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Can I have $Q = R = I$ as covariance matrices for a kalman filter?

0

$begingroup$

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

share|cite|improve this question

edited Mar 25 at 10:37

Daniel Mårtensson

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  


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  $begingroup$
  @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

0

$begingroup$

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

share|cite|improve this question

edited Mar 25 at 10:37

Daniel Mårtensson

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

share|cite|improve this question

edited Mar 25 at 10:37

Daniel Mårtensson

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419

$endgroup$

share|cite|improve this question

edited Mar 25 at 10:37

Daniel Mårtensson

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419

share|cite|improve this question

edited Mar 25 at 10:37

Daniel Mårtensson

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419

asked Mar 25 at 10:10

Daniel MårtenssonDaniel Mårtensson

993419