22-Mec-A3 System Analysis and Control · Undated paper

Question 4 of 6: Absolute stability of a hydraulic servo by Routh’s criterion

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 16-Mec-A3 System Analysis and Control, National Exam, May 2019 sitting. Three hours, closed book, Casio or Sharp approved calculator only. Six questions; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied with the paper. Every one of the six questions is worked below.

Reference texts. Ogata, Modern Control Engineering, 5th ed. (Chs. 5–8); Nise, Control Systems Engineering, 8th ed. (Chs. 4, 6, 8, 10); Dorf & Bishop, Modern Control Systems, 13th ed.; Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed.

Question 4: Absolute stability of a hydraulic servo by Routh’s criterion (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. The block diagram of the paper, in which the force error $F_e(s)=F_r(s)-A\,P(s)$ drives a spool valve of transfer function $(As+K_1)/(Ms^2+cs+k)$ to produce a flow $Q_o(s)$; the load flow $Q_L(s)$ is subtracted, and the net flow charges the compressible fluid volume through $1/\left[(V/\beta)s+L\right]$ to give the cylinder pressure $P(s)$. The pressure is fed back through the piston area $A$.

Symbol	Meaning
$M,\ c,\ k$	spool mass, viscous damping and centring-spring rate
$A$	piston area (also the flow gain of the spool)
$K_1$	steady-state flow gain of the valve
$V,\ \beta$	trapped fluid volume and bulk modulus (so $V/\beta$ is the fluid capacitance)
$L$	leakage coefficient across the piston

Find. The inequality among $M,\ c,\ k,\ A,\ K_1,\ V/\beta$ and $L$ that guarantees absolute stability of the closed loop.

[Figure not reproduced: Question 4 block diagram: hydraulic position servo with pressure feedback through the piston area A, redrawn from the exam paper. The load flow Q L (s) is a disturbance and does not enter the characteristic equation. See the official exam paper.]

Approach. The disturbance $Q_L$ affects only the numerator of the response, so the characteristic equation comes from the loop gain alone: form $1+G(s)H(s)=0$, clear the denominators to obtain a cubic in $s$, and apply the Routh criterion to that cubic.

1. Write the loop gain. Travelling once round the loop, from the summing junction through the valve, through the fluid capacitance, and back through the area $A$, $$G(s)H(s)=\frac{As+K_1}{Ms^2+cs+k}\cdot\frac{1}{\dfrac{V}{\beta}s+L}\cdot A$$
2. Form the characteristic equation. Setting $1+GH=0$ and multiplying out both denominators, $$\left(Ms^2+cs+k\right)\left(\frac{V}{\beta}s+L\right)+A\left(As+K_1\right)=0$$
3. Expand into a cubic. Writing $a\equiv V/\beta$ for brevity and collecting powers of $s$: $$M a\,s^3+\left(ML+ca\right)s^2+\left(cL+ka+A^2\right)s+\left(kL+AK_1\right)=0$$ Note where the piston area enters twice: $A^2$ lifts the $s$ coefficient (pressure feedback acts like extra damping), while $AK_1$ lifts the constant term (it also raises the loop gain). These two effects pull in opposite directions, which is the heart of the question.
4. Apply Routh to the cubic. For $a_3s^3+a_2s^2+a_1s+a_0=0$ the array reduces to two conditions: every coefficient positive, and the $s^1$ entry $\left(a_2a_1-a_3a_0\right)/a_2$ positive. Since $M$, $a=V/\beta$, $c$, $k$, $L$, $A$ and $K_1$ are all physically positive, the coefficient condition is automatic and only the product test binds: $$\boxed{\left(ML+\frac{cV}{\beta}\right)\left(cL+\frac{kV}{\beta}+A^2\right)>\frac{MV}{\beta}\left(kL+AK_1\right)}$$
5. Read the design message. Solving the same inequality for the valve gain gives an explicit ceiling, $$K_1<\frac{\left(ML+ca\right)\left(cL+ka+A^2\right)-MakL}{MaA},\qquad a=\frac{V}{\beta}$$ so the servo is stable up to a maximum flow gain and unstable above it. Everything that raises the left-hand side stabilises the loop: more spool damping $c$, more leakage $L$ (deliberate cross-port bleed is a classic hydraulic fix), and a larger piston area through the $A^2$ term. Everything that raises the right-hand side destabilises it: a heavier spool $M$, a larger trapped volume $V$ or a softer fluid (small $\beta$, for instance from entrained air), and a higher valve gain $K_1$.
6. Note the limiting case. With zero leakage, $L=0$, the condition collapses to $ca\left(ka+A^2\right)>MaAK_1$, i.e. $c\left(ka+A^2\right)>MAK_1$; if in addition the spool damping is small the inequality fails and the servo oscillates. This is the familiar result that a tightly sealed, lightly damped hydraulic actuator on a large trapped volume is prone to sustained hunting.

Quantity	Result
Loop gain	$GH=\dfrac{A(As+K_1)}{(Ms^2+cs+k)\left[(V/\beta)s+L\right]}$
Characteristic equation	$Ma\,s^3+(ML+ca)s^2+(cL+ka+A^2)s+(kL+AK_1)=0$, $a=V/\beta$
Coefficient conditions	Satisfied automatically for positive physical parameters
Binding stability condition	$(ML+ca)(cL+ka+A^2)>Ma(kL+AK_1)$
Maximum valve gain	$K_1<\left[(ML+ca)(cL+ka+A^2)-MakL\right]/(MaA)$
Stabilising parameters	$c$, $L$, $A^2$, $\beta$ (increase); $M$, $V$, $K_1$ (decrease)

---