Usually when I see the Kalman filter in lectures, the state equation is
$$\mathbf{x}_{k+1} = A_k \mathbf{x}_k + B_k a_k + r_k^{(s)}$$
and the measurement equation is
$$z_k = H \cdot \mathbf{x}_k + r_k^{(m)}$$
where $r_k^{(m)}, r_k^{(s)}$ is normal distributed noise.
Then, the covariance in t...
I've added an explanation for $C_k^{(r_s)}$. For $Q$ and $G$ I can't add one, because I don't know it. I hoped that somebody would be familiar with this notation, as the Kalman filter is pretty well-known and somebody might have seen this notation before.
Obviously, $Q$ and $G$ are matrices. I could also add which shape they have to have, but I guess that doesn't help.
@PeterK. What I named $C_k^{(r_s)}$ is named $Q$ by greg.czerniak.info/guides/kalman1 (and probably other sources, too). I just think $C_k^{(r_s)}$ ($C$ for covariance, $k$ for the $k$-th step, $r_s$ for "random" and "system") makes more sense.
@PeterK. My state equation and my measurment equation are the same as eq. 4.1 and eq. 4.2 from cs.unc.edu/~tracker/media/pdf/SIGGRAPH2001_CoursePack_08.pdf . The covariance prediction is the same as in (4.10). There is no missunderstanding on my side (or at least not only on my side) how to model the system.
The system model you use in the link does not include a $G$ matrix. The question you are asking assumes the system model uses a $G$ matrix. Is that not a point of confusion?
The $G$ matrix is obtained in the same way the $A$, $B$, and $C$ matrices are obtained: you decide how you want to model your system and make selections for those system parameters based on that. Selecting a $G$ that is not $I$ (identity) usually means that you know that only some of your states are driven by process noise. Usually the states not driven by process noise are just driven by the state update matrix, $A$ here.
> The system model you use in the link does not include a $G$ matrix. The question you are asking assumes the system model uses a $G$ matrix. Is that not a point of confusion?
Ah, yes, you're right. I missed that. It is actually in the slides. Thank you! Now I think I could answer the real question myself. (But the second part of your answer is what I was looking for. I only couldn't phrase it properly as I missed that the state equations were different)
@PeterK. Thank you! Nice answer! (Because of the notation: I am learning about the Kalman filter in 3 different lectures simultaneously. To make things worse, I can only learn from the slides as I can only attend one of the three lectures. All lecturers use different, but similar notation. So I chose to take a mixture of the notations which I could remember best. As I don't know which textbooks are well-known in this area with typical notation, I chose to take the notation I am used to.)
Discussion between Peter K. and Marti…
Discussion between Peter K. and Marti…