Question 8 of 8: PID Controller 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. 98-Phys-B5 Control, National Examination December 2017
— a three-hour closed-book examination with one double-sided handwritten formula/notes
sheet permitted and an approved calculator. Questions 1 and 2 are compulsory; candidates choose
three of the remaining six (3–8). Every question is nonetheless answered in full below so
the paper remains a complete study resource. All eight questions carry equal value (20 marks
each).
Reference texts. N. S. Nise, Control Systems Engineering, 7th ed.
(root locus, Routh–Hurwitz, frequency-response lag/lead design, Nyquist stability,
state-space representation, steady-state errors); K. Ogata, Modern Control Engineering,
5th ed. (frequency-response compensator design, controllability/observability, PID pole
placement).
Question 8: PID Controller Design by Pole Placement with Pole–Zero Cancellation (20 marks)
Find. $\zeta$, $\omega_n$; $K_p,K_i,K_d$ (one of two solution sets, with
justification); the PID controller's two real zeros.
Approach. Convert the PO/settling-time specs to $\zeta,\omega_n$. The
closed-loop characteristic equation is a CUBIC (3rd order, from the controller's integrator);
match its quadratic factor to the desired 2nd-order pair and let the THIRD (real) pole $p_3$ be
a free parameter. The pole–zero cancellation condition (one PID zero coincides with
$p_3$) supplies the missing equation, which turns out to be a cubic in $p_3$ with a trivial root
at $0$ and two genuine candidates — choose the one giving a physically implementable
(positive) derivative gain.
Step 1 — $\zeta$ and $\omega_n$ from the specs.
$$\zeta=\frac{-\ln(0.10)}{\sqrt{\pi^2+\ln^2(0.10)}}\approx\boxed{0.591},\qquad
\omega_n=\frac{4}{\zeta\times1}\approx\boxed{6.766\ \text{rad/s}}.$$
Step 2 — set up the pole-placement + cancellation equations.
The forward path is $\big(K_ds^2+K_ps+K_i\big)\cdot5/\big[s(s^2{+}10s{+}15)\big]$, giving the
closed-loop characteristic cubic
$$s^3+(10{+}5K_d)s^2+(15{+}5K_p)s+5K_i=0.$$
Writing this as $(s^2+2\zeta\omega_ns+\omega_n^2)(s+p_3)$ and matching coefficients:
$$K_d=\frac{p_3+2\zeta\omega_n-10}{5},\quad K_p=\frac{\omega_n^2+2\zeta\omega_np_3-15}{5},\quad
K_i=\frac{\omega_n^2p_3}{5}.$$
Pole–zero cancellation requires $s=-p_3$ to be a root of the PID numerator
$K_ds^2+K_ps+K_i$, i.e. $K_dp_3^2-K_pp_3+K_i=0$. Substituting the three expressions above turns
this into a CUBIC in $p_3$ alone:
$$0.2p_3^3-2.0p_3^2+3.0p_3=0\ \Rightarrow\ p_3=0,\ 1.838,\ 8.162.$$
$p_3=0$ is degenerate ($K_i=0$, no integral action — defeats the purpose of PID and is
discarded).
Step 3 — choose the physically implementable solution set.
$$p_3=1.838:\quad K_p=9.097,\ K_i=16.83,\ K_d=\boxed{-0.0325}\ \ (\text{negative — rejected}).$$
$$p_3=8.162:\quad K_p=19.22,\ K_i=74.74,\ K_d=+1.232\ \ (\text{positive — PHYSICALLY
IMPLEMENTABLE}).$$
The first set demands a slightly NEGATIVE derivative gain, which is atypical for a standard PID
implementation (it would act like extra phase lag rather than lead) and is not the intended
design. Choosing $\boxed{p_3=8.162}$:
$$\boxed{K_p\approx19.22,\qquad K_i\approx74.74,\qquad K_d\approx1.232.}$$
Step 4 — verify and identify the PID's two real zeros.
Rebuilding the closed-loop characteristic cubic with these gains and solving numerically gives
poles at exactly $-8.162$ (real, cancelling) and $-4\pm j5.458$ — matching
$-\zeta\omega_n\pm j\omega_n\sqrt{1{-}\zeta^2}=-4\pm j5.458$ exactly. The PID numerator
$K_ds^2+K_ps+K_i=1.232s^2+19.22s+74.74$ factors as
$$\boxed{G_c(s)=K_p+\frac{K_i}{s}+K_ds=\frac{1.232(s+8.162)(s+7.430)}{s}},$$
so the controller's two real zeros are $\boxed{-8.162}$ (the one that cancels the closed-loop's
third pole) and $\boxed{-7.430}$ (the surviving zero, which shapes the closed-loop response
alongside the dominant complex pair).
