# Conditions for controllability canonical form to be non-observable but detectable

I have a system in controllability canonical form: $\dot{x} = \begin{bmatrix} 0 & 1 & 0 & ... & 0 \\ 0 & 0 & 1 & ... & 0 \\ . & . & . & ... & . \\ . & . & . & ... & . \\ 0 & 0 & 0 & ... & 1 \\ -a_1 & -a_2 & -a_3 & ... & -a_{n} \end{bmatrix}x + \begin{bmatrix} 0 \\ . \\ . \\ . \\ 0 \\ 1 \end{bmatrix}u$ And $y = \begin{bmatrix} c_1 & 1 & 0 & ... & 0 \end{bmatrix}x$.

In order for the system to be unobservable but detectable, what are the implied conditions for $c_1$?

I have tried to convert the systems back to the transfer function, in order to show that the denominator or the nominator is a prime polynomial, but unsuccessfully.

Do you have any ideas?

The above system is unobservable and detectable if

1) $\quad s=-c_1$ is a root of the polynomial $s^n+a_ns^{n-1}+\cdots+a_2s+a_1$

2) $\qquad c_1>0$

Let $x:=[x_1\cdots x_n]^T$ and $X_i(s)$ the Laplace transform of $x_i(t)$. Since the system is in the controllable canonical form we have from the ODEs describing the system (ignoring initial conditions)

$$sX_i(s) =X_{i+1}(s) \qquad, i=1,\cdots,n-1 \qquad\qquad (1)\\ sX_n(s)=-a_1X_1(s)-\cdots-a_nX_n(s)+U(s)\quad\quad (2)$$ From (1) we have that

$$X_i(s)=s^{i-1}X_1(s) \qquad,i=1,\cdots,n\qquad\qquad(3)$$

For example we can write $X_3(s)=sX_2(s)=s\cdot sX_1(s)=s^2X_1(s)$. Replacing (3) in (2) we obtain

$$X_1(s)=\frac{U(s)}{s^n+a_ns^{n-1}+\cdots+a_2s+a_1}$$

Since $y=c_1x_1+x_2$ we have that $$Y(s)=c_1X_1(s)+X_2(s)=(c_1+s)X_1(s)$$ and therefore

$$\frac{Y(s)}{U(s)}=\frac{s+c_1}{s^n+a_ns^{n-1}+\cdots+a_2s+a_1}$$

The uncontrollable and unobservable modes of a system appear in the form of pole-zero cancellations in the transfer function. Since the system is already in the controllable canonical form there are no uncontrollable modes. Therefore the system is unobservable if we have pole-zero cancellations i.e. the zero $-c_1$ must also be a pole of the system (a root of the polynomial $s^n+a_ns^{n-1}+\cdots+a_2s+a_1$). For the system to be detectable, the unobservable modes must be stable and therefore we must have $-c_1<0$.

• You are the KING! – user1415066 Dec 2 '16 at 10:29

You only need to check the following for detectability. $$Av = \lambda v \quad \text{with} \quad \mathbf{Re}(\lambda) \geq 0 \implies cv \neq 0$$