The delay a chain adds

The edge of a chain found that stages each inside their own edge are not, together, inside theirs: the stages share one error budget. Four ×10 stages from a 1 MHz amplifier are 1 per cent from ideal at 7.06 kilohertz rather than the single stage’s 14.2, and the chain’s edge moves as the number of stages to the power minus a half, because each stage’s error grows as the square of the frequency. It closed on the chain in time: a cascade’s step settles to a tolerance more slowly than any stage’s, and settling time has its own law.

It has, and it is not the frequency edge’s law turned upside down. The magnitude response that edge was read from is a statement about how much of a sine gets through, and it is blind to when. A chain of stages delays as well as filters, and the delay adds up linearly with the number of stages while the filtering adds up as the square root. A step sees both. The result is that the chain’s settling time grows faster than its bandwidth shrinks, and a chain designed to the frequency rule is late.

The chain, and its step in closed form #

The chain is the amplifier chain of the earlier essay with its gain divided out: N identical stages, each a single pole at the closed-loop bandwidth, a hundred kilohertz for a ×10 stage from a 1 MHz amplifier, so a time constant of 1.59 microseconds. Each stage’s step response is $1 - e^{-t/\tau}$, and a chain of N of them has a step response with a closed form that turns up far from circuits:

$$y_N(t) = 1 - e^{-t/\tau}\sum_{k=0}^{N-1}\frac{(t/\tau)^k}{k!}.$$

That is the cumulative distribution of the Erlang distribution — the time for N independent exponential waits to finish — and it is the chain’s step because each stage’s impulse response is an exponential delay and the chain’s is their convolution. The settling time to a tolerance is the time after which $1 - y_N$ stays below it; the response rises without overshoot, so that is a single crossing and it is bisected.

The second route is to march the chain: N first-order lags in a row, each one’s output the next one’s input, stepped forward by fourth-order Runge–Kutta from rest, with the settling time read off the last stage’s trace. At one, four and sixteen stages the march and the closed form agree to two parts in a thousand, which is the march’s own resolution. Above sixteen the closed form is the only practical route, and it has to be evaluated in logarithms: at a thousand stages $e^{-t/\tau}$ and $(t/\tau)^k/k!$ underflow and overflow long before their product does, and summed directly they report a thousand stages settling in 745 time constants — before their own delay, which no chain can do.

The frequency rule, and how far short it falls #

One stage settles within 1 per cent in $\ln 100 = 4.61$ time constants, 7.33 microseconds. Two take 6.64, four 10.0, sixteen 26.7. The frequency edge’s rule for the same chain says that N stages stay within 1 per cent of flat up to a frequency smaller than one stage’s by $\sqrt{((1.01)^{2/N} - 1)/((1.01)^2 - 1)}$, which for sixteen stages is a factor of 4.02. Read as a time — a slower edge settles proportionally later — it would put sixteen stages at 4.02 times one stage’s settling. They take 5.81.

The gap grows with the chain. At four stages the frequency rule says 2.01 and the step needs 2.18; at sixteen, 4.02 against 5.81; and the hero figure’s long chains show where it ends up. A thousand and twenty-four stages take 1,100 time constants to settle, where the frequency rule would allow about 148: the settling time is heading not for $\sqrt N$ times one stage’s but for N time constants, the chain’s delay.

The delay the magnitude cannot see #

The step responses show the two things a chain does to an edge. Each added stage slides the edge one time constant to the right — the centre of each curve is at N time constants — and each widens it, so that the 10–90 per cent rise goes from 2.20 time constants for one stage to 10.2 for sixteen. The widening is what the frequency edge measured: a slower rise is a narrower band. The slide is what it could not measure, because a pure delay has a magnitude of exactly one at every frequency and changes no magnitude response at all.

The delay that is a centre of mass found that a filter’s group delay at zero frequency is exactly the centroid of its impulse response. Each stage here contributes one time constant of that centroid, so N stages contribute N, and the centroid of the chain’s impulse response sits at $N\tau$ whatever else happens. The settling time cannot be less than about that, because the response cannot have settled before most of its impulse response has arrived. That is a floor linear in N, and the frequency rule’s $\sqrt N$ has nothing to say about it.

