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

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.
19 min read 7 figures Two routes to a numberEvery model has an edgeThe constants decide

Assumes: Where the current comes back · A ladder is not a line

Two tracks running side by side on a board are coupled by two things, and it is worth resisting the temptation to treat them as one. A mutual capacitance between them injects a current into the quiet track proportional to the rate of change of voltage on the active one, and that current splits and runs towards both ends. A mutual inductance between them injects a voltage into the quiet track proportional to the rate of change of current, and a voltage source in a line drives its two ends in opposite directions.

So at the near end of the quiet track — the end beside where the signal was launched — the two arrive with the same sign and add. At the far end they arrive with opposite signs and subtract.

That difference is not a detail of sign conventions. It is the reason far-end crosstalk behaves in a way that catches people out, and the reason a stripline has none at all.

Two tracks, and a far end that cancels exactly when the field is all in one materialcomputed 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.100n1µ10µ100µ1m10m100m1M10Mfrequency (hertz)voltage on the quiet track, as a fraction of the drivethe near end — the two addthe far end is 7.5e-19 of the drive — below this axis entirelya section is 1° long hereCm/Ct0.05Lm/Lt ÷ Cm/Ct1one-way delay667.1 psnear end at 10.1 MHz1.06e-3far end at 10.1 MHz3.47e-19backward coefficient0.02500a section is 1° at50.0 MHzsolved, then checked — twelve coupled sectionsthe far end nulls at a ratio of one
Fig. 1 Two coupled ladders, twelve sections each, a hundred millimetres long and terminated in fifty ohms at all four ends. The mutual capacitance is five per cent of the total and the mutual inductance is equal to it — a homogeneous dielectric. The near end of the quiet track carries a thousandth of the drive; the far end carries nothing at all, and the caption says how much nothing.

The model, and why it is a ladder

Each track is twelve sections of series inductance and shunt capacitance, sized so that the pair has the right characteristic impedance and the right one-way delay: fifty ohms and 667 picoseconds for a hundred millimetres in a dielectric of relative permittivity four.

The coupling is stamped as two things. A capacitance between the corresponding nodes of the two ladders — and the shunt capacitance to ground is reduced by that amount, so that the track’s own impedance does not move when the coupling does, which is what a coupling means and what a model that simply adds CmC_m gets wrong. And a mutual inductance on every corresponding pair of series inductors, using the solver’s own coupling element, which is an off-diagonal in the matrix rather than a transformed pair.

That element carries a trap the network library names by hand and this figure had to avoid: a coupling refers to the inductor objects, not to node names, because there is no such thing as coupling between two places. Building the ladder twice and coupling the first build’s inductors would have referred to elements that are not in the network being solved. It is why the netlist is constructed by one function that returns both the elements and nothing else.

Two proportionalities with one constant

The claim the figure exists to make is not about a value. It is about which combination each end responds to.

Writing kC=Cm/Ctk_C = C_m/C_t and kL=Lm/Ltk_L = L_m/L_t for the two coupling fractions, the solved network at ten megahertz gives

ratio kL/kCk_L/k_C near end far end
0.4 7.40×10−47.40\times10^{-4} 3.17×10−43.17\times10^{-4}
0.7 8.98×10−48.98\times10^{-4} 1.59×10−41.59\times10^{-4}
0.9 1.00×10−31.00\times10^{-3} 5.29×10−55.29\times10^{-5}
1.0 1.06×10−31.06\times10^{-3} 3.47×10−193.47\times10^{-19}
1.1 1.11×10−31.11\times10^{-3} 5.29×10−55.29\times10^{-5}
1.4 1.27×10−31.27\times10^{-3} 2.11×10−42.11\times10^{-4}
2.0 1.59×10−31.59\times10^{-3} 5.29×10−45.29\times10^{-4}

Divide the near-end column by kC+kLk_C + k_L and every row gives 1.048×10−21.048\times10^{-2}. Divide the far-end column by ∣kC−kL∣|k_C - k_L| and every row gives 1.048×10−21.048\times10^{-2}.

The same constant. The figure asserts all three of those: that the near end is proportional to the sum with one constant across the slider, that the far end is proportional to the difference with one constant, and that the two constants are the same to five parts in a thousand.

That is what makes the fourth row a cancellation rather than a small number. The far end is not a weaker coupling; it is the identical coupling with a minus sign in it, and when the two mechanisms are equal there is nothing left. 3.5×10−193.5\times10^{-19} of the drive is zero to the last bits of a double, in a quantity that is 10−310^{-3} one row up.

Why the two couplings would ever be equal

They are equal whenever the field is entirely in one material, and that is not a coincidence — it is a consequence of the two coupling matrices being inverses of one another for a lossless homogeneous structure. Both are determined by the same geometry through the same integrals, and in a uniform dielectric the electric and magnetic pictures give the same numbers.

