The sections a wavelength needs

A ladder is not a line built a chain of series inductors and shunt capacitors — the construction every textbook uses to introduce a transmission line, sharing out a line’s inductance and capacitance — and asked it to do what a line does with a step. It found that the LC ladder does the right things in roughly the right places and gets the numbers wrong for a very long time. Forty sections still ring by five per cent on every plateau; the ringing falls as the −0.54 power of the section count; and reaching one per cent would take, by extrapolation, something of the order of a thousand sections.

That essay measured the LC ladder in the time domain, against a step, and a step is the hardest possible test of an LC ladder. This essay asks the same LC ladder a frequency-domain question, one sine wave at a time, and gets an answer that is exact, closed-form, and much kinder. It also explains why the step is so unkind.

A step against twenty sections #

The time-domain picture is worth having in front of the argument. Twenty sections carry the inductance and capacitance of a metre of fifty-ohm cable with a delay of 4.83 nanoseconds. The staircase they produce has the right plateaus in the right order, with ringing on each and an arrival that starts early. Nothing in that picture says which part of the LC ladder’s failure is fundamental and which is a matter of how many sections were used. The frequency domain separates the two cleanly.

One section, and the phase it imposes #

A single section — an inductance $L_s$ in series, then a capacitance $C_s$ to ground — has a transmission matrix whose entries can be written down by inspection. A wave travelling down an endless chain of identical sections picks up the same phase at every section, and that phase, the chain’s Bloch phase θ, follows from half the trace of the matrix:

$$\cos\theta = 1 - \frac{\omega^2 L_sC_s}{2}, \qquad \theta = 2\arcsin\frac{\omega}{\omega_c}, \qquad \omega_c = \frac{2}{\sqrt{L_sC_s}}.$$

A true line with the same inductance and capacitance per section imposes $\omega\sqrt{L_sC_s} = 2\omega/\omega_c$ per section, which is the first term of the arcsine’s series. So below the cutoff $\omega_c$ the chain is a line whose phase per section is too large by a factor of $\arcsin(x)/x$, with $x = \omega/\omega_c$. At the cutoff the arcsine reaches a right angle and the phase per section reaches half a turn; above it $\cos\theta$ is less than minus one, the phase is complex, and the chain attenuates rather than propagates. A ladder is a line below a frequency and a low-pass filter above it, and the frequency is set by one section.

The group delay — the delay of an envelope, and the one that decides when the energy of a pulse arrives — is the derivative of the phase with respect to frequency, and the arcsine’s derivative gives it at once: too long by a factor of $1/\sqrt{1 - x^2}$. The group delay error is always the larger of the two, by about three to one at small $x$.

The chain, solved #

The figure solves the twenty-section chain as a network at each frequency, between fifty-ohm terminations at both ends, reads the phase of its output, and turns it into a delay. Beside it are the two closed forms: the phase-delay error $\arcsin(x)/x - 1$ and the group-delay error $1/\sqrt{1 - x^2} - 1$.

The cutoff is 1.32 gigahertz, which is $N/(\pi T)$ with $N$ sections and a total delay $T$ — twenty over π times 4.83 nanoseconds. The chain’s delay is one per cent too long at 318 megahertz, where $x$ is 0.241, and its group delay at 185 megahertz, where $x$ is 0.140. Well below a tenth of the cutoff the solved delay follows the Bloch phase to two parts in a thousand, and the figure checks that; closer to the cutoff the solved curve ripples about the closed form.

The ripple belongs to the terminations, and it is the second thing a frequency-domain view makes visible. A ladder’s own impedance is not fifty ohms at every frequency. Looked into from its inductor end, an endless chain of these sections presents $Z_0\sqrt{1 - x^2}$, and from its capacitor end $Z_0/\sqrt{1 - x^2}$; both are fifty ohms only at direct current. Fifty-ohm terminations are therefore slightly mismatched to the chain everywhere else, the chain reflects a little at each end, and those small reflections go back and forth and put a ripple on the delay. They are the frequency-domain face of the ringing on the time-domain plateaus. And the impedance’s departure is the group delay’s departure — $1/\sqrt{1 - x^2}$ again — so a ladder is one per cent out in impedance at the same frequency it is one per cent out in group delay.

Doubling the sections doubles every frequency on the axis. Forty sections have their cutoff at 2.63 gigahertz and are one per cent slow in delay at 637 megahertz and in group delay at 370. Eighty sections move those to 5.27 gigahertz, 1.27 and 739 megahertz. There is no convergence rate to fit here, because the whole error is a fixed function of frequency over cutoff and the cutoff is in exact proportion to the number of sections. What was an uncertain power law in the time domain is a change of scale in the frequency domain.

