Can I have $Q = R = I$ as covariance matrices for a kalman filter? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Partial differentiation of vector to find Jacobian (extended Kalman filter)Kalman Filter to determine position and attitude from 6DOF IMU (accelerometer + gyroscope)How to estimate variances for Kalman filter from real sensor measurements without underestimating process noise.How to handle the noise covariance matrices in a basic Kalman Filter setup?Extended Kalman Filter destablizingHow to determine the transition probability in Sequential Importance Sampling (SIS) for Particle FilterKalman filter using accelerometer and system dyanamical modelDo I understand these expressions correctly (Kalman filter)?How to obtain kalman filter?LQG with bias rejection for quadcopter attitude control
Does Parliament need to approve the new Brexit delay to 31 October 2019?
Slither Like a Snake
Match Roman Numerals
Program that generates brainfuck code that outputs given text
How are presidential pardons supposed to be used?
Can withdrawing asylum be illegal?
What information about me do stores get via my credit card?
He got a vote 80% that of Emmanuel Macron’s
Who or what is the being for whom Being is a question for Heidegger?
Why is superheterodyning better than direct conversion?
Didn't get enough time to take a Coding Test - what to do now?
Semisimplicity of the category of coherent sheaves?
Single author papers against my advisor's will?
Did the new image of black hole confirm the general theory of relativity?
Wolves and sheep
Windows 10: How to Lock (not sleep) laptop on lid close?
Make it rain characters
"... to apply for a visa" or "... and applied for a visa"?
Movie about afterlife I think? Large towers with clothing and food?
How long does the line of fire that you can create as an action using the Investiture of Flame spell last?
Python - Fishing Simulator
How to colour the US map with Yellow, Green, Red and Blue to minimize the number of states with the colour of Green
What does the torsion-free condition for a connection mean in terms of its horizontal bundle?
Is there a trick to getting spices to fix to nuts?
Can I have $Q = R = I$ as covariance matrices for a kalman filter?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Partial differentiation of vector to find Jacobian (extended Kalman filter)Kalman Filter to determine position and attitude from 6DOF IMU (accelerometer + gyroscope)How to estimate variances for Kalman filter from real sensor measurements without underestimating process noise.How to handle the noise covariance matrices in a basic Kalman Filter setup?Extended Kalman Filter destablizingHow to determine the transition probability in Sequential Importance Sampling (SIS) for Particle FilterKalman filter using accelerometer and system dyanamical modelDo I understand these expressions correctly (Kalman filter)?How to obtain kalman filter?LQG with bias rejection for quadcopter attitude control
$begingroup$
Assume that we have no noise in our system. We using a low pass filter to filer away some peaks in the measurements.
But our goal is just to estimate the state $X_k$. Can we set the $Q_k$ and $R$ to the identity matrix $I$? Or do we need to compute $Q_k$ and $R$?
We can assume that we have no noise from our process.
We know $A, B, C, X_0, P_0, U_k, Y_k, H$ but not $Z_k, W_k$
control-theory optimal-control linear-control kalman-filter
$endgroup$
|
show 2 more comments
$begingroup$
Assume that we have no noise in our system. We using a low pass filter to filer away some peaks in the measurements.
But our goal is just to estimate the state $X_k$. Can we set the $Q_k$ and $R$ to the identity matrix $I$? Or do we need to compute $Q_k$ and $R$?
We can assume that we have no noise from our process.
We know $A, B, C, X_0, P_0, U_k, Y_k, H$ but not $Z_k, W_k$
control-theory optimal-control linear-control kalman-filter
$endgroup$
$begingroup$
Why do you have to filter out peaks if there is no noise acting on the system?
$endgroup$
– Kwin van der Veen
Mar 25 at 13:10
$begingroup$
@KwinvanderVeen It can be disturbance from the microcontroller. I know that $R$ can be found by $R = cov(Y_k)$ when $Y_k$ is steady state. But how about $Q$?
$endgroup$
– Daniel Mårtensson
Mar 25 at 14:12
$begingroup$
Q is the prior variance ( or sometimes they call it the initial variance ) of the state. If you know everything else, it can be estimated using prediction error decomposition but that's a time-series methodology. ( see Andrew Harvey's blue book ) .I'm not familiar with how they would do it in control field which could be something totally different.
$endgroup$
– mark leeds
Mar 26 at 5:30
$begingroup$
@markleeds it's very difficult to find Q in reality for an unknow process ? Can I then set $Q=0$?
