Stability of Two Characteristic Equations (Routh-Hurwitz): 23-Mechatronics-A1 Systems Dynamics and Controls, Undated paper, Q1

Question 1: Stability of Two Characteristic Equations (Routh-Hurwitz)

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. Two closed-loop characteristic equations in the variable $p$ (equivalent to the Laplace variable $s$): (a) a 4th-order polynomial, (b) a 5th-order polynomial, both with real coefficients.

Find. Whether each system is stable or unstable using the Routh-Hurwitz criterion, treating any root on or to the right of the imaginary axis ("limited stability") as instability.

Approach. Build the Routh array for each characteristic equation from its coefficients and read the sign pattern of the first column — the number of sign changes equals the number of right-half-plane (RHP) roots.

Part (a) — Routh array for $p^4+4p^3+8p^2+8p+3=0$.

Row Col 1 Col 2 Col 3

$p^4$ 1 8 3

$p^3$ 4 8 0

$p^2$ $(4{\cdot}8-1{\cdot}8)/4=6$ $(4{\cdot}3-1{\cdot}0)/4=3$ 0

$p^1$ $(6{\cdot}8-4{\cdot}3)/6=6$ 0 0

$p^0$ $(6{\cdot}3-6{\cdot}0)/6=3$ 0 0
Part (a) — read the first column. First column: $1,\,4,\,6,\,6,\,3$ — all positive, zero sign changes, so no roots lie in the RHP. $$\boxed{\text{System (a): STABLE}}$$

Row	Col 1	Col 2	Col 3
$p^4$	1	8	3
$p^3$	4	8	0
$p^2$	$(4{\cdot}8-1{\cdot}8)/4=6$	$(4{\cdot}3-1{\cdot}0)/4=3$	0
$p^1$	$(6{\cdot}8-4{\cdot}3)/6=6$	0	0
$p^0$	$(6{\cdot}3-6{\cdot}0)/6=3$	0	0

Part (b) — Routh array for $p^5+2p^4+3p^3+8p^2+p+4=0$.

Row	Col 1	Col 2	Col 3
$p^5$	1	3	1
$p^4$	2	8	4
$p^3$	$(2{\cdot}3-1{\cdot}8)/2=-1$	$(2{\cdot}1-1{\cdot}4)/2=-1$	0
$p^2$	$(-1{\cdot}8-2({-1}))/{-1}=6$	$(-1{\cdot}4-2{\cdot}0)/{-1}=4$	0
$p^1$	$(6({-1})-({-1}){\cdot}4)/6=-1/3$	0	0
$p^0$	4	0	0

Part (b) — read the first column. First column: $1,\,2,\,-1,\,6,\,-1/3,\,4$ — sign changes $+\!\to\!-$, $-\!\to\!+$, $+\!\to\!-$, $-\!\to\!+$: four sign changes, so four roots lie in the RHP (confirmed by direct root-solving: a real root at $-2.28$ plus two RHP-crossing complex pairs). $$\boxed{\text{System (b): UNSTABLE (4 RHP roots)}}$$

System	Order	1st-column sign changes	RHP roots	Stability
(a)	4	0	0	Stable
(b)	5	4	4	Unstable

Concept & Theory

Topic: The Routh-Hurwitz stability criterion.

Key relations: Routh array rows are built alternately from the even-power and odd-power coefficients of the characteristic polynomial; each subsequent entry equals (cross-product of the two rows above)/(pivot of the row directly above).

Why this works: the number of sign changes in the completed array's first column equals the number of characteristic-equation roots strictly in the right half of the $s$-plane, so stability can be certified — positive entries, no sign change — without ever solving for the roots directly.

Common pitfall: stopping at "no negative entries" instead of counting sign changes specifically; and forgetting that a zero first-column entry or an entire zero row (marginal/"limited" stability, roots exactly on the imaginary axis) must be reported as instability here, per the question's own definition.

Source: Nise, Control Systems Engineering, Ch. 6 (Routh-Hurwitz criterion); Ogata, Modern Control Engineering, §5-6.

Question 1 of 6: Stability of Two Characteristic Equations (Routh-Hurwitz)

Question 1: Stability of Two Characteristic Equations (Routh-Hurwitz)