RMS measurement error with square wave - Page 1
Forum section: electronics / projects

author: taydin
author_rank: Frequent Contributor
timestamp: February 24, 2020, 08:04:52 pm
content: Hello guys. I'm generating a square wave from my signal genarator that is 100 Hz and 2 Vpp. The square wave is bipolar and the positive and negative peaks have the same magnitude. Then I attach this signal to my scope and make the following three measurements:1) Duty cycle is 10%2) Duty cycle is 50%3) Duty cycle is 90%And I'm measuring 1 Vrms at the scope. No problems so far.Then I do the same three measurements with my Keysight U1282A true rms multimeter. With 50% duty cycle, the measurement is bang on. But with both 10% and 90%, the measurement is about 0.6 Vrms. With 20% and 80%, the measurement is about 0.8 Vrms. So what gives?PS: Please don't get hung up on the usage of the "duty cycle" here, I'm just trying to explain the relative duration of the positive part and the negative part of the square wave.

author: taydin
author_rank: Frequent Contributor
timestamp: February 24, 2020, 08:05:43 pm
content: I have tried this with other multimeters I have and all exhibit about the same measurement error.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 24, 2020, 08:13:10 pm
content: A square wave with other than 50% duty cycle comtains a DC part. For the RMS readings there is sometime a confusion in showing only the AC part or the combined AC and DC value. So both readings can be correct just different things. 1 V für AC +DC and 0.6 V for the AC coupled part. For DMMs it is normal to show the AC coupled RMS, so without the DC.

author: macboy
author_rank: Super Contributor
timestamp: February 24, 2020, 08:15:45 pm
content: Your scope is set to DC coupled (which is fine) but the multimeters are all using AC coupling. Set the scope to AC coupled input and look at what the RMS measurements become in that case. Note that the waveform shifts with repect to ground because it isn't equal positive vs negative duration.Some higher end multimeters can use DC coupled input when measuring AC (also sometimes called AC+DC mode). This should give more accurate results in this case.

author: taydin
author_rank: Frequent Contributor
timestamp: February 24, 2020, 08:19:07 pm
content: Thanks a lot for the responses guys. This makes perfect sense. I vaguely remember that the U1282A can handle AC+DC, but not sure in what context. Time to read up on the manual.

author: vk6zgo
author_rank: Super Contributor
timestamp: February 25, 2020, 12:58:42 am
content: Quote from: Kleinstein on February 24, 2020, 08:13:10 pmA "square wave" with other than a 50% duy cycle is no longer a square wave ------- It is a rectangular wave.ln any case, "duty cycle" is not an appropriate term for a rectangular waveform which is bipolar.Duty cycle applies to rectangular waves which return to zero volts for some proportion of each cycle.("unipolar")A bipolar rectangular wave is able to produce power for the whole cycle if applied to a resistive load, a unipolar rectangular wave cannot.In each case, the waveform's RMS voltage is the equivalent DC voltage which would be necessary to dissipate the same power in a resistive load.A RMS meter which cannot distnguish between unipolar & bipolar waveforms will get at least one of them wrong.

author: SiliconWizard
author_rank: Super Contributor
timestamp: February 25, 2020, 01:23:49 pm
content: Quote from: macboy on February 24, 2020, 08:15:45 pmYes.Refer to what RMS means (integral).https://en.wikipedia.org/wiki/Root_mean_squareFor a "square" signal with equal positive and negative amplitude, whatever the "duty cycle" (if we can call it that), we'll get the exact same value as this is the square root of the integral of f(t)^2 divided by the time period, so the duty cycle doesn't matter: if your amplitude is A (signal going from +A to -A), then f(t)^2 = A^2 for all t, and thus the RMS value is |A|. Of course this is for an ideal "square" wave (a step function.)If the negative and positive phases have different absolute amplitude (which may happen depending on the coupling mode), which we can also translate to a non-zero DC offset, then of course the result is completely different.

author: bson
author_rank: Unknown
timestamp: February 26, 2020, 12:06:49 am
content: Look at the definition of standard deviation:$$\sigma = \sqrt{\frac{1}{N}{\sum_{i=1}^N{(x_i-\bar{x})}^2}}$$Then look at RMS:$$rms = \sqrt{\frac{1}{N}{\sum_{i=1}^N{x_i}^2}}$$The difference is that std dev subtracts the mean, so is deviation around the mean.RMS is implicitly deviation around 0.When you have a DC component your measurement is no longer centered around 0.