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*Subject*: Re: [amsat-bb] doppler shift*From*: Franklin Antonio <antonio@xxxxxxxxxxxx>*Date*: Wed, 30 Sep 1998 16:07:14 -0700*In-Reply-To*: <000701bdecbe$704a49a0$bff6a5ce@grandsdtrand>

At 03:05 PM 9/30/98 , Mark Flanagan wrote:

Am I correct in stating that the doppler shift of a receive frequency starts higher than the actual frequency as it is approaching then continues lower as it going away?

Yes. Just like cars going around a racetrack. The vroooom sound is higher in pitch when the car is coming toward you, and lower as it is going away.

Does the frequency appear to stabilize when the distance from the satellite to the qth becomes constant, even if it is briefly?

We have to be careful with use of words. You said "when the distance from satellite to the qth becomes constant, even if it is briefly". I think I know what you meant, but to be sure, I'll state it in my own words. Doppler shift is caused by rate of change of distance. If the distance is not changing, then the doppler shift is zero. This happens when a satellite (or race car) reaches the point of closest approach. Of course a low-earth-orbit satellite doesn't stop at the point of closest approach. It zips thru really quickly. But at that one point, the doppler shift is zero for an instant.

You used the words "appear to stabilize", and that raises another issue. When you listen, you notice the rate of change of frequency. At the point of closest approach, the doppler shift goes thru zero, but the rate of change of doppler shift is really high at that time. So I wouldn't describe this point as a point where anything "appears to stabilize". In fact, the time around closest approach is the time when doppler shift is changing most rapidly. As with the race car, the received frequency appears stable when the satellite is far away (either coming or going), but as it zips past you, things change rapidly.

The formula for the frequency shift is df = - f * dr/dt / c where f is the transmitted frequency, and c is the speed of light, and dr/dt is the rate of change of distance vs time. The only thing to remember from this formula is that things are simply proportional. Proportional to f, and proportional to dr/dt.

Hope that was helpful.

**References**:**doppler shift***From:*Mark Flanagan

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