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

The near end that stops growing

Near-end crosstalk is a length only until the coupled run's round trip equals the edge's rise time; past that it is a constant. Solved mode by mode and returned to time, a 200 ps edge on 100 mm of coupled track gives a plateau of 12.51 mV per volt — Kb, a quarter of the sum of the two couplings, on the driven line's half-volt — held for the 1.33 ns round trip. The knee is at v·tr/2: 7 mm for a 100 ps edge, 75 mm for 1 ns. Past it a longer run adds width and not height, and the pulse's area keeps growing as Kb times the round trip.
16 min read 5 figures Every model has an edgeTwo routes to a numberMeasured, not tabulated

Assumes: The far end that cancels · The staircase in time

The far end that cancels measured the two ways a pair of parallel tracks couple — a mutual capacitance injecting a current, a mutual inductance injecting a voltage — and found that at the quiet track’s near end they add while at its far end they subtract, with one constant in front of both. How wide a null is then measured how quickly the far end’s cancellation comes back as the two couplings part.

Both worked in the frequency domain, at a steady sinusoid, which is the natural place to see the two mechanisms as matrix entries. But crosstalk is almost always met as a time-domain event — a digital edge on one track and a glitch on its neighbour — and the frequency-domain picture leaves out the one property of the near end that makes it predictable in practice. A near-end coupling coefficient, quoted as a number, sounds like a gain: twice the coupled length, twice the crosstalk. That is true only for short runs. Beyond a length set by the edge itself, the near end stops growing altogether, and the length goes into the pulse’s width instead.

This essay measures that in time, on an exact model of the coupled pair, and finds where the knee is, what sets the plateau and what keeps growing after it.

Near-end crosstalk from a 200 ps edge, 100 mm of coupled track against 10 mmcomputed by solving, not by drawing, the pair solved mode by mode and returned to time from its harmonics. Over 100 mm the near end rises with the edge, holds a plateau of 12.51 mV per volt for twice the line's delay, 1.33 ns, and falls. Over 10 mm the round trip is shorter than the edge and the pulse never reaches the plateau: 8.34 mV. The plateau is Kb, a quarter of the sum of the two couplings, on the 0.5 V the driven line carries: 12.50 mV.05101501234time, nsquiet line's near end, mV per volt of driveKb: 12.50 mV10 mmsolved, then checked — modes, then harmonicsplateau for 2Td
Fig. 1 Near-end crosstalk from a 200 ps edge. Over 100 mm the quiet line’s near end rises with the edge, holds a plateau of 12.51 mV per volt of drive for twice the line’s delay, 1.33 ns, and falls. Over 10 mm the round trip is shorter than the edge and the pulse never reaches the plateau: 8.34 mV. The plateau is Kb, a quarter of the sum of the two couplings, on the 0.5 V the driven line carries: 12.50 mV. The slider moves the coupled length.

The pair, solved mode by mode

The two tracks are each a 50-ohm line in a dielectric of relative permittivity four, so a signal travels at half the speed of light, 150 millimetres a nanosecond. Their coupling is a mutual capacitance of five per cent of each track’s capacitance and a mutual inductance set by a ratio to it — equal in a homogeneous dielectric, larger on a microstrip whose field is partly in air. All four ends are terminated in 50 ohms, and one track is driven from a 50-ohm source, so the driven line carries half the source voltage.

The earlier measurements solved this as a ladder network: twelve sections of series inductance and shunt capacitance on each track, stamped with their mutual elements. That is the right model for a matrix and a poor one for a time-domain march, because every section adds poles and a long run needs many of them. So here the pair is solved exactly instead. With all four ends terminated in the same resistance, the even and odd modes — the two tracks moving together and in opposition — separate completely: the transformation that diagonalises the coupled line also diagonalises its terminations. Each mode is then a single uniform line with its own impedance and its own speed, from the same per-length inductances and capacitances the ladder network uses, and its two ends are given by the chain matrix of a uniform line in closed form. The quiet track’s near and far ends are the even mode minus the odd.

