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Re: [Amps] Measuring RF Power

To: "amps@contesting.com" <amps@contesting.com>
Subject: Re: [Amps] Measuring RF Power
From: David Kirkby <david.kirkby@onetel.net>
Date: Sat, 26 Mar 2005 12:44:50 +0000
List-post: <mailto:amps@contesting.com>
R. Measures wrote:

>>
>> I think Gary is right, and you are wrong, and I have rigorously proved
>> that.
>>
>> In any case, it would be pointless for me to try to answer your question
>> when nothing  is not defined properly.
>
>
> What does "nothing is not defined properly" mean? 

Interpreting the unintentional double-negative correctly,  it means 
everything is defined properly. But I did correct myself when I said in 
a short post "I mean nothing (P, I or E) are defined properly of 
course."  Yes I originally made a typo - I accept that. When I know I am 
wrong I admit my mistakes, but this was simply an unintentional 
double-negative. That is a mistake I guess most of us have made at some 
point in our lives.

Since the discussion is about mean, peak and RMS values, without you 
having stated whether P, I or E are mean, RMS or peak values, it would 
have been impossible to answer. (Also, E is usually reserved for 
electric field, not voltage).

> As I understand this matter, for AC, V-peak ^2 / R = P-peak,

Yes, I agree with you there.

> V-rms^2 / R = P-rms.

That is *not* true.

It will not take you to a minute to skip all the maths and take a look 
at the last graph at

http://www.g8wrb.org/useful-stuff/rmspower.html

You wil see a graph with 3 coloured lines, near the bottom.

* The green line shows a simple sine wave, which we have all seen many 
times. It has the usual relationship Vrms=Vpeak/1.414 or 
Vrms=0.707*Vpeak. (Sorry, the first half is hidden by the blue line, but 
I think you get what I mean).

* The blue graph shows a rectified sine wave, which you would get as the 
output voltage from an ideal full wave bridge rectifier, if you applied 
a sine wave to the input of the bridge. Like the green sine wave, this 
has a relationship that Vrms=0.707*Vpeak.

* The red graph shows the power at any time.  It is *not* sinusoidal in 
shape, nor is it like the rectified sine wave, hence  you can *not* 
assume the relationship that peak=1.414 x RMS.

There is a small difference in the shape of the red and blue lines. That 
is why there is a sqrt(3/2)=22.5% difference between RMS and mean values 
of power.

Prms=1.225*Pmean  - Assuming a sine wave voltage and current.

I hope that explains it. If you can come back with a technical 
explanation of where my logic is flawed, then let me know, but please 
don't come back with something silly like "bananas" as you did before.

Others have read the page at

http://www.g8wrb.org/useful-stuff/rmspower.html

and agree that it is technically correct.

RMS power is not a particularly useful quantity, but it can be computed, 
and it can be shown it is not the same as mean power for a sinusoidal 
input voltage.

If I am wrong, give me some convincing evidence and I'll admit it. I am 
not the sort of person who will try to justify something I know is 
wrong, but in this case, nobody has a shred of evidence to disprove me, 
whereas I have provided a detailed proof, starting from the mathematical 
definition of RMS.

You are the only person who seems to be arguing this Rich, so at least 
back it up.

-- 
Dr. David Kirkby, 
G8WRB

Please check out http://www.g8wrb.org/ 
of if you live in Essex http://www.southminster-branch-line.org.uk/



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