Sections per wavelength #

That proportionality means the natural unit is not the number of sections but the number of sections per wavelength of the line being imitated. A line of delay $T$ is $fT$ wavelengths long at frequency $f$, so $N$ sections give $N_\lambda = N/(fT)$ sections per wavelength, and substituting the cutoff gives the whole of the result in one line:

$$x = \frac{\omega}{\omega_c} = \frac{\pi}{N_\lambda}.$$

Every error above is a function of how many sections the ladder has per wavelength and of nothing else — not of the impedance, not of the length, not of the dielectric.

For a delay within one per cent an LC ladder needs 13.0 sections per wavelength. For a group delay within one per cent — and so, at the same frequency, an impedance within one per cent — it needs 22.4. A tenth of a per cent needs 40.6 and 70.3; ten per cent, 4.6 and 7.5. Both counts grow as the reciprocal square root of the error, because both errors are quadratic in $x$ to leading order: $\arcsin(x)/x - 1 \approx x^2/6$ and $1/\sqrt{1-x^2} - 1 \approx x^2/2$.

The figure holds the closed form to a solved chain at the point that matters most. At the frequency where twenty sections should be one per cent out in group delay, the solved chain between fifty-ohm terminations is out by 1.09 per cent; the extra nine hundredths is the terminations’ ripple at that frequency.

The practical rule most often quoted is ten sections per wavelength. It is 1.72 per cent slow in phase delay and 5.33 per cent slow in group delay, and five per cent out in impedance. That is a respectable description of a line for many purposes and a poor one for anything that depends on when a pulse’s energy arrives or on what impedance a source sees, and the rule is usually quoted without saying which.

Why a step converges so slowly #

The frequency-domain answer is kind and exact, and the time-domain one was neither. The reason is the step, not the LC ladder.

An ideal step contains every frequency, with an amplitude falling only as the reciprocal of frequency. Whatever number of sections is chosen, a fixed fraction of the step’s content lies above that LC ladder’s cutoff, where the LC ladder transmits nothing, and a further fraction lies between a tenth of the cutoff and the cutoff, where the LC ladder is a line with a large and frequency-dependent delay error. The part above the cutoff is removed, which smears the edge; the part just below it arrives late by different amounts at different frequencies, which rings. Doubling the sections moves the cutoff up an octave, and the step still has content above it, falling by only half. No finite LC ladder ever passes a step correctly, and the error falls slowly because the step’s spectrum falls slowly.

That is the mechanism behind both of the exponents a ladder is not a line fitted on that chain of inductors and capacitors, and it predicts their direction without predicting their values. An LC ladder is a low-pass filter whose cutoff rises in proportion to the section count, so the edge it produces should narrow roughly in proportion to one over the section count; the measured exponent was −0.73 rather than −1, because the edge a dispersive filter produces near its cutoff is shaped by the group delay’s rise as well as by the cutoff itself. The plateau ringing is the Gibbs phenomenon of a filter with a sharp edge, and a sharp edge’s overshoot does not fall with the edge’s frequency at all; what falls is the part of it that has not yet decayed by the time the next plateau is measured.

Both fitted exponents are drawn above as the time domain measured them. Neither is a property of the LC ladder alone; each is the LC ladder’s fixed frequency-domain error integrated against the spectrum of an ideal step, and a different input — an edge with a finite rise time — would give different exponents from the same closed form. That is why the plateau fit carried a residual of nearly nine per cent and resisted extrapolation: it was fitting a curve whose shape depends on how much of a slowly falling spectrum lies near each LC ladder’s cutoff, and that fraction does not change as a clean power of the section count.

The step is also the reason the rule of ten per wavelength is not wrong so much as unanchored. Ten per wavelength of what frequency is the whole question, and for a step there is no such frequency. For a real edge there is one: an edge of 10–90 per cent rise time $t_r$ has most of its content below about $0.35/t_r$. A metre of cable carrying a 100-picosecond edge needs to be a line up to about 3.5 gigahertz, which is 16.9 wavelengths of that cable, and for one per cent of group delay at that frequency the LC ladder needs 22.4 sections for each: 379 sections. For a one-nanosecond edge, 38. The time-domain extrapolation of a thousand sections was an answer for an edge of zero rise time, which no physical source produces.

The frequency-domain reading also joins this LC ladder to the rest of the field. The staircase in time is the step a real line delivers, which is what the LC ladder was imitating; the delay that is not one number found a lossy line dispersive in exactly the way this LC ladder is — a delay that depends on frequency — for a physical rather than a constructional reason; and several sections, and the band they buy uses sections of line deliberately, where each one is exact and the band is set by how they are combined. The compression of frequency near a cutoff is the same compression the corner that moved found in a bilinear-transformed filter near half its sample rate.

