I've been reading about how a non-minimum phase system with a Right-Half Plane Zero and Right-Half Plane Pole can become impossible to control IF the RHP Pole becomes faster than the RHP Zero.

I was curious if one can show such a system is uncontrollable via the Controllability Matrix.

I was starting with the following system.

$$ G(s)=\frac{s-z}{(s+p)(s-p)}$$

However, I was tripping up on how to convert that to State Space Form, so I can get the A and B matrices and then obtain the Controllability Matrix.

Curious if anyone had insights. Thanks!

  • \$\begingroup\$ you might already know this, but if not it is a little tangential, but maybe will be helpful for insight into zeros within the frequency domain approach. In the closed loop expression (1) you can alter the poles of G, but the zeros are going to stay right where they are, although the compensator can add additional zeros. (2) A RHP zero (s-z) is equivalent to [(s-z)/(s+z)](s+z), and the part in brackets is an all-pass which approximates a delay. \$\endgroup\$
    – Pete W
    Commented Feb 12, 2021 at 16:25

1 Answer 1


The state space system will have \$A \in \mathbb{R}^{2 \times 2}\$ and other matrices of appropriate size, so it will be

$$ \dot{x} = Ax + B u$$

The poles of the system will only depend on the eigenvalues of the matrix \$A\$. And since


We have that \$ \det(sI-A) = (s+p)(s-p) = 0\$. Writing that in the controllable canonical form we have

$$ \dot{x} = \begin{bmatrix} 0 & 1 \\ -p^2 & 0\end{bmatrix} x + \begin{bmatrix} 0 \\ 1\end{bmatrix} u$$

The transfer function of that state space equation will be

$$ C(sI-A)^{-1}B$$


$$(sI-A)^{-1} = \begin{bmatrix} -s/(-s^2 + p^2) & s/(s^3 - sp^2) \\ p^2/(s^2 - p^2) & s/(s^2 - p^2) \end{bmatrix} $$

we will have

$$(sI-A)^{-1}B = \begin{bmatrix} s/(s^3 - sp^2) \\ s/(s^2 - p^2) \end{bmatrix} = \begin{bmatrix} 1 \\ s \end{bmatrix} \frac{1}{s^2 - p^2} $$

Now it should be very simple to determine the \$C\$ matrix

$$C = \begin{bmatrix} -z & 1\end{bmatrix}.$$

All that being said, the non-minimum phase should not be a problem if you assume that you know the state \$x\$ and your model is correct \$(A,B,C)\$. As far as I know, non-minimum phase is a limitation of classical control and PID. While State-space control has different drawbacks (like needing a good model of the plant, fast estimators and dealing with nonlinearities "explicitly".)


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