Interval Polynomials and Robust Performance

Abstract

In the previous chapters, control systems with discretized nonlinear characteristics were considered. However, the problems of these nonlinear control systems are generally difficult to analyze. In this chapter, the case where a nonlinear characteristic can be treated as a set of linear characteristics (in other words, interval parameters) is considered. This concept is based on the validity of Aizerman’s conjecture. It is assumed that the transfer function (and the characteristic polynomial) is expressed by interval parameters. The control performance of these interval systems is discussed from the viewpoint of the roots area of a characteristic polynomial. The robust stability condition for discrete interval systems is derived using Sturm’s theorem (the division algorithm), which has been applied to the Routh-Hurwitz criterion.

Appendices

Appendix A: Kharitonov Rectangles

With respect to continuous-time systems, consider the following characteristic polynomials [7]:

$$\begin{aligned} &\tilde{F}(s)=\tilde{a}_0s^n+\tilde{a}_1s^{n-1}+ \cdots +\tilde{a}_{n-1}s+\tilde{a}_n, \\ &~~~~\tilde{a}_k \in [a_k^-, a_k^+],~~k=0,1,2, \cdots ,n. \end{aligned}$$

(6.64)

Considering the imaginary axis, i.e., s=jω, Eq. (6.64) can be given as

$$ \tilde{F}(j\omega) =[P^-(\omega),P^+(\omega)]+j[Q^-(\omega),Q^+(\omega)]. $$

(6.65)

When n is odd,

$$\begin{aligned} &P^-(\omega)=a_n^--a_{n-2}^+\omega^2+\cdots+(-1)^{(n-1)/2}a_1^+\omega^{n-1}\\ &P^+(\omega)=a_n^+-a_{n-2}^-\omega^2+\cdots+(-1)^{(n-1)/2}a_1^-\omega^{n-1}\\ &Q^-(\omega)=a_{n-1}^-\omega-a_{n-3}^+\omega^3+\cdots-(-1)^{(n-1)/2}a_0^+\omega^n\\ &Q^+(\omega)=a_{n-1}^+\omega-a_{n-3}^-\omega^3+\cdots-(-1)^{(n-1)/2}a_0^-\omega^n. \end{aligned}$$

On the other hand, when n is even,

$$\begin{aligned} &P^-(\omega)=a_n^--a_{n-2}^+\omega^2+\cdots-(-1)^{n/2}a_0^+\omega^n\\ &P^+(\omega)=a_n^+-a_{n-2}^-\omega^2+\cdots-(-1)^{n/2}a_0^-\omega^n\\ &Q^-(\omega)=a_{n-1}^-\omega-a_{n-3}^+\omega^3+\cdots+(-1)^{n/2}a_1^+\omega^{n-1}\\ &Q^+(\omega)=a_{n-1}^+\omega-a_{n-3}^-\omega^3+\cdots+(-1)^{n/2}a_1^-\omega^{n-1}. \end{aligned}$$

For a third-order system, (6.65) can be written directly as follows:

$$\tilde{F}(s)=[a_0^-,a_0^+]s^3+[a_1^-,a_1^+]s^2+[a_2^-,a_2^+]s +[a_3^-,a_3^+]. $$

Therefore,

$$ \tilde{F}(j\omega) =[a_3^-,a_3^+]-[a_1^-,a_1^+]\omega^2 +j\{[a_2^-,a_2^+]\omega-[a_0^-,a_0^+]\omega^3\}. $$

(6.66)

When considering ω>0, the following vertices are obtained:

$$\begin{aligned} &P^-(\omega)=a_3^--a_1^+\omega^2\\ &P^+(\omega)=a_3^+-a_1^-\omega^2\\ &Q^-(\omega)=a_2^-\omega-a_0^+\omega^3\\ &Q^+(\omega)=a_2^+\omega-a_0^-\omega^3. \end{aligned}$$

Of course, when ω<0, the following vertices are obtained:

$$\begin{aligned} &P^-(\omega)=a_3^--a_1^+\omega^2\\ &P^+(\omega)=a_3^+-a_1^-\omega^2\\ &Q^-(\omega)=a_2^+\omega-a_0^-\omega^3\\ &Q^+(\omega)=a_2^-\omega-a_0^+\omega^3 \end{aligned}$$

although they are conjugated with each other.

