Question 7 of 8: Controller Design by Pole Placement — PI + Dynamic Rate Feedback
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 17-Phys-B5 Systems and Control, National Exam, May
2019 — 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.
(transient-response specifications, root locus, Routh–Hurwitz, frequency-response and lead
compensator design, state-space representation, steady-state errors); K. Ogata, Modern
Control Engineering, 5th ed. (root-locus construction, PID/rate-feedback compensator design,
controllability/observability, Mason's Gain Formula on signal-flow graphs).
Question 7: Controller Design by Pole Placement — PI + Dynamic Rate Feedback
(20 marks)
Figure Q7.1 — single summing junction; PI controller in the forward
path; process $1/(s^2+8s+5)$; DYNAMIC rate-feedback path $(1+\tau_d s)$ multiplying $Y(s)$ back
into the same junction.
Given. $G_{PI}(s)=K_p(1+1/(\tau_i s))$; process $1/(s^2+8s+5)$; feedback
signal $Y(s)(1+\tau_d s)$ subtracted from $R(s)$ at a SINGLE summing junction.
Find. $G_{cl}(s)$ and $Q(s)$; $\zeta,\omega_n,K_{dc}$; $K_p,\tau_i,\tau_d$.
Approach. Write $E(s)=R(s)-Y(s)(1+\tau_ds)$, close the loop algebraically, and
note the forward path's own integrator already guarantees $e_{ss}(step)=0$ (so $K_{dc}=1$
automatically). Place the dominant complex pair by matching the char. equation at the target
$s=-\zeta\omega_n\pm j\omega_d$, then choose $\tau_i$ so the resulting third pole coincides with
the numerator's own zero at $s=-1/\tau_i$, cancelling it out of the transfer function.
Part 1) — closed-loop TF and $Q(s)$.
$Y(s)=\dfrac1{s^2+8s+5}K_p\!\left(1+\dfrac1{\tau_is}\right)\!\big[R(s)-Y(s)(1+\tau_ds)\big]$.
Clearing denominators and collecting $Y(s)$:
$$\boxed{G_{cl}(s)=\frac{K_p(\tau_is+1)}
{\tau_is^3+(K_p\tau_d\tau_i+10\tau_i)s^2+(K_p\tau_d+K_p\tau_i+8\tau_i)s+K_p}}$$
$$Q(s)=s^3+(K_p\tau_d+10)s^2+\left(K_p+\frac{K_p\tau_d}{\tau_i}+8\right)s+\frac{K_p}{\tau_i}=0
\ \ (\text{dividing through by }\tau_i).$$
At $s=0$, $G_{cl}(0)=K_p/K_p=1$ for ANY $K_p,\tau_i,\tau_d$ — the loop's own integrator
(from the PI controller's $1/\tau_is$ term) guarantees $e_{ss}(step)=0$ structurally.
Part 2) — $\zeta,\omega_n,K_{dc}$. $PO=5\%\Rightarrow\zeta=0.6901$;
$T_{settle(2\%)}=1$ s $\Rightarrow\omega_n=4/(\zeta\times1)=5.796$ rad/s.
$$\boxed{\zeta=0.6901,\quad\omega_n=5.796\ \text{rad/s},\quad K_{dc}=1\ \text{(automatic)}.}$$
Part 3) — pole placement with cancellation. Target dominant pair
$s=-\zeta\omega_n\pm j\omega_n\sqrt{1-\zeta^2}=-4.000\pm j4.1948$. Substituting into $Q(s)=0$
gives two linear equations in $K_p,K_p\tau_d$ for any trial $\tau_i$; searching for the $\tau_i$
that ALSO places the third root exactly at $s=-1/\tau_i$ (the pole-zero cancellation condition)
converges to
$$\boxed{\tau_i=1.4633\ \text{s},\quad K_p=33.596,\quad \tau_d=0.02034\ \text{s}}$$
(a small POSITIVE rate-feedback constant). Checking: $Q(s)$ at these
values factors as $(s+4-4.1948j)(s+4+4.1948j)(s+0.6834)$, and $1/\tau_i=0.6834$ EXACTLY matches
the third pole — confirming the cancellation, so the closed-loop TRANSFER FUNCTION reduces
to the pure 2nd-order dominant-pole form. Simulating the full closed loop directly (no
cancellation assumed) gives $PO=5.00\%$ and $T_{settle(2\%)}=1.00$ s — the design hits
both specs essentially exactly.
