: Ken Javor[SMTP:ken.ja...@emccompliance.com]
Sent: Tuesday, October 30, 2001 6:20 PM
To: umbdenst...@sensormatic.com; emc-p...@majordomo.ieee.org;
dmck...@corp.auspex.com
Subject: Re: The Trouble with Convention, The Final Chapter
Following your logic, I was just following
warranted. The proper response is to do the job right and so notify the
customer so that he imposes the correct requirements even handedly.
--
From: umbdenst...@sensormatic.com
To: emc-p...@majordomo.ieee.org, dmck...@corp.auspex.com,
ken.ja...@emccompliance.com
Subject: RE: The Trouble
The point! We have all missed the point! :-)
I do not dispute the science.
The question was not what is the correct science, rather, what is
expected by the FCC (or any other spectrum authority) for successful
processing of the submittal. It became apparent that the science did not
match
Subject: RE: The Trouble with Convention
Date: Tue, Oct 23, 2001, 8:18 AM
Chris,
I don't believe we are addressing math proofs in this situation. Just as
the free space impedance of 377 ohms (51.5 dB) does not apply to the
reactive near field but is specified by ETSI for conversion from dBuV
Subject: RE: The Trouble with Convention
Date: Tue, Oct 23, 2001, 7:37 AM
Don,
The mathematical proofs to verify that 20log(D) is a valid method to
calculate the change in dBuV for a voltage signal with duty cycle D are
mathematically incorrect. There is no sanity check.
Multiplying D times V
The important thing is, first average the quantities, then convert to dB.
Ever seen folks doing video averaging on a log-scaled analyzer display?
Sure you have. And it's wrong. How wrong?
Take two samples, 100 dBq and 25 dBq. Sum their amplitudes in dB (100dBq +
25dBq= 125dbq) and divide by
For the FCC calculations, I can understand ...
E = I*R = 1uA*377 ohms = 377*10^-6 Volts
dBuV = 20log(377*10^-6V/1uV) = 51.5 dBuV
Assume you measure XuA's and you want to
convert to dBuV's.
E = I*R = XuA*377ohms = X*377*10^-6 Volts
and ...
dBuV = 20log[X*377*10^-6 Volts/1uV]
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