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

[Didier]

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?

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
> 
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