$endgroup$
– Daniel Mårtensson
Mar 26 at 13:24
$begingroup$
Hi: I would think, even in the control field, you should give it a prior ( in bayesian framework ) or estimate it ( classical framework ). Setting it to zero says that the state has no prior variance so, no, I don't think that's the approach to take. Hopefully someone else can chime in. But check out Harvey's blue book because, if everything else is known, it's not that hard to estimate it.
$endgroup$
– mark leeds
Mar 27 at 16:59
|
show 2 more comments
$begingroup$
Assume that we have no noise in our system. We using a low pass filter to filer away some peaks in the measurements.
But our goal is just to estimate the state $X_k$. Can we set the $Q_k$ and $R$ to the identity matrix $I$? Or do we need to compute $Q_k$ and $R$?
We can assume that we have no noise from our process.
We know $A, B, C, X_0, P_0, U_k, Y_k, H$ but not $Z_k, W_k$
control-theory optimal-control linear-control kalman-filter
$endgroup$
Assume that we have no noise in our system. We using a low pass filter to filer away some peaks in the measurements.
But our goal is just to estimate the state $X_k$. Can we set the $Q_k$ and $R$ to the identity matrix $I$? Or do we need to compute $Q_k$ and $R$?
We can assume that we have no noise from our process.
We know $A, B, C, X_0, P_0, U_k, Y_k, H$ but not $Z_k, W_k$
control-theory optimal-control linear-control kalman-filter
control-theory optimal-control linear-control kalman-filter
edited Mar 25 at 10:37
Daniel Mårtensson
asked Mar 25 at 10:10
Daniel MårtenssonDaniel Mårtensson
993419
993419
$begingroup$
Why do you have to filter out peaks if there is no noise acting on the system?
$endgroup$
– Kwin van der Veen
Mar 25 at 13:10
$begingroup$
@KwinvanderVeen It can be disturbance from the microcontroller. I know that $R$ can be found by $R = cov(Y_k)$ when $Y_k$ is steady state. But how about $Q$?
$endgroup$
– Daniel Mårtensson
Mar 25 at 14:12
$begingroup$
Q is the prior variance ( or sometimes they call it the initial variance ) of the state. If you know everything else, it can be estimated using prediction error decomposition but that's a time-series methodology. ( see Andrew Harvey's blue book ) .I'm not familiar with how they would do it in control field which could be something totally different.
$endgroup$
– mark leeds
Mar 26 at 5:30
$begingroup$
@markleeds it's very difficult to find Q in reality for an unknow process ? Can I then set $Q=0$?
$endgroup$
– Daniel Mårtensson
Mar 26 at 13:24
$begingroup$
Hi: I would think, even in the control field, you should give it a prior ( in bayesian framework ) or estimate it ( classical framework ). Setting it to zero says that the state has no prior variance so, no, I don't think that's the approach to take. Hopefully someone else can chime in. But check out Harvey's blue book because, if everything else is known, it's not that hard to estimate it.
$endgroup$
– mark leeds
Mar 27 at 16:59
|
show 2 more comments
$begingroup$
Why do you have to filter out peaks if there is no noise acting on the system?
$endgroup$
– Kwin van der Veen
Mar 25 at 13:10
$begingroup$
@KwinvanderVeen It can be disturbance from the microcontroller. I know that $R$ can be found by $R = cov(Y_k)$ when $Y_k$ is steady state. But how about $Q$?
$endgroup$
– Daniel Mårtensson
Mar 25 at 14:12
$begingroup$
Q is the prior variance ( or sometimes they call it the initial variance ) of the state. If you know everything else, it can be estimated using prediction error decomposition but that's a time-series methodology. ( see Andrew Harvey's blue book ) .I'm not familiar with how they would do it in control field which could be something totally different.
$endgroup$
– mark leeds
Mar 26 at 5:30
$begingroup$
@markleeds it's very difficult to find Q in reality for an unknow process ? Can I then set $Q=0$?
$endgroup$
– Daniel Mårtensson
Mar 26 at 13:24
$begingroup$
Hi: I would think, even in the control field, you should give it a prior ( in bayesian framework ) or estimate it ( classical framework ). Setting it to zero says that the state has no prior variance so, no, I don't think that's the approach to take. Hopefully someone else can chime in. But check out Harvey's blue book because, if everything else is known, it's not that hard to estimate it.