Example 6.5

Consider the following (monic) interval polynomial (the highest coefficient a ₀=1):

$$\tilde{F}(s) =s^3+[1.6,2.4]s^2+[1.6,2.4]s+[0.8,1.2]. $$

Obviously, when ω>0,

$$\begin{aligned} &P^-(\omega)=0.8-2.4\omega^2\\ &P^+(\omega)=1.2-1.6\omega^2\\ &Q^-(\omega)=1.6\omega-\omega^3\\ &Q^+(\omega)=2.4\omega-\omega^3. \end{aligned}$$

And, when ω<0,

$$\begin{aligned} &P^-(\omega)=0.8-2.4\omega^2\\ &P^+(\omega)=1.2-1.6\omega^2\\ &Q^-(\omega)=2.4\omega-\omega^3\\ &Q^+(\omega)=1.6\omega-\omega^3. \end{aligned}$$

Figure 6.15(a) shows the trace of Kharitonov rectangles when ω: −ω _c →ω _c . From the figure, one can see that the stability of the interval system is determined by only the four vertices of the rectangular array. The robust stability condition is necessary and sufficient for interval (uncertain) systems. In this example, a view of the root areas is shown in Fig. 6.15(b).

Fig. 6.15

Kharitonov rectangles and roots area

Appendix B: Roots of Polynomials and Sturm’s Theorem

The basic concept of Sturm’s theorem is given as follows [6, 13]. Let f(x) be a “real” polynomial.⁴ Denote it by f ₀(x) and its derivative f′(x) by f ₁(x). Applying Euclid’s division algorithm, the following sequence is obtained:

$$\begin{aligned} &f_0(x)=q_1(x)f_1(x)-f_2(x), \\ &f_1(x)=q_2(x)f_2(x)-f_3(x), \\ &~~\vdots \\ &f_{n-2}(x)=q_{n-1}(x)f_{n-1}(x)-f_n(x), \\ &f_{n-1}(x)= q_n(x)f_n, \end{aligned}$$

(6.67)

where f _i (x) is of degree lower than that of f _i−1(x) for 1≤i≤n. The signs of the remainders are negated from those in the algorithm. Note that the divisor f _k that yields a zero remainder is the greatest common divisor of f(x) and f′(x). The sequence f ₀,f ₁,⋯,f _n is called a Sturm sequence for the polynomial f.

Sturm’s Theorem

The number of distinct (simple) zeros of a polynomial f(x) in x∈(x ₁,x ₂) is equal to the excess of the number of changes of sign in the sequence f ₀(x ₁), ⋯, f _n−1(x ₁), f _n (x ₁) over the number of changes of sign in the sequence f ₀(x ₂), ⋯, f _n−1(x ₂), f _n (x ₂).

In other expressions, if the number of sign changes that cross the zero of f ₀(x)/f ₁(x) for x:x ₁→x ₂ is defined as N(x ₁,x ₂) and the number of sign changes of sequence f ₀(x ₂), ⋯, f _n−1(x ₂), f _n (x ₂) is defined as V(x), we obtain the relationship

$$ N(x_1,x_2)=V(x_1)-V(x_2). $$

(6.68)

Here, f(x) can be multiplied by a positive constant or a factor involving x provided that the factor remains positive in x∈(x ₁,x ₂). Moreover, there is no common divisor between f ₀(x) and f ₁(x), because the zeros of f(x) are distinct.

Example 6.6

(Real Roots)

Consider the following polynomial to apply Sturm’s theorem [16]:

$$f(x)=x^3-1.5x^2-0.16x-0.24. $$

The first step polynomials f ₀ and f ₁ are given as:

$$\begin{aligned} &f_0(x)=x^3-1.5x^2-0.16x-0.24,\\ &f_1(x)=3x^2+3x-0.16. \end{aligned}$$

Therefore, the results of the division algorithm are obtained as follows:

$$\begin{aligned} &f_2(x)=0.607x+0.213,\\ &f_3(x)=0.844. \end{aligned}$$

Table 6.1 lists the signs of each Sturm sequence, and the number of sign changes is given in Table 6.2. From these tables, we obtain, for example,

$$\begin{aligned} &N(-2,2)=V(-2)-V(2)=3,\\ &N(-1,2)=V(-1)-V(1)=2,\\ &N(0,1)=V(0)-V(1)=1. \end{aligned}$$

Table 6.1 Sign changes of sequence f _i (i=1,2,3,4)

Table 6.2 The number of sign changes N(x ₁,x ₂)

This result is clear from Fig. 6.16.