Two routes to the quiet line: the distributed pair solved by modes, and a 48-section ladder network of coupled inductors and capacitors. computed by solving, not by drawing, at an inductive-to-capacitive coupling ratio of 1.6 so that the far end is not zero. Lines: the modes, exact for a uniform lossless pair. Dots: the ladder network, solved by nodal analysis with its coupled inductors as matrix entries. They agree to 2.3e-5 below 30 MHz and part only where a 2 mm section stops being short against the wavelength. The time-domain figures use the modes.
Fig. 2 Two routes to the quiet line, at an inductive-to-capacitive coupling ratio of 1.6 so that the far end is not zero. Lines: the pair solved by modes, exact for a uniform lossless pair. Dots: a 48-section ladder network of coupled inductors and capacitors, solved by nodal analysis. They agree to 2.3 × 10⁻⁵ below 30 MHz and part only where a 2 mm section stops being short against the wavelength.

The two routes share nothing but the per-length values. One is a sum of two closed-form transmission lines; the other a 192-element network solved as a matrix. Below 30 megahertz they agree to 2.3 parts in 100,000, and above that the ladder network departs, as the sections a wavelength needs says a ladder network must once its sections stop being short. The modal solution has no sections to run out of, which is why the time-domain figures here use it.

Time comes back from frequency through harmonics. The drive is a periodic trapezoid — a 40-nanosecond period, so long against every line drawn that the edge’s own return half a period later never reaches the window — and its harmonics are exact, a product of two sinc functions. Each harmonic is passed through the modal solution and the results are summed. The 0-to-100-per-cent rise time is the edge’s one parameter.

A plateau, held for the round trip

The near-end waveform over a long run has a shape that the frequency-domain coefficient does not hint at. It rises with the edge, reaches a plateau, holds it, and falls — and the plateau’s length is exactly twice the line’s one-way delay.

The reason is the way the coupling is collected. As the edge travels down the driven track, every millimetre it passes injects a small backward wave into the quiet track — a capacitive current and an inductive voltage that add for a wave travelling backward. That backward wave travels to the quiet track’s near end at the same speed. The near end therefore receives, at any instant, the contributions from a stretch of the driven line whose length is the distance the edge covers in half the time — and once the edge has fully risen, each of those contributions is a fixed amount per millimetre of edge front. The sum is constant as long as the edge is somewhere on the line, which is from the moment it enters until it reaches the far end; the last backward contributions then take another delay to return. So the plateau starts when the edge has fully risen and lasts for the round trip, 1.33 nanoseconds for 100 millimetres.

Its height is the backward coupling coefficient times the voltage on the driven line:

Vnear=Kb⋅Vline,Kb=14(CmC+LmL).V_{\text{near}} = K_b \cdot V_{\text{line}}, \qquad K_b = \frac14\left(\frac{C_m}{C} + \frac{L_m}{L}\right).

With both couplings at five per cent KbK_b is 0.025, and on the 0.5 volts the driven line carries per volt of drive that is 12.50 millivolts. The modal solution’s plateau is 12.51.

Why the two mechanisms add going backward

The plateau’s height is a sum, and the sum has a direction in it that is worth spelling out, because it is the whole difference between the two ends. At each point along the run the passing edge does two things to the quiet track. Its rising voltage pushes a current through the mutual capacitance into the quiet track, and that current divides equally, half flowing toward each end. Its rising current induces a voltage along the quiet track through the mutual inductance, and that voltage, appearing in series, drives a current around the quiet track’s loop — toward the near end and away from the far end, in opposite senses at the two ends.

So at the near end the capacitive half and the inductive contribution have the same sign and add, and at the far end they have opposite signs and subtract. That is the frequency-domain result of the far end that cancels, read as two currents choosing directions. What the time domain adds is the bookkeeping of when. Contributions heading backward meet the oncoming edge’s later contributions and are collected over a stretch of line, which is why the near end becomes a plateau; those heading forward ride along with the edge and arrive all together at the far end, which is why the far end, when it does not cancel, arrives as a single pulse at the moment the edge does.

