How to prove that difference equation is stable when its roots are inside the unit circle.

Question

What is the proof for the result that a process governed by difference equations is stable if all of its poles are inside the unit circle?

Motivation for the question, take a difference equation:

$$y[n] = b_1x[n]+a_1y[n-1]+a_2y[n-2]...$$

Perform Z-transform (taking a short-cut):

$$\frac{1}{1-a_1z^{-1}-a_2z^{-2}...}$$

Setting the characteristic equation (the denominator) to zero, and solving for z, the process is stable if the absolute value of z < 1. Question: Why does the procedure work?

Answer 1

One way you can try to see it is through an analogy with the Laplace transform...

$$e^{-at}\stackrel{\mathcal{L}}\longleftrightarrow \frac{1}{s+a}$$

Where $a\in\mathbb{C}$. If $\Re\{a\}>0$, the exponential decays and the system is stable. If $\Re\{a\}=0$, there is no damping and the system is marginally stable. $\Re\{a\}<0$, the exponential grows infinitely and the system is unstable.

Therefore, the absolutely stable region is mapped as the left half of the s-Plane.

Now, for the Z-Transform:

$$a^n\stackrel{\mathcal{Z}}\longleftrightarrow \frac{z}{z-a}$$

We know that asequence is stable if it is absolutely summable. To simplify the analysis, this time, let us consider $a\in\mathbb{R}$. If $a<1$, $\sum_{n=0}^{\infty}a^{n}<\infty$ decays and converges. If $a=1$, in this case we have the discrete unit step, whose pole lay over the unit circle, characterizing marginal stability. If $a>1$, $\sum_{n=0}^{\infty}a^{n}$ increases indefinitely and does not converge.

Therefore, the absolute stable region is mapped inside the unit circle on the z-Plane.

You can also understand this through the relationship between the Region of Convergence of the Z-Transform ($ROC$) and the unit circle. For that, please refer to this answer on a question about region of convergence.

Answer 2

My own attempt at an answer (definitely very hard around the edges):

Consider a sequence:

$$b+a_0z+a_1z^2+a_3z^3... = \lim_{c \rightarrow \infty} c$$

Now, if z is less than one, some a has to be infinite. Therefore, the process is only bounded if z > 1. I reversed the z operator so the solution is inverted.