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 $\tau=\omega t$:
$$f(t)=\frac{2}{\pi} + \sum_{k=1}^{\infty} \frac{4}{1-4k^2} \cos(2 k \omega t).$$