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 > > > 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), > > The

Re: Is This Right?

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: umbdenst...@sensormatic.com >To: emc-p...@majordomo.ieee.org >Subject: RE: Is This Right? >Date: Fri, Oct 19, 2

RE: Is This Right?

[mailto:umbdenst...@sensormatic.com] Sent: 19 October 2001 14:38 To: emc-p...@majordomo.ieee.org 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 > Then dB

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 > 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 =