12/31/2019 Transformation Matrix Polar To Cartesian
I am dealing with a time series in polar coordinates and I am applying the Kalman filter for predictions. The time series is related with the satellite orbite.
However my prediction and estimation for the variance are expressed in polar coordinates [r,theta].
I know how to convert my prediction in cartesian coordinates with the function
Convert the Cartesian coordinates defined by corresponding entries in matrices x and y to polar coordinates theta and rho. Cartesian to Cylindrical Coordinates. Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. Spherical-polar coordinates. 1.1 Specifying points in spherical-polar coordinate s. To specify points in space using spherical-polar coordinates, we first choose two convenient, mutually perpendicular reference directions (i and k in the picture).
But I do not know how to deal with the variance since it is not a linear operator.
I provide you my data in order if you can help me with the transformation:
And the variance matrix for the first prediction is :
I would like to know how to obtain this matrix in cartesian coordinates for the first prediction. thanks!
JPVJPV
3 Answers
This had me perplexed as well. I think I found the answer: the formula provided above,
,
follows from the most general form of error propagation. The formula is correct, provided you're OK with making a few assumptions, in particular that you are OK with linearizing the transformation.
See https://en.wikipedia.org/wiki/Propagation_of_uncertainty#Non-linear_combinations. This section has a subsection, 'Caveats and Warnings,' which I think is worth approaching with an open mind (so you don't end up biased :P ).
Beepboop bebopBeepboop bebop
In developing tracking filters for radar systems, I've used the following technique:
1) Determine the polar to Cartesian rotation matrix as
2) Perform the following matrix multiplication to obtain the covariance matrix in Cartesian coordinates, Pcart:
FrankFrank
The best way to accomplish this is to find the Jacobian of the function Fhat = Jacobian[f(r,theta)]. If the variance matrix in spherical is R(polar), then P(Cart) = Fhat*R*Fhat'. Using a Rotation matrix gives you the wrong answer, as it simply rotates the Cartesian covariance into another 'rotated' Cartesian system. See the Appendix 18.B in my book 'Bayesian Estimation and Tracking: A Practical Guide' for a complete derivation of this formula and how to use it.
Tony HaugTony Haug
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However my prediction and estimation for the variance are expressed in polar coordinates [r,theta].
I know how to convert my prediction in cartesian coordinates with the function
But I do not know how to deal with the variance since it is not a linear operator.
I provide you my data in order if you can help me with the transformation:
And the variance matrix for the first prediction is :
I would like to know how to obtain this matrix in cartesian coordinates for the first prediction. thanks!
JPVJPV
3 Answers
This had me perplexed as well. I think I found the answer: the formula provided above,
,
follows from the most general form of error propagation. The formula is correct, provided you're OK with making a few assumptions, in particular that you are OK with linearizing the transformation.
See https://en.wikipedia.org/wiki/Propagation_of_uncertainty#Non-linear_combinations. This section has a subsection, 'Caveats and Warnings,' which I think is worth approaching with an open mind (so you don't end up biased :P ).
Beepboop bebopBeepboop bebop
In developing tracking filters for radar systems, I've used the following technique:
1) Determine the polar to Cartesian rotation matrix as
2) Perform the following matrix multiplication to obtain the covariance matrix in Cartesian coordinates, Pcart:
FrankFrank
The best way to accomplish this is to find the Jacobian of the function Fhat = Jacobian[f(r,theta)]. If the variance matrix in spherical is R(polar), then P(Cart) = Fhat*R*Fhat'. Using a Rotation matrix gives you the wrong answer, as it simply rotates the Cartesian covariance into another 'rotated' Cartesian system. See the Appendix 18.B in my book 'Bayesian Estimation and Tracking: A Practical Guide' for a complete derivation of this formula and how to use it.
Tony HaugTony Haug
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