WEBVTT NOTE Doppler and the Shape of a Satellite Pass 1 00:00:00.000 --> 00:00:08.241 One pass, one recording. A low Earth orbit satellite is visible for eight to fifteen minutes. 2 00:00:08.241 --> 00:00:20.913 In that time it moves fast enough that its Doppler shift is obvious on the display, no measurement equipment required, just a wide enough view. 3 00:00:20.913 --> 00:00:39.522 The exercise: record the whole pass without touching the tuning, then look at what you captured. Setting up. Pick a target. The NOAA weather satellites at 137 MHz are ideal, strong, continuous, and predictable. 4 00:00:39.522 --> 00:00:53.434 The ISS packet downlink at 145.825 MHz works too. Get a pass prediction. Any tracking app; a pass reaching more than 40° elevation gives the clearest result. 5 00:00:53.434 --> 00:01:11.423 Tune to the satellite's nominal frequency and set a span of at least 20 kHz so the whole Doppler excursion stays on screen. Do not touch the dial for the entire pass. The point is to see the signal move. 6 00:01:11.423 --> 00:01:26.059 What you will see. The waterfall shows a signal that starts high in frequency, sweeps downward through the nominal frequency, and ends low. An S-shaped curve lying on its side. 7 00:01:26.059 --> 00:01:36.453 Three things to read off it: The total excursion. At 137 MHz a LEO satellite shifts about ±3 kHz, roughly 6 kHz peak to peak. 8 00:01:36.453 --> 00:01:53.251 The shift scales with frequency, so the same orbit at 435 MHz gives about ±10 kHz, which is why UHF satellite work needs continuous tuning and VHF often does not. The steepest point is closest approach. 9 00:01:53.251 --> 00:01:57.658 Doppler shift depends on the rate of change of range. 10 00:01:57.658 --> 00:02:12.793 It is largest when the satellite is coming straight at you or straight away, and passes through zero at the moment of closest approach, where the curve crosses the nominal frequency. 11 00:02:12.793 --> 00:02:23.104 That crossing is TCA, and you can read it off your recording to within a few seconds. The slope tells you the pass geometry. 12 00:02:23.104 --> 00:02:35.911 A high overhead pass produces a sharp, almost vertical transition through zero, the range changes fast. A low pass on the horizon produces a gentle slope. 13 00:02:35.911 --> 00:02:41.898 You can distinguish a 10° pass from an 80° one from the waterfall alone. 14 00:02:41.898 --> 00:02:52.094 The fading. Look closely at the signal's amplitude during the pass. It will not be steady. Expect regular deep fades, several per minute. 15 00:02:52.094 --> 00:03:04.224 Two causes, both from the satellites lesson: Spin modulation, the satellite is rotating, and its linearly polarised antenna sweeps past your polarisation and back. 16 00:03:04.224 --> 00:03:19.927 Faraday rotation, the ionosphere rotates the polarisation of the signal passing through it, by an amount that varies with the path. Both make a linearly polarised receiving antenna fade deeply and unpredictably. 17 00:03:19.927 --> 00:03:32.058 This is the observation that justifies a circularly polarised antenna, and having watched a NOAA image get shredded by polarisation fading you will not forget why. 18 00:03:32.058 --> 00:03:45.633 Turning it into a measurement. With the recording in hand: Measure the peak upward and peak downward offsets. Halve the difference, that is the one-way Doppler magnitude. 19 00:03:45.633 --> 00:04:15.739 Compute the satellite's line-of-sight velocity at the start of the pass: v equals ( delta f) divided by (f) times c A 3 kHz shift at 137 MHz gives (3000) divided by (137 times 10 to the power of 6 ) times 3 times 10 to the power of 8 is approximately 6, 600 m/s, which is close to LEO orbital velocity, because at acquisition the satellite is moving almost directly toward you. 20 00:04:15.739 --> 00:04:30.512 You have just measured the orbital speed of a spacecraft with a thirty dollar receiver. That is a fair demonstration that the physics in these lessons is not decorative. Check yourself. 21 00:04:30.512 --> 00:04:42.330 Where in the pass is the Doppler shift zero? You observe a total Doppler excursion of 20 kHz peak to peak on a 435 MHz satellite. Is that plausible? 22 00:04:42.330 --> 00:04:50.715 Your recording shows the signal fading deeply about every eight seconds. What is that, and what fixes it? 23 00:04:56.715 --> 00:05:09.713 At closest approach (TCA), where the range stops decreasing and starts increasing, so the rate of change of range is momentarily zero. 24 00:05:09.713 --> 00:05:25.524 Yes, about ±10 kHz at 435 MHz, since Doppler shift scales with frequency and a LEO gives roughly ±3 kHz at 137 MHz. Spin modulation, and possibly Faraday rotation. 25 00:05:25.524 --> 00:05:34.254 A circularly polarised antenna is insensitive to the orientation and largely removes both.