$endgroup$
– mark leeds
Mar 27 at 16:59
$begingroup$
Why do you have to filter out peaks if there is no noise acting on the system?
$endgroup$
– Kwin van der Veen
Mar 25 at 13:10
$begingroup$
Why do you have to filter out peaks if there is no noise acting on the system?
$endgroup$
– Kwin van der Veen
Mar 25 at 13:10
$begingroup$
@KwinvanderVeen It can be disturbance from the microcontroller. I know that $R$ can be found by $R = cov(Y_k)$ when $Y_k$ is steady state. But how about $Q$?
$endgroup$
– Daniel Mårtensson
Mar 25 at 14:12
$begingroup$
@KwinvanderVeen It can be disturbance from the microcontroller. I know that $R$ can be found by $R = cov(Y_k)$ when $Y_k$ is steady state. But how about $Q$?
$endgroup$
– Daniel Mårtensson
Mar 25 at 14:12
$begingroup$
Q is the prior variance ( or sometimes they call it the initial variance ) of the state. If you know everything else, it can be estimated using prediction error decomposition but that's a time-series methodology. ( see Andrew Harvey's blue book ) .I'm not familiar with how they would do it in control field which could be something totally different.
$endgroup$
– mark leeds
Mar 26 at 5:30
$begingroup$
Q is the prior variance ( or sometimes they call it the initial variance ) of the state. If you know everything else, it can be estimated using prediction error decomposition but that's a time-series methodology. ( see Andrew Harvey's blue book ) .I'm not familiar with how they would do it in control field which could be something totally different.
$endgroup$
– mark leeds
Mar 26 at 5:30
$begingroup$
@markleeds it's very difficult to find Q in reality for an unknow process ? Can I then set $Q=0$?
$endgroup$
– Daniel Mårtensson
Mar 26 at 13:24
$begingroup$
@markleeds it's very difficult to find Q in reality for an unknow process ? Can I then set $Q=0$?
$endgroup$
– Daniel Mårtensson
Mar 26 at 13:24
$begingroup$
Hi: I would think, even in the control field, you should give it a prior ( in bayesian framework ) or estimate it ( classical framework ). Setting it to zero says that the state has no prior variance so, no, I don't think that's the approach to take. Hopefully someone else can chime in. But check out Harvey's blue book because, if everything else is known, it's not that hard to estimate it.
$endgroup$
– mark leeds
Mar 27 at 16:59
$begingroup$
Hi: I would think, even in the control field, you should give it a prior ( in bayesian framework ) or estimate it ( classical framework ). Setting it to zero says that the state has no prior variance so, no, I don't think that's the approach to take. Hopefully someone else can chime in. But check out Harvey's blue book because, if everything else is known, it's not that hard to estimate it.
$endgroup$
– mark leeds
Mar 27 at 16:59
|
show 2 more comments
0
active
oldest
votes
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161599%2fcan-i-have-q-r-i-as-covariance-matrices-for-a-kalman-filter%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161599%2fcan-i-have-q-r-i-as-covariance-matrices-for-a-kalman-filter%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Why do you have to filter out peaks if there is no noise acting on the system?
$endgroup$
– Kwin van der Veen
Mar 25 at 13:10
$begingroup$
@KwinvanderVeen It can be disturbance from the microcontroller. I know that $R$ can be found by $R = cov(Y_k)$ when $Y_k$ is steady state. But how about $Q$?
$endgroup$
– Daniel Mårtensson
Mar 25 at 14:12
$begingroup$
Q is the prior variance ( or sometimes they call it the initial variance ) of the state. If you know everything else, it can be estimated using prediction error decomposition but that's a time-series methodology. ( see Andrew Harvey's blue book ) .I'm not familiar with how they would do it in control field which could be something totally different.
$endgroup$
– mark leeds
Mar 26 at 5:30
$begingroup$
@markleeds it's very difficult to find Q in reality for an unknow process ? Can I then set $Q=0$?
$endgroup$
– Daniel Mårtensson
Mar 26 at 13:24
$begingroup$
Hi: I would think, even in the control field, you should give it a prior ( in bayesian framework ) or estimate it ( classical framework ). Setting it to zero says that the state has no prior variance so, no, I don't think that's the approach to take. Hopefully someone else can chime in. But check out Harvey's blue book because, if everything else is known, it's not that hard to estimate it.
$endgroup$
– mark leeds
Mar 27 at 16:59