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What will be stability if we have only one single pole at origin in s domain?? and what will be the case for multiple poles at origin in s domain?

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A system with simple distinct poles on the imaginary axis (and note that the origin is on the imaginary axis) and no poles in the right half-plane is called marginally stable. If you have poles with multiplicity greater than $1$ on the imaginary axis, or if there are poles in the right half-plane, then the system is unstable.

For discrete-time systems, the same is true if you replace "imaginary axis" by "unit circle".

Transients in marginally stable systems do not decay, but neither do they grow without bounds. In practice, we usually want to avoid marginally stable systems. An ideal integrator is an example of a marginally stable system.

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    $\begingroup$ I would like to expand a little with regards to stability. There are many definitions in control theory for stability, such as BIBO, asymptotic, exponential and Lyapunov. For simple poles on the imaginary axis (and no right half plane poles) only satisfies Lyapunov stability and not any of the other mentioned stability criteria. And all mentioned criteria fail when you have a non-simple pole on the imaginary axis. It can be noted that you can have multiple poles on the imaginary axis, as long as they are all simple, and still be Lyapunov stable. $\endgroup$ – fibonatic Sep 28 '19 at 11:20
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    $\begingroup$ @abtj: It's just as I said, the origin $s=0$ is on the imaginary axis, so a pole at $s=0$ is on the imaginary axis. And, obviously, not all poles on the imaginary axis are necessarily at the origin. $\endgroup$ – Matt L. Sep 28 '19 at 11:20
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    $\begingroup$ @abtj imaginary axis just means a pole with zero real part; its imaginary part can be anything. This is also true for the origin (the origin has both zero real and zero imaginary part). $\endgroup$ – fibonatic Sep 28 '19 at 11:22
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    $\begingroup$ @abtj A simple digital resonator can have two poles on the z-plane's unit circle and not be unstable. Such a resonator would instead be marginally stable. $\endgroup$ – Richard Lyons Sep 28 '19 at 16:12
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    $\begingroup$ At the risk of stating the obvious, it is not about the number of poles on the unit circle, but about the multiplicity of the poles. Separate simple poles on the unit circle, no matter how many, result in a marginally stable system, poles on the circle with multiplicity greater than $1$ cause the system to be unstable. $\endgroup$ – Matt L. Sep 29 '19 at 7:35

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