Absolute stability analysis in the z-plane

Absolute stability asks only whether the system is stable or unstable. A closed-loop discrete-time system with pulse transfer function $C(z)/R(z) = G(z)/[1+GH(z)]$ is stable if all roots of the characteristic equation

$$P(z) = 1 + GH(z) = 0$$

lie inside the unit circle $|z| < 1$. A simple pole on $|z| = 1$ makes the system critically stable. Any pole outside the circle, or a repeated pole on it, makes the system unstable.

Factoring a high-order $P(z)$ is tedious, so two direct tests are used:

1. Jury stability test, applied directly in the z-plane.
2. Bilinear transformation + Routh–Hurwitz: substitute $z = (w+1)/(w-1)$. This maps the inside of the unit circle onto the left half of the w-plane, so the ordinary Routh test can be used.

Jury stability test

Write $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$. The system is stable if and only if all of these hold:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. In the Jury table, $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, … down to the row with three elements.

The rows of the Jury table are built as:

$$b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1} \end{vmatrix},\qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1} \end{vmatrix}$$

Example 1

Test $P(z) = z^3 - 1.3z^2 - 0.08z + 0.24 = 0$. Here $a_0 = 1$, $a_1 = -1.3$, $a_2 = -0.08$, $a_3 = 0.24$.

The condition $P(1) > 0$ fails, so the system is unstable. Factoring confirms this: the roots are $z = 1.2, 0.5, -0.4$, and $z = 1.2$ lies outside the unit circle.

Example 2: stable range of gain

Take $G(z) = \dfrac{K(0.3679z + 0.2642)}{(z-1)(z-0.3679)}$ with unity feedback. The characteristic equation is:

$$z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0$$

• $|a_2| < 1$: $0.3679 + 0.2642K < 1$, so $K < 2.392$
• $P(1) = 0.6321K > 0$, so $K > 0$
• $P(-1) = 2.7358 - 0.1037K > 0$, so $K < 26.38$

Answer: The system is stable for $0 < K < 2.392$. At $K = 2.392$ it is critically stable and oscillates.