Fourier transforms


By Tom Hardcastle (P2477) on Monday, May 15, 2000 - 11:18 pm :

Hi. It looks like to follow some of the discussions here I'm going to have to find out what a Fourier transformation is.

In fact, I don't know anything called Fourier. If it can be done briefly, can someone give me a description and examples of uses. Otherwise, any recommendations of easily available textbooks would be appreciated.


By Dan Goodman (Dfmg2) on Tuesday, May 16, 2000 - 01:38 am :
OK, here is a quick intro to Fourier transforms. I'm going to assume a reasonable knowledge of complex numbers. I.e. What is a complex number and exponentials of complex numbers. If you don't know about complex numbers, you'll be able to understand the first part, but not the second.

First of all, a Fourier series is a simpler notion, so we'll start with that. If f(x) is a nice(ish) function that is periodic with period 2L (i.e. f(2L)=f(0)), then we can write:

f(x)=(1/2) a0 + n=1 ( an cos(nπx/L)+ bn sin(nπx/L)

for some coefficients an and bn . In fact, we can work out these coefficients from f:

an =(1/L) 0 2L f(x)cos(nπx/L)dx

bn =(1/L) 0 2L f(x)sin(nπx/L)dx

(I might be out by a constant factor here, I'm not sure).

Why would anyone do this? Well, there are various reasons. Firstly, you can single out frequency components by finding the Fourier series, and use this to find the effect of high pass filters on sound signals for instance (a high pass filter removes high frequencies from a sound signal, which can be used for (e.g.) removing hiss from tape). It's also useful in physics quite a lot for solving partial differential equations with boundary conditions (e.g. Laplace's equation).

A Fourier transformation (FT) is a related notion. The Fourier transformation of a function f(x) is F(ω)=(1/2π) - - e-iωx f(x)dx

The inverse Fourier transform of F(ω) is

f(x)=(1/2π) - eiωx F(ω)dω

The Fourier inversion theorem states (roughly) that the inverse Fourier transform of the Fourier transform of a function is the original function. Actually, it is not quite the original function, it is the original function where it is continuous, and it is the midpoint of the upper and lower values of the function at a discontinuity. For periodic functions, the inverse FT of an FT simplifies to the Fourier series above. The inverse FT of the FT is sometimes called the Fourier representation of the function.

Fourier transforms have various properties which can be deduced from the definition, one important one is the following. If g(x)=f'(x) the derivative of f, then the FT of g is G(ω)=(iω)F(ω). This can help us to solve differential equations. For instance, if we want to solve:

y''(x)+ay'(x)+by(x)=f(x)

Then we take the FT and get

(- ω2 )Y(ω)+aiωY(ω)+bY(ω)=F(ω)

Which we can easily solve for Y(ω). Now we just take the inverse FT of Y(ω) and we get the original function y(x) which satisfies the differential equation.

There are lots of uses of Fourier transforms, in the other discussion for instance (Money) I mentioned that the Fourier transform of position space is momentum space in Quantum Physics. Someone has just made a slightly more in depth post about this.

So there you go, Fourier in a nutshell. I've just finished revising it for my exams as it happens, so if anyone notices any big mistakes, PLEASE TELL ME! Did you follow all that?

p.s. If you are keen, you might like to try and prove some of the properties of FTs, for instance if g(x)=f(x+a) then G(ω)= e-iωa F(ω).