D
Didier
Wed, Dec 24, 2008 5:14 PM
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@febo.com
[mailto:time-nuts-bounces@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 10log(p1/p0) or
20log(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:
- dB's are dB's. 3dB represents the doubling of a power ratio,
6dB represents the doubling of a voltage ratio.
- 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@febo.com To unsubscribe,
go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.
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@febo.com
> [mailto:time-nuts-bounces@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@febo.com To unsubscribe,
> go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
MD
Magnus Danielson
Wed, Dec 24, 2008 5:48 PM
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
-----Original Message-----
From: time-nuts-bounces@febo.com
[mailto:time-nuts-bounces@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 10log(p1/p0) or
20log(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:
- dB's are dB's. 3dB represents the doubling of a power ratio,
6dB represents the doubling of a voltage ratio.
- 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@febo.com To unsubscribe,
go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.
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@febo.com
>> [mailto:time-nuts-bounces@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@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@febo.com
> To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
D
Didier
Wed, Dec 24, 2008 8:23 PM
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@febo.com
[mailto:time-nuts-bounces@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
dB, but refering to the first issue, when you combine two
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
-----Original Message-----
From: time-nuts-bounces@febo.com
[mailto:time-nuts-bounces@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
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:
- dB's are dB's. 3dB represents the doubling of a power ratio,
6dB represents the doubling of a voltage ratio.
- 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@febo.com To unsubscribe, go to
https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.
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@febo.com
> [mailto:time-nuts-bounces@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@febo.com
> >> [mailto:time-nuts-bounces@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@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@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@febo.com To unsubscribe,
> go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
MD
Magnus Danielson
Wed, Dec 24, 2008 8:47 PM
6dB means the power is four times the power of one oscillator, how does the
power go up by 4 when combining two oscillators?
When you add the two signals together. A linear addition give twice the
amplitude and 4 times the power assuming the same impedance, but to
achieve this you need an active amplifier. A passive adder will
naturally not produce more power out than put into it, so it dampens the
signal, but will also dampen the noise applied to the input, so the same
relative gain remains.
Cheers,
Magnus
Didier skrev:
> 6dB means the power is four times the power of one oscillator, how does the
> power go up by 4 when combining two oscillators?
When you add the two signals together. A linear addition give twice the
amplitude and 4 times the power assuming the same impedance, but to
achieve this you need an active amplifier. A passive adder will
naturally not produce more power out than put into it, so it dampens the
signal, but will also dampen the noise applied to the input, so the same
relative gain remains.
Cheers,
Magnus
CH
Chuck Harris
Thu, Dec 25, 2008 4:44 AM
Hi Didier,
I presumed you at one time knew the difference between voltage
ratio, and power ratio dB, but your question gave the impression
that you might have needed a tiny nudge to refresh your memory.
If both oscillators are in lock step, you obviously can combine
them using an appropriately wound transformer to gain a 3dB increase
in power. But it would seem that the noise, being random and
independent, should add constructively as often as destructively.
I don't know the answer to your question... I did once, but that was
30 years ago... but I get the feeling that the noise power is going
to add up as some fractional power giving you a net improvement
in signal to noise... probably close to 3dB, but not quite 3dB...
I just don't remember anymore.
-Chuck Harris
Didier wrote:
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
Hi Didier,
I presumed you at one time knew the difference between voltage
ratio, and power ratio dB, but your question gave the impression
that you might have needed a tiny nudge to refresh your memory.
If both oscillators are in lock step, you obviously can combine
them using an appropriately wound transformer to gain a 3dB increase
in power. But it would seem that the noise, being random and
independent, should add constructively as often as destructively.
I don't know the answer to your question... I did once, but that was
30 years ago... but I get the feeling that the noise power is going
to add up as some fractional power giving you a net improvement
in signal to noise... probably close to 3dB, but not quite 3dB...
I just don't remember anymore.
-Chuck Harris
Didier wrote:
> 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
DJ
Didier Juges
Thu, Dec 25, 2008 2:29 PM
Chuck,
That was exactly my point. I have used phase combining of power amplifiers
to increase power and I know it improves broadband noise performance (noise
figure) even though that is usually not why we combine power amplifiers. My
last phase-combined amplifier combines two 100W Ku band TWTs for a very
compact airborne 200W satellite transmitter (much shorter than if I had been
using a 200W TWT, which was a requirement in order to fit in a small
diameter unit)
For power amplifiers, the improvement in noise figure (broad band,
uncorrelated noise) is about 3dB, but I am not sure it applies when you
combine two similar oscillators that may have a similar noise pattern. For
instance, combining TWTs does not improve power supply ripple induced
sidebands of course (when both tubes operate from the same supply). I would
expect that combining low noise osacillators will improve the random noise
by close to 3dB, but not all oscillator perturbations are broadband
uncorrelated. Particularly, the supply induced, temperature and load pulling
variations will not improve at all, limiting the benefits of the technique
for low noise oscillators.
