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I have read in some dsp texts that when ROC includes unit circle,system is stable

But i am bit confused in difference between stability and marginal stability depending Upon ROC

Especially i am confused by part c of below dsp.se question

Z transform stability

Where in part C, ROC is 1/3 < |z| < 1 and analysis answer of part c is written as "non-causal & marginally stable" . I am confused by marginal stability. why it is marginally stable?If our ROC is changed to 1/3 < |z| < 1.1, will now it becomes stable or will it still remains marginally stable?

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  • $\begingroup$ By which part are you confused? Can you try to more precisely point out what you need help with? $\endgroup$
    – Marcus Müller
    Commented Feb 25, 2020 at 21:03

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Marginally stable means that an otherwise stable system has one or more simple poles on the unit circle (in discrete time), or on the imaginary axis (in continuous time).

The consequence of that is that transients don't decay, but they also don't grow without bounds. Marginally stable systems are unstable in the bounded-input bounded-output (BIBO) sense.

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  • $\begingroup$ By simple poles, you mean non repeated poles? $\endgroup$
    – DSP_CS
    Commented Feb 26, 2020 at 14:16
  • $\begingroup$ @Man: No, single distinct poles, not multiple poles. $\endgroup$
    – Matt L.
    Commented Feb 26, 2020 at 14:57
  • $\begingroup$ If we had ROC as 1/3 < |z| < 1.1, will it become fully stable? $\endgroup$
    – DSP_CS
    Commented Feb 26, 2020 at 16:34
  • $\begingroup$ @Man: Yes, in that case it can't be marginally stable (I would have hoped that that had become clear in my answer). $\endgroup$
    – Matt L.
    Commented Feb 26, 2020 at 17:45
  • $\begingroup$ will it not be stable in this new case?because now roc includes unit circle?but you are saying it will be marginally stable $\endgroup$
    – DSP_CS
    Commented Feb 28, 2020 at 14:01

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