AC - DC estimator

[TooEagerToLearn]

Hello again,

I am trying to do a spectral analysis one 2 AC signals and have another question, so I decided to start another post because it is about a different vi. I am using RMS.vi to find the RMS value of current and voltage. How is that different from AC-DC estimator? Thanks!

[Bob_Schor]

To answer the difference between RMS and AC/DC estimator, consider a 1 v DC level with a 0.5v square wave riding on top of it.  The DC level is 1v, and the AC level (obtained after subtracting the DC) is sqrt(sqr(0.5)+sqr(0.5)) = sqrt(0.5) = 0.707.

The RMS is sqrt(sqr(1.5)+sqr(0.5)) = sqrt(2.5) = 1.581.  Note RMS is not DC + AC.

BS

[TooEagerToLearn]

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Bob_Schor wrote:

To answer the difference between RMS and AC/DC estimator, consider a 1 v DC level with a 0.5v square wave riding on top of it.  The DC level is 1v, and the AC level (obtained after subtracting the DC) is sqrt(sqr(0.5)+sqr(0.5)) = sqrt(0.5) = 0.707.

The RMS is sqrt(sqr(1.5)+sqr(0.5)) = sqrt(2.5) = 1.581.  Note RMS is not DC + AC.

BS

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THank you for that response Bob! Couple follow up questions/comments.

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To answer the difference between RMS and AC/DC estimator, consider a 1 v DC level with a 0.5v square wave riding on top of it.  The DC level is 1v, and the AC level (obtained after subtracting the DC) is sqrt(sqr(0.5)+sqr(0.5)) = sqrt(0.5) = 0.707.

When you say 0.5V riding on 1V DC level, Does that mean RMS of AC is 1V and Vpeak will be sqrt(2) * 1V = 1.414V ? Or Vpeak is 1 + 0.5 = 1.5V ? And what does it mean when you say AC level? I thought that RMS of an AC signal is the DC level which would be LESS than AC Level.

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The RMS is sqrt(sqr(1.5)+sqr(0.5)) = sqrt(2.5) = 1.581.  Note RMS is not DC + AC.

I am not sure how you go this? Seems like you have added 0.5 to 1 and subtracted 0.5 from 1. Vpk-pk in this case is 2V?

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[Bob_Schor]

To answer the difference between RMS and AC/DC estimator, consider a 1 v DC level with a 0.5v square wave riding on top of it.  The DC level is 1v, and the AC level (obtained after subtracting the DC) is sqrt(sqr(0.5)+sqr(0.5)) = sqrt(0.5) = 0.707.

When you say 0.5V riding on 1V DC level, Does that mean RMS of AC is 1V and Vpeak will be sqrt(2) * 1V = 1.414V ? Or Vpeak is 1 + 0.5 = 1.5V ? And what does it mean when you say AC level? I thought that RMS of an AC signal is the DC level which would be LESS than AC Level.

Sorry, I'm not an engineer, more a math/science type, so my terminology may be faulty.  What I meant was the sum of a constant 1V (some might call this the "bias") and a square wave whose peak is +0.5v and trough is -0.5v.  I tend to think in "amplitude" -- I'd express a sinusoid as "bias + amplitude * sin (2pi*f*t - phase).  By "AC Level", I meant the RMS of the AC (time-varying) part of the signal, i.e. of the square wave.  Sorry for the sloppy language.

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The RMS is sqrt(sqr(1.5)+sqr(0.5)) = sqrt(2.5) = 1.581.  Note RMS is not DC + AC.

I am not sure how you go this? Seems like you have added 0.5 to 1 and subtracted 0.5 from 1. Vpk-pk in this case is 2V?

The peak of the DC + square wave is 1.5, the trough is 0.5.  RMS is sqrt (sum(squares)).  Peak-to-trough voltage is (1 + 0.5) - (1 - 0.5) = 1v, same as peak-to-peak of square wave (DC doesn't enter into peak-to-peak calculations, but does influence RMS).

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BS

[billko]

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@TooEagerToLearn wrote:

Hello again,

I am trying to do a spectral analysis one 2 AC signals and have another question, so I decided to start another post because it is about a different vi. I am using RMS.vi to find the RMS value of current and voltage. How is that different from AC-DC estimator? Thanks!

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What is the actual point of the discussion?  Did you end up with different results for each and want to know which is correct for what you are trying to do?  (I'm not familiar with the AC-DC Estimator.)

I'm just asking because I thought maybe Bob was kind of going down the wrong path with this discussion.