--
> From: 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
>
>
ed is what counts. For those still in the dark
> about 20 log () or 10 log (), my suggestion is to go straight to the source
> -- ask the FCC what they specify for your situation.
>
>
> Don Umbdenstock
> Sensormatic
>
>
>> --
>> From: Ken Javor[SMTP:k
ompliance.com]
> Sent: Monday, October 29, 2001 2:41 PM
> To: umbdenst...@sensormatic.com; emc-p...@majordomo.ieee.org;
> dmck...@corp.auspex.com
> Subject: The Trouble with Convention, The Final Chapter
>
> In the face of all the responses I and others gave last week showin
rp.auspex.com,
ken.ja...@emccompliance.com
>Subject: The Trouble with Convention
>Date: Mon, Oct 22, 2001, 4:46 PM
>
> Similarly, it appears the same issue of convention is the basis of certain
> FCC clauses, for example, the reporting of the output of an averaging
> detector as called for by 15.2
nettest.com
>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
> react
Mr. Maxwell is right, yesterday's mailing from Mr. Umbdenstock is incorrect
- I was just losing interest in the issue.
--
>From: "Chris Maxwell"
>To: , ,
,
>Subject: RE: The Trouble with Convention
>Date: Tue, Oct 23, 2001, 7:37 AM
>
> Don,
>
>
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 two,
t; Sent: Tuesday, October 23, 2001 8:37 AM
> To: umbdenst...@sensormatic.com; emc-p...@majordomo.ieee.org;
> dmck...@corp.auspex.com; ken.ja...@emccompliance.com
> Subject: RE: The Trouble with Convention
>
> Don,
>
> The mathematical proofs to verify that 20log(D
.@sensormatic.com [SMTP:umbdenst...@sensormatic.com]
> Sent: Monday, October 22, 2001 5:47 PM
> To: emc-p...@majordomo.ieee.org; dmck...@corp.auspex.com;
> ken.ja...@emccompliance.com
> Subject: The Trouble with Convention
>
>
> Pursue the right question and one might rec
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]
Separating
Pursue the right question and one might receive a meaningful answer.
How do you convert from dBuA to dBuV when measuring a 50 kHz signal at 10
meters? How do you convert from linear terms to log terms when addressing
the output of an averaging detector where the limit is field strength in
units
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