Circuit Response
every model has an edge, and the figure carries it
Lines, where a wire has a length

A ladder is not a line

A transmission line is usually introduced as the limit of a chain of inductors and capacitors as the number of sections goes to infinity. That is true, and it gives entirely the wrong impression of how close a finite chain gets. Forty sections still ring through every plateau by five per cent, and extrapolating the fitted convergence, reaching one per cent would need about nine hundred and sixty.
16 min read 7 figures Two routes to a numberEvery model has an edgeThe constants decide

The wave picture of the previous essay is a construction: a step launches, travels, reflects, comes back, reflects again, and the voltage anywhere is the superposition of everything that has arrived so far. It is not a solve. Nothing in it goes through the matrix that every other answer on this site comes out of.

So it needs a second route, and one is available that could hardly be more different. Replace the line with N inductor-capacitor sections and march the resulting lumped network forward in time with the trapezoidal rule — the same integrate that draws every step response in the transients field, and one that has never heard of a wave.

20 inductor-capacitor sections, against the line they are meant to becomputed by solving, not by drawing by the trapezoidal rule over 2,600 steps. The ladder reaches two per cent of full scale at 0.86 delays, before the wave picture says anything can have arrived, and its plateaus are wrong by up to 0.079 V. Neither is a small correction to the wave answer; they are what a network of 20 poles does when asked to be a delay.00.50011.50202468time, in propagation delaysvolts at the far endthe wave answer20 sectionsone delaysolved, then checked — a ladder against a latticeplateaus 0.08 V out at 20 sections
Fig. 1 Twenty inductor-capacitor sections against the staircase they are meant to be. The ladder has to produce the delay, the characteristic impedance and the reflection coefficients; nothing in it knows any of them. What it produces is recognisably the staircase, with ringing on every plateau and an edge that begins before the wave picture says anything can have arrived. The slider is the number of sections.

What the ladder gets right

The agreement is real and it is worth stating before the disagreements, because the disagreements are the point of the page and it would be easy to leave the wrong impression.

The ladder reaches roughly the right plateaus, in roughly the right places, in the right order. It overshoots into an open circuit and undershoots into a low resistance. Its long-term limit is the resistive divider, because at direct current the inductors are shorts and the capacitors are opens and the ladder is a resistive divider. Everything qualitative about the staircase is reproduced by a network that contains no delay, no impedance and no reflection coefficient, which is a genuinely striking thing and the reason the construction is taught.

What it does not do is get the numbers right, and the interesting part is by how much and how slowly that improves. It also gets one thing wrong in kind rather than in degree, and that one is worth naming immediately because no number of sections fixes it: the ladder’s far end starts moving straight away. There is no delay in a lumped network — every node is connected to every other through some path of finite impedance, so a change anywhere appears everywhere at once, attenuated but not delayed. Adding sections makes the early movement smaller and never makes it zero.

Two errors, and they behave differently

Two quantities are measured, and separating them is what makes the convergence legible.

The plateau error is taken on the flat stretches between arrivals, where the wave picture says the voltage is constant. The ladder is not constant there; it is ringing from the last edge, because a chain of N sections is a low-pass filter with N poles and an abrupt input excites all of them.

The rise time is the 10–90% transition of the first arrival, which the wave picture puts at zero. A lumped chain has no delay line in it, only that filter, and its edge is smeared over something like the reciprocal of the filter’s corner.

They are measured at five, ten, twenty and forty sections, and the comparison begins after the first arrival — before it the two disagree completely and correctly, since the lattice says the far end is at zero and the ladder says it is at something small and rising. Averaging that in would hide the very difference the figure exists to show.

What it would take for a ladder to be a linecomputed by solving, not by drawing at 4 section counts. The plateau error falls as N^-0.54 and the edge's rise time as N^-0.73, the second with a worst residual of 0.92%. Extrapolating the first, a ladder within 1% of the wave answer on its plateaus needs about 961 sections. The extrapolation is drawn dashed because that is what it is.-3-2-10101001ksections in the ladderlog₁₀ of the errorplateau error, N^-0.54rise time ÷ delay, N^-0.731%961 sections, extrapolatedmeasured to heresolved, then checked — two fits, residuals statedmeasured to 40, extrapolated to 961
Fig. 2 The two errors against the number of sections, on logarithmic axes, with the fitted power laws drawn through them and the extrapolation drawn dashed. The plateau error falls as N^−0.54 with a worst residual of 8.7%; the rise time falls as N^−0.73 with a worst residual of 0.9%. The slider is the error to reach, and the marked edge is the extrapolated section count.

