Your eqn (2) is in error. This is how it works. 10 log (aV2^2/V2^2) = 10 log (a) + 10 log (V2/V2)^2 = 10 log (a) + 20 log (V2/V2) = 10 log (a)
---------- >From: [email protected] >To: [email protected] >Subject: RE: Is This Right? >Date: Fri, Oct 19, 2001, 8:37 AM > > > 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. 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.
