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I just came from a class on Fourier Transformations as applied to signal processing and sound. It all seems pretty abstract to me, so I was wondering if there were any physical systems that would behave like a Fourier transformation.

That is, if given a wave, a purely physical process that would "return" the Fourier transform in some meaningful way. Like, you gave it a sound wave and you would see, "Oh, there are a lot of components of frequency 1kHz...a few of frequency 10kHz...some of 500Hz..."

I've seen things happening where, if you put sand on a speaker, the sand would start to form patterns on the speakers that are related to the dominant wavelengths/fundamental frequencies of the sound. Is this some sort of natural, physical fourier transform?

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You seem to be referring to Chladni figures. – user172 Nov 8 '10 at 5:49
I added some remarks on your last paragraph where the phenomena are not only related to a Fourier approach. Greets. – Robert Filter Dec 13 '10 at 8:46
up vote 31 down vote accepted

Your ear is an effective Fourier transformer.

An ear contains many small hair cells. The hair cells differ in length, tension, and thickness, and therefore respond to different frequencies. Different hair cells are mechanically linked to ion channels in different neurons, so different neurons in the brain get activated depending on the Fourier transform of the sound you're hearing.

A piano is a Fourier analyzer for a similar reason.

A prism or diffraction grating would be a Fourier analyzer for light. It spreads out light of different frequencies, allowing us to analyze how much of each frequency is present in a given source.

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I didn't even think of prisms/light! That's pretty cool. Could you elaborate on how a Piano is a Fourier analyzer? How would you "read" it? – Justin L. Nov 4 '10 at 7:36
make a sound near the piano. some strings will vibrate more and others less. the more a string vibrates, the more of its fundamental frequency is in the sound. – Mark Eichenlaub Nov 4 '10 at 8:18
To elaborate on the diffraction grating part, once you have the 2D Fourier transform of a beam by passing it through a grating, you can chop off parts of the transformed beam and pass it through a second grating to filter out parts of the spectrum of the beam. I saw this done once by using an old 35mm slide as a diffraction grating for a laser beam, and you could make details on the slide go away by chopping off the high-frequency components of the transformed beam. – Chris Granade Nov 4 '10 at 13:20
I think it is more accurate to say that the ear is a sort of xylophone. To be pedantic, I think the "frequency domain" is a distinct concept from the "Fourier transform". – nibot Nov 4 '10 at 23:16

I see that two examples in optics have been mentioned, a diffraction grating by Mark Eichenlaub, and a lens by sigoldberg1. I would like to elaborate a bit, because there is a subtle difference between the two.

On the one hand, a diffraction grating separates out light of different frequencies, i.e. colors, transforming them into different positions. This is analogous to how the 1-D Fourier transform works on a sound or electrical signal.

On the other hand, a lens takes the 2-D Fourier transform of a monochromatic light beam. Monochromatic means there is only one frequency or color. You might wonder why the Fourier transform isn't a single peak, if there's only one frequency in the light. That's because a lens transforms spatial frequencies (roughly equivalent to angles, but I will explain them below) into positions. Mathematically, this is the same Fourier transform, but it transforms an entirely different property of the light.

It took me the longest time to understand exactly what spatial frequencies were. I finally got it one day, reading chapter 4, "Fourier Optics", of Saleh and Teich's Fundamentals of Photonics, which I whole-heartedly recommend if you don't understand my explanation here.

You probably learned in your class that you can write any signal as a Fourier series, which is the sum of frequency components, each with their own amplitude and phase. The Fourier transform is a sort of continuous version of that. Well, you can express any monochromatic light beam as the sum of many plane waves traveling at different angles, all with the same frequency, but each with their own amplitude and phase. Is this starting to sound familiar? Each of these plane waves is a spatial frequency. Just as with the step from Fourier series to Fourier transform, you can make the step from discrete spatial frequencies to 2-D Fourier transform.

A positive lens focuses each of these spatial frequencies to a separate point. For example, here are illustrations of two different plane waves being focused by a lens.

Plane wave hitting a lens face on

When a plane wave hits a lens face on, the lens focuses it to a point on the optical axis, at the focal distance of the lens.

Plane wave hitting a lens at an angle

However, when the plane wave is incident at an angle, the focus is still at the focal distance, but displaced from the optical axis. This is how a lens transforms "angle" to position.

1-D Fourier transform using a lens

So if you have a more complicated beam made up of many spatial frequencies (which, as I said, are just plane waves), they are all focused to separate points in the same plane on the other side of the lens, one focal distance away. This is why we say it computes the 2-D Fourier transform of a monochromatic light beam. In fact, in the 1970s, when computers weren't so fast, people actually experimented with using lenses to calculate Fourier transforms instantaneously!

