[time-nuts] second step : I need a clarification
John Miles
jmiles at pop.net
Sat Jan 9 02:00:09 UTC 2010
>
> this is very clear explanation / 2E-11 * 10E9 = (2*10)E(-11 +
> 9) = 20E-2 = 2E-1 = 0.2 Hz_
>
> thanks very much .
> now the first step is clear .
>
> assuming a timebase at 10MHz with a short term stability of 2E-11 the
> 10MHz should be near 10.000.000.000. +/- 2 . ( or better floating
> inside +/- 2)
>
> if this is correct I do not understand why driving my specrum analyzer
> the result at 10GHz is a residual FM near 2 Hz .
> driving the same analyzer with 10MHz OCXO I do not see residual FM ( or
> is very little ) The spectrum analyzer is hp 8566b .
> Is there a well know explanation or I I make some mistakes ?
How are you looking at 2 Hz sidebands with an 8566B?
The 8566's 10->100 MHz reference loop is 300 Hz wide, so it does benefit
from a clean 10 MHz source.
-- john, KE5FX
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