A delay that adds and a spread that adds in quadrature #

The Erlang distribution has a mean of $N\tau$ and a standard deviation of $\sqrt N\,\tau$, and for many stages it becomes a normal distribution with those two numbers. That turns the chain’s step into something that can be read off a table.

The rise over $\sqrt N$ climbs from 2.197 time constants for one stage — the single exponential’s $\ln 9$ — to 2.563 at a thousand stages, which is twice the 90th percentile of a normal distribution, 2 × 1.2816. And the settling time beyond the delay, divided by $\sqrt N$, falls from 3.605 for one stage to 2.372 at a thousand, approaching 2.326, the normal distribution’s 99th percentile. So for a long chain the 1 per cent settling time is very nearly $N\tau + 2.326\sqrt N\,\tau$: a delay that adds stage by stage and a spread that adds in quadrature, exactly as the times of N independent waits do.

That is the whole law, and it explains the hero figure’s shape. For a few stages the spread term dominates and the settling time grows roughly as $\sqrt N$; for many the delay dominates and it grows as N; in between it follows neither, and a power law fitted across one to sixteen stages returns whatever exponent the mixture gives.

Why the spread is a square root #

The square root in the spread has the same origin as every other square root in measurement. Each stage’s impulse response is an exponential delay of mean $\tau$ and variance $\tau^2$, and the chain’s impulse response is the distribution of the sum of N such delays. Means add, so the chain’s centroid is at $N\tau$; variances add, so its spread is $\sqrt N\,\tau$. The window the square-root law has found that averaging N readings divides the noise by $\sqrt N$ only when the readings are independent, and here the condition holds by construction, because the stages are buffered and each one’s delay is its own.

The normal limit then follows from the central limit theorem, and the hero figure shows how slowly it arrives. At sixteen stages the 1 per cent time is 26.7 time constants against the normal estimate’s $16 + 2.326 \times 4 = 25.3$; at a thousand it is 1,100 against 1,098. The approach is slow because an exponential delay is skewed, with a long tail on the late side, and the skew of a sum of N falls only as $1/\sqrt N$. A tight tolerance is read in exactly that tail, which is why the tolerance figure’s exponents are lowest where the tail matters most.

The exponent depends on the tolerance #

The exponent is not a property of the chain alone. Settling to ten per cent, a chain’s time grows as N to the 0.803; to one per cent, 0.638; to a part in ten thousand, 0.489; to a part per million, 0.410. A loose tolerance is reached on the rising edge, which the delay moves directly, so the delay’s linear growth shows strongly. A tight tolerance is reached deep in the tail, and a single stage’s tail is already long — $\ln(1/\varepsilon)$ time constants, 13.8 for a part per million — so the chain’s extra delay is a smaller fraction of the total and the exponent falls. Over one to sixteen stages, only near a part in ten thousand does the time-domain exponent happen to equal the frequency edge’s half.

A boundary is a model and a tolerance found that every edge is computed from a fraction of error nobody states, and that the edge’s dependence on the tolerance is its own exponent. The chain in time adds a second dependence: the tolerance does not only move the edge, it changes how the edge scales with the number of stages. A specification that quotes settling “to within 1 per cent” and one that quotes “to 16 bits” are not the same measurement scaled.

What a designer has to give each stage #

The useful form of all this is a bandwidth budget. If a chain of N stages must settle as fast as one stage would, every stage’s bandwidth must rise by the ratio of the chain’s settling time to one stage’s.

To settle within 1 per cent in one stage’s time, four stages need 2.18 times the bandwidth each and sixteen need 5.81. The frequency rule, which keeps the chain flat to 1 per cent, asks for 2.01 and 4.02. A designer who gives each stage the bandwidth the flatness rule demands and expects the chain to settle in one stage’s time is 45 per cent late at sixteen stages. At a tolerance of 0.01 per cent the same budget asks for less — 1.73 and 3.83 — because the single stage’s own tail is long and the chain’s delay a smaller part of it.

In the example the earlier essay used, a single ×10 stage from a 1 MHz amplifier settles to 1 per cent in 7.33 microseconds. Four such stages, to settle in the same time, need amplifiers of 2.18 MHz; sixteen need 5.81 MHz. A data acquisition chain of input buffer, gain stage, filter and driver is four stages, and its settling budget is two and a fifth times what any one of them would need alone.