Fig. 6.16

Cubic curve and Sturm sequences (f(x)=x ³+1.5x ²−0.16x−0.24)

Complex Roots Problem

The Sturm sequence can be extended to a complex roots problem. Consider the following circle as a closed contour in the complex plane:

$$ z=\varrho\mathrm{e}^{j\phi}+x_0+jy_0,~~~(\phi:-\pi\to\pi), $$

(6.69)

where ϱ, (x ₀,y ₀), and ϕ are the radius, the center, and the angle of rotation of a specified circle, respectively. The problem of a sectorial area in the complex plane can be considered in a large circle ∂Γ, as shown in Fig. 6.3(a).⁵

Figures 6.17(a), (b) show examples of the circular contour in a complex plane. The circular contour (6.36) can also be written as the following rational function:

$$ z=\varrho\cdot\frac{1+jv}{1-jv}+x_0+jy_0 =\frac{A+Bv}{1-jv}, $$

(6.70)

where

$$v=\tan\left(\frac{\phi}{~2~}\right) $$

and A and B are complex constants written as A=ϱ+x ₀+jy ₀ and B=y ₀+j(ϱ−x ₀). Clearly, the relationship between variables ϕ and v is

$$\begin{aligned} &\phi=-\pi;~~v=-\infty,\\ &\phi=0;~~~~~v=0,\\ &\phi=+\pi;~~v=+\infty. \end{aligned}$$

Fig. 6.17

Circular contours (f(z)=z ²+z+1, x ₀=−0.5, (a) y ₀=0.1, ϱ=1.5, (b) y ₀=0.866, ϱ=0.3)

Consider the following characteristic polynomial in general:

$$ F(z)=a_0z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_n. $$

(6.71)

If (6.69) is substituted into (6.71), a numerator polynomial with complex coefficients is obtained:

$$ \Phi(jv)=(1-jv)^n F(jv) =P(v)+jQ(v). $$

(6.72)

Here, polynomials P(v) and Q(v) can be written as

$$\begin{aligned} &P(v)=a_{0,0}v^n+\cdots +a_{0,n-1}v+a_{0,n}, \\ &Q(v)=b_{0,0}v^n+\cdots +b_{0,n-1}v+b_{0,n}. \end{aligned}$$

(6.73)

Therefore, argument change 2μπ for polynomial F(z) becomes (2μ−n)π for Φ(jv) by adding change −nπ in the argument of (1−jv)ⁿ. When P/Q (or −Q/P) is considered, the number of sign changes that cross zero for v:−∞→+∞ is n−2μ.

The coefficients in (6.73) are calculated by

$$ \Phi(jv) =\sum_{k=0}^n a_k(A+Bv)^{n-k}(1-jv)^k $$

(6.74)

from (6.69). This equation is expanded as

and thus we can obtain

where

denotes a combination symbol.

By setting f ₀(v)=P(v) and f ₁(v)=Q(v), the following division algorithm can be executed:

$$\begin{aligned} &f_{2\kappa-2}(v)=q_{2\kappa-1}(v)f_{2\kappa-1}(v)-f_{2\kappa}(v), \\ &~~~~~~~~\kappa=1,2, \cdots ,n. \end{aligned}$$

(6.75)

If f ₀(v) and f ₁(v) are of the n-th order for v, then f ₂(v),f ₃(v),⋯,f _2n are expressed as:

$$\begin{aligned} &f_2(v)=a_{1,1}v^{n-1}+ \cdots +a_{1,n} \\ &f_3(v)=b_{1,1}v^{n-1}+ \cdots +b_{1,n} \\ &~~~~~~\cdots \\ &f_{2n-2}(v)=a_{n-1,n-1}v+a_{n-1,n} \\ &f_{2n-1}(v)=b_{n-1,n-1}v+b_{n-1,n} \\ &f_{2n}=a_{n,n}. \end{aligned}$$

