Q Method and Wahba's Problem - Revision history

Yash: /* Q-Method */

2018-02-19T12:50:07Z

‎Q-Method

Yash: /* Q-Method */

2018-02-19T12:47:37Z

‎Q-Method

Yash: /* Q-Method */

2018-02-19T12:08:40Z

‎Q-Method

Yash at 12:07, 19 February 2018

2018-02-19T12:07:53Z

Yash at 12:01, 19 February 2018

2018-02-19T12:01:50Z

Yash at 03:14, 10 February 2018

2018-02-10T03:14:52Z

Yash at 03:14, 10 February 2018

2018-02-10T03:14:24Z

Yash: /* Q-Method */

2018-02-10T03:13:30Z

‎Q-Method

Yash at 03:11, 10 February 2018

2018-02-10T03:11:54Z

Yash at 03:11, 10 February 2018

2018-02-10T03:11:32Z

@@ Line 31: / Line 31: @@
 <math>g(\bar{q}) = \bar{q}^TKq</math><br \>
 Where K is a 4x4 Matrix given by
-[[File:Equation33.png|frame|center]]
+[[File:Equation77.png|frame|center]]
 Wherein,<br \>
 <math>B = \sum w_k(v_{kb}v_{ki}^T)</math><br \>

@@ Line 25: / Line 25: @@
 We can see that in order to minimize J, we need to maximize g(R). This problem can be written in a better and more elegant way by representing rotation matrix R in terms of the quaternion
 [[File:Equation29.png|frame|center]]
-R can be written in terms of q as
+R can be written in terms of q as <math>R = (q_4^2-q^Tq)I+2qq^T-2q_4[q  \times]</math><br \>
 Where [q x] is the cross product matrix associated with q
 [[File:Equation31.png|frame|center]]

@@ Line 33: / Line 33: @@
 Where K is a 4x4 Matrix given by
 [[File:Equation33.png|frame|center]]
-Wherein,
+Wherein,<br \>
 <math>B = \sum w_k(v_{kb}v_{ki}^T)</math><br \>
 <math>S = B+B^T</math><br \>

@@ Line 34: / Line 34: @@
 [[File:Equation33.png|frame|center]]
 Wherein,
-[[File:Equation34.png|frame|center]]
+<math>B = \sum w_k(v_{kb}v_{ki}^T)</math><br \>
 Now we want to maximise g. However, noting that quaternion elements are not independent, the constraint <math>\bar{q}^Tq = 1</math> must also be satisfied. So we use the lagrangian multiplier method and differentiate with respect to <math>\bar{q}</math> we get <math>K\bar{q} = \lambda \bar{q}</math>. So now, we have reduced the problem to an eigenvalue problem for the matrix K.
 Since, K is a 4x4 matrix, it can have at most 4 different eigenvalues. Substituting <math>K\bar{q} = \lambda \bar{q}</math> in <math>g(\bar{q}) = \bar{q}^TKq</math>, we get

@@ Line 35: / Line 35: @@
 Wherein,
 [[File:Equation34.png|frame|center]]
-Now we want to maximise g. However, noting that quaternion elements are not independent, the constraint [[File:Equation35.png]] must also be satisfied. So we use the lagrangian multiplier method and differentiate with respect to [[File:Equation36.png]] we get [[File:Equation37.png]] So now, we have reduced the problem to an eigenvalue problem for the matrix K.
+Now we want to maximise g. However, noting that quaternion elements are not independent, the constraint <math>\bar{q}^Tq = 1</math> must also be satisfied. So we use the lagrangian multiplier method and differentiate with respect to <math>\bar{q}</math> we get <math>K\bar{q} = \lambda \bar{q}</math>. So now, we have reduced the problem to an eigenvalue problem for the matrix K.
-Since, K is a 4x4 matrix, it can have at most 4 different eigenvalues. Substituting [[File:Equation37.png]] in [[File:Equation38.png]], we get
+Since, K is a 4x4 matrix, it can have at most 4 different eigenvalues. Substituting <math>K\bar{q} = \lambda \bar{q}</math> in <math>g(\bar{q}) = \bar{q}^TKq</math>, we get
 [[File:Equation39.png|frame|center]]
 Now we want to maximise g. So we will chose the eigenvector corresponding to the maximum eigenvalue.

@@ Line 8: / Line 8: @@
 [[File:Equation72.png|frame|center]]
 In this expression, J is the loss function that has to be minimized, k is the counter for the
-N observations, <math<w_k</math> is weight assigned to the kth measurement, <math>v_{kb}</math> is the matrix consisting of the measured
+N observations, <math>w_k</math> is weight assigned to the kth measurement, <math>v_{kb}</math> is the matrix consisting of the measured
 components in the body frame, and <math>v_{ki}</math> is the matrix consisting of components in the inertial
 frame as determined using appropriate mathematical models. This loss function is a

@@ Line 4: / Line 4: @@
 for each of the N vectors. It is clear that this set of equations will be overdetermined for N > 2,
 and hence the equation, in general, cannot be satisfied for each k = 1,2..N. Hence,
-we wish to find a solution for R_{bi} that in some way minimizes the overall error for
+we wish to find a solution for <math>R_{bi}</math> that in some way minimizes the overall error for
 the N vectors. <ref>http://www.dept.aoe.vt.edu/~cdhall/courses/aoe4140/attde.pdf</ref><br \>
 [[File:Equation72.png|frame|center]]
 In this expression, J is the loss function that has to be minimized, k is the counter for the
-N observations, w_k is weight assigned to the kth measurement, v_{kb} is the matrix consisting of the measured
+N observations, <math<w_k</math> is weight assigned to the kth measurement, <math>v_{kb}</math> is the matrix consisting of the measured
-components in the body frame, and v_{ki} is the matrix consisting of components in the inertial
+components in the body frame, and <math>v_{ki}</math> is the matrix consisting of components in the inertial
 frame as determined using appropriate mathematical models. This loss function is a
 sum of the squared errors for each vector measurement. If the measurements and
 mathematical models are perfect, then J = 0, since the first equation on this page will be satisfied for all N vectors. If there are any errors or noisy measurements, then J > 0. The smaller
-we make J, the better the approximation of R_{bi}.
+we make J, the better the approximation of <math>R_{bi}</math>.
 == Q-Method ==