The inductive half depends on the quiet track having a return to complete its loop, and the loop’s area is set by how the return current flows under the pair. On a board with a continuous plane directly beneath, the return hugs each track and the mutual inductance is small; across a gap in the plane the return detours and the mutual inductance grows, which is the mechanism where the current comes back measures for a single track. The near end on a run that crosses a split plane is therefore larger than the per-length values of the unbroken stack-up predict, by however much the detour adds to the inductive term.

The knee is half the edge

A short run cannot reach the plateau, because the edge is still rising when it arrives at the far end: the backward contributions from a partial edge return before the full edge has been collected. The near end then peaks at a fraction of the plateau equal to the round trip over the rise time.

Near-end crosstalk peak against coupled length, for edges of 100 ps, 300 ps and 1 ns. computed by solving, not by drawing. Each rise time gives a line rising in proportion to the length and then a plateau at 12.50 mV per volt, Kb on the driven line's half of the source. The knee is where the round trip equals the rise time, at v·tr/2: 7 mm for 100 ps, 22 mm for 300 ps, 75 mm for 1000 ps. Past it a longer run adds width, not height.
Fig. 3 Near-end peak against coupled length, for edges of 100 ps, 300 ps and 1 ns. Each rises in proportion to the length and then stops at 12.50 mV per volt. The knee is where the round trip equals the rise time, at v·tr/2: 7 mm for 100 ps, 22 mm for 300 ps and 75 mm for 1 ns.

So the near end is a length up to the saturation length Lsat=v tr/2L_{\text{sat}} = v\,t_r/2, and a constant past it. Below the knee the measured peaks are proportional to length to within eight per cent, and above twice the knee they are the plateau to within three per cent, for all three edges. The knee moves with the rise time and the plateau does not: 7 millimetres for a 100-picosecond edge, 22 for 300 picoseconds, 75 for a nanosecond.

That reverses the intuition a coupling coefficient creates. On a board with 100-picosecond edges every parallel run longer than about seven millimetres produces the same near-end crosstalk, so shortening a 100 millimetre run to 50 buys nothing at the near end. Only a run shorter than the knee is improved by being shortened, and for fast edges that is a very short run indeed. The lengths that matter are set by the edge, not by the layout.

It also says why a slow driver helps the near end only on short runs. A one-nanosecond edge on a 20-millimetre run is well below its 75-millimetre knee and produces about a quarter of the plateau; the same run with a 100-picosecond edge is past its knee and produces all of it. The near end on long runs is indifferent to the edge rate, which is the property the staircase in time found in a line’s reflections: a feature whose height is set by impedances and whose duration is set by delays.

Height saturates; area does not

The plateau is a statement about height. What a victim circuit actually responds to is not always height, and the pulse’s other dimension behaves differently.

The near-end pulse's height and its area against length, for a 200 ps edge. computed by solving, not by drawing. The height stops at Kb once the run passes 15 mm. The area — the volt-seconds the quiet line's near end receives — does not stop: it is Kb times the round-trip time at every length, 0.83 mV·ns at 5 mm and 53.3 mV·ns at 320 mm. A receiver that responds to the peak sees saturation; one that integrates — a slow input, a charge-sensitive node — sees the length.
Fig. 4 The near-end pulse’s height (dark) and its area (pale) against length, for a 200 ps edge. The height stops at Kb once the run passes 15 mm. The area — the volt-seconds the quiet line’s near end receives — is Kb times the round-trip time at every length, 0.83 mV·ns at 5 mm and 53.3 mV·ns at 320 mm.

