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22-Elec-A2 Systems and Control · May 2016

Question 5 of 8: 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, May 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 candidates choose 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) — time response and second-order specs (Ch. 4), block reduction (Ch. 5), Routh–Hurwitz (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID and lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10–11), state space, controllability/observability (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling. All step responses, block diagrams, pole–zero maps, root loci, the Bode sketches and the Nyquist plot 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$; the figures on the exam paper (step responses, root loci, Bode plots) are read graphically, so the values below carry normal chart-reading tolerance.

Question 5 — PID design by pole placement [5 + 5 + 10]

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.

Given. forward controller $G_c=K_p(s+K_i)/s$, process $1/(s^2+7s+5)$, feedback $H=K_ds+1$.

Find. $Q(s)$; the target $(\zeta,\omega_n)$; and $K_p,K_d,K_i$.

R(s)−+P+IKp(1+Ki/s)Process1/(s2+7s+5)Y(s)Kd s + 1Figure Q5.1 — P+I forward control with rate (Kd s+1) feedback
Figure Q5.1: P+I forward control with rate ($K_ds+1$) feedback.

Part (1) — closed-loop transfer function

  1. Forward and loop. $G_f=\dfrac{K_p(s+K_i)}{s(s^2+7s+5)}$, $H=K_ds+1$, so $\dfrac{Y}{R}=\dfrac{K_p(s+K_i)}{s(s^2+7s+5)+K_p(K_ds+1)(s+K_i)}$.
  2. Characteristic equation. $Q(s)=s^3+(7+K_pK_d)s^2+(5+K_p+K_pK_dK_i)s+K_pK_i=0$.

Part (2) — target second-order model

  1. Damping from overshoot. $PO=10\%\Rightarrow\zeta=\boxed{0.591}$.
  2. Natural frequency from settling. $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=1\Rightarrow\zeta\omega_n=4\Rightarrow\omega_n=4/0.591=\boxed{6.77\text{ rad/s}}$. Dominant poles $-4\pm j5.46$.
Q5 — desired CL poles: dominant pair + 3rd pole (=-Ki) cancels zero-50-55ReIm
Q5.2: desired closed-loop poles — dominant pair plus a third real pole at $-K_i$ that cancels the controller zero.

Part (3) — controller gains

  1. Desired characteristic polynomial. $(s^2+8s+45.8)(s+K_i)=s^3+(K_i+8)s^2+(8K_i+45.8)s+45.8K_i$ (third pole at $-K_i$ cancels the zero at $-K_i$).
  2. Match the $s^0$ term. $K_pK_i=45.8K_i\Rightarrow \boxed{K_p=45.8}$ ($=\omega_n^2$).
  3. Match the $s^2$ term. $7+K_pK_d=K_i+8\Rightarrow K_pK_d=K_i+1$.
  4. Match the $s^1$ term. Substituting $K_pK_d=K_i+1$ gives $8K_i+45.8=5+45.8+(K_i+1)K_i\Rightarrow K_i^2-7K_i+5=0\Rightarrow K_i=6.19$ or $0.81$. Choose $\boxed{K_i=6.19}$ so the third pole at $-6.19$ stays outside the dominant pair (the $0.81$ root would dominate and spoil the design).
  5. Rate gain. $K_d=\dfrac{K_i+1}{K_p}=\dfrac{7.19}{45.8}=\boxed{0.157}$.
QuantityValue
$\zeta$ / $\omega_n$$0.591$ / $6.77$ rad/s
$K_p$$45.8$
$K_i$$6.19$
$K_d$$0.157$