The phase a delay never stops adding

The area that feedback only moves measured Bode’s sensitivity integral on a three-pole amplifier and found it exact: the logarithm of the device sensitivity $|1/(1 + T)|$, integrated over linear frequency, has as much area above zero as below it — 6.74 kilohertz-nepers each side at 34.9° of margin, net zero to eight parts in a billion. The theorem behind it needs only two things of the loop: that it falls at least two poles’ worth at high frequency, and that nothing in it is unstable when the loop is cut. A pure delay satisfies both. It adds no poles and no zeros, and $e^{-sT}$ is analytic everywhere, so a loop with a delay in it must still balance.

But a delay does something no pole does. A pole’s phase saturates — it contributes ninety degrees and then stops — while a delay’s phase grows without limit, $360\,f\,T$ degrees at frequency $f$. Above crossover, where every other contribution to the loop’s phase has settled, the delay keeps turning it. That is the area essay’s last open question: whether the repayment, with a phase that never stops turning, breaks into a comb. It does, and the comb turns out to do the opposite of what a comb of small peaks suggests.

Delays are not exotic in feedback. A digital controller takes time to compute; a sensor at the end of a cable reports what happened a cable’s length ago; an amplifier’s input stage and a power stage’s driver each contribute a fraction of a microsecond that a three-pole model leaves out. Each is, to the loop, a phase that keeps growing with frequency.

The same loop, ten microseconds late #

The loop is the area essay’s: a gain of $10^5$, poles at 10 Hz, 4 kHz and 2 MHz, closed for a gain of a hundred, crossing over at 5.73 kHz with 34.9° of margin. The netlist solves one frequency at a time, and an integral that follows a comb to tens of megahertz needs a hundred thousand frequencies, so the integrals here use the loop’s closed form — $A_0\beta$ over its three poles — after holding it against the netlist’s cut loop at four frequencies, where the two agree to a part in a billion. The delay multiplies both alike.

With 10 µs of delay the crossover does not move — a delay changes no magnitude — but the margin falls from 34.9° to 14.3°, since 10 µs at 5.73 kHz is 20.6 degrees. The sensitivity’s peak rises from 5.70 dB to 13.53 dB, and above the peak the curve no longer settles: it swings above and below zero with a period of 100 kHz, the reciprocal of the delay. And the integral, taken to 50 MHz on a grid with sixty steps a tooth, balances: 7.144 kHz·Np below zero and 7.144 above, net $9\times10^{-5}$ hertz-nepers.

The slider runs the delay through 0, 2, 5 and 10 µs. The margin falls through 34.9°, 30.7°, 24.6° and 14.3°, the peak rises through 5.70, 6.80, 8.76 and 13.53 dB, and at every setting the areas agree. The theorem holds exactly as stated. Bode’s integral is not a statement about phase margin or about how a loop’s phase behaves above crossover. It is a statement about analyticity, and a delay is analytic.

A comb, and why #

Far above crossover the loop gain is small and the logarithm of the sensitivity is, to second order in it,

$$\ln\left|\frac{1}{1 + T}\right| \approx -\operatorname{Re} T.$$

Without a delay, T’s phase settles towards −270° as the three poles take effect, its real part settles to one sign, and the sensitivity approaches one from one side. With a delay, T’s phase goes on turning, a full cycle every $1/T_d$, and its real part changes sign twice a cycle.

Above a hundred kilohertz the computed sensitivity is $-\operatorname{Re} T$ to within a twentieth of $|T|$, and it crosses zero seven times between 40 and 400 kHz — twice per period of the delay. The envelope is the loop gain’s own magnitude, falling at forty decibels a decade between the second pole and the third, so the teeth are a fifth of a decibel at 40 kHz and a few thousandths by 300 kHz. Each tooth above zero is followed by one below, and the obvious guess is that they cancel, so that the comb contributes nothing and the main peak must carry the whole repayment.

The peak overpays, and the comb refunds #

The guess is wrong in an instructive direction. Split the area three ways: the desensitisation proper, below zero up to the sensitivity’s first zero crossing; the main peak, the stretch above zero after it; and the comb, everything beyond.

Without a delay the split is trivial: the main peak repays the desensitisation exactly, 6,744.8 Hz·Np each, and there is no comb. With 5 µs the desensitisation is 6,602.9 Hz·Np, ending at 4.24 kHz rather than 4.48; the main peak is 6,843.0, 103.6 per cent of it; and the comb nets −240.2. With 10 µs the desensitisation is 6,484.5, the main peak 6,937.1, 107.0 per cent, and the comb −452.6.

