Question 8 of 8: PID design by pole placement with pole–zero cancellation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, December 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, $\zeta$–resonant-peak and phase-margin 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), lead/lag and PID design (Ch. 9–11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations, pole placement and frequency-domain compensator design. All block diagrams, root loci, pole–zero maps, Bode and Nyquist plots, closed-loop magnitude curves and step responses below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$, the phase-margin chart uses $\Phi_m\approx100\,\zeta$, 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 from the exact transfer functions given in the paper; the supplied plots are used only to corroborate $\zeta$, $\Phi_m$ and the crossover frequencies. Where a printed figure’s hand-read tick differs from the exact value, the exact value governs and the gap is flagged.
Question 8 — PID design by pole placement with pole–zero cancellation [20]
Figure Q8.1 — parallel-PID controller $K_p+K_i/s+K_d s$ ahead of the positioning process $5/(s^2+10s+15)$, unit feedback.
Given. $P(s)=5/(s^2+10s+15)$ (poles $-1.84,-8.16$); $PO=10\%$, $T_{settle(\pm2\%)}=1$ s. Find. $\zeta,\omega_n$, then $K_p,K_d,K_i$. Approach. map the transient specs to a dominant pair, write the third-order closed-loop characteristic, and force a controller zero to cancel the non-dominant pole.
Closed-loop characteristic. $1+G_cP=0$ gives $s(s^2+10s+15)+5(K_ds^2+K_ps+K_i)=0$, i.e. $$s^3+(10+5K_d)s^2+(15+5K_p)s+5K_i=0.$$
Impose the desired poles. Write the target as $(s^2+8s+45.8)(s+p_3)$ and match: $$10+5K_d=8+p_3,\quad 15+5K_p=45.8+8p_3,\quad 5K_i=45.8\,p_3.$$
Cancellation condition. Requiring a controller zero at $-p_3$ ($K_dp_3^2-K_pp_3+K_i=0$) and substituting the three relations reduces to $$p_3^2-10p_3+15=0\ \Rightarrow\ p_3=8.16\ \text{or}\ 1.84.$$ These are exactly the plant poles — the PID zero cancels a plant pole.
Choose the solution. $p_3=1.84$ gives $K_d=-0.03\lt0$ (a non-physical negative rate gain and a right-half-plane controller zero), so it is rejected. Take $\boxed{p_3=8.16}$, which cancels the fast plant pole and leaves the third closed-loop pole far to the left for clean dominance: $$\boxed{K_p=19.2,\quad K_d=1.23,\quad K_i=74.7.}$$
Factored controller. $G_c(s)=\dfrac{K_ds^2+K_ps+K_i}{s}=\dfrac{1.23\,(s^2+15.6s+60.6)}{s}=\dfrac{1.23\,(s+8.16)(s+7.43)}{s}$ — two real zeros at $\boxed{-8.16\ \text{and}\ -7.43}$. The zero at $-8.16$ cancels the closed-loop pole there, leaving the dominant pair $-4\pm j5.46$ ($\zeta=0.59,\omega_n=6.77$) as designed.
Q8 closed-loop pole–zero map: dominant pair $-4\pm j5.46$ ($\zeta=0.59$), third pole $-8.16$ cancelled by the controller zero at $-8.16$; residual zero at $-7.43$.
Dominant-model unit-step response: $10\%$ overshoot and $\pm2\%$ settling at $1.0$ s, meeting both specifications.
