NivaarExam PrepOfficial exam papers ↗

22-Elec-A2 Systems and Control · December 2016

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, 07-Elec-A2 Systems & Control, December 2016, 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 ζ–overshoot and ζ–resonant-peak design plots are supplied). 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, Bode, Nyquist, gain/phase margins (Ch. 10), lead–lag and PID design (Ch. 9–11), state space and Mason’s rule (Ch. 3, 5); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and second-order correlations. All block diagrams, root loci, pole–zero maps, Bode sketches, Nyquist plots and the 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Questions 4, 6 and 7 are read from printed Bode/response plots, so those values carry normal chart-reading tolerance; the s-domain results (Q1, Q2, Q3, Q5, Q8) are exact.

Question 8 — Series-PID design by pole placement [20]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

−+R(s)Kp(1+Ki/s)(Kd s+1)series PIDP(s)1/(s^2+10s+15)Y(s)Figure Q8.1 - series-PID positioning loop
Figure Q8.1 — series-PID positioning loop, $C(s)=K_p(1+K_i/s)(K_d s+1)$, process $1/(s^2+10s+15)$.

Item (1) — closed-loop TF and characteristic equation

  1. Open loop. $C(s)=\dfrac{K_p(s+K_i)(K_d s+1)}{s}$, so $L(s)=\dfrac{K_p(s+K_i)(K_d s+1)}{s(s^2+10s+15)}$.
  2. Closed loop (unity feedback): $T(s)=\dfrac{K_p(s+K_i)(K_d s+1)}{s(s^2+10s+15)+K_p(s+K_i)(K_d s+1)}$.
  3. Characteristic equation. Expanding, $$\boxed{Q(s)=s^3+(10+K_pK_d)s^2+\big(15+K_p(1+K_iK_d)\big)s+K_pK_i=0.}$$

Item (2) — target $\zeta,\omega_n$

  1. Damping. $PO=10\%\Rightarrow\zeta=\dfrac{-\ln0.1}{\sqrt{\pi^2+\ln^2 0.1}}=\boxed{0.591}$.
  2. Natural frequency. $T_{settle(2\%)}=\dfrac{4}{\zeta\omega_n}=0.9\Rightarrow\zeta\omega_n=4.44$, so $\omega_n=\dfrac{4.44}{0.591}=\boxed{7.52}$ rad/s. Desired pair: $s^2+8.89s+56.5$.

Items (3)–(4) — gains by pole placement

Approach. require $Q(s)=(s^2+2\zeta\omega_n s+\omega_n^2)(s+K_i)$, so the third pole is at $-K_i$ and cancels the closed-loop zero at $-K_i$; match coefficients.

Q8 - CL poles: dominant pair (zeta=0.59,wn=7.52) + pole -1.84 cancels zero -1.84-10-50-55ReIm
Closed-loop poles: dominant pair ($\zeta=0.59,\ \omega_n=7.52$) plus the real pole at $-K_i$ which cancels the $(s+K_i)$ zero; the remaining zero at $-1/K_d$ is pushed far left (chosen solution) so the response is nearly ideal second order.
  1. Match the $s^0$ term. $K_pK_i=\omega_n^2K_i\Rightarrow \boxed{K_p=\omega_n^2=56.5}$.
  2. Match $s^2$ and $s^1$. $10+K_pK_d=K_i+8.89$ and $15+K_p(1+K_iK_d)=56.5+8.89K_i$. Eliminating $K_d$ gives the promised quadratic $$K_i^2-10K_i+15=0\ \Rightarrow\ K_i=8.16\ \text{or}\ 1.84.$$
  3. Two candidate designs. $K_i=8.16$: $K_d=0.125$ (leftover zero $-1/K_d=-8.02$); $K_i=1.84$: $K_d=0.0128$ (leftover zero $-1/K_d=-77.8$). Both give $K_p=56.5$ and exact cancellation of the third pole.
  4. Choice (item 4). After the $(s+K_i)$ cancellation the closed loop is $T(s)=\dfrac{K_pK_d(s+1/K_d)}{s^2+8.89s+56.5}$; the leftover zero at $-1/K_d$ shapes the transient. For $K_i=1.84$ that zero sits at $-77.8$ — far from the dominant pair — so the actual response is almost exactly the target second-order model. For $K_i=8.16$ the zero at $-8.0$ is close enough to add noticeable extra overshoot. $$\boxed{K_p=56.5,\quad K_i=1.84,\quad K_d=0.0128.}$$
QuantityValue
$\zeta,\omega_n$$0.591,\ 7.52$ rad/s
$K_p$$56.5$
Chosen $K_i,K_d$$1.84,\ 0.0128$ (zero at $-77.8$)
Rejected set$K_i=8.16,\ K_d=0.125$ (zero at $-8.0$, more overshoot)
$e_{ss(step)}$$0\%$ (integral action, Type 1)
Back to the paper →