The area under the near-end pulse is Kb⋅Vline⋅2TdK_b \cdot V_{\text{line}} \cdot 2T_d at every length, short or long, to within one per cent. Below the knee the pulse is short and low; above it the pulse is at full height and its width grows with the length. Either way the volt-seconds are proportional to the coupled length.

Which of the two matters depends on what is listening. A fast logic input that switches on a threshold responds to height, and for it near-end crosstalk saturates. A slow input, a node with a capacitor on it, a reference that integrates what it is given, or a receiver whose noise margin is a time-and-voltage area responds to the volt-seconds, and for it the near end grows with the run at any length. The same pair of tracks is a fixed problem for one receiver and a length-proportional one for another, and the frequency-domain coefficient, which grows linearly with length at low frequency, is a statement about the second kind.

The plateau is the sum, whatever the proportion

The far end subtracts the two couplings and can be cancelled by making them equal; the near end adds them, so it should not care how the total is divided.

The saturated near end against the ratio of the two couplings, capacitive coupling held at 5%. computed by solving, not by drawing, 150 mm with a 200 ps edge. The plateau is (Cm/C + Lm/L)/4 on the driven line's 0.5 V at every ratio — 9.38 at 0.5, 11.26 at 0.8, 12.51 at 1, 14.08 at 1.25, 16.27 at 1.6, 18.79 at 2 mV per volt — a straight line through the sum. The near end adds the two mechanisms whatever their proportion, which is why it cannot be cancelled by balancing them the way the far end can.
Fig. 5 The saturated near end, 150 mm with a 200 ps edge, against the ratio of inductive to capacitive coupling, the capacitive held at 5 per cent. The plateau is (Cm/C + Lm/L)/4 on the driven line’s 0.5 V at every ratio — 9.38 mV per volt at 0.5, 12.51 at 1, 16.27 at 1.6 and 18.79 at 2 — a straight line through the sum.

Across a fourfold range of the ratio between the two mechanisms, the saturated near end is a quarter of their sum on the driven line’s voltage, to within three per cent: 9.38 millivolts per volt when the inductive coupling is half the capacitive, 18.79 when it is twice. The figure is a straight line because the near end has no cancellation in it to exploit.

That is the practical asymmetry between the two ends. How wide a null is found the far end vanishing in a homogeneous dielectric and returning steeply as the dielectric becomes inhomogeneous, so the far end is something a stack-up can be designed to suppress. The near end can only be reduced by reducing the couplings themselves — moving the tracks apart, or bringing a plane closer — because both mechanisms push it the same way. On a stripline buried in a single dielectric, the far end is cancelled and the near end is the whole of the crosstalk; that is why stripline crosstalk budgets are written almost entirely in terms of near-end saturation.

What the plateau is used for

A designer budgeting crosstalk on a bus does not need a simulation of every run to use these results; the saturated near end gives a bound that holds for every run past the knee. With five per cent couplings the bound is 2.5 per cent of the line voltage. A 3.3-volt signal swinging fully on a 50-ohm line, driven from a matched source, puts 1.65 volts on the line and at most 41 millivolts on its neighbour’s near end, however long the parallel run. Two aggressors, one on each side, put at most twice that, since near-end contributions from different lines add.

If that is too much, the bound says where the margin comes from. Halving the couplings halves it, and couplings fall roughly as the square of the spacing-to-height ratio once the tracks are a few dielectric heights apart. Terminating the quiet line’s near end matters too: the plateau is measured into 50 ohms, and a quiet line whose near end is driven by a low-impedance output reflects the backward wave toward its far end, where it arrives a delay later at full size — the resistor at the wrong end is the same argument for a signal rather than for crosstalk. Shortening the run, the first instinct, helps only below the knee, and for a 100-picosecond edge that means below seven millimetres.

The area result adds one caution to the budget. A run long enough to be well past its knee still delivers volt-seconds in proportion to its length, so a victim whose sensitivity is to energy rather than to peak — a sampling node, an analog reference sharing a route with a clock — is not protected by saturation, and for it the run length does matter.

