Jury test: z⁴−1.2z³+0.07z²+1.5z−0.06

We apply the Jury test to $P(z) = a_0z^4 + a_1z^3 + a_2z^2 + a_3z + a_4$, where $a_0 = 1$, $a_1 = -1.2$, $a_2 = 0.07$, $a_3 = 1.5$, $a_4 = -0.06$, and $n = 4$.

Coefficient conditions

1. $|a_4| < a_0$: $0.06 < 1$, satisfied.
2. $P(1) = 1 - 1.2 + 0.07 + 1.5 - 0.06 = 1.31 > 0$, satisfied.
3. $(-1)^4P(-1) = 1 + 1.2 + 0.07 - 1.5 - 0.06 = 0.71 > 0$, satisfied.

Jury table

$$b_k = \begin{vmatrix} a_4 & a_{3-k} \\ a_0 & a_{k+1} \end{vmatrix} = a_4a_{k+1} - a_0a_{3-k},\qquad c_k = \begin{vmatrix} b_3 & b_{2-k} \\ b_0 & b_{k+1} \end{vmatrix}$$ $$\begin{aligned} b_3 &= a_4^2 - a_0^2 = 0.0036 - 1 = -0.9964 \\ b_2 &= a_4a_3 - a_0a_1 = -0.09 + 1.2 = 1.11 \\ b_1 &= a_4a_2 - a_0a_2 = -0.0042 - 0.07 = -0.0742 \\ b_0 &= a_4a_1 - a_0a_3 = 0.072 - 1.5 = -1.428 \end{aligned}$$ $$\begin{aligned} c_2 &= b_3^2 - b_0^2 = 0.9928 - 2.0392 = -1.0464 \\ c_1 &= b_3b_2 - b_0b_1 = -1.1060 - 0.1060 = -1.2120 \\ c_0 &= b_3b_1 - b_0b_2 = 0.0739 + 1.5851 = 1.6590 \end{aligned}$$

Remaining conditions

4. $|b_3| > |b_0|$: $0.9964 > 1.428$ is false.
5. $|c_2| > |c_0|$: $1.0464 > 1.6590$ is false.

Conclusion

The first three conditions hold, but the table conditions fail. Hence the system is unstable.

Check by root finding: the roots are $z = 1.0084 \pm j0.8572$ (with $|z| = 1.3235$), $z = -0.8568$ and $z = 0.0400$. Two complex poles lie outside the unit circle. This also shows why checking only $P(1)$ and $P(-1)$ is not enough.