Question 4 of 8: PID with rate feedback by pole–zero cancellation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 07-Elec-A2 Systems & Control, May 2014, 3 hours, closed book (approved calculator plus one signed 8.5 × 11" formula sheet). Questions 1 and 2 are compulsory; answer three of the remaining six (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 and stability (Ch. 6), steady-state error and error constants (Ch. 7), root locus (Ch. 8), PID and lag/lead design (Ch. 9–11), frequency response, Nyquist and gain/phase margins (Ch. 10–11), state space, controllability/observability and pole placement (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of dominant-poles modelling and Mason’s rule; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, Bode/Nyquist plots and signal-flow graphs below are redrawn as inline figures.
Reading the exam figures. Q1 supplies an open-loop Bode plot and Q6 supplies open-loop Bode plots of the process. Where a transfer function is given exactly in the text (Q1’s system is the same plant as Q2, stated exactly there), every graphical reading is also confirmed analytically and the exact model governs. Q6 is a pure graph-reading problem (no closed form given); its numeric reads are approximate and are flagged accordingly. Note the printed exam mislabels the header of the last two pages “May 2013” — this is the May 2014 sitting throughout.
Question 4 — PID with rate feedback by pole–zero cancellation [5 + 5 + 10]
[Figure not reproduced: Figure Q4.1 (redrawn) — P+I forward path, inner rate ($K_d s$) feedback, unity outer loop. See the official exam paper.]
Part A — Closed-loop transfer function [5]
Approach. Close the inner rate loop first, cascade the PI controller, then close the unity outer loop.
Inner rate loop. Around $P(s)=\dfrac{2}{(s+0.1)^2}$ with feedback $K_d s$: $$\frac{P}{1+P\,K_d s}=\frac{2}{(s+0.1)^2+2K_d s}.$$
Forward path (add PI). With $C(s)=\dfrac{K_p s+K_i}{s}$, $$G_{fwd}(s)=\frac{2(K_p s+K_i)}{s\,[(s+0.1)^2+2K_d s]}.$$
Close the unity loop. $G_{cl}=\dfrac{G_{fwd}}{1+G_{fwd}}$: $$\boxed{G_{cl}(s)=\frac{2(K_p s+K_i)}{s^3+(0.2+2K_d)s^2+(0.01+2K_p)s+2K_i}.}$$ The open loop has a free integrator ($1/s$ from the PI), so the system is Type 1 — $e_{ss\%}=0$ to a step is guaranteed. The closed-loop zero is at $s=-K_i/K_p$.
Part B — Dominant-pole model [5]
Damping from overshoot. $\zeta=\dfrac{-\ln(0.10)}{\sqrt{\pi^2+\ln^2(0.10)}}=\boxed{0.591}$.
Natural frequency from settling time. $T_s=\dfrac{4}{\zeta\omega_n}=5$ s gives $\zeta\omega_n=0.8$, so $\omega_n=\dfrac{0.8}{0.591}=\boxed{1.353\ \text{rad/s}}$ ($\omega_n^2=1.831$).
Model. Dominant poles $s=-0.8\pm j1.09$; $G_m(s)=\dfrac{1.831}{s^2+1.6s+1.831}$.
Part C — Controller gains by pole–zero cancellation [10]
Approach. Force the cubic characteristic polynomial to factor as $(s^2+2\zeta\omega_n s+\omega_n^2)(s+p_3)$ with the third pole $p_3$ placed exactly on the closed-loop zero $-K_i/K_p$, so numerator and third factor cancel and $G_{cl}=G_m$.
Design pole–zero map: dominant pair $-0.8\pm j1.09$; the third closed-loop pole at $-0.00625$ cancels the closed-loop zero at the same location.
Match coefficients. Expanding $(s^2+1.6s+1.831)(s+p_3)$ and equating to the characteristic polynomial: $$0.2+2K_d=1.6+p_3,\quad 0.01+2K_p=1.831+1.6p_3,\quad 2K_i=1.831\,p_3.$$
Unity numerator match. For $G_{cl}=G_m$ the surviving numerator gain must be $\omega_n^2$: $2K_p=\omega_n^2\Rightarrow K_p=\dfrac{1.831}{2}=\boxed{0.916}$.
Third pole. Substituting $2K_p=\omega_n^2$ into the $s^1$ match gives $p_3=\dfrac{0.01}{2\zeta\omega_n}=\dfrac{0.01}{1.6}=\boxed{0.00625}$ — a slow pole that cancels the zero $-K_i/K_p$.
Remaining gains. From the $s^0$ and $s^2$ matches: $$K_i=\tfrac{1}{2}\omega_n^2 p_3=\boxed{0.00572},\qquad K_d=\tfrac{1}{2}(1.6+p_3-0.2)=\boxed{0.703}.$$ Check: $K_i/K_p=0.00572/0.916=0.00625=p_3$ ✓ (zero exactly cancelled).
