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<math>x_{\mathrm{square}}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin{\left ((2k-1)2\pi ft \right )}\over(2k-1)} </math>
<math>x_{\mathrm{square}}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin{\left ((2k-1)2\pi ft \right )}\over(2k-1)} </math>


That's alot of numbers seemingly out of the blue, at first observance. I bet your questions are:
1. What is a Fourier Transform? <br>
2.How do I get one? <br>
3.If I've got one, what can I do with it?
Unfortunately, the Fourier Transform isn't a Transformer. If it was, you would have seen it in the movie that came out lately. Even if
it was a transformer, I bet it would look like a stately old English professor instead of a metal beast, since it always follows the rules.
One way to explain a Fourier Transform
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt



Revision as of 18:55, 8 October 2007

An Introduction to the Fourier Transform

A Fourier Transform is a representation of a function using a large number of sinusoids added together to create it.

For example, a square wave could be represented by: xsquare(t)=4πk=1sin((2k1)2πft)(2k1)

That's alot of numbers seemingly out of the blue, at first observance. I bet your questions are:

1. What is a Fourier Transform?
2.How do I get one?
3.If I've got one, what can I do with it?

Unfortunately, the Fourier Transform isn't a Transformer. If it was, you would have seen it in the movie that came out lately. Even if it was a transformer, I bet it would look like a stately old English professor instead of a metal beast, since it always follows the rules.

One way to explain a Fourier Transform X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt

</math>

Suppose that we have some function, say β(t), that is nonperiodic and finite in duration.
This means that β(t)=0 for some Tα<|t|

Now let's make a periodic function γ(t) by repeating β(t) with a fundamental period Tζ. Note that limTζγ(t)=β(t)
The Fourier Series representation of γ(t) is
γ(t)=k=αkej2πfkt where f=1Tζ
and αk=1TζTζ2Tζ2γ(t)ej2πktdt
αk can now be rewritten as αk=1Tζβ(t)ej2πktdt
From our initial identity then, we can write αk as αk=1TζB(kf)
and γ(t) becomes γ(t)=k=1TζB(kf)ej2πfkt
Now remember that β(t)=limTζγ(t) and 1Tζ=f.
Which means that β(t)=limf0γ(t)=limf0k=fB(kf)ej2πfkt
Which is just to say that β(t)=B(f)ej2πfktdf

So we have that [β(t)]=B(f)=β(t)ej2πftdt
Further 1[B(f)]=β(t)=B(f)ej2πfktdf

Some Useful Fourier Transform Pairs

[α(t)]=1αf(ωα)

[c1α(t)+c2β(t)] =(c1α(t)+c2β(t))ej2πftdt
=c1α(t)ej2πftdt+c2β(t)ej2πftdt
=c1α(t)ej2πftdt+c2β(t)ej2πftdt=c1A(f)+c2B(f)


[α(tγ)]=ej2πfγA(f)
[α(t)*β(t)]=A(f)B(f)
[α(t)β(t)]=A(f)*B(f)
Some other usefull pairs can be found here: Fourier Transforms

A Second Approach to Fourier Transforms