[Amps] Measuring RF Power

[R.Measures]

On Mar 25, 2005, at 2:00 PM, David Kirkby wrote:

> R.Measures wrote:
>
>>
>>>
>>> You can not find RMS power by multiplying RMS voltage by RMS current.
>>>
>>
>> I disagree. .
>>
>
> Rich, I think you are wrong.  Here's  my reasoning - I may well be 
> wrong, as maths never was a good point of mine.
>
> You need to understand calculus to properly work out the RMS of an 
> arbitrary waveform, as you probably know.
> Hence this explanation will really only be of use to someone who knows 
> integration. But I can't see a simple way of computing the RMS value 
> of the power, without it. People will just argue, but if you go back 
> to first principles, you should get the right answer.

The right answer is that RMS current x RMS potential = AC P
>
> This is the definition of RMS
>
> rms = Sqrt [ 1/T *  integral_of_waveform^2 ]
>
> where the period you integrate over must be a complete cycle (2 Pi 
> radians, or an integer multiple of it).
>
> Here I run Mathematica (a maths program from http://www.wolfram.com/) .
>
> This for a simple sine wave, with a peak value of 1.
> I integrate sin(t)^2 over 2 Pi radians, then take the square root, 
> which is what the definition of rms is.
>
> Mathematica 5.1 for Sun Solaris (UltraSPARC)
> Copyright 1988-2004 Wolfram Research, Inc.
> -- Motif graphics initialized --
>
> In[1]:= Sqrt[1/(2 Pi) Integrate[Sin[t]^2, {t, 0, 2 Pi}]]
>
>           1
> Out[1]= -------
>        Sqrt[2]
>
> It gives an answer of 1/sqrt(2), the approximate value of which is
>
> In[2]:= N[%1]
>
> Out[2]= 0.707107
>
> So, the rms value of a sine wave is 0.707 times the peak.
>
> Now assume for simplicity the resistor is 1 Ohm. Hence the current 
> will follow the voltage exactly and will be sin(t) too. Hence the 
> power is varies as a sin(t)^2.  I shows this below as sin(t) * sin(t), 
> but take the square of that (effectively sin(t)^4), since the 
> definition of RMS requires you square up the waveform.
>
>
> In[6]:= Sqrt[1/(2 Pi) Integrate[ (Sin[t] Sin[t])^2, {t,0,2 Pi}] ]
>
>
>             3
>        Sqrt[-]
>             2
> Out[6]= -------
>           2
> I doubt that is formatted too well, but it is saying the answer is 
> sqrt(3/2) / 2. The numerical value of which is:
>
> In[7]:= N[%6]
>
> Out[7]= 0.612372
>
> So for a peak voltage of one volt, the rms value of the voltage is 
> 0.707 V, but the RMS value of the power is 0.612 W.
>
>
> So we have
> 1) RMS value of a sinusoidal voltage of 1 V peak = 0.707 V.
> 2) RMS value of a sinusoidal current of 1 V peak = 0.707 A
> 2) RMS value of the power (which will not be sinusoidal) is 0.613 W.
>
> Note the instantaneous power will not be sinusoidal, but sin(t)^2. 
> That is sort of like a rectified sine wave, but it is not *exactly* 
> the same shape.
>
> -- 
> Dr. David Kirkby, G8WRB
>
> Please check out http://www.g8wrb.org/ of if you live in Essex 
> http://www.southminster-branch-line.org.uk/
>
>
>
>
>

Richard L. Measures, AG6K, 805.386.3734.  www.somis.org