I wasn't going to weigh in on this but...  what was presented by Mr. 
Umbdenstock is equivalent to saying that since 2 + 2 = 4, then 2 x 2 = 4.
It is tautological.  The decibel scale is a power ratio.  If a signal has a
particular duty cycle then it is the total power that is affected by the
duty cycle ratio.  If something is on 100% and then you reduce the on-time
to 50%, clearly you consume half the previous POWER.

dB = 10 log (P1/P2)

Let "a" be the duty cycle ratio, with 0<a<1, so that P1 = aP2.

Then dB = 10 log (aP2/P2) = 10 log (a).  QED.

----------
>From: [email protected]
>To: [email protected], [email protected]
>Subject: RE: duty cycle correction factors
>Date: Thu, Oct 18, 2001, 12:26 PM
>

>
> Stuart,
>
> Duty cycle in 15.231 is related to a voltage ratio, therefore  20 log(duty
> cycle) will provide the correct factor.
>
> Demonstrate it to yourself.  Start with a given value (say 100V), multiply
> this by some duty cycle (say 15% or .15).  Convert the result to dB.  This
> is your reference result.  Now take 20 log of a duty cycle (.15).  Convert
> your given value (100V) to dB.  Add the numbers together, duty cycle dBs to
> the given value dBs, and behold -- the same answer as the reference result.
>
> Best regards,
>
> Don
>
>> ----------
>> From:  Stuart Lopata[SMTP:[email protected]]
>> Reply To:  Stuart Lopata
>> Sent:  Thursday, October 18, 2001 12:00 PM
>> To:  emc
>> Subject:  duty cycle correction factors
>>
>>
>> Part 15.231 devices use a duty cycle correction factor to adjust peak
>> readings.  The duty cycle represents the fractional on-time over a given
>> period of time (that must be under some limit).  Anyways, given this
>> fractional time, d, how do you make the conversion to dB?
>>
>> 10log(d) or 20log(d)?
>>
>> There have been some misinterpretations, since the readings are made at a
>> span of zero hertz (voltage readings).  Normally, a reduction in voltage
>> would use the 20log scale.  However, since the duty cycle does not
>> represent
>> a scale down (it represents the off-time versus on-time), the 10log scale
>> seems more appropriate.
>>
>> I have seen conflicting documents, so would like your professional
>> opinions!
>>
>> Thanks,
>>
>> Stuart Lopata
>>
>>
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