Controllable realisation of $\frac{s^4+1}{4s^4+2s^3+2s+1}$ is both controllable and observable? 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: 

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

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

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?
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.