[time-nuts] Basic question regarding comparing two frequencies
Bob Camp
lists at rtty.us
Mon Jul 26 16:38:42 UTC 2010
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
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