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?

As I know, maybe I m wrong, the duty cycle only changes the rise and fall time, changing the form of the wave, not the voltage value. Duty cycle is a ratio of rise time versus fall time.

 

Duty cycle is the % on time or (time on)/(period).

The RMS value will vary with the DC offset, the offset to peak value and the duty cycle.

Another way of looking at the RMS value of a waveform is What DC voltage would be applied to a restive load that would diddipate the same power as the waveform of interest.

 

Hi,

The best bet is to go back to the definition. The name RMS actually is an acronym for what we have to calculate to get the RMS value of any waveform.

RMS stands for "the Root of the Mean of the Square".
Since this is written out sort of backwards, we calculate using that backwards so first we square the wave, then take the mean, then find the square root.

To square any wave in time simply means we start with a general wave:
v(t)

then square it:
v(t)^2

then find the mean (Tp is the total period):
integrate(v(t)^2,t)/Tp

then take the square root.

Now with a square or rectangular wave the 'on' time is constant, so we are dealing with a voltage that is constant for part of the time. This is depicted as:
y=V

So you would square that, then find the mean, then take the square root, and that would be the RMS value for a rectangular wave which is all above zero.

Now the reason i gave the definition from the acronym was because this changes with offset. The wave would actually be:
y=V+V(0)

where V(0) is the offset.

There are shortcuts around that help calculate this stuff for various waves like triangle, sine, etc. But once you run into a special case you need to apply the definition.

For a quick example, if you have a sine wave with zero offset the shortcut is to divide the peak value by the square root of 2. That shortcut is only valid though for the sine wave with zero offset though, so dont try to apply it to any other type of wave except a 50 percent duty cycle square wave with no offset (all above zero).
For example, for a sine wave with peak voltage Vpk and offset v0 we get:
Vrms=sqrt(Vpk^2+2*v0^2)/sqrt(2)

so you can see that the offset v0 changes the result.

 

Hi,

The best bet is to go back to the definition. The name RMS actually is an acronym for what we have to calculate to get the RMS value of any waveform.

RMS stands for "the Root of the Mean of the Square".
Since this is written out sort of backwards, we calculate using that backwards so first we square the wave, then take the mean, then find the square root.

This is not the definition, but rather just a useful descriptor of how to calculate the value that follows from the definition.

The definition is that the effective voltage of a waveform is the DC voltage that, if applied to a resistive load, would deliver the same average power to that load as the waveform being considered. Applying this definition yields the result that the effective voltage is the square root of the mean of the squared-voltage of the waveform.

 

Well i would call that the mathematical expression but it does serve as a definition because that is a direct result of equating the power of an unknown current wave in a resistor R to the power caused by a DC current in the same resistor value R.

 

Well i would call that the mathematical expression but it does serve as a definition because that is a direct result of equating the power of an unknown current wave in a resistor R to the power caused by a DC current in the same resistor value R.

If A is defined as one thing, we can't then go and say that it is a direct result of something else. Either it is what it is because we defined it that way or it is what it is because it is a consequence of something else. We can't have it both ways.

 

This is not the definition, but rather just a useful descriptor of how to calculate the value that follows from the definition.

The definition is that the effective voltage of a waveform is the DC voltage that, if applied to a resistive load, would deliver the same average power to that load as the waveform being considered. Applying this definition yields the result that the effective voltage is the square root of the mean of the squared-voltage of the waveform.

The TS in post #1 says: "My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS? "

He refers to the definition of RMS, not the definition of "effective".

In post #23 when MrAl says: "The best bet is to go back to the definition. ", it's reasonable to assume that he's referring to the same definition the TS referred to, namely the definition of RMS, which he explains in detail.

The word "effective" is not used in this thread until you used it in post #24.

It just so happens that the definition of RMS, which is different than the definition of "effective", also gives the "DC voltage that, if applied to a resistive load, would deliver the same average power to that load as the waveform being considered."

 

The TS in post #1 says: "My question is, if my multimeter reading is correct, how would I get to that value by the definition of RMS? "

He refers to the definition of RMS, not the definition of "effective".

In post #23 when MrAl says: "The best bet is to go back to the definition. ", it's reasonable to assume that he's referring to the same definition the TS referred to, namely the definition of RMS, which he explains in detail.

The word "effective" is not used in this thread until you used it in post #24.

It just so happens that the definition of RMS, which is different than the definition of "effective", also gives the "DC voltage that, if applied to a resistive load, would deliver the same average power to that load as the waveform being considered."

Okay, then how does going back to the definition of RMS have anything to do with getting the value shown on the multimeter unless and until you show how the two are related?

 

Okay, then how does going back to the definition of RMS have anything to do with getting the value shown on the multimeter unless and until you show how the two are related?

What are you referring to when you say "...the two..."?

 

Duty cycle is the % on time or (time on)/(period).

The RMS value will vary with the DC offset, the offset to peak value and the duty cycle.

