Routh Array and Routh–Hurwitz Criterion of Stability (20 marks): 22-Elec-A2 Systems and Control, Undated paper, Q1

Question 1: Routh Array and Routh–Hurwitz Criterion of Stability (20 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. A hydraulic flow loop under parallel-form PID control, assumed to close as a unity-feedback loop.

Quantity	Symbol	Value
Controller (parallel PID)	$G_{PID}(s)$	$K_p\left(1+\dfrac{1}{\tau_i s}+\tau_d s\right)$
Integral time	$\tau_i$	0.25 s
Derivative time	$\tau_d$	0.10 s
Hydraulic process	$G(s)$	$\dfrac{10}{(s+1)^2}$

Find. (1) the closed-loop transfer function $G_{cl}(s)=Y(s)/R(s)$ as a ratio of polynomials in $K_p$; (2) the critical gain(s) $K_{crit}$ at marginal stability and the resulting oscillation frequency $\omega_{osc}$; (3) the range of $K_p$ giving stable operation.

Figure S1.1 - Unity-feedback loop for Question 1: PID controller ahead of the hydraulic process.

Approach. Put the controller over a common denominator, form the open-loop product $L=G_{PID}G$, close the loop, then apply the Routh–Hurwitz criterion to the resulting cubic characteristic polynomial.

Write the controller as a single rational function. Placing the three parallel terms over the common denominator $s$ gives$$G_{PID}(s)=K_p\frac{\tau_d s^2+s+1/\tau_i}{s}=K_p\frac{0.1s^2+s+4}{s}.$$
Form the open-loop transfer function. Multiplying by the process and clearing the factor 10 through the numerator,$$L(s)=G_{PID}(s)G(s)=\frac{10K_p(0.1s^2+s+4)}{s(s+1)^2}=\frac{K_p\left(s^2+10s+40\right)}{s(s+1)^2}.$$The controller therefore contributes a pole at the origin and a pair of complex zeros at $s=-5\pm j3.873$.
Close the loop. With unity feedback $G_{cl}=L/(1+L)$, and since $s(s+1)^2=s^3+2s^2+s$,$$\boxed{G_{cl}(s)=\frac{K_p\left(s^2+10s+40\right)}{s^3+(2+K_p)s^2+(1+10K_p)s+40K_p}}$$
Build the Routh array for $Q(s)=s^3+(2+K_p)s^2+(1+10K_p)s+40K_p$:$$\begin{array}{c|cc}s^3 & 1 & 1+10K_p\\s^2 & 2+K_p & 40K_p\\s^1 & c_1 & 0\\s^0 & 40K_p & \end{array}\qquad c_1=\frac{(2+K_p)(1+10K_p)-40K_p}{2+K_p}.$$
Marginal stability sets the $s^1$ element to zero. Expanding the numerator of $c_1$,$$(2+K_p)(1+10K_p)-40K_p=10K_p^2-19K_p+2=0,$$a quadratic in $K_p$ — so there are two critical gains, not one:$$K_p=\frac{19\pm\sqrt{281}}{20}\;\Rightarrow\;\boxed{K_{crit,1}=0.1118,\qquad K_{crit,2}=1.7882}$$
Get each oscillation frequency from the auxiliary polynomial. At marginal stability the $s^2$ row forms $A(s)=(2+K_p)s^2+40K_p=0$, so $\omega_{osc}=\sqrt{40K_p/(2+K_p)}$:$$\omega_{osc,1}=\sqrt{\frac{40(0.1118)}{2.1118}}=1.456\ \text{rad/s},\qquad \omega_{osc,2}=\sqrt{\frac{40(1.7882)}{3.7882}}=4.345\ \text{rad/s}.$$Substituting each gain back into $Q(s)$ confirms a root pair exactly on the imaginary axis ($\pm j1.4555$ with a third root at $-2.112$; $\pm j4.3453$ with a third root at $-3.788$).
Read off the stable range. The first-column entries $1$, $2+K_p$ and $40K_p$ are positive for every $K_p\gt 0$, so stability is governed entirely by the sign of $c_1$. The upward parabola $10K_p^2-19K_p+2$ is positive outside its roots, hence$$\boxed{0\lt K_p\lt 0.1118\quad\text{or}\quad K_p\gt 1.7882}$$The safe region is a union of two intervals: the loop is stable at low gain, goes unstable in the band $0.1118\lt K_p\lt 1.7882$, and re-stabilises above it. This is conditional stability, produced by the two complex controller zeros pulling the locus back into the left half-plane at high gain.

Result	Value
Closed-loop transfer function	$\dfrac{K_p(s^2+10s+40)}{s^3+(2+K_p)s^2+(1+10K_p)s+40K_p}$
Lower critical gain	$K_{crit,1}=0.1118$
Oscillation frequency at $K_{crit,1}$	$\omega_{osc,1}=1.456$ rad/s
Upper critical gain	$K_{crit,2}=1.7882$
Oscillation frequency at $K_{crit,2}$	$\omega_{osc,2}=4.345$ rad/s
Stable range of $K_p$	$0\lt K_p\lt 0.1118$ or $K_p\gt 1.7882$

Topic: Routh–Hurwitz stability of a PID loop, and conditional stability.

Key relations: Routh $s^1$ element $c_1=\dfrac{a_2a_1-a_0a_3}{a_2}$ for $a_3s^3+a_2s^2+a_1s+a_0$; auxiliary polynomial $A(s)=a_2s^2+a_0$ gives $\omega_{osc}=\sqrt{a_0/a_2}$.

Why this works: A sign change in the Routh first column counts one right-half-plane root. Because the derivative term makes the $s^2$ coefficient itself depend on $K_p$, the $s^1$ element becomes quadratic in $K_p$ and can change sign twice, so the stability boundary is crossed twice.

Common pitfall: Assuming there is a single $K_{crit}$ and reporting only one root of the quadratic. Equally, taking $\omega_{osc}=\sqrt{a_0}$ and forgetting the $s^2$ pivot $a_2$ — the auxiliary polynomial keeps that coefficient.

Source: Nise, Control Systems Engineering, 8th ed., Ch. 6 (Stability), and Ogata, Modern Control Engineering, 5th ed., Sec. 5–6 (Routh criterion, PID action).

Question 1 of 8: Routh Array and Routh–Hurwitz Criterion of Stability (20 marks)

Question 1: Routh Array and Routh–Hurwitz Criterion of Stability (20 marks)