A stripline is homogeneous. The conductors are buried between two planes with dielectric on all sides, the field is entirely in that dielectric, and kC=kLk_C = k_L exactly. A stripline has no far-end crosstalk.

A microstrip is not. One side of the track faces the board and the other faces air, so part of the electric field is in a material of permittivity one and part is in a material of permittivity four. The capacitive coupling is weakened by the air path more than the inductive coupling is — magnetic permeability is the same in both — so kC<kLk_C < k_L and the far end appears.

That gives the essay’s sharpest statement. What arrives at the far end is a measurement of how much of the field is in air, not of how close the tracks are. Moving them apart reduces both couplings and reduces the far end proportionally; burying them changes the ratio and can remove it entirely. The two repairs are not the same, and only one of them has a limit.

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.
Fig. 2 The same pair with the inductive coupling forty per cent larger than the capacitive one, which is roughly a microstrip. The near end has moved by twenty per cent and the far end has gone from nothing to two parts in ten thousand of the drive — an effect that exists only because the field is in two materials.

What each end does with frequency, and where the model stops

Both couplings are derivatives, so both rise a decade per decade: measured across the axis the slope is 1.00 decades per decade for the near end, which is what a dV/dtdV/dt and a dI/dtdI/dt do to a sinusoid.

There is a second regime the figure deliberately does not reach, and saying why is the honest part of this essay. On a real pair of lines the near-end coupling stops rising once the signal has lasted longer than a round trip — the reflections that build it up have all arrived — and settles at the backward coefficient

kb=14(CmCt+LmLt)k_b = \frac{1}{4}\left(\frac{C_m}{C_t} + \frac{L_m}{L_t}\right)

which is 0.025 here. A round trip is 1.33 ns, so that saturation happens above about 750 MHz.

A twelve-section ladder cannot be trusted there. Each section is 8.33 mm long and is one degree of phase at 50 MHz by the limits field’s own criterion, so the model’s ceiling is fifteen times below the frequency at which the interesting saturation begins. Reaching it would need several hundred sections, which is the same finding this field’s ladder essays already record: a lumped ladder approaches a line slowly, and the number of sections needed goes up with the frequency being asked about.

So the figure’s axis stops at 50 MHz and the backward coefficient appears in the panel as a number rather than as a level on the plot. What is measured here is the low-frequency regime, where the proportionalities are clean and the cancellation is exact; what is quoted is where the other regime begins.

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.
Fig. 3 The same pair with the inductive coupling only four tenths of the capacitive one, which is the direction a track over a slotted or perforated plane moves in. At 10.1 MHz the near end is 7.40×10⁻⁴ of the drive and the far end 3.17×10⁻⁴ — and the far end’s sign is the opposite of the microstrip case above, because the difference it is made of has changed direction. A quantity that can be cancelled can also be reversed, which is something a magnitude specification cannot express.
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.
Fig. 4 Seven tenths, between the two. The far end has halved to 1.59×10⁻⁴ while the near end has risen from 7.40×10⁻⁴ to 8.98×10⁻⁴. The two ends move in opposite directions as the ratio changes, which is a sum and a difference behaving as a sum and a difference — and it is why no single number describes the coupling between two tracks.

What the near end does that the far end does not

The two ends differ in a second way that the proportionalities above do not show, and it matters when the drive is an edge rather than a sinusoid.

The near-end coupling is a sum of contributions arriving from every point along the pair, each delayed by the round trip to that point. So for a step it is spread out: it begins when the edge is launched and lasts for a full round trip — 1.33 ns here — at a level set by the backward coefficient. A long pair gives a long, low pulse of near-end crosstalk, and making the pair longer makes it last longer without making it taller.

The far-end coupling is a difference of contributions that all arrive at the same instant, because each one travels with the edge. So for a step it is a single narrow spike, one rise time wide, and its height grows in proportion to the length. Making the pair longer makes it taller.

That is the practical asymmetry and it is not visible on a frequency axis. Doubling the coupled length leaves near-end crosstalk at the same amplitude and doubles far-end crosstalk, so a design that is comfortable at fifty millimetres can fail at two hundred for a reason that has nothing to do with either coupling changing.

Where the far end is what fails

A clock beside a data line on a backplane. Backplane traces are usually stripline, and the absence of far-end crosstalk there is the reason long parallel runs are tolerable at all. The same layout on the surface of the board is a different problem, and the difference is a permittivity rather than a spacing.

A ribbon cable. Wires in air with a return conductor somewhere have very unequal coupling fractions, and the far-end term is large. It is the reason a ribbon cable is used with alternate ground conductors — which changes both couplings and their ratio at once.

