What does it mean to Fourier transform a function, i.e. #hat(f)(omega) = mathcalF_(omega)[f(x)]#?

1 Answer

Answer:

The Fourier transform is a way of converting one function to another function (details to follow). It is useful for solving differential equations. I'm not an expert in this, but I believe it's also useful for signal processing, which is at the heart of communication technology (see Wikipedia's page )

Explanation:

Given a "sufficiently nice" function #f(x)# (details on what "sufficiently nice" means will not be delved into), defined on the entire real line #RR#, the Fourier transform of #f(x)#, denoted #hat{f}(omega)#, is defined by the formula:

#hat{f}(omega)=\int_{-\infty}^{\infty}f(x)e^{i \omega x}\ dx#, where #i# is the imaginary unit (#i^{2}=-1#).

Note 1: Sometimes people put an extra #-2pi# in the power of #e#. Alternatively, sometimes people put an extra #1/sqrt{2pi}# in front of the integral.

Note 2: With respect to the given improper integral, the #omega# is a constant. However, since we can (hopefully) do this for many values of #omega#, we ultimately think of #omega# as the variable for the "new function" #hat{f}#. Typically #omega# is a real variable, though #hat{f}(omega)# is a complex number (in general).

Note 3: Technically speaking, then, the Fourier transform is the (more general kind of) function that maps #f# to #hat{f}#. Sometimes this is written as #mathcal{F}(f)=hat{f}# (in other words, #mathcal{F}# is the symbol for this function that maps #f# to #hat{f}#).

As an example, let #f(x)=e^{-|x|}#. Then, not worrying about convergence issues, and using Euler's formula #e^{i theta}=cos(theta)+i sin(theta)#, we get:

#\hat{f}(omega)=\int_{-\infty}^{0}e^{x}e^{i omega x}\ dx+\int_{0}^{\infty}e^{-x}e^{i omega x}\ dx#

#=\int_{-\infty}^{0}(e^{x}cos(omega x)+ie^{x}sin(omega x))\ dx+\int_{0}^{\infty}(e^{-x}cos(omega x)+ie^{-x}sin(omega x))\ dx#.

Now (via Wolfram Alpha or tables of integrals),

#\int (e^{x}cos(omega x)+ie^{x}sin(omega x))\ dx=i/(i-omega)e^{x}cos(omega x)-1/(i-omega)e^{x}sin(omega x)+C#

and

#\int (e^{-x}cos(omega x)+ie^{-x}sin(omega x))\ dx=-1/(1-i omega)e^{-x}cos(omega x)-i/(1-i omega)e^{-x}sin(omega x)+C#

Since #cos(0)=1# and #sin(0)=0# and #e^{x}->0# as #x->-\infty#, this leads to

#hat{f}(omega)=i/(i-omega)+1/(1-i omega)#

#=(i(-i-omega))/((i-omega)(-i-omega))+(1+i omega)/((1-i omega)(1+i omega))#

#=(1-i omega)/(1+omega^2)+(1+i omega)/(1+omega^2)=2/(1+omega^2)#