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In the realm of LTI systems (perhaps even in general systems, I am not sure) it is so that if a system is asymptotically stable then it is also BIBO stable. Is that the case when a system is marginally stable too? Meaning that if it is, then it too is BIBO stable.

For an LTI system to be BIBO stable we simply need that for any bounded input, the response does not exceed some finite bound. According to my understanding of marginal stability it is so that there does exist inputs that lead to both bounded and unbounded outputs. As such is it so in general that marginally stable systems ARE NOT BIBO stable?

I would like to say that marginally stable systems are NOT BIBO stable. As an example I can give you a system with transfer function $H(s)=1/s$ and give you as input $x(t)=u(t)$ where $u(t)$ is the heaviside function a.k.a the unit step function. That would yield unbounded outputs no?

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  • $\begingroup$ I think that marginally "stable" systems have an impulse response that is bounded. Putting single poles directly on the unit circle results in a marginally stable system. But "BIBO" means that the output is bounded to some finite value given any input that is bounded to a given finite value. If you have a pole located at $z=1$, that is a marginally stable system. The impulse response is a step function and bounded forever. But if you inputted to that system a step function (which is bounded), the output is a ramp and would grow forever. Not BIBO. $\endgroup$ Oct 16, 2022 at 18:19
  • $\begingroup$ Looks like you answered your question with the Analog analog of the digital answer I put in my comment. We both have integrators. $\endgroup$ Oct 16, 2022 at 18:20

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That's exactly it, I can just confirm the answer stated in your question:

marginally stable systems are not BIBO-stable.

Another example is a system with two complex conjugate poles on the imaginary axis (considering continuous time here). It's basically an oscillator. If excited with an input with constant amplitude at the same frequency as the system's oscillation frequency, the output's envelope will grow linearly. Hence, the output signal is unbounded even though the input is bounded.

Also take a look at this related question and its answers.

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  • $\begingroup$ Interesting, could you give a concrete example of a transfer function and an input that would give this result? $\endgroup$
    – NoName123
    Oct 16, 2022 at 18:36
  • $\begingroup$ So with poles like that we just have an oscillator, so for example a sinusoid. If we do a convolution of that with say $x(t)=4$ the result will just be an unbounded ramp. Is this a good example of what you were talking about? $\endgroup$
    – NoName123
    Oct 16, 2022 at 18:43
  • $\begingroup$ @NoName123: It's the example I gave in my answer: a sinusoidal input at the frequency where the poles are located, assuming that the poles are on the imaginary axis (or on the unit circle in discrete time). $\endgroup$
    – Matt L.
    Oct 16, 2022 at 18:50
  • $\begingroup$ But a sinusoid does not have constant amplitude? $\endgroup$
    – NoName123
    Oct 16, 2022 at 18:51
  • $\begingroup$ @NoName123: Well, depends on what you mean by amplitude. A signal $A\sin(\omega_0t)$ has a constant amplitude $A$. It doesn't need to be a constant signal. $\endgroup$
    – Matt L.
    Oct 16, 2022 at 18:53

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