Jury stability criterion

Remarks

The proof rests on the Schur–Cohn reduction. Write $P^*(z) = z^n P(1/z)$, so that $\abs{P^*(z)} = \abs{P(z)}$ on the unit circle because the coefficients are real. When $\abs{a_0} < a_n$, Rouché's theorem shows that $P$ is Schur stable if and only if the polynomial $(a_n P(z) - a_0 P^*(z))/z$ of degree $n - 1$ is Schur stable. The next row of the table lists the coefficients of this reduced polynomial in reverse order and with the opposite sign, so the table repeats the reduction until degree $2$. The conditions 1 and 2 are necessary, since a real polynomial with $a_n > 0$ and no roots in $[1, \infty)$ or $(-\infty, -1]$ has the stated signs at $\pm 1$. For $n = 2$ the criterion reads $\abs{a_0} < a_2$ and $\abs{a_1} < a_0 + a_2$. For instance $z^2 - z + \tfrac{1}{2}$ passes: $\tfrac{1}{2} < 1$ and $1 < \tfrac{3}{2}$. The polynomial $z^2 - \tfrac{5}{2} z + 1 = (z - 2)(z - \tfrac{1}{2})$ fails condition 1, since $P(1) = -\tfrac{1}{2}$.

Going deeper

For a student

The criterion is the discrete-time analogue of the Routh–Hurwitz stability criterion, which tests whether roots lie in the left half-plane. Combined with the stability theorem for linear recurrences, it decides from the coefficients whether every solution of a linear recurrence tends to zero. A degree $n$ polynomial needs $n + 1$ conditions in total: three simple ones and $n - 2$ from the table. When a parameter enters the coefficients, the conditions give the region of parameters where the system is stable.