Jury test: 2z⁴+7z³+10z²+4z+1

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

$P(z) = 2z^4 + 7z^3 + 10z^2 + 4z + 1$, so $n = 4$, $a_0 = 2$, $a_1 = 7$, $a_2 = 10$, $a_3 = 4$, $a_4 = 1$.

Conditions 1 to 3

• $|a_4| = 1 < a_0 = 2$: satisfied
• $P(1) = 2 + 7 + 10 + 4 + 1 = 24 > 0$: satisfied
• $(-1)^4P(-1) = 2 - 7 + 10 - 4 + 1 = 2 > 0$: satisfied

Jury table

$$\begin{aligned} b_0 &= a_4a_1 - a_0a_3 = 1(7) - 2(4) = -1 \\ b_1 &= a_4a_2 - a_0a_2 = 1(10) - 2(10) = -10 \\ b_2 &= a_4a_3 - a_0a_1 = 1(4) - 2(7) = -10 \\ b_3 &= a_4a_4 - a_0a_0 = 1 - 4 = -3 \end{aligned}$$ $$\begin{aligned} c_0 &= b_3b_1 - b_0b_2 = (-3)(-10) - (-1)(-10) = 20 \\ c_1 &= b_3b_2 - b_0b_1 = (-3)(-10) - (-1)(-10) = 20 \\ c_2 &= b_3b_3 - b_0b_0 = 9 - 1 = 8 \end{aligned}$$

Condition 4

• $|b_3| = 3 > |b_0| = 1$: satisfied
• $|c_2| = 8 > |c_0| = 20$: not satisfied

Conclusion

One Jury condition fails, so the system is unstable.

Check: the roots of $P(z)$ are $z = -1.526 \pm j1.078$ (magnitude 1.868, outside the unit circle) and $z = -0.224 \pm j0.305$ (magnitude 0.379, inside). Two roots lie outside the unit circle, which confirms the result.

Answer: Unstable, because $|c_2| = 8 < |c_0| = 20$.