2

Given matrices $Q$ and $R \succ 0$, the Kalman filter problem is

$$ \begin{array}{ll} \underset {P \succeq0, K} {\text{minimize}} & \operatorname{tr} (P) \\ \text{subject to} & (A - K C) P (A - K C)^T + K R K^T + Q = P \end{array} $$

I am trying to learn about methods to convert the above optimization to have a convex optimization form.

Method 1 - Relaxation: This is a minimization problem, so we can perform the tight relaxation:

$$ (A-KC) P (A-KC)^T + KRK^T + Q \preceq P$$

We can then follow a solution suggested by @KBS:

  1. Multiply the Riccati inequality on both sides by $P^{-1}$: $$ (P^{-1}A-P^{-1}KC) P (P^{-1}A-P^{-1}KC)^T + P^{-1}KRK^TP^{-1} + P^{-1}QP^{-1} \preceq P^{-1}$$
  2. Denote $X:= P^{-1}, Y= XK$ to obtain: $$ (XA-YC) X^{-1} (X A-YC)^T + YRY^T + XQX \preceq X$$
  3. Apply the Schur complement to this inequality: \begin{align} \begin{bmatrix} X & XA - YC & Y & X\\ (XA - YC)^T & X & 0& 0\\ Y^T & 0 & R^{-1}&0\\ X& 0 & 0 & Q^{-1} \end{bmatrix} \succeq 0 \end{align}

All remains is to formulate the objective function $\operatorname{tr} (Z)$ where $Z=X^{-1}$. Since it is a minimization, we can take the tight relaxation $Z\succeq X^{-1}$ and write it as the LMI \begin{align} \begin{bmatrix} Z & I\\ I & X \end{bmatrix} \succeq 0. \end{align}

Method 2: Precision matrix The Riccati equation can be written using the precision matrix \begin{align} P^{-1}&= (APA^T+Q)^{-1} + C^TR^{-1}C, \end{align} I believe it is possible, but I am not sure how to convert this to an LMI.

Method 3: ?? Are there more methods to convert this optimization into a convex optimization?

As a remark, I know how to formulate Kalman as an LMI using the controllability Gramian without the Riccati equation, but I am trying to summarize all methods that utilize the standard Riccati.

Morad
  • 690
  • 2
    Perform a congruence transformation with respect to $X=P^{-1}$ on the Riccati inequality, use the change of variables $U=KX$ and use the Schur complement on the nonlinear terms. – KBS Mar 06 '25 at 07:05
  • 1
    There is no relaxation here, all the transformations are tight. Instead of step 4, you can just apply the Schur complement on $XQX$ as $Q$ is not a decision variable, it is a known constant matrix. For the cost, you can minimize trace($Z$) where $\begin{bmatrix} Z & I\I & X\end{bmatrix}\succeq0$. – KBS Mar 06 '25 at 15:28
  • Looks fine to me. Did you try numerically to see if it works? – KBS Mar 07 '25 at 17:49
  • 1
    It is just a change of variables related to the observer gain. It does not have any particular meaning other than you can construct the observer gain from it. – KBS Mar 09 '25 at 12:41
  • 1
    Better to ask another question for that and clearly define all the variables and the properties they have. – KBS Mar 09 '25 at 13:26
  • 1
    Then, I would recommend to improve the structure of your posts into sections. – KBS Mar 09 '25 at 21:48
  • @KBS done! btw, is there a good literature? I could only find the Cavelry arxiv booklet but it only uses the Gramians for H2 estimation – Morad Mar 10 '25 at 09:10

0 Answers0