I think I've found the rub here... What Ken says below is mathematically correct.
What Don says below what's below is also mathematically correct. It's the application of the math (and the absence of, or assumed position of, a parenthesis or two) that is getting in the way. Assuming voltage (V) across a load, R, the power through the load is V**2/R. (I hope we can all agree on this.) Now, if the VOLTAGE were scaled by a factor "a", then power would be: (aV)**2/R. In this case, the "a" directly scales the voltage, so it gets squared when you calculate power. So, If you were to convert this to dB, you would get a 20log(a) term, which is what Don did. Mathematically correct, but doesn't have anything to do with duty cycle. The original question asked about a duty cycle. In this case, the voltage level is not changed or scaled. It is simply turned on and off with a duty cycle "a". It's probably best to think of this situation in terms of power being defined as energy/time. Consider one on/off cycle with a period of "T". Assume that the voltage is "on" during a*T and off during the rest of the cycle. So, in this case, ON TIME is being scaled by "a". The energy "E" delivered in one cycle (assuming square waves) is equal to the instantaneous power multiplied by the time. When V is on, Eon = (a*T)(V**2/R) When V is off, Eoff=0 So, the energy for the entire cycle------- E = Eon + Eoff = (a*T)*(V**2/R) The average power would be E / T which would yield a*(V**2)/R. Now, to convert to dB, we need to assume a reference power. We then take 10 times the log of the power ratio. This is the basic definition of dB, so it should lead us in the right direction. Let the reference power be V**2/R. ---which is what the power would be if V were on all of the time. dB = 10log((a*(V**2)/R) / ((V**2)/R)) everything but "a" cancels out. >>>>>> dB=10log(a) assuming--- 0<a<1, constant load, square wave for voltage----- Q.E.D. Note that this only works for square waves, if the waveform were sinusoidal or otherwise screwy shaped, the energy calculation would be different, possibly yielding an entirely different power formula. If only world peace were as straightforward as Math. Stay compliant guys. Chris Maxwell | Design Engineer - Optical Division email [email protected] | dir +1 315 266 5128 | fax +1 315 797 8024 NetTest | 6 Rhoads Drive, Utica, NY 13502 | USA web www.nettest.com | tel +1 315 797 4449 | > -----Original Message----- > From: Ken Javor [SMTP:[email protected]] > Sent: Friday, October 19, 2001 5:07 PM > To: [email protected]; [email protected]; > [email protected] > Subject: Re: duty cycle closure > > > No. You aren't applying the rule correctly. As I stated earlier: > > log a*b = log a + log b > log b^n = n log b > Combining, it is clear that > log (a*b^n) = log a + n log b. > > ---------- > >From: [email protected] > >To: [email protected], [email protected] > >Subject: duty cycle closure > >Date: Fri, Oct 19, 2001, 2:33 PM > > > > > > > Going back to fundamentals -- > > > > Given a = duty cycle = average power > > and define "^2" = "squared" > > > > Then P[ave] = a P[ref] > > > > P = V^2/R > > > > V[ave]^2/R = aV[ref]^2/R, the Rs cancel leaving > > > > V[ave]^2 = aV[ref]^2 > > > > 10 log(V[ave]^2) = 10 log (aV[ref]^2), which is equivalent to > > > > 20 log (V[ave]) = 20 log (aV[ref]) > > > > = 20 log (a) + 20 log (V[ref]) > > > > In the last equation one sees the duty cycle isolated as "20 log > (a)" when > > referring to power in terms of voltage. > > > > Best regards, > > > > Don Umbdenstock > > Sensormatic > > > > > >> ---------- > ------------------------------------------- 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.
