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Esran

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## Homework Statement

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].

## Homework 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]

## 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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