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

$$H(s)=\frac{s^4+1}{4s^4+2s^3+2s+1}$$

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?

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  • $\begingroup$ Why should a system not be both controllable and observable? $\endgroup$ – SampleTime Oct 28 '17 at 14:06
  • $\begingroup$ Well I thought that when a system is in controllable canonical form then it must be controllable, but unobservable. $\endgroup$ – john melon Oct 28 '17 at 14:07
  • $\begingroup$ Why do you think that your original $H(s)$ is not controllable? $\endgroup$ – Kwin van der Veen Oct 28 '17 at 18:27
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    $\begingroup$ A state space model in the controllable canonical form is indeed controllable, however it could be unobservable it doesn't have to be. $\endgroup$ – Kwin van der Veen Oct 28 '17 at 20:17
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    $\begingroup$ If you don't have pole-zero cancellation in your transfer function, then every controllable realization is a minimal realization, thus an observable one. (and vice versa) $\endgroup$ – polfosol Oct 31 '17 at 13:53

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