Question 2 of 8: Error Constants, DC Gain and 2nd-Order Dominant-Pole Model at $K_{op}=3.0$
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 2: Error Constants, DC Gain and 2nd-Order Dominant-Pole Model at
$K_{op}=3.0$ (20 marks, compulsory)
Given. $G_{open}(s)=0.6(s+10)/[s(s^2+2s+3)]$ (Type 1, one pole at the
origin); $G_{cl}(s)=0.6(s+10)/[(s+1.827)(s^2+0.173s+3.284)]$ — independently reproduced from
Question 1's $G_{cl}(s)$ at $K_p=3$, which cross-checks this data exactly.
Find. 1) $K_{pos},K_v,K_a$ and the three steady-state errors. 2) $G_{cl}(0)$.
3) $K_{dc},\zeta,\omega_n,G_m(s)$ for the 2nd-order dominant-pole model. 4) $PO$,
$T_{settle(\pm2\%)}$, $T_{rise(0-100\%)}$.
Approach. Error constants come directly from the appropriate
$\lim_{s\to0}s^kG_{open}(s)$; the dominant-pole model keeps the lightly-damped complex pair
($\text{Re}=-0.0865$) and drops the far real pole ($-1.827$, over $21\times$ farther from the
imaginary axis), matching $K_{dc}$ to the true DC gain; the step-response specs follow from the
standard second-order formulas.
Part 1) — error constants. $G_{open}(s)$ is Type 1 (a single pole
at $s=0$), so
$$K_{pos}=\lim_{s\to0}G_{open}(s)=\infty\ \Rightarrow\ e_{ss(step)}\%=0\%,$$
$$K_v=\lim_{s\to0}sG_{open}(s)=\frac{0.6\times10}{3}=\boxed{2}\ \Rightarrow\
e_{ss(ramp)}=\frac{1}{K_v}=\boxed{0.5\ \text{V/V}},$$
$$K_a=\lim_{s\to0}s^2G_{open}(s)=0\ \Rightarrow\ e_{ss(parab)}=\infty.$$
Part 2) — closed-loop DC gain. $$G_{cl}(0)=\frac{0.6\times10}
{1.827\times3.284}=\frac{6.0}{6.001}\approx\boxed{1.000}$$ — consistent with
$e_{ss(step)}=0$ found in Part 1 (a Type 1 loop always tracks a step exactly, so the
closed-loop DC gain must be unity).
Part 3) — 2nd-order dominant-pole model. The complex pair
$s^2+0.173s+3.284$ gives directly
$$\omega_n=\sqrt{3.284}=\boxed{1.812\ \text{rad/s}},\qquad
\zeta=\frac{0.173}{2\omega_n}=\boxed{0.0477}.$$
Matching the model's DC gain to the true system's (Part 2) gives $K_{dc}=G_{cl}(0)=1.0$,
so
$$\boxed{G_m(s)=\frac{3.284}{s^2+0.173s+3.284}}.$$
Part 4) — unit-step response specs. With $\zeta=0.0477$,
$\omega_n=1.812$ rad/s:
$$PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=\boxed{86.1\%},\qquad
T_{settle(\pm2\%)}=\frac{4}{\zeta\omega_n}=\boxed{46.2\ \text{s}}.$$
For the rise time, $\theta=\cos^{-1}\zeta=1.523$ rad and
$\omega_d=\omega_n\sqrt{1-\zeta^2}=1.810$ rad/s, so
$$T_{rise(0-100\%)}=\frac{\pi-\theta}{\omega_d}=\boxed{0.894\ \text{s}}$$
(this very light damping is expected: $K_{op}=3$ is close to the $K_{crit}=3.75$ found in
Question 1, so the closed loop is nearly marginally stable, matching the huge overshoot and
settling time.)