What the model leaves out

The lines are lossless. A real track’s resistance and its dielectric’s loss attenuate the backward contributions from the far part of the run more than those from the near part, so the plateau slopes downward along its length instead of holding flat, and the edge on the driven line slows as it travels, which moves the knee outward along the run. The cable that hides two things measures what loss conceals in a single line, and the same concealment applies to the coupled pair.

Every end is terminated in exactly the line’s impedance. That is what lets the modes separate, and it is also not quite true of a coupled pair, whose even and odd modes have impedances slightly above and below 50 ohms; a 50-ohm termination is therefore slightly mismatched to each mode, and small reflections ride on the plateau. They are present in the modal solution and visible in the figures as the plateau’s last decimal.

The coupling is uniform along the run. Real routes enter and leave their parallel section through bends and vias, where the coupling tapers; a tapered entry spreads the near-end rise over a longer time and lowers it slightly below the plateau on runs near the knee. Only two conductors are modelled: a third track beside the victim adds its own near end, and a grounded guard trace between them changes both couplings, which the guard that is part of the return treats as a return-path question.

The number to keep

Near-end crosstalk in time is a plateau of Kb=(Cm/C+Lm/L)/4K_b = (C_m/C + L_m/L)/4 times the driven line’s voltage — 12.51 millivolts per volt here, for five per cent couplings — held for the run’s round trip. It is reached once the run is longer than half the distance the edge travels in its rise time, v tr/2v\,t_r/2: seven millimetres for a 100-picosecond edge, 75 for a nanosecond. Past that a longer run adds width and not height, while the pulse’s area keeps growing as KbK_b times the round trip. The plateau adds the two couplings in any proportion, so unlike the far end it cannot be cancelled by balancing them.

Still open: the far end in time, the rise time that is too fast, and the receiver that integrates

The far end as a derivative. In an inhomogeneous dielectric the far end does not cancel, and in time it should behave as the opposite of the near end: a pulse as wide as the edge, arriving after one delay, with a height proportional to the run length and inversely to the rise time. Whether that law holds, and where it stops holding, is the companion measurement.

The edge faster than the modes are matched. When even and odd modes travel at different speeds, a long enough run separates them by more than an edge, and both ends change character. For the far end that is the length where its growth must stop.

The integrating receiver. A victim with a capacitive input filters the near-end pulse, and its response is set by area for long time constants and height for short ones. Marching a receiver’s input network on the modal solution would turn the area result into a voltage at a decision point.

Part 3 on crosstalk

One argument about Crosstalk, and one of 4 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

The null is a V and not a bowl: one per cent of ratio error is 5.0e-3 of the near end. computed by solving, not by drawing. The same twelve coupled sections read at 10.0 MHz, with the ratio of the two couplings swept across the null rather than sat at one setting. The far end divided by the near end is |1 − r|/(1 + r) at every point — a straight-sided V through zero, first order in the departure with a coefficient of one half, and not a rounded minimum with a flat bottom. So there is no tolerance band: a ratio one per cent off gives 4.98e-3 of the near end and ten per cent off gives 4.76e-2, and the exchange rate between them is fixed. The upper trace is the near end over the same sweep, which moves by 11 per cent while the lower one moves through 14 decades — the two ends are the same coupling read as a sum and as a difference, which is why one of them has a zero in it and the other cannot. How wide a null is Part 2 — A far end at 3.5×10⁻¹⁹ of the drive is a statement about arithmetic until somebody asks how far the two couplings may differ before it comes back. The answer has no flat bottom in it: the far end divided by the near end is |1−r|/(1+r) exactly, so the null is a V and a ratio one per cent off returns 4.98×10⁻³ of the near end. On the axis a board is built to that is a difference of 0.0081 between the two modes' effective permittivities, out of 3.99 — two parts in a thousand, and 0.675 picoseconds of mode skew over a hundred millimetres. Both ends of the quiet line, 100 mm of microstrip-like coupling, edge 200 ps. computed by solving, not by drawing. The near end (pale) is the saturated plateau, 16.28 mV per volt. The far end (dark) is a single pulse arriving with the edge after one delay, 0.676 ns, as wide as the edge and 25.16 mV per volt deep — negative, because the inductive coupling exceeds the capacitive. Its depth is the coupling difference, the delay and the edge's slope multiplied: 24.96 predicted. The far end that is a derivative Part 4 — Where the two couplings are unequal the far end does not cancel, and in time it is the derivative of the edge: a single pulse as wide as the rise time, arriving after one delay, its height the coupling difference times the delay times the edge's slope. On 100 mm of microstrip-like coupling a 200 ps edge gives −25.16 mV per volt against 24.96 predicted, and the height falls as one over the rise time with a fitted slope of −1.000 while the saturated near end sits still. It grows with length without a knee until the two modes separate by a whole edge — 249 mm for a 50 ps edge — and then stops at half the driven line's voltage.

