Absolute Stability of a Hydraulic System (Routh's Criterion): 23-Mechatronics-A1 Systems Dynamics and Controls, Undated paper, Q4

Question 4: Absolute Stability of a Hydraulic System (Routh's Criterion)

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. A closed-loop hydraulic-servo block diagram: forward path $(As+K_1)/(Ms^2+cs+k)$ converts a force error $F_e(s)$ to a flow $Q_o(s)$; a summing junction combines it with the load-flow disturbance $Q_L(s)$; the net flow drives plant $1/[(V/\beta)s+L]$ to output pressure $P(s)$; $P(s)$ feeds back through gain $A$ to the input summing junction. $M,c,k,A,K_1,V,\beta,L$ are all positive physical constants (mass, damping, stiffness, valve/feedback gain, valve flow-gain, chamber volume, fluid bulk modulus, leakage coefficient).

Find. The algebraic relationship among the parameters that keeps the closed loop absolutely stable.

Figure — hydraulic servo loop: force-error stage $(As+K_1)/(Ms^2+cs+k)$, load-flow disturbance $Q_L(s)$, chamber/leakage plant $1/[(V/\beta)s+L]$, pressure feedback gain $A$.

Check

The printed summing-junction signs at the $Q_o(s)/Q_L(s)$ node are read as $Q_o(s)-Q_L(s)$ (load flow subtracted as a disturbance). This assumption does not affect the stability result below: $Q_L(s)$ is an additive disturbance that never closes a feedback path, so it cannot appear in the closed-loop characteristic equation regardless of its sign at that junction.

Approach. Since $Q_L(s)$ does not appear in the loop's own feedback path, the closed-loop pole locations are set entirely by $1+A\,G_1(s)G_2(s)=0$ for the $F_r(s)\to P(s)$ loop; build that characteristic equation and apply Routh's criterion to the resulting cubic.

Form the loop transfer function and characteristic equation. $$G_1(s)G_2(s)=\frac{As+K_1}{(Ms^2+cs+k)\left[(V/\beta)s+L\right]},\qquad (Ms^2+cs+k)\left[(V/\beta)s+L\right]+A(As+K_1)=0$$
Expand and collect by powers of $s$. $$\frac{MV}{\beta}s^3+\left(LM+\frac{Vc}{\beta}\right)s^2 +\left(A^2+Lc+\frac{Vk}{\beta}\right)s+(AK_1+Lk)=0$$ Call the four coefficients $a_0,a_1,a_2,a_3$ (all manifestly positive, since every physical constant above is positive).
Routh's criterion for a general cubic $a_0s^3+a_1s^2+a_2s+a_3$. The array reduces to row $s^1$: $(a_1a_2-a_0a_3)/a_1$, row $s^0$: $a_3$; with $a_0,a_1,a_2,a_3 >0$ already guaranteed, the only condition that can fail is $a_1a_2>a_0a_3$. $$\boxed{\text{Absolute stability} \iff \left(LM+\frac{Vc}{\beta}\right) \left(A^2+Lc+\frac{Vk}{\beta}\right) > \frac{MV}{\beta}\,(AK_1+Lk)}$$

Coefficient	Expression
$a_0$	$MV/\beta$
$a_1$	$LM+Vc/\beta$
$a_2$	$A^2+Lc+Vk/\beta$
$a_3$	$AK_1+Lk$
Stability condition	$a_1a_2 > a_0a_3$ (shown above)

Concept & Theory

Topic: Routh's criterion applied to a symbolic (parametric) 3rd-order characteristic equation.

Key relations: for $a_0s^3+a_1s^2+a_2s+a_3$ with all coefficients positive, the Routh array collapses to the single shortcut condition $a_1a_2>a_0a_3$.

Why this works: the $s^1$ row entry $(a_1a_2-a_0a_3)/a_1$ is the only array entry whose sign is not already forced positive by positive coefficients alone; a sign change there signals a right-half-plane pole pair, so requiring it to stay positive is necessary and sufficient for a stable cubic with positive coefficients.

Common pitfall: dropping the $A^2$ term the feedback path itself contributes to $a_2$ (the SAME gain $A$ also appears once, linearly, in $a_3$ — easy to carry into only one of the two coefficients); and forgetting that $Q_L(s)$ is a disturbance, not part of the denominator, so it never enters the characteristic equation.

Source: Ogata, Modern Control Engineering, §5-6; Nise, Control Systems Engineering, §6.3 (Routh-Hurwitz for parametric systems).

Question 4 of 6: Absolute Stability of a Hydraulic System (Routh's Criterion)

Question 4: Absolute Stability of a Hydraulic System (Routh's Criterion)