^2}\sin(n\pi)-\frac{2\sqrt{L}}{n \pi}\cos(n\pi)$.

$\cos(n\pi)$ will be $(-1)^n$ and $\sin(n\pi)$ will be 0. And so we have

$a_n=(-1)^{n+1}\frac{2\sqrt{L}}{n \pi}.$

$g(x)=\sum_n(-1)^{n+1}\frac{2}{n \pi}\sin(\tfrac{n\pi x}{L})$

Again we see that an infinite number of frequencies are represented in the signal but that their amplitude falls off at higher frequency.

Vectors and Fourier Series: Part 1

When I was first presented with Fourier series, I mostly viewed them as a bunch of mathematical tricks to calculate a bunch of coefficients. I didn't have a great idea about why we were calculating these coefficients, or why it was nice to always have these sine and cosine functions. It wasn't until later that I realized that I could apply much of the intuition I had for vectors to Fourier series. This post will be the first in a series of three that develop this intuition:

• Part 1: Developing tools from vectors
• Part 2: Using these tools for Fourier series
• Part 3: A few examples using these tools

We can start with the abstract notion of a vector. We can think about a vectors as just an arrow that points in some direction with some length. This is a nice geometrical picture of a vectors, but it is difficult to use a picture to do a calculation. We want to turn our geometrical picture into the usual algebraic picture of vectors.

$\vec r=a\hat x+b\hat y$

Choosing a Basis

We will need to develop some tools to do this. One tool we will need is a basis. In our algebraic picture, choosing a basis means that we choose to describe our vector, $\vec r$, in terms of the components in the $\hat x$ and $\hat y$ direction. As usual, we need our basis vectors to be linearly independent, have unit length, and we need one basis vector per dimension (they span the space).

Projection Operators

Now the question becomes: how can we calculate the components in the different directions? The way we learn to do this with vectors is by projection. So, we need projection operators. Things that eat our vector, $\vec r$, and spit out the components in the $\hat x$ and $\hat y$ directions. For the example vector above, this would be:

\begin{aligned}\text{Proj}_x(\vec r)&=a,\\\text{Proj}_y(\vec r)&=b.\end{aligned}

We want these projection operators to have a few properties, and as long as they have these properties, any operator we can cook up will work. We want the projection operator in the $x$ direction to only pick out the $x$ component of our vector. If there is an $x$  component, the projection operator should return it, and if there is no $x$ component, it should return zero.

Great! Because we have used vectors before, we know that the projection operators are dot products with the unit vectors.

\begin{aligned}\text{Proj}_x(\vec r) &= \hat x\cdot\vec r=a\\\text{Proj}_y(\vec r) &= \hat y\cdot\vec r = b\end{aligned}

We can also check that these projection operators satisfy the properties that we wanted our operators to have.

So, now we have a way to take some arbitrary vector—maybe it was given to us in magnitude and angle form—and break it into components for our chosen basis.

Vector Rebuilder

The last tool we want is a way of taking our components and rebuilding our vector in our chosen basis. I don't know of a mathematical name for this, so I'm going to call it a vector rebuilder. We know how to do this with our basis unit vectors:

$\vec r = a\hat x+b\hat y = \text{Proj}_x(\vec r) \hat x+\text{Proj}_y(\vec r)\hat y = \sum_{e=x,y}\text{Proj}_e(\vec r)\hat e.$

So, to recap, we have developed three tools:

• Basis: chosen set of unit vectors that allow us to describe any vector as a unique linear combination of them.
• Projection Operators: set of operators that allow us to calculate the components of a vector along any basis vector.
• Vector Rebuilder: expression that gives us our vector in terms of our basis and projection operators.

This may seem silly or overly pedantic, but during Part 2, I'll (hopefully) make it clear how we can develop these same tools for Fourier analysis and use them to gain some intuition for the goals and techniques used.