# Observability of a system

Suppose we are given the following system: $$$$\dot{x} = Ax \\ y = Cx$$$$ We are given the following statements:

$$1.$$ $$A \in R^{n\text{x}n}$$ and $$C \in R^{m\text{x}n}$$ are exponentially stable in the sense of Lyapunov and the system is observable.

$$2.$$ The modes of the system are $$e^{\lambda_1t}$$,$$e^{\lambda_2t}$$, ..., $$e^{\lambda_nt}$$ with $$\lambda_i \neq\lambda_j$$ if $$i \neq j$$.

Now consider the following sysem: $$$$\ddot{x} = Ax \\ y = Cx$$$$ I want to determine whether this new system is controllable or not.

My attempt:

The new system has modes $$e^{\sqrt{\lambda_1}t}$$,$$e^{\sqrt{\lambda_2}t}$$, ..., $$e^{\sqrt{\lambda_n}t}$$.

We can construct the observability gramian as follows:

$$$$W_o(t_1,t_2) = \int_{t_0}^{t_1}\Phi(\tau,t_0)C(\tau)C(\tau)^{T}\Phi(\tau,t_0)^{T}$$$$

Now we know that for the first system $$\Phi(\tau,t_0)$$ is $$n$$x$$n$$ diagonal matrix with elements $$e^{\lambda(\tau-t_0)}$$ and $$W_o$$ has rank $$n$$. On the other hand, for the second system we have elements $$e^{\sqrt{\lambda}(\tau-t_0)}$$ and I cannot figure out how I can prove $$W_o$$ has rank n or not.

First of all, you missed a lot of eigenvalues in your approach (the system has $$2n$$ eigenvalues) and your approach into the problem is way more complicated than it should.

Let $$X=(x,\dot x)$$, then we have that

$$\dot{X}=\begin{bmatrix}0 & I\\ A & 0\end{bmatrix}X,\quad y=[C\quad 0]X.$$

We also have that

$$\begin{bmatrix}0 & I\\ A & 0\end{bmatrix}^{2i}=\begin{bmatrix}A^i & 0\\ 0 & A^i\end{bmatrix}$$ and $$\begin{bmatrix}0 & I\\ A & 0\end{bmatrix}^{2i+1}=\begin{bmatrix}0 & A^i\\ A^{i+1} & 0\end{bmatrix}.$$

So, $$CA^{2i}=[CA^i\quad 0]$$ and $$CA^{2i+1}=[0\quad CA^i]$$. Therefore, the observability matrix is given by

$$\begin{bmatrix} C & 0\\ 0 & C\\ CA & 0\\ 0 & CA\\ \vdots\\ CA^{n-1} & 0\\ 0 & CA^{n-1}\\ \end{bmatrix}.$$

Clearly, if the system $$(A,C)$$ is observable, then the above matrix is full rank, meaning that the considered system is also observable.

• Yeah sorry the modes should be $e^{\sqrt{\lambda_1}t}$,$e^{-\sqrt{\lambda_1}t}$,$e^{\sqrt{\lambda_2}t}$,$e^{-\sqrt{\lambda_2}t}$ ..., $e^{\sqrt{\lambda_n}t}$,$e^{-\sqrt{\lambda_n}t}$ right?
– eet
Commented Sep 20, 2022 at 19:04
• @eet Yes this is correct. The characteristic polynomial is actually $\det(s^2I-A)$.
– KBS
Commented Sep 20, 2022 at 19:08