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Consider a 1D point-mass moving along an axis. A force $u$ is applied as control. There is no gravity or other forces involved. The system can be described in state space equations as:

$$\begin{align} A &= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \\ B &= \begin{bmatrix} 0 \\ 0 \\ \dfrac{1}{M} \end{bmatrix} \\ C &= \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \\ D &= [0] \end{align}$$

The shown system is controllable, but not observable. Not even structurally observable and most certainly not fully observable. Thus, it should be impossible to construct an observer for this system.

However, if I know the initial state of the system, I can compute the full state at every time, i.e. by integrating the system's output. How does this fall in line with the concept of observability? How would I incorporate the initial state into the equations?

I can't find the error in my train of thought, but I am certain there is one. Do I misunderstand observability?

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Observability means that you can estimate the complete state using only the output, without knowing the initial state. In other words, you have to figure out where you are without knowing where you were initially.

A more practical reason why this rarely works is that when you are limited by non-perfect sensors and non-zero sampling time, taking the integral of the acceleration will cause growing errors in your estimates of positions and velocity. Thus, even if you know the initial state you will "lose track" of it over time.

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  • $\begingroup$ Hm. I thought that is what people call "fully observable", as in: given the sequences of input and output you can reconstruct x in finite time. How does "observable" and "fully observable" differ then? $\endgroup$ – FirefoxMetzger Dec 31 '16 at 21:40
  • $\begingroup$ I'm not aware of "non-fully observable". I guess it could refer to a case where some states are observable and some aren't. $\endgroup$ – Daniel Nilsson Jan 1 '17 at 12:24

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