RMS Voltage/Duty-Cycle

bernardomarques

Joined Nov 14, 2019
7
Hi,

I have a 100Hz 1V (peak-to-peak) square wave on a function generator, I have my true RMS multimeter plugged into it and I meausure around 0.5 V_RMS. So far so good!
Now I change it’s duty-cycle to 10% and I measure around 0.3 V_RMS.

Does this make sense?

My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS?

MrAl

Joined Jun 17, 2014
6,643
Hi,

I have a 100Hz 1V (peak-to-peak) square wave on a function generator, I have my true RMS multimeter plugged into it and I meausure around 0.5 V_RMS. So far so good!
Now I change it’s duty-cycle to 10% and I measure around 0.3 V_RMS.

Does this make sense?

My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS?
Hi,

Where is your zero voltage point on each wave?

bernardomarques

Joined Nov 14, 2019
7
Hi,

Where is your zero voltage point on each wave?
On my first wave the 0V was right in the middle, so the positive and the negative values were symmetrical.

Now on the second one, the 10% duty-cycled wave, my teacher suggested that the 0 voltage point would change.

Something to do with some default setting of the function generator.

I thought it would stay the same but apperently it didn’t.

Thank you.

WBahn

Joined Mar 31, 2012
24,854
Hi,

I have a 100Hz 1V (peak-to-peak) square wave on a function generator, I have my true RMS multimeter plugged into it and I meausure around 0.5 V_RMS. So far so good!
Now I change it’s duty-cycle to 10% and I measure around 0.3 V_RMS.

Does this make sense?

My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS?
By doing the math.

On the first waveform is appears that the DC was right in the middle of the wave, which is what you would expect if the output is AC-coupled.

So assume that it is AC-coupled (and what does this mean mathematically) and do the math.

Without even needing a calculator you can determine that your measured result is very close to what you would expect.

Now you take a shot at deriving the expected result.

MrAl

Joined Jun 17, 2014
6,643
On my first wave the 0V was right in the middle, so the positive and the negative values were symmetrical.

Now on the second one, the 10% duty-cycled wave, my teacher suggested that the 0 voltage point would change.

Something to do with some default setting of the function generator.

I thought it would stay the same but apperently it didn’t.

Thank you.
Hi,

That makes sense because many function generators even the old style analog ones had a zero offset adjustment where you can change the zero point of the sine, square, or triangle.
You could for example have the square wave all below zero or all above zero or anywhere in between.

A key point is that if the wave is entirely below or above zero by some non zero amount the RMS value will change dramatically because there is a significant DC component which has it's own RMS value that adds to the total.

bernardomarques

Joined Nov 14, 2019
7
By doing the math.

On the first waveform is appears that the DC was right in the middle of the wave, which is what you would expect if the output is AC-coupled.

So assume that it is AC-coupled (and what does this mean mathematically) and do the math.

Without even needing a calculator you can determine that your measured result is very close to what you would expect.

Now you take a shot at deriving the expected result.
Ok! I did assume that it was AC coupled but like I said before my teacher suggested that 0V point had now changed when the duty-cycle was inserted. Something to do with the default setting of the function generator (BK 4055B).

Would calculating the V_avg (assuming AC coupling) help?

RBR1317

Joined Nov 13, 2010
502
Something to do with the default setting of the function generator (BK 4055B).
If you can't trust the calibration of the function generator, does your lab have a DC-coupled scope available to determine what the function generator is actually doing?

The Electrician

Joined Oct 9, 2007
2,750
Hi,

I have a 100Hz 1V (peak-to-peak) square wave on a function generator, I have my true RMS multimeter plugged into it and I meausure around 0.5 V_RMS. So far so good!
Now I change it’s duty-cycle to 10% and I measure around 0.3 V_RMS.

Does this make sense?

My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS?
What is the manufacturer and model of your multimeter?

bernardomarques

Joined Nov 14, 2019
7
What is the manufacturer and model of your multimeter?
It’s a UNI-T 139C.

bernardomarques

Joined Nov 14, 2019
7
Ok! I did assume that it was AC coupled but like I said before my teacher suggested that 0V point had now changed when the duty-cycle was inserted. Something to do with the default setting of the function generator (BK 4055B).

Would calculating the V_avg (assuming AC coupling) help?
Is this a valid way to do it?

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MrAl

Joined Jun 17, 2014
6,643
Hello again,

If the signal is AC coupled then it may not matter what you set the offset to. It only matters when the offset causes the wave to clip either on the high peak or the low peak.
For example, let's say the max output peak adjustment is +10v and -10v with the offset at 0v.
If you set it for +5v and -5v and set the offset to +1v, then the output goes from -4v up to +5v. But if you set the offset for +6v then the output should go from +1v to +11v but because the max is +10v it will only go from +1v to +10v so we lose 1v due to clipping which of course means the RMS voltage changes from what it would have been if it could have gone from +1v to +11v.

Now when AC coupled and set to go plus and minus 5v with the offset set to +6v we still lose the upper end even though the capacitor used for the AC coupling causes the wave to be centered at 0v.

So yeah it should be checked with a scope to make sure the wave is what you think it should be.

WBahn

Joined Mar 31, 2012
24,854
Ok! I did assume that it was AC coupled but like I said before my teacher suggested that 0V point had now changed when the duty-cycle was inserted. Something to do with the default setting of the function generator (BK 4055B).

Would calculating the V_avg (assuming AC coupling) help?
Yes, the 0 V point of the waveform, relative to the min and max voltages of the waveform has shifted because the AC coupling removes the DC component, which means that the average voltage of the waveform is zero.

So where does the 0 V point need to be on the waveform in order for the DC component to be 0 V?

bernardomarques

Joined Nov 14, 2019
7
Yes, the 0 V point of the waveform, relative to the min and max voltages of the waveform has shifted because the AC coupling removes the DC component, which means that the average voltage of the waveform is zero.

So where does the 0 V point need to be on the waveform in order for the DC component to be 0 V?

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WBahn

Joined Mar 31, 2012
24,854
Is this a valid way to do it?
It looks like you have your second waveform going from -0.7 V to +0.9 V. I thought your waveform was 1.0 V peak-to-peak.

bernardomarques

Joined Nov 14, 2019
7
It looks like you have your second waveform going from -0.7 V to +0.9 V. I thought your waveform was 1.0 V peak-to-peak.
-0.1 (sorry my writing can be confusing sometimes)

KeepItSimpleStupid

Joined Mar 4, 2014
3,640
TRMS meters can sometimes do DC+AC and AC readings.

VICTORR

Joined Mar 18, 2019
4
Hi,

I have a 100Hz 1V (peak-to-peak) square wave on a function generator, I have my true RMS multimeter plugged into it and I meausure around 0.5 V_RMS. So far so good!
Now I change it’s duty-cycle to 10% and I measure around 0.3 V_RMS.

Does this make sense?

My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS?
I think it make sense because the rms value is 0.707 the first value. I mean 5V x 0.707 = 0.3535V. I hope this help.

WBahn

Joined Mar 31, 2012
24,854
I think it make sense because the rms value is 0.707 the first value. I mean 5V x 0.707 = 0.3535V. I hope this help.
You need to revisit these relationships.

To get the RMS value of one waveform, why would you multiply the RMS value of some other waveform by 1/√2?

You are also implying that the duty cycle has no influence on the result since you didn't feel it necessary to take it into account. Well, if that's true, then should you do the same thing for duty cycles of 20%, 25%, 45%, and even 50%? But that would mean that the RMS voltage of a 50% duty cycle waveform is 0.707 times the RMS voltage of a 50% duty cycle waveform. See the problem?