Cherish (mareserinitatis) wrote,
Cherish
mareserinitatis

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Fourier Transforms - pt. 1

Now that I have a few minutes to breathe, I can get going on the posts I had planned to write on Fourier Transforms (no, not Transformers).

Hopefully you've read all the background on Fourier series. (If not, go to the bottom of the post and click on the words "fourier series" in the tags. It will bring you to the relevant posts in reverse chronological order.) They're nifty things, and rather integral to the idea of the Fourier Transform.

Let's begin with our definition of the general Fourier series:

(1).

The coefficients Dn were defined as

(2).

Now, let's stick (2) into (1). The result is

(3).

Now we're going to go through and make some changes about the assumptions we made. What we want to do is to find a way to represent a general signal by either it's frequency content or as the original function. To do that, we'll have to make some changes to (3).

To begin, we assumed that we were trying to use our Fourier series to represent a periodic function. That is, our function f(x) would repeat a pattern over a given period T. We could also say that f(x) = f(x + T).

But what if that isn't true? What if what we have is just some general signal and not something that repeats in a well-behaved manner?

The first thing we can do is to imagine that we have a signal that is periodic, but the period goes from -∞ to +∞. Therefore, rather than integrating over the period To, we will integrate from -∞ to +∞.

Our equation looks like

(4).

Secondly, we assumed that we were dealing with discrete harmonics, which is what the indicated. What if we want to look at every possible frequency? In this situation, we can assume that our n becomes insignificantly small. This means we can chuck our n and just use ω. (Realistically, we can't look at every possible frequency. Our data will provide some constraints, but we'll get into that later. For now, we'll pretend it's possible to look at all of them.)

(5)

This creates a problem, though. What about that summation?

This goes back to our treatment of n: the distance between frequencies or harmonics becomes infinitesimally small. Hopefully, you'll remember that Riemann sums can be used to approximate the area under a function. As the width of the columns of the Riemann sum become infinitesimally small, the sum will become an integral (at the limit). So in this case, our summation will actually become an integral with limits of -∞ to +∞.

(6)

Oh dear. The mathematicians are howling because we've now entered the realm of BAD MATH. We have an integral with no infinitesimal at the end.

Fortunately, we still have one unexamined term: 1/To. That will save us!

If we're looking at the reciprocal of the period, and we changed our period to be infinity, then the reciprocal of that period will be...wait for it...an infinitesimal! However, rather than using it in its present form, we're going to put it in terms of our frequency.

Remember that

(7).

If we take the infinitesimal of ω, we get

(8).

And putting this into (6) and moving the 1/2π term outside the integral leaves us with

(9).

I know what you're thinking: "This gets me what?"

What this has done is given us a way to change the signal into a form that represents it as a function of frequency as well as a way to change the frequency dependent representation back into the original signal.

So if you look at the term in brackets, you see that we are eliminating the dependence on x because that is the variable of integration. That leaves us with a function in terms of ω simply as a result of the fact that there aren't more variables in there. The term in brackets is the Fourier Transform (or FT as shorthand), which we will set equal to F(ω).

(10)

Some folks prefer to make the left-hand side a function of i ω or, actually, j ω.

J?!

Electrical engineers use j to denote the square root of negative one because i is used as the symbol for current. The reason for using j ω rather than simply ω is that this combination can be replaced by s, which (in most cases) gives you the Laplace Transform, an exceptionally nifty device for analyzing simple circuits.

Now what about outside of the brackets?

If we set the term in the brackets equal to F(ω), we get

(11),

which we call the Inverse Fourier Transform (IFT).

So that's one way to get the Fourier Transform and it's inverse. The FT will take a signal and give you a function that is dependent only on frequency. Taking the IFT of that function will, hypothetically, give you back your original signal.

In terms of symbols, one might say that



and

.

This means that the left-hand side will equal the transform (either Fourier or Inverse Fourier) of the function in the curly brackets.

Using them, however, will take a bit of practice, and will have to wait for another post.
Tags: fourier series, fourier transforms, math
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