> > 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 _______________________________________________ time-nuts mailing list -- [email protected] To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts and follow the instructions there.
