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We know in Laplace Transform, if the poles lie on the left of $j\omega$ axis, we can say the system is stable. Similarly can we comment on the stability based on poles position in $\mathcal Z$-transform domain?

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    $\begingroup$ inside the unit disc. $\endgroup$ – percusse Oct 29 '17 at 17:15
  • $\begingroup$ If a visual representation is more to your liking, think of the convertion of s to z, where the left-hand side of the analog axis becomes a circle, due to the tan() treatment. So, as both the comments and the answer say, that's where the ROC lies. Search for a paper from Constantinides, "Spectral transformations for digital filters", last time I saw it it was freely available (to read, at the very least). $\endgroup$ – a concerned citizen Mar 30 '18 at 11:10
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Short Answer:

All the poles of a causal (right-sided) and stable LTI system must be inside the unit circle whereas all the poles of an acausal (left-sided) and stable LTI system must be outside the unit circle.

Explanation:

First, consider the following right-sided (causal) exponential sequence $h[n] = a^n u[n]$ whose $\mathcal{Z}$-Transform is: $$H(z) = \sum_{n=0}^{\infty} a^n z^{-n} = \frac{1}{1 - a z^{-1}} ~~~,~~~ |z| > |a|.$$

As can be seen for a right-sided (causal) sequence the ROC of $H(z)$ extends outward from the pole to the infinity.

Now, consider a causal LTI system with right-sided impulse response $h[n]$ whose transfer function can be expressed in terms of ratio of polynomials:

$$H(z) = \sum_{n=0}^{\infty} h[n]z^{-n} = \frac{P(z)}{Q(z)} = K \frac{\prod_{k=0}^{M-1} (1 - d_z z^{-1})}{\prod_{k=0}^{N-1} (1 - d_p z^{-1})} $$

Where $d_p$ are the poles of the transfer function at which $H(z)$ diverges. Hence, associated with the $\mathcal{Z}$-Transform $H(z)$, is a region of convergence (ROC) which is singly connected region that does not include any poles inside. Which implies that ROC for the causal system extends outwards from the largest pole.

On the other hand, the BIBO stability of LTI system requires that $|y[n]|<\infty$ for all bounded inputs $|x[n]| < B$, which translates to $h[n]$ be absolutely summable as follows:

$$ |y[n] = \sum_{k} h[k] x[n-k] | < \infty \implies \sum_{k=0}^{\infty} |h[k]| < \infty$$

This means, for stable system, at $|z| = 1$, the $\mathcal{Z}$ transform $$ H(z) = \sum_{k=0}^{\infty} h[k] z^{-k} < \sum_{k=0}^{\infty} |h[k] z^{-k}| = \sum_{k=0}^{\infty} |h[k]| < \infty$$

converges! Thus for a BIBO stable LTI system, its ROC must include the unit-circle $|z|=1$ inside.

Combinig these two concepts yields: for a causal LTI system to be stable its ROC must include the unit circle and it should extend outwards from the largest pole to infinity. The corollary of this fact is that the largest pole of the stable and causal LTI system is inside the unit circle. Hence we conclude that all of the poles of the stable and causal systems are inside the unit circle.

An opposite complementary conclusion occurs for the stable but acausal LTI systems (whose $h[n]$ includes left-sided terms like $-a^nu[-n-1]$) for which the ROC is inwards from the smallest pole, and must include the unit circle (for stability), and therefore that all of its poles must be out of the unit circle; i.e., it's smallest pole is larger than $|z|=1$.

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