The numbers

The plateau error is 0.163 V at five sections, 0.128 V at ten, 0.079 V at twenty and 0.055 V at forty, against a first-arrival amplitude of 1.667 V. Fitting a straight line through the logarithms gives an exponent of −0.54.

The rise time, as a fraction of the propagation delay, is 0.443, 0.261, 0.157 and 0.096. The exponent is −0.73.

Both fits are quoted with their worst residuals, which is the only honest way to state a convergence rate. The rise-time fit’s is 0.92% — four points that lie on a line to within one per cent, which is a fit worth extrapolating. The plateau fit’s is 8.7%, which is a fit worth quoting and treating with suspicion at any distance from the measured range.

Extrapolating the plateau fit to one per cent gives about nine hundred and sixty sections. That number is drawn dashed and labelled as an extrapolation, and the reason for the care is exactly the 8.7% residual: an exponent uncertain in the second digit, extrapolated across a decade and a half, gives a section count uncertain by a factor of two either way. The right reading is “of order a thousand”, not “961”.

5 inductor-capacitor sections, against the line they are meant to becomputed by solving, not by drawing by the trapezoidal rule over 2,600 steps. The ladder reaches two per cent of full scale at 0.67 delays, before the wave picture says anything can have arrived, and its plateaus are wrong by up to 0.163 V. Neither is a small correction to the wave answer; they are what a network of 5 poles does when asked to be a delay.00.50011.50202468time, in propagation delaysvolts at the far endthe wave answer5 sectionsone delaysolved, then checked — a ladder against a latticeplateaus 0.16 V out at 5 sections
Fig. 3 Five sections. The plateau error is 0.163 V — a tenth of the arrival amplitude — and the edge takes almost half a propagation delay to rise. This is what the phrase “a line is a ladder of many sections” looks like when “many” is a number one would actually build, and it is why an artificial delay line made of discrete parts has a bandwidth rather than a delay.

What the comparison had to be careful about

Three decisions in the measurement change the answer, and each is worth stating because a different choice would give a different exponent and there would be no way to tell from the number alone.

Where the plateaus are sampled. The comparison takes the middle two-fifths of each interval between arrivals, from 0.3 to 0.7 of the way along. Sampling nearer the edges would include the transition itself, which is not a plateau in either model, and would report an error that is mostly about timing rather than about level.

When the comparison starts. After 1.3 propagation delays, which is to say after the first arrival has finished. Before it the two models disagree completely and correctly, and including that stretch would put a large constant into every error and flatten the exponent towards zero.

How many time steps. Two thousand six hundred over eight propagation delays, which is about three hundred and twenty-five per delay and comfortably finer than the fastest feature at forty sections. The trapezoidal rule’s own error falls by a factor of four for every halving of the step, which the transients field measures directly across five step sizes, so it is straightforward to confirm that the integration is not what is being measured here.

None of the three is a free parameter in the sense of having been tuned. Each was chosen once, for a stated reason, and the reason in every case is that the alternative measures something other than the quantity named.

Why the convergence is so slow

Neither exponent is one, and neither is one half, and the reason for that is worth a paragraph because it says something about what is being approximated.

A ladder of N sections is a low-pass filter whose corner is at roughly N/πT_d. An ideal delay line passes every frequency and delays them all equally; the ladder passes frequencies up to its corner and stops. So the error is essentially the energy in the step above the ladder’s corner, and for a step — whose spectrum falls as 1/f — that energy falls slowly with the corner frequency.

There is a second effect on top of it, which is that the ladder does not merely truncate the spectrum, it also disperses it: the components near the corner are delayed differently from those well below it, so the edge does not simply become gentler, it acquires the ringing visible on every plateau. That ringing is a Gibbs-like phenomenon and its envelope decays much more slowly than its frequency content suggests.

The measured exponents are therefore properties of the interaction of those two effects over the range tested, rather than a clean asymptotic law, and the honest way to present them is as a fit over a stated range with a stated residual — which is what the figure does.