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Interesting. I was once briefly involved in a project that had to do precise wavefront sensing (for adaptive optics). I never thought about the Fourier approach because we instead had an array of lenses, each focusing on their own CCD. This allowed spatial sampling of the wavefront rather than Fourier sampling. In hindsight, thought, we did have a "first order" detector hooked up to a tilting mirror that was, I suppose, finding the strongest Fourier component. – Mark Eichenlaub Nov 4 '10 at 18:44
A while ago now (late 90's), I heard some cool talks on using these Fourier properties of lenses to do cryptography. By putting a phase mask in the focal plane of a lens looking at some picture or object of interest, you can turn the image formed by that lens into what looks like random static, because you change the way all those spatial Fourier components add up. You can undo the encryption by doing the same thing in reverse- sending the encrypted image through a lens and applying the right phase mask gets the original picture back. It was cool, but I don't know if it went anywhere. – Chad Orzel Nov 5 '10 at 16:50
I've never heard of these being called "spatial frequencies" - it's kind of a misleading name, IMO, but I think I see what you're getting at. If I understand you correctly, it's basically a Fourier transformation from 2D momentum space into 2D position space. – David Z Nov 7 '10 at 7:53
@David Zaslavsky, exactly. I call them "spatial frequencies" because that's what Saleh and Teich call them and that's where I first learned about them ;-) – ptomato Nov 7 '10 at 20:56

Remember the double slit experiment? The interference pattern is the Fourier transform of the hole(s). This boggled my mind when I first learned it. In the limit where the screen is far from the mask, the rays of light actually physically compute the Fourier transform (see Fraunhofer diffraction).

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This is also really cool! I didn't touch on it in my answer, but as you say, propagation over long distances also computes the Fourier transform. This is why laser beams are often Gaussian-shaped - because the Gaussian is its own Fourier transform, so it stays the same over long propagation distances! – ptomato Nov 5 '10 at 0:17
Hmm, I hadn't thought of that! Does this mean that higher order (Hermite-Gauss, Laguerre-Gauss, etc) are also their own Fourier transforms? – nibot Nov 5 '10 at 0:23

This relates to one of the coolest examples ever - lenses and how you see.

Very, very, roughly, when light interacts with a macroscopic object the result is the Fourier transform of the shape of the object, contained in the scattered light. A lens basically computes the Fourier transform of some of the scattered light. The Fourier transform is its own inverse, i.e. applied twice we get the identity. So using the lens in your eye, your retina sees the shape of the object! Same for cameras, obviously. I'll try to find some references and edit later to include them.

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In imaging microscopy the 2D-Fourier Transform is a useful tool to analyse the images and improve the S/N ratio.

I had your same opinion at the beginning but then my boss, who actually is a FTT entusiast, was able to convince me to the actual usefulness (comic) of the technique!

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Do imaging microscopy and 2D Fourier Transforms happen naturally? – Justin L. Nov 4 '10 at 17:19
Justin: Yes. For instance, in crystallography, it is common to expose a crystal to a beam of electrons and to measure the diffraction pattern, which is the Fourier transform of the crystal's density pattern. One problem is that, when the diffraction pattern is captured by photographic film, only the amplitude information is recorded but the phase information is lost. (There are algorithms to get around this.) – nibot Nov 5 '10 at 0:25

I want to add something to the speaker issue and show that the Fourier transformation is not enough to explain it (as maybe stated implicitly by the answers provided so far).

In a first approximation you can describe the eigenfunctions of a plate subject to some mechanical (e.g. acoustic excitation) by the Helmholtz equation

$\Delta\psi(x,y) + \mathbf{k}^2\psi(x,y) = 0$ for ${x,y}\in \Omega = {(x,y)\in(-L_x...L_x,-L_y...L_y)}$

Solutions to this equation are plane waves if, and now comes the point that was not discussed so far, you have some rectangular boundary with Dirichlet conditions, say

$\psi \equiv 0$ for $(x,y) \in \partial\Omega$

Then, you really have (the m'th) resonances/eigenfrequencies at some $k_x*L_x = m$ and analogously for $y$.

But if the boundaries are somehow different, you get other eigenfunctions like Bessel Functions for a spherical system. The situation gets totally messi if you have a shape that is not integrable like for dynamical billiards. Then you can observe chaotic eigenfunctions.

So, to sum up, it is not enough to know the Fourier transformation to explain such wave phenomena.



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Expounding on the 1st part of @Einchenlaub's answer (and describing my favourite example),

The Cochlea in the ear (in which the hair follicles are present) is a beautiful example of how the fourier transform is carried out physically.
The changing diameter of the cochlea's tube causes different frequencies to resonate in different parts of the cochlea, so hair follicles in those regions are thus most sensitive to a particular frequency. The signals coming out of all the hair cells are the fourier transform of the incoming sound (if you organize the signals by position in the cochlea).

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