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The Bancroft method allows to obtain a direct solution of the receiver position and the clock offset, without requesting any "a priori" knowledge for the receiver location.
{{Article Infobox2
|Category=Fundamentals
|Authors=J. Sanz Subirana, J.M. Juan Zornoza and M. Hernández-Pajares, Technical University of Catalonia, Spain.
|Level=Advanced
|YearOfPublication=2011
|Title={{PAGENAME}}
}}
The Bancroft method allows obtaining a direct solution of the receiver position and the clock offset, without requesting any "a priori" knowledge for the receiver location.


  insert Appendix D here.
 
==Raising and resolution==
 
Let <math>PR^j</math> the prefit-residual of satellite-<math>j</math>, computed from equation (1)
 
::<math>
R^j=\rho^j+c(\delta t-\delta t^j)+T^j+\hat{\alpha}\, I^j+TGD^j+\mathcal{M}^j+{\boldsymbol \varepsilon}^j
\qquad \mbox{(1)}</math>
 
 
after removing all model terms not needing the a priory knowledge of the receiver position:<ref group="footnotes">The tropospheric and ionospheric terms,  <math>T^j</math> and <math>\hat{\alpha} \,I^j</math>, can not be included, because the need to consider the satellite-receiver ray. Off course, after an initial computation of the receiver coordinates, the method could be iterated using the ionospheric and tropospheric corrections to improve the solution.</ref>
 
::<math>
PR^j\equiv R^j +c\,\delta t^j-TGD^j
\qquad \mbox{(2)}</math>
 
 
 
Thence, neglecting the tropospheric and ionospheric terms, as well as the multipath and receiver noise, the equation (3)
 
::<math>
\begin{array}{r}
R^j-D^j\simeq \sqrt{(x^j-x)^2+(y^j-y)^2+(z^j-z)^2}+c\,\delta t\\[0.3cm]
j=1,2,...,n~~~~ (n \geq 4)\\
\end{array}
\qquad \mbox{(3)}</math>
 
can be written as:
 
::<math>
PR^j = \sqrt{(x^j-x)^2+(y^j-y)^2+(z^j-z)^2}+c \, \delta t
\qquad \mbox{(4)}</math>
 
 
 
Developing the previous equation (4), it follows:
 
::<math>\left[{x^j}^2+{y^j}^2+{z^j}^2-{PR^j}^2 \right]-2 \left[x^j x+y^j y+z^j z-{PR^jc\,\delta t} \;  \right] + \left[x^2+y^2+z^2-(c\,\delta t)^2 \right]=0 \qquad \mbox{(5)}</math>
 
 
Then, calling <math>{\mathbf r}=[x,y,z]^T</math> and considering the inner product of
Lorentz <ref group="footnotes">
<math>
\left \langle{\mathbf a},{\mathbf b}\right \rangle={\mathbf a}^{t} \; {\mathbf M} \; {\mathbf b}=
\left[
\begin{array}{c}
a_1,a_2,a_3,a_4
\end{array}
\right]
\left(
\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&-1\\
\end{array}
\right)
\left[
\begin{array}{c}
b_1\\
b_2\\
b_3\\
b_4
\end{array}
\right]
</math>
</ref>
the previous equation (5) can be expressed in a more compact way as:
 
::<math>\frac{1}{2} \left  \langle
\left[
\begin{array}{c}
{\mathbf r}^j\\
PR^j\\
\end{array}
\right],
\left[
\begin{array}{c}
{\mathbf r}^j\\
PR^j\\
\end{array}
\right]
\right \rangle
-
\left  \langle
\left[
\begin{array}{c}
{\mathbf r}^j\\
PR^j\\
\end{array}
\right],
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
\right \rangle
+
\frac{1}{2} \left  \langle
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right],
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
\right \rangle
=0
\qquad \mbox{(6)}</math>
 
 
 
The former equation can be raised for every satellite (or prefit-residual <math>PR^j</math>).
 
