I've seen the relative questions. I'm studying a textbook and I'm confused as to how it connects bibo stability with asymptotic stability for an LTI system.
For asymptotic stability we check if the system's response goes to zero for zero input. For bibo stability if the input is finite the output must be finite for zero initial conditions.
The way I see it the output consists of two terms, the zero state response and the zero input response. Each of the stability checks above has to do with one of the terms.
If a system is asymptotically stable the zero input response tends to zero as time goes to infinity. How can this imply that the zero state response is finite (bibo stability) if it is not taken into account (initial conditions are set to 0) when we check asymptotic stability?
Best Answer
BIBO stability refers to the property that a bounded input applied to a system leads to a bounded output. However, the inital conditions actually doesn't matter. For LTI systems, BIBO stability is normally checked by considering the transfer function, where no initial conditions occur. It is just a transfer behavior. However, when you formulate BIBO stability in the time domain, then the initial conditions occur explicitly.
Asymptotic stability refers to the stability of an equilibrium point (it is a stability concept w.r.t. equilibria). Therefore, actually you can not speak from zero input response. In the case of exponential stability, which is a special case of asymptotic stability, you can conlude that the state or the output (as one component of the state) of the system (no matter if linear or nonlinear) is bounded for a bounded input. (However, one has to take care of the region of exponential stability in the state space. When exponential stability holds globally then there is no problem. Additionally one has to take care that the input function $g(x)$ in $\dot x = f(x) + g(x)u$ is Lipschitz continuous)
What you are seemingly looking for is stability of the origin (not asymptotic stability, not BIBO stability). In the case of an $n$th order LTI system you have this e.g. when you have $n-1$ eigenvalues with negative real part and only one single eigenvalue with zero real part. (A system with stable equilibrium is not BIBO, since the input is integrated for all times, therefore not bounded.)