2
$\begingroup$

It is known that a system marginally stable if and only if the real part of every pole in the system's transfer-function is non-positive, one or more poles have zero real part, and all poles with zero real part are simple roots (i.e. the poles on the imaginary axis are all distinct from one another).[Wikipedia].

My question is based on the definition, a system with state space representation (A,B,C), where $A = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$ that has a non simple root at origin, is considered unstable? why?

$\endgroup$
1
$\begingroup$

Suppose that

$$\mathrm A = \mathrm O_2 \qquad \qquad \mathrm b = \begin{bmatrix} 0\\ 1\end{bmatrix} \qquad \qquad \mathrm c = \begin{bmatrix} 1 & 1\end{bmatrix}$$

and that the initial condition is $\mathrm x_0 := (x_{10}, x_{20})$. Hence, the states are

$$x_1 (t) = x_{10} \qquad \qquad \qquad x_2 (t) = x_{20} + \int_0^{t} u (\tau) \,\mathrm{d}\tau$$

and the output signal is

$$y (t) = (x_{10} + x_{20}) + \int_0^{t} u (\tau) \,\mathrm{d}\tau$$

Suppose that we start from zero initial conditions and that the input signal is constant, say, $u = 1$. Hence, the output signal is given by

$$y (t) = \int_0^{t} u (\tau) \,\mathrm{d}\tau = t$$

Thus, even though the LTI system is internally marginally stable (the zero eigenvalue has two $1 \times 1$ Jordan blocks, rather than one $2 \times 2$ Jordan block), it is not BIBO (bounded input, bounded output) stable, as a bounded input can produce an unbounded output.

$\endgroup$
  • 3
    $\begingroup$ You are right about the fact that system is not BIBO stable. But I don't think the part of your answer about "internally marginally stable (i.e., no eigenvalues with positive real parts)" is true. (Take A = [0 1;0 0] for example) $\endgroup$ – user356937 Jul 31 '16 at 5:33
  • $\begingroup$ @user356937 You are absolutely correct. The system is internally marginally stable because the zero eigenvalue only has $1 \times 1$ Jordan blocks. I will edit my answer. $\endgroup$ – Rodrigo de Azevedo Jul 31 '16 at 10:55
  • $\begingroup$ Thanks, I was looking for this answer. btw, can u give me a reference too? And also, do you think that simple root is not opposite of repetitive root? $\endgroup$ – user356937 Jul 31 '16 at 20:54
  • $\begingroup$ @user356937 Take a look at lecture 9 of books.google.com/books?id=tvd4ILdJUQoC $\endgroup$ – Rodrigo de Azevedo Jul 31 '16 at 21:18
0
$\begingroup$

Since not all the eigenvalues of $A$ are negative, then your system is not asymptotically stable. However, even having all the eigenvalues on the complex axis you system can be unstable in the sense BIBO (Bounded Input / Bounded Output) but marginally stable (if all the zero eigenvalues are simple). Check the answer from @Rodrigo for more details.

$\endgroup$
  • 2
    $\begingroup$ but for A = [0 1; 0 0] all eigenvalues are non-positive and the system is unstable. $\endgroup$ – user356937 Jul 31 '16 at 5:35
  • $\begingroup$ You are totally right with your comment. Reviewing my notes of Linear systems, a deep observation has to be made indeed. The answer from @Rodrigo about the "internal" concept (looking at the multiplicity of the zero eigenvalue) has to be introduced in order to address correctly your example. $\endgroup$ – user51196 Aug 1 '16 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.