ASEN 5070 Statistical Orbit Determination I Fall 2012 Professor Jeffrey S. Parker

CCARColorado Center for

Astrodynamics Research

University of ColoradoBoulder 1

ASEN 5070Statistical Orbit Determination I

Fall 2012

Professor Jeffrey S. ParkerProfessor George H. Born

Lecture 21: Exam 2 Debrief and More Fun




Homework 9 due this week.◦ Make sure you spend time studying for the exam

Homework 10 out Thursday.◦ Give you a small reprieve to focus on HW9.

Announcements




Quiz 17 Review




Quiz 17 Review




Quiz 17 Review




Quiz 17 Review

The matrix of partials of one observation relative to the state parameters is identical to the other matrix.




Quiz 17 Review




Quiz 17 Review




Quiz 17 Review




Quiz 17 Review




Due this Thursday

HW#9




HW#9 Tip Try building x-hat from the data given online. If you can get

that to work then you’ll have a better chance of getting your own x-hat to match the solutions!

Grab the accumulated matrices HTWH and HTWY. Try computing inv(HTWH+P0bar)*HTWY




HW#9 Tip Your x-hat should match to at least 1 digit of precision in

each parameter (hopefully 3). It will not be identical!◦ Different integrator◦ Different tolerance◦ Different computer◦ Different inverter

inv(HTWH+P0bar) is very poorly conditioned (e-34 I believe)

Matlab’s “inv” function will not produce the right answer.




HW#9 Tip inv(HTWH+P0bar) is very poorly conditioned (e-34 I

believe)

R = chol( HTWH+P0bar) Inv(R) is also poorly conditioned, but only e-1.

This is far better.

If RTR = (HTWH+P0bar), what is inv( RTR )?




Overall, the class did well. Most everyone grasped the concepts.

Nobody got 100% - so don’t worry if your grade was lower than 90. (curve TBD)

Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief

i.e., High Precision but low accuracy!




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief

Only guarantees a nonnegative definite!




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief




Exam 2 Debrief

4x3




Exam 2 Debrief




Exam 2 Debrief

[3x4]*[4x3] = [3x3] (hint: it’s always nxn)




Exam 2 Debrief




Exam 2 Debrief

1. one observation vector includes 4 independent pieces of information. We only need 3 pieces of information.




Exam 2 Debrief




Exam 2 Debrief

Then Phi, A, y, H-tildex-hat, P, x-bar, P-bar




Exam 2 Debrief




Exam 2 Debrief

X* (the reference trajectory)x-bar (the a priori deviation, nominally zero)P-bar (the a priori covariance)Y_i (the observations)omega and sigma (though that’s specific to this problem and it’s okay if you didn’t include that!




Exam 2 Debrief




Quick Break

Next up: Stuff.

◦ Prediction Residual◦ Givens◦ Householder◦ Future: Process Noise, Smoothing



University of ColoradoBoulder

The Prediction Residual













This would be especially important in the case of the EKF




Next: Orthogonal transformations: Givens, Householder






















So how do we select Q?

Givens, Householder, many methods

Choices from here




Givens




Givens




Givens




Givens




Givens




Givens

Apply the rotation across the matrix, converting it into a triangular matrix




Givens




Givens




Accuracy Comparison for Batch and Givens – Finite Precision Computer

Consider 1 11 11 1

H

Machine precision is such that

The normal matrix is given by

2

3 33 3 2

TH H

; exact solution

our computer will drop the and23 3

3 3 2TH H

To order , hence itis singular

0TH H

Notice that a vector of Observations is not needed.Why?





Consequently, the Batch Processor will fail to yield a solution. Note that thisIllustrates the problem with forming HTH, i.e. numerical problems are amplified.

The Cholesky decomposition yields: 33

30 0

R

R is singular and will not yield a solution for . x̂





Use the Givens transformation to determine R 1st zero element (2,1) of H 0 1 1

0 1 10 0 1 1 1

C SS C





Use the Givens transformation to determine R 1st zero element (2,1) of H 0 1 1

0 1 10 0 1 1 1

C SS C

1 1x 2 1x

22 21 2

12

xSx x

12 21 2

12

xCx x

3 23 3

1 2 1 2 0 1 1 2 2 2 21 2 1 2 0 1 1 0 0

0 0 1 1 1 1 1

Note that the magnitude of the columns of [H y] are unchanged.





Next zero element (3,1) 122

x 2 1x ,

3 23 3

1 2 1 2 0 1 1 2 2 2 21 2 1 2 0 1 1 0 0

0 0 1 1 1 1 1

1 11 2 3

S

2 2 23 6

C ,

2 6 0 1 3 2 2 2 2 3 3 30 1 0 0 0 0 0

1 11 3 0 2 6 0 2 3





Next zero element (3,2) 1 0x 223

x , 1S 0C ,

3 3 31 0 0 3 3 30 0 1 0 0 0 2 30 1 0 0 00 2 3

2 6 0 1 3 2 2 2 2 3 3 30 1 0 0 0 0 0

1 11 3 0 2 6 0 2 3





The Givens transformations yield

3 3 3

203

R

Which will yield a valid solution for ̂xIn fact

3 0 3 3 3

2 23 3 03 3

TR R

2

3 33 3 2





Which is the exact solution result for . Hence, for this example the orthogonal transformations would yield the correct solution. However, the estimation error covariance matrix would be incorrect because our computer would drop the

TH H

2 term.




Givens used rotations to null values until R became upper-triangular

Householder uses reflections to accomplish the same goal

74

Other Orthogonal Transformations




Homework 9 due this week.◦ Make sure you spend time studying for the exam

Homework 10 out Thursday.◦ Give you a small reprieve to focus on HW9.

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ASEN 5070 Statistical Orbit Determination I Fall 2012 Professor Jeffrey S. Parker

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