Lecture 28 - CCAR - University of Colorado Boulder
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Transcript Lecture 28 - CCAR - University of Colorado Boulder
ASEN 5070: Statistical Orbit Determination I
Fall 2014
Professor Brandon A. Jones
Lecture 28: Orthogonal Transformations
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Lecture quiz due at 5pm
Exam 2 – Friday, November 7
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Minimum Variance
Conventional Kalman Filter
Extended Kalman Filter
Prediction Residuals
Handling Observation Biases
Numeric Considerations in the Kalman
Batch vs. CKF vs. EKF
Effects of Uncertainties on Estimation
Potter Square-Root Filter
Cholesky Decomposition w/ Forward and
Backward substitution
Singular Value Decomposition Methods
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Least Squares via Orthogonal Transformations
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Recall the least squares cost function:
By property 4 on the previous slide and Q an orthogonal
matrix:
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The method for selecting R defines a
particular algorithm
◦ Givens Transformations (Section 5.4)
◦ Householder Transformation (Section 5.5)
◦ Gram-Schmidt Orthogonalization
Not in the book and we won’t cover it
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LS Solution via Givens Transformations
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Consider the desired result
To achieve this, we select the Givens matrix such that
We then use this transformation in the top equation
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We do not want to add non-zero terms to the
previously altered rows, so we use the
identity matrix except in the rows of interest:
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After applying the transformation, we get:
Repeat for all remaining non-zero elements
in the third column
What if the term is already 0 ?
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Need to find the orthogonal matrix Q to yield a
matrix of the form of the RHS
Q is generated using a series of Givens
transformations G
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We select G to get a zero for the term in red:
To achieve this, we use:
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We select G to get a zero for the term in red:
To achieve this, we use:
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We select G to get a zero for the term in red:
To achieve this, we use:
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We select G to get a zero for the term in red:
To achieve this, we use:
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We select G to get a zero for the term in red:
To achieve this, we use:
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We select G to get a zero for the term in red:
To achieve this, we use:
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We select G to get a zero for the term in red:
To achieve this, we use:
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We now have the required Q matrix (for this
conceptual example):
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Givens Transformations – An New Example
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Consider the case where:
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We then have the matrices needed to solve
the system:
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Batch vs. Givens
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Consider the case where:
The exact solution is:
After truncation:
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Well, the Batch can’t handle it. What about
Cholesky decomposition?
Darn, that’s singular too.
Let’s give Givens a shot!
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Hence, Givens transformations give us a solution
for the state
Home Exercise: Why is this true?
Note: R is not equal to H !
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Still a problem w/ P !
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Givens uses a sequence of rotations to
generate the R matrix
Instead, Householder transformations use a
sequence of reflections to generate R
◦ See the book for details
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