Okay, it’s not as glamorous as a luxury yacht filled with supermodels, but it can help you:

• Decompose any signal and reduce the noise
  
• Identifying Cycles, Trends, and Forecasts Time Series 
• Extract features for Machine Learning Models
• Manage Risks
• Detecting Trading Opportunities via Frequency Clusters
• And much more ...
  

Basically, if you’re building a trading bot, this is one of the tools you’ll want in your arsenal. In this article, I’ll cover the basics, but stay tuned for a follow-up where we’ll explore its full potential—and the common pitfalls to avoid.

For those who aren’t math enthusiasts, don’t worry! I’ll make everything as intuitive as possible, so you can apply it like a true signal-processing wizard in your own projects.

The world is beautiful, isn’t it? (Oh, and if something isn’t clear, ask me! I may not have the answer, but at least it’ll boost my SEO. Thanks!😁)

 

What is Fourier Analysis?

Analysis, that branch of mathematics dealing with functions, took a revolutionary turn thanks to a certain Joseph Fourier (yes, a French guy—cocorico! 🐓). He studied the famous series that now bear his name. His idea? To decompose periodic functions into a sum of cosines and sines. A brilliant approach that, among other things, helped solve the heat equation and allowed us to deeply explore a multitude of periodic signals—like those fancy brain wave tracings on an EEG.

But what does this have to do with stock prices, you ask? Absolutely nothing. That little historical detour was just for my own amusement!

That being said, even though stock prices and financial "signals" are not periodic (at least not on a large scale), Fourier’s second major discovery is actually useful to us here. Indeed, as early as 1810, he demonstrated that any function, whether periodic or not, could be decomposed into a sum of trigonometric functions—yes, those same cosines and sines! More precisely, we state:

For any function integrable over ℝ:

• Let $f$ be such that the integral of its absolute value is finite, meaning:  $\int_{\mathbb{R}} |f(x)| \, dx < +\infty$
• For the math geeks, this is closely related to absolute convergence.
• For everyone else, it just means that the area under the curve, counted in absolute value (as if we flattened everything to be positive), is finite.
• And for the rest of the world… let’s just say it applies to any "normal" function.

Now, such a function $f$ can be transformed as follows:

$\hat{f}(\xi) = \int_{\mathbb{R}^n} f(x) e^{-2i\pi \xi \cdot x} \, dx$

 

This is what we call the Fourier transform. And we can retrieve $f(x)$ using the inverse transform:

$f(x) = \int_{\mathbb{R}^n} \hat{f}(\xi) e^{2i\pi \xi \cdot x} \, d\xi$

 

Alright, let’s take it slow! No need to panic.

The formula we really care about here is the inverse Fourier transform. Why? Because it reveals something amazing: if we take a well-behaved function $f$, we can express it as a sum of complex exponential functions.

And these complex exponentials, thanks to Euler’s formula, can be rewritten as:

$e^{-2i\pi \xi \cdot x} = \cos(2\pi \xi \cdot x) - i \sin(2\pi \xi \cdot x)$

 

Which, after some calculations, gives us:

$f(x) = \int_{\mathbb{R}^n} \left( \text{Re}(\hat{f}(\xi)) \cos(2\pi \xi \cdot x) - \text{Im}(\hat{f}(\xi)) \sin(2\pi \xi \cdot x) \right) d\xi$

 

In simpler terms, $f(x)$ is a sum (okay, a continuous sum, but still a sum) of sine and cosine signals. In other words, we’ve just managed to break down a non-periodic function into a superposition of periodic signals. Pretty cool, right? Here is a comparison to help you break it down :

Magnificent? Absolutely. But buckle up, because this is just the warm-up before the real fireworks begin.

 

But how can we make sense of it?

The key to interpretation lies precisely in this decomposition. To fully grasp its usefulness in finance, let's explore the intuition behind this transform.

Imagine a function $f(x)$ that depends on $x$, a distance, where $f$ is an aperiodic function describing heat distribution along a metal bar. Let’s assume that $f(x)$ is measured in Kelvin and $x$ in meters. To the American units club: the rest of us are waiting for you to invent 'light-years per ounce' as the next standard !

Now, let’s apply the Fourier transform to decompose this signal into a sum of periodic signals. And when we talk about periodic signals, we’re not just talking about periods (yes, I know…pretty logic right) but also about frequencies, which are simply the inverse of the period. This means we are decomposing our temperature function into different frequency components. But what does that really mean?

Let’s take a closer look. In the exponential term $e^{-2i\pi \xi \cdot x}$, the exponent $-2i\pi\xi \cdot x$ must be dimensionless. This implies that $\xi$, the frequency, must have units of $m^{-1}$ (inverse meters). This means it is a spatial frequency —compared to a time frequency that represents the inverse of time itself!😍

Next, the integral tells us that the temperature in Kelvin can be expressed as a sum of terms $\hat{f}(\xi) e^{2i\pi \xi \cdot x} \, d\xi$. Since the exponential function is dimensionless and $d\xi$ has units of $m^{-1}$, it follows that $\hat{f}(\xi)$ must have units of Kelvin × meters so that the entire expression remains in Kelvin. In other words, $\hat{f}(\xi)$ represents the amplitude of the contribution of the spatial frequency $\xi$ to the temperature.

A crucial point: this decomposition is unique for each spatial frequency. It is logically impossible to have two waves with the same frequency of $2 \, m^{-1}$.

 

Let’s Crunch Some Numbers to Make It Clearer

Now, suppose that for our metal bar, we obtain $\hat{f}(\xi) = 5 \, K \times m$ for the frequencies $\xi = 1 \, m^{-1}$ and $\xi = 2 \, m^{-1}$. This means that these two spatial frequencies contribute equally to the heat distribution in our bar. However, this equality in amplitude does not mean that the temperature variation is the same:

• For $\xi = 1 \, m^{-1}$, the temperature variation is $\hat{f}(\xi) = 5 \, K \times m$ over 1 meter, meaning a difference of 5 Kelvin per meter.
• For $\xi = 2 \, m^{-1}$, the temperature variation is also $\hat{f}(\xi) = 5 \, K \times m$, but over only 50 cm (1/2 meter), meaning the temperature changes occur much more rapidly.

Quick Suprise Test: 

What if we found an amplitude of $3 \, K \times m$ for a frequency $\xi = 5 \, m^{-1}$?