asen 5070 statistical orbit determination i fall 2012 professor jeffrey s. parker
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ASEN 5070 Statistical Orbit Determination I Fall 2012 Professor Jeffrey S. Parker Professor George H. Born Lecture 21: Exam 2 Debrief and More Fun. Announcements. Homework 9 due this week. Make sure you spend time studying for the exam Homework 10 out Thursday. - PowerPoint PPT PresentationTRANSCRIPT
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ASEN 5070Statistical Orbit Determination I
Fall 2012
Professor Jeffrey S. ParkerProfessor George H. Born
Lecture 21: Exam 2 Debrief and More Fun
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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
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Quiz 17 Review
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Quiz 17 Review
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Quiz 17 Review
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Quiz 17 Review
The matrix of partials of one observation relative to the state parameters is identical to the other matrix.
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Quiz 17 Review
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Quiz 17 Review
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Quiz 17 Review
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Quiz 17 Review
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Due this Thursday
HW#9
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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
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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.
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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 )?
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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
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
i.e., High Precision but low accuracy!
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
Only guarantees a nonnegative definite!
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
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Exam 2 Debrief
4x3
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Exam 2 Debrief
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Exam 2 Debrief
[3x4]*[4x3] = [3x3] (hint: it’s always nxn)
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Exam 2 Debrief
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Exam 2 Debrief
1. one observation vector includes 4 independent pieces of information. We only need 3 pieces of information.
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Exam 2 Debrief
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Exam 2 Debrief
Then Phi, A, y, H-tildex-hat, P, x-bar, P-bar
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Exam 2 Debrief
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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!
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Exam 2 Debrief
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Quick Break
Next up: Stuff.
◦ Prediction Residual◦ Givens◦ Householder◦ Future: Process Noise, Smoothing
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The Prediction Residual
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The Prediction Residual
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The Prediction Residual
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The Prediction Residual
This would be especially important in the case of the EKF
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Next: Orthogonal transformations: Givens, Householder
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So how do we select Q?
Givens, Householder, many methods
Choices from here
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Givens
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Givens
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Givens
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Givens
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Givens
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Givens
Apply the rotation across the matrix, converting it into a triangular matrix
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Givens
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Givens
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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?
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Accuracy Comparison for Batch and Givens – Finite Precision Computer
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
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Accuracy Comparison for Batch and Givens – Finite Precision Computer
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
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Accuracy Comparison for Batch and Givens – Finite Precision Computer
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.
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Accuracy Comparison for Batch and Givens – Finite Precision Computer
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
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Accuracy Comparison for Batch and Givens – Finite Precision Computer
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
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Accuracy Comparison for Batch and Givens – Finite Precision Computer
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
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CCARColorado Center for
Astrodynamics Research
University of ColoradoBoulder
Accuracy Comparison for Batch and Givens – Finite Precision Computer
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.
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CCARColorado Center for
Astrodynamics Research
University of ColoradoBoulder
Givens used rotations to null values until R became upper-triangular
Householder uses reflections to accomplish the same goal
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Other Orthogonal Transformations
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CCARColorado Center for
Astrodynamics Research
University of ColoradoBoulder 75
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.
The End