How do you find the fourier expansion coefficients?

[Vitani11]

Homework Statement

I need to expand this piecewise function f(x) = h for a<x<L and f(x) = 0 for 0<x<a. I am told that this is a square wave so a_o and a_n in the expansion are 0 (odd function). Therefore I only need to worry about b_n. The limits on the integral are from a to L, but what about the coefficient? Is it 1/(L-a)? Also for the sine term which is inside the integrand - is this just (L-a) in replace of the L in the general formula?That's what I did and I want to make sure.

Homework Equations

b_n = 1/L∫f(x)sin(nπx/L)dx where the limits are from -L to L in general.
I have b_n = 1/(L-a)∫hsin(nπx/(L-a))dx where the limits are from a to L.

The Attempt at a Solution

I've found the Fourier expansion to be 4hL/π(L-a) ∑sin(nπx/(L-a))

 

[Ken Miller]

It looks like you are stating some things that aren't true, and plugging some values into formulas without really understanding how the formulas work.
First questions:
1) Can you verify for yourself whether this is a square wave? What is the definition of a square wave? Does it matter?
1a) Is this really a square wave? It is a rectangular wave, yes, but square? I think many (most?) would say that a square wave has equal amounts of on/off (high/low, etc.). 1b) Does the answer to 1a) matter?
2) Is this an odd function? Draw it out, and use the definition of odd function.
3) Do the answers to these questions affect your calculations?
4) If the limits of integration are generally (-L, L), how many periods of the waveform would be included?

 

[eys_physics]

Odd function means $f(-x)=-f(x)$. How do you know that is odd? It is not obvious from the information you give.