[Math] BIBO stability with positive eigenvalue

control theorydynamical systems

Suppose I have the transfer function of a 2nd order Linear Time-Invariant system and there are only two poles, one positive and one negative, can I conclude that the system is not BIBO (Bounded Input Bounded Output) stable? Is there a theorem that links the poles of a transfer function to the BIBO stability?

Best Answer

BIBO stability states that when the system starts in the origin at $t=0$ and a bounded input $u(t)$ is applied, such that $|u(t)|<a\ \forall\, t>0$, with $a$ some positive constant, then the system output also remains bounded (there exists some constant $b$ such that $|y(t)|<b$). This basically comes down to that the impulse response of the system should always be bounded. This implies that all poles should have a negative real part for continuous LTI systems.

However if we consider a state space model representation of a system,

$$ \left\{ \begin{align} \dot{x} & = A\, x + B\, u \\ y & = C\, x + D\, u \end{align}\right. $$

then the state matrix $A$, does not have to be Hurwitz. Namely if the system is controllable then all the eigenvalues of $A$ would correspond to the poles of the transfer function. But if unstable modes of the system are not controllable, then they can't be disturbed out their equilibrium at the origin. For example,

$$ A = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 0 \end{bmatrix}, $$

is BIBO stable, even though $A$ has an eigenvalue of $1$. However I do have to note that only controllable (and observable) modes of a system are visible in transfer functions. So if the poles of a transfer function all have a negative real part then it will be BIBO stable; if not then it is not BIBO stable.