Primer on Fourier TransformsDate: 01/27/2003 at 09:16:28 From: David Subject: Fourier Transform Hi, 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! David Date: 01/27/2003 at 12:28:57 From: Doctor Roy Subject: Re: Fourier Transform Hi, 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 concepts. 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 is: 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 function. 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 removed. - Doctor Roy, The Math Forum http://mathforum.org/dr.math/ 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 place. Thanks again for your help, Yours, David |
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