# Transfer function with cancellable zero pole and controllability

I have a transfer function (From Ogata's Modern Control Engineering)

$$\frac{s+2.5}{(s+2.5)(s-1)}$$

and the theory says the system has a pole zero cancellation and is uncontrollable.

They said that a state space rep of this transfer function has A and B matrix of:

$$A = \begin{bmatrix} 0 & 1 \\ 2.5 & -1.5 \end{bmatrix} \ \text{and} \ B = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

Using rank(ctrb(A,B)) in MATLAB, the result is not equal to the dimension of the state space so it not controllable.

So I got curious and used tf2ss in MATLAB and got another state space rep:

$$A = \begin{bmatrix} -1.5 & 2.5 \\ 1 & 0 \end{bmatrix} \ \text{and} \ B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Using rank(ctrb(A,B)) in MATLAB, I got a value equal to the dimension of the state space so it is controllable.

What have I misconceived? (and could someone teach me on how to make matrices in Markdown for the above?)

• Your A matrix has a typo, the 2 should be 2.5 and to learn about matrix formatting in MathJax take a look at this. – fibonatic Jul 22 '18 at 13:33
• Yea but the rank of the controllability matrix still remains 2 for the result I got from MATLAB.... – aldo Jul 22 '18 at 13:41

## 2 Answers

Normally state space models who are equivalent to the same transfer function are also equivalent to each other, such that there exists a similarity transformation between them. However if the considered transfer function has pole zero cancellations then an equivalent state space model would be a rank deficient controllability- or observability matrix, but it is not fixed which of the two has to be rank deficient (those ranks are conserved under a similarity transformation). But the lowest rank of either the controllability- or observability matrix will be the same for every state space model which is equivalent to the transfer function. So when checking pole zero cancellation you have to check for both controllability and observability.

PS: One might argue that there still exists a similarity transform between those state space models. Only the transform should be rank deficient itself (not invertible) but still satisfy $\hat{A}\,T = T\,A$ and $\hat{C}\,T = C$.

• Wow you're right; I checked the observability of my MATLAB result and it indeed is rank deficient. So it's a tradeoff then? i.e. if I want to be able to control the states then I have to sacrifice the possibility of observing one of the states, and similarly for wanting to observe all the states? – aldo Jul 22 '18 at 14:10
• @aldo When you want to build a controller you need both observability and controllability in order to completely shape your dynamics, so there is not really a tradeoff if in both cases you can't completely shape your dynamics. – fibonatic Jul 22 '18 at 14:29

Here is the derivation for your problem - Note the two states and the trick used to get the second state 