Hi I believe what they do is:
DSB modulate the 5 MHz with 500 Hz to get 5.0005 and 4.9995 MHz Filter out the 4.9995 MHz with a crystal filter or by using an I/Q modulator (I believe Austron did the I/Q thing rather than the filter). Divide the result by 5 to get 1.0001 MHz Mix the 1.0001 with an incoming 1 MHz from the DUT Look at the 100 Hz beat note out of the mixer. That all (of course) assumes you have 1 MHz out of the DUT in the first place. Otherwise there's a divide the DUT to 1 MHz step in there as well. Bob -----Original Message----- From: time-nuts-boun...@febo.com [mailto:time-nuts-boun...@febo.com] On Behalf Of Peter Vince Sent: Monday, July 26, 2010 10:32 AM To: Discussion of precise time and frequency measurement Subject: Re: [time-nuts] Basic question regarding comparing two frequencies Sorry Bert, I don't follow the last part about the 100Hz - can you explain further please? (and is that 100.00 or 100.01 Hz?) Peter On 26 July 2010 14:27, <ewkeh...@aol.com> wrote: > Hi, > ten years ago not having a super counter I copied the input circuit of > the Austron 2110 that using an XOR gate mixes 5 MHz with 500 Hz getting > 5.0005 MHz. It is devided down to 1.0001 Mhz which in turn is mixed in 74 HC 74 > D F/F giving 100 Hz, that most counters are able to count at high > resolution. Still use it today. May be a time-nuts project. > Bert Kehren _______________________________________________ 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.
