I am trying to find the controllable realization of the following transfer function:


I approach this by first using polynomial division to assure that $H(s)$ is strictly proper. This yields:

$$H(s)=1/4+\frac{-1/8\cdot s^3-1/8\cdot s+3/16}{s^4+1/2\cdot s^3+1/2\cdot s+1/4}$$

From this I can extract the controllable canonical form where the matrices A, B, C and D are:

$$A=\begin{bmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 1\\-1/4 & -1/2 & 0 & -1/2\end{bmatrix}$$

$$B=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$$ $$C=\begin{bmatrix}3/16 &-1/8 &0 &-1/8\end{bmatrix}$$ $$D=1/4$$

From this I am confused because when I construct the observability and controllability matrices I found that the system is both controllable and observable (both matrices have full rank). This seems like a contradiction. Can it be possible? What am I doing wrong?

  • 2
    $\begingroup$ Controllability and obsrvabilty are not mutually exclusive, indeed having both is usually good $\endgroup$
    – user28715
    Oct 28 '17 at 16:02
  • $\begingroup$ Ok thank you! Its just that I thought that in the canonical controllable realization the system cannot also be observable. But I guess this is false. How can I ensure that my system is both observable an controllable? $\endgroup$
    – john melon
    Oct 28 '17 at 16:11
  • 1
    $\begingroup$ You did the tests. You can try doing the same tests on a different canonical form and should get the same answer. The system doesn’t mind how we write it down. State space expressions can be realized by more than one way for a particular system. If the system is both, it should be both. If you can find a counter example, it would be interesting to see it $\endgroup$
    – user28715
    Oct 28 '17 at 16:23

Controllability is a property of $A,B$ and observability is a property of $A,C$ matrices.

Transfer functions don't exhibit this property since if there is such a problem then you will not get a nonzero transfer function as there will be no possibility of control so nothing to transfer or nothing to observe. Hence, this is also a state space property.

The main theorem says that a state space realization is minimal if and only if it is both observable and controllable.

These representations allows one to quickly see, if exists, the unobservable and uncontrollable parts of the system. Such representations are obtained via the procedure called Kalman decomposition.

Here you have only constructed the controllable canonical form and already found out that the system is also observable. Thus the represantation is minimal.

  • $\begingroup$ A transfer function can be said to be "uncontrollable"/"unobservable" if you have pole zero cancellation. Namely if a pole can be cancelled with a zero then the denominator will be reduced in order and thus the system will require less initial conditions. However does changes nothing about the Bode or Nyquist diagram. $\endgroup$
    – fibonatic
    Oct 29 '17 at 6:15
  • $\begingroup$ @fibonatic Transfer functions form an equivalence class just like rational numbers so that can't be related to system properties as you can always also add spurious factors. Also you implicitly assume zero initial conditions for transfer functions anyways. $\endgroup$
    – percusse
    Oct 29 '17 at 9:48

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