> > 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?
The original intention of the calculations. The first relation, dB = 10 log (aP2/P2) = 10 log (a) is a "power" relation. The second relation, dB = 10 log (aV2^2/V2^2) = 20 log (a) is a "voltage relation. " Equating the two is invalid since you're trying to equate two different concepts. Doesn't mean anything. - Doug McKean ------------------------------------------- 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.
