What is the differance between mean value response and RMS? - Page 1
Forum section: electronics / beginners

author: ali6x944
author_rank: Frequent Contributor
timestamp: February 19, 2016, 12:29:51 pm
content: Hi guys,I own a UNI-T UT33C for a long time, and I never had a problem with it, the multimeter is measuring every thing properly , except AC voltage measurement , its seems to not respond properly to DC voltages by not detecting them , instaed the the reading is double the DC voltage aplyed to it, i read the manual and i saw this that is diplaying the mean value responce instaed of pk-pk or pk or RMS...So,What is the differance between mean value response and RMS?

author: ade
author_rank: Unknown
timestamp: February 19, 2016, 03:57:19 pm
content: Mean responding meters work only for sinusoidal waveforms. If the signal is anything other than a sine wave (e.g., a square wave, DC) then the measurements will be wrong.A "True RMS" meter will work on non-sinusoidal waveforms.

author: TimFox
author_rank: Super Contributor
timestamp: February 19, 2016, 04:53:49 pm
content: In general, an AC voltmeter can be built to respond to one of these variables:"Average responding" measures the mean absolute value of the voltage."Peak responding" measures either the peak-to-peak value (or sometimes the peak value above zero)."True RMS" measures the square root of the mean value of the voltage squared, over a suitable averaging time for the "mean" operation.Many (if not most) AC voltmeters now sold respond to the average value, but display the result (meter or digital display) applying a calibration for the RMS value of a true sine wave. This is fine for a known sinusoidal waveform, but will give an incorrect RMS value for a square wave or random noise signal (as examples). For both of those cases, if you know what the waveform is, there is a calculable correction factor from the displayed reading to the RMS value. Also, some (if not most) AC meters are capacitor-coupled, and will not measure the DC component.It gets messy if you really need the RMS value and the waveform is some unknown combination of sine wave, DC, random noise, and sine wave harmonics. For that, you need the true RMS meter.

author: helius
author_rank: Super Contributor
timestamp: February 19, 2016, 05:22:20 pm
content: Of course, just because you have a True RMS meter doesn't mean you can accurately measure noise amplitude. They all have limits on the frequency and crest factor they can detect.

author: ali6x944
author_rank: Frequent Contributor
timestamp: February 20, 2016, 08:42:55 am
content: Is there a mathimatical way to convert the reading to RMS

author: rs20
author_rank: Super Contributor
timestamp: February 20, 2016, 09:46:14 am
content: Quote from: ali6x944 on February 20, 2016, 08:42:55 amIf you know theshapeof the waveform you're measuring, and you have an accurate model of your multimeter, then yes.If you assume that the multimeter has a linear response (i.e., doubling the input signal doubles the measured value), then that calculation will be a simple scaling factor.But then again, if you have the tools required to do all this calibration (i.e., an oscilloscope), then you should just be making your RMS measurements directly with the oscilloscope.If you assume that your multimeter is not just linear, but a theoretically perfect averaging multimeter (a remarkable assumption to make given that that's as difficult and expensive as making a true RMS meter), then the conversion factors can be calculated.Finally, I would point out that people are, in general, overly obsessed with getting True RMS readings, as if True RMS is the One True Reading. One example of a specific case where RMS is appropriate is power dissipationin a resistor-- the power dissipated in a resistor is equal to Vrms^2 / R, because it's the square of the instantaneous voltage that determines the instantaneous power, and if you write out the integration that calculates average power over time, it looks similar to the integration that generates the RMS reading. If you can't explainwhyTrue RMS is the right measurement to make like this, then you're likely wasting your time making the wrong measurement.

author: ade
author_rank: Unknown
timestamp: February 20, 2016, 09:48:07 am
content: Simplistically, if you know the shape of the input signal then you can compute a correction factor.In a sine wave, the difference between the mean value and the true RMS value is a factor of 1.11. So a mean responding meter typically works by measuring the average, multiplying this value by 1.11, then displaying it as an RMS value.For a square wave, the mean value is equal to the true RMS value. But the meter assumes the input was a sine wave so it multiplies the mean by 1.11 before displaying the (incorrectly converted) RMS result. So to get the true RMS you need to divide whatever is displayed by 1.11.

author: Dave
author_rank: Super Contributor
timestamp: February 20, 2016, 01:29:00 pm
content: Quote from: ade on February 20, 2016, 09:48:07 amThis applies for the mean of thefull-wave rectifiedsine. The mean of the plain old sine is 0.The math behind this is that the mean of the full-wave rectified sine is \( \frac{2}{\pi} \cdot V_{peak} \) (about 0.637 * Vpeak) and the RMS value is \( \frac{1}{\sqrt{2}} \cdot V_{peak} \) (0.707 * Vpeak).\( \frac{0.707}{0.637} = 1.11 \)

author: ade
author_rank: Unknown
timestamp: February 20, 2016, 07:38:40 pm
content: Hence I said "simplistically".To really understand the math behind it, we'd have to delve more into the definition of RMS and why/when it is used instead of peak or rectified average.