The Global Positioning System (GPS) - a type of Global Navigation Satellite System (GNSS) - uses trilatera- tion to estimate the position of a receiver with respect to the position of satellites, as depicted in the image below*. 1 4- 3 2 Three satellites would normally be sufficient for localisation but a fourth satellite allows for corrections due to clock errors in the receiver. With tuned clocks in satellites and receiver, the distance between satellite and the user / receiver is given by dk = ctk = c(Tk - Tu). Here c = 299792.458 km/s is the speed of light, and the signal flight time t = Tk – T₁ is the difference between the respective clock times T₁ and T₁ (in seconds) on satellite and the receiver. Errors in the receiver clock make this formula unsuitable, however. Instead we use the usual 2-norm distance formula modified with a clock correction term. Using Carte- sian coordinates in units of kilometres, the distance between satellite at (ak, yk, 2k) and the user/receiver at (x, y, z) is d = √(x − xk)² + (y − Yk )² + (z − zk)² + ct, where t (in seconds) incorporates the clock differences. k = 1,2,3,4. With apparent distance dk = ctk and measurements (k, Yk, 2k, tk), the values (x, y, z, t) can be deter- mined by solving the system of equations √√(x − xk)² + (y — Yk)² + (z − zk)² + c(t − tk) = 0, k = 1,2,3,4. (1) Newton's method, in its multi-function, multi-variable form, provides an ideal solution method for this. Taking f(x, y, z, t)=√√(x − xk)² + (y − Yk)² + (z − zk)² + c(t − tk), we solve the system f = 0 in the iterative, linearised Newton form, k = 1,2,3,4, (2) J(m) (m) = f(m) (3) Recalling that the Jacobian matrix J(m) is a matrix of partial derivatives, the form of ƒ in equation (2) makes for unappealing partial derivatives, especially with computational errors in mind. 1. A different version of fk can be obtained by modifying equation (1), leading to the more palatable partial derivatives, afı дх =2(x-x1), an მყ =2(y-y1), af Əz afı =2(2-21), =2c² (t1-t). Ət Determine f(x, y, z, t). 2. Construct the Jacobian matrix J(m) and the function vector f(m). 3. With the correction vector defined as c(m) = x(m+1) = x(m) and x being the ordered list (x, y, z, t), describe the algorithmic steps required to solve for the user / receiver position.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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--- Please answer the three questions provided in the attached image ---

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The Global Positioning System (GPS) - a type of Global Navigation Satellite System (GNSS) - uses trilatera-
tion to estimate the position of a receiver with respect to the position of satellites, as depicted in the image
below*.
1
4-
3
2
Three satellites would normally be sufficient for localisation but a fourth satellite allows for corrections
due to clock errors in the receiver.
With tuned clocks in satellites and receiver, the distance between satellite and the user / receiver is
given by dk = ctk = c(Tk - Tu). Here c = 299792.458 km/s is the speed of light, and the signal flight time
t = Tk – T₁ is the difference between the respective clock times T₁ and T₁ (in seconds) on satellite and
the receiver. Errors in the receiver clock make this formula unsuitable, however.
Instead we use the usual 2-norm distance formula modified with a clock correction term. Using Carte-
sian coordinates in units of kilometres, the distance between satellite at (ak, yk, 2k) and the user/receiver
at (x, y, z) is
d = √(x − xk)² + (y − Yk )² + (z − zk)² + ct,
where t (in seconds) incorporates the clock differences.
k = 1,2,3,4.
With apparent distance dk = ctk and measurements (k, Yk, 2k, tk), the values (x, y, z, t) can be deter-
mined by solving the system of equations
√√(x − xk)² + (y — Yk)² + (z − zk)² + c(t − tk) = 0,
k = 1,2,3,4.
(1)
Newton's method, in its multi-function, multi-variable form, provides an ideal solution method for this.
Taking
f(x, y, z, t)=√√(x − xk)² + (y − Yk)² + (z − zk)² + c(t − tk),
we solve the system f = 0 in the iterative, linearised Newton form,
k = 1,2,3,4,
(2)
J(m) (m) = f(m)
(3)
Recalling that the Jacobian matrix J(m) is a matrix of partial derivatives, the form of ƒ in equation (2)
makes for unappealing partial derivatives, especially with computational errors in mind.
1. A different version of fk can be obtained by modifying equation (1), leading to the more palatable
partial derivatives,
afı
дх
=2(x-x1),
an
მყ
=2(y-y1),
af
Əz
afı
=2(2-21),
=2c² (t1-t).
Ət
Determine f(x, y, z, t).
2. Construct the Jacobian matrix J(m) and the function vector f(m).
3. With the correction vector defined as c(m) = x(m+1) = x(m) and x being the ordered list (x, y, z, t),
describe the algorithmic steps required to solve for the user / receiver position.
Transcribed Image Text:The Global Positioning System (GPS) - a type of Global Navigation Satellite System (GNSS) - uses trilatera- tion to estimate the position of a receiver with respect to the position of satellites, as depicted in the image below*. 1 4- 3 2 Three satellites would normally be sufficient for localisation but a fourth satellite allows for corrections due to clock errors in the receiver. With tuned clocks in satellites and receiver, the distance between satellite and the user / receiver is given by dk = ctk = c(Tk - Tu). Here c = 299792.458 km/s is the speed of light, and the signal flight time t = Tk – T₁ is the difference between the respective clock times T₁ and T₁ (in seconds) on satellite and the receiver. Errors in the receiver clock make this formula unsuitable, however. Instead we use the usual 2-norm distance formula modified with a clock correction term. Using Carte- sian coordinates in units of kilometres, the distance between satellite at (ak, yk, 2k) and the user/receiver at (x, y, z) is d = √(x − xk)² + (y − Yk )² + (z − zk)² + ct, where t (in seconds) incorporates the clock differences. k = 1,2,3,4. With apparent distance dk = ctk and measurements (k, Yk, 2k, tk), the values (x, y, z, t) can be deter- mined by solving the system of equations √√(x − xk)² + (y — Yk)² + (z − zk)² + c(t − tk) = 0, k = 1,2,3,4. (1) Newton's method, in its multi-function, multi-variable form, provides an ideal solution method for this. Taking f(x, y, z, t)=√√(x − xk)² + (y − Yk)² + (z − zk)² + c(t − tk), we solve the system f = 0 in the iterative, linearised Newton form, k = 1,2,3,4, (2) J(m) (m) = f(m) (3) Recalling that the Jacobian matrix J(m) is a matrix of partial derivatives, the form of ƒ in equation (2) makes for unappealing partial derivatives, especially with computational errors in mind. 1. A different version of fk can be obtained by modifying equation (1), leading to the more palatable partial derivatives, afı дх =2(x-x1), an მყ =2(y-y1), af Əz afı =2(2-21), =2c² (t1-t). Ət Determine f(x, y, z, t). 2. Construct the Jacobian matrix J(m) and the function vector f(m). 3. With the correction vector defined as c(m) = x(m+1) = x(m) and x being the ordered list (x, y, z, t), describe the algorithmic steps required to solve for the user / receiver position.
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