Question 7 of 8: Root Locus, Critical Gain, and a 5% Overshoot Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B5 Control, National Examination May 2017
— 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
Nyquist stability, state-space representation, steady-state errors); K. Ogata, Modern
Control Engineering, 5th ed. (root-locus and PID compensator design, controllability/
observability, Mason's Gain Formula on signal-flow graphs).
Question 7: Root Locus, Critical Gain, and a 5% Overshoot Design (20 marks)
Figure Q7.2 — root locus for $G(s)=10/[(s+2)(s+4)(s+7)]$: real-axis
branches (poles $-2,-4$ merging at the breakaway point $-2.88$; pole $-7$ moving left), the
complex branches through the $j\omega$-axis crossing at $K_{crit}=59.4$, and the design pole
at $K_{op}\approx3.82$.
Given. $G(s)=10/[(s+2)(s+4)(s+7)]$, unity feedback, proportional gain $K_p$.
Find. 1) Asymptote angles, breakaway point, centroid, $j\omega$-crossing
($\omega_{osc}$, $K_{crit}$). 2) $K_{op}$ for $PO\approx5\%$ and the resulting
$T_{settle(\pm5\%)}$, $T_{rise(0-100\%)}$, $e_{ss(step\%)}$. 3) Commentary on the dominant-pole
model's accuracy.
Approach. Apply the standard root-locus rules (real-axis segments, asymptote
angles/centroid, breakaway via $dK_p/ds=0$) to the open-loop poles $-2,-4,-7$; find the
$j\omega$-crossing via Routh–Hurwitz on the characteristic polynomial. For the design,
invert the overshoot formula for $\zeta$, find where the $\zeta$-line intersects the locus by
the angle condition, then read $K_{op}$ and the transient specs off the standard 2nd-order
formulas.
Part 1) — asymptotes, centroid, breakaway.
Three poles, no zeros $\Rightarrow$ 3 asymptotes at
$\theta=\dfrac{(2k{+}1)180^\circ}{3}=\boxed{60^\circ,180^\circ,300^\circ}$, centroid
$\sigma_a=\dfrac{-2-4-7}{3}=\boxed{-4.33}$. Real-axis locus exists on $(-4,-2)$ and
$(-\infty,-7)$ (odd pole/zero count to the right). Breakaway: with
$f(s)=(s+2)(s+4)(s+7)=s^3+13s^2+50s+56$, solve $f'(s)=3s^2+26s+50=0\Rightarrow
s=-2.88\ \text{or}\ -5.79$; only $s=-2.88$ lies ON the locus segment $(-4,-2)$ (the other root,
$-5.79$, falls in $(-7,-4)$ where the locus does NOT exist — an even pole count to the
right — so it is rejected): $\boxed{\text{breakaway}=-2.88}$.
Part 1) — $j\omega$-crossing via Routh–Hurwitz.
Char. eq. $s^3+13s^2+50s+56+10K_p=0$. Routh's $s^1$ row vanishes at
$K_p=\tfrac{13(50)-(56+10K_p)}{13}=0\Rightarrow\boxed{K_{crit}=59.4}$; the auxiliary equation
from the $s^2$ row, $13s^2+(56+594)=0$, gives $s^2=-50\Rightarrow
\boxed{\omega_{osc}=\sqrt{50}\approx7.07\ \text{rad/s}}$ (matching the orange markers in the
sketch, where the third, real closed-loop pole sits at exactly $s=-13$).
Part 2) — damping ratio and dominant pole for $PO\approx5\%$.
Inverting $PO=100e^{-\zeta\pi/\sqrt{1-\zeta^2}}=5$ gives $\boxed{\zeta\approx0.690}$
($\theta=\cos^{-1}\zeta\approx46.4^\circ$ off the negative real axis). Solving the angle
condition (sum of angles from the three poles $=180^\circ\pmod{360^\circ}$) numerically along
this $\zeta$-line locates the dominant closed-loop pole at
$$s_{dom}\approx-2.315\pm j2.427\quad(\omega_n=|s_{dom}|\approx3.354\ \text{rad/s}).$$
Evaluating $K_p=-f(s_{dom})/10$ at that point gives $\boxed{K_{op}\approx3.82}$ (the third closed-loop pole from the resulting cubic lands at
$s\approx-8.37$, real, as marked in green above).
Part 2) — step-response estimates. Using $\sigma=\zeta\omega_n=2.315$
and $\omega_d=2.427$: $T_{settle(\pm5\%)}\approx3/\sigma\approx\boxed{1.30\ \text{s}}$;
$T_{rise(0-100\%)}=(\pi-\cos^{-1}\zeta)/\omega_d\approx\boxed{0.96\ \text{s}}$ (exact closed
form for a pure 2nd-order, zero-free response); and with $G(0)=10/(2\cdot4\cdot7)=0.1786$,
$$e_{ss(step)\%}=\frac{100}{1+K_{op}G(0)}=\frac{100}{1+3.82(0.1786)}\approx\boxed{59.5\%}$$
(a large residual error, since this is Type 0 proportional-only control — no
integrator to drive it to zero).
Part 3) — dominant-pole model vs. the actual response.
The third closed-loop pole ($-8.37$) is only $8.37/2.315\approx3.6\times$ farther from the
imaginary axis than the dominant pair's real part — short of the usual
$\ge5\times$–$10\times$ separation needed for the 2nd-order estimate to be accurate. A
direct time-domain simulation of the actual (3rd-order) closed loop at $K_{op}$ confirms this:
the true peak overshoot is only $\approx4.7\%$ (slightly under the nominal 5%) and the response
re-enters the $\pm5\%$ band by $t\approx0.85$ s — noticeably FASTER than the
$1.30$ s dominant-pole estimate. $\boxed{\text{The third pole's own decay measurably
speeds up settling and slightly trims the overshoot versus the pure 2nd-order prediction.}}$
