Top-down controller design by pole matching: 22-Elec-A2 Systems and Control, May 2015, Q3

Question 3 — Top-down controller design by pole matching [7 + 5 + 8]

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. $G(s)=\dfrac{10}{s(s+2)}$ (one free integrator $\Rightarrow$ Type 1, so a step error of 0 is automatic); $G_c(s)=\dfrac{a_1s+a_0}{b_1s+1}$.

Find. $G_{cl}(s)$ and $Q(s)$; the desired cubic; and $a_1,a_0,b_1$.

Part 1 — Closed-loop transfer function [7]

Forward path. $G_cG=\dfrac{10(a_1s+a_0)}{(b_1s+1)\,s(s+2)}$.
Close the unity loop. $G_{cl}=\dfrac{G_cG}{1+G_cG}=\dfrac{10(a_1s+a_0)}{(b_1s+1)s(s+2)+10(a_1s+a_0)}$.
Characteristic equation. Expanding the denominator,$$Q(s)=b_1s^3+(2b_1+1)s^2+(2+10a_1)s+10a_0=0.$$

Part 2 — Desired characteristic equation [5]

Dominant damping. $PO=5\%\Rightarrow\zeta=0.690$.
Dominant natural frequency. $T_{settle}=\dfrac{4}{\zeta\omega_n}=4$ s $\Rightarrow\zeta\omega_n=1$, so $\omega_n=\dfrac{1}{0.690}=1.449$ rad/s ($\omega_n^2=2.10$). Dominant pair: $-1\pm j1.05$.
Third pole. Ten times farther left than the dominant real part ($-1$): $s=-10$.
Desired cubic. $(s^2+2s+2.10)(s+10)$:$$Q_{des}(s)=s^3+12s^2+22.1s+21=0.$$ (The Type-1 plant already gives $e_{ss}(\text{step})=0$.)

Part 3 — Matching and controller identification [8]

Approach. Divide $Q(s)$ by $b_1$ to make it monic, then equate coefficients with $Q_{des}$.

$s^2$ coefficient. $\dfrac{2b_1+1}{b_1}=12\Rightarrow2+\dfrac1{b_1}=12\Rightarrow\boxed{b_1=0.1}$.
$s^0$ coefficient. $\dfrac{10a_0}{b_1}=21\Rightarrow10a_0=2.1\Rightarrow\boxed{a_0=0.21}$.
$s^1$ coefficient. $\dfrac{2+10a_1}{b_1}=22.1\Rightarrow2+10a_1=2.21\Rightarrow\boxed{a_1=0.021}$.
Controller nature. The zero is at $-a_0/a_1=-10$ and the pole at $-1/b_1=-10$ — they coincide. Hence $G_c(s)=\dfrac{0.021(s+10)}{0.1(s+10)}=0.21$: the compensator collapses to a $\boxed{\text{pure Proportional controller, }K_p=0.21}$. Structurally it is the degenerate limit of a lead/lag network (pole/zero ratio = 1); the controller pole supplies the desired third pole at $-10$ while its zero cancels it, leaving a clean second-order dominant response.

Check: the third pole never appears. Because the compensator zero cancels its pole at $-10$, the realised closed loop is exactly $G_{cl}(s)=\dfrac{2.1}{s^2+2s+2.1}$ — second order, DC gain 1, poles $-1\pm j1.05$. This is the ideal outcome (no lingering third pole), which is why a plain proportional gain meets all three specifications here.

Quantity	Result
$Q(s)$	$b_1s^3+(2b_1+1)s^2+(2+10a_1)s+10a_0$
Desired cubic	$s^3+12s^2+22.1s+21$
$b_1$ / $a_0$ / $a_1$	$0.1$ / $0.21$ / $0.021$
Controller	$G_c=0.21$ (pure P; pole–zero cancel at $-10$)

Concept & Theory

Topic. Controller synthesis by characteristic-equation matching.

Key relations. Dominant model $s^2+2\zeta\omega_ns+\omega_n^2$ with $\zeta\omega_n=4/T_s$; augment with a non-dominant pole; equate monic coefficients with the plant+controller characteristic polynomial. (Nise Ch. 9; Ogata §10.)

Why this works. Three controller unknowns give three characteristic-polynomial coefficients, so any monic cubic target is reachable. The specific numbers here happen to force a coincident pole and zero, which is not an error but the design collapsing to its simplest realisation.

Common pitfall. Reporting a “lead/lag” from the raw form without checking the pole/zero ratio — a ratio of 1 means proportional. Also, do not expect a third closed-loop pole once the cancellation is recognised.

Source. Nise 7e, §9.2–9.5.

Question 3 of 8: Top-down controller design by pole matching

Question 3 — Top-down controller design by pole matching [7 + 5 + 8]

Part 1 — Closed-loop transfer function [7]

Part 2 — Desired characteristic equation [5]

Part 3 — Matching and controller identification [8]