# How to derive stability of analogue and digital systems

I'm trying to find out why digital systems are stable iff poles oh their transfer function are inside the unit circle (or are there any others conditions?).

I understand analogous systems are stable iff all poles of their transfer function are negative 'cause of inverse Laplace transform, where $\mathcal{L}^{-1} \left \{ \frac {1}{s+a} \right \} = e^{-at} \cdot \mathcal{1} (t)$. (by the $\mathcal{1} (t)$ i mean heaviside function). Is that right?

But how to derive conditions for stability of discrete system?

What others ways for derivating stability of discrete and continuous systems exist?

• What is meant by the stability of a signal? Stability of a system, which is asked for in the last two sentences of your question, is something that is often discussed in the literature, and is indeed the question that Matt has answered in the last paragraph of his answer. The first part of his answer discusses whether a signal is bounded, that is, if there exists a constant $M$ such that $|x(n)|\leq M$ for all $n\geq 0$, and the example shows that the sequence defined by $x(n) = a^n, n \geq 0$ is unbounded if $|a| > 1$. Please don't use stable signal as a name for this notion. – Dilip Sarwate May 12 '13 at 13:39
• You're right, I'm sorry. I was in hurry and I learn DSP only short time, so it didn't sound weird to me, 'cause I'm still not used to these notions. – user50222 May 12 '13 at 13:56
• One additional minor suggestion: in English, the wording is analog system or analogue system or continuous-time system (as contrasted with digital or discrete-time system) and not analogous system as in your title. – Dilip Sarwate May 12 '13 at 14:15
• For $|a|<1$ the signal is not only bounded but also absolutely summable, which is a much stronger condition. Only an absolutely summable sequence, if seen as a system's impulse response, corresponds to a stable system. BIBO stability basically means: a bounded input signal filtered by an absolutely summable impulse response results in a bounded output signal. An impulse response which is only bounded but not absolutely summable results in an unstable system. An example of such a system is an accumulator with $h(n)=\sigma(n)$, where $\sigma(n)$ is the step function. – Matt L. May 12 '13 at 19:10

In the discrete-time case you can use the $\mathcal{Z}$-transform:

$$\mathcal{Z}\{x(n)\}=X(z)=\sum_{n=0}^{\infty}x(n)z^{-n}\tag{1}$$

Equation (1) is the unilateral $\mathcal{Z}$-transform which is useful for causal signals which are zero for $n<0$. Now consider the signal $x(n)=a^n$, $n\ge 0$:

$$X(z)=\sum_{n=0}^{\infty}a^nz^{-n}$$

This series converges for $|a|<1$ and in this case $X(z)$ is given by:

$$X(z)=\frac{1}{1-az^{-1}}=\frac{z}{z-a}\tag{2}$$

From (2) you can see that $a$ is the pole of $X(z)$, and for $x(n)$ to be stable (if seen as a system's impulse response), and for $X(z)$ to exist, $|a|<1$ must hold, i.e. the pole must be inside the unit circle of the complex $z$-plane. This example is analogous to the continuous-time example in your question.

Also take a look at the concept of BIBO-stability (bounded-input-bounded-output stability), which leads to the same conclusion: for continuous time signals/systems the poles must be in the strict left half of the $s$-plane, and for discrete-time signals/systems the poles must be strictly inside the unit circle of the $z$-plane. The condition on the time-domain signals is that they be absolutely integrable in the continuous case, and absolutely summable in the discrete-time case:

$$\int_{-\infty}^{\infty}|x(t)|dt <\infty\\ \sum_{n=-\infty}^{\infty}|x(n)| <\infty$$