Suppose we are given the following system: \begin{equation} \dot{x} = Ax \\ y = Cx \end{equation} 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: \begin{equation} \ddot{x} = Ax \\ y = Cx \end{equation} 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:

\begin{equation} 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} \end{equation}

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.

Please correct me if I made a mistake in my attempt.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – eet
    Commented Sep 20, 2022 at 19:04
  • 1
    $\begingroup$ @eet Yes this is correct. The characteristic polynomial is actually $\det(s^2I-A)$. $\endgroup$
    – KBS
    Commented Sep 20, 2022 at 19:08

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