If you take the $H(s) = \mathcal{L}[h(t)]$ the poles are on the imaginary axis so the system should be marginally stable, but is it?

  • $\begingroup$ The system is supposed to be causal so $h(t) = 0, \forall t < 0$. So yes, I meant $sin(t)u(t)$, sorry. $\endgroup$ – LaplaAlFourier Dec 16 '20 at 7:26
  • $\begingroup$ Okay, thanks for clarifying. $\endgroup$ – Matt L. Dec 16 '20 at 10:52

The Laplace transform of the given impulse response, i.e., the transfer function of the corresponding system, has two poles on the imaginary axis. According to the definition of marginal stability, that system is marginally stable.

It is, however, not BIBO stable, because given an input with a frequency equal to the frequency of the poles will result in an unbounded response. For a causal continuous-time LTI system to be BIBO stable, all its poles must be strictly in the left half-plane, i.e., their real parts must be negative.


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.