The cost in complexity for combining oscillators is much more significant
than for amplifiers. Amplifiers only need phase matching (electrical
length). Oscillators need to be phase locked to begin with.
Merry Christmas to all.
Didier KO4BB
-----Original Message-----
From: time-nuts-bounces@febo.com [mailto:time-nuts-bounces@febo.com] On
Behalf Of Chuck Harris
Sent: Wednesday, December 24, 2008 10:45 PM
To: Discussion of precise time and frequency measurement
Subject: Re: [time-nuts] New topics (was Re: He isaTime-Nut
Troublemaker....)
Hi Didier,
I presumed you at one time knew the difference between voltage ratio, and
power ratio dB, but your question gave the impression that you might have
needed a tiny nudge to refresh your memory.
If both oscillators are in lock step, you obviously can combine them using
an appropriately wound transformer to gain a 3dB increase in power. But it
would seem that the noise, being random and independent, should add
constructively as often as destructively.
I don't know the answer to your question... I did once, but that was 30
years ago... but I get the feeling that the noise power is going to add up
as some fractional power giving you a net improvement in signal to noise...
probably close to 3dB, but not quite 3dB...
I just don't remember anymore.
-Chuck Harris
Didier wrote:
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
Chuck,
That was exactly my point. I have used phase combining of power amplifiers
to increase power and I know it improves broadband noise performance (noise
figure) even though that is usually not why we combine power amplifiers. My
last phase-combined amplifier combines two 100W Ku band TWTs for a very
compact airborne 200W satellite transmitter (much shorter than if I had been
using a 200W TWT, which was a requirement in order to fit in a small
diameter unit)
For power amplifiers, the improvement in noise figure (broad band,
uncorrelated noise) is about 3dB, but I am not sure it applies when you
combine two similar oscillators that may have a similar noise pattern. For
instance, combining TWTs does not improve power supply ripple induced
sidebands of course (when both tubes operate from the same supply). I would
expect that combining low noise osacillators will improve the random noise
by close to 3dB, but not all oscillator perturbations are broadband
uncorrelated. Particularly, the supply induced, temperature and load pulling
variations will not improve at all, limiting the benefits of the technique
for low noise oscillators.
The cost in complexity for combining oscillators is much more significant
than for amplifiers. Amplifiers only need phase matching (electrical
length). Oscillators need to be phase locked to begin with.
Merry Christmas to all.
Didier KO4BB
-----Original Message-----
From: time-nuts-bounces@febo.com [mailto:time-nuts-bounces@febo.com] On
Behalf Of Chuck Harris
Sent: Wednesday, December 24, 2008 10:45 PM
To: Discussion of precise time and frequency measurement
Subject: Re: [time-nuts] New topics (was Re: He isaTime-Nut
Troublemaker....)
Hi Didier,
I presumed you at one time knew the difference between voltage ratio, and
power ratio dB, but your question gave the impression that you might have
needed a tiny nudge to refresh your memory.
If both oscillators are in lock step, you obviously can combine them using
an appropriately wound transformer to gain a 3dB increase in power. But it
would seem that the noise, being random and independent, should add
constructively as often as destructively.
I don't know the answer to your question... I did once, but that was 30
years ago... but I get the feeling that the noise power is going to add up
as some fractional power giving you a net improvement in signal to noise...
probably close to 3dB, but not quite 3dB...
I just don't remember anymore.
-Chuck Harris
Didier wrote:
> 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
_______________________________________________
time-nuts mailing list -- time-nuts@febo.com To unsubscribe, go to
https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.
CH
Chuck Harris
Mon, Dec 29, 2008 3:10 PM
Didier,
I wonder what would happen if you took N, uncorrelated and well isolated,
10 MHz sources and added them together?
-Chuck Harris
The cost in complexity for combining oscillators is much more significant
than for amplifiers. Amplifiers only need phase matching (electrical
length). Oscillators need to be phase locked to begin with.
Merry Christmas to all.
Didier KO4BB
Didier,
I wonder what would happen if you took N, uncorrelated and well isolated,
10 MHz sources and added them together?
-Chuck Harris
>
> The cost in complexity for combining oscillators is much more significant
> than for amplifiers. Amplifiers only need phase matching (electrical
> length). Oscillators need to be phase locked to begin with.
