Perhaps I oversimplified. The definitions may be conditioned by what the FCC is looking for. And in general, I have always tested my understandings for a sanity check, not as a proof.
So, going back to the origins of the question, in some sections the FCC refers to an averaging detector, and a preference to use duty cycle with peak detection to provide the "averaging detector" reading. The FCC commented that they preferred math over averaging detectors due to linearity issues (per comments on a submission). So let's test the understanding: Given a 100uV signal measured by the peak detector in my spectrum analyzer. Given a 15 % duty cycle. The FCC would call this a signal equivalent to an averaging detector output of 15uV, 100 x .15 = 15 uV. If I wanted to simplify the handling of factors, I would apply the formula 10 log (P1/P2) or 10 log (V1^2/V2^2) = 10 log (V1/V2)^2 = 2*10 log (V1/V2) or in general, 20 log ("V"). The signal converted to dB would be 20 log (15) or 23.5dB If I want to simplify the handling of factors, I would apply the formula to the given value, 20 log (100) or 40 dB. If I apply the test to Ken's formula 10 log (a) = 10 log (.15) we have -8.2dB. As we are multiplying in linear terms, that means we are adding in log terms. 40 + (-8.2) = 31.8 dB If we apply the formula 20 log (.15) we have -16.5 dB. 40 + (-16.5) = 23.5 dB, which compares to 23.5 dB above. There is a piece missing somewhere as demonstrated when a test is applied. Don Umbdenstock > ---------- > From: Ken Javor[SMTP:ken.ja...@emccompliance.com] > Sent: Thursday, October 18, 2001 3:37 PM > To: umbdenst...@sensormatic.com; emc-p...@majordomo.ieee.org; > stu...@timcoengr.com > Subject: Re: duty cycle correction factors > > 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: umbdenst...@sensormatic.com > >To: emc-p...@majordomo.ieee.org, stu...@timcoengr.com > >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:stu...@timcoengr.com] > >> 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 > >> > >> > >> ------------------------------------------- > >> This message is from the IEEE EMC Society Product Safety > >> Technical Committee emc-pstc discussion list. > >> > >> Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/ > >> > >> To cancel your subscription, send mail to: > >> majord...@ieee.org > >> with the single line: > >> unsubscribe emc-pstc > >> > >> For help, send mail to the list administrators: > >> Michael Garretson: pstc_ad...@garretson.org > >> Dave Heald davehe...@mediaone.net > >> > >> For policy questions, send mail to: > >> Richard Nute: ri...@ieee.org > >> Jim Bacher: j.bac...@ieee.org > >> > >> All emc-pstc postings are archived and searchable on the web at: > >> No longer online until our new server is brought online and the old > >> messages are imported into the new server. > >> > > > > ------------------------------------------- > > This message is from the IEEE EMC Society Product Safety > > Technical Committee emc-pstc discussion list. > > > > Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/ > > > > To cancel your subscription, send mail to: > > majord...@ieee.org > > with the single line: > > unsubscribe emc-pstc > > > > For help, send mail to the list administrators: > > Michael Garretson: pstc_ad...@garretson.org > > Dave Heald davehe...@mediaone.net > > > > For policy questions, send mail to: > > Richard Nute: ri...@ieee.org > > Jim Bacher: j.bac...@ieee.org > > > > All emc-pstc postings are archived and searchable on the web at: > > No longer online until our new server is brought online and the old > > messages are imported into the new server. > > > ------------------------------------------- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. 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