You are missing the fact you have your brackets wrong in: dB = 10 log (aV2^2/V2^2) = 10 log (aV2/V2)^2 = 20 log (a) ? Eq. (2) should be dB = 10 log (a (V2^2/V2^2)) = 10 log a ((V2/V2)^2) = 10 log (a) Eq. (2)
-----Original Message----- From: [email protected] [mailto:[email protected]] Sent: 19 October 2001 14:38 To: [email protected] Subject: RE: Is This Right? More to the proof discussion launched by the duty cycle question, given > 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). Eq. > (1) > If 10 log (P1/P2) = 10 log (V1^2/V2^2) = 10 log (V1/V2)^2 = 20 log (V1/V2), Then does it follow that, dB = 10 log (aV2^2/V2^2) = 10 log (aV2/V2)^2 = 20 log (a) ? Eq. (2) If this is true, then duty cycle "a" = 10 log (a) from Eq. (1) and = 20 log (a) from Eq. (2) What am I missing? Don > ---------- > From: UMBDENSTOCK, DON > Sent: Thursday, October 18, 2001 5:03 PM > To: UMBDENSTOCK, DON; [email protected]; [email protected]; > 'Ken Javor' > Subject: RE: duty cycle correction factors > > 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:[email protected]] > Sent: Thursday, October 18, 2001 3:37 PM > To: [email protected]; [email protected]; > [email protected] > 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: [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 > >> > >> > >> ------------------------------------------- > >> 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: > >> [email protected] > >> with the single line: > >> unsubscribe emc-pstc > >> > >> For help, send mail to the list administrators: > >> Michael Garretson: [email protected] > >> Dave Heald [email protected] > >> > >> For policy questions, send mail to: > >> Richard Nute: [email protected] > >> Jim Bacher: [email protected] > >> > >> 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: > > [email protected] > > with the single line: > > unsubscribe emc-pstc > > > > For help, send mail to the list administrators: > > Michael Garretson: [email protected] > > Dave Heald [email protected] > > > > For policy questions, send mail to: > > Richard Nute: [email protected] > > Jim Bacher: [email protected] > > > > 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: [email protected] with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Michael Garretson: [email protected] Dave Heald [email protected] For policy questions, send mail to: Richard Nute: [email protected] Jim Bacher: [email protected] 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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