Question 8 of 8: Series-PID design by pole placement
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, May 2017, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula sheet; a Laplace-transform table and the standard $\zeta$–overshoot and $\zeta$–resonant-peak design plots are supplied on pages 2–3). Questions 1 and 2 are compulsory; a complete paper is five questions, so a candidate chooses three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — Routh–Hurwitz (Ch. 6), root locus (Ch. 8), steady-state error and static error constants (Ch. 7), frequency response, Nyquist, gain/phase margins (Ch. 10), PID and compensator design (Ch. 9–11), state space, controllability/observability and Mason’s rule (Ch. 3, 5, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations and pole-placement. All block diagrams, root loci, pole–zero maps, Nyquist plots, step responses and the state-realization diagrams below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Every result here is derived analytically in the $s$-domain and cross-checked; the design charts are used only to corroborate $\zeta$ from $PO$.
Question 8 — Series-PID design by pole placement [20]
Figure Q8.1 — series-PID loop, controller $K_p(1+K_i/s)(K_d s+1)$ driving the process $1/(s^2+7s+8)$, unity feedback.
Item (1) — closed-loop TF and characteristic equation
Open loop. $G_c(s)=\dfrac{K_p(s+K_i)(K_d s+1)}{s}$, so $L(s)=\dfrac{K_p(s+K_i)(K_d s+1)}{s(s^2+7s+8)}$ and $G_{cl}=\dfrac{L}{1+L}$ with numerator $K_p(s+K_i)(K_d s+1)$.
Natural frequency. $T_{settle(\pm2\%)}=\dfrac{4}{\zeta\omega_n}=2$ s $\Rightarrow \zeta\omega_n=2$, so $\omega_n=\dfrac{2}{0.517}=\boxed{3.87\ \text{rad/s}}$. The dominant pair is $-2\pm j\omega_d$, $\omega_d=\omega_n\sqrt{1-\zeta^2}=3.31$ rad/s.
Item (3) — pole placement and gains
Approach. the third pole is set to $-K_i$ so the closed-loop zero at $s=-K_i$ cancels it. Match $Q(s)$ to $(s^2+4s+\omega_n^2)(s+K_i)$ with $\omega_n^2=14.97$.
Derivative gains. $K_d=\dfrac{K_i-3}{14.97}$: for $K_i=5.56$, $K_d=+0.171$; for $K_i=1.44$, $K_d=-0.104$.
Item (4) — choice and commentary
Select the physical set. The branch $K_i=1.44$ gives a negative derivative gain $K_d=-0.104$ (a non-minimum-phase RHP zero at $+9.6$ and a controller that is not a proper lead) — reject it. Choose $$\boxed{K_p=14.97,\quad K_i=5.56,\quad K_d=0.171.}$$
Model vs actual. The third pole at $-K_i=-5.56$ is cancelled by the controller zero at $-K_i$, so it does not appear in the response; the remaining closed-loop zero at $-1/K_d=-5.85$ is far from the dominant pair, so its effect is small. The actual response therefore tracks the second-order model closely — slightly faster rise and a touch more overshoot than the ideal because a finite left-half-plane zero always adds a little lead. (The rejected set would have placed an uncancelled RHP zero, badly degrading the transient.)
Q8 closed-loop poles at the chosen gains: dominant pair $-2\pm j3.31$ ($\zeta=0.52,\omega_n=3.87$) and a third pole at $-5.56$ cancelled by the controller zero; a far zero at $-5.85$ remains.
