The sections an edge needs

A ladder is not a line drove a chain of LC sections with an ideal step and found it disappointing. Forty sections still ring through every plateau by five per cent, the convergence is slow, and reaching one per cent would take, by extrapolation, something of the order of a thousand sections. The sections a wavelength needs then asked the same chain a frequency-domain question and got an exact and much kinder answer: thirteen sections per wavelength for one per cent of phase, twenty-two for the group delay of a pulse. It closed with a promise it did not check. A chain’s failure with a step is a property of the step. Given an edge with a finite rise time, the number of sections needed is finite and computable, and it is a few tens of sections per metre of fast cable rather than a thousand.

That sentence is right in its first half and misleadingly simple in its second. There is a number, and it can be measured directly. It is not per metre, or per rise, or per wavelength at the edge’s frequency, and any of those rules would give a wrong answer by a growing factor on a longer line. The rule that holds comes straight from the group delay the earlier essay already wrote down, and it holds across two orders of magnitude of length within a few per cent.

Taking the reflections away #

The earlier chain ran between a 10 Ω source and an open end, because that is the bounce diagram it was being compared with, and every plateau had a reflection in it. Here both ends are matched, with a half capacitor at each end and the chain’s own impedance as source and load. With no reflections to speak of, whatever the far end shows is the chain’s own effect on a passing edge, which is what a designer substituting sections for a line needs to know.

The edge is a raised cosine with a stated 10–90 per cent rise. It has a definite start and end and no energy at the corner, and in contrast with a linear ramp it has no sudden changes of slope to excite anything for reasons unrelated to the question. The march is a leapfrog on the section voltages and currents, in units where the characteristic impedance is one and the edge’s rise is one, with a time step well under a section’s delay. Everything is therefore a function of two numbers: the line’s delay in rises, $T_d/t_r$, and the number of sections.

A netlist of the same chain, solved for a step by the ordinary trapezoidal integration and convolved with the edge’s own slope, gives a second route. On twenty sections and ten rises the two agree on the overshoot to a hundredth, 17.5 per cent against 17.5.

What an edge shows that a step does not #

Before the main measurement, it is worth seeing why the step was so unkind.

The chain has a cut-off at $2/\tau$ for a section delay $\tau$, and a group delay that rises towards it without limit. A step has energy at every frequency up to that cut-off. So the frequencies near the cut-off arrive very late and at their full amplitude, and they appear behind the edge as a ring whose frequency is the cut-off. More sections raise the cut-off and move the ring, but they do not remove it, since a step has energy at the new cut-off too. Asking how many sections make a step’s overshoot small is a question with a discouraging answer because the overshoot is largely a shape determined by the cut-off’s existence rather than by its value. It behaves like the Gibbs overshoot of a truncated Fourier sum, and for the same reason.

A finite edge is different in kind. Its spectrum has a top, so past some number of sections the cut-off is above anything the edge contains. Below that point, as the figure shows with twenty sections, the edge rings a good deal anyway, at 17.5 per cent, and the chain is not yet adequate. But the ring is now something the count can actually remove.

The count, against the length #

So the measurement: for a stated overshoot, bisect the number of sections, at each length.

For one per cent of overshoot: 6 sections on a line one rise long, 17 on two, 62 on five, 183 on ten, 520 on twenty, 2,048 on fifty and 5,760 on a hundred. The slope over the last decade is 1.50. For five per cent the numbers are lower, 54 on ten rises and 1,648 on a hundred, and the slope is 1.48. A power law like this over two decades, with the exponent very close to a simple fraction, is begging for a one-line reason.

