[time-nuts] Basic question regarding comparing two frequencies

[Bob Camp]

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-bounces at febo.com [mailto:time-nuts-bounces at 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,  <EWKehren at 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 at febo.com
To unsubscribe, go to
https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
and follow the instructions there.