# Position of poles and Stability in $z$ domain

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?

• inside the unit disc. – percusse Oct 29 '17 at 17:15
• 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). – a concerned citizen Mar 30 '18 at 11:10

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.

The explanation:

Consider a causal LTI system with right-sided impulse response $h[n]$ whose transfer function can be expressed in terms of ratio of polynomials so that its $\mathcal{Z}$-Transform is

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

Where $d_p$ are the poles of the transfer function at which the sum diverges hence shall be avoided. Therefore associated with the $\mathcal{Z}$-Transform $H(z)$ is a region of convergence (ROC) which is a single connected circular region on the z-plane that cannot contain any poles inside.

Consider a right-sided (causal) exponential sequence $h[n] = a^n u[n]$ whose $\mathcal{Z}$-Transform is: $$H_a(z) = \sum_{n=0}^{\infty} a^n z^{-n} = \frac{1}{1 - a z^{-1}} ~~~,~~~\text{ ROC } |z| > |a|$$

As can be seen for a right-sided (causal) sequence the ROC extends outward from the pole to the infinity. Hence for the transfer function $H(z)$ of a causal system with $N$ poles, the ROC must extend outside from the largest pole to the infinity.

On the other hand, the BIBO stability of LTI system requires that $|y[n]|<\infty$ for all bounded inputs $|x[n]| < B$. By defining the output as a convolution sum, this translates to

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

This means that a BIBO stable causal LTI system's impulse response $h[n]$ is (must be) absolutely summable. Which means that for $|z| = 1$ 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 stable system, its ROC must include the unit-circle $|z|=1$.

Combinig these two concepts yields the following: 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 conclusion occurs for the stable but acausal LTI systems for which the ROC is inwards from the smallest pole and includes the unit circle and therefore that all of its poles must be out of the unit circle.