*I'm rewriting the question after I deleted the previous one because I wasn't clear enough. Hope it's fair

Given a system of the general form:

\begin{align} x_{[k+1]} &= A\,x_{[k]} + B\,u_{[k]} \\ y_{[k]} &= C\,x_{[k]} + D\,u_{[k]} \end{align}

I would like to know how I should place the poles for the closed-loop observer system $A−L\,C$. I know that the observer has to be faster than the real system poles so the poles of $A-LC$ should be more close to zero than $A+B\,K$. But I don't know about any other constraint of the position. If there isn't any other bound why we don't place the poles at 0 and make $A-L\,C$ converge in one step?


2 Answers 2


For LTI systems you can design the observer and state feedback separately, due to certainty equivalence. So the observer does not have to be faster then the state feedback in order to ensure stability.

Often a system will have input constraints (such as saturation), which will indirectly place constraints on $K$. However for your observer model there are no such constraints, so it is possible to place the observer poles closer to the origin then the state feedback poles.

However placing the observer poles much closer to the origin gives less and less performance gains, since the error dynamics will eventually be dominated by the slowest pole (usually from the state feedback). Another side effect of placing the observer poles closer to the origin is that the start up transient often has a bigger overshoot compared with slower observer poles.

Another important aspect which can be taken into consideration is the effects of noise/disturbances on the system, because a fast observer might amplify those more than a slower observer.

  • $\begingroup$ I think you're referring to a continuous system, while I'm referring to a discrete system. $\endgroup$ Commented May 18, 2018 at 9:20
  • $\begingroup$ @TommasoBendinelli Certainty equivalence holds for both continues as discrete time. $\endgroup$
    – fibonatic
    Commented May 18, 2018 at 9:58
  • $\begingroup$ Ah okey, and why the slowest pole from the state feedback influence the error dynamics? I thought the error dynamics was just influenced by the poles of (A-LC) that if (A,C) is observable can be freely assign $\endgroup$ Commented May 18, 2018 at 13:19

What you're referring to is called a deadbeat observer. The problem with these deadbeat observers/controllers is that there is a bandwidth/sensitivity trade off which basically means that raising the bandwidth will improve the transient response but will result in higher noise sensitivity.

It is easier to see in SISO systems where you can compute the close loop sensitivity and complementary sensitivity transfer functions, look at their Bode plots and see the noise sensitivity raises as the bandwidth gets higher.

You can read about this "waterbed effect" here (it's control oriented but it's basicallt the same):



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