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Primer on Fourier Transforms

Date: 01/27/2003 at 09:16:28
From: David
Subject: Fourier Transform


I have been learning about the Fourier Transform with reference to 
sound waves. I understand that the Fourier transform can be used to 
'separate out' or identify the component sine waves of an individual 
sound. But what I want to know is, how does the Fourier Transform 
extract these individual sine waves? Are there any examples that show 
the maths involved?

Thanks for any help!


Date: 01/27/2003 at 12:28:57
From: Doctor Roy
Subject: Re: Fourier Transform


Thanks for writing to Dr. Math.

It is not possible to really deeply understand Fourier series without
the complicated math.

However, it is possible to get a very good intuitive grasp of the

So, here is a simple primer on Fourier transforms. The idea behind the 
Fourier transform is very simple: Can we represent a rectangular
function, or a square wave, with a weighted sum of sinusoids?

At first glance, we would think not. After all, rectangular functions
have discontinuities and sinusoids do not have discontinuities. But
Fourier found a way to represent a square wave using only sinusoids.
This was the basis for his Doctoral dissertation. Of course, he didn't
get it. His problem was that he couldn't explain how the
discontinuities formed (the jumps in the graph of a square wave). It
wasn't until a decade later that he found a "fix." He defined the
concept of equal "almost everywhere." The idea is that the "energy,"
defined in a certain way, of both the sinusoid version and the
original square wave is the same. The energy of both functions at the
discontinuities is zero, so they don't matter. So we have a method of
representing square waves (and really almost any "normal" function)
with sinusoids.

There is one property of sinusoids that is important. They can be
defined by their frequency. So, Fourier's method was to represent each
square wave as a weighted sum (so each frequency component had a
different amplitude) of sinusoids.

This leads to the next great property: orthogonality. Sinusoids, which 
are a multiple of a fundamental frequency (or fundamental harmonic), 
are "orthogonal" to each other. What does this mean? I assume you are 
familiar with the x-y plane. The x and y axes are orthogonal, or 
perpendicular to each other. So, any point on the x-y plane can be 
described by the x and y components separately, and we can calculate 
many useful quantities very easily because of the orthogonality. If 
the bases were not orthogonal, we wouldn't be able to calculate many 
things as easily.

For example, let's imagine a "crazy" system, where the x axis is one
of the bases and the other basis is the line x = y.  Notice that we
can still describe every point on the line by these two components.
But this description has its limitations.

So, in a system of sinusoids, all multiples of a fundamental frequency
are orthogonal to each other. So the Fourier "transform" actually
DOES transform signals from one orthogonal system to another
orthogonal system.

And here's the way each frequency component is separated:

Each sinusoid is orthogonal to each other. That means we can figure 
out "how much" of each sinusoid is present in a function. For example, 
consider the x-y example above. If we have the point x = 2 and y = 5, 
it is very easy to find out how much "x" and how much "y" is present. 
However, it is much harder to find out how much "x" and how much 
"x = y" there is in this point. The orthogonality means that by 
calculating the Transform integral, we find exactly "how much" of a
particular sinusoid is present. We don't get any energy or weight from
another sinusoid because of the orthogonality.

Here's a "simple" example:

Let's say we had a flat signal of amplitude 1. The Fourier Transform 

    a_k = integral( 1 * exp(-j*w*k*t) dt)

The exp(-j*w*t) is the sinusoid. Since it is a complex exponential 
(has a complex number since j = sqrt(-1)), we can write it as:

   exp(-j*w*t) = cos(w*t) - j*sin(w*t)

Here w is the fundamental frequency. If we pick a particular k, we
pick which frequency we are calculating for. The integral calculates
how much "weight" or "energy" is found for a particular frequency in a

So, if we had a periodic function, like a square wave, we would notice
that certain frequencies would  show up more (multiples of the 
frequency of the square wave itself) and others would not (non-
multiples). A function like a flat signal of unit amplitude contains 
all frequencies.

This is about the simplest explanation possible without either
understanding the mathematics or taking a course. The above is really
a couple weeks of Fourier theory at the introductory level at a
university compressed into a few paragraphs and with most of the math

- Doctor Roy, The Math Forum 

Date: 01/28/2003 at 09:30:19
From: David
Subject: Thank you (Fourier Transform)

Hello Dr Roy,

Thank you very much for your insight into the Fourier Transform. Your 
description of the transform in relation to square waves gave me a new 
understanding. I am now delving into the maths of the transform and am 
hoping that it will soon click. However, you are right when you say 
that taking a course would be beneficial. Learning the Fourier 
Transform from complicated maths books alone is quite a slow process. 
I am considering taking a course in DSP if I am lucky enough to gain a 

Thanks again for your help,

Associated Topics:
College Calculus

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