After the tracking process in a Software GPS receiver, I need to compute the pseudoranges between the receiver and the satellite (without having in mind the clock corrections). In the tracking process I obtain the Code Phase in chips for each satellite, which actually is the "pseudorange". So I compute it as:

Code_Phase/1023(chips/msec)*1E-3(msec)*3E8(speed light)

Where I obtain a certain value in meters. For example:

92.57/1023*1E-3*299792458 = 2.71E4 m

Am I computing the pseudorange correctly? or Am I forgetting something important? Since I compare this method with other I know it's correct, and both pseudoranges are quite different (for example in the given example, by this other method I obtain 2.17E7 m).

  • \$\begingroup\$ If anybody else is wondering, that big number is the speed of light. You need to multiply it by the time the signal needs to reach you, which indeed depends on the phase. \$\endgroup\$ – Vladimir Cravero Jan 2 '17 at 18:32
  • \$\begingroup\$ I've corrected the number, I've changed to 3E8, which is more well-known. I don't understand how to compute the time the signal needs to reach me, can you explain it? thanks \$\endgroup\$ – Pep Jan 2 '17 at 18:41
  • \$\begingroup\$ I do not think this is the right place, your question is not really about electronics engineering and I am fairly sure there is plenty of resources already online. Have you tried to search a bit? \$\endgroup\$ – Vladimir Cravero Jan 2 '17 at 18:45
  • \$\begingroup\$ I've posted here since there are some other questions related to GPS, but maybe is not the right place. I've already read some books, but still I cannot solve the problem. \$\endgroup\$ – Pep Jan 2 '17 at 19:36
  • \$\begingroup\$ Does your Code_Phase wrap modulo 1023? It should be extended to the left with the bit-counter of the navigation message to give you space-vehicle transmission time (Note that this leaves an ambiguitiy of 20 which is resolved by clock-data-recovery). If you do so, you will never find that 92.57 chips are in flight between you and the SV, which indeed make for only 27km distance (your formula is correct). \$\endgroup\$ – Andreas Jan 2 '17 at 19:57

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