Question 6 of 6: Feedback around a pure transport delay

Question 6: Feedback around a pure transport delay (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 whose entire plant is a one-second transport delay.

Find. The admissible range of $K$ and the corresponding phase margin, the steady-state step-tracking error as a function of $K$, the unit step response at $K = 1$, and a redesigned controller giving zero step error while keeping 6 dB of gain margin.

Approach. Exploit the fact that a delay is all-pass: its magnitude is identically one and all of its effect is in the phase. That decouples the magnitude and phase conditions completely, so every margin can be written in closed form without a Padé approximation.

1. Part (a) — separate magnitude from phase. On the imaginary axis $$|L(j\omega)| = K\,\bigl|e^{-j\omega}\bigr| = K, \qquad \angle L(j\omega) = -\omega\ \text{rad} = -57.296\,\omega\ \text{degrees}.$$ The magnitude is flat and the phase falls without bound. Routh's criterion is unavailable here because $e^{-s}$ is transcendental and the characteristic equation is not a polynomial; the frequency-domain margins are the natural tools instead.
2. Locate the phase crossover. The phase reaches $-180^\circ$ when $\omega\theta = \pi$, so with $\theta = 1\ \text{s}$ $$\omega_{pc} = \frac{\pi}{\theta} = \pi = 3.1416\ \text{rad/s}.$$ This frequency does not depend on $K$ at all, which is a direct consequence of the delay being all-pass.
3. Impose the gain-margin specification. The gain margin is the reciprocal of the loop magnitude at the phase crossover, $$\mathrm{GM} = \frac{1}{|L(j\omega_{pc})|} = \frac{1}{K},$$ and requiring at least 6 dB — taken, as is conventional, as a factor of two — gives $1/K \ge 2$: $$\boxed{\;0 \lt K \le 0.5\;}$$
4. Determine the corresponding phase margin. The phase margin is measured at the gain crossover, where $|L| = 1$. But $|L(j\omega)| = K \le 0.5$ at every frequency, so the magnitude curve never reaches 0 dB and there is no gain crossover: $$\boxed{\;\text{No gain crossover exists for } K \lt 1, \text{ so the phase margin is infinite.}\;}$$ This is the graded observation. A loop whose magnitude is uniformly below unity cannot be destabilised by any amount of additional phase lag, so no finite margin can be quoted; the gain margin is the only meaningful measure for this plant.
5. Part (b) — write the error transfer function. For unity feedback, $$E(s) = \frac{R(s)}{1 + L(s)} = \frac{R(s)}{1 + K e^{-s}} .$$ The delay has unit gain at DC, $e^{-s}\big|_{s=0} = 1$, so $L(0) = K$.
6. Apply the final-value theorem. With $R(s) = 1/s$, $$e_{ss} = \lim_{s\to0} s\cdot\frac{1/s}{1 + Ke^{-s}} = \frac{1}{1 + K},$$ $$\boxed{\;e_{ss} = \frac{1}{1+K}\;}$$ The loop is Type 0, so a proportional controller can only shrink the error, never remove it; at the largest gain allowed by (a), $K = 0.5$, the residual error is still $1/1.5 = 0.667$, i.e. two-thirds of the step.
7. Part (c) — expand the closed-loop response at $K = 1$. With $K = 1$ the closed-loop transfer function is $$T(s) = \frac{e^{-s}}{1 + e^{-s}} = e^{-s}\bigl(1 - e^{-s} + e^{-2s} - e^{-3s} + \cdots\bigr) = \sum_{n=1}^{\infty}(-1)^{n+1} e^{-ns},$$ a geometric series that converges for $|e^{-s}| \lt 1$, i.e. in the right half plane.
8. Invert term by term. Each term $e^{-ns}/s$ inverts to a unit step delayed by $n$ seconds, so the unit step response is $$\boxed{\;y(t) = \sum_{n=1}^{\infty}(-1)^{n+1}\,u(t-n) = \begin{cases}0, & 2m \le t \lt 2m+1\\ 1, & 2m+1 \le t \lt 2m+2\end{cases}\;}$$ The output is a square wave alternating between 0 and 1 every second, of period 2 s, which never settles. This is exactly what the margin analysis predicts: at $K = 1$ the gain margin is $1/K = 1$, that is 0 dB, so the loop sits precisely on the stability boundary and sustains an undamped oscillation at $\omega_{pc} = \pi\ \text{rad/s}$ — a period of $2\pi/\pi = 2\ \text{s}$, matching the square wave exactly.
Question 6(c): unit-step response at K = 1. The output is a square wave that switches between 0 and 1 every second and never settles - the loop is marginally stable.
9. Part (d) — choose the controller structure. Zero steady-state error to a step requires the loop to be Type 1, and since the plant supplies no pole at the origin the controller must. The simplest choice that adds nothing else is a pure integrator, $$C(s) = \frac{K}{s}, \qquad L(s) = \frac{K e^{-s}}{s}.$$
10. Locate the new phase crossover. The integrator adds a constant $-90^\circ$, so $$\angle L(j\omega) = -90^\circ - 57.296\,\omega = -180^\circ \;\Longrightarrow\; \omega_{pc}' = \frac{\pi/2}{\theta} = 1.5708\ \text{rad/s},$$ half the previous value — the price paid for the integrator.
11. Size the gain for 6 dB. The magnitude at that frequency is no longer flat, $|L(j\omega)| = K/\omega$, so $$\mathrm{GM} = \frac{1}{|L(j\omega_{pc}')|} = \frac{\omega_{pc}'}{K} = \frac{\pi/2}{K} \ge 2 \;\Longrightarrow\; \boxed{\;C(s) = \frac{K}{s}, \qquad 0 \lt K \le \frac{\pi}{4} = 0.7854\;}$$
12. Report the margins of the redesign. The gain crossover of the new loop satisfies $K/\omega = 1$, so $\omega_{gc}' = K$, and $$\mathrm{PM}' = 180^\circ - 90^\circ - 57.296K = 90^\circ - 57.296K .$$ At the limiting gain $K = \pi/4$ this evaluates to $$\mathrm{PM}' = 90^\circ - 45^\circ = 45^\circ,$$ a textbook-comfortable margin. The redesigned loop therefore delivers zero steady-state step error, 6 dB of gain margin and $45^\circ$ of phase margin simultaneously — and unlike part (a) it now has a finite phase margin, because the integrator makes the magnitude frequency-dependent and a genuine gain crossover exists.