Jury stability test explained

The Jury stability test is an algebraic test for the absolute stability of a discrete-time (sampled-data) system. It works directly in the z-plane. Using only the coefficients of the characteristic polynomial $P(z)$, it tells whether all roots lie inside the unit circle $|z| = 1$, without solving for the roots. It does for the z-plane what the Routh–Hurwitz test does for the s-plane.

Procedure

Write the characteristic equation as

$$P(z) = a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0, \qquad a_0 > 0$$

Form the Jury table:

Each even row is the row above it written in reverse. The elements are $2\times2$ determinants:

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

Stability conditions

The system is stable if and only if all of these hold:

1. $|a_n| < a_0$
2. $P(z)\big|_{z=1} > 0$
3. $(-1)^nP(z)\big|_{z=-1} > 0$
4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, …, $|q_2| > |q_0|$

There are $n+1$ conditions in total. Check conditions 1–3 first. If any of them fails, the system is unstable and no table is needed. For $n = 2$ no table is needed at all.

Example

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

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

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

The system is stable for $0 < K < 2.392$.

Features

• It works directly with z-domain coefficients, so no bilinear transformation is needed.
• It also gives the range of a gain $K$ for stability, as in the example.
• If a condition becomes an equality (for example $P(1) = 0$), a root lies on the unit circle and the system is critically stable.
• It shows only absolute stability, not relative stability (damping, margins).