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I tried to solve this myself, but I'm forced to beg for help.

I have a very short waveform, too short to resolve accurately on my scope. When I expand the waveform, the RMS of one complete pulse is 6.6 Vrms, 0.000 001 2 seconds long.

What would the RMS be if the time period was extended to 0.000 254 seconds long (with the same single pulse)?

Seems like it should be a simple calculation, but I can't seem to make it work. I searched the web for a calculator, but no joy.

The answer appears to be as follows...

A 6.6 Vrms 0.0000012 second pulse, in a 0.0000012 second (833,333.33 Hz) time period would be 6.6 Vrms because the pulse fills the time period 100%.

A 6.6 Vrms 0.0000012 second pulse in a 0.000254 second (3,3937 Hz) time period should be approx 0.399 Vrms.

Is this correct?

ocrdu
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William Mingin
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    Could you show us the waveform? Did you use a specific circuit to make this waveform? – Christianidis Vasilis Jan 26 '21 at 16:16
  • Both area and period are stretching by the same factor, so RMS value doesn't change. – across Jan 26 '21 at 16:19
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    The RMS of any repetitive waveform doesn't depend on time. – Andy aka Jan 26 '21 at 16:21
  • I suspect that you are using RMS calculation for something where it doesn't mean much. It's only meaningful on a repetitive waveform. If the pulse is 1.2 μs long, what is the periodic time? – Transistor Jan 26 '21 at 16:58
  • Yeah if you could generate some graphics showing the zoomed in view that would help give a better answer. My take on the comments so far is that they don't understand your question. Some pictures would help clarify. Either pictures or sketches or whatever you can manage. – user57037 Jan 26 '21 at 17:26
  • Note that in engineering we use milli (m)and micro (μ or u) so that we don't need so many decimal places. '1.2 μs' and '254 ms' is so much easier to read. – Transistor Jan 26 '21 at 19:07
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    Is the signal actually periodic at the slower rate but you need to zoom in so much for the scope to measure the value? – Justin Jan 26 '21 at 19:44

4 Answers4

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I have a very short waveform, too short to resolve accurately on my scope.

So, you don't actually know what the shape of the waveform is with any accuracy, since you can't see it. How would you calculate the RMS value of a signal you know nothing about.

When I expand the waveform, the RMS of one complete pulse is 6.6 Vrms, 0.000 001 2 seconds long.

So, now somehow you are able to resolve the waveform accurately? Where did the Vrms value come from now? Are you perhaps talking about a literal square pulse that has 6.6Vpp amplitude and 12μs length?

Nothing about the question makes much sense to me.

The answer appears to be as follows... A 6.6 Vrms 0.0000012 second pulse, in a 0.0000012 second (833,333.33 Hz) time period would be 6.6 Vrms because the pulse fills the time period 100%.

The RMS value is always measured over some time period, and you're only converting between two values by scaling them with time. The original RMS value must also be attached to a time period. It has no meaning otherwise if the signal is non-periodic, like a single pulse would be.

An aperiodic signal with 6.6Vrms calculated over 1.2μs interval will maintain its rms value if you "translate" it to... the same time interval. That's by definition - nothing much to see here.

A 6.6 Vrms 0.0000012 second pulse in a 0.000254 second (3,3937 Hz) time period should be approx 0.399 Vrms.

A 0.000'254s=254μs period is equivalent to the frequency of 1/254μs=3937Hz, not "3,3937". It's much better to type using reasonable metric magnitude suffixes and avoid too many decimals, whether zero or otherwise. It's too easy to have typos otherwise.

A 6.6Vrms aperiodic signal measured over a 12μs interval, with zero value outside of the interval, will measure \$6.6\cdot(12/254)=0.31{\rm\,V}\$ over the interval of 254μs, not 0.4V as you stated.

Kuba hasn't forgotten Monica
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A single pulse (over infinite time) has an RMS value of zero.

Stefan Wyss
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The RMS value of a sampled signal (coming from a scope) is similar to a standard deviation with no mean, so if you wanted to export the data to a math package and take the standard deviation.

enter image description here
Source: https://en.wikipedia.org/wiki/Standard_deviation

Now set \$\mu\$ (the mean) to zero and you get... RMS

enter image description here
Source: https://en.wikipedia.org/wiki/Root_mean_square

But this won't make very much sense, because the RMS value will depend on the sampling and the window size, if the window size is increase, the value will change.

This is because of the integration, and the scale is changing which might explain why your not getting meaningful results.

This doesn't happen for periodic signals, as integrating over longer periods of time does not change the RMS value.

I personally would stick with peak to peak to describe a short pulse.

Voltage Spike
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The RMS value is the height of an rectangle with the same area as that between signal and zero line over the given sample time. If your signal is periodic, the sample time does not matter unless it's an integer multiple of the period length.

For non-periodic signals, you have to tell RMS multiplied with the sample time —the area of the rectangle— instead. It is only useful if you always apply the same noise level to tell the start and end of the signal.

Janka
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