# Fourier coefficients of a double frequency signal

I have a doubt regarding the fourier coefficients of a signal. I have a signal generator of sin(wt). After a single phase bridge rectifier of 4 diodes, I obtain in the output a signal equal to abs(sin(x)). As this signal is even, I know that bn coefficients of the Fourier Series decomposition will be zero. Anyway, I don't know what I should do to obtain the fundamental harmonic (in this case a2, as the frequency is twice in the output). I have two options, but I don't know which one, if any, is the valid one:

1) The first way I find to obtain it is by applying the definition, and separating the absolute value of the sine in two parts, obtaining: $$\frac{2}{2\pi}[\int_0^\pi sin wt *cos 2wt dw - \int_{\pi}^{2\pi} sin wt *cos 2wt dw]$$

2) As the frequency in the output is the double, compute the following: $$\frac{2}{\pi}\int_0^\pi sin wt *cos 4wt dw$$ Considering that now the expression would impy $$cos(2nwt)$$ due to the double frequency in the output.

Which one would be the valid one? Any help is appreciated!

• Too much hand-waving. I can't follow what it is you're trying to do, or why you think that the second expression is in any way equivalent to the first. Try sketching the graphs of their components to see what I mean. – Dave Tweed May 19 '17 at 2:13
• The fundamental frequency is $\small 2\omega$ but in the second expression you have $\small 4\omega$, which is the 2nd harmonic. Also, the integrals should be wrt $\small \omega t$, not $\small \omega$ – Chu May 19 '17 at 8:36