Let
f(
z) be a complex polynomial. The process is as follows: • Compute the polynomials P_0(y) and P_1(y) such that f(iy)=P_0(y)+iP_1(y) where
y is a
real number. • Compute the
Sylvester matrix associated to P_0(y) and P_1(y). • Rearrange each row in such a way that an odd row and the following one have the same number of leading zeros. • Compute each
principal minor of that matrix. • If at least one of the minors is negative (or zero), then the polynomial
f is not stable.
Example Let f(z) = az^2 + bz + c (for the sake of simplicity we take real coefficients) where c\neq 0 (to avoid a root in zero so that we can use the Routh–Hurwitz theorem). First, we have to calculate the real polynomials P_0(y) and P_1(y): f(iy) = -ay^2 + iby + c = P_0(y) + iP_1(y) = -ay^2 + c + i(by). Next, we divide those polynomials to obtain the generalized
Sturm chain: \begin{align} P_0(y) &= \left( \tfrac{-a}{b} y \right)P_1(y)+c, &&\implies P_2(y) = -c, \\[4pt] P_1(y) &= \left( \tfrac{-b}{c} y \right)P_2(y), &&\implies P_3(y) = 0, \end{align} and the
Euclidean division stops. Notice that we had to suppose different from zero in the first division. The generalized Sturm chain is in this case \Bigl( P_0(y), P_1(y), P_2(y) \Bigr) = (c-ay^2, by ,-c). Putting y=+\infty, the sign of (c-ay^2) is the opposite sign of and the sign of is the sign of . When we put (y=-\infty), the sign of the first element of the chain is again the opposite sign of and the sign of is the opposite sign of . Finally, has always the opposite sign of . Suppose now that is Hurwitz-stable. This means that w(+\infty)-w(-\infty)=2 (the degree of ). By the properties of the function , this is the same as w(+\infty)=2 and w(-\infty)=0. Thus, and must have the same sign. We have thus found the
necessary condition of stability for polynomials of degree 2.
Routh–Hurwitz criterion for second, third and fourth-order polynomials For a second-order polynomial P(s) = a_2s^2 + a_1s + a_0 = 0, all coefficients must be positive: a_i > 0 \quad \text{for} \quad i = 0, 1, 2. For a third-order polynomial P(s) = a_3s^3 + a_2s^2 + a_1s + a_0 = 0, all coefficients must be positive: a_i > 0 \quad \text{for} \quad i = 0, 1, 2, 3 and also we need a_2a_1 - a_3a_0 > 0. For a fourth-order polynomial P(s) = a_4s^4 + a_3s^3 + a_2s^2 + a_1s + a_0 = 0, all coefficients must be positive: a_i > 0 \quad \text{for} \quad i = 0, 1, 2, 3, 4 and also we need \begin{align} a_3a_2 - a_4a_1 & > 0\\ a_3a_2a_1 - a_4a_1^2 - a_3^2a_0 & > 0. \end{align} (When this is derived you do not know all coefficients should be positive, and you add a_3a_2 > a_1.) In general the Routh stability criterion states a polynomial has all roots in the open left half-plane if and only if all first-column elements of the Routh array have the same sign. All coefficients being positive (or all negative) is necessary for all roots to be located in the open left half-plane. That is why here a_n is fixed to 1, which is positive. When this is assumed, we can remove a_3a_2 > a_1 from fourth-order polynomial, and conditions for fifth- and sixth-order can be simplified. For fifth-order we only need to check that \Delta_2>0, \Delta_4>0 and for sixth-order we only need to check \Delta_3>0, \Delta_5>0 and this is further optimised in
Liénard–Chipart criterion. Indeed, some coefficients being positive is not independent with principal minors being positive, like a_2 > 0 check can be removed for third-order polynomial.
Higher-order example A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain. For an th-degree polynomial whose all coefficients are the same signs D(s) = a_ns^n + a_{n-1}s^{n-1} + \cdots + a_1s + a_0 the table has rows and the following structure: \begin{matrix} a_n & a_{n-2} & a_{n-4} & \dots \\ a_{n-1} & a_{n-3} & a_{n-5} & \dots \\ b_1 & b_2 & b_3 & \dots \\ c_1 & c_2 & c_3 & \dots \\ \vdots & \vdots & \vdots & \ddots \end{matrix} where the elements b_i and c_i can be computed as follows: \begin{align} b_i &= \frac{a_{n-1} \times a_{n-2i} - a_n \times a_{n-(2i+1)}}{a_{n-1}} \\[4pt] c_i &= \frac{b_1 \times a_{n-(2i+1)} - a_{n-1} \times b_{i+1}}{b_1} \end{align} When completed, the number of sign changes in the first column will be the number of roots whose real part are non-negative. \begin{matrix} 0.75 & 1.5 & \ 0 \ & \ 0 \ \\ -3 & 6 & 0 & 0 \\ 3 & 0 & 0 & 0 \\ 6 & 0 & 0 & 0 \end{matrix} In the first column, there are two sign changes (, and ), thus there are two roots whose real part are non-negative and the system is unstable. The characteristic equation of an example
servo system is given by: b_0s^4 + b_1s^3 + b_2s^2 + b_3s + b_4 = 0 For which we have the following table: \begin{matrix} b_ 0 & b_2 & \quad b_4 \quad & \quad 0 \quad \\[4pt] b_1 & b_3 & 0 & 0 \\[4pt] \frac{b_1b_2 - b_0b_3}{b_1} & \frac{b_1b_4-b_ 0 \times 0}{b_1} = b_4 & 0 & 0 \\ \frac{(b_1b_2 - b_0b_3)b_3 - b_1^2b_4}{b_1b_2 - b_0b_3} & 0 & 0 & 0 \\[4pt] b_4 & 0 & 0 & 0 \end{matrix} for stability, all the elements in the first column of the Routh array must be positive when b_0 > 0. And the conditions that must be satisfied for stability of the given system as follows: s^4+6s^3+11s^2+6s+200=0 We have the following table : \begin{matrix} 1 & 11 & 200 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 20 & 0 & 0 \\ -19 & 0 & 0 & 0 \\ 20 & 0 & 0 & 0 \end{matrix} there are two sign changes. The system is unstable, since it has two right-half-plane poles and two left-half-plane poles. The system cannot have jω poles since a row of zeros did not appear in the Routh table. ==See also==