# Does the stability of an LTI system depend on the input?

Consider a system $$\G(s) = \frac{1}{s}\$$. It is marginally stable, as it has its only pole at $$\s=0\$$. However, if we apply a step input, the output is $$\ tu(t) \$$, which turns out to be unstable. But, the stability or instability of a system should not depend on the nature of the input. If it has a single pole at $$\ s= 0\$$, it should remain marginally stable, no matter what the input is.

So, how do we reconcile these conflicting notions?

• It is marginally stable. If you apply an impulse, the response does not go to infinity. – Chu Jan 16 at 8:13
• Marginal stability does not imply bibo stability, so there is nothing contradictory here – AVK Jan 16 at 16:36
• @Chu, what if we apply a step – ShiS Jan 17 at 11:05
• The input has nothing to do with the system's relative stability. Stability is a system property. In this case there's a pole at the origin, so it's on the boundary. – Chu Jan 17 at 15:28