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Relating to my previous question, I'm trying to understand how the LS method work for multi-constellation GNSS trilateration/multilateration. I understand for 2 constellations, trilateration require at least 5 pseudorange equations to account for the different receiver clock offsets for different constellations.

What is the minimum number of pseudorange equations per constellation? I.e, will one equation for one constellation be enough as long as the total number of equations meet the minimum number required for LS to work?

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With a few assumption about your position (i.e. on the surface of the Earth), you can locate yourself with three timing codes from the same constellation. However, a fourth satellite's timing code (from the same constellation) makes calculation more convenient because of the clock precision and synchronization requirements. This also give you the "absolute" time with high precision.

In relation with your previous question, you can locate yourself once you have 4 satellites from the same constellation. However, you still need at least 2 satellites from another constellation in order to add useful information to your multi-constellation system because you need to manage the time offset (which is a combination of intersystem time offset and device time delays).

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  • Thanks for your comment! I was also thinking the same thing. But I don't really understand why at least 2 satellites from another constellation (and not 1) is needed to account for the inter-system offset. Shouldn't the existence of an additional parameter (say r_glo that I used in my previous question) in one satellite already be enough? How will an extra equation with r_glo help?
    – Shawn Lim
    Jan 4, 2022 at 13:53
  • The time offset is partly due to intersystem offset and partly due to the receiver device, so you cannot get it in a simple way as far as I know. Therefore you use the information from the first satellite of another constellation in order to measure the time offset, but then you cannot reuse this information to derive the distance.
    – radouxju
    Jan 4, 2022 at 15:03
  • I see. Mathematically this makes sense as well. Using 4 satellites of the first constellation and 1 satellite of a second constellation would be practically useless because the first 4 pseudorange equations will yield the positional coordinates and then just use it to solve for the second constellation time offset.
    – Shawn Lim
    Jan 4, 2022 at 15:34

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