(6.76)

Here, each coefficient can be given by the following sequential operations:

$$\begin{aligned} &a_{1,p}=b_{0,p} \left( \frac{a_{0,0}}{b_{0,0}} \right)-a_{0,p}, \\ &b_{1,p}=a_{1,p+1} \left( \frac{b_{0,0}}{a_{1,1}} \right)-b_{0,p}, \\ &~~~~~(p=1,2, \cdots ,n) \\ &~~~~~\cdots \\ &a_{q,p} =b_{q-1,p} \left( \frac{a_{q-1,q-1}}{b_{q-1,q-1}} \right)-a_{q-1,p}, \\ &b_{q,p} =a_{q,p+1} \left( \frac{b_{q-1,q-1}}{a_{q,q}} \right)-b_{q-1,p}, \\ &~~~~~(p=q, \cdots ,n) \\ &~~~~~\cdots \\ &a_{n,n} =b_{n-1,n} \left( \frac{a_{n-1,n-1}}{b_{n-1,n-1}} \right)-a_{n-1,n}, \\ &~~~~~(a_{q,n+1}=0). \end{aligned}$$

(6.77)

If the number of sign changes which cross the zero of f ₀(v)/f ₁(v) for v:v ₁→v ₂ is expressed as N(v ₁,v ₂) and the number of sign changes of sequence f ₀(v),f ₁(v),⋯,f _2n is expressed as V(v), the following relationship is obtained:

$$ N(v_1,v_2)=V(v_1)-V(v_2). $$

(6.78)

Since the condition is N(−∞,+∞)=n−2μ,

$$ V(-\infty)-V(+\infty)=n-2\mu $$

(6.79)

is obtainable.

On the other hand, the following expressions are derived from (6.76):

$$\begin{aligned} &\lim_{v\to -\infty}\frac{f_0(v)}{f_1(v)} =\frac{a_{0,0}}{b_{0,0}}, \\ &\lim_{v\to -\infty}\frac{f_1(v)}{|v|f_2(v)} =-\frac{b_{0,0}}{a_{1,1}}, \\ &~~~~~~~\vdots \\ &\lim_{v\to -\infty}\frac{f_{2n-1}(v)}{|v|f_{2n}(v)} =-\frac{b_{n-1,n-1}}{a_{n,n}}, \\ &\lim_{v\to +\infty}\frac{f_0(v)}{f_1(v)} =\frac{a_{0,0}}{b_{0,0}}, \\ &\lim_{v\to +\infty}\frac{f_1(v)}{|v|f_2(v)} =\frac{b_{0,0}}{a_{1,1}}, \\ &~~~~~~~\vdots \\ &\lim_{v\to +\infty}\frac{f_{2n-1}(v)}{|v|f_{2n}(v)} =\frac{b_{n-1,n-1}}{a_{n,n}}, \end{aligned}$$

(6.80)

The condition of (6.78) corresponds to observing whether the following ratios (ratios to the polynomial of different orders) are negative or not:

$$\begin{aligned} &\lim_{v \rightarrow +\infty} \frac{f_1(v)}{|v|f_2(v)} =\frac{b_{0,0}}{a_{1,1}}, \\ &\lim_{v \rightarrow +\infty} \frac{f_3(v)}{|v|f_4(v)} =\frac{b_{1,1}}{a_{2,2}}, \\ &~~~~~~~\vdots \\ &\lim_{v \rightarrow +\infty}\frac{f_{2n-1}(v)}{|v|f_{2n}(v)}=\frac{b_{n-1,n-1}}{a_{n,n}}, \end{aligned}$$

(6.81)

because ratios to the polynomial of same order (i.e., $\frac{a_{0,0}}{b_{0,0}}, \frac{a_{1,1}}{b_{1,1}},\ldots$) are canceled based on (6.79). Suppose that the number of negative ratios is N and the number of positive ratios is P. Thus, P−N=n−2μ can be obtained from (6.79). Since P+N=n, N=μ is given.