Chains this subject already contains #

The law applies wherever a signal passes through stages that each delay it, and several of the subject’s own objects are such chains. A filter built as a cascade of sections is one: the corner error a filter hides in its sections found the sections’ individual errors cancelling in the magnitude, and in time the sections’ delays do not cancel at all. A converter’s input path is another: a buffer, a gain stage, an anti-alias filter and a driver, each comfortably fast alone. Settling to half a least significant bit of sixteen bits is a tolerance of 7.6 parts per million, where one stage needs 11.8 time constants, two need 14.5, four need 19.0 and eight need 26.5 — 1.61 times one stage’s time for four stages, a smaller factor than at 1 per cent because the tail dominates, and still a factor nobody gets by reading each stage’s data sheet alone.

A stage that rings is the case the law does not cover, and it is where the cliff before the fastest settling lives: a second-order stage’s settling time jumps discontinuously with its damping, because the last excursion outside the tolerance band appears or disappears whole. A chain of such stages inherits a cliff from each, and where the cliffs of different stages fall decides the chain’s settling as much as their delays do.

The delay inside a loop #

A chain’s delay matters most when the chain sits inside a feedback loop, because there the delay is not a lateness but a phase. Each stage contributes $\arctan(f/f_b)$ of lag at a frequency $f$, so N stages contribute N times that, and far below their bandwidth the total is very nearly $2\pi f N\tau$ — the phase of a pure delay of $N\tau$. A loop that crosses unity at a tenth of the stages’ bandwidth loses 5.7 degrees of margin to each stage, 23 to four of them and 91 to sixteen. What is left at crossover measured a loop’s margin as the phase remaining where its gain reaches one; a chain in the loop takes that phase linearly in its length, where the flatness it costs grows only as the square root.

So the two readings of a chain diverge furthest where it matters most. Measured as a flat band, sixteen stages are four times narrower than one. Measured as a phase at a fixed frequency, they are sixteen times worse. A designer who reads a chain’s magnitude response and judges it fast enough to sit inside a loop has read the half of the chain that grows slowly.

What the closed form assumes #

Every stage is a single real pole with no overshoot, identical to the others and buffered from them. Real amplifier stages have second poles and overshoot, and an overshooting stage’s step settles later than its time constant suggests while also being partly corrected by the next; a chain of overshooting stages can settle faster or slower than this law, according to whether the overshoots reinforce. Stages of unequal bandwidth have a delay equal to the sum of their time constants and a spread equal to the root of the sum of their squares, which is the same law with the slowest stages carrying most of the spread.

And the settling here is linear. A stage that slews on a large step adds a delay proportional to the step’s size, a nonlinear term on top of the linear delay; how long a large step takes to settle measured it for one amplifier, and the step that is too big found where it begins. In a chain only the stages that see a large step slew, which is usually the last, since the earlier ones carry the signal before the gain.

The number to keep #

A chain of N identical one-pole stages has a step that is the Erlang distribution’s, with a delay of N time constants and a spread of $\sqrt N$. Its 1 per cent settling time is 4.61 time constants for one stage, 10.0 for four and 26.7 for sixteen, and tends to $N\tau + 2.326\sqrt N\,\tau$. Over one to sixteen stages it grows as N to the 0.638 at 1 per cent, against the frequency edge’s half; for sixteen stages to settle as fast as one, each needs 5.81 times the bandwidth, where the flatness rule gives 4.02.

Still open: unequal stages, a stage that overshoots, and errors that cancel #

Unequal stages. A chain whose stages have different bandwidths has a delay that is the sum of its time constants and a spread that is the root of their squares. Which stage to make fastest, for a fixed total of bandwidth, is a budget question with a closed answer in the normal limit and an Erlang-like one for a few stages, and the two may not agree.

A stage that overshoots. A second-order stage with a little overshoot settles a single step later than its rise suggests, but in a chain its overshoot is filtered by the next stage. Whether a chain of slightly underdamped stages settles faster than the same chain of single poles of equal bandwidth is the measurement that would decide how the chain’s stages should be compensated.

Errors that cancel. A chain that alternates stages whose amplitude errors have opposite signs can be flatter than any of its stages, which the frequency edge would reward. In time the delays do not cancel; whether a flatter chain also settles faster is the comparison that would say which of the two edges a designer can trade against the other.