PID Controller Design by Pole Placement: 17-Phys-B5 Systems and Control, May 2016, Q5

Question 5: PID Controller Design by Pole Placement (20 marks)

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 path $G_c(s)G_p(s)=K_p(1+K_i/s)\cdot\dfrac{1}{s^2+7s+5}$; feedback path $H(s)=K_ds+1$ (tachometer/rate feedback wrapped around the plant output).

Find. The closed-loop transfer function and characteristic equation; the target $\zeta,\omega_n$; and the controller gains $K_p,K_d,K_i$ that place the poles as specified.

Approach. Combine $G_c(s)$ and $G_p(s)$ into one forward-path fraction, close the loop with $H(s)=K_ds+1$, then match the resulting cubic's coefficients to $(s^2+2\zeta\omega_ns+\omega_n^2)(s+K_i)$ — the pole-zero-cancellation constraint folds directly into matching the constant term.

Part 1) — closed-loop TF and characteristic equation. $$G_c(s)G_p(s)=\frac{K_p(s+K_i)}{s(s^2+7s+5)},\qquad \frac{Y}{R}=\frac{G_cG_p}{1+G_cG_pH}=\frac{K_p(s+K_i)}{s(s^2+7s+5)+K_p(s+K_i)(K_ds+1)}.$$ Expanding the denominator gives the Characteristic Equation $$\boxed{Q(s)=s^3+(K_dK_p+7)s^2+(K_dK_iK_p+K_p+5)s+K_iK_p=0.}$$
Part 2) — target $\zeta,\omega_n$. $PO=10\%\Rightarrow \zeta=\dfrac{-\ln0.10}{\sqrt{\pi^2+\ln^20.10}}=\boxed{0.591}$; the $\pm2\%$ settling-time formula $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=1$ s gives $\omega_n=4/(0.591\times1)=\boxed{6.77\ \text{rad/s}}$.
Part 3) — pole placement with the cancellation constraint. Requiring $Q(s)\equiv(s^2+2\zeta\omega_ns+\omega_n^2)(s+K_i)$ and matching coefficients term by term: the CONSTANT term gives $K_iK_p=\omega_n^2K_i\Rightarrow\boxed{K_p=\omega_n^2=45.8}$ immediately (the $K_i$ cancels, which is exactly the pole-zero-cancellation condition doing its job). The $s^2$ and $s^1$ coefficients then give two simultaneous equations in $K_d,K_i$: $$K_dK_p+7=K_i+2\zeta\omega_n,\qquad K_dK_iK_p+K_p+5=2\zeta\omega_nK_i+\omega_n^2,$$ which reduce to a quadratic in $K_i$ with two admissible positive roots; the smaller, better-conditioned solution is $$\boxed{K_p=45.8,\qquad K_d=0.0395,\qquad K_i=0.807.}$$ Substituting back, the closed-loop poles are exactly $-4.00\pm j5.458$ (matching $\zeta\omega_n=4.00$, $\omega_d=\omega_n\sqrt{1-\zeta^2}=5.458$) and a real pole at $-0.807=-K_i$, confirming the intended cancellation. (A second valid root of the same quadratic, $K_d=0.157$, $K_i=6.19$, places the same dominant pair with the cancelling pole pushed further left; either is a defensible design.)

Final results — Question 5
Quantity	Value
$Q(s)$	$s^3+(K_dK_p{+}7)s^2+(K_dK_iK_p{+}K_p{+}5)s+K_iK_p$
$\zeta$, $\omega_n$	$0.591$, $6.77$ rad/s
$K_p$	$45.8$
$K_d$	$0.0395$
$K_i$	$0.807$
Resulting poles	$-4.00\pm j5.46$ and $-0.807$

Topic: PID/rate-feedback pole placement with a deliberate pole-zero cancellation.

Key relations: matching a characteristic polynomial to $(s^2+2\zeta\omega_ns+\omega_n^2)(s+p_3)$ term by term is the standard way to place BOTH the dominant complex pair and a specified real pole simultaneously; the constant-term match is what enforces $p_3=K_i$.

Why this works: the integral controller contributes a zero at $s=-K_i$; choosing the closed-loop's third pole to sit at exactly that same location makes that mode's residue vanish in the transfer function, leaving a response that behaves, to a very good approximation, like the pure second-order dominant pair — the entire point of the “pole-zero cancellation” instruction.

Common pitfall: forgetting that matching THREE polynomial coefficients against three unknowns is generically NONLINEAR here (the $K_dK_iK_p$ cross-term), so more than one algebraically valid $(K_d,K_i)$ pair can exist; both should be checked for positivity and a sensible-magnitude cancelling pole before choosing one.

Source: Nise, Control Systems Engineering, Ch. 9 (PID design by pole placement); Ogata, Modern Control Engineering, Ch. 10.

Question 5 of 8: PID Controller Design by Pole Placement

Question 5: PID Controller Design by Pole Placement (20 marks)