I got this system state representation:

$$\begin{align} \frac{dx}{dt} &= \begin{bmatrix}-3&0&0\\0&\alpha-2&0\\\alpha&0&2+\alpha\end{bmatrix}x+\begin{bmatrix}6\\2\\1\end{bmatrix}u\\ y &= \begin{bmatrix}0&0&1\end{bmatrix}x \end{align} $$

I shall find the biggest possible range of values for the system to be BIBO(Bounded Input Bounded Output) stable.

My eigenvalues are $-3, \alpha-2, \alpha +2$.
So my system is asymptotically stable for all values of $\alpha < -2$.

The eigenvalues of the system matrix A are the poles of the transfer function of the system. And if the poles have a negative real part, the system is BIBO stable.
So my system is also BIBO stable for all values of $\alpha < -2$.

Is my guessing here correct?

  • $\begingroup$ If I'm not wrong, BIBO, stable and asymptotically stable are equivalent for LTI systems. So you are right. $\endgroup$ – Shiyu Oct 8 '11 at 14:48
  • $\begingroup$ They are pretty equivalent, but I think not exactly. I am a bit skeptical because asymptotically stability and BIBO stability are 2 different tasks in my exercise... $\endgroup$ – madmax Oct 8 '11 at 16:54
  • 2
    $\begingroup$ BIBO includes pure imaginary eigenvalues, asymptotical stability does not allow them since the responses are either constant or sustained oscillations. $\endgroup$ – user13838 Oct 8 '11 at 20:18

As the name BIBO (Bounded Input-Bounded Output) suggests, pure oscillations or constant trajectories are allowed. Consider the following simpler cases for the autonomous systems (with initial condition $x(0) = \pmatrix{6&2&1}^T$ if you like):

  1. $$\dot x = \pmatrix{0&&\\&0&1\\&-1&0}x$$
  2. $$\dot x = \pmatrix{0&&\\&-1&1\\&0&-1}x$$
  3. $$\dot x = \pmatrix{-1&&\\&-1&1\\&0&-1}x$$

A simple $e^{At}$ calculation reveals that the trajectories are all bounded for all initial conditions in $\mathbb{R}^3$ but only 3. is asymptotically stable.

Hence, the answer for the system given in the question is : If $\alpha\leq-2$ then BIBO stability. If further, $\alpha<-2$ then asymmptotical stability. In the case where $\alpha=-2$ you get poles on the imaginary axis, hence oscillations. Here you have the so-called marginal stability.


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