How to minimize RMS calculation error due to sampling?

Question

Say I have a sinusoidal signal of 49.93 Hz, which is sampled at 16 kHz. A period of the input signal would then contain 16000/49.93 = 320.448 samples. Naturally, the number of samples has to be an integer, so it is rounded to 320 samples. Say the RMS value of the continuous input wave is 100 V (volts). When I calculate the RMS value of the 320 samples, the result is 99.91407 V. The error introduced is then 99.91407V - 100V = -0.08593V.

What would be the best way to minimize this error?

Answer 1

There's many ways to approach this, but the classical "estimation of power in a system having data on harmonic oscillations" would be too just apply a quadrature mixture to baseband to it and take the RMS of the result.

Things would be easy if you knew the phase of your oscillation. If you knew the phase, you could multiply with an amplitude-one oscillation of the same phase, and get a new oscillation at twice the frequency, and at 0 frequency. You'd then eliminate the higher frequency component through filtering, and just look at the amplitude of the 0-frequency component.

Problem: you don't know the phase, usually, nor do you know the frequency that exactly.

There's two possible approaches: in the first, you just rely on the fact that analytical signals to real harmonic signals have constant complex envelope, and in the second you just recover phase and frequency.

first approach

Calculate a cos and a -sin of the same frequency with amplitude 1/2, multiply each with the observed signal, apply a low-pass filter to eliminate anything above say, half the original frequency (and especially at twice the original frequency). You get two new signals, one from the cosine-branch and one from the sine-branch of your processing. Use the cosine branch as real part and the sine branch as imaginary part.

Notice how the amplitude of the two parts depends on the phase of your original signal, but that the magnitude of the complex number that they form together is constant.

So, you just take the magnitude square of the complex signal you formed and get the square off your amplitude. Apply the mean and the root of it to that.

second approach

You could also just train an oscillator to have the same phase and frequency as the observed signal, and use that to multiply, then filter away higher frequency components to get rid of the double frequency.

That would be a Phase-Locked Loop (PLL) approach.

Answer 2

Here are some more ideas.

1. If your frequency is not drifting to much you can easily choose a better frame size. In your example you get and error of about -60dB for a frame size of 320 or 321. The "correct" value would be 320.45. If you simply use $320+321 = 641$ the error drops below -80dB. That would be an effective frame size of 320.5. If you use $11*320+9*321=6409$ the error drops below -110 dB. Obviously you need to adapt the frame size as the frequency changes but that's not that hard to do.
2. The easiest way is to simply square the signal and lowpass filter it. The squared signal has the (desired) RMS at DC plus a second spectral line at twice the signal frequency. The lowpass filter removes this line and allows dialing in the trade-off between accuracy and "agility", i.e. how fast it reacts to changes. For example, a 6th order butterworth at 5 Hz will reduce the error to a whopping -160dB but it takes about 2 seconds to get there.