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In its chapter on Kalman filters, my DSP book states, seemingly out of the blue, that the stationary Kalman filter for a system

$$\begin{cases} x(t+1) &= Ax(t) + w(t) \\ y(t) &= Cx(t) + v(t) \end{cases}$$

has the predictor

$$\hat{x}(t+1|t) = (A-A\bar{K}C)\hat{x}(t|t-1) + A\bar{K}y(t)$$

and stationary state vector covariance and Kalman gain

$$\bar{P} = A\bar{P}A^T - A\bar{P}C^T ( C\bar{P}C^T + R )^{-1}C\bar{P}A^T + Q$$ $$\bar{K} = \bar{P}C^T(C\bar{P}C^T+R)^{-1}$$

where $Q$ and $R$ denote the covariances of the input noise $w$ and measurement noise $v$, respectively.

I can't see how to arrive at this from the minimum variance predictor. Could someone explain it to me, or point me to a resource that derives the expression? This is the time-variant minimum-variance filter, which I can derive:

$$\hat{x}(t+1|t) = (A-K(t)C)\hat{x}(t|t-1) + K(t)y(t)$$ $$P(t+1|t) = A\left(P(t|t-1)-P(t|t-1)C^T(CP(t|t-1)C^T + R)^{-1}CP(t|t-1)\right)A^T+Q$$ $$K(t) = AP(t|t-1)C^T(CP(t|t-1)C^T+R)^{-1}$$

I'm just unsure about how to go from here to the stationary filter above.

Update: I can see that substituting $\bar{P} = P(t+1|t) = P(t|t-1)$ and $K(t)=A\bar{K}$ into the time-variant filter results in the stationary filter, but why multiply with $A$? Is this just a symptom of an unfortunate choice of notation, meaning that either $K$ or $\bar{K}$ doesn't really denote the Kalman gain?

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  • $\begingroup$ No, it's not possible to "see" the predictor from the equations for the system. I think it would be better if you read a text book on Kalman filters instead of asking us to derive it for you (which would be just regurgitating something from a text book). Optimal Filtering by Anderson and Moore might be a good place to start. It's derived in chapter 5, if I remember correctly. $\endgroup$ – Lorem Ipsum Dec 4 '11 at 15:24
  • $\begingroup$ @yoda: Thanks. My question was if someone could point me to a better resource than the text book that my course recommends, so that's an answer. $\endgroup$ – Andreas Dec 4 '11 at 15:55
  • $\begingroup$ @yoda: By the way, in case I was unclear: I'm not asking for a derivation from the state-space system, but from the minimum variance Kalman filter. I've updated the question to make it clearer that I can derive a time-invariant Kalman filter, just not the stationary one. $\endgroup$ – Andreas Dec 4 '11 at 16:05
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    $\begingroup$ Which text are you getting the above from? If someone has access to it, it may be useful so we can see the full context. $\endgroup$ – Jason R Dec 28 '11 at 22:17
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Your derivations are correct.

$\bar P = P(t|t-1)$ and $K(t) = A \bar K$

Is this your confusion:

  1. Why didn't they have the term $t|t-1$ in the Kalman Gain and Covariance Matrix Expressions?
  2. How can this be "stationary" when your derivation shows that it is time varying?

  1. Bad choice of notation on the book's part

Let's look at the expression: $ \bar P = A\bar PA^T - A\bar P C^T(C\bar P C^T + R)^{-1}C\bar PA^T + Q $. The fact that $\bar P $ is a function of itself shows a recursive relationship. In other words, it uses its past values. So, it is NOT the same for all time instants - It changes at every iteration.

  1. Misunderstanding of the word "stationary".

When the author of the book said "stationary" he/she did not mean that $P$ and $K$ have the same value at all time instants. Instead, the author wanted to emphasise that the expressions for those to values are the same for all statistical realisations. Stationarity is a statistical concept which means that the Statistics of the system is same all the time. Look at the expressions for $\bar P$ and $\bar K$ again. They depend only on \

  • Previous values of themselves
  • Transition matrices $A$ and $C$ which are deterministic and in your case time-invariant ($A$ and $C$ are the same at all times)
  • $Q$ and $R$ which are the noise covariance matrices. These 2 matrices describe the statistics of the noises and are the same in all realisations and time instances.

The Kalman gain, $K$, and the state covariance Matrix $P$ will have the same value for all realisations of this random process. (Side Note: None of these 2 terms depend on the measurements, $y$. So they can be computed beforehand.)


Conclusion:

The "time-variant" equations you derived were equivalent to the ones in the book. Besides, the notational differences, there was a slight misunderstanding on your part regarding what changes and what doesn't.

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    $\begingroup$ I don't remember what the problem I had was when I asked the question, but now it makes sense. Thanks! $\endgroup$ – Andreas Mar 20 '13 at 17:39

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