Jury test: z³−1.1z²−0.1z+0.2

Jury stability test (Ogata form)

For $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$, all roots lie inside the unit circle if and only if:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, and so on, for the rows of the Jury table

The table entries are

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

Coefficients

$P(z) = z^3 - 1.1z^2 - 0.1z + 0.2$, so $n = 3$, $a_0 = 1$, $a_1 = -1.1$, $a_2 = -0.1$, $a_3 = 0.2$.

Conditions 1 to 3

• $|a_3| = 0.2 < a_0 = 1$: satisfied
• $P(1) = 1 - 1.1 - 0.1 + 0.2 = 0$: not greater than 0. The equality means there is a root exactly at $z = 1$.
• $(-1)^3P(-1) = -(-1 - 1.1 + 0.1 + 0.2) = 1.8 > 0$: satisfied

Jury table (for completeness)

$$\begin{aligned} b_0 &= a_3a_1 - a_0a_2 = 0.2(-1.1) - 1(-0.1) = -0.12 \\ b_1 &= a_3a_2 - a_0a_1 = 0.2(-0.1) - 1(-1.1) = 1.08 \\ b_2 &= a_3a_3 - a_0a_0 = 0.04 - 1 = -0.96 \end{aligned}$$

• $|b_2| = 0.96 > |b_0| = 0.12$: satisfied

Conclusion

All conditions hold except $P(1) > 0$, which is exactly zero. So one root lies on the unit circle at $z = 1$ and the others lie inside. The system is critically (marginally) stable, not asymptotically stable.

Check by factorising:

$$P(z) = (z - 1)(z^2 - 0.1z - 0.2) = (z - 1)(z - 0.5)(z + 0.4)$$

The roots are $z = 1$, $0.5$ and $-0.4$, which confirms the result.

Answer: Critically (marginally) stable, with a root at $z = 1$ (because $P(1) = 0$).