The reason is in the Bloch relation the earlier essay measured. A section of delay $\tau$ has a phase $2\arcsin(\omega\tau/2)$, against a line’s $\omega\tau$, and the leading error in that is $\omega^3\tau^3/24$. Summed over N sections of $\tau = T_d/N$, the total phase error across the chain is

$$\Delta\phi = \frac{\omega^3 T_d^3}{24\,N^2}.$$

The size of the disturbance to the edge depends on this phase at the frequencies the edge contains. The edge’s highest important frequency is of the order of $1/t_r$, so $\Delta\phi \sim (T_d/t_r)^3/24N^2$. Holding it fixed, for a fixed overshoot, requires

$$N \propto \left(\frac{T_d}{t_r}\right)^{3/2}.$$

That is a simple result and it has a sharp consequence. For one per cent the constant is close to 5.8: 5.8 at one rise against 6 measured, 183 at ten against 183, and 5,800 at a hundred against 5,760. So a single formula is good to a few per cent over two orders of magnitude of length, and it was derived in two lines. The factor between 1 per cent and 5 per cent is 3.4, so in this regime overshoot goes as about the inverse square of the count, as the phase error does.

Not per rise, and not per wavelength #

The number of sections per rise is $N/(T_d/t_r)$, and this law says it rises as the square root of the length. One rise needs 6 per rise, ten need 18.3, a hundred need 57.6. A rule stating some number of sections per rise is therefore right for one length. A rule of thumb in circuit simulation says to use about ten sections per rise of the fastest edge, and that is right for a line about three rises long at one per cent, and it is optimistic by a factor of six on a hundred rises.

The contrast with the earlier essay’s frequency-domain answer is instructive. There thirteen sections per wavelength gave one per cent of phase, and that is a local statement about error per unit of length, a fractional error in delay. It is true, and it does not care about the length, because it is a fraction. What an edge cares about is an absolute phase error, in radians, at the top of its spectrum. A fractional error of one per cent is a tiny absolute one on a short line and a large one on a long line. That difference, between a fractional phase error and an absolute one, is exactly the extra half power.

The delay a chain adds found a similar structure for cascaded amplifiers in time: a delay adds as the number of stages and a spread as the root, so that long chains are unlike short ones. Here the mechanism is different but the moral is similar. A discrete thing’s cumulative error, on something long, has a different law from what any one section suggests.

Three things leaving at different rates #

Overshoot is not the only thing a chain can get wrong, and the three obvious readings of an edge do not leave together.

The half-height delay is far the easiest: 1.52 per cent out at 28 sections, 0.586 at 56, 0.0577 at 112. A chain gets its delay right long before it gets its edge right, because a delay reads the chain’s low-frequency behaviour, which converges immediately. So a delay line built from sections, tested with a time-of-flight measurement, can read correctly when the shape it delivers is badly wrong. At 28 sections the delay is within 1.5 per cent and the rise is 38.6 per cent long, with 14 per cent of overshoot behind it.

The rise is the most sensitive at low counts: 153 per cent long at ten sections and 89 at sixteen. It crosses the overshoot near 80 sections, and by 112 it is the smaller of the two at 0.616 per cent. At the largest count it reads slightly negative, −0.34 per cent, with the overshoot still at 0.692. A little dispersion actually sharpens the middle of an edge, at the price of ringing behind it, in the same way a filter’s small overshoot does, and the corner that says nothing about an edge measures a related fact. So at high counts the 1 per cent overshoot figure is the more demanding of the two. A person checking the chain by its rise would have stopped sooner than one checking its overshoot.

A note of honesty about the overshoot’s own slope on this figure: at ten rises it is 1.38 at the largest counts, which is less than the 2 the phase error suggests. The ring is still separating from the edge at these counts, and the overshoot is in transition. Across lengths at a fixed overshoot the law is clean, but the local slope of overshoot against count at one length has not reached its asymptote.

In metres of cable #

Putting the law into the units a bench uses shows what the promise of “a few tens of sections per metre” was worth.

A nanosecond edge on a metre of cable, five rises long, asks for 62 sections. That is the earlier essay’s “a few tens”, and on a metre it holds. On three metres it asks for 338, more than five times as many on three times the length, and on ten metres for 2,048. A 100 ps edge on one metre, fifty rises long, needs the same 2,048. The two are the same line in its own units, since only the ratio matters.

So any lumped model of a piece of cable has to be sized for the edge and the length together. A circuit simulator’s lossless transmission line avoids the whole problem by doing the delay exactly. A chain of sections is useful where the line has something that varies along its length or a nonlinear load partway along, or where the simulator has no good line element, and the fixed rule that a person typically brings to that has a right answer on one line only. The load spread along the track used a chain of real capacitive loads, and the same count now says how many separate inputs are enough for such a bus to look continuous to a given edge.

