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*Subject*: Re: [amsat-bb] Doppler freq. shift*From*: Paul Williamson <kb5mu@xxxxxxxxx>*Date*: Tue, 27 Apr 2004 21:49:31 -0700*In-Reply-To*: <001d01c42c6f$cf8701a0$0300a8c0@hamcomputer>

>I need some help understanding the Doppler freq shift on the sat's >downlinks. My understanding of Doppler is the freq rises as the object >approaches and decreases as it moves away. That sounds almost right, but it's stated in a confusing way. Let me try to clarify. This would be easier with a blackboard or a napkin. Try to visualize. When the object is approaching (range is decreasing) the Doppler shift is a frequency increase, proportional to the speed of closure. Likewise, when the object is receding (range is increasing), the Doppler shift is a frequency decrease, again proportional to the rate of change of the range. We say "range-rate" as a shorthand for this speed along the line from you to the object. Now think about the geometry of a LEO pass. Let's make it an overhead pass for simplicity. At the beginning of the pass, the satellite is near the horizon, and rushing toward you. It turns out that the range-rate is highest at this moment. As the satellite gets closer to you, it rises in the sky, and more and more of its orbital speed is perpendicular to the line between you and the satellite. That is, the satellite is moving faster across the sky but not getting closer as fast as before. Because the range-rate is now lower, the Doppler shift is proportionately lower. When the satellite is exactly overhead, it is moving across the sky but not getting any closer to you or any further away. That is, the range is momentarily constant. The Doppler shift is thus zero, and you hear the downlink on the exact frequency it was transmitted on. After the satellite passes through overhead, it begins to move away from you. More and more of the orbital speed is parallel to the line between you and the satellite, until it reaches a maximum value as the satellite disappears over the horizon. So during this time, the Doppler shift (now negative) increases to a maximum. So, the Doppler shift is always falling. It starts out at a big positive value, goes through zero as the satellite goes overhead, and then ends up at a big negative value at LOS. That's why you see the frequency decreasing throughout the pass. For "overhead" you can substitute "time of closest approach" for a non-overhead pass, and everything above is still true. It's just a little harder to visualize that way. I specified LEO because with a high elliptical orbit it is possible to get other Doppler curves, because the rotation of the Earth becomes a significant factor. A LEO satellite always describes a beautiful S-curve in frequency. The Doppler shift is big but nearly constant near the ends of the pass, and changes quickly but smoothly in the middle, passing through zero at the time of closest approach. Plotting this curve from observations and finding the slope of the steepest part of the curve is a good way to measure the orbit; it's possible to derive your own Keplerian elements from several of these observations. Any clearer now? 73 -Paul kb5mu@amsat.org ---- Sent via amsat-bb@amsat.org. Opinions expressed are those of the author. Not an AMSAT member? Join now to support the amateur satellite program! To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**Follow-Ups**:**RE: Doppler freq. shift***From:*David Smith

**References**:**Doppler freq. shift***From:*Bill Bruno

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