Question 1 of 6: Proportional-integral design for a double-lag plant

Question 1: Proportional-integral design for a double-lag plant (25 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. A unity-feedback loop in which the controller drives the plant and a disturbance is subtracted at the plant output.

Find. The gains $K_p$ and $K_i$ that meet the margin specification at the largest practical crossover, the shapes of the two closed-loop magnitude responses, and the steady-state output for constant reference and disturbance.

[Figure not reproduced: Question 1 feedback loop, redrawn from the exam page. The disturbance d is SUBTRACTED at the plant output: y = P(s)u - d. See the official exam paper.]

Approach. Write the loop transfer function as $C(s)P(s)$ with the PI zero placed explicitly, use the phase condition to locate the largest crossover that still leaves $45^\circ$ of margin, then use the magnitude condition $|L(j\omega_{gc})| = 1$ to fix the gain; finally read the two closed-loop transfer functions off the block diagram and take their DC limits.

1. Part (a) — write the loop gain in zero-pole form. Factoring the proportional gain out of the controller exposes the single design freedom that a PI controller really has, the location of its zero: $$C(s) = K_p\,\frac{s + z}{s}, \qquad z = \frac{K_i}{K_p},$$ so the open-loop transfer function is $$L(s) = C(s)P(s) = \frac{16K_p\,(s+z)}{s\,(s+2)^2}.$$ The loop is Type 1: one pole at the origin supplied by the integrator, and two plant poles at $s = -2$.
2. Write the phase condition. Along the imaginary axis the three factors contribute independently, $$\angle L(j\omega) = \arctan\!\frac{\omega}{z} - 90^\circ - 2\arctan\!\frac{\omega}{2},$$ so the phase margin is $$\mathrm{PM}(\omega_{gc}) = 180^\circ + \angle L(j\omega_{gc}) = 90^\circ + \arctan\!\frac{\omega_{gc}}{z} - 2\arctan\!\frac{\omega_{gc}}{2}.$$ Only the first two terms are at the designer's disposal; the last term is the plant's own phase lag and grows without bound as the crossover is pushed out.
3. Establish the ceiling on the crossover. The zero can contribute at most $+90^\circ$, and it does so only in the limit $z \to 0$, which is the same as switching the integrator off. Substituting that best case gives $$90^\circ + 90^\circ - 2\arctan\!\frac{\omega}{2} \ge 45^\circ \;\Longrightarrow\; \arctan\!\frac{\omega}{2} \le 67.5^\circ,$$ $$\boxed{\;\omega_{gc} \lt 2\tan 67.5^\circ = 4.828\ \text{rad/s}\;}$$ No PI controller whatever can hold $45^\circ$ of margin beyond this frequency, so "as large as feasible" means approaching 4.83 rad/s without letting the integral action collapse.
4. Choose the zero, then solve for the crossover. Pushing $z$ toward zero buys crossover but destroys the low-frequency disturbance rejection that the integrator was added for, so the design rule adopted here is the standard one: place the PI zero a decade below the intended crossover, $z = \omega_{gc}/10$, which recovers $\arctan 10 = 84.29^\circ$ of the available $90^\circ$ while leaving the integral action fully effective. The phase condition then closes on a single unknown: $$90^\circ + 84.289^\circ - 2\arctan\!\frac{\omega_{gc}}{2} = 45^\circ \;\Longrightarrow\; \arctan\!\frac{\omega_{gc}}{2} = 64.645^\circ,$$ $$\boxed{\;\omega_{gc} = 2\tan 64.645^\circ = 4.220\ \text{rad/s}, \qquad z = \frac{K_i}{K_p} = 0.4220\ \text{rad/s}\;}$$ This sits at 87 % of the theoretical ceiling.
