Mixing
Kalman Filter applied to linear discrete time process and interpretation of the estimated covariance matrix
$begingroup$
I want to have a deeper understanding of the discrete time Kalman Filter.
As a part of this I have modeled a forced, damped, mass spring system numerically in the Jupyter Notebook available here:
https://colab.research.google.com/drive/14c3-moLnPlbWArnT5I0_MEW6zytYW6Mv
The model of the system is:
$ddot{x} = -k (x - x_0) - mu v + f sin(5t)$
Where:
$x$ is the spring position, unit $[mathrm{m}]$.
$k$ is the spring constant, unit $[mathrm{N/m}]$.
$mu$ is the viscous friction, unit $[mathrm{Ns/m}]$.
$x_0$ is the spring equilibrium position, unit $[mathrm{m}]$.
$f$ is the external force amplitude, unit $[mathrm{N}]$.
$ddot{x}$ is the double derivative of $x$ with respect to time, or $frac{mathrm{d}^2 x}{mathrm{d} t^2}$.
The question is: To my understanding this process should be optimally estimated by the Kalman filter. My understanding is that this implies that the filter estimate $hat{x}$ should be between $x pm sigma_x$ with a probability of 0.682.
In my numerical experiments the position estimate follows this relation perfectly, but the velocity estimate does not at all. I am struggling to understand the exact cause of this. The notebook explains each step of the numerics, and shows how the distribution of the position estimate and velocity estimate differ. You can edit and comment in the notebook to test changes and explain where things go wrong.
Any help to shed light on this would be greatly appreciated.
covariance kalman-filter
asked Jan 18 at 12:18
lysgaardlysgaard
62
$endgroup$
add a comment |
$begingroup$
I want to have a deeper understanding of the discrete time Kalman Filter.
As a part of this I have modeled a forced, damped, mass spring system numerically in the Jupyter Notebook available here:
https://colab.research.google.com/drive/14c3-moLnPlbWArnT5I0_MEW6zytYW6Mv
The model of the system is:
$ddot{x} = -k (x - x_0) - mu v + f sin(5t)$
Where:
$x$ is the spring position, unit $[mathrm{m}]$.
$k$ is the spring constant, unit $[mathrm{N/m}]$.
$mu$ is the viscous friction, unit $[mathrm{Ns/m}]$.
$x_0$ is the spring equilibrium position, unit $[mathrm{m}]$.
$f$ is the external force amplitude, unit $[mathrm{N}]$.
$ddot{x}$ is the double derivative of $x$ with respect to time, or $frac{mathrm{d}^2 x}{mathrm{d} t^2}$.
The question is: To my understanding this process should be optimally estimated by the Kalman filter. My understanding is that this implies that the filter estimate $hat{x}$ should be between $x pm sigma_x$ with a probability of 0.682.
In my numerical experiments the position estimate follows this relation perfectly, but the velocity estimate does not at all. I am struggling to understand the exact cause of this. The notebook explains each step of the numerics, and shows how the distribution of the position estimate and velocity estimate differ. You can edit and comment in the notebook to test changes and explain where things go wrong.
Any help to shed light on this would be greatly appreciated.
covariance kalman-filter
asked Jan 18 at 12:18
lysgaardlysgaard
62
$endgroup$
$begingroup$
I found the problem. The issue was coming from having an incorrect state model matrix $A$. The problem was thus purely from a bug in the numerical implementation. The numerical example has been updated to be correct. So others seeing this question in the future can have it as reference.
$endgroup$
– lysgaard
Jan 21 at 12:37
add a comment |
$begingroup$
I want to have a deeper understanding of the discrete time Kalman Filter.
As a part of this I have modeled a forced, damped, mass spring system numerically in the Jupyter Notebook available here:
https://colab.research.google.com/drive/14c3-moLnPlbWArnT5I0_MEW6zytYW6Mv
The model of the system is:
$ddot{x} = -k (x - x_0) - mu v + f sin(5t)$
Where:
$x$ is the spring position, unit $[mathrm{m}]$.
$k$ is the spring constant, unit $[mathrm{N/m}]$.
$mu$ is the viscous friction, unit $[mathrm{Ns/m}]$.
$x_0$ is the spring equilibrium position, unit $[mathrm{m}]$.
$f$ is the external force amplitude, unit $[mathrm{N}]$.
$ddot{x}$ is the double derivative of $x$ with respect to time, or $frac{mathrm{d}^2 x}{mathrm{d} t^2}$.
The question is: To my understanding this process should be optimally estimated by the Kalman filter. My understanding is that this implies that the filter estimate $hat{x}$ should be between $x pm sigma_x$ with a probability of 0.682.
In my numerical experiments the position estimate follows this relation perfectly, but the velocity estimate does not at all. I am struggling to understand the exact cause of this. The notebook explains each step of the numerics, and shows how the distribution of the position estimate and velocity estimate differ. You can edit and comment in the notebook to test changes and explain where things go wrong.