Example 6.7

(Quadratic Equation)

As a simple example, consider the following quadratic polynomial:

$$ F(z)=z^2+z+1. $$

(6.82)

First, a specified circle is assumed to be x ₀=−0.5, y ₀=0.1, and ϱ=1.5, as shown in Fig. 6.17(a). Then, numerator polynomials P(v) and Q(v) written as

$$\Phi(jv)=(1-jv)^2F(jv)=P(v)+jQ(v) $$

are given as

$$\begin{aligned} &P(v)=f_0(v)=-2.99v^2+2.99,\\ &Q(v)=f_1(v)=0.3v^2+3.02v-0.3. \end{aligned}$$

By setting f ₀(v)=P(v), f ₁(v)=Q(v), the following results are obtained from (6.76):

$$\begin{aligned} &f_0(v)=-2.99v^2+2.99, \\ &f_1(v)=0.30v^2+3.02v-0.30, \\ &f_2(v)=-30.1v-5.98, \\ &f_3(v)=-2.96v-0.30, \\ &f_4=2.93. \end{aligned}$$

(6.83)

Since the sequence of fractions, (6.81), is given as

$$\frac{b_{0,0}}{a_{1,1}}=\frac{0.30}{-30.1}<0, ~~~~~~\frac{b_{1,1}}{a_{2,2}}=\frac{-2.96}{2.93}<0, $$

the number of roots in the specified circle is μ=2.

The discriminating procedure given above can be interpreted as follows. The specified contour encircles two roots of the quadratic equation. As a result, the mapping curve is as shown in Fig. 6.18(a). Thus, the argument change of Φ(jv) becomes 4π−2π=2π by adding a change of −2π to the argument of (1−jv)². Figure 6.18(b) shows the Φ curve for this case.

Fig. 6.18

Mapping f(jv) and Φ(jv) for the specified circle

On the other hand, when a circle is chosen as x ₀=−0.5, y ₀=0.8, and ϱ=0.4, as shown in Fig. 6.17(b), numerator polynomials P(v) and Q(v) are given by

$$\begin{aligned} &P(v)=f_0(v)=-0.27v^2+0.27,\\ &Q(v)=f_1(v)=0.64v^2+0.10v+0.64. \end{aligned}$$

The Sturm sequence is given as

$$\begin{aligned} &f_0(v)=-0.27v^2+0.27, \\ &f_1(v)=0.64v^2+0.10v-0.30, \\ &f_2(v)=-0.04v-0.54, \\ &f_3(v)=8.09v-0.64, \\ &f_4=0.54. \end{aligned}$$

(6.84)

Since the sequence of fractions, (6.81), is given as

$$\frac{b_{0,0}}{a_{1,1}}=\frac{0.64}{-0.04}<0, ~~~~~~\frac{b_{1,1}}{a_{2,2}}=\frac{8.09}{0.54}>0, $$

the number of roots in the specified circle is μ=1.

Example 6.8

(Cubic Equation)

Consider the following cubic polynomial:

$$ F(z)=z^3+2z^2+2z+1. $$

(6.85)

A specified circle is chosen as x ₀=−0.5, y ₀=0.1, and ϱ=1.5. Then, numerator polynomials P(v) and Q(v) written as

$$\Phi(jv)=(1-jv)^3F(jv)=P(v)+jQ(v) $$

are given as

$$\begin{aligned} &P(v)=-0.599v^3-9.03v^2-0.303v+5.95,\\ &Q(v)=-2.96v^3+0.603v^2+9.06v+0.899. \end{aligned}$$

By setting f ₀(v)=P(v), f ₁(v)=Q(v), the following results are obtained from (6.76):

$$\begin{aligned} &f_0(v)=-0.599v^3-9.03v^2-0.303v+5.95, \\ &f_1(v)=-2.96v^3+0.603v^2+9.06v+0.899, \\ &f_2(v)=9.152v^2+2.1364v-5,7681, \end{aligned}$$

(6.86)

$$\begin{aligned} &f_3(v)=-1.294v^2-7.1945v-0.899, \\ &f_4(v)=48.749v+12.127, \\ &f_5(v)=6.8726v+0.899, \\ &f_6=-5.7497. \end{aligned}$$

(6.87)

Since the sequence of fractions, (6.81), is given as

$$\frac{b_{0,0}}{a_{1,1}}=\frac{-2.96}{9.152}<0, ~~~~\frac{b_{1,1}}{a_{2,2}}=\frac{-1.294}{48.749}<0, ~~~~\frac{b_{2,2}}{a_{3,3}}=\frac{6.8726}{-5.7497}<0, $$

the number of roots in the specified circle is μ=3.

