Given a system describing a 2-vehicles platoons that can be described as such :

  • $v_k$ the speed of each vehicle
  • $s$ the gap between the vehicles.

We can describe the state of the system as :

  • $\dot{s} = v_0 - v_1$
  • $\dot{v_k} = u_k$

Hence, writing the state of the system as $x = (v_0, s, v_1)^T$ yields the following state equation : $$\dot{x} = Ax + Bu$$ with : $A = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$, $B = \mathbb{1}$.

The eigenvalues of $A$ are all $0$, so the system should be marginally stable.

However, it seems to me that taking any $v_1 \neq v_0$ as initial conditions gives $s \to \infty$ so $x$ isn't bounded, so it isn't marginally stable.

Where am I wrong ?

  • $\begingroup$ Shouldn't it be: $$B=\begin{pmatrix}1&0\\0&0\\0&1\end{pmatrix}$$ $\endgroup$
    – fibonatic
    Commented Jan 16, 2019 at 11:08

1 Answer 1


Your system is in fact not marginally stable. It can be verified with a condition stated here on wikipedia

If the system is in state space representation, marginal stability can be analyzed by deriving the Jordan normal form: if and only if the Jordan blocks corresponding to poles with zero real part are scalar is the system marginally stable.

You can find the Jordan normal form with the MATLAB function $\texttt{jordan()}$, which yields

$$\begin{bmatrix} 0&1&0\\0&0&0\\0&0&0 \end{bmatrix}$$

The corresponding Jordan blocks are $\begin{bmatrix} 0&1\\0&0 \end{bmatrix}$ and $0$. Clearly, the first Jordan block is not a scalar but a 2x2 matrix.

Therefore, the system is not marginally stable.

You can also see it by finding the eigenvalues and corresponding eigenvectors, it turns out that one of the eigenvalues has multiplicity 2, which violates the condition under the link above.

and all poles with zero real part are simple roots


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