Hello,

How can I find the peak-to-peak voltage of a square wave produced by a function generator when I only have a True-RMS multimeter?

I'm new at this so if someone could dumb it down for me, I'd appreciate it.

Is it possible? Thanks.

 

For a square-wave, RMS is the peak voltage.

Or do you need it for duty cycles other than 50%?

ETA: RMS - Peak Calculator for different waves

 

and hence the peak-to-peak voltage would be twice the RMS reading.

 

For a square-wave, RMS is the peak voltage.

Or do you need it for duty cycles other than 50%?

ETA: RMS - Peak Calculator for different waves

No 50% duty is all I need. It says positive square wave. I forgot to mention that it is a +/- square wave equally bipolar. So if it was 10Vpp, it would be +5, -5V...if that makes sense.

Is duration (T) one period?

Thank you.

 

and hence the peak-to-peak voltage would be twice the RMS reading.

The reading of a multimeter is RMS, but since it's meant to give the RMS value of a sine wave, wouldn't it screw something up when its a square wave?

 

Yes, T is usually the period, or \(\frac{1}{f}\) where f is frequency.

 

The reading of a multimeter is RMS, but since it's meant to give the RMS value of a sine wave, wouldn't it screw something up when its a square wave?

If the meter is TRUE - RMS and marked that way, then it will give you the correct RMS voltage of any waveform (up to the frequency ability of the meter)

If the meter cost under $40-$50 or so, it is probably only True RMS for sinewaves, and the actual voltage of a different waveform is anybody's guess in the case of an Sine RMS only meter.

 

The RMS value is what constant DC voltage will give the same heating effect as the signal of arbitrary wave form. Hence mathematically, you have to compute the integral of the square of the amplitude and then take the square root. So when you square the negative portion, it is the same value as the positive part. Hence the RMS is the same as if the square wave was a constant DC signal of half the peak-to-peak voltage.

 

If the meter is TRUE - RMS and marked that way, then it will give you the correct RMS voltage of any waveform (up to the frequency ability of the meter)

If the meter cost under $40-$50 or so, it is probably only True RMS for sinewaves, and the actual voltage of a different waveform is anybody's guess in the case of an Sine RMS only meter.

http://www.amprobe.com/amprobe/usen/Multimeters/Industrial-Multimeters-/AM-270.htm?PID=73125

I'll have to look at the datasheet for my meter then.

Thank you.

 

That meter is a True RMS meter, meaning it will give you the RMS value of any waveform.

It may also give you the peak voltage if you set it to peak record.

 

The RMS value is what constant DC voltage will give the same heating effect as the signal of arbitrary wave form. Hence mathematically, you have to compute the integral of the square of the amplitude and then take the square root. So when you square the negative portion, it is the same value as the positive part. Hence the RMS is the same as if the square wave was a constant DC signal.

I got the first part about the heating. (Actually learned that in school!). Second part not so much..Let's see. I'll read it again.

 

Draw a square wave centered above and below 0V. So in your example of 10V p-p, you end up with 5V above and 5V below.
Now flip the negative portion to the positive side. You end up with a DC voltage equal to 5V, half the peak-to-peak value.

A true RMS reading voltmeter would give the same reading as the 5V DC voltage if they were to have the same heating effect.

 

Hello,

How can I find the peak-to-peak voltage of a square wave produced by a function generator when I only have a True-RMS multimeter?

I'm new at this so if someone could dumb it down for me, I'd appreciate it.

Is it possible? Thanks.

Be careful. Most modern "true RMS meters" do some kind of integral approxination and are only accurate on signals with a crest factor less than 3. The phrase "True RMS" must be taken with a grain of salt, unless it is a thermal meter and they are rare and very expensive.

 

Be careful. Most modern "true RMS meters" do some kind of integral approxination and are only accurate on signals with a crest factor less than 3. The phrase "True RMS" must be taken with a grain of salt, unless it is a thermal meter and they are rare and very expensive.

The crest factor of a square-wave is 1, so that should not be a problem for the meter.

 

The other problem is bandwidth limitation. The leading edges of a square wave have very high frequency content. It would be like using a scope with a bandwidth of about 50 kHz to look at a square wave. The vertical edges would be "leaning" more like a triangle wave.

 

I'm going to have to try that with my meter sometime. It's a true RMS with min, max and average (Fluke 89 IV). I know that mathematically a 10Vpp square wave of 50% duty cycle should read 5Vrms, min = -5V, max = 5V and average = 0V but it would be interesting to verify.

 

Note however that the frequency range is quite limited: accuracy is only specified from 50Hz into the audio range, and degrades considerably above 500Hz. For square waves, the effective frequency range may be reduced as thy contain a large content of higher harmonics which may fall outside the meter bandwidth. See data sheet: http://content.amprobe.com/DataSheets/AM270.pdf

This meter cannot therefore really be expected to give accurate results for square waves above the lower audio range.