Where this goes next

Two tracks, and a far end that cancels exactly when the field is all in one material. computed by solving, not by drawing, on 12 coupled sections of a 100 mm pair terminated in 50 Ω at all four ends. A mutual capacitance injects a current proportional to dV/dt and splits it towards both ends of the quiet track; a mutual inductance injects a voltage proportional to dI/dt and drives the two ends in opposite directions. So the near end goes as Cm/Ct + Lm/Lt and the far end as their difference, with the same constant in front of both — measured here as 1.048e-2 either way, over a slider that moves the ratio by five times. The consequence is that the far end is not a smaller effect but a cancellation: at a ratio of one it is 7.52e-19 of the drive, which is zero to the last bits of a double, while the near end is 1.048e-3. That is why a stripline has no far-end crosstalk and a microstrip has some — what shows up there measures the field that is in air, not the spacing. The model is lumped and stops where it says: a section is one degree long at 50.0 MHz. The far end that cancels Two mechanisms couple two parallel tracks: a mutual capacitance injecting a current and a mutual inductance injecting a voltage. They add at the near end of the quiet track and subtract at the far end, with the same constant in front of both — measured here as 1.048 times ten to the minus two either way, across a slider that moves their ratio by five times. So the far end is not a smaller effect: when the two couplings are equal it is 3.5 times ten to the minus nineteen of the drive, which is zero to the last bits of a double. The null is a V and not a bowl: one per cent of ratio error is 5.0e-3 of the near end. computed by solving, not by drawing. The same twelve coupled sections read at 10.0 MHz, with the ratio of the two couplings swept across the null rather than sat at one setting. The far end divided by the near end is |1 − r|/(1 + r) at every point — a straight-sided V through zero, first order in the departure with a coefficient of one half, and not a rounded minimum with a flat bottom. So there is no tolerance band: a ratio one per cent off gives 4.98e-3 of the near end and ten per cent off gives 4.76e-2, and the exchange rate between them is fixed. The upper trace is the near end over the same sweep, which moves by 11 per cent while the lower one moves through 14 decades — the two ends are the same coupling read as a sum and as a difference, which is why one of them has a zero in it and the other cannot. How wide a null is A far end at 3.5×10⁻¹⁹ of the drive is a statement about arithmetic until somebody asks how far the two couplings may differ before it comes back. The answer has no flat bottom in it: the far end divided by the near end is |1−r|/(1+r) exactly, so the null is a V and a ratio one per cent off returns 4.98×10⁻³ of the near end. On the axis a board is built to that is a difference of 0.0081 between the two modes' effective permittivities, out of 3.99 — two parts in a thousand, and 0.675 picoseconds of mode skew over a hundred millimetres. A 1 V step onto 1.00 m of 50 Ω line into an open circuit. computed 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. 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.

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Characteristic impedanceCrosstalkElectrical lengthLumped approximationMutual capacitanceMutual inductancePropagation delayRise timeTransmission lineVerification