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EDIT: I failed to mention that the system's inverse also needs to be causal and stable. I cannot wrap my mind around on how when a system and its inverse are both causal and stable and LTI IIR it is safe to say that the system is also minimum phase.

So the causality for a system means that $$h[n]=0, \mbox{for all } n<0,$$ but what does the stability mean when talking about a system? That there are no poles inside the unit circle? and if so how does it turn out that when a system is causal and stable it's also min phase.

Also intuition-wise when is a system inverse causal and stable?

I should clarify that I'm talking about discrete time.

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but what does the stability mean when talking about a system?

It means that all poles are INSIDE the unit circle.

and if so how does it turn out that when a system is causal and stable its also min phase.

Sorry, you got this wrong. Causal, stable and LTI does NOT imply minimum phase. A simple counter examples is a one-sample delay. It's causal, stable and LTI but definitely not minimum phase.

Minimum phase requires that all zeros are also INSIDE the unit circle as well.

Update

When you invert a system the poles become the zeros and the zeros become the poles. A minimum phase has both all zeros and poles inside the unit circle, that means that poles and zeros of the inverse will also be inside the unit circle and hence it's also minimum phase.

A non-minimum phase system has zeros outside the unit circle. If you invert the system, this zero turns into a pole that's outside the unit circle and hence the inverse is unstable (or non-causal).

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  • $\begingroup$ I don't know if this changes anything but I should say the proof that I saw was talking about an I I R system, so causal stable LTI and IIR $\endgroup$
    – Nyquist-er
    May 7 at 15:25
  • $\begingroup$ It does not change anything. It just means that it has poles other than $z=0$ $\endgroup$
    – Hilmar
    May 7 at 15:45
  • $\begingroup$ But it is true that when a system and its inverse are both causal and stable then the system is min-phase though right? I probably should change my question to that $\endgroup$
    – Nyquist-er
    May 7 at 15:47
  • $\begingroup$ Yes, that's correct.. $\endgroup$
    – Hilmar
    May 7 at 16:33

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