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Author Topic: Average Power Computation - MEAN or RMS? - Audience Area  (Read 11508 times)
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"What method of computation should be applied to a Power Waveform Trace to obtain the Average Power of the sample set under consideration?"




This is the Audience Area where comments can be made regarding the debate in the Participant Area.
   
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The confusion lies in that when you look at voltage or current, the power associated with these parameters is proportional to the square  of the measurement.  That's why you make RMS measurements on AC voltage-alone or current-alone waveforms to "rationalize" them down to DC equivalents.  It's a way of giving you a measure of the true power capability inherent in an AC voltage or current waveform.  Alternatively, voltage times current is in itself a measurement that is the product of two variables and is therefore inherently "squared."   

When it comes to a "power waveform," that is an abstraction derived from two other real-world variables.  The power waveform already factors in the "squared" factor associated with the voltage and current waveforms.  Therefore it makes no sense at all to "square power" and then take the mean value of that and then calculate the square root.  The only thing that you have to do is average out the power over an entire periodic cycle.

When you want to calculate energy over a cycle, you integrate the power waveform over that cycle to get your energy calculation.  You do not integrate over the square of the power waveform and then take the square root.  Again, that makes no sense.

MileHigh
   
Group: Guest

When you want to calculate energy over a cycle, you integrate the power waveform over that cycle to get your energy calculation.


The problem is when you do the integration, the negative area cancels the positive area.  The arguement is that negative area does not mean negative energy because energy cannot be negative.  What Lawrence wants to find is the apparent power and doing the root mean square method to the power wave form is not the correct algorithm.  Taking the mean power with respect to coordinate is an algorithm for real power.  Taking the mean power with respect to area is the apparent power.
   

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It's not as complicated as it may seem...
The problem is when you do the integration, the negative area cancels the positive area.  
True, and this is precisely what the scope will do if such a wave form is the case. So, there is no "problem" actually.

Quote
The argument is that negative area does not mean negative energy because energy cannot be negative.
In terms of the integral, indeed the energy IS negative, and would in this case be going back to the source. The scope would happily show this, and it would be correct.

Quote
What Lawrence wants to find is the apparent power and doing the root mean square method to the power wave form is not the correct algorithm.  Taking the mean power with respect to coordinate is an algorithm for real power.  Taking the mean power with respect to area is the apparent power.
Don't speculate about what Lawrence wants to find. Watch the debate, then perhaps we will all know what he wants.

.99
   
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I want to ask you something Poynt.

If I give you a power waveform P(t).  Find the phase differential.  Can you do this?
   

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It's not as complicated as it may seem...
I want to ask you something Poynt.

If I give you a power waveform P(t).  Find the phase differential.  Can you do this?


If it is from a sinusoidal source, then yes it is possible to determine what the phase differential was.

.99
   
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 ;D Looks like I have underestimated mr. Tseung
   

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It's not as complicated as it may seem...
Participants,

Each of you have well established your stance on the issue.

However, as of yet your statements can be considered as opinion rather than fact.

What is missing is substantiation in support of your respective stances.

Poynt99 has provided an internet quote of Roy Lewallen as a form of substantiation. While this quote tells us what Average Power is not, it fails to provide a definition of what Average Power is nor does it discuss the use of MEAN in a calculation.

Roy does offer a clue when he states in his summary:
"* The RMS values of voltage and current are useful because they can be used to calculate the average power."

The debate still needs a verifiable definition for "Average Power"

Moderator


I would encourage the moderator to do the best he can with what is being presented by both participants as substantiation.

One factor in deciding on the moderator for this debate, was his current knowledge of electrical theory. Most if not all of the arguments put forth thus far are quite readily vetted by those with the background this moderator possesses.

It is hoped that the moderator is not asking the participants to derive the fundamentals of algebra, Ohm's law, and the like, as this would be exremely counter-productive and of no value in this debate. It is understood that the participants and moderator alike are adequately versed in the basics of math etc.

In the spirit efficiency and to retain a sense of focus, I encourage the moderator to utilize his background (and other technical resources if necessary) in the topic at hand to effectively resolve this debate.

.99
   
Group: Guest
For an example comparison of RMS vs. MEAN for a nonsinusoidal periodic waveform see the attachments.

Notice that Running RMS and Running MEAN are included along with final RMS and final MEAN calculations.

Cheers,

Harvey

EDIT: Added phase shifted version with Cos(θ) inclusion. Notice the reduction in Average Power and the negative instantaneous power.

« Last Edit: 2011-02-05, 01:32:04 by Harvey »
   

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It's not as complicated as it may seem...
Good stuff.

Thanks Harvey.

.99
   
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Quote

Using the MEAN of the instantaneous power automatically handles the phase angle problem by removing the reactive power from the total, thereby leaving the Average Power, or Real Power.