 
If four measurements are available, thence, the following matrix can be written, containing all the available information on satellite coordinates and pseudoranges (every row corresponds to a satellite):
 
::<math>
{\mathbf B}=
\left(
\begin{array}{cccc}
x^1&y^1&z^1&PR^1\\
x^2&y^2&z^2&PR^2\\
x^3&y^3&z^3&PR^3\\
x^4&y^4&z^4&PR^4\\
\end{array}
\right)
\qquad \mbox{(7)}</math>
 
 
Then, calling:
 
::<math>\Lambda=
\frac{1}{2}
\left  \langle
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right],
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
\right \rangle
\; ,
\;
{\mathbf 1}=
\left[
\begin{array}{c}
1\\
1\\
1\\
1\\
\end{array}
\right]
\; ,
\;
{\mathbf a}=
\left[
\begin{array}{c}
a_1\\
a_2\\
a_3\\
a_4\\
\end{array}
\right]
\;
\mbox{being}
\;
\;
a_j=
\frac{1}{2}
\left  \langle
\left[
\begin{array}{c}
{\mathbf r}^j\\
PR^j\\
\end{array}
\right],
\left[
\begin{array}{c}
{\mathbf r}^j\\
PR^j\\
\end{array}
\right]
\right \rangle
\qquad \mbox{(8)}</math>
 
 
 
The four equations for pseudorange can be expressed as:
 
::<math>{\mathbf a} -{\mathbf B}\,{\mathbf M} \left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
+\Lambda \; {\mathbf 1}=0\;\;,\;\;\;\; \mbox{being} \;\;\;\;\;\;
{\mathbf M}=\left(
\begin{array}{cccc}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&-1\\
\end{array}
\right)
\qquad \mbox{(9)}</math>
 
 
from where:
 
::<math>
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
={\mathbf M} {\mathbf B}^{-1} (\Lambda \; {\mathbf 1} + {\mathbf a})
\qquad \mbox{(10)}</math>
 
 
 
Then, taking into account the following equality
 
:: <math>\langle {\mathbf M}{\mathbf g},{\mathbf M}{\mathbf h} \rangle=\langle {\mathbf g},{\mathbf h} \rangle \qquad \mbox{(11)}</math>,
 
 
and that
 
::<math>
\Lambda=
\frac{1}{2}
\left  \langle
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right],
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
\right \rangle
\qquad \mbox{(12)}</math>, 
 
 
from the former expression (10), one obtains:
 
::<math>
\left \langle {\mathbf B}^{-1} {\mathbf 1}, {\mathbf B}^{-1} {\mathbf 1} \right \rangle \Lambda^2+ 2\left [  \left \langle {\mathbf B}^{-1} {\mathbf 1}, {\mathbf B}^{-1} {\mathbf a} \right \rangle -1 \right ] \Lambda +  \left \langle {\mathbf B}^{-1} {\mathbf a}, {\mathbf B}^{-1} {\mathbf a} \right \rangle =0
\qquad \mbox{(13)}</math>
 
 
The previous expression (13) is a quadratic equation in <math>\Lambda</math> (note that matrix <math>{\mathbf B}</math> and the vector <math>\mathbf a </math> are also known) and provides two solutions, that introduced in expression (10) provides the searched solution:
 
::<math>
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
\qquad \mbox{(14)}</math>.
 
The other solution is far from the earth surface.
 
==Generalisation to the case of <math>n</math>-measurements:==
 
 
If more than four observations are available, the matrix <math>{\mathbf B}</math> is not square. However, multiplying by <math>{\mathbf B}^T</math>, one obtains (Least Squares solution):
 
::<math>
{\mathbf B}^T{\mathbf a} -{\mathbf B}^T {\mathbf B}\,{\mathbf M} \left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
+\Lambda \; {\mathbf B}^T {\mathbf 1}=0
\qquad \mbox{(15)}</math>
 
 
where:
 
::<math>
\left[
\begin{array}{c}
{\mathbf r}\\
c\,\delta t\\
\end{array}
\right]
={\mathbf M} ({\mathbf B}^T {\mathbf B})^{-1}{\mathbf B}^T(\Lambda \; {\mathbf 1} + {\mathbf a})
\qquad \mbox{(16)}</math>
 
 
and then:
 