Any help to shed light on this would be greatly appreciated.
covariance kalman-filter
asked Jan 18 at 12:18
lysgaardlysgaard
62
$endgroup$
I want to have a deeper understanding of the discrete time Kalman Filter.
As a part of this I have modeled a forced, damped, mass spring system numerically in the Jupyter Notebook available here:
https://colab.research.google.com/drive/14c3-moLnPlbWArnT5I0_MEW6zytYW6Mv
The model of the system is:
$ddot{x} = -k (x - x_0) - mu v + f sin(5t)$
Where:
$x$ is the spring position, unit $[mathrm{m}]$.
$k$ is the spring constant, unit $[mathrm{N/m}]$.
$mu$ is the viscous friction, unit $[mathrm{Ns/m}]$.
$x_0$ is the spring equilibrium position, unit $[mathrm{m}]$.
$f$ is the external force amplitude, unit $[mathrm{N}]$.
$ddot{x}$ is the double derivative of $x$ with respect to time, or $frac{mathrm{d}^2 x}{mathrm{d} t^2}$.
The question is: To my understanding this process should be optimally estimated by the Kalman filter. My understanding is that this implies that the filter estimate $hat{x}$ should be between $x pm sigma_x$ with a probability of 0.682.
In my numerical experiments the position estimate follows this relation perfectly, but the velocity estimate does not at all. I am struggling to understand the exact cause of this. The notebook explains each step of the numerics, and shows how the distribution of the position estimate and velocity estimate differ. You can edit and comment in the notebook to test changes and explain where things go wrong.
Any help to shed light on this would be greatly appreciated.
covariance kalman-filter
covariance kalman-filter
asked Jan 18 at 12:18
lysgaardlysgaard
62
asked Jan 18 at 12:18
lysgaardlysgaard
62
asked Jan 18 at 12:18
lysgaardlysgaard
62
asked Jan 18 at 12:18
lysgaardlysgaard
62
asked Jan 18 at 12:18
lysgaardlysgaard
62
62
$begingroup$
I found the problem. The issue was coming from having an incorrect state model matrix $A$. The problem was thus purely from a bug in the numerical implementation. The numerical example has been updated to be correct. So others seeing this question in the future can have it as reference.
$endgroup$
– lysgaard
Jan 21 at 12:37
add a comment |
$begingroup$
I found the problem. The issue was coming from having an incorrect state model matrix $A$. The problem was thus purely from a bug in the numerical implementation. The numerical example has been updated to be correct. So others seeing this question in the future can have it as reference.
$endgroup$
– lysgaard
Jan 21 at 12:37
$begingroup$
I found the problem. The issue was coming from having an incorrect state model matrix $A$. The problem was thus purely from a bug in the numerical implementation. The numerical example has been updated to be correct. So others seeing this question in the future can have it as reference.
$endgroup$
– lysgaard
Jan 21 at 12:37
$begingroup$
I found the problem. The issue was coming from having an incorrect state model matrix $A$. The problem was thus purely from a bug in the numerical implementation. The numerical example has been updated to be correct. So others seeing this question in the future can have it as reference.
$endgroup$
– lysgaard
Jan 21 at 12:37
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
Your Kalman Filter seems to be tracking the position and velocity fairly well, so I don't think it is broken.
Have you looked at the offdiagonal elements of your covariance matrix? If these are nonzero, that would mean there is correlation between the position and velocity errors.
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
$endgroup$
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your Kalman Filter seems to be tracking the position and velocity fairly well, so I don't think it is broken.
Have you looked at the offdiagonal elements of your covariance matrix? If these are nonzero, that would mean there is correlation between the position and velocity errors.
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
$endgroup$
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
add a comment |
$begingroup$
Your Kalman Filter seems to be tracking the position and velocity fairly well, so I don't think it is broken.
Have you looked at the offdiagonal elements of your covariance matrix? If these are nonzero, that would mean there is correlation between the position and velocity errors.
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
$endgroup$
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
add a comment |
$begingroup$
Your Kalman Filter seems to be tracking the position and velocity fairly well, so I don't think it is broken.
Have you looked at the offdiagonal elements of your covariance matrix? If these are nonzero, that would mean there is correlation between the position and velocity errors.
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
$endgroup$
Your Kalman Filter seems to be tracking the position and velocity fairly well, so I don't think it is broken.
Have you looked at the offdiagonal elements of your covariance matrix? If these are nonzero, that would mean there is correlation between the position and velocity errors.
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
answered Jan 18 at 13:41
SRVfender01SRVfender01
112
112
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
add a comment |
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
$begingroup$
The problem was found to be an incorrect model matrix, $A$. I have updated the numerical code to be correct. Thank you for the help.
$endgroup$
– lysgaard
Jan 21 at 12:38
add a comment |
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$begingroup$
I found the problem. The issue was coming from having an incorrect state model matrix $A$. The problem was thus purely from a bug in the numerical implementation. The numerical example has been updated to be correct. So others seeing this question in the future can have it as reference.
$endgroup$
– lysgaard
Jan 21 at 12:37