1
$\begingroup$

I have the following misunderstanding regarding observability and observer.

Observability of a linear dynamical system is the same as the "recoverability" of the initial condition $x(0)$. A system is observable if $x(0)$ is recoverable through output and input information in finite time.

On the other hand, an observer is a linear dynamical system whose state $\widehat x$ approximates $x$ through tuning of the system parameter, specifically the Luenberger gain $L$.

Since an observer does not require the initial condition $x(0)$ AT ALL, therefore what does an observer have to do with observability?

I dug this question deeper and found a reference that said to the effect: suppose a system $\dot x = Ax + Bu$ is observable, then there exists some transformation $z= Tx$ such that $\dot z = \widetilde A z+ \widetilde B y + \widetilde N u, y = \widetilde{C}z$. This transformed system "kind of" looks like an observer. Then perhaps we can "add'' an input of the form $L (y - \widehat y)$ to it (and change the variables to estimated variables) so we get an observer

$$\dot {\widehat z} = \widetilde A \widehat {z}+ \widetilde B \widehat {y} + \widetilde N \widehat {u} + L(y - \widehat y), \widehat y = C \widehat z$$

I find this not to be very well-motivated (almost like pulling something out of thin air). It also does not say anything about the recoverability of $x(0)$.

I think if $x(0)$ was somehow used in the construction of the observer, then I can better see the necessity of observability. Is there an argument that closes this gap?

$\endgroup$
3
  • 1
    $\begingroup$ Observability of a system is not required for using an observer. The name is misleading, "reference controlled synthesizer" would be closer to the truth. $\endgroup$ Commented Apr 15 at 13:53
  • $\begingroup$ If the initial condition is known, you can just run a (properly initialized) copy of your system and get the same trajectory as the real one. From this point of view, state estimation is possible if you can reconstruct from observations the original mismatch between the actual state and the initialization of your observer. $\endgroup$ Commented Apr 17 at 13:34
  • $\begingroup$ A classic counterexample to the arguments given in the comments would be a trolley system such that only velocity is measured and the position is observed. No observer exists that the position estimation converges to the actual position in the presence of noise in this case, because detectability is lost. An observer is useful if the initial condition is not known beforehand and/or there is noise in the measurements. If you can measure/determine everything perfectly, there is no need for an observer in the first place. $\endgroup$ Commented Oct 2 at 16:48

1 Answer 1

Reset to default
1
$\begingroup$

When you write the error dynamics between the actual states and observed states, $\dot{e}:=x-\hat{x}$, you obtain the dynamical system $\dot{e}:=(A-LC)e$. Therefore, observability (or detectability as a weaker condition) is required for the existence of an observer gain $L$ such that the error between the actual states and observed states to go to $0$. Otherwise, there is no guarantee that the observer actually observes the states.

$\endgroup$
2
  • $\begingroup$ What if the pair $(A,C)$ is not detectable while $A$ is stable? :) $\endgroup$ Commented Apr 22 at 15:27
  • 1
    $\begingroup$ @KBS If $A$ is stable, $(A,C)$ must be detectable by definition. Consider the extreme case $C=0$ and $A$ is stable. In this case I can use the states of any stable system (with appropriate dimensions) as an observer since in the limit the estimated and actual states would be the same, which is 0. $\endgroup$ Commented Oct 2 at 16:42

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.