Find the Fourier series of a 'broken' function which is periodic in 2L

[knowlewj01]

Homework Statement

Sketch the function:

$f(x) = \begin{Bmatrix} \frac{-x}{a} & 0 \leq x \leq a \\ \frac{x-L}{L-a} & a \leq x \leq L$

where f(x) is an odd function and is periodic in 2L.
And a is a constant less than L/2

Find the Fourier series for the function f(x).

Homework Equations

The Attempt at a Solution

I have attached a rough sketch of the function, its not perfect but good enough to get the general idea. We can see that the function is odd and Discontinuous at f(a) and f(a+L)

the general formula for the Fourier series is:

$\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n cos\left(\frac{n\pi x}{L}\right) + b_n sin\left(\frac{n\pi x}{L}\right)$

we are told that the function is odd, hence all the a_n terms must be 0.

so the Fourier series is in the form:

$\sum_{n=1}^{\infty} b_n sin(\frac{n\pi x}{L})$

to find the b_n terms use the formula:


$b_n = \frac{1}{2L} \int_{-L}^L f(x) sin(\frac{n\pi x}{L}) dx$

odd function * odd function = even function, so:

$b_n = \frac{1}{L} \int_0^L f(x) sin(\frac{n\pi x}{L}) dx$


so.. do i then split this up into 2 separate integrals:

$b_n = \frac{1}{L}\left[ \int_0^a \left(\frac{-x}{a}\right) sin(\frac{n\pi x}{L}) dx + \int_a^L \left(\frac{x-L}{L-a}\right) sin(\frac{n\pi x}{L})\right]$

am i able to do this?

the thing that is confusing me the most is wether i am using the right values for L (ie, not sure wether its supposed to be 1/L , 2/L , 1/2L etc)

I followed this through and ended up with a horrible expression containing terms in:

$sin\left(\frac{n \pi a}{L}\right)$

and

$cos\left(\frac{n \pi a}{L}\right)$


thanks

 

[vela]

Looks good so far.

The period T is 2L, so the angular frequency is ω = 2π/T = 2π/2L = π/L. The argument of the sines and cosines is nωx, which is equal to what you have.

 

[knowlewj01]

I've worked the integrals through and ended up with the solution:

$f(x) = \sum_{n=1}^{\infty} \frac{-2L^2}{an^2 \pi^2 (L-a)} sin\left(\frac{n \pi a}{L}\right) sin\left(\frac{n \pi x}{L}\right)$

I'm not sure if this is right. I've never seen any examples where the sin and cos terms havn't become 0 or (-1)^n.

 

[knowlewj01]

Also, I have never seen that way of finding the factor outside the integration

$\omega = \frac{2 \pi}{T}$

does that work every time? just use T = period.

 

[vela]

Looks good. Here's a plot of your series using the first ten terms with L=1 and a=0.2.

 

[knowlewj01]

looks right then, thanks. How did you make that curve? Matlab, or Mathematica or something? You just reminded me that i need to get a copy of one of them :)

 

[vela]

Also, I have never seen that way of finding the factor outside the integration

$\omega = \frac{2 \pi}{T}$

does that work every time? just use T = period.

Yes, this relationship between the angular frequency and period always holds.

looks right then, thanks. How did you make that curve? Matlab, or Mathematica or something? You just reminded me that i need to get a copy of one of them :)

I used Mathematica.

 

[knowlewj01]

Thanks for Your help