The same step through all three, at order 5Overshoot measured off each curve: Butterworth 12.8%, Chebyshev 12.4%, Bessel 0.8%. The family with the flat delay barely overshoots; the other two, whose delay varies by tens of per cent, overshoot by more than ten and are hard to tell apart — magnitude flatness does not decide it.00.500101234time (milliseconds)output, for a 1 V step inButterworth 12.8%Chebyshev 12.4%Bessel 0.8%solved, then checked — residues, checked by integrationovershoot follows the delay, not the magnitude
Fig. 4 A filter’s step response from the filters field, where the same mechanism appears without a line anywhere. A steeper skirt buys attenuation and pays for it in ringing and delay variation, and a ladder pretending to be a delay line is paying exactly that price without having asked for the attenuation.

Where the ringing sits, and the other place it appears

The ringing on the plateaus has a frequency, and identifying it turns a visual impression into a number.

An N-section ladder representing a line of total delay T_d has section inductance L′ℓ/N and section capacitance C′ℓ/N, so each section’s own resonant frequency is N/(2πT_d) — twenty sections on a 4.83 ns line puts it at about 660 MHz. The chain’s cutoff is twice that, at N/(πT_d), and the ringing visible on the plateaus is the chain ringing near its own cutoff.

Two predictions follow and both are visible on the slider. Doubling the sections doubles the ringing frequency, so the oscillations on the plateaus become finer as sections are added rather than smaller — which is exactly why the plateau error, measured as a worst departure, falls so much more slowly than the eye expects. And the ringing’s decay is set by the terminations rather than by the ladder, so into an open circuit it barely decays at all.

The same mechanism appears in a field of this collection that has nothing to do with lines. What a steep skirt costs measures the ringing a filter’s step response acquires as its skirt steepens, and finds the same trade: a sharper cutoff in frequency is a longer ring in time, and no arrangement of poles avoids it. A ladder pretending to be a delay line is paying that price without having asked for the attenuation, which is the least favourable position on the trade it is possible to occupy.

What this says about the lumped model generally

The conclusion is stronger than “the lumped model is approximate”, and it is the reason this field exists rather than being an appendix to Kirchhoff’s own frequency.

A limit that converges as N^−0.54 is not a practical approximation. There is no section count that a board or a simulator would actually use at which the ladder is a good description of a line. The wave picture is therefore not a convenient summary of a lumped one; the lumped one is a bad approximation to it, and the two are different objects that happen to have the same limit.

That reverses the order in which the two are usually introduced, and the reversal is the useful part. A reader who learns the ladder first and the line second carries an intuition that the line is a refinement. A reader who measures the convergence carries the opposite one, which is closer to correct: above the frequency where a circuit is electrically large, the line is the description and the lumped network is the thing that needs justifying.

A 1 V step onto 1.00 m of 50 Ω line into an open circuitcomputed by solving, not by drawing as a sum of 81 arrivals. The source drives 0.8333 V into the line immediately — set by 10 Ω against the line's 50 Ω, and not by the load, which it cannot yet know about. One delay of 4.83 ns later the far end reaches 1.6666 V. The staircase settles at 0.999990 V, which is what the resistive divider gives.00.50011.5002468time, in propagation delaysvoltsthe divider: 1.0000 V0.833 V: the source against the linethe far endthe near endfirst arrival, 4.83 nssolved, then checked — a series against a dividernothing at the far end before 4.83 ns
Fig. 5 The wave picture the ladder is being measured against, from the previous essay. Its own credentials are established separately: the reflections sum to the resistive divider the circuit becomes, and the far end sits at exactly zero until one delay has elapsed. Neither of those checks involves a ladder, so the comparison on this page is between two independently checked answers rather than between an answer and a reference.

What a ladder is good for

None of this makes the ladder useless, and it is worth saying what it is actually for.

It is a real circuit. An artificial delay line built from discrete inductors and capacitors is a manufactured component, and everything on this page is a description of how it behaves rather than a criticism of it. Its bandwidth is its corner frequency, its delay is N√(LC), and its ringing is a specification.

It is how a simulator without a line model copes. A network simulator that has transmission-line elements uses them. One that does not can approximate with sections, and the measurements here say what that costs — which is a legitimate thing to want to know, and an argument for using enough sections that the answer is at least of the right shape.

It shows where the impedance comes from. The characteristic impedance of the ladder is √(L/C) per section, which is the same √(L′/C′) the continuous line has, and watching a chain of components produce it is a better explanation of why a cable has a resistance that dissipates nothing than any amount of prose about fields.