Routh-Hurwitz Criterion

Finally, the relationship between the above results based on Sturm’s theorem and the classical Routh-Hurwitz criterion is presented. In general, a characteristic equation for continuous-time systems is written as

$$ F(s)=a_0s^n+a_1s^{n-1}+\cdots+a_{n-1}s+a_n=0, $$

(6.88)

where s is the Laplace transform variable and a _i (i=1,2,⋯,n) are real coefficients. The stability of a control system having characteristic equation (6.88) is discriminated by a contour on the imaginary axis and a large half-contour in the right half-plane (RHP) or left half-plane (LHP) and its mapping in the F-plane.

On the imaginary axis s=jω (ω:−∞→∞),

$$ F(j\omega)=P(\omega)+jQ(\omega). $$

(6.89)

The real and imaginary parts of (6.89) are given as follows.

• (1) When n=1,5,9,⋯,

  $$\begin{aligned} &P^{(1)}(\omega)=a_1\omega^{n-1}-a_3\omega^{n-3}+a_5\omega^{n-5}-\cdots,\\ &Q^{(1)}(\omega)=a_0\omega^{n}-a_2\omega^{n-2}+a_4\omega^{n-4}-\cdots. \end{aligned}$$

• (2) When n=2,6,10,⋯,

  $$\begin{aligned} &P^{(2)}(\omega)=-a_0\omega^{n}+a_2\omega^{n-2}-a_4\omega^{n-4}+\cdots,\\ &Q^{(2)}(\omega)=a_1\omega^{n-1}-a_3\omega^{n-3}+a_5\omega^{n-5}-\cdots. \end{aligned}$$

• (3) When n=3,7,11,⋯,

  $$\begin{aligned} &P^{(3)}(\omega)=-a_1\omega^{n-1}+a_3\omega^{n-3}-a_5\omega^{n-5}+\cdots,\\ &Q^{(3)}(\omega)=-a_0\omega^{n}+a_2\omega^{n-2}-a_4\omega^{n-4}+\cdots. \end{aligned}$$

• (4) When n=4,8,12,⋯,

  $$\begin{aligned} &P^{(4)}(\omega)=a_0\omega^{n}-a_2\omega^{n-2}+a_4\omega^{n-4}-\cdots,\\ &Q^{(4)}(\omega)=-a_1\omega^{n-1}+a_3\omega^{n-3}-a_5\omega^{n-5}+\cdots. \end{aligned}$$

If the following polynomials are considered:

$$\begin{aligned} &f_0(\omega)=a_0\omega^n-a_2\omega^{n-2}+a_4\omega^{n-4}-\cdots \\ &f_1(\omega)=a_1\omega^{n-1}-a_3\omega^{n-3}+a_5\omega^{n-5}-\cdots, \end{aligned}$$

(6.90)

the four cases of polynomials obviously become

$$\begin{aligned} &P^{(1)}(\omega)=f_1(\omega),~~~Q^{(1)}(\omega)=f_0(\omega)\\ &P^{(2)}(\omega)=-f_0(\omega),~~~Q^{(2)}(\omega)=f_1(\omega)\\ &P^{(3)}(\omega)=-f_1(\omega),~~~Q^{(3)}(\omega)=-f_0(\omega)\\ &P^{(4)}(\omega)=f_0(\omega),~~~Q^{(4)}(\omega)=-f_1(\omega). \end{aligned}$$

Based on the above premise, the following division algorithm is executed:

$$\begin{aligned} &f_{k-1}(\omega)=q_k(\omega)f_k(\omega)-f_{k+1}(\omega), \\ &~~~~~~~~k=1,2, \cdots ,n. \end{aligned}$$

(6.91)

