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Re: [Amps] Measuring RF Power

To: amps@contesting.com
Subject: Re: [Amps] Measuring RF Power
From: Bill Fuqua <wlfuqu00@uky.edu>
Date: Tue, 22 Mar 2005 15:41:50 -0500
List-post: <mailto:amps@contesting.com>
The following is an example that I presented some time ago on AMPS. Hope it 
is useful again.
The biggest problem is that for some reason the registration of the values 
in the columns do not line up with some email programs.

73
Bill wa4lav







>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     Volts     Volts Squared      Power into 50 ohms Watts
>0              0           0                             0
>45           0.7071    0.5                          0.01
>90            1           1                             0.02
>135         0.7071     0.5                          0.01
>180          0           0                             0
>225       -0.7071     0.5                          0.01
>270         -1           1                             0.02
>335         -.7071    0.5                           0.01
>
>totals        0          4                             0.08
>average of totals    0.5                          0.01
>
>
>Square Root of average voltage squared is .7071 volts RMS
>
>
>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          Volts          VoltsSquared               Power into 50 
>ohms       Power Squared Watts
>0                  0                 0                                0 
>                               0
>45               0.7071          0.5                             0.01 
>                         0.0001
>90                1                 1                               0.02 
>                             0.0004
>135             0.7071          0.5                             0.01 
>                         0.0001
>180              0                  0                               0 
>                              0
>225             -.7071           0.5                             0.01 
>                          0.0001
>270             -1                  1                               0.02 
>                              0.0004
>335             -.7071           0.5                             0.01 
>                           0.0001
>totals            0                 4                              0.08 
>                              0.0012
>average of
>totals                              0.5                              0.01 
>                                0.00015
>Square Root of
>average                           0.7071                          N/A 
>                           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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