Where the lump begins #

The same count says where a piece of track can be treated as no sections at all. The boundary that is a rise time found that a track stops being a node when its delay reaches a fraction of the edge’s rise, and that essay’s boundary is a statement about reflections at mismatched ends. Here the ends are matched and the question is what the chain does in passing. At a delay of half a rise the law asks for two sections, and a single section is close to adequate, which is the sense in which a short track is a lump: one L and one C, with no error a measurement would notice. Between that and a few rises the count rises from a handful to some dozens, and the lumped description becomes a chain rather than a single element.

A chain as a real delay line #

The other use of these sections is real hardware: an LC artificial line is an ordinary way to get a delay on a board without having a metre of cable. Here the numbers are severe. For a delay of ten rises at one per cent, 183 sections means 183 inductors and 183 capacitors, and that is why a real delay line of that kind has its delay-to-rise ratio specified as a figure of merit, and it is typically only a few for a compact design. A real line is often also deliberately designed with mutual coupling between its inductors, the bridged or constant-k forms, since coupling flattens a section’s group delay. The phase error’s cubic coefficient is then smaller by a factor that design can choose, so the constant 5.8 falls substantially and the exponent stays at 3/2, since that came from the form of the accumulation. Flat delay bought with more delay makes a similar exchange in a filter, spending parts to reduce a delay’s frequency dependence.

The law also explains why a very high ratio of delay to rise is so expensive and is generally made with some actual transmission medium instead. A ratio of a hundred in lumped parts is 5,760 sections, and in coaxial cable it is a length. Physical distribution solves this problem at no cost, and that is the plainest possible statement of what a distributed element is for.

What the march leaves out #

The chain has no loss, and a real cable has skin and dielectric loss, which limit the far end’s bandwidth in their own way and would reduce the overshoot on a long line in the same way the loss that hides a mismatch reduces a reflection. So on a real cable the count needed for fidelity to the real line may be smaller than this. The match is exact at the frequency where the chain’s impedance has its low-frequency value. A chain’s image impedance moves with frequency as $\sqrt{1 - (\omega\tau/2)^2}$, so a resistive termination has a small reflection at high frequency. The half capacitors at each end reduce that, though they do not remove it. At the counts that hold 1 per cent, this is a small effect, but at low counts it contributes some of what is on the figure here.

What to carry #

A matched chain of LC sections needs about $5.8\,(T_d/t_r)^{3/2}$ of them to keep an edge within one per cent of overshoot: 6 on one rise, 183 on ten, 5,760 on a hundred, measured. A section’s phase error goes as the cube of frequency and the inverse square of the count, and the total accumulates over the line, so sections per rise rise as the root of the length rather than staying fixed. The delay is right long before the edge is. In cable a 1 ns edge asks for 62 sections on a metre and 338 on three. An ideal step rings at every count because it has energy at every cut-off. The promise of a few tens of sections per metre is true on about one metre of one edge.

Still open: loss along the chain, a bridged section, and a load partway down #

Sections with loss. A real cable’s loss rises with frequency and rounds the far-end edge, and it would reduce both the overshoot of a chain and the fidelity demanded of one. Adding a series resistance to each section gives a lossy chain whose count against the real line can be measured, and it would say whether a lossy line’s model needs fewer sections per unit of length.

A bridged section. Coupling between adjacent inductors flattens the group delay, and the constant 5.8 should fall by a factor the coupling can choose, while the exponent should not move. Whether a coupling of a stated value reduces the count by as much as the textbook’s analysis of one section suggests, once an entire edge is what is being judged, is a march with one more element in it.

A load partway along. A tap on a chain is a discontinuity in a chain’s impedance at that frequency, and the load spread along the track measured buses of equal taps on real lines. How the tap’s reflection compares with the chain’s own dispersion, on an edge at a given number of sections, would say when a bus model’s section count is limited by the taps and when by the sections between them.