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What are the necessary(must) conditions for stability in z domain? I am sure about one(ROC must include unit circle) Is there any other such condition which states that there shouldn't be any poles in ROC?

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    $\begingroup$ Poles are not allowed in the ROC in any way. ROC means region of convergence, that is, an area of the complex plane where the transform converges. Convergence means the transform takes finite values. Poles, by definition, are points of the complex plane where the transform takes an infinite value. $\endgroup$ – GKH Mar 4 '20 at 17:20
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The one and only condition for BIBO stability of a 1D discrete-time system, in the z-domain, is that its transfer functions's ROC (region of convergence) should include the unit circle : $|z| =1$. Therefore, it's a necessary and sufficient condition for BIBO stability of a 1D SISO system.

There are no other conditions (to my knowledge).

EDIT [based on comments] : By definition of ROC (region of convergence) there cannot be any poles inside the ROC region. This's related not with stability but with convergence. And also, causality is not related with stability too.

Properties of ROC is a different question to be answered.

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  • $\begingroup$ This only applies if you're allowing non-causal systems though, because a pole that's outside the unit circle has to be associated with the non-causal portion of the system response for it to be stable. $\endgroup$ – TimWescott Mar 4 '20 at 23:20
  • $\begingroup$ conditions for BIBO stability for more than 1D discrete-time system?? $\endgroup$ – engr Mar 5 '20 at 4:56
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    $\begingroup$ @TimWescott I might be missing something. If the ROC includes the unit circle then there CANNOT be any poles on or outside the unit circle for a causal system since the ROC extends outward from but not including the outermost pole. Right? So this does apply to causal systems and your restriction then would not be correctly stated (?) $\endgroup$ – Dan Boschen Mar 5 '20 at 12:37
  • $\begingroup$ @DanBoschen You're right -- I rarely think of stability in terms of the ROC, which is how I got myself tied in a knot. $\endgroup$ – TimWescott Mar 5 '20 at 14:40
  • $\begingroup$ @TimWescott same here which is why I bothered to ask. And why I participate in Stack Exchange, it helps untangle my ever tangling knots :) $\endgroup$ – Dan Boschen Mar 5 '20 at 16:52
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Here is one viewpoint in evaluating system stability.

[1] System z-plane pole(s) lie outside the unit circle: System impulse response increases, with time, toward ±infinity. System frequency response does not exist. System is unstable.

[2] System z-plane pole(s) lie on the unit circle: System impulse response remains non-zero and finite for all time. System frequency response exists but contains infinite-magnitude value(s). System is conditionally stable.

[3] System z-plane pole(s) lie inside the unit circle: System impulse response decreases, with time, toward zero. System frequency response exists and contains no infinite-magnitude value(s). System is stable.

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  • $\begingroup$ All of the three above conditions are assuming "causal" system? $\endgroup$ – engr Mar 5 '20 at 10:24
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    $\begingroup$ That is an excellent viewpoint, and I also like to think about stability in terms of both time, frequency, and complex frequency: ROC includes $|z| =1$ $\leftrightarrow$ Fourier sum converges $\leftrightarrow$ Frequency response exists $\leftrightarrow$ impulse response is absolutely summable $\leftrightarrow$ LTI system is BIBO stable. $\endgroup$ – GKH Mar 5 '20 at 20:11
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    $\begingroup$ @GKH. Your words make very good sense to me. $\endgroup$ – Richard Lyons Mar 6 '20 at 8:45
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ROC cant include poles because the impulse response wont converge if it includes poles and hence the system will be unstable. Also, the ROC must include the unit circle because if it includes the unit circle, the Fourier transform is defined since the unit circle represents convergence of FFT.

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  • $\begingroup$ Hey @Anushka, it's not a curve, it's a region of the complex plane. ROC curve is another, irrelevant term. $\endgroup$ – GKH Mar 5 '20 at 20:07
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    $\begingroup$ Oh yes...my bad. I mixed detection theory term with dsp term $\endgroup$ – DSP Novice Mar 5 '20 at 23:44

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