So if the user specified the frequency of the unknown periodic waveform, would this be possible? Is this specified frequency the same as the frequency of the fundamental component?Just based on the information you have given I would say no. You cannot determine the frequency of the fundamental component. You did not say the input signal was band-limited by an anti-aliasing filter so you're just SOL.
If you know the frequency of the fundamental component you have a shot. You also have to be careful that high frequencies don't alias for lower frequencies. That is why you need an analog anti-aliasing filter ahead of ANY digital process you plan to apply.So if the user specified the frequency of the unknown periodic waveform, would this be possible? Is this specified frequency the same as the frequency of the fundamental component?
But frequency is important nonetheless. For example, if the signal frequency is 60Hz and your sampling frequency is 60Hz too, you'll be sampling all peaks (and overestimate RMS) or all zeroes (and underestimate). There are other frequency effects too.There is not a frequency term in the RMS formula.
So how do I proceed if this information is known?If you know the frequency of the fundamental component you have a shot.
This is the RMS of the sampled waveform, not necessairly the RMS of its fundamental component, right?Over one periodSquare the samples
Take the mean
Take the square root
Right. And if you sample this for 1000 seconds, the result will be different than if you sample it for 1500 seconds.What you are describing only occurs when the sample rate and the frequency of the unknown signal are coherent (phase locked). The OP indicated that this is not the case. If I sample a 60 Hz sine wave at a 60 Hz rate I will eventually get enough samples to accurately describe its RMS value if, and only if, the two clocks are not coherent. The time period in this case is the reciprocal of the difference in their frequency. For instance: 60.000 Hz minus 60.001 Hz = .001HZ or 1000 seconds.
No doubts. This is becuase errors are independent and autocorrelations are zero.Lets look at from a different point of view. Suppose the signal being sampled was Gaussian noise. The longer you sample, the more accurately you can approximate what the RMS value is.
Any agreement?
If the ANALOG ANTI_ALIASING FILTER that I've mentioned on multiple occasions is used there will only be the fundamental.This is the RMS of the sampled waveform, not necessairly the RMS of its fundamental component, right?
by Aaron Carman
by Aaron Carman
by Jake Hertz
by Jake Hertz