[time-nuts] Frequency error -> Parabolic Time Error?

Brooke Clarke brooke at pacific.net
Sat Apr 9 18:48:49 EDT 2005


Hi Tom:

Now suppose the reference oscillator is a Cesium type, does that 
automatically mean that df is zero?

Have Fun,

Brooke

Tom Van Baak wrote:

>Yes, the curve is described by a quadratic formula,
>same as any parabola. The time, T, (phase) of a
>reference oscillator is given by:
>
>T = t0 + f0 * t + 1/2 * df * t^2.
> -- where t0 is the initial time offset (phase),
> -- and f0 is the initial frequency offset,
> -- and df is the frequency drift per unit time.
>
>This can give rise to several different looking shapes
>depending on the initial conditions.
>
>/tvb
>
>----- Original Message -----
>From: "Brooke Clarke" <brooke at pacific.net>
>To: "Discussion of precise time and frequency measurement"
><time-nuts at febo.com>
>Sent: Saturday, April 09, 2005 15:18
>Subject: [time-nuts] Frequency error -> Parabolic Time Error?
>
>
>  
>
>>Hi:
>>
>>Suppose that my reference oscillator is offset from 10.0 MHz by some
>>small amount.  If I plot the time interval between the reference
>>oscillator and a GPS 1 PPS what will the plot look like over a long time
>>period, a parabola?
>>
>>Have Fun,
>>
>>Brooke
>>
>>--
>>w/Java http://www.PRC68.com
>>w/o Java http://www.pacificsites.com/~brooke/PRC68COM.shtml
>>http://www.precisionclock.com
>>
>>
>>
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>>    
>>
>
>
>
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>  
>

-- 
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