4
\$\begingroup\$

I have the calculated the transfer function of system one $$ G_{1}(s) = \frac{-(s-2)}{(s+1)^2} $$ and of system two $$ G_{2}(s) = \frac{s+2}{s-2} $$ Now I have to check if the hole system $$ G(s) = G_1(s)\cdot G_2(s)=\frac{-(s-2)}{(s+1)^2}\cdot \frac{(s+2)}{(s-2)} = \frac{-(s+2)}{(s+1)^2} $$ is BIBO and asymptotically stable. The question is, if a pole is shorten out by the mutiplication with the other system, is it still BIBO and asymptotically stable? If I look at the hole system G(s) it is both, it has it poles with Re < 0 and the final value theorm is proofing a asymptotic stability.

But G2(s) is not asymptotically or BIBO stable. Is G(s) still asymptotically and BIBO stable or can it not be it, because one sub system is not asymptotically/BIBO stable?

\$\endgroup\$
  • 2
    \$\begingroup\$ Only in theory can a zero exactly cancel a pole. \$\endgroup\$ – Chu Oct 19 at 16:33
4
\$\begingroup\$

Bibo stability is all about systems external stability which is determined by applying the external input with zero initial condition (transfer function in other words) so if you check bibo stability of G(s) ,it would be bibo stable

But

Asymptotic stability is all about systems internal stability which can be determined by applying the non zero initial condition and no external input ,and if two system as in your example are cascaded if any one of them is unstable(or both) then its modes will not decay and reaches up to infinity (in your example) after infinite time ,so G(s) would be asymptotic unstable

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you for your answer, I understand it now. So in two systems can stabilze themself to be BIBO stable, if one of these isn't? \$\endgroup\$ – tugmn Oct 19 at 15:58
  • \$\begingroup\$ @tugmn I don't understand your question ,bibo stability of overall system (in your example) doesn't mean they stablize themselves rather we cannot determine its unstability by input output test (transfer function) because your overall system is either uncontrollable or unobservable or both and in Cascade system configuration (only) an unstable system is enough to make overall system unstable.in other words bibo stability is necessary condition but not sufficient for overall system to be stable in sense of asymptotic \$\endgroup\$ – user215805 Oct 19 at 16:05
  • \$\begingroup\$ I am sorry, my English is not good. My idea was wrong, my question was if you could just cascade the system to BIBO stabilze it, but as you said it is not enough for a asymptotoc stability and therefore makes no sense to do it. So BIBO stability is just a necessary condition, and does not say much about the system, just that the output is never +/-infinity for a bounded input. \$\endgroup\$ – tugmn Oct 19 at 16:31
  • \$\begingroup\$ Yeah you got that ! Bibo stability tells about stability of system correctly when system is controllable as well as observable (no pole zero cancellation) otherwise it doesn't \$\endgroup\$ – user215805 Oct 19 at 16:42
3
\$\begingroup\$

A system with a cancelled pole and zero may still be unstable (just not in the BIBO sense) -- it can be mathematically shown that a system with such a cancellation may still misbehave due to its response to an initial condition - this is in contrast to a stable system with all poles in the LHP, where the contribution of the initial condition to the output decays to zero. This violates asymptotic stability, but doesn't address the BIBO condition.

Furthermore, in a practical analogue realization, it is very difficult to precisely cancel a pole with a zero due to physical variation of components; your system may end up being unstable, with a pole very close to a zero in the realization.

| improve this answer | |
\$\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.