Integration by parts (Fourier Series)

[mrjoet]

Hi,

I'm trying to work out the Fourier series coefficients for a triangular wave form.

So far I have got as far as:

$\displaystyle C_n = \frac{1}{T} \left( \int_0^\frac{T}{4} \frac{4At}{T}e^{-j2π\frac{n}{T}t} dt + \int_\frac{T}{4}^\frac{3T}{4} \left( 2A - \frac{4At}{T} \right) e^{-j2π\frac{n}{T}t} dt + \int_\frac{3T}{4}^T \left( \frac{4At}{T} - 4A \right) e^{-j2π\frac{n}{T}t} dt \right)$

I try integration by parts but seem to get lost, I know I should end up with:

$\displaystyle C_n = \frac{A}{π^2 n^2} \left( 2e^{-jπ\frac{n}{2}} - 1 - 2e^{-jπ\frac{3n}{2}} + e^{-j2πn} \right)$

Is anyone able to show how to get to the answer? Thanks in advance!

 

[Cubist]

Quite a monster! If I was doing this I'd split up terms like this...

$\displaystyle \int_\frac{T}{4}^\frac{3T}{4} \left( 2A - \frac{4At}{T} \right) e^{-j2π\frac{n}{T}t} dt$

in the following way...

$\displaystyle =\int_\frac{T}{4}^\frac{3T}{4} 2A e^{ct} dt - \int_\frac{T}{4}^\frac{3T}{4} \frac{4At}{T} e^{ct} dt \,$ where $\displaystyle c=-j2π\frac{n}{T}$

$\displaystyle =2A \int_\frac{T}{4}^\frac{3T}{4} e^{ct} dt - \frac{4A}{T} \int_\frac{T}{4}^\frac{3T}{4} t e^{ct} dt$

...so that the whole integration is broken down into two basic integration problems

$\displaystyle \int e^{ct} dt \,$ and $\displaystyle \, \int t e^{ct} dt$

Obviously it will be a bit messy/ long so be careful as you go. If you're still struggling please post your work. (I have not checked that the answer can be obtained from your equation)

 

[mrjoet]

Thanks for the tip! I'll have another go this weekend and post my work!