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I don't understand why the following makes sense:

Given a second-order mass damper system in continuous time:

$H(s) = \frac{1}{ms^{2}+cs}$

Its inverse $H^{-1}(s)$ is unrealizable as a transfer function $G(s) = \frac{Y(s)}{U(s)}$ has a state space realization if and only if the degree of $Y(s)$ is less than or equal to the degree of $U(s)$.

However, the zero-order hold (ZOH) equivalent of $H(s)$ given by:

$H(z^{-1}) = \frac{z^{-1}(b_{0}+b_{1}z^{-1})}{1+a_{1}z^{-1}+a_{2}z^{-2}}$

has a realizable inverse $H^{-1}(z^{-1})$ as its degree in the denominator and numerator is equal.

How does this make sense?

I have read that when designing a disturbance observer, we need a $Q(z)$ filter such that $Q(z)G^{-1}(z)$ is realizable as $G^{-1}(z)$ will not be.

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You give a transfer function in $z^{-1}$, but for the purposes of making a state-space model, you need a transfer function in $z$ (more on that later). So your $H$ becomes $$H(z) = \frac{b_0 z + b_1}{z^2 + a_1 z + a_2}$$ When you invert that it's improper, which shows that you need to know information one time step into the future. That's not possible (or, if it is, you should take up life as a stock broker).

(Note that you don't need state space to know this: you can just write the corresponding difference equation for your $H^{-1}$ and you'll see that it's asking for future values of the input to affect the current value of the output.)

Appendix

The reasoning for doing it in $z$ is because the time-domain state-space realization is

$$\begin{split} x_k = A x_{k-1} + B u_k \\ y_k = C x_{k-1} + D u_k\end{split}$$

For $D = 0$, the resulting transfer function is strictly proper in $z$ (but not in $z^{-1}$); for $D \ne 0$ the resulting transfer function in $z$ has equal orders in the numerator and denominator.

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  • $\begingroup$ I could add delay to be proper but yet still not be invertible- to be invertible a system must also have all zeros inside the unit circle (a minimum phase system); anything else (maximum phase or linear phase) has zeros outside the unit circle which become poles when inverted and therefore unstable (for causal systems). Agree? $\endgroup$ – Dan Boschen Dec 13 '19 at 20:48
  • $\begingroup$ Adding delay puts a pole at $z = 0$, which is about as inside of the unit circle as you can get. And even if the resulting system were unstable, it's going to be used for an observer, so it could conceivably be tractable to stabilize. $\endgroup$ – TimWescott Dec 13 '19 at 21:01
  • $\begingroup$ Ah I see- so stability isn't a factor here in determining invertibility; good point- thanks. I was envisioning systems where the degree of the numerator is the same as the denominator (regardless if the delay pole is inside the circle, was thinking of the other poles), which isn't invertible when stability is the concern but I get your point that in this case it would/could be stabilized in the overall system. $\endgroup$ – Dan Boschen Dec 13 '19 at 21:05
  • $\begingroup$ If the inverse were unstable it would certainly make life harder -- but you would like the system to be causal if you want to realize it in the real world. $\endgroup$ – TimWescott Dec 14 '19 at 0:43
  • $\begingroup$ Yes that all makes sense - thanks! $\endgroup$ – Dan Boschen Dec 14 '19 at 0:48

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