Another way of looking at the RMS value of a waveform is What DC voltage would be applied to a restive load that would diddipate the same power as the waveform of interest.

I ignored, as many other things in the life, that duty cycle is the % of on time. But left more doubts than clear ideas. 1- that applies only for square wave forms or applies for and triangle and/or sawtooth. If applies for triangle and sawtooth 2- what is the on time, rise or fall?

 

I ignored, as many other things in the life, that duty cycle is the % of on time. But left more doubts than clear ideas. 1- that applies only for square wave forms or applies for and triangle and/or sawtooth. If applies for triangle and sawtooth 2- what is the on time, rise or fall?

Hi,

Sometimes the 'on' time of a triangle or sawtooth is whatever time the wave is non zero but for a continuous triangle or sawtooth the 'duty cycle' would be considered 100 percent. These kinds of waves often require going back to the actual calculation of RMS which is the "Root of the Mean of the Square" talked about previously. There are no short cuts for some waves that's what you have to realize now.

If you dont like integrating in:
RMS=sqrt(integrate(v(t)^2,t,0,T)/T)

then what you can do is divide the wave up into small pieces and calculate the individual quantities then sum them and get the RMS value. This might be approximate but it is another method. We can cover that if you like.

 

Okay, then how does going back to the definition of RMS have anything to do with getting the value shown on the multimeter unless and until you show how the two are related?

Hello again,

You mean we have to go back to the equivalent definition of effective before we can measure anything in real life and calculate the RMS value?

Try this:
Do a Google search of "Definition of RMS Voltage".
You dont have to search for "Definition of Effective Voltage".

But ultimately i think what you are trying to do is subsume the definition of RMS voltage or current with the definition of 'effective' voltage or current and that is understandable. That is more or less the physical view as compared to the mathematical view. However, if we are going to go back to basics then cant we also then subsume that definition with an even more general definition and say that's the real basis of RMS and also Effective?
The more general definition would include everything physical not just electrical where electrical is just a part of the entire physical realm.

Many people that ask about RMS values however just want to know about the Root of the Mean of the Square and how to calculate it.

 

Hello again,

You mean we have to go back to the equivalent definition of effective before we can measure anything in real life and calculate the RMS value?

Try this:
Do a Google search of "Definition of RMS Voltage".
You dont have to search for "Definition of Effective Voltage".

But ultimately i think what you are trying to do is subsume the definition of RMS voltage or current with the definition of 'effective' voltage or current and that is understandable. That is more or less the physical view as compared to the mathematical view. However, if we are going to go back to basics then cant we also then subsume that definition with an even more general definition and say that's the real basis of RMS and also Effective?
The more general definition would include everything physical not just electrical where electrical is just a part of the entire physical realm.

Many people that ask about RMS values however just want to know about the Root of the Mean of the Square and how to calculate it.

you mean effective and RMS are not the same terms? That s it? if so, What are the differences?

 

Hi,

I knew this was going to happen

Start with the average power in a constant resistance R:
\(
Pavg=(\frac{1}{T}){\int_{0}^{T}{\mathrm{i}\left( t\right) }^{2}\,R\,dt}
\)

and the power in the same resistor R using a DC current:
\(
P={I}_{eff}^{2}\,R
\)
(note we call the current the 'effective' current with the presumption that it will be shown to be true later in the discussion)

Equate the two:
\(
(\frac{R}{T}){\int_{0}^{T}{\mathrm{i}\left( t\right) }^{2}dt}={I}_{eff}^{2}\,R
\)

Solve for \({I}_{eff}\):
\(
{I}_{eff}=\sqrt{(\frac{1}{T}){\int_{0}^{T}{\mathrm{i}\left( t\right) }^{2}dt}}

\)

and that is what we call the RMS value of the current because we take the root of the mean of the square of the current wave.
So Irms is the (square) root of the mean of the square (of the current)

\(
{I}_{RMS}=\sqrt{(\frac{1}{T}){\int_{0}^{T}{\mathrm{i}\left( t\right) }^{2}dt}}
\)

"Ieff" is the physical phenomenon where the power heating is observed to be the same for the DC current as the AC current when the DC current is the effective current of the AC current wave.
"Irms" is the mathematical definition of the RMS current.
The mathematical expressions are the same for both. There are other ways to calculate Irms but the integral expression is the most concise and mathematically exact.

A side note is that the final expression does not contain the resistance R used in the calculation. That is because the value of R does not matter we use the same R for both measurements.

An experiment would go as follows...
Find two resistors the same value and same power rating.
Apply the AC current to be measured to the first resistor. Measure the temperature of the first resistor call it T1.
Apply an adjustable DC current to the second resistor. Measure the temperature call it T2.
Adjust the DC current such that T2=T1 after waiting some time between adjustments.
After some time has passed when T2=T1 the DC current applied is the same as the effective current of the AC wave.

The RMS current stated in pure text is:
Irms=sqrt[(1/T)*integrate(i(t)^2,t,0,T)]