So the delay does three things at once, and they all point the same way. It shrinks the desensitisation, because the extra lag brings the loop’s phase closer to −180° just below crossover, where $|1 + T|$ is then smaller than $|T|$, and the sensitivity reaches one at a lower frequency. It makes the main peak overpay the shrunken debt, because the margin is smaller and the peak taller. And the comb, whose teeth below zero slightly outweigh those above, hands the excess back as further desensitisation — 7 per cent of it at 10 µs, spread over hundreds of kilohertz.

Two more numbers explain why the total areas rose when the desensitisation fell. The area below zero in the hero figure, 7,144 Hz·Np, is the desensitisation plus every negative tooth of the comb; the area above, also 7,144, is the main peak plus every positive tooth. The comb adds to both sides, and the sides still balance. A designer reading the hero figure’s two equal numbers as “the delay increased the desensitisation” would have it backwards: the useful part, below crossover, went down, and what went up was the comb’s churn far above the band the loop was built for.

Why the comb’s net is below zero #

The comb’s teeth alternate, so its net is the small remainder of a large cancellation, and the remainder has a sign for a reason that can be read off the numbers. Far above crossover the comb is $-\operatorname{Re} T$ to first order — the second-order term, half the real part of $T^2$, contributes less than six Hz·Np of the 453 — and $T$ is a rotating arrow whose length falls at forty decibels a decade. An alternating series whose terms shrink is dominated by its first term, and here the first term is the tooth just after the main peak.

With 10 µs of delay the main peak ends at 26.7 kHz, far above the crossover, because the delay keeps $|1 + T|$ below one for a long stretch past it. The tooth that follows, from 26.7 to 50 kHz, lies below zero and contributes −354 Hz·Np on its own. Every tooth after it is smaller than its predecessor, and the next hundred kilohertz add only −95; everything above 150 kHz adds less than four. So the comb’s net is essentially its first tooth, and that tooth is below zero because the main peak, which ends where $T$’s real part changes sign, is necessarily followed by a stretch where it has the other sign.

That turns the overpayment into something a designer can picture. The main peak with a delay is wider as well as taller than the same loop’s without one — it runs from 4.06 kHz to 26.7 kHz — and the extra width carries more area than the shrunken desensitisation owes. The first tooth of the comb then takes back most of the excess, and the rest of the comb takes back the remainder in ever smaller instalments, each alternating in sign.

What a microsecond of delay costs, in the loop’s own currency #

The delay figures above can be compressed into rules that do not need the integral. The margin falls by $360\,f_c\,T_d$ degrees, 2.06 degrees per microsecond at this crossover, so the delay a loop can tolerate is its margin divided by that rate: 16.9 µs for 34.9°. The peak follows the margin, a decibel and a half above the margin-implied value, so a stated peak is a stated margin is a stated delay. And the desensitisation, the part of the area that does the loop’s job, falls by about 0.4 per cent per microsecond here — small, because the delay’s phase is small at the low frequencies where the desensitisation lives, and growing as the delay approaches its limit.

Read the other way, a fixed delay fixes how high the crossover may go. A digital controller with 10 µs of computation loses twenty degrees at 5.73 kHz, forty at 11.5 and all of a forty-five-degree loop’s margin at 12.5. Halving the delay doubles every one of those frequencies, which is why the latency of a control computation, rather than its sample rate, is so often the number that limits a digitally controlled loop’s bandwidth.

What the margin says about the peak, and what it misses #

At the crossover $|T| = 1$, so $|1 + T| = 2\sin(\varphi/2)$ for a margin φ, and the sensitivity there is $1/(2\sin(\varphi/2))$ exactly. The peak can only be higher. With no delay it is 5.70 dB against 4.45 at crossover; with 5 µs, 8.76 against 7.42; with 10 µs, 13.53 against 12.11; with 15 µs, 24.73 against 23.24. The margin reads the peak from below by about 1.2 to 1.5 dB across the whole range, and both go to infinity together at 16.92 µs, where the delay has used up all 34.9° — 34.9 divided by 360 times 5.73 kHz.

That near-constant gap is reassuring and is the reason margin is a usable design number for a delayed loop: a designer who sizes the margin for a stated peak will be wrong by about a decibel and a half, in the safe direction only if the gap is added. The delay does not hide anything from the margin that a pole would not. Not every loop is so obliging: Stable, and unstable with less gain found one whose margins could not see that turning its gain down breaks it. This loop’s magnitude falls monotonically, so it crosses unity once whatever the delay, and its one phase margin is the whole account of its stability. What the margin cannot see is the comb, which sits well below the peak and changes nothing about stability, and the shrinking of the desensitisation, which is a statement about the loop’s low-frequency behaviour that a single phase reading at crossover could not make.

