Fourier Series of Full Wave Rectifier

[thercias]

Homework Statement

Determine the Fourier series for the full-wave rectifier defined as
f(t) = sinωt for 0 < ωt < pi
-sinωt for -pi < ωt < 0

Homework Equations

The Attempt at a Solution

This looks like an even function, so bm = 0
Ao = 1/pi∫sinωt from 0 to pi
= 1/pi(-cos(ωt))/ω) from 0 to pi
= 2/piω

An = 2/pi∫sin(ωt)cos(nt) from 0 to pi (because the function is even)
=2/pi∫(0.5(sin(ωt-nt)+0.5(sin(ωt+nt)) from 0 to pi
=-1/pi(cos(ωt-nt)/(ω-n) + cos(nt+ωt)/(n+ω)) from 0 to pi
= -(cos(pi(ω-n))-1)/(n+ω) -(cos(pi(ω+n))-1)/(pi(n+ω))

I'm stuck at this part. I don't know how to simplify those or what that equals to and I've been looking around for a very long time trying to figure it out...any help?

 

[vanhees71]

First of all you should set \tau=\omega t. Then your calculations become right. Further you must specify with respect to which integral you integrate. Of course you are right, that it's an even function and thus it's a pure cosine series.

Then you have
a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} \mathrm{d} \tau f(\tau) \cos(n \tau)=\frac{2}{\pi} \int_0^{\pi} \mathrm{d} \tau \sin \tau \cos(n \tau)
and
f(\tau)=\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(n \tau).

 

[thercias]

ok, using that way I got

Ao = 2/pi

and An = -(cos(pi(1-n))-1)/pi(1-n) -(cos(pi(1+n))-1)/(pi(1+n))
my problem with this part is knowing what the cosine part equals out to. it seems to oscillate back and fourth from -1 to +1 as n increases
so it becomes
-(((-1)^n+1)-1)/pi(1-n)-(((-1)^n+1)-1)/pi(1+n)
this doesn't feel right though, am i doing something wrong?

 

[vanhees71]

Your coefficients are correct. Simplified and written in LaTeX makes it a bit better readable:
a_n=\frac{2}{\pi(1-n^2)}[1+\cos(n \pi)].
There's only an apparent problem at n=1, but a direct calculation of the integral gives a_1=0. Thus you have for k \in \mathbb{N}_0
a_{2k}=\frac{4}{\pi(1-4 k^2)}, \quad a_{2k+1}=0.
The Fourier series thus is
f(\tau)=\frac{2}{\pi} + \sum_{k=1}^{\infty} \frac{4}{1-4k^2} \cos(2 k \tau)
or substituting back \tau=\omega t:
f(t)=\frac{2}{\pi} + \sum_{k=1}^{\infty} \frac{4}{1-4k^2} \cos(2 k \omega t).