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. $\endgroup$ Dec 16, 2020 at 7:26
  • $\begingroup$ Okay, thanks for clarifying. $\endgroup$
    – Matt L.
    Dec 16, 2020 at 10:52

1 Answer 1


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.


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