Inside GNSS Media & Research

MAR-APR 2018

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www.insidegnss.com M A R C H / A P R I L 2 0 1 8 Inside GNSS 29 in the east/west direction. Before moving on, although we used the example of a geostationary satellite, the exact same effect applies to non-geostationary orbits as well. e main difference is that the satellite positions in Figures 1 and 2 would not necessarily be directly above the user, and the distance between the user and satellites, projected into the equato- rial plane (which is shown in Figures 1 and 2), will vary with time as satellites move along their orbits. e good news is that regardless of the orbit, the meth- od of compensation is the same. Simple Solution To remove the discrepancy between the measured and computed signal paths, we need to compute the ECEF position of the satellite at the time of transition in the ECEF frame at the time of signal reception. Fortunately, this is easily accomplished by realizing that the two coordinate frames are related by a rotation about the z-axis. Mathematically, we can write where is a position vector at the subscripted time (or frame), and R 3 (ω e . (t r – t t )) is the rotation matrix about the z-axis by the angle subtended by the Earth rotated during signal propagation. Applying the transformation in (1) yields the position of the yellow satel- lite in Figure 2, which allows for the proper computation of the (orange) user position. e astute reader might be wonder- ing how the propagation time is com- puted. is can be found by iterating to a solution: first, assume an initial distance between the user and satel- lite (e.g., 70 milliseconds); then com- pute the satellite position using this assumed distance (for Earth rotation compensation); use the approximate user position to re-compute the range to the satellite; and finally use this range to compute the satellite position. e accuracy of the user position in the iteration is not typically a problem. e reason is because, even with a posi- tion error of 10 kilometers, the worst- case propagation time error would be 33.3 μs (i.e., 10 km / 3e8 m/s). Multiply- ing this by Earth rotation rate (~7.3e-5 rad/s) yields an angular error of about 2.4 nanoradians. Even over an orbital radius of 26,000 kilometers (assuming a MEO orbit), the orbital error is less than a decimeter. en, of course, aer the first epoch, the position error is typically several orders of magnitude smaller making the effect of user posi- tion error negligible. Summary is article has shown why Earth rota- tion needs to be accounted for when computing satellite coordinates for GNSS applications. e compensation is simple but crucial steps for obtaining the highest possible positioning accu- racies. GNSS SOLUTIONS

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