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
> >>
> >>
> >> -------------------------------------------
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