This appears to be the missing piece . . . > ---------- > From: Ken Javor[SMTP:[email protected]] > Sent: Thursday, October 18, 2001 4:39 PM > To: [email protected]; [email protected]; > [email protected] > Subject: FW: duty cycle correction factors > > There is another issue which is different which may be what the FCC is > after > (I didn't read the referenced part 15 paragraph). FCC calls it pulse > desensitization. It was what the mil and aerospace world used to call a > broadband signal. If a signal is in the passband of a receiver for less > time than it takes to charge the IF bandpass filter, then the filter > output > is the average of the input. For instance, if the signal lasted one tenth > of a filter time constant, then the potential (Volts) that the filter > charges to is one tenth of the peak level that was fed into the filter. > This is very important when the test instrumentation uses a different > bandwidth than the real world victim protected by the requirement. Take > broadcast TV as an example. FCC/CISPR requires a 120 kHz bandwidth, but > TV > uses 6 MHz. An interference signal lasting 1 us with a 10% duty cycle (on > 1 > us, off 9 us) would be averaged by the CISPR measurement and would appear > at roughly 10% the level it would present to the TV receiver. In this > case, > the pulse desensitization factor is calculated as 20 log (duty cycle), > because we are talking about a coherent signal where the voltage is > proportional to the duty cycle, There is one last very important concept > here and that is duty cycle itself. Duty cycle is not an absolute but is > relative to the filter time constant. We could imagine a signal lasting > 10 > us and having a 10% duty cycle (10 us on, 90 us off) and the CISPR > receiver > and the TV would report exactly the same level. Duty cycle is relative to > receiver bandwidth. > > ---------- > From: "Ken Javor" <[email protected]> > To: [email protected] , [email protected] , > [email protected] > Subject: Re: duty cycle correction factors > Date: Thu, Oct 18, 2001, 2:37 PM > > > 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. > > >
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