The third of those is worth one more sentence, because it is the reason the ladder route was built at all rather than a second analytic model being found. A chain of ideal inductors and capacitors contains nothing that dissipates: every element stores energy and gives it back. Yet the ladder, driven from a source through a ten-ohm resistance, initially draws exactly the current that a fifty-ohm resistor would — and it does so for as long as the far end has not yet answered, which for twenty sections is roughly one propagation delay.

A network of pure reactances presenting a resistance is the single most counter-intuitive statement in this field, and the ladder demonstrates it in a form that can be solved by the same matrix as every other circuit in the collection. The line model asserts it as a property of Z₀; the ladder produces it out of components whose individual behaviour contains nothing of the sort.

10.0 cm of track, solved as a lumped circuit and as a lineThe two agree to 0.030% at 3.97 MHz, where the track is one degree long, and to 30.1% at 143 MHz, where it is a tenth of a wavelength. Above that the lumped model is not approximately right; it is describing a different object.1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz
Fig. 6 The same piece of track as a two-element lumped circuit and as a line, from the limits field. That figure is this page’s argument in the frequency domain and at N = 1: below the crossover the two agree, above it they describe different objects, and everything on this page is a measurement of how much of the gap a larger N closes.

The measurement stops at forty, and why

The section counts are five, ten, twenty and forty, and the reason the sequence does not continue is arithmetic rather than editorial.

An integration costs roughly the cube of the section count per time step, because each step solves a matrix whose size is proportional to the number of sections. Forty sections take about three seconds for one trace; eighty take about twenty-five; a hundred and sixty would take three minutes. A figure that a reader can drag is regenerated at every position of its slider at build time, so a cost of three minutes per frame is not a figure.

The honest consequence is that the extrapolation to a thousand sections spans a decade and a half beyond anything measured, and that is why it is drawn dashed and quoted as an order of magnitude. It would be entirely possible to run the eighty-section case once, offline, and add a fifth point; what would not be possible is to have that point present in an interactive figure without making every build a quarter of an hour longer.

That trade — the range of a measurement against the cost of regenerating it — is a constraint this collection meets often and states rarely, and it is worth stating here because the extrapolation is the page’s most quotable number and the least well supported.

What it would take for a ladder to be a linecomputed by solving, not by drawing at 4 section counts. The plateau error falls as N^-0.54 and the edge's rise time as N^-0.73, the second with a worst residual of 0.92%. Extrapolating the first, a ladder within 5% of the wave answer on its plateaus needs about 49 sections. The extrapolation is drawn dashed because that is what it is.-3-2-10101001ksections in the ladderlog₁₀ of the errorplateau error, N^-0.54rise time ÷ delay, N^-0.735%49 sections, extrapolatedmeasured to heresolved, then checked — two fits, residuals statedmeasured to 40, extrapolated to 49
Fig. 7 The same fit, asked for five per cent rather than one. Forty-nine sections — a number one might actually build, and only a little beyond the range measured, so the extrapolation is a short one. The contrast with the thousand needed for one per cent is the shape of a power law with an exponent near a half: each further decade of accuracy costs a hundredfold in sections.

Where this goes next

00.50011.5002468time, in propagation delaysvoltsthe divider: 1.0000 V0.833 V: the source against the linethe far endthe near endfirst arrival, 4.83 nssolved, then checked — a series against a dividernothing at the far end before 4.83 ns The staircase in time A source driving a metre of cable does not know what is on the far end of it for 4.83 nanoseconds. What it drives into during that time is decided by the cable's characteristic impedance and nothing else — and when the far end finally answers, the answer comes back as a staircase whose limit is the resistive divider the circuit was going to be all along. 1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz Kirchhoff's own frequency The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board. passband deviationdecibels, peak to trough below 0.8 f_cButterworth0.443 dBChebyshev0.500 dBBessel1.882 dBattenuation at three times the cornerdecibels downButterworth47.7 dBChebyshev64.0 dBBessel28.3 dBgroup-delay variation across the passbandper cent, slowest against fastestButterworth48.0%Chebyshev49.0%Bessel0.1%solved, then checked — nine measurements, three networksevery number here moves with the order What a steep skirt costs A filter's order buys attenuation at a known rate — twenty decibels per decade per pole, and no arrangement of components changes it. What varies between families is how quickly the slope is reached, and the currency it is paid for in is delay: the steepest of the three distorts delay eight hundred times more than the gentlest.