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Fourier series representation for F(t) = 0 or sin(wt) [depending on range]

by Esran
Tags: fourier series, sine
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Esran
#1
Oct9-10, 02:11 PM
P: 71
1. The problem statement, all variables and given/known data

Obtain the Fourier series representing the function [tex]F(t)=0[/tex] if [tex]-2\pi/w<t<0[/tex] or [tex]F(t)=sin(wt)[/tex] if [tex]0<t<2\pi/w[/tex].

2. Relevant equations

We have, of course, the standard equations for the coefficients of a Fourier expansion.

[tex]a_{n}=2/\tau\int^{1/2\tau}_{-1/2\tau}F(t')cos(nwt')dt'[/tex]
[tex]b_{n}=2/\tau\int^{1/2\tau}_{-1/2\tau}F(t')sin(nwt')dt'[/tex]

3. The attempt at a solution

Clearly, in the first part of the interval we're working with the integrals inside the equations for our coefficients go to zero, so we need focus only on the region [tex]0<t<2\pi/w[/tex]. The curious thing is, the problem did not mention a period, and when we assume a period of [tex]4\pi/w[/tex], each of our coefficients vanish. Should I instead assume a period of [tex]2\pi/w[/tex] and integrate over the region [tex]-\pi/w<t<\pi/w[/tex]?

Here is what I have:

For a_{n}: integral_0^((2 pi)/w)1/2 pi w cos(n w x) sin(w x) dx = (pi sin^2(pi n))/(1-n^2)

For b_{n}: integral_0^((2 pi)/w)1/2 pi w sin(w x) sin(n w x) dx = (pi sin(2 pi n))/(2 (n^2-1))

Once again, as you can see, the coefficients vanish (since n is an integer). What is going wrong here?
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vela
#2
Oct9-10, 05:33 PM
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The ω that appears in F(t) isn't the same as the frequency that appears in the sine and cosines. As you said, the period of the function is T=4π/ω, so the fundamental frequency of the Fourier expansion is ω0=2π/T=ω/2.

Also, note that the integration results you wrote down are only valid when n2-1≠0. You need to treat cases like that separately.
Esran
#3
Oct9-10, 06:44 PM
P: 71
Oh yeah. The fundamental frequency. Thanks!


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