>
> Merry Christmas to all.
>
> Didier KO4BB
>
D
Didier
Mon, Dec 29, 2008 4:38 PM
I think you will get a narrow (how narrow will depend on how close they are
in frequency and how many there are) band of noise centered around 10 MHz.
Didier
-----Original Message-----
From: time-nuts-bounces@febo.com
[mailto:time-nuts-bounces@febo.com] On Behalf Of Chuck Harris
Sent: Monday, December 29, 2008 9:11 AM
To: Discussion of precise time and frequency measurement
Subject: Re: [time-nuts] New topics (was Re: He isaTime-Nut
Troublemaker....)
Didier,
I wonder what would happen if you took N, uncorrelated and
well isolated, 10 MHz sources and added them together?
-Chuck Harris
The cost in complexity for combining oscillators is much more
significant than for amplifiers. Amplifiers only need phase
(electrical length). Oscillators need to be phase locked to
Merry Christmas to all.
Didier KO4BB
I think you will get a narrow (how narrow will depend on how close they are
in frequency and how many there are) band of noise centered around 10 MHz.
Didier
> -----Original Message-----
> From: time-nuts-bounces@febo.com
> [mailto:time-nuts-bounces@febo.com] On Behalf Of Chuck Harris
> Sent: Monday, December 29, 2008 9:11 AM
> To: Discussion of precise time and frequency measurement
> Subject: Re: [time-nuts] New topics (was Re: He isaTime-Nut
> Troublemaker....)
>
> Didier,
>
> I wonder what would happen if you took N, uncorrelated and
> well isolated, 10 MHz sources and added them together?
>
> -Chuck Harris
>
>
> >
> > The cost in complexity for combining oscillators is much more
> > significant than for amplifiers. Amplifiers only need phase
> matching
> > (electrical length). Oscillators need to be phase locked to
> begin with.
> >
> > Merry Christmas to all.
> >
> > Didier KO4BB
> >
>
> _______________________________________________
> time-nuts mailing list -- time-nuts@febo.com To unsubscribe,
> go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
CH
Chuck Harris
Mon, Dec 29, 2008 7:08 PM
That's what I would expect. There would, of course be a fair bit
of low frequency beating and the like, but if kept to a long window, it
should average out to a bell curve that is exactly centered on the average
frequency of the references.
Now, I guess the companion question is: Of what use would it be?
-Chuck Harris
Didier wrote:
I think you will get a narrow (how narrow will depend on how close they are
in frequency and how many there are) band of noise centered around 10 MHz.
Didier
That's what I would expect. There would, of course be a fair bit
of low frequency beating and the like, but if kept to a long window, it
should average out to a bell curve that is exactly centered on the average
frequency of the references.
Now, I guess the companion question is: Of what use would it be?
-Chuck Harris
Didier wrote:
> I think you will get a narrow (how narrow will depend on how close they are
> in frequency and how many there are) band of noise centered around 10 MHz.
>
> Didier
MD
Magnus Danielson
Mon, Dec 29, 2008 10:14 PM
That's what I would expect. There would, of course be a fair bit
of low frequency beating and the like, but if kept to a long window, it
should average out to a bell curve that is exactly centered on the average
frequency of the references.
Which I would guess would be adrift, so it would be kind of hard.
Actually, I wonder if it for two unsynchronized clocks would look like
an amplitude or phase modulated signal with a sine modulation. This
would cause a very typical sine-modulation curve on the histogram, and
with gaussian bell shaped outer edges. You would need to lock them up
for this hollow shape to fall out and it all become just a bell-shape.
Now, I guess the companion question is: Of what use would it be?
If they are free-running, I think the functionality is fairly unuseful.
If you lock them up, well, that's another issue.
Cheers,
Magnus
Chuck Harris skrev:
> That's what I would expect. There would, of course be a fair bit
> of low frequency beating and the like, but if kept to a long window, it
> should average out to a bell curve that is exactly centered on the average
> frequency of the references.
Which I would guess would be adrift, so it would be kind of hard.
Actually, I wonder if it for two unsynchronized clocks would look like
an amplitude or phase modulated signal with a sine modulation. This
would cause a very typical sine-modulation curve on the histogram, and
with gaussian bell shaped outer edges. You would need to lock them up
for this hollow shape to fall out and it all become just a bell-shape.
> Now, I guess the companion question is: Of what use would it be?
If they are free-running, I think the functionality is fairly unuseful.
If you lock them up, well, that's another issue.
Cheers,
Magnus