Question 1 of 8: Servo-Positioning Signal-Flow Graph — Closed-Loop and Disturbance TFs, 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 Exams May 2018
— 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 and PID/rate-feedback compensator design,
controllability/observability, Mason's Gain Formula on signal-flow graphs).
Question 1: Servo-Positioning Signal-Flow Graph — Closed-Loop and Disturbance TFs,
Stability Range (20 marks, compulsory)
Figure Q1.1 — the diagram is a SIGNAL-FLOW GRAPH (Mason's Gain
Formula), not a chain of summing boxes: every circle is a node whose value is the SUM of its
incoming branches. The PI controller and amplifier feed node $P$ (armature input); $T_{dist}(s)$
injects with gain $-1$ at the node between Armature and Load; the tachometer-style path
$-0.05$ feeds back from the node after Load to node $P$, entirely inside the dashed
"Motor and Robotic Arm Dynamics" box; the outer $-1$ path closes from $\Omega_{load}(s)$ back to
the first node.
Given. $G_c(s)=K_p(1+10/s)=K_p(s+10)/s$ ($\tau_i=0.1$ s); amplifier
gain $0.5$; armature $4/(s+1)$; local tachometer feedback $-0.05$ (wraps the armature+load pair
only); load $10/(s+1)$; gearbox $1/100$; outer feedback $-1$ from $\Omega_{load}(s)$.
Find. 1) $G_{cl}(s)$ as a polynomial ratio in $K_p$. 2) $G_d(s)$ as a
polynomial ratio in $K_p$. 3) The stable range of $K_p$, $K_{crit}$, $\omega_{osc}$.
Approach. Write the node equations of the signal-flow graph directly (each
node = sum of incoming branch values) and solve the resulting linear system symbolically for
$\Omega_{load}(s)$ as a function of $\Omega_{ref}(s)$ and $T_{dist}(s)$; differentiate to isolate
$G_{cl}$ and $G_{dist}$. Apply Routh–Hurwitz to the shared characteristic polynomial for
part 3.
Part 1) — node equations and $G_{cl}(s)$. Label the nodes $N_1$
(controller input), $N_2$ (controller output), $P$ (armature input), $Q$ (node after Armature,
where $T_{dist}$ injects), $S$ (node after Load, source of the $-0.05$ feedback). Then:
$$N_1=\Omega_{ref}-\Omega_{load},\quad N_2=G_c(s)N_1,\quad P=0.5N_2-0.05S,$$
$$Q=\frac{4}{s+1}P-T_{dist},\quad S=\frac{10}{s+1}Q,\quad \Omega_{load}=\frac{S}{100}.$$
Eliminating $N_1,N_2,P,Q,S$ (six linear equations, six unknowns) and collecting the
$\Omega_{ref}$-coefficient with $T_{dist}=0$ gives
$$\boxed{G_{cl}(s)=\frac{\Omega_{load}(s)}{\Omega_{ref}(s)}
=\frac{K_p(s+10)}{5s^3+10s^2+(15+K_p)s+10K_p}}.$$
As a check, at $K_p=3$ this reduces to $G_{cl}(s)=3(s+10)/(5s^3+10s^2+18s+30)$, whose
denominator factors as $5(s+1.827)(s^2+0.173s+3.284)$ — EXACTLY the closed-loop TF quoted
as given data in Question 2, confirming the signal-flow-graph reading above is the one the
paper intends.
Part 2) — $G_d(s)$. Collecting instead the $T_{dist}$-coefficient
(with $\Omega_{ref}=0$) from the same six equations gives
$$\boxed{G_d(s)=\frac{\Omega_{load}(s)}{T_d(s)}
=\frac{-(s^2+s)}{10s^3+20s^2+(30+2K_p)s+20K_p}}$$
— exactly twice the denominator of $G_{cl}(s)$ (both transfer functions share the same
closed-loop characteristic equation, as they must, since they come from the same feedback
loop).
Part 3) — Routh–Hurwitz stability range. The characteristic
equation is $5s^3+10s^2+(15+K_p)s+10K_p=0$. The Routh array is
$$\begin{array}{c|cc} s^3 & 5 & 15+K_p\\ s^2 & 10 & 10K_p\\
s^1 & \dfrac{10(15+K_p)-5(10K_p)}{10}=15-4K_p & 0\\ s^0 & 10K_p \end{array}$$
Every first-column entry must be positive: $15-4K_p>0\Rightarrow K_p<3.75$, and
$10K_p>0\Rightarrow K_p>0$. So
$$\boxed{0\lt K_p\lt3.75\ \text{for stable operation}}.$$
At $K_p=K_{crit}=3.75$ the $s^1$ row vanishes; the auxiliary equation from the $s^2$ row,
$10s^2+10K_{crit}=0\Rightarrow s^2=-3.75$, gives the marginal poles
$s=\pm j\sqrt{3.75}=\pm j\sqrt{15}/2$, i.e.
$$\boxed{K_{crit}=3.75,\qquad \omega_{osc}=\frac{\sqrt{15}}{2}\approx1.937\ \text{rad/s}}.$$
(Confirmed directly: at $K_p=3.75$ the cubic factors exactly as
$\tfrac54(s+2)(4s^2+15)=0$, roots $s=-2,\ \pm j1.9365$.)
