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?

 

This might help -

https://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-pulse-and-square-waveforms/

Regards, Dana.

 

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?

 

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.

 

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.

 

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.

 

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?

 

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?

 

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?

 

What is the manufacturer and model of your multimeter?

It’s a UNI-T 139C.

 

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?

 

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.

 

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?

 

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?

 

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.

 

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)

 

TRMS meters can sometimes do DC+AC and AC readings.

 

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.

 

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?

 

Go back to Calculus:

https://en.wikipedia.org/wiki/Root_mean_square