5. Apply the magnitude condition to fix the gain. At the crossover the loop magnitude must be unity. With $\omega_{gc} = 4.2205$ and $z = 0.42205$, $$|P(j\omega_{gc})| = \frac{16}{\omega_{gc}^2 + 4} = \frac{16}{21.813} = 0.7335, \qquad |C(j\omega_{gc})| = K_p\,\frac{\sqrt{\omega_{gc}^2 + z^2}}{\omega_{gc}} = 1.00499\,K_p,$$ and setting the product to one, $$\boxed{\;K_p = \frac{1}{0.7335 \times 1.00499} = 1.357, \qquad K_i = z\,K_p = 0.5725\;}$$ so $C(s) = (1.357s + 0.5725)/s$.
6. Confirm the design. Evaluating the loop directly at the design frequency gives $|L(j4.2205)| = 1.000$ and $\angle L(j4.2205) = -135.00^\circ$, i.e. exactly $45.0^\circ$ of phase margin. The closed-loop characteristic polynomial $s(s+2)^2 + 16K_p s + 16K_i = s^3 + 4s^2 + 25.70s + 9.160$ has roots $$s = -1.812 \pm j4.589, \qquad s = -0.3763,$$ all in the left half-plane, so the design is stable. The slow real pole at $-0.376$ is the integrator's own root pulled off the origin by the PI zero — it is the price of the Type-1 behaviour and it dominates the long tail of the step response.
7. Compare with the obvious alternative. A designer who instead cancels one plant pole ($z = 2$) obtains $L = 16K_p/[s(s+2)]$, whose phase margin condition $90^\circ - \arctan(\omega/2) = 45^\circ$ forces $\omega_{gc} = 2$ rad/s and $K_p = 0.354$, $K_i = 0.707$. That is a legitimate design, but its bandwidth is less than half of the one above, so it fails the “as large as feasible” part of the specification. Cancellation is attractive because the algebra is easy, not because it is fast.
8. Part (b) — identify the two closed-loop transfer functions. From the block diagram, $e = r - y$ and $y = P(s)C(s)e - d$, so $$y = \underbrace{\frac{L}{1+L}}_{T(s)}\,r \;-\; \underbrace{\frac{1}{1+L}}_{S(s)}\,d , \qquad L(s) = C(s)P(s).$$ $T$ is the complementary sensitivity and $S$ the sensitivity; they satisfy $S + T = 1$ at every frequency, which is what makes the two sketches mirror images about the 0 dB line.
9. Sketch the magnitudes. Because $L$ has one pole at the origin, $|S(j\omega)| \approx \omega/(4K_i) $ at low frequency — a $+20$ dB/decade climb out of zero — while $|T| \approx 1$ (0 dB). Near $\omega_{gc}$ the two curves cross at about $-6$ dB and $S$ peaks; the standard estimate for a $45^\circ$ margin is $$M_s \approx \frac{1}{2\sin(\mathrm{PM}/2)} = \frac{1}{2\sin 22.5^\circ} = 1.31 \;(2.32\ \text{dB}).$$ Beyond the crossover $L$ dies away, so $T \to L$ and rolls off at $-40$ dB/decade (relative degree two), while $S \to 1$, i.e. flat at 0 dB — at high frequency the loop is simply too slow to fight the disturbance and passes it straight through.

10. Part (c) — take the DC limits. For constant inputs the final-value theorem applies to each channel separately. The integrator makes $|L(j\omega)| \to \infty$ as $\omega \to 0$, hence $$T(0) = \lim_{s\to 0}\frac{L}{1+L} = 1, \qquad S(0) = \lim_{s\to 0}\frac{1}{1+L} = 0,$$ and therefore $$\boxed{\;y(\infty) = T(0)\,r_0 - S(0)\,d_0 = r_0\;}$$ The steady-state output equals the reference regardless of the size of the constant disturbance, and regardless of $K_p$ and $K_i$. The integral action, not the gain, is what earns this: it holds a non-zero control signal at zero error, and that offset is exactly what cancels $d_0$. Had the disturbance entered at the plant input instead, the same argument would still give $y(\infty) = r_0$, but the required control offset would be $u(\infty) = d_0 + $ the term needed to hold the output, a distinction worth checking on every sitting because the two block diagrams look almost identical at a glance.