The impairment no equaliser can remove, because it is driven by somebody else's data. Why near-end and far-end coupling behave so differently, why stripline has none of the second kind at all, and what the three-widths rule is actually worth.
Loss and reflection are both deterministic functions of the victim's own data. The pulse response tells you exactly what the channel will do to any bit pattern, so a filter designed against that pulse response can subtract the damage — which is what every equaliser in SerDes_Equalisation does.
Crosstalk is driven by a different lane's data, which the victim's receiver has never seen and cannot predict. To that receiver it is indistinguishable from noise, so it enters the budget as a variance to be added in quadrature rather than as a tap to be cancelled. No amount of equalisation removes it, and the only places to fix it are the geometry and the routing.
A receiver that can observe the aggressor can cancel it, and some modern transceivers do exactly that for the near-end crosstalk from their own transmitter, which is the one aggressor whose data is known locally. That is genuinely effective and it is also the narrow case: the aggressors on a backplane connector belong to other devices, and nothing on the victim's die knows what they are sending.
There is also a vocabulary trap worth clearing. Near-end crosstalk is measured at the end of the victim nearest the aggressor's transmitter, and far-end at the other end. The names describe where you measure, not where the coupling happens, and the two behave so differently that keeping them straight matters.
Two adjacent lines are coupled both capacitively, through the mutual capacitance $C_m$ between them, and inductively, through the mutual inductance $L_m$. Both couplings inject current into the victim; the difference is direction. Capacitive coupling injects current that flows both ways along the victim. Inductive coupling drives a loop, which sends current one way and returns it the other.
Towards the near end the two add; towards the far end they subtract. That gives the two coupling coefficients:
$$K_b = \frac{1}{4}\left(\frac{L_m}{L} + \frac{C_m}{C}\right), \qquad K_f = -\frac{1}{2}\left(\frac{L_m}{L} - \frac{C_m}{C}\right).$$
Everything distinctive about crosstalk follows from that one sign. The near-end coefficient is a sum of two positive quantities and cannot vanish. The far-end coefficient is a difference, and it vanishes whenever the two ratios are equal — which happens whenever the dielectric surrounding the pair is uniform.
The claim on the previous slide is checkable. The field solver is given two identical cross-sections: a pair of coupled striplines buried in uniform laminate, and a pair of coupled microstrips with laminate below and air above. It computes the capacitance matrix with the dielectrics present and again with them replaced by vacuum, and obtains the inductance matrix from the second. Nothing in that procedure knows what answer is expected.
The stripline result is exact in the sense that matters: the two ratios are equal because the medium is uniform, and the solver confirms it to six decimal places. The microstrip magnitudes deserve more caution. An open microstrip has fields extending indefinitely into the air above it, and this solver encloses its cross-section in a grounded box; the enclosure perturbs the air-side field and overstates the difference between the two modal velocities. The sign, the mechanism and the contrast with stripline are sound. The far-end coupling in decibels is not a number to budget from, which is why the budget on slide 09 is built from the near-end term on an inner layer.
In a homogeneous medium every mode travels at the same speed, $c/\sqrt{ \varepsilon_r}$, and that forces $LC = \mu\varepsilon$ as matrices, not merely as scalars. The mutual and self terms are then locked in the same ratio and the difference in $K_f$ is identically zero. It is a consequence of the medium, not a property of the geometry, which is why it survives any spacing, any trace width and any asymmetry as long as the dielectric is uniform.
The two kinds of crosstalk also scale differently with the length over which the lines run together, and this is what decides which one a given design has to worry about.
Near-end crosstalk builds up while the aggressor's edge travels, but the contribution from a point further along arrives later and later, spread over a window two one-way delays wide. Once the coupled section is long enough that the backward wave from the far end is still arriving when the edge has finished, making it longer adds nothing more — it simply extends the duration. The saturation length is the distance the edge covers in half its own rise time.
Far-end crosstalk behaves quite differently. Each element of coupled length adds a contribution that arrives at the same instant as all the others, because it travels with the aggressor's edge, so the total grows in proportion to length — and it is easy to stop there, because that is where the standard first-order formula stops. It cannot be the whole story, since a linear growth would eventually predict a crosstalk larger than the aggressor that caused it.
What actually happens is that the far-end voltage is the difference between the even and odd modes arriving at slightly different times, $v_{fext}(t) = \tfrac{1}{2}[v(t-\tau_e) - v(t-\tau_o)]$. While the modal delay difference is small compared with the rise time this difference is proportional to it, which reproduces the linear formula. Once the delay difference exceeds the rise time the two modes arrive fully separated, and the far-end crosstalk saturates at half the swing. The model computes the bounded form and reports the linear one alongside it, so the point at which the familiar expression stops being usable is visible rather than assumed.
