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?

  • $\begingroup$ The answer to both questions is yes. For the second question, it can be proved in different ways. One of which is the inverse Laplace transform $\endgroup$ – polfosol Mar 14 '17 at 13:39
  • $\begingroup$ @polfosol but then if BIBO stability is linked to the poles of the transfer function, which is the difference between simple stability (i.e., all poles with non-positive real part) and BIBO stability? $\endgroup$ – cholo14 Mar 14 '17 at 13:48
  • $\begingroup$ For LTI systems, it is safe to say that all types of stability are equivalent $\endgroup$ – polfosol Mar 14 '17 at 13:51
  • $\begingroup$ @polfosol This is not entirely correct. Namely Lyapunov stability does not have the exact same constraints are BIBO stability. This is because Lyapunov stability also allows poles with zero real part. And when using a state space model it is also possible to have a system which is BIBO but not Lyapunov stable (see the example in my answer). $\endgroup$ – Kwin van der Veen Mar 14 '17 at 14:12
  • $\begingroup$ @fibonatic I am not unfamiliar with Lyapunov stability. When I say it is safe, I mean from a practical point of view. $\endgroup$ – polfosol Mar 14 '17 at 14:18

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.

  • $\begingroup$ Stated otherwise: 1) if I am considering transfer functions, I have BIBO stability if and only if all the poles have non-positive real part; 2) if I am considering a state space model, the system can be BIBO stable even if there are some positive eigenvalues of A. Are statemets 1 and 2 correct? $\endgroup$ – cholo14 Mar 14 '17 at 13:58
  • $\begingroup$ @cholo14 Kinda sorta. Mathematically speaking, your second statement seems correct. But from a physical point of view (which is the whole point of stability analysis) that's impossible. $\endgroup$ – polfosol Mar 14 '17 at 14:03
  • 1
    $\begingroup$ @cholo14 If you want to be more mathematically correct then "non-positive real part" should be "negative real part", since this first also includes zero real part, but that is not BIBO stable. And for state space model you do have to specify that only the uncontrollable eigenvalues may have non-negative real parts. $\endgroup$ – Kwin van der Veen Mar 14 '17 at 14:09
  • $\begingroup$ @fibonatic a pole with 0 real part would not lead to BIBO stability because the impulse response wouldn't be absolutely integrable from -inf to +inf, even if bounded, correct? $\endgroup$ – cholo14 Mar 14 '17 at 14:16
  • $\begingroup$ @cholo14 I do not remember the official mathematical definition from the top of my head, but what you stated seems correct. However I often like to think of an example. For example an integrator ($1/s$) is not BIBO stable, because the response to a step will become a ramp, which in not bounded, or a undamped mass spring system exited at its resonance frequency will also blow up and thus not be bounded. $\endgroup$ – Kwin van der Veen Mar 14 '17 at 14:41

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.