Segmenting a line for a simulator #

The same arithmetic answers the question a circuit simulator poses every time a track is modelled as a chain of lumped segments, which is still the commonest way to put an interconnect into a netlist that has no transmission-line element. The question is how many segments, and the answer from this essay is: enough that the highest frequency of interest sees twenty-two per wavelength if timing matters and thirteen if only phase does.

Take a thirty-centimetre track in a laminate with a relative permittivity of 4, whose delay is about two nanoseconds, carrying a signal whose content matters to five gigahertz. At five gigahertz the track is ten wavelengths long, so one per cent of group delay needs 224 segments and one per cent of phase delay 130. A model built with the ten-segments-per-wavelength rule would use a hundred, and its edges would arrive five per cent late at the top of the band — about a hundred picoseconds on a two-nanosecond track — with a ripple on the impedance the driver sees. A model built with ten segments in total, which is not unusual, has its cutoff at 1.6 gigahertz and simply does not pass the upper two thirds of the band.

That is not a statement about any particular simulator; it is a statement about the chain the simulator is given. The number that should be chosen first is the highest frequency, and the section count follows from it. A segment count chosen first and a frequency inferred afterwards is how a model comes to be quietly low-pass filtered at a frequency nobody picked. The symptom in a simulation is an eye that looks slightly better than the board’s, because the model has removed the highest harmonics along with their crosstalk and their reflections, and nothing in the waveform says that a filter was added rather than a line modelled.

The cutoff as something to design with #

The cutoff that makes an LC ladder a poor line makes it a good filter, and the construction is old enough to have a name. A chain of identical inductor–capacitor sections is a constant-k low-pass filter, and its cutoff is exactly $2/\sqrt{L_sC_s}$: the frequency at which the Bloch phase per section reaches half a turn. Below it the filter passes with a phase delay the arcsine gives; above it, attenuates by an amount that grows with every section.

Seen from that side, the ringing a ladder is not a line measured on the plateaus of that chain of inductors and capacitors is the step response of a steep low-pass filter, and what a steep skirt costs is the essay about what that response costs in delay distortion. The two descriptions — a line that fails above a frequency, a filter that works below one — are one object, and which description is appropriate depends on whether the signal’s content lies well below the cutoff or reaches it. An artificial delay line is the first use, a constant-k filter the second, and the same chain of parts is either, according to the signal put through it.

What this settles, and where the LC ladder still earns its place #

An LC ladder is a line to a stated accuracy below a stated frequency, and both are exact. The cutoff is $2/\sqrt{L_sC_s}$ of one section, the delay error is $\arcsin(x)/x - 1$, the group delay and impedance errors are $1/\sqrt{1 - x^2} - 1$, and $x$ is π over the sections per wavelength. That is the whole of the lumped approximation’s frequency-domain edge.

The number to quote is sections per wavelength at the highest frequency that matters, and the accuracy to quote with it is the group delay’s rather than the phase delay’s for anything carrying pulses. Thirteen per wavelength is a one-per-cent line for a sine wave’s phase, twenty-two for a pulse’s timing.

The LC ladder’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.

And an LC ladder is not only a model. The via that is a piece of line found a single pad–barrel–pad via behaving as one section of exactly this kind, delaying an edge by $\sqrt{LC}$ and reflecting a doublet that grows as the edge approaches the via’s own delay. A via is an LC ladder of one section, and the error it makes is this essay’s error at $N = 1$. Kirchhoff’s own frequency put a frequency on when a piece of copper stops being a node; this puts a frequency on when a chain of nodes starts being a line.

Still open: an LC ladder with loss, an LC ladder with half-sections at its ends, and the edge through the ladder #

Loss in each section. A real line has series resistance and shunt conductance, and a ladder built to imitate one puts a resistor in each section. Above the frequency where the section’s reactance exceeds its resistance the Bloch phase above is unchanged in form; below it the chain is diffusive, and the delay that is not one number measured what that does to a real line. Whether a lossy LC ladder converges on a lossy line at the same number of sections per wavelength, or needs more because loss makes the low-frequency end dispersive as well, is a distinct question.

Half-sections at the ends. The ripple here comes from terminating a chain whose own impedance is $Z_0\sqrt{1-x^2}$ at one end and $Z_0/\sqrt{1-x^2}$ at the other in a fixed fifty ohms. The classical remedy is to end the chain with half a section, which makes the two ends’ impedances equal and moves the mismatch to fourth order. Solving the same chain with half-sections would say whether the terminations’ ripple — the frequency-domain face of the time-domain ringing — falls with them, and by how much.

The edge through the LC ladder. The sections-per-wavelength count above uses a rule of thumb for an edge’s highest frequency. Passing a raised-cosine edge of known rise time through a LC ladder in the frequency domain and measuring the transmitted rise time and overshoot against a line’s would replace the rule with a measurement, and would test whether 379 sections for a 100-picosecond edge on a metre of cable is the right order.