A connector. A connector is a short region where the geometry changes and the two couplings scale differently, so a pair that is balanced along its whole length gets a far-end contribution concentrated in a few millimetres. It is a small length and a large ratio change, and the product is what matters.

The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres.
Fig. 5 The other geometric quantity this field measures, and the one that decides the couplings’ size. Where the return current runs sets the loop area, and the loop area sets the mutual inductance — so the crossover this figure computes is upstream of both numbers in the essay above.
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.
Fig. 6 Twice, the far edge of the slider and further into air than an ordinary microstrip reaches. The near end is 1.59×10⁻³ and the far end 5.29×10⁻⁴ — a third of the near end, rather than the nothing it was at unity. Over the whole slider the near end runs monotonically from 7.40×10⁻⁴ to 1.59×10⁻³ while the far end falls to zero at one and rises again beyond it. That shape is what a cancellation looks like; an attenuation has no zero in it.

Two repairs and what each one costs

The proportionalities give the repairs directly, and they are worth separating because they are usually offered as one piece of advice.

Increase the spacing. Both kCk_C and kLk_L fall roughly as the inverse square of the separation for tracks over a plane, so both ends improve together and the ratio is nearly unchanged. It costs board area, it has diminishing returns, and it cannot remove the far end because it cannot make the difference of two shrinking numbers vanish.

Change the ratio. Burying the pair, or adding a coverlay, or putting a plane above them as well as below, moves the electric field out of the air and takes kCk_C towards kLk_L. It costs a layer. What it buys is the far-end null, and the null is exact rather than asymptotic — which is why it is the repair for a long parallel run and spacing is the repair for a short one.

The third thing people do, adding a grounded track between the two, is a mixture of both: it reduces both couplings, and it changes the ratio in a direction that depends on where the guard track’s vias are. It is the least predictable of the three and the most often recommended.

A coupled pair as a two-port, at k = 0.2. computed by solving, not by drawing. Each port driven in turn with the other open, four solves, and the four impedance parameters read out. The diagonal terms measure each winding's own inductance — 10.0000 mH and 40.0000 mH against 10 and 40 — and both transfer terms measure the mutual inductance, 4.0000 mH against k√(L₁L₂) = 4.0000. The two transfer terms agree to 1.41e-16, which is reciprocity — a property of the device rather than of the measurement, and the first thing a coupling stamped into the wrong row would break. Neither of this site's two standing checks can see it: a coupling adds no current and dissipates nothing.
Fig. 7 A coupled pair described the other way. Four impedance parameters recovered by four solves is the complete description of two coupled windings, and a coupled pair of tracks is the same object with a different geometry — which is why the same solver element carries both.

What this essay does not claim

That the couplings are computed from a geometry. They are not. Cm/CtC_m/C_t and Lm/LtL_m/L_t are given as inputs, and everything above is a statement about a netlist with those values in it. Getting them from a stack-up is a field problem and belongs to whichever site owns electromagnetics; what is owned here is what a netlist with a stated coupling does.

That the far-end null survives loss. It does not exactly. A lossy line has a small phase difference between the two mechanisms and the null becomes a minimum rather than a zero. The model here is lossless, which is why the number is 10−1910^{-19} rather than something physical, and the honest reading is that the null is exact in the model and deep in reality.

That twelve sections is a good model of a line. It is a good model below a few tens of megahertz for this length, and the figure says so on its own axis. This field’s ladder essays measure exactly how bad it is above that.

That near-end crosstalk is the lesser problem. It is usually the larger number and it is frequently the easier one, because it appears at the driver’s end where the source impedance is low and the victim is often being driven rather than listened to. Which end matters is a property of the system rather than of the geometry.

The number the sum gives

One more reading of the same pair of proportionalities is worth having, because it says how large the whole effect is before any of the geometry is known.

Add the two ends: the near end goes as kC+kLk_C + k_L and the far end as kC−kLk_C - k_L, so their sum is 2kC2k_C and their difference is 2kL2k_L. Two measurements at the two ends of one quiet track therefore recover the two coupling fractions separately, with no field solution and no knowledge of the stack-up — which is the same shape of construction the magnetics field uses to get a coupling coefficient from two series measurements.

That is the useful experiment on a real board. Drive one track, measure both ends of its neighbour, and the sum and difference of what is measured are the capacitive and inductive couplings. If the far end reads zero, the pair is homogeneous; if it reads half the near end, the inductive coupling is three times the capacitive one, and nothing else needs to be known to say so.

What the cancellation depends on, and where it is used

The far end vanishing at a ratio of one is a statement about a model as much as about a pair of tracks. A ladder is not a line is the measurement of what a lumped chain is worth as a stand-in for a line, and it is the reason the axis here stops at fifty megahertz. One number from two measurements is where the coupling element itself is checked, by a route neither of this site’s standing verifications can supply. The receiver that is a branch is the same lattice with a third