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In a gps receiver there is an inevitable time delay in the signal travelling in the antenna lead to the circuitry of the receiver. In a typical 5 metre lead the delay is about 15 nanoseconds, which, added on to the time for the signal to travel from the satellite to the antenna, would mean a serious inaccuracy if it was not compensated for. How is this compensation achieved?

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Typically they have short or no leads - but note that a uniform extra delay through the system isn't a problem, it just means that the GPS tells you where the antenna is not where the decoder is.

This is magnified in situations where GPS is re-broadcast indoors. Again, it gives you the location of the external antenna not the decoder.

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    \$\begingroup\$ In precision timing applications, receivers have an explicit configuration setting that compensates for the antenna cable delay, so that the 1pps output has that delay removed. But as you say, they're still reporting the location of the antenna. \$\endgroup\$
    – Dave Tweed
    Commented Jan 27, 2015 at 13:10
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    \$\begingroup\$ I added the extra to my question before I saw the comment by pjc50, but the reference I quote confirms that some installations have a configurable allowance for the cable delay. Patch antennas attached to the Rx do indeed have very short leads, but I have a couple of puck antennas with 5 metre leads to the Rx. \$\endgroup\$ Commented Jan 27, 2015 at 14:44
  • \$\begingroup\$ is the link dead ? \$\endgroup\$
    – Blup1980
    Commented Jul 25, 2019 at 7:31
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There is usually no need for any compensation at all. The GPS receiver measures the location of the phase center of the antenna. What happens between the phase center of the antenna and the receiver affects all of the satellite signals equally and hence will not affect the position calculation. If a precise time is needed, then the propagation delay of the cable is compensated for in the timing outputs of the receiver.

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Here is a quotation from a source that bears out what I think:

The Global Positioning System has a clever, effective solution to this problem. Every satellite contains an expensive atomic clock, but the receiver itself uses an ordinary quartz clock, which it constantly resets. In a nutshell, the receiver looks at incoming signals from four or more satellites and gauges its own inaccuracy. In other words, there is only one value for the "current time" that the receiver can use. The correct time value will cause all of the signals that the receiver is receiving to align at a single point in space. That time value is the time value held by the atomic clocks in all of the satellites. So the receiver sets its clock to that time value, and it then has the same time value that all the atomic clocks in all of the satellites have. The GPS receiver gets atomic clock accuracy "for free."

When you measure the distance to four located satellites, you can draw four spheres that all intersect at one point. Three spheres will intersect even if your numbers are way off, but four spheres will not intersect at one point if you've measured incorrectly. Since the receiver makes all its distance measurements using its own built-in clock, the distances will all be proportionally incorrect.

The receiver can easily calculate the necessary adjustment that will cause the four spheres to intersect at one point. Based on this, it resets its clock to be in sync with the satellite's atomic clock. The receiver does this constantly whenever it's on, which means it is nearly as accurate as the expensive atomic clocks in the satellites.

I am still convinced that this is correct and that there are just a lot of various ways of presenting it.

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Asking the manufacture of the receiver or reading their docs is a good place to start. One manufacturer of computer server GPS cards comes up with an answer of 1.2 ns per 1 ft of cable length. Many customers might need several hundred feet of antenna cable to get from a server machine room to a commercial building rooftop, so it is a real issue.

Under typical conditions, the expected cable and amplifier delays are negligible. The following formula is used to calculate the cable delay:

  • D = ( L * C ) / V

Where:

  • D = Cable delay in nanoseconds
  • L = Cable length in feet
  • C = Constant derived from velocity of light: 1.016
  • V = Nominal velocity of propagation expressed as decimal, i.e. %66 = 0.66 Value is provided by cable manufacturer.

When using LMR- 400 or an equivalent coax cable, this formula equates to approximately 1.2 nanoseconds of delay per every foot of cable. To calculate the Offset value (cable delay), multiply the length of the entire cable run by “1.2” and then enter this value into the Offset field.

Examples of LMR–400 coax cable delays:

100 feet of cable = 120 nanoseconds of cable delay

200 feet of cable = 240 nanoseconds of cable delay

300 feet of cable = 360 nanoseconds of cable delay

(Above quote from Spectracom - Antenna Cable Length Considerations)

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  • \$\begingroup\$ Welcome to the site :-) "it is a real issue" Just to be clear, this calculation is related to the 1PPS offset (as mentioned in this comment from Dave Tweed) and not to any positional accuracy issue, isn't it? (Of course, the GPS position of the antenna and not of the receiver, will be decoded if the two are not co-located.) Thanks. \$\endgroup\$
    – SamGibson
    Commented Jul 25, 2019 at 1:21
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Sorry, Harry, but you're dead wrong in your answer.

The entire point of the GPS system is that the Rx clocks do NOT have to be as accurate as the satellites. As a matter of fact, Rx clock is irrelevant to the accuracy of the system. GPS works by detecting the difference in reported clock times from the satellites. From these differences (it takes the comparison between several satellite clocks to do the job), the relative position of the receiver can be calculated. That's the first part.

The second part consists of each satellite transmitting, and the receiver receiving, ephemeris data, which allows the receiver to tell exactly where each satellite is (in Earth coordinates, not in absolute terms), and from that derive the receiver location.

If the position of the receiver were determined by a single comparison of two satellites' data, Rx clock would have to be very precise. However, using several comparisons (which is needed anyways) the Rx clock errors can be compensated for. See, as a start, the Wiki article on "pseudorange".