So the average power is real power.   Why didn't anyone says it earlier. ;D 

Good battle. Thanks.
   

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It's not as complicated as it may seem...
So the average power is real power.   Why didn't anyone says it earlier. ;D 

Good battle. Thanks.


I did even before the debate began. I guess you weren't paying attention  :P

http://www.overunityresearch.com/index.php?topic=538.msg10288#msg10288

.99
   
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Quote
With the greatest respect for Lawrence, I find in favor of declaring Poynt99 the winner of this debate.

While it may be possible to use a math setting on the scopes to multiply the RMS values of voltage and current this would only work for purely resistive loads where no phase angle exists. In order to accurately calculate the Average Power using the RMS method the phase angle must be included in the calculation.

Using the MEAN of the instantaneous power automatically handles the phase angle problem by removing the reactive power from the total, thereby leaving the Average Power, or Real Power.

I just want to mention again, that you cannot multiply the RMS values of voltage and current for non-linear loads.  I demonstrated this with a brute-force analysis in an earlier posting.  This applies for both sinusoidal and arbitrary waveform excitation.  The LTJT and many other circuits make use of diodes or LEDs, which are non-linear components.

That's one of my pet peeves where you see experimenters using LEDs all the time and attempting to make measurements on them with basic equipment.  By the same token your eyes simply cannot make anything more than a subjective estimate of the brightness of the LEDs, nor can your eyes account for the flashing.

If you are doing an experiment you are better off working with resistors and a true-RMS multimeter if you are trying to measure your output power.  In addition, you should try to understand what the limitations of the true-RMS metering is like for your particular multimeter.  From what I have seen for the cheaper ones there are no specifications provided by the manufacturers.  If you are serious about measuring output power and you don't have or can't afford a DSO then you can look for alternative and cheaper ways to do these measurements.  That should be part of the fun!

MileHigh
   
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I just want to mention again, that you cannot multiply the RMS values of voltage and current for non-linear loads.  I demonstrated this with a brute-force analysis in an earlier posting.   This applies for both sinusoidal and arbitrary waveform excitation.  The LTJT and many other circuits make use of diodes or LEDs, which are non-linear components.


MH: could you link to this please?  Thanks,

Humbugger
   
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It's not as complicated as it may seem...
I want to ask you something Poynt.

If I give you a power waveform P(t).  Find the phase differential.  Can you do this?


Again, assuming that the power trace is an undistorted sinusoid, yes.

Here's how, just for you Gibbs.  ;)


If P+ is the power trace positive peak, and P- the negative peak, the phase angle of the voltage and current can be computed as follows:

PHASE = ARCCOS [(P+ + P-) / (P+ - P-)]


Let's try a couple examples:

If P+ is 17.07W and P- is -2.93W, we have:
PHASE = ARCCOS [(17.07 - 2.93) / (17.07 + 2.93)]
PHASE = ARCCOS (14.14/20)
PHASE = ARCCOS (0.707)
PHASE = 45º

If P+ is 4W and P- is -1W, we have:
PHASE = ARCCOS [(4 - 1) / (4 + 1)]
PHASE = ARCCOS (3/5)
PHASE = ARCCOS (0.600)
PHASE = 53.13º

.99
   
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I'm impressed.

I wonder how you found out the mean power related to real power (or like Harvey said, it takes care of the phase).
   

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It's not as complicated as it may seem...
I'm impressed.

I wonder how you found out the mean power related to real power (or like Harvey said, it takes care of the phase).

If we were to draw a horizontal line representing the MEAN (DC component) of a sine wave, it would bisect the wave form into equal top and bottom halves. There is a relationship between the wave form average and the phase angle.

By using the positive and negative peak values as I showed, we are effectively finding the wave form MEAN (but normalized), then taking the ARCCOS of it to obtain the phase angle.

Makes sense?

.99
   
Group: Guest
Voila!

http://www.overunityresearch.com/index.php?topic=538.msg10326#msg10326

Hi MH,

Feel free to re-work those spread sheets and put in a nonlinear load like the LED. All you would need to do is plug in a voltage threshold before the current begins flowing. The easiest way to do that would be to set the current to zero in the forward bias direction for any voltage value below the threshold and then set it at the LED operating current (to show a series limiting resistor) for the duration of the forward bias. Then for all the negative voltage, set the current to zero for the reverse bias direction.

Thus by just altering the data in the Current column in the first spread sheet, you can map the effect that waveform would have on a nonlinear LED load with current protection.

If you make changes in the second sheet, then you have to account for the fact that a phase shift exists - I forget if I made that Capacitive or Inductive . . . you can figure it out.

Question: Since an LED and Series Resistor is present across the secondary winding does the inductance of the winding have a phase shift for the output section of the FLEET device?  ;)



   
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