Question 1 of 8: PID + Hydraulic Process — Closed-Loop TF and Routh–Hurwitz Stability Range
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 17-Phys-B5 Systems and Control, National Exam, May
2019 — a three-hour closed-book examination with one double-sided handwritten formula/notes
sheet permitted and an approved calculator. Questions 1 and 2 are compulsory; candidates choose
three of the remaining six (3–8). Every question is nonetheless answered in full below so
the paper remains a complete study resource. All eight questions carry equal value (20 marks
each).
Reference texts. N. S. Nise, Control Systems Engineering, 7th ed.
(transient-response specifications, root locus, Routh–Hurwitz, frequency-response and lead
compensator design, state-space representation, steady-state errors); K. Ogata, Modern
Control Engineering, 5th ed. (root-locus construction, PID/rate-feedback compensator design,
controllability/observability, Mason's Gain Formula on signal-flow graphs).
Question 1: PID + Hydraulic Process — Closed-Loop TF and Routh–Hurwitz Stability
Range (20 marks, compulsory)
Find. $G_{cl}(s)$ as a function of $K_p$; $K_{crit}$ and $\omega_{osc}$; the
stable range of $K_p$.
Approach. Combine $G_{PID}(s)G(s)$ into a single rational function of $K_p$,
close the unit-feedback loop, then apply the Routh array to the resulting characteristic
polynomial.
Combine controller and process. $G_{PID}(s)=K_p\left(1+\dfrac4s+0.1s\right)
=K_p\dfrac{0.1s^2+s+4}{s}$. Multiplying by $G(s)=10/(s+1)^2$, the $0.1\times10=1$ factor cancels
cleanly:
$$G_{PID}(s)G(s)=K_p\,\frac{s^2+10s+40}{s(s+1)^2}.$$
Close the loop. $G_{cl}(s)=\dfrac{G_{PID}G}{1+G_{PID}G}$. With
$s(s+1)^2=s^3+2s^2+s$,
$$\boxed{G_{cl}(s)=\frac{K_p(s^2+10s+40)}{s^3+(K_p+2)s^2+(10K_p+1)s+40K_p}}.$$
Routh array. Characteristic equation $s^3+(K_p+2)s^2+(10K_p+1)s+40K_p=0$:
$$\begin{array}{c|cc} s^3 & 1 & 10K_p+1\\ s^2 & K_p+2 & 40K_p\\
s^1 & \dfrac{(K_p+2)(10K_p+1)-40K_p}{K_p+2} & 0\\ s^0 & 40K_p \end{array}$$
The $s^1$-row numerator is $10K_p^2-19K_p+2$. Setting it to zero (marginal stability):
$$K_p=\frac{19\pm\sqrt{19^2-4(10)(2)}}{20}=\frac{19\pm\sqrt{281}}{20}
\ \Rightarrow\ \boxed{K_{crit,1}=0.1118,\quad K_{crit,2}=1.7882}.$$
Because this row is QUADRATIC in $K_p$ (not linear), there are two critical gains and the system
is only conditionally stable, a pattern typical of closely-spaced pole/zero pairs (root-locus zero near the process pole here).
$\omega_{osc}$ at each crossing. The auxiliary equation from the $s^2$ row,
$(K_p+2)s^2+40K_p=0$, gives $\omega_{osc}=\sqrt{40K_p/(K_p+2)}$ at each $K_{crit}$. Substituting
(confirmed by direct root-finding on the full cubic at each $K_p$):
$$\boxed{\omega_{osc,1}=1.4555\ \text{rad/s at }K_p=0.1118,\qquad
\omega_{osc,2}=4.3453\ \text{rad/s at }K_p=1.7882.}$$
Stable range. Both first-column entries $K_p+2$ and $40K_p$ are positive for
all $K_p\gt0$, so stability is governed purely by the sign of $10K_p^2-19K_p+2$, a upward-opening
parabola that is positive outside its roots and negative between them:
$$\boxed{0\lt K_p\lt0.1118\ \ \text{OR}\ \ K_p\gt1.7882\ \ \text{(stable)};\qquad
0.1118\lt K_p\lt1.7882\ \ \text{(unstable)}.}$$
Confirmed: at $K_p=0.5$ (inside the "gap") the closed-loop poles are
$-2.85,\ 0.177\pm2.64j$ — genuinely unstable.
