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[AMPS] RMS Power

To: <amps@contesting.com>
Subject: [AMPS] RMS Power
From: Mike" <W4EF@dellroy.com (Mike)
Date: Thu, 28 Jun 2001 07:34:54 -0700
Hi Billy, 

For a continuous wave RF signal across a resistive load, the average power
delivered to the load will be equal to the 1/4 * Epk-pk^2 / R. Not sure what
you mean by a symetrically modulated. If your talking about an two-tone
modulated SSB signal, the average power delivered to the load will be 
1/2 the peak "envelope" power or 1/8 * Epk-pk^2 / R. If your talking about
an AM signal 100% modulated by a single tone, the average power delivered
to the load will be 1/4 * Epk-pk^2/R*(1/4+1/16+1/16) = 1/4 * Epk-pk^2/R * 
(0.375). 
Or in other words, the average power will be 37.5% of the peak power
for the 100% modulated AM signal. Perhaps this is where your friend was
seeing the the 1/3 power reading on the Bird 43 (37.5% +/- 5% full scale error
could account for a 1/3 power reading). 

BTW, the term RMS (root-mean-square) derives from the mathematical 
operation performed on the voltage or current waveform used when 
determining the average power associated with the waveform. You square
the waveform, average the waveform over the period of the lowest frequency
component, and then take the square root. This value can then be used 
to calculate the average power associated with the waveform (P = Erms^2/R)
independent of the waveform complexity. 
 

73 de Mike, 
W4EF................................................................................................


----- Original Message ----- 
From: "Billy Ward" <billydeanward@hotmail.com>
To: <amps@contesting.com>
Sent: Wednesday, June 27, 2001 9:16 PM
Subject: [AMPS] RMS Power


> 
> Hey guys,
> 
> I am having quite a debate with a group of hams and cb'ers about the subject 
> of RMS power. I have been a Ham for almost 44 years and have never had 
> anyone deny that there is such a thing as RMS Power.  I am aware that there 
> are other amplifier manufacturers on this reflector and I would like some 
> input on how to calculate RMS Power of a linear amplifier. I have heard or 
> seen in a book somewhere that Average power is equal to the RMS voltage 
> times the RMS current.  It seems to me that if one is measuring RMS voltage 
> and RMS current, the product would be RMS power but it also seems that the 
> reference that I read that said otherwise was one of the main respectable 
> ARRL OR Orr handbooks.
> 
> One of the gang that I have had the discussion with says that when you 
> measure a modulation envelope peak to peak symetrically modulated signal 
> across a dummy load with a calibrated scope and calculate the power using 
> Esquared/R = P, that the Bird 43 will show 1/3 of that value when looking at 
> the dead key instead of 1/4. He therefore thinks that the Bird is a POS. I 
> happen to like the Bird 43 and have a lot of respect for it. Has anyone here 
> got a comment on that?
> 
> It has been years since I had everything set up on the bench to make these 
> measurements accurately but I would swear that when I used to frequently 
> make these type of tests that I came up with carriers that were 1/4 the 
> modulated peaks.  I know that the modulation envelope is double the size 
> during 100% modulation and that calls for 4 times the power of the dead key 
> value.  I also am aware that the carrier does not actually change in size 
> and that the modulation is a mixing process.
> 
> Billy
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