Past the critical delay, the integral counts the growth #

A loop with more than 16.92 µs in it is unstable. Its sensitivity is no longer analytic in the right half-plane — $1/(1 + T)$ has poles there, at the closed-loop poles that crossed — and the integral’s premise fails. It does not fail into nonsense. For a loop that is stable when cut and unstable when closed, the integral of $\ln|1/(1 + T)|$ over frequency in hertz is minus half the sum of the real parts of the unstable closed-loop poles.

Newton’s method on $1 + L(s)e^{-sT_d} = 0$, started near the crossover, finds the pair. At 12, 14 and 16 µs its real part is −3,039, −1,716 and −512 per second, and the net area is zero to the integration’s accuracy. At 17 µs the pair has crossed: its real part is +46.1 per second and the net area is −46.9 Hz·Np. At 20 µs, +1,547.7 and −1,547.7; at 25 µs, +3,535.1 and −3,535.1. Two routes — a root of a characteristic equation in the complex plane and an area on the frequency axis — give the same number to a part in a thousand once the pair is clear of the axis, and to a hertz-neper at 17 µs, where the pole is so close to the axis that the sensitivity is a needle the grid only just follows.

So the integral does not merely fail for an unstable loop; it measures how unstable. At 20 µs the oscillation that builds up grows by a factor of e every 646 µs, and the sensitivity’s logarithm has lost exactly that growth rate, in hertz-nepers, from its balance. It is the same identity what is left at crossover reads as a margin, carried past the point where a margin stops being a number.

Where a real loop’s delay comes from #

The delay here is a pure $e^{-sT}$, which a loop meets in two common forms. One is a digital controller’s computation: the sample is taken, the arithmetic runs, the correction is applied one computation later, and the loop has a delay of that computation time in it whatever the sample rate. Half a clock against a pole priced a converter’s share of it as 180 degrees times the crossover over the clock, which is this essay’s $360\,f_c\,T_d$ at half a clock. The other is transport: a sensor at the end of a cable, a remote voltage sense on a long supply line, a mechanical or thermal path whose response arrives late. Neither adds a pole, and both, by this essay’s measurements, cost the loop margin at a rate set by the crossover frequency — 2.06 degrees per microsecond here — and turn its above-crossover sensitivity into a comb at the reciprocal of the delay.

A third form is worth separating because it is not a pure delay. A sampled loop with a zero-order hold behaves, well below its sample rate, like a continuous loop with half a sample of delay, and that is a useful first estimate of what sampling costs. But near half the sample rate the two part: the hold’s response folds back and ends, where a delay’s comb runs on. A sampled loop’s integral is taken over a finite band, and that changes its conclusions more than the half-sample estimate suggests. The band a sampled loop repays in takes that finite band as its subject.

What it leaves out #

Only a pure delay. Real delays come with dispersion — a cable’s delay depends a little on frequency, a filter’s group delay a great deal — and a delay whose phase grows less than linearly makes a comb whose teeth spread apart. The balance still holds; the split between peak and comb would change.

And only this loop. A loop with more margin to start with can afford more delay before the peak climbs; a loop with less has less. The 1.2 to 1.5 dB gap between peak and margin-implied value is this loop’s, and a loop with two poles close to crossover, or a resonance near it, would have a different gap.

Still open: the sampled loop, the bound on the peak, and a delay the loop can predict #

The sampled loop’s integral. A loop closed through a sampler and hold has a sensitivity defined only up to half the sample rate, and its Bode integral is taken over that finite band. The comb here repaid part of the debt far above crossover; a sampled loop has nowhere far above crossover to put anything, and the question is how much higher its peak must be as a result, and whether the finite band puts a floor under the peak that no design can go beneath.

A bound on the peak from the band. Half to the amplifier, half to the resistors split a loop’s sensitivity between its device and its divider at every frequency; the continuous integral says the areas of the first balance and says nothing about how high the peak must be, because the repayment can always be spread thinner over a wider band. Any constraint that confines the repayment — a finite band, a loop gain required to be small above some frequency — turns the equality into a lower bound on the peak, and computing that bound for this loop would say how close its ordinary compensation comes to the best possible.

A predictor in the loop. A Smith predictor removes a known delay from the loop’s characteristic equation by modelling it, so the margin no longer falls with the delay. It cannot remove the delay from the loop’s response to disturbances, and whether its sensitivity still carries the comb — and where the predictor’s own modelling error puts the peak — is the practical form of the question this essay asked.