Rise time acts on the two mechanisms in opposite directions, which is a genuinely counter-intuitive result and a useful one.
Far-end crosstalk is proportional to the rate of change of the aggressor, so halving the rise time doubles it. Near-end crosstalk, once saturated, does not depend on rise time at all — its amplitude is fixed by $K_b$ alone. What the rise time changes is the saturation length: a faster edge saturates in a shorter distance, so a shorter coupled section reaches full amplitude.
The practical consequence is that moving to a faster technology on the same board makes far-end crosstalk worse in direct proportion while leaving near-end amplitude where it was. On an inner layer, where the far-end term is zero, a rise-time reduction therefore costs very little crosstalk; on an outer layer it costs a great deal. That is another entry in deck 05's list of reasons to bury the fast pairs.
Layout guidelines are usually written as a multiple of the trace width: keep neighbours three widths away, or five. The rule is convenient because the router knows the trace width. It is also the wrong variable, because what sets the coupling is the spacing relative to the dielectric height, and the two move independently whenever the stackup changes.
A grounded trace between two signals is an appealing idea and a frequently disappointing one. The solver is given three conductors rather than two, so the guard is treated as a conductor with charge of its own rather than as a wall, and the comparison is run three ways: no guard, a guard tied to the planes, and a guard left floating.
A guard trace only works if it is grounded along its length, stitched to the planes often enough that it cannot support a standing wave. Grounded at its two ends and nowhere else, it is a resonator: near its resonances it is an efficient path from aggressor to victim rather than a barrier. Stitching vias every tenth of a wavelength at the highest frequency of interest is the usual requirement, and at fourteen gigahertz on FR-4 that is roughly every four millimetres — a lot of vias for a benefit that the same board area spent on spacing would deliver more reliably.
A real lane has neighbours on both sides, above and below, and through the connector. Their data is uncorrelated with each other's and with the victim's, so their contributions add in power rather than in amplitude: four equal aggressors give twice the amplitude of one, not four times.
The quantity the standards define is integrated crosstalk noise, which weights each aggressor's coupling by the spectrum of the transmitted signal and the response of a reference receiver and reports a single root-mean-square voltage. The version computed here keeps the essential structure — uncorrelated sources adding in power, producing a variance that sits beside receiver noise in the budget — without the full weighting of the published method.
The arrangement costed below is the one this series actually works against: a stripline lane with four neighbours at three trace widths, which is a routed lane pitch rather than the one-dielectric-height separation used earlier to demonstrate the coefficients. Only near-end coupling is counted, because on an inner layer the far-end coupling is identically zero — which is this deck's central result, arriving in the budget.
Because crosstalk adds in quadrature with the other noise terms and cannot be equalised away, its effect on the budget is direct: it reduces the ratio of signal to noise that the whole design rests on. Expressing that reduction in decibels makes it comparable with the remedies priced in the equalisation deck, where changing the board bought one and a half to two and a half decibels and more silicon bought a quarter of a decibel.
Deck 17 runs the same crosstalk figure through a channel operating margin calculation and finds that varying it across the range a real connector spans moves the verdict by several decibels — more than any of the receiver parameters the standard specifies. Crosstalk is not a secondary term in a compliance calculation; on a dense connector it is frequently the dominant one.
| Quantity | Expression | Worth remembering |
|---|---|---|
| Near-end coefficient | $K_b=\frac{1}{4}(L_m/L + C_m/C)$ | A sum; cannot vanish |
| Far-end coefficient | $K_f=-\frac{1}{2}(L_m/L - C_m/C)$ | A difference; vanishes in a uniform dielectric |
| Near-end saturation length | $\ell_{\text{sat}} = v\,t_r/2$ | Longer coupling adds duration, not amplitude |
| Far-end amplitude | $\tfrac12\min(\Delta\tau/t_r,\,1)$ | Linear in length while $\Delta\tau < t_r$, where it equals $K_f\,\ell\,t_{pd}/t_r$; saturates at half the swing thereafter |
| Stripline far-end | zero | The modal velocities are equal, so $\Delta\tau$ is identically zero at any length |
| Many aggressors | $\sqrt{\sum V_i^2}$ | Uncorrelated sources add in power |
| Guard trace | — | Needs stitching every tenth of a wavelength or it is a resonator |
| Equalisation | — | Does nothing; the fix is geometry |
Deck 6 of eleven in Signal Integrity & High-Speed Digital Design. Every figure on this page is computed by si_models/deck06 and embedded as data; nothing is typed in by hand.
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