wujing1129/aae6102_lab

This is my AAE6102 coursework, welcome discussion <^_^>

Introduction

The primary purpose of this Lab experiment is to perform single-point positioning using satellite broadcast ephemeris data and receiver observations for a given epoch. The Least-Square Method has been used for estimating the receiver's coordinate and the receiver's clock offset.

The receiver's initial position is X0 = -2694685.473, Y0 = -4293642.366, Z0 = 3857878.924 (WGS 84 XYZ in meters), and the initial clock offset of the receiver is zero.

Instruction

The usage of the source code:

1. Download the zip file and uncompressed.
2. Set the current folder of Matlab to spp folder.
3. Type *main* in command line windows.

The Frame of Codes

1. readrcv.m: Read the rcvr.dat file and store the data in rcv struct.
2. readeph.m: Read the eph.dat file and store the data in Eph_broadcast struct.
3. eph2satxyz.m: Convert the orbit parameters to ECEF coordinate (The earth rotation has been considered, and the relativity effect also be considered).
4. XYZ2BLH.m
5. XYZ2NEU.m
6. main.m: Using the Least-Square method to calculate the receiver's coordinate and clock offset.

Results

The ECEF position and the clock offset of satellites.

svid	X (m)	Y (m)	Z (m)	Clock offset (s)
5	-8855545.4737	-22060115.1685	-11922133.1713	0.00018907
6	-8087233.0607	-16945956.5996	18816198.5368	-8.3932e-08
17	-21277119.8022	-7467119.5023	14287505.0468	-0.0002049
30	-17713900.9391	-19797464.3067	19213.2028	-1.0041e-05
10	9027646.532	-12319233.163	21737387.4107	3.3248e-05
23	-19452319.1118	-16750376.9085	-6918519.6344	1.036e-05
22	-13649531.2661	8229504.3157	21122957.8349	0.00022268
26	6162910.1746	-25286774.5207	-3541191.2494	0.00028099

The iteration of this LS problem.

Iterations	Position X,Y,Z changed (m)	Receiver clock changed (s)	Updated position ECEF (m)
1	dx = -5737.933 (m) dy = 1099.090 (m) dz = -2610.774 (m)	dtr = 0.001733 (s)	X = -2700423.406332 Y = -4292543.276071 Z = 3855268.150447
2	dx = -0.130 (m) dy = -0.100 (m) dz = 0.105 (m)	dtr = -0.000000 (s)	X = -2700423.536456 Y = -4292543.375690 Z = 3855268.255359
3	dx = -0.000 (m) dy = 0.000 (m) dz = 0.000 (m)	dtr = 0.000000 (s)	X = -2700423.536456 Y = -4292543.375690 Z = 3855268.255359

Reference

1. ARINC Research Corporation. (2000). GPS Interface Control Document ICD-GPS-200 (IRN-200C-004): Navstar GPS Space Segment and Aviation User Interfaces.

aae6102_lab's Issues

来你的github 逛逛。

我在中南看到很多人在做brdc-ppp很难理解为啥这样做。你有兴趣研究吗？