Question 6 of 6: Nyquist analysis and margins for a delayed integrator

Question 6: Nyquist analysis and margins for a delayed integrator (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 with a proportional controller and a plant consisting of a one-second transport delay in series with an integrator.

Find. The polar (Nyquist) plot of the open-loop transfer function at $K = 1$, the gain and phase margins, the gain giving a $50^\circ$ phase margin, and the steady-state output for a step reference with a ramp disturbance.

Approach. A transport delay is all-pass, so magnitude and phase separate cleanly: the magnitude comes entirely from the integrator and the phase is the integrator's fixed $-90^\circ$ plus a term growing linearly with frequency. That separation makes both crossover frequencies available in closed form, with no Padé approximation and no Routh array (which is unavailable anyway, the characteristic equation being transcendental).

1. Part (a) — write the frequency response. With $K = 1$, $$L(j\omega) = \frac{e^{-j\omega}}{8j\omega}, \qquad |L(j\omega)| = \frac{1}{8\omega}, \qquad \angle L(j\omega) = -90^\circ - \omega\ \text{rad} = -90^\circ - 57.30^\circ\omega .$$ The delay contributes no attenuation at all — that is what “all-pass” means — but it rotates the phasor by an angle proportional to frequency.
2. Describe the locus. As $\omega$ increases from $0^{+}$ the magnitude falls monotonically as $1/(8\omega)$ while the phase winds clockwise without limit, so the plot is an inward spiral. At $\omega \to 0^{+}$ it starts infinitely far down the negative imaginary axis; it crosses successive axes at $\omega = \pi/2,\ \pi,\ 3\pi/2, \dots$; and it converges on the origin as $\omega \to \infty$. Unlike a rational transfer function, it does not approach the origin along a fixed asymptote — it circles in.
3. Part (b) — find the phase crossover and the gain margin. The first crossing of the negative real axis is where the phase reaches $-180^\circ$: $$-90^\circ - \omega\,\frac{180^\circ}{\pi} = -180^\circ \;\Longrightarrow\; \omega_{pc} = \frac{\pi}{2} = 1.571\ \text{rad/s}.$$ There $|L| = 1/(8\times 1.5708) = 0.07958$, so $$\boxed{\;\mathrm{GM} = \frac{1}{|L(j\omega_{pc})|} = 8\cdot\frac{\pi}{2}\cdot 1 = 4\pi = 12.57 \;(21.98\ \text{dB}) \text{ at } \omega_{pc} = 1.571\ \text{rad/s}\;}$$ Equivalently the loop stays stable until $K$ reaches $4\pi$, and the delay alone sets that limit: without it the loop would be stable for every positive gain.
4. Find the gain crossover and the phase margin. Unity magnitude requires $1/(8\omega) = 1$, so $$\omega_{gc} = \frac{1}{8} = 0.125\ \text{rad/s}, \qquad \angle L(j\omega_{gc}) = -90^\circ - 0.125\times 57.30^\circ = -97.16^\circ,$$ $$\boxed{\;\mathrm{PM} = 180^\circ - 97.16^\circ = 82.84^\circ \text{ at } \omega_{gc} = 0.125\ \text{rad/s}\;}$$ The margin is generous because the crossover sits far below the frequency at which one second of delay matters: at $0.125$ rad/s the delay has eaten only $7.2^\circ$.
5. Part (c) — design for a $50^\circ$ margin. Raising $K$ moves the crossover to the right but leaves the phase curve untouched, so first find where the phase equals $-130^\circ$: $$-90^\circ - \omega\,\frac{180^\circ}{\pi} = -130^\circ \;\Longrightarrow\; \omega\,\frac{180^\circ}{\pi} = 40^\circ \;\Longrightarrow\; \omega_{gc} = \frac{40\pi}{180} = 0.6981\ \text{rad/s}.$$ Then choose $K$ so that this frequency is the crossover, $|L(j\omega_{gc})| = K/(8\omega_{gc}) = 1$: $$\boxed{\;K = 8\,\omega_{gc} = 8 \times 0.6981 = 5.585\;}$$ This is comfortably below the stability limit $4\pi = 12.57$, as it must be, and the new crossover is 5.6 times the old one — the delay only becomes expensive close to $\omega_{pc}$.
6. Part (d) — write the two output channels. Reading the block diagram with $e = r - y$ and $y = P(s)Ke - d$, $$y = \frac{L}{1+L}\,r - \frac{1}{1+L}\,d = T(s)\,r - S(s)\,d, \qquad S(s) = \frac{8s}{8s + Ke^{-s}} .$$
7. Take the reference term. The loop is Type 1 (the plant integrator), so $L(s) \to \infty$ as $s \to 0$ and $T(0) = 1$. A step of magnitude $r_0$ therefore contributes exactly $r_0$ to the steady-state output, with zero position error.
8. Take the disturbance term. The ramp has $D(s) = d_0/s^{2}$, so by the final-value theorem $$\lim_{t\to\infty} y_d(t) = -\lim_{s\to 0} s\,S(s)\,\frac{d_0}{s^{2}} = -d_0\lim_{s\to 0}\frac{S(s)}{s} = -d_0\lim_{s\to 0}\frac{8}{8s + Ke^{-s}} = -\frac{8d_0}{K}.$$ The limit is finite — not zero, and not divergent — because $S(s)$ has exactly one zero at the origin, which cancels one power of $s$ from the ramp but not two.
9. Combine and interpret. $$\boxed{\;y(\infty) = r_0 - \frac{8\,d_0}{K}\;}$$ so with the part-(c) gain $K = 5.585$ the offset is $1.432\,d_0$. Two readings earn the marks. First, a single integrator in the loop tracks a step perfectly but can only hold a constant error against a ramp; rejecting the ramp entirely would need a second integrator, which this loop cannot afford because the delay would then make it unstable at any gain. Second, the error falls only as $1/K$, and $K$ is capped at $4\pi$ by the delay, so the best attainable steady-state offset is $8d_0/(4\pi) = 0.637\,d_0$ — a hard limit imposed by the transport lag, not by the choice of controller structure.