[time-nuts] New topics (was Re: He isaTime-Nut Troublemaker....)

Didier didier at cox.net
Wed Dec 24 20:23:49 UTC 2008


6dB means the power is four times the power of one oscillator, how does the
power go up by 4 when combining two oscillators?

Didier 

> -----Original Message-----
> From: time-nuts-bounces at febo.com 
> [mailto:time-nuts-bounces at febo.com] On Behalf Of Magnus Danielson
> Sent: Wednesday, December 24, 2008 11:48 AM
> To: Discussion of precise time and frequency measurement
> Subject: Re: [time-nuts] New topics (was Re: He isaTime-Nut 
> Troublemaker....)
> 
> Didier skrev:
> > Chuck,
> > 
> > I am quite familiar with how to calculate a voltage or 
> power ratio in 
> > dB, but refering to the first issue, when you combine two 
> oscillators, 
> > does the noise improve by 3dB?
> 
> If you look back at my previous post, when combining the two 
> outputs, the noise raises by 3 dB, but the signal strength 
> raises by 6 dB, giving a net effect of lowering the noise by 
> 3 dB relative the signal strength.
> 
> You will find this described in many places for transistors. 
> It occurs for instance in the MAT-0x series of datasheets 
> among several other places.
> 
> Cheers,
> Magnus
> 
> > Didier
> > 
> >> -----Original Message-----
> >> From: time-nuts-bounces at febo.com
> >> [mailto:time-nuts-bounces at febo.com] On Behalf Of Chuck Harris
> >> Sent: Wednesday, December 24, 2008 10:02 AM
> >> To: Discussion of precise time and frequency measurement
> >> Subject: Re: [time-nuts] New topics (was Re: He is aTime-Nut
> >> Troublemaker....)
> >>
> >> Didier wrote:
> >>
> >>>> Square root of 2 is about 1,414 or about 3,01 dB.
> >>> I am always confused when considering noise, is it 
> 10*log(p1/p0) or 
> >>> 20*log(p1/p0)?
> >> A moment's reflection on why the 10 log  vs. 20 log, might help.
> >>
> >> The conversion from a power ratio to dB is:
> >>
> >> dB = 10 log (P1/P2)
> >>
> >> Remember that Power = VxV/R, so:
> >>
> >> P1/P2 = (V1xV1)/(V2xV2), the R's cancelling.
> >>
> >> So,
> >>
> >> dB = 10 log [(V1^2)/V2^2)] or, 10 log[(V1/V2)^2]
> >>
> >> If we want to express this as a ratio of voltages, rather than a 
> >> ratio of powers (there's a pun in there somewhere ;-), we need to 
> >> take the square root of (V1/V2)^2 outside of the log.
> >>
> >> To do this, we need to remember that log[X^2] = 2 log X, so:
> >>
> >> dB = 10 log[(V1/V2)^2] = 20 log[V1/V2]
> >>
> >> A couple of things to note:
> >>
> >> 1) dB's are dB's.  3dB represents the doubling of a power ratio,
> >>     6dB represents the doubling of a voltage ratio.
> >> 2) Convention says that if -dB's are loss, and +dB's are gain, but 
> >> that
> >>     is just convention.
> >>
> >> -Chuck Harris
> >>
> >> _______________________________________________
> >> time-nuts mailing list -- time-nuts at febo.com To unsubscribe, go to 
> >> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> >> and follow the instructions there.
> > 
> > 
> > _______________________________________________
> > time-nuts mailing list -- time-nuts at febo.com To unsubscribe, go to 
> > https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> > and follow the instructions there.
> 
> _______________________________________________
> time-nuts mailing list -- time-nuts at febo.com To unsubscribe, 
> go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.





More information about the Time-nuts_lists.febo.com mailing list