::<math>
\begin{array}{r}
\left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1} \right \rangle \Lambda^2+ 2\left [  \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a} \right \rangle -1 \right ] \Lambda +\\[0.3cm]
+ \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a} \right \rangle =0
\end{array}
\qquad \mbox{(17)}</math>
 
 
==Notes==
<references group="footnotes"/>


[[Category:Fundamentals]]
[[Category:Fundamentals]]
[[Category:GNSS Measurements Modelling]]

Latest revision as of 11:12, 7 July 2014


FundamentalsFundamentals
Title Bancroft Method
Author(s) J. Sanz Subirana, J.M. Juan Zornoza and M. Hernández-Pajares, Technical University of Catalonia, Spain.
Level Advanced
Year of Publication 2011

The Bancroft method allows obtaining a direct solution of the receiver position and the clock offset, without requesting any "a priori" knowledge for the receiver location.


Raising and resolution

Let [math]\displaystyle{ PR^j }[/math] the prefit-residual of satellite-[math]\displaystyle{ j }[/math], computed from equation (1)

[math]\displaystyle{ R^j=\rho^j+c(\delta t-\delta t^j)+T^j+\hat{\alpha}\, I^j+TGD^j+\mathcal{M}^j+{\boldsymbol \varepsilon}^j \qquad \mbox{(1)} }[/math]


after removing all model terms not needing the a priory knowledge of the receiver position:[footnotes 1]

[math]\displaystyle{ PR^j\equiv R^j +c\,\delta t^j-TGD^j \qquad \mbox{(2)} }[/math]


Thence, neglecting the tropospheric and ionospheric terms, as well as the multipath and receiver noise, the equation (3)

[math]\displaystyle{ \begin{array}{r} R^j-D^j\simeq \sqrt{(x^j-x)^2+(y^j-y)^2+(z^j-z)^2}+c\,\delta t\\[0.3cm] j=1,2,...,n~~~~ (n \geq 4)\\ \end{array} \qquad \mbox{(3)} }[/math]

can be written as:

[math]\displaystyle{ PR^j = \sqrt{(x^j-x)^2+(y^j-y)^2+(z^j-z)^2}+c \, \delta t \qquad \mbox{(4)} }[/math]


Developing the previous equation (4), it follows:

[math]\displaystyle{ \left[{x^j}^2+{y^j}^2+{z^j}^2-{PR^j}^2 \right]-2 \left[x^j x+y^j y+z^j z-{PR^jc\,\delta t} \; \right] + \left[x^2+y^2+z^2-(c\,\delta t)^2 \right]=0 \qquad \mbox{(5)} }[/math]


Then, calling [math]\displaystyle{ {\mathbf r}=[x,y,z]^T }[/math] and considering the inner product of Lorentz [footnotes 2] the previous equation (5) can be expressed in a more compact way as:

[math]\displaystyle{ \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right] \right \rangle - \left \langle \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle + \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle =0 \qquad \mbox{(6)} }[/math]


The former equation can be raised for every satellite (or prefit-residual [math]\displaystyle{ PR^j }[/math]).


If four measurements are available, thence, the following matrix can be written, containing all the available information on satellite coordinates and pseudoranges (every row corresponds to a satellite):

[math]\displaystyle{ {\mathbf B}= \left( \begin{array}{cccc} x^1&y^1&z^1&PR^1\\ x^2&y^2&z^2&PR^2\\ x^3&y^3&z^3&PR^3\\ x^4&y^4&z^4&PR^4\\ \end{array} \right) \qquad \mbox{(7)} }[/math]


Then, calling:

[math]\displaystyle{ \Lambda= \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle \; , \; {\mathbf 1}= \left[ \begin{array}{c} 1\\ 1\\ 1\\ 1\\ \end{array} \right] \; , \; {\mathbf a}= \left[ \begin{array}{c} a_1\\ a_2\\ a_3\\ a_4\\ \end{array} \right] \; \mbox{being} \; \; a_j= \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}^j\\ PR^j\\ \end{array} \right] \right \rangle \qquad \mbox{(8)} }[/math]


The four equations for pseudorange can be expressed as:

[math]\displaystyle{ {\mathbf a} -{\mathbf B}\,{\mathbf M} \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] +\Lambda \; {\mathbf 1}=0\;\;,\;\;\;\; \mbox{being} \;\;\;\;\;\; {\mathbf M}=\left( \begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-1\\ \end{array} \right) \qquad \mbox{(9)} }[/math]


from where:

[math]\displaystyle{ \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] ={\mathbf M} {\mathbf B}^{-1} (\Lambda \; {\mathbf 1} + {\mathbf a}) \qquad \mbox{(10)} }[/math]


Then, taking into account the following equality

[math]\displaystyle{ \langle {\mathbf M}{\mathbf g},{\mathbf M}{\mathbf h} \rangle=\langle {\mathbf g},{\mathbf h} \rangle \qquad \mbox{(11)} }[/math],


and that

[math]\displaystyle{ \Lambda= \frac{1}{2} \left \langle \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right], \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \right \rangle \qquad \mbox{(12)} }[/math],


from the former expression (10), one obtains:

[math]\displaystyle{ \left \langle {\mathbf B}^{-1} {\mathbf 1}, {\mathbf B}^{-1} {\mathbf 1} \right \rangle \Lambda^2+ 2\left [ \left \langle {\mathbf B}^{-1} {\mathbf 1}, {\mathbf B}^{-1} {\mathbf a} \right \rangle -1 \right ] \Lambda + \left \langle {\mathbf B}^{-1} {\mathbf a}, {\mathbf B}^{-1} {\mathbf a} \right \rangle =0 \qquad \mbox{(13)} }[/math]


The previous expression (13) is a quadratic equation in [math]\displaystyle{ \Lambda }[/math] (note that matrix [math]\displaystyle{ {\mathbf B} }[/math] and the vector [math]\displaystyle{ \mathbf a }[/math] are also known) and provides two solutions, that introduced in expression (10) provides the searched solution:

[math]\displaystyle{ \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] \qquad \mbox{(14)} }[/math].

The other solution is far from the earth surface.

Generalisation to the case of [math]\displaystyle{ n }[/math]-measurements:

If more than four observations are available, the matrix [math]\displaystyle{ {\mathbf B} }[/math] is not square. However, multiplying by [math]\displaystyle{ {\mathbf B}^T }[/math], one obtains (Least Squares solution):

[math]\displaystyle{ {\mathbf B}^T{\mathbf a} -{\mathbf B}^T {\mathbf B}\,{\mathbf M} \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] +\Lambda \; {\mathbf B}^T {\mathbf 1}=0 \qquad \mbox{(15)} }[/math]


where:

[math]\displaystyle{ \left[ \begin{array}{c} {\mathbf r}\\ c\,\delta t\\ \end{array} \right] ={\mathbf M} ({\mathbf B}^T {\mathbf B})^{-1}{\mathbf B}^T(\Lambda \; {\mathbf 1} + {\mathbf a}) \qquad \mbox{(16)} }[/math]


and then:

[math]\displaystyle{ \begin{array}{r} \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1} \right \rangle \Lambda^2+ 2\left [ \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf 1}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a} \right \rangle -1 \right ] \Lambda +\\[0.3cm] + \left \langle ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a}, ({\mathbf B}^T {\mathbf B})^{-1} {\mathbf B}^T{\mathbf a} \right \rangle =0 \end{array} \qquad \mbox{(17)} }[/math]


Notes

  1. ^ The tropospheric and ionospheric terms, [math]\displaystyle{ T^j }[/math] and [math]\displaystyle{ \hat{\alpha} \,I^j }[/math], can not be included, because the need to consider the satellite-receiver ray. Off course, after an initial computation of the receiver coordinates, the method could be iterated using the ionospheric and tropospheric corrections to improve the solution.
  2. ^ [math]\displaystyle{ \left \langle{\mathbf a},{\mathbf b}\right \rangle={\mathbf a}^{t} \; {\mathbf M} \; {\mathbf b}= \left[ \begin{array}{c} a_1,a_2,a_3,a_4 \end{array} \right] \left( \begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-1\\ \end{array} \right) \left[ \begin{array}{c} b_1\\ b_2\\ b_3\\ b_4 \end{array} \right] }[/math]