Jury test: z⁴−1.2z³+0.07z²+0.3z−0.08

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

$P(z) = z^4 - 1.2z^3 + 0.07z^2 + 0.3z - 0.08$, so $n = 4$, $a_0 = 1$, $a_1 = -1.2$, $a_2 = 0.07$, $a_3 = 0.3$, $a_4 = -0.08$.

Conditions 1 to 3

• $|a_4| = 0.08 < a_0 = 1$: satisfied
• $P(1) = 1 - 1.2 + 0.07 + 0.3 - 0.08 = 0.09 > 0$: satisfied
• $(-1)^4P(-1) = 1 + 1.2 + 0.07 - 0.3 - 0.08 = 1.89 > 0$: satisfied

Jury table

$$\begin{aligned} b_0 &= a_4a_1 - a_0a_3 = (-0.08)(-1.2) - 0.3 = -0.2040 \\ b_1 &= a_4a_2 - a_0a_2 = (-0.08)(0.07) - 0.07 = -0.0756 \\ b_2 &= a_4a_3 - a_0a_1 = (-0.08)(0.3) + 1.2 = 1.1760 \\ b_3 &= a_4^2 - a_0^2 = 0.0064 - 1 = -0.9936 \end{aligned}$$ $$\begin{aligned} c_0 &= b_3b_1 - b_0b_2 = (-0.9936)(-0.0756) - (-0.2040)(1.1760) = 0.3150 \\ c_1 &= b_3b_2 - b_0b_1 = (-0.9936)(1.1760) - (-0.2040)(-0.0756) = -1.1839 \\ c_2 &= b_3^2 - b_0^2 = 0.9872 - 0.0416 = 0.9456 \end{aligned}$$

Condition 4

• $|b_3| = 0.9936 > |b_0| = 0.2040$: satisfied
• $|c_2| = 0.9456 > |c_0| = 0.3150$: satisfied

Conclusion

All the Jury conditions are satisfied, so all the roots lie inside the unit circle and the system is stable.

Check by factorising: $P(z) = (z - 0.8)(z - 0.5)(z + 0.5)(z - 0.4)$. The roots are $0.8$, $0.5$, $-0.5$ and $0.4$, all with $|z| < 1$.

Answer: Stable. All conditions hold, with $|b_3| = 0.9936 > 0.2040$ and $|c_2| = 0.9456 > 0.3150$.