# RMS of absolute-valued sinusoid

A signal $$x(t)=\left|\sin(2\pi(1000)t)\right|$$ ($$\left|x\right|$$ is the absolute value function) is fed to an ideal low-pass filter with cutoff frequency of $$1500\,\mathrm{Hz}$$ to produce an output $$y(t)$$, what is the RMS voltage of $$y(t)?$$

My initial belief is that it will be zero since the absolute value doubles the frequency of the signal and so the frequency of $$y(t)$$ will be $$2000\,\mathrm{Hz}$$ which is beyond the cut-off frequency of the ideal low pass filter.

However, upon reflecting on the Fourier transform of $$y(t)$$, I believe it will have a DC component with zero frequency which will give it an RMS voltage. Moreover, $$y(t)$$ is a fully rectified signal which, when passed through low pass filters in reality (i.e. capacitors) will yield a smooth DC signal.

How do I find the RMS voltage of the resulting $$y(t)?$$ Is it really nonzero? Will it be just the same as the RMS voltage of $$x(t)?$$

Your reasoning is correct. The signal $x(t)=|\sin(2\pi f_0t)|$ is periodic with period $T/2$, where $T=1/f_0$. Consequently, it has components at frequencies $2k/T$, $k=0,1,\ldots$, i.e., at $0$Hz, $2000$Hz, $4000$Hz, etc. Since the ideal low pass filter has a cut-off frequency of 1500Hz, all harmonics are completely suppressed, and only the DC component appears at the output.
I'm sure that you know how to compute the DC component of $x(t)$ (which, if positive, equals its RMS value).