Then, f ₂(ω),f ₃(ω),⋯,f _n are obtained as follows:

$$\begin{aligned} &f_0(\omega)=a_{0,0}\omega^n+a_{0,2}\omega^{n-2} +a_{0,4}\omega^{n-4}+\cdots \\ &f_1(\omega)=a_{1,1}\omega^{n-1}+a_{1,3}\omega^{n-3} +a_{1,5}\omega^{n-5}+\cdots \\ &f_2(\omega)=a_{2,2}\omega^{n-2}+a_{2,4}\omega^{n-4}+\cdots \\ &f_3(\omega)=a_{3,3}\omega^{n-3}+a_{3,5}\omega^{n-5}+\cdots \\ &~~~~~~\cdots \\ &f_{n}=a_{n,n}. \end{aligned}$$

(6.92)

For an easier understanding, polynomials f ₀(ω) and f ₁(ω) are also written in (6.92). Here, a _0,0=a ₀, a _0,2=−a ₂, a _0,4=a ₄, ⋯ and a _1,1=a ₁, a _1,3=−a ₃, a _1,5=a ₅, ⋯. Moreover, each coefficient can be given by the following sequential operations:

$$\begin{aligned} &a_{2,p}=a_{1,p+1}\left(\frac{a_{0,0}}{a_{1,1}}\right)-a_{0,p} \\ &(p=2,4,\cdots) \\ &~~~~~~\cdots \\ &a_{q,p} =a_{q-1,p+1}\left(\frac{a_{q-2,q-2}}{a_{q-1,q-1}}\right)-a_{q-2,p} \\ &(q=3,4,\cdots,~~p=q,q+2,\cdots) \\ &~~~~~~\cdots \\ &a_{n,n} =a_{n-1,n+1}\left(\frac{a_{n-2,n-2}}{a_{n-1,n-1}}\right)-a_{n-2,n}. \end{aligned}$$

(6.93)

In regard to the four cases, the following Sturm sequences are obtained.

• (1) In this case, f ₀(ω)/f ₁(ω)=Q ⁽¹⁾(ω)/P ⁽¹⁾(ω) should be calculated (when considering a large contour in the RHP). Since a _0,0=a ₀, a _0,2=−a ₂, a _0,4=a ₄,⋯, a _1,1=a ₁, a _1,3=−a ₃, a _1,5=a ₅,⋯, the following coefficients (Routh’s series) are obtained:

  $$\begin{aligned} &a_{2,2}=-a_3\left(\frac{a_0}{a_1}\right)+a_2,~~~ a_{2,4}=a_5\left(\frac{a_0}{a_1}\right)-a_4,~~~ ~~\cdots\\ &a_{3,3}=a_{2,4}\left(\frac{a_1}{a_{2,2}}\right)+a_3,~~~ a_{3,5}=a_{2,6}\left(\frac{a_1}{a_{2,2}}\right)-a_5,~~~\cdots\\ &a_{4,4}=a_{3,5}\left(\frac{a_{2,2}}{a_{3,3}}\right)-a_{2,4}\\ &~~~~~\cdots. \end{aligned}$$

• (2) In this case, f ₀(ω)/f ₁(ω)=−P ⁽²⁾(ω)/Q ⁽²⁾(ω) is assumed to be calculated (when considering a large contour in the RHP). Since a _0,0=−a ₀, a _0,2=a ₂, a _0,4=−a ₄,⋯, a _1,1=a ₁, a _1,3=−a ₃, a _1,5=a ₅,⋯, the following series are obtained:

  $$\begin{aligned} &a_{2,2}=-a_3\left(\frac{a_0}{a_1}\right)+a_2,~~~~ a_{2,4}=a_5\left(\frac{a_0}{a_1}\right)-a_4,~~~ ~~\cdots\\ &a_{3,3}=a_{2,4}\left(\frac{a_1}{a_{2,2}}\right)+a_3,~~~ a_{3,5}=a_{2,6}\left(\frac{a_1}{a_{2,2}}\right)-a_5,~~~\cdots\\ &a_{4,4}=a_{3,5}\left(\frac{a_{2,2}}{a_{3,3}}\right)-a_{2,4}\\ &~~~~~\cdots. \end{aligned}$$

