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I want to simulate a system whose impulse response is like the following:

$$h(t) = e^{-at} \sin(t)$$

The graph of which should look similar to the plot below:

example impulse response

I want to simulate the output for this system in response to an arbitrary input signal. I believe I can use control theory to do this. Is the impulse response sufficient to derive a state space model, and how do I do it? Under what conditions is the impulse response sufficient?

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    $\begingroup$ The response of the zero initial state system to an arbitrary input is $h * u$, you can derive a state space model, but it ignores unobservable/uncontrollable dynamics (since the impulse response is from a zero initial state. I have no idea what you mean by 'sufficient'. $\endgroup$ – copper.hat Nov 2 '14 at 17:36
  • $\begingroup$ What are "unobservable dynamics"? By sufficient, I mean "Is knowing the impulse response enough to come up with a state space model". $\endgroup$ – statusfailed Nov 2 '14 at 17:39
  • $\begingroup$ Like I wrote, you can generally come up with a state space model, but it cannot model unobservable/uncontrollable dynamics (they don't appear in the transfer function). There is a canonical decomposition theorem (or similar name) which shows this decomposition for LTI systems. $\endgroup$ – copper.hat Nov 2 '14 at 17:40
  • $\begingroup$ There is no magic here. If it can't be controlled, it can't appear in the transfer function, and if it can't be observed it can't appear in the transfer function. Think of an entirely independent, unconnected to any input or output, system with dynamics $\dot{x} = x$. This is unstable, and if considered as part of the overall system, could be catastrophic, but will not appear in the transfer function. $\endgroup$ – copper.hat Nov 2 '14 at 17:44
  • $\begingroup$ To use a bad analogy, if someone hides a infinitesimally small and light time bomb in my car (so I can't control or observe it), then from my perspective (the driver) the transfer function is the same as before, but clearly the overall system is different now. $\endgroup$ – copper.hat Nov 2 '14 at 17:49

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