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[AMPS] RMS POWER

To: <amps@contesting.com>
Subject: [AMPS] RMS POWER
From: billydeanward@hotmail.com (Billy Ward)
Date: Fri, 29 Jun 2001 04:31:05 -0000
Yeah, I guess your right Bill.  ENOUGH!  I have not made myself clear and it 
is probably my fault.  The below is standard stuff that I knew and have been 
teaching for 30 years.  It still does not answer my question.  I will 
re-group and later on, if I have not figured out what I am trying to figure 
out, I will find a better way to ask the question.

Thanks Guys,

Billy


>From: William Fuqua <Wlfuqu00@pop.uky.edu>
>To: <amps@contesting.com>, <billydeanward@hotmail.com>
>Subject: [AMPS] RMS POWER
To: <amps@contesting.com>
>Date: Thu, 28 Jun 2001 13:21:47 -0400
>
>OK enough...
>
>RMS has one meaning, definition or procedure if you want to call it that.
>
>R   take square root
>         of
>M   the average (mean)
>         of
>S    the square of a number of data points
>
>  It may look complicated  as an equation or an integral expression 
>(calculus)
>but it is SIMPLE!!!!
>
>For example...
>
>
>Take one cycle of a 1 volt sine wave and divide into at least 8 equally
>spaced points starting at zero degrees.
>Calculate instantaneous voltage, voltage squared and power into 50 ohms
>load at each point.
>Do not use 360 degrees because it is the same point as 0 degrees. You may
>do this for any number
>of full cycles and you may want to do this on a spread sheet  for each
>degree (ie 360 points) or what ever.
>
>Phase   Voltage         Voltage Squared Power into 50 ohms
>                                                         Watts
>0               0               0                       0
>45              .7071           .5                      .01
>90              1               1                       .02
>135             .7071           .5                      .01
>180             0               0                       0
>225             -.7071          .5                      .01
>270             -1              1                       .02
>335             -.7071          .5                      .01
>
>totals          0               4                       .08
>average of totals               .5                      .01
>Square Root of average                  .7071
>
>
>The RMS value of the signal  is .7071 times the peak (1 Volt)= .7071
>volts  GEE look at above bold and underlined.
>
>What is the power of this signal into a 50 ohm load?
>.7071 volts squared divided by 50 ohms=  .01 watts   Look again  above at
>the average power bold.
>
>It is all very consistent with calculations using RMS voltage or
>Current  and resistance and power.
>
>Now what happens when we take RMS of the Power??
>
>
>Phase   Voltage         Voltage Squared Power into 50 ohms      Power 
>Squared
>                                                         Watts
>0               0               0                       0
>     0
>45              .7071           .5                      .01
>     .0001
>90              1               1                       .02
>     .0004
>135             .7071           .5                      .01
>     .0001
>180             0               0                       0
>     .0
>225             -.7071          .5                      .01
>     .0001
>270             -1              1                       .02
>     .0004
>335             -.7071          .5                      .01
>     .0001
>totals          0               4                       .08
>     .0012
>average of
>totals               .5                      .01                     .00015
>Square Root of
>average                  .7071
>0.01225  RMS power ?????
>
>OK above is your RMS power  for  what ever good it is....
>And it does not relate to any of the other numbers you may use such as
>voltage, current.  resistance or power. But does give you 22.5% bigger
>number in this example.
>73
>Bill wa4lav
>
>
>

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