Fourier Series

  • #1
I understand what the Fourier Theorem means, as well as how it behaves, I just don't understand how the math actually pans out or in what order to do what.

I'm gonna start off with what I know.

[tex]f(x) = \frac{a_0}{2} \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})[/tex]

while,
[tex] a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx[/tex]

[tex]a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos({nx})\ dx[/tex]

[tex]b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin({nx})\ dx[/tex]

This is of course only the case with periodic functions. Depending on how the graph looks it is possible to derive [tex]f(x)[/tex], the spatial period, and maybe even its tendency to be even or odd.

(Even/odd meaning whether or not the beginning of the wavelength is at the origin. If it is/does, it's odd and only contains cosine terms, if not and it behaves more like a sine wave (highest amplitude at origin) than it is even, thus not containing any cosine terms.)

if we are given [tex]f(x)[/tex], all we do is find [tex]a_{0}[/tex], [tex]a_{n}[/tex] and [tex] B_{n}[/tex] then plug into the first equation. Is this right? I am getting absurdly long answers doing this, and as far as I can tell I can't find any way of finding out whether I'm headed in the right direction or not.

Here is one of the problems I'm having trouble with:

[tex]f(x) = A\cos({\frac{\pi x}{\lambda}})[/tex], find the fourier series (it is assumed the function is periodic on the interval [tex][0,2\lambda][/tex])
 
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Answers and Replies

  • #2
Office_Shredder
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The very important, first latex image isn't working. Neither is the image labelled (2)

Try
[tex]f(x) = \frac{A_0}{2} \sum_{n=1}^{\infty}(A_n}\cos{k_nx} + B_{n}\sin{k_nx})[/tex]

and

[tex] = [\frac{\sin(n k x)}{(n k)^2} - \frac{x\cos(n k x)}{n k}]_{-\lambda/2}^{lambda/2}[/tex]

Did I get those right? I'm not a latex expert, I just fixed up some underscores and hoped for the best :D
 
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  • #3
Thanks OS..formatting is working now, still need help though.
 
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  • #4
459
5
NutriGrainKiller said:
[tex]f(x) = \frac{a_0}{2} \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})[/tex]
You are missing a plus sign in the middle. It should be
[tex]f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n}\cos{nx} + b_{n}\sin{nx})[/tex]

NutriGrainKiller said:
if we are given [tex]f(x)[/tex], all we do is find [tex]a_{0}[/tex], [tex]a_{n}[/tex] and [tex] B_{n}[/tex] then plug into the first equation. Is this right?
Yes, that is the way you do it.

NutriGrainKiller said:
I am getting absurdly long answers doing this, and as far as I can tell I can't find any way of finding out whether I'm headed in the right direction or not.

Here is one of the problems I'm having trouble with:

[tex]f(x) = A\cos({\frac{\pi x}{\lambda}})[/tex], find the fourier series (it is assumed the function is periodic on the interval [tex][0,2\lambda][/tex])
This is because the formulas for the coefficients you have stated only hold true for functions whose period is [tex]2\pi[/tex]. The more general formulas are:

[tex] a_{0} = \frac{2}{T} \int_{t_0}^{t_0 + T} f(x) dx[/tex]

[tex]a_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(x) \cos({n{\omega}_0 x})\ dx[/tex]

[tex]b_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(x) \sin({n{\omega}_0 x})\ dx[/tex]

where [tex]{\omega}_0 = \frac{2\pi}{T}[/tex] and [tex]\int_{t_0}^{t_0 + T} [/tex] means that the integral is over any particular period. In your example, the period is [tex]2\lambda[/tex] so [tex]{\omega}_0 = \frac{2\pi}{2\lambda} = \frac{\pi}{\lambda}[/tex] and your integral should look something like this: [tex]\int_{0}^{2\lambda}[/tex].
 
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  • #5
459
5
Also, the function you have given in your example is very weird because the function and its fourier series is the same. You can't decompose your function any further.
 

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