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" <[EMAIL PROTECTED]> To: "Discussion of precise time and frequency measurement" <[email protected]> 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
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