When is a discrete time transfer function unrealizable?

Question

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.

Answer 1

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.