  In either case (1) or (2), the top of Routh’s series can be written as

  $$\begin{aligned} &a_{1,1}=a_1 \\ &a_{2,2}= \begin{vmatrix} a_1 & a_3\\ a_0 & a_2 \end{vmatrix} \diagup a_1 \\ &a_{3,3}= \begin{vmatrix} a_1 & a_3 & a_5\\ a_0 & a_2 & a_4\\ 0 & a_1 & a_3 \end{vmatrix} \diagup \begin{vmatrix} a_1 & a_3\\ a_0 & a_2 \end{vmatrix} \\ &a_{4,4}= \begin{vmatrix} a_1 & a_3 & a_5 & a_7\\ a_0 & a_2 & a_4 & a_6\\ 0 & a_1 & a_3 & a_5\\ 0 & a_0 & a_2 & a_4 \end{vmatrix} \diagup \begin{vmatrix} a_1 & a_3 & a_5\\ a_0 & a_2 & a_4\\ 0 & a_1 & a_3 \end{vmatrix} \\ &~~~\cdots \\ &a_{n,n}=a_n,~~~~(\mathrm{because}~~a_m=0~~\mathrm{for}~~m>n). \end{aligned}$$

  (6.94)

  Each term of (6.94) corresponds to a Hurwitz determinant and its principal minors.

  However, in regard to the remaining two cases, the following Sturm’s sequences are calculated.

• (3) In this case, if f ₀(ω)/f ₁(ω)=Q ⁽³⁾(ω)/P ⁽³⁾(ω) is calculated as above for (1), the following series are obtained:

  $$\begin{aligned} &a_{2,2}=a_3\left(\frac{a_0}{a_1}\right)-a_2,~~~~ a_{2,4}=-a_5\left(\frac{a_0}{a_1}\right)+a_4,~~~ ~~\cdots\\ &a_{3,3}=-a_{2,4}\left(\frac{a_1}{a_{2,2}}\right)-a_3,~~~ a_{3,5}=-a_{2,6}\left(\frac{a_1}{a_{2,2}}\right)+a_5,~~~\cdots\\ &a_{4,4}=a_{3,5}\left(\frac{a_{2,2}}{a_{3,3}}\right)-a_{2,4}\\ &~~~~~\cdots. \end{aligned}$$

• (4) In this case, if f ₀(ω)/f ₁(ω)=−P ⁽⁴⁾(ω)/Q ⁽⁴⁾(ω) is calculated as above for (2), the following series are obtained:

  $$\begin{aligned} &a_{2,2}=a_3\left(\frac{a_0}{a_1}\right)-a_2,~~~~ a_{2,4}=-a_5\left(\frac{a_0}{a_1}\right)+a_4,~~~ ~~\cdots\\ &a_{3,3}=-a_{2,4}\left(\frac{a_1}{a_{2,2}}\right)-a_3,~~~ a_{3,5}=-a_{2,6}\left(\frac{a_1}{a_{2,2}}\right)+a_5,~~~\cdots\\ &a_{4,4}=a_{3,5}\left(\frac{a_{2,2}}{a_{3,3}}\right)-a_{2,4}\\ &~~~~~\cdots. \end{aligned}$$

In either case (3) or (4), the top of Routh’s series can be written as

$$\begin{aligned} &a_{1,1}=-a_1 \\ &a_{2,2}=- \begin{vmatrix} a_1 & a_3\\ a_0 & a_2 \end{vmatrix} \diagup a_1 \\ &a_{3,3}=- \begin{vmatrix} a_1 & a_3 & a_5\\ a_0 & a_2 & a_4\\ 0 & a_1 & a_3 \end{vmatrix} \diagup \begin{vmatrix} a_1 & a_3\\ a_0 & a_2 \end{vmatrix} \\ &a_{4,4}=- \begin{vmatrix} a_1 & a_3 & a_5 & a_7\\ a_0 & a_2 & a_4 & a_6\\ 0 & a_1 & a_3 & a_5\\ 0 & a_0 & a_2 & a_4 \end{vmatrix} \diagup \begin{vmatrix} a_1 & a_3 & a_5\\ a_0 & a_2 & a_4\\ 0 & a_1 & a_3 \end{vmatrix} \\ &~~~\cdots \\ &a_{n,n}=-a_n. \end{aligned}$$

(6.95)

Each term of (6.95) corresponds to a Hurwitz determinant and its principal minors. Although it is different from (6.94) in sign, the number of sign changes is invariant (i.e., zero).