Jury test: z³+2.1z²+1.44z+0.32

We apply the Jury test to $P(z) = a_0z^3 + a_1z^2 + a_2z + a_3$, where $a_0 = 1$, $a_1 = 2.1$, $a_2 = 1.44$, $a_3 = 0.32$, and $n = 3$.

Conditions on the coefficients

1. $|a_3| < a_0$: $0.32 < 1$, satisfied.
2. $P(1) = 1 + 2.1 + 1.44 + 0.32 = 4.86 > 0$, satisfied.
3. $(-1)^3P(-1) > 0$: $P(-1) = -1 + 2.1 - 1.44 + 0.32 = -0.02$, so $(-1)^3P(-1) = 0.02 > 0$, satisfied.

Jury table

$$b_k = \begin{vmatrix} a_3 & a_{2-k} \\ a_0 & a_{k+1} \end{vmatrix}$$ $$\begin{aligned} b_2 &= a_3^2 - a_0^2 = 0.1024 - 1 = -0.8976 \\ b_1 &= a_3a_2 - a_0a_1 = 0.32(1.44) - 2.1 = -1.6392 \\ b_0 &= a_3a_1 - a_0a_2 = 0.32(2.1) - 1.44 = -0.768 \end{aligned}$$

4. $|b_2| > |b_0|$: $0.8976 > 0.768$, satisfied.

Conclusion

All the Jury conditions are satisfied, so the system is stable.

Check by factoring: $z = -0.5$ is a root, since $-0.125 + 0.525 - 0.72 + 0.32 = 0$. Then $P(z) = (z+0.5)(z^2 + 1.6z + 0.64) = (z+0.5)(z+0.8)^2$. The roots are $-0.5, -0.8, -0.8$, all inside $|z| = 1$. The margin is small: $(-1)^3P(-1)$ is only 0.02 because two roots sit at $-0.8$, close to $z = -1$.