Mohinder S. Grewal - Global Navigation Satellite Systems, Inertial Navigation, and Integration

(2.39)

with unknown 4 × 1 state vector

We can rewrite the four equations in matrix form as

or

(2.40)

where

Then, we premultiply both sides of Eq. (2.40)by M −1:

If the rank of M , the number of linear independent columns of the matrix M , is less than 4, then M will not be invertible.

Differentiate Eq. (2.21)with respect to time without C b.

(2.41)

Differentiate Eq. (2.41)with respect to

(2.42)

where .

In classical navigation geometry, the components (3 × 3) of this unit vector are often called direction cosine. It is interesting to note that these components are the same as the position linearization shown in Eqs. (2.26a)and (2.26b).

Equations (2.42)and (2.26b)will be used in GPS/INS tightly coupled implementation as measurement equations for pseudoranges and/or delta pseudoranges in chapters 11 and 12 in the extended Kalman filters. Equation (2.27)will be used in integrity determination of GNSS satellites in Chapter 9 and from Eq. (2.41),

(2.43)

where

	=	range rate (known)
ρ r	=	range (known)
( x, y, z )	=	satellite positions (known)
( , , )	=	satellite rates (known)
X , Y , Z	=	user position (known from position calculations)
( , , )	=	user velocity (unknown)

For three satellites, Eq. (2.43)becomes

(2.44)

Equation (2.44)becomes

(2.45)

(2.46)

where

However, if the rank of N is <3, N will not be invertible.

Refer to Appendix B for coordinate system definitions and to Section B.3.10 for satellite orbit equations.

1 2.1 Which of the following coordinate systems is not rotating?North–east–down (NED)East–north–up (ENU)Earth‐centered, Earth‐fixed (ECEF)Earth‐centered inertial (ECI)Moon‐centered, moon fixed

2 2.2 Show that the 3 × 3 identity matrix. (Hint: ).

3 2.3 Rank VDOP, HDOP, and PDOP from smallest (best) to largest (worst) under normal conditions:VDOP ≤ HDOP ≤ PDOPVDOP ≤ PDOP ≤ HDOPHDOP ≤ VDOP≤PDOPHDOP ≤ PDOP ≤ VDOPPDOP ≤ HDOP ≤ VDOPPDOP ≤ VDOP ≤ HDOP

4 2.4 UTC time and the GPS time are offset by an integer number of seconds (e.g. 16 seconds as of June 2012) as well as a fraction of a second. The fractional part is approximately.0.1–0.5 s1–2 ms100–200 ns10–20 ns

5 2.5 Derive equations (2.41)and (2.42).

6 2.6 For the following GPS satellites, find the satellite position in ECEF coordinates at t = 3 seconds. (Hint: See Appendix B.) Ω0 and θ0 are given below at time t0 = 0:Ω0 (°)θ0 (°)(a)32668(b)2634

7 2.7 Using the results of the previous problem, find the satellite positions in the local reference frame. Reference should be to the COMSAT facility in Santa Paula, California, located at 32.4° latitude, −119.2° longitude. Use coordinate shift matrix S = 0. (Refer to Section B.3.10.)

8 2.8 Given the following GPS satellite coordinates and pseudoranges:SatelliteΩ0 (°)θ0 (°)ρ (m)1326682.324 × 1072263402.0755 × 10731461982.1103 × 1074862712.3491 × 107Find the user's antenna position in ECEF coordinates.Find the user's antenna position in locally level coordinates referenced to 0° latitude, 0° longitude. Coordinate shift matrix S = 0.Find the various DOPs.

9 2.9 Given two satellites in north and east coordinateswith pseudorangesand starting with an initial guess of xest, yest, find the user's antenna position.

10 2.10 A satellite position at time t = 0 is specified by its orbital parameters as Ω0 = 92.847°, θ0 = 135.226°, α = 55°, R = 26 560 000 m.Find the satellite position at one second, in ECEF coordinates.Convert the satellite position from (a) with user atfrom WGS84 (ECEF) to ENU coordinates with origin at