Partner: Sania Ghazvini Abhari
GPS, Conditioning, and Nonlinear Least
Squares
There are three available satellites, which transmit signals from a
position in space to GPS receivers on earth. The receivers determine the
(x,y,z) coordinates of the satellite position. When the signal is collected, the t time, which is the difference between
the time the signal was transmitted and received, is determined. The speed
of the signal is the speed of light so c*t gives the distance
the satellite is from the receiver. Three satellites is the same as three
known spheres with two points of
intersection. Since the clocks on earth are not as accurate as the atomic
clocks onboard
the satellites, we must fix this difference between the accurate
and inaccurate time by adding an extra satellite d.
Using this equation:
(x-A1)²+(y-B1)²+(z-C1)²=[(c(t1-d)]²
Where 1,2,3 denotes the satellite with parameters A1, B1, C1, t1.
(x-A2)²+(y-B2)²+(z-C2)²=[(c(t2-d)]²
(x-A3)²+(y-B3)²+(z-C3)²=[(c(t3-d)]²
(x-A4)²+(y-B4)²+(z-C4)²=[(c(t4-d)]²
By expanding and contracting the radii of the circles by the right
common amount, a common intersection between the spheres can be found.
Part 1
The quadratic formula was used to solve the system of equations. We used
determinants to solve for d and then found x,y,z in terms of d.
The satellite positions are (Ai,Bi,Ci,ti)
Satellite 1: (15600,7540,20140,.07074)
Satellite 2: (18760,2750,18610,.07220)
Satellite 3: (17610,14630,13480,.07690)
Satellite 4: (19170,610,18390,.07242)
The correct answer: (x,y,z)= (-41.77271,-16.78919,6370.0596) and
d=-.003202 seconds.
The second solution was: (x,y,z) = (-39.7478,-134.2741,-9413.6) and
d=0.1852 seconds
Part1 Code
Part 3
Using the same satellites from Part 1, we solved for x,y,z,d using
Multivariable Newton'€™s Method.
Initial vector (x0,y0,z0,d0)= (0,0,6370,0)
The solution we got matched the solution of Part 1 with (x,y,z)=
(-41.77271,-16.78919,6370.0596) and
d=-.003202 seconds.
Part3 Code
Newton's Method Code in 4 equations
Part 4
We defined four new satellite positions with parameters Ai,Bi,Ci,ti in
sperical coordinates (ρ,φ,θ) where:
Ai= ρiCosφiCosθ
Bi= ρiCosφiSinθ
Ci= ρSinφ
and ρ=26570 km
0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ 2π
where each satillate has the range: Ri= √(Ai² + Bi² +
(Ci-6370)².
The travel time for each satellite is: ti=d+Ri/c
We changed the times by Δt=+10-8 and
Δt=-10-8 and used them with the same satellites to find a
nex,y,z,d to compare with our actual solution and then we found the Error
Magnifcation facter EMF:
EMF=(Δx,Δy,Δz)/c(Δt1...Δtm)
Our least EMF was: 3.336x10-6 which was an outlier,
so the next least was: 1.139
Our greatest EMF was: 3.206, which is the condition number
Our average EMF was: 2.06
The maximum position error in kilometers is: 0.00739 kilometers
Part4 Code
Part 5
We did the same thing in Part 5 as in Part 4 except that we
changed the θ's and φ's to be within 5% of each other.
Our least EMF was: 3.336x10-6 which was an outlier,
so the next least was: 4.006
Our greatest EMF was: 273.7, which is the condition number
Our average EMF was: 137.2
The maximum position error in kilometers is: 0.012 kilometers
Part5 Code
Part 6
For this part, we did the same thing as in Part 4, but we added four
more satellites to have a total of 8.
Our least EMF was: 3.336x10-6 which was an outlier,
so the next least was: .619
Our greatest EMF was: 1.656, which is the condition number
Our average EMF was: .878
The maximum position error in kilometers is: 0.00368 kilometers
Part6 Code
Newton Code for 8 equations
Conclusion
Since the smallest EMF was .619 from Part 6 when we used 8 satellites,
the more satesllites used, the less amount of error. The maximum GPS error
using any of these parts is .012 kilometers using bunched together
satellites, while the least GPS error is .00368 kilometers using 8 unbunched
satellites.
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