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#### aNxello

##### New member

- Feb 20, 2013

- 13

So I'm stuck in yet another problem, I've been going through some Fourier Series problems (including one that I already asked about around here, and go great help!) and I've managed to get past most, but I got once again stuck on one.

So for this one we got an alternating current of the form

\(\displaystyle i(t) = A\sin(\omega t)\)

that is passed through a half-wave rectifier, which the current when it is flowing in the positive direction only.

So I assumed to do a half-range sine Fourier Series

I approached it by taking the range and re-defining it, realizing that the period was \(\displaystyle 2\frac{\pi }{\omega}\) so from \(\displaystyle -\frac{\pi }{\omega}\) to \(\displaystyle \frac{\pi }{\omega}\)

So since it is only on the positive side I made it so

\(\displaystyle f(t)=\begin{Bmatrix}A\sin(\omega t) & 0 \leq t\leq \frac{\pi}{\omega } \\

0 & -\frac{\pi}{\omega } \leq t \leq 0

\end{Bmatrix}\)

However the problem also told me the answer, but the coefficients I get are nothing like the answer, this is the answer:

\(\displaystyle i_{out} = \frac{A}{\pi } + \frac{A}{2}\sin(\omega t )- \frac{2A}{\pi}\sum \frac{\cos[(n+1)\omega t ]}{n(n+1)}\)

There's also a second part, which is what would happen if the same current went through a full-wave rectifier, and for that the solution is:

\(\displaystyle i_{out} = \frac{2A}{\pi } -\frac{4A}{\pi}\sum \frac{\cos(n \omega t)}{n^{2}-1}\)

I really hope you guys can help, this is driving me crazy