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Let's say I have a standard system \begin{align} x(t+1)&=Ax\\ y(t) &=Cx(t) \end{align}

As you can see $B=0$, so the system is not controllable.

For the steady state Kalman filter I'd say that if $A$ is Hurwitz then it converges.

But what if $A$ is say, $\mathbf I_n$ (i.e. eye(n)), what can I say about the optimality of the SS. filter varying $C$?

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In this setting you are considering an observation problem. Controllability does not play a role. In geenral, the condition under which you can obtain a convergent estimate is that the couple $(A,C)$ is detectable, i.e. that you can design a gain matrix $L$ so that $A+LC$ is Hurwitz. Under this hypothesis the eigenvalues of $A+LC$ will be with negative real-part and they are assigned in a optimal way, in the kalman sense. The kalman gains depend on $C$, therefore as you vary $C$ you should vary the gains accordingly to obtain optimality

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  • $\begingroup$ you are right, the observability indeed guarantees that the kalman gain converges (the algebraic riccati equation has at least one positive definite solution).... but without controllability I can't conclude that the steady state filter will be stable. It depends on the eigenvalues of the ss filter's matrix A, which I fear I can't tell without actual numbers in the problem. if A,B was observable (or A stable) then the ss filter would be stable. do you agree with me? I tought a day about it. $\endgroup$ – user3149593 Jun 7 '16 at 17:44
  • $\begingroup$ Note that I used Hurwitz instead of ''stable'', in a discrete time system the condition should be that the magnitude of the eigenvalues is less than 1... Forgive me this abuse of notation, but I hope it makes sense. $\endgroup$ – user3149593 Jun 7 '16 at 17:47
  • $\begingroup$ Well, if A is not Hurwitz but $A+LC$ is, then your estimate converges. If the state diverges then obviously so does your estimate, infact I dont know how much sense does it make to estimate unstable systems. $\endgroup$ – LJSilver Jun 8 '16 at 8:28
  • $\begingroup$ Converges in the sense that the error goes to zero $\endgroup$ – LJSilver Jun 8 '16 at 8:28
  • $\begingroup$ But obviously the steady state will depend on A $\endgroup$ – LJSilver Jun 8 '16 at 8:29

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