Question 2 of 8: Frequency Response and Root Locus — Verifying Question 1
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 2: Frequency Response and Root Locus — Verifying Question 1
(20 marks, compulsory)
Given. Open-loop transfer function (from Q1)
$K_pF(s)=K_p\dfrac{s^2+10s+40}{s(s+1)^2}$; poles at $s=0,-1,-1$; zeros at $s=-5\pm j\sqrt{15}
\approx-5\pm3.873j$.
Find. Gain-margin verification of $K_{crit}$ from $G(j\omega)$; full root-locus
geometry; confirmation against Question 1.
Approach. Part 1 uses the classical gain-margin construction
($K_{crit}=1/|F(j\omega_{pc})|$ at the phase-crossover frequency where $\angle F(j\omega)=-180^\circ$). Parts 2–3 use standard root-locus
construction rules on $K_pF(s)$.
Check
Figure Q2.1's printed break frequencies (a system
rolling off from 0 dB at $\omega=1$, steepening at $\omega=10$) do not match
$G(s)=10/(s+1)^2$ at all (a clean double real pole at $\omega=1$ only). Rather than read
unreliable numbers off that figure, the SAME phase-crossover/gain-margin technique the figure is
meant to illustrate is applied analytically to $F(j\omega)$ directly below, which is mathematically
equivalent to what a clean version of the plot would show.
Part 1) — gain-margin check.
$F(j\omega)=\dfrac{(j\omega)^2+10j\omega+40}{j\omega(j\omega+1)^2}$. Evaluating at the two
$\omega_{osc}$ values found in Question 1:
$$F(j1.4555)=-8.9408+0j\ \ (\angle=-180^\circ)\ \Rightarrow\ K_{crit,1}=1/8.9408=0.1118,$$
$$F(j4.3453)=-0.5592+0j\ \ (\angle=-180^\circ)\ \Rightarrow\ K_{crit,2}=1/0.5592=1.7882.$$
Both EXACTLY reproduce Question 1's Routh–Hurwitz values — the frequency response
confirms the critical gains.
Part 2a) — asymptotes and centroid. $n=3$ poles, $m=2$ zeros
$\Rightarrow n-m=1$ asymptote, at angle $(2(0)+1)\times180^\circ/1=180^\circ$ (i.e. the single
excess branch heads to $-\infty$ along the negative real axis; with only one asymptote a
"centroid" isn't a separate construction point).
Part 2b) — real-axis segments and breakaway point. The segment
$(-1,0)$ has one pole (at $s=0$) to its right — odd — so it lies on the locus.
Solving $dK/ds=0$ for $K(s)=-s(s+1)^2/(s^2+10s+40)$ gives one real, positive-$K$ root:
$$\boxed{s_{break}=-0.3527,\quad K=0.00404}$$
(confirmed by direct root tracking: two real branches, one leaving $s=0$ and one leaving the
double pole at $s=-1$, meet at $s=-0.3527$ and break into a complex-conjugate pair immediately
above $K=0.004$).
Part 2c) — imaginary-axis crossings. These are exactly the
Routh-derived boundaries: the complex branch pair crosses into the right half-plane at
$K_p=K_{crit,1}=0.1118$ ($\omega_{osc}=1.4555$), continues unstable, then crosses back into the
LEFT half-plane at $K_p=K_{crit,2}=1.7882$ ($\omega_{osc}=4.3453$) — a conditionally stable
locus that re-enters the left half-plane at high gain, ultimately approaching the finite zeros
$-5\pm3.873j$ as $K_p\to\infty$.
Figure Q2.2 (sketch) — root locus of $K_p(s^2+10s+40)/[s(s+1)^2]$.
× = open-loop poles, ◯ = open-loop zeros, orange dots = breakaway point and the two
imaginary-axis crossings.