So, no, the Rx clock most definitely does not have to be as accurate as the satellite clocks. It's worth pointing out that there are two very different uses for GPS - position and time determination. The article you linked to discusses antenna configurations for both uses, and I'm afraid you missed the distinction.

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    \$\begingroup\$ Well, GPS receiver clocks might not be as accurate, in the normal sense of the word, but when the GPS is locked, they are by definition synchronised to a few 10s of ns. Has to be, if position is accurate to a few m. \$\endgroup\$
    – tomnexus
    Commented Jan 27, 2015 at 15:21
  • \$\begingroup\$ I have just read the Wikipedia article on Pseudorange, and I think it substantially says the same as I have, with the proviso that maybe the Rx clock is not actually corrected, but that a calculation of its error is used instead. This is like the use of a chronometer on a ship for navigation. The essential property is that it ticks steadily and with a constant period. Whether it gains or loses is no matter, as long as the rate is known so that it can be allowed for. \$\endgroup\$ Commented Jan 27, 2015 at 15:23
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    \$\begingroup\$ It's a little misleading to say that the receiver clock is completely irrelevant to the accuracy of the solution. While its absolute error will be measured and compensated for in the process of solving the equations, its short-term stability is extremely important, because it is the timebase against which all of the satellite signals are correlated. \$\endgroup\$
    – Dave Tweed
    Commented Jan 27, 2015 at 16:13
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I think I have worked out how, although I haven't been able to find the question answered elsewhere, so I am posting this hoping it will be verified or corrected by experts.

The clock in the receiver (Rx) has to be as accurate as those in the satellites. If the Rx clock is slow, it will make the distances from the satellites seem longer, because it will be further behind than it should be, and will measure extra time. Alternatively if it is fast, it will be in advance and measure less time, putting the Rx closer than its true position. Either of these situations causes a spread of positions calculated by triangulation from pairs of satellites. This is compensated for by a statistical analysis of the measured times and positions, and adjusting the Rx clock to to give the greatest consistency in the set of calculated position, and this process brings the Rx clock into synchronism with the satellites' clocks.

The delay in the cable is the equivalent of the clock being extra slow, so it is automatically taken account of by the process I have described. The Rx clock is made to be fast on those in the satellites by the time of the delay in the cable.

I have given a presentation on GPS, and attach two images of slides which were the basis of my explanation. I do not have references for the original source of these images.

How fix is madeAdjusting the receiver's clock

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    \$\begingroup\$ The antenna cable delay affects all of the satellite signals by the same amount. Since the position algorithm works on the differences between satellite signal phases, rather than their absolute phases, the cable delay cancels out. But the receiver ends up calculating the position of the antenna rather than the position of the receiver itself. \$\endgroup\$
    – Dave Tweed
    Commented Jan 27, 2015 at 13:14
  • \$\begingroup\$ I'm sorry to disagree with the comments that say the fact that all the signals are delayed the same amount cancels out the effect. I'm not an expert in the maths involved, but I believe this extra delay will cause a spread of calculated positions, which can be adjusted for statistically. \$\endgroup\$ Commented Jan 27, 2015 at 14:40
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    \$\begingroup\$ No, the extra delay does not cause a "spread". It really is the same for all signals. I have worked extensively with GPS in precision timing, positioning and interferometry applications, and I can tell you that it "just works". \$\endgroup\$
    – Dave Tweed
    Commented Jan 27, 2015 at 14:45
  • \$\begingroup\$ I am sorry Dave Tweed, I did not read your comment carefully enough before I leapt in to add my own, so I apologise for my misguided remarks. I am obviously not an expert, and I would be glad for your views on the additional material, on which I base my understanding (or mis-understanding) in my question, which I have put there because I can't put images in these comments. \$\endgroup\$ Commented Jan 27, 2015 at 15:02
  • \$\begingroup\$ @Harry I struggled with this for a long time, flipping between the delay idea and the spread-of-positions idea. It helps to imagine a 1 km long antenna cable, so it's forced to decide. \$\endgroup\$
    – tomnexus
    Commented Jan 27, 2015 at 15:24
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The GPS Pseudo Ranging process calculate and compensate the transmission time from GPS satellite to receiver by generate a replica ranging code then align the code phases, thus eliminate the transmission time taken from satellite to receiver. This allows the discipline of receiver clock to match GPS master clock on board of satellite possible.
The cable delay, no matter how much, is included in this process. Thus it needs no corrections.

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  • \$\begingroup\$ No, it's not - the receiver clock simply runs slow by the delay introduced by the antenna cable. Thought experiment: what if the cable were 1 km long. 10 km? How would the receiver know the cable length? It wouldn't see anything change. \$\endgroup\$
    – tomnexus
    Commented Jan 24, 2018 at 8:10
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Position is calculated using Time Difference of Arrival (TDOA) at the receiver. Therefore, the receiver must be able to measure accurate time interval between arriving signals, but does not need to be absolute accurate over a long period of time. Since all signals from all satellites travel through the same antenna cable, the same delay is added to all these signals, but the time interval between the arriving signals is not affected by the cable length. Therefore, position calculation does not care for cable length, but only if you need the time synch pulse when a reference signal is received, then you have to configure a cable time delay.

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  • \$\begingroup\$ Welcome to the site. While what you are saying here is true, you are not saying anything which several other people did not already explained several years before you. Positive contributions to the site would be similarly accurate answers to questions which do not already have an accurate answer, either because they are new or those which enduring value which no one has yet figured out how to answer. \$\endgroup\$ Commented Dec 11, 2020 at 18:52

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