4
$\begingroup$

I have a system which has the following transfer function $$H(s)=\frac{\beta + s}{s^{2} + 2\alpha s + \beta^{2}}$$ where $\beta = \sqrt{\omega^{2} + \alpha^{2}}$ and $\alpha>0$. The degree of numerator is less than the degree of denominator, moreover the zeros of $s^{2} + 2\alpha s + \beta^{2}$ are: $$s_1=-\alpha+i\omega,\qquad s_2=-\alpha-i\omega$$ If I remember correctly the courses I took at university many years ago, the system should be stable because the zeros of the denominator have a negative real part.

But what kind of stability are we talking about? (Asymptotic, global ...) Could someone help me by adding some more details and possibly references that may be useful to review this topic better?

$\endgroup$
2
  • $\begingroup$ What sort of stability matters to you? Provided $\alpha$ is positive and big enough, it should be stable for most definitions of stability. $\endgroup$
    – Peter K.
    Jul 9 at 18:22
  • $\begingroup$ The definition I remember working with most of the time with these problems is bounded input bounded output. $\endgroup$ Jul 9 at 20:06

2 Answers 2

3
$\begingroup$

Laplace-domain transfer functions describe linear systems*. If a system is linear, then its local stability properties are always the same as its global stability properties.

So if it's stable at all then it exhibits both asymptotic and global stability. By the same token, if it's unstable at all, then there's no choice of initial values that have some local stable attractor.

* Note that I am not talking about real-world systems that can be reasonably described by transfer functions -- all real-world systems are nonlinear if you push them hard enough; yet there's a whole lot of them that can be profitably treated as linear for the sake of analyzing their behavior. So if you've computed a transfer function for a real system then you have -- with or without realizing it -- taken a nonlinear system and made an approximate, linear, description of it. The linear system you describe is globally stable or globally unstable -- the nonlinear system you started with may not be.

$\endgroup$
3
$\begingroup$

Assuming that the given transfer function describes a causal system, it is true that the given system is stable because all its poles lie in the left half-plane. This type of stability referred to here is BIBO stability (bounded-input, bounded-output stability). BIBO-stability is also called external stability because it only describes the relationship between signals observable at the inputs and outputs of the system. This is different from a system's internal (asymptotic) stability. A system can be BIBO-stable but internally (asymptotically) unstable. BIBO-stability says nothing about the internal behavior of a system. A complete system characterization is given by the state-variable description, which is an internal description of the system, revealing whether a system is asymptotically stable.

An example of a BIBO-stable but asymptotically unstable system is the following causal system consisting of two concatenated sub-systems [1]:

$$H(s)=H_1(s)H_2(s)=\frac{1}{s+1}\tag{1}$$

with

$$H_1(s)=\frac{s-1}{s+1}\quad\textrm{and}\quad H_2(s)=\frac{1}{s-1}$$

Clearly, the total system is BIBO-stable because there are no poles on or to the right of the imaginary axis, but the system $H_2(s)$ is unstable, and hence, the total system is internally (asymptotically) unstable. This cannot be inferred from the transfer function $(1)$ because of the pole-zero cancellation. Only a state-variable description of the system could reveal its internal instability.


[1] B.P. Lathi and R. Green, Linear Systems and Signals, 3rd ed., p. 200.

$\endgroup$

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.