Routh–Hurwitz ranges of a parameter: 22-Mec-A3 System Analysis and Control, May 2014, Q2

Question 2: Routh–Hurwitz ranges of a parameter (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four third-order characteristic polynomials in which the single parameter $\alpha$ occupies a different coefficient position in each case:

Case	Characteristic equation	Position of α
(a)	$s^3+s^2+s+\alpha=0$	constant term $a_0$
(b)	$s^3+s^2+\alpha s+1=0$	$s$ coefficient $a_1$
(c)	$s^3+\alpha s^2+s+1=0$	$s^2$ coefficient $a_2$
(d)	$\alpha s^3+s^2+s+1=0$	leading coefficient $a_3$

Find. For each case, the complete range of real $\alpha$ that places every root strictly in the open left half-plane (i.e. the range for which the system is asymptotically stable).

Approach. Build the Routh array once for the general cubic $a_3s^3+a_2s^2+a_1s+a_0$, extract the two independent conditions that its first column carries, then substitute each case in turn; a root-location check by direct factorisation confirms every boundary.

Form the general Routh array. For $a_3s^{3}+a_2s^{2}+a_1s+a_0=0$ the array is $$\begin{array}{c|cc} s^{3} & a_3 & a_1\\ s^{2} & a_2 & a_0\\ s^{1} & \dfrac{a_2a_1-a_3a_0}{a_2} & 0\\ s^{0} & a_0 & \end{array}$$ All four first-column entries must share the same sign (conventionally all positive), which for a cubic collapses to the compact Hurwitz statement
State the two stability conditions. Every coefficient must be positive and the middle-row test must hold: $$\boxed{\;a_3,a_2,a_1,a_0 > 0 \quad\text{and}\quad a_2a_1 > a_3a_0\;}$$ The first condition is necessary but not sufficient; the product condition is what actually detects the pair of complex roots crossing the imaginary axis.
Case (a): $\alpha$ in the constant term. Here $a_3=a_2=a_1=1$ and $a_0=\alpha$. Positivity demands $\alpha > 0$; the product test gives $1\times 1 > 1\times\alpha$, i.e. $\alpha < 1$. Combining, $$\boxed{\;0 < \alpha < 1\;}$$ At $\alpha=1$ the polynomial factors as $(s+1)(s^{2}+1)$, confirming a pair of roots exactly on the imaginary axis at $s=\pm j1$; at $\alpha=0$ a root sits at the origin.
Case (b): $\alpha$ multiplying $s$. Now $a_1=\alpha$ and $a_0=1$, so positivity requires $\alpha > 0$ and the product test requires $1\times\alpha > 1\times 1$. The second is the binding condition: $$\boxed{\;\alpha > 1\;}$$ There is no upper limit — increasing the $s$ coefficient adds damping without limit. At $\alpha = 1$ the polynomial is again $(s+1)(s^{2}+1)$.
Case (c): $\alpha$ multiplying $s^{2}$. With $a_2=\alpha$, positivity gives $\alpha > 0$ and the product test gives $\alpha\times 1 > 1\times 1$, so once more $$\boxed{\;\alpha > 1\;}$$ Physically this is the same statement as case (b): the $s^{2}$ coefficient is the sum of the pole magnitudes, so it too must be large enough to keep the complex pair damped.
Case (d): $\alpha$ as the leading coefficient. A negative $\alpha$ makes the coefficients change sign, so no root pattern can be stable; for $\alpha > 0$ the product test reads $1\times 1 > \alpha\times 1$. Hence $$\boxed{\;0 < \alpha < 1\;}$$ Note that $\alpha\to 0$ is a degenerate limit: the polynomial drops to second order ($s^{2}+s+1$, stable), but the third root has escaped to infinity, so the strict inequality is retained.
Verify by root computation. Substituting representative values and factoring numerically confirms every range: $\alpha=0.5$ in (a) gives roots $-0.6478,\ -0.1761\pm j0.8607$ (all stable); $\alpha=1.5$ in (a) gives $-1.2041,\ 0.1020\pm j1.1115$ (unstable), and the same pattern of confirmation holds for (b), (c) and (d).

The four cases together make the pedagogical point of the question: the same cubic can be destabilised by making any one coefficient too small relative to the others, but which direction is "too small" depends on where the parameter sits. Parameters in $a_1$ or $a_2$ have lower bounds; parameters in $a_0$ or $a_3$ have upper bounds.

Case	Characteristic equation	Stable range of α
(a)	$s^3+s^2+s+\alpha=0$	0 < α < 1
(b)	$s^3+s^2+\alpha s+1=0$	α > 1
(c)	$s^3+\alpha s^2+s+1=0$	α > 1
(d)	$\alpha s^3+s^2+s+1=0$	0 < α < 1

Topic. Routh–Hurwitz parametric stability analysis of a cubic.

Key relations. Routh array first column $\{a_3,\ a_2,\ (a_2a_1-a_3a_0)/a_2,\ a_0\}$; cubic Hurwitz criterion $a_i>0$ and $a_2a_1>a_3a_0$; number of right-half-plane roots equals the number of sign changes in the first column.

Why this works. The Routh array is a compact form of the Sturm/Hurwitz determinant test: each row is generated by a cross-multiplication that is equivalent to a Euclidean division of the polynomial by its "even" and "odd" parts. Sign changes in the first column count the roots with positive real part, so demanding no sign changes forces every root into the left half-plane. When a parameter appears linearly, the conditions become a set of linear inequalities whose intersection is the stable interval; the boundaries of that interval are exactly the parameter values at which roots sit on the imaginary axis, so they can always be checked by factoring an auxiliary polynomial.

Common pitfall. Students routinely stop at "all coefficients positive", which misses case (a)'s upper bound entirely, and they forget that a negative leading coefficient in case (d) is not rescued by dividing through — dividing by a negative number flips every coefficient sign and the positivity test fails anyway. A second trap is quoting a closed interval; the endpoints put roots on the $j\omega$ axis and are marginally stable, not stable.

Source. Ogata, Modern Control Engineering, 5th ed., §5.6 (Routh's stability criterion); Dorf & Bishop, Modern Control Systems, 13th ed., Ch. 6.2.

Question 2 of 6: Routh–Hurwitz ranges of a parameter

Question 2: Routh–Hurwitz ranges of a parameter (25 marks)