Lag-controller design in the frequency domain: 22-Elec-A2 Systems and Control, May 2015, Q6

Question 6 — Lag-controller design in the frequency domain [15 + 5]

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. Uncompensated Type 0, $K_{pos}=G(0)=10$; targets $e_{ss}=2\%\Rightarrow K_{pos,c}=49$, $\Phi_m=45^\circ$ at $\omega_{cp}=4$ rad/s. At $\omega=4$: $|G(j4)|=3.43$, $\angle G(j4)=-111.1^\circ$.

Find. $a_0,a_1,b_1$, $G_c(s)$, and the compensated step specs.

Part 1 — Lag-controller calculation [15]

Approach. The DC gain sets the error constant; the values at the target crossover ($|G_cG|=1$, $\angle G_cG=-135^\circ$) then fix the two time constants.

DC gain for the error spec. $K_{pos,c}=G_c(0)\,G(0)=a_0(10)=49\Rightarrow\boxed{a_0=4.9}$.
Requirements at $\omega_{cp}=4$. For $|G_cG(j4)|=1$ with $|G(j4)|=3.43$: $|G_c(j4)|=\dfrac{1}{3.43}=0.292$. For $\angle G_cG(j4)=-135^\circ$ (i.e. $PM=45^\circ$) with $\angle G(j4)=-111.1^\circ$: $\angle G_c(j4)=-23.9^\circ$ — a genuine lag contribution.
Solve for $a_1,b_1$. With $G_c(j4)=\dfrac{a_0+j4a_1}{1+j4b_1}$, imposing the magnitude and phase above gives $$\boxed{a_1=2.59,\qquad b_1=9.80.}$$
Controller. $$\boxed{G_c(s)=\frac{2.59\,s+4.9}{9.80\,s+1}}$$ — zero at $-a_0/a_1=-1.89$, pole at $-1/b_1=-0.102$. The pole is nearer the imaginary axis than the zero, confirming a lag network. DC gain $4.9$ (raises $K_{pos}$ to 49); high-frequency gain $a_1/b_1=0.264$ (the mid-band attenuation that pulls the crossover down to 4 rad/s).

Uncompensated (blue) vs lag-compensated (red) open loop: the lag lifts the low-frequency gain for the $2\%$ error while attenuating near crossover, moving $\omega_{cp}$ to $4$ rad/s with $45^\circ$ phase margin.

Lag pole–zero: pole $-0.10$ inside, zero $-1.89$ outside.

Part 2 — Compensated step estimates [5]

Damping. $\zeta\approx\dfrac{\Phi_m}{100}=0.45$ (or exactly $0.456$).
Overshoot. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}\approx\boxed{20\%}$.
Settling time. Taking $\omega_n\approx\omega_{cp}=4$ rad/s, $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}\approx\dfrac{4}{0.45(4)}=\boxed{2.2\ \text{s}}$.
Steady-state error. By construction $e_{ss}=\dfrac{1}{1+K_{pos,c}}=\dfrac{1}{50}=\boxed{2\%}$.

Quantity	Uncompensated	Compensated
$K_{pos}$ / $e_{ss}$	$10$ / $9.1\%$	$49$ / $2\%$
$\Phi_m$ / $\omega_{cp}$	$23.5^\circ$ / $8.83$	$45^\circ$ / $4$ rad/s
$PO$ / $T_{settle}$	$\approx58\%$ / —	$\approx20\%$ / $2.2$ s
Controller	$G_c(s)=\dfrac{2.59s+4.9}{9.80s+1}$ (lag)

Concept & Theory

Topic. Frequency-domain lag-compensator design.

Key relations. $K_{pos}=\lim_{s\to0}G_cG$; at crossover $|G_cG|=1$ and $\angle G_cG=-180^\circ+\Phi_m$; lag DC gain $a_0$ sets the error constant, high-frequency gain $a_1/b_1$ sets the attenuation. (Nise Ch. 11; Ogata Ch. 9.)

Why this works. A lag raises the low-frequency gain to meet the steady-state error while attenuating the magnitude near the old crossover, so the new crossover falls to a lower frequency where the plant phase is more favourable — recovering phase margin for a transient spec while fixing the error.

Common pitfall. Placing the lag corners too near crossover adds excess phase lag; here the design deliberately accepts $\approx24^\circ$ to hit the specified $\omega_{cp}=4$ and $\Phi_m=45^\circ$ exactly, so the corners sit close to crossover.

Source. Nise 7e, §11.3–11.4.

Question 6 of 8: Lag-controller design in the frequency domain

Question 6 — Lag-controller design in the frequency domain [15 + 5]

Part 1 — Lag-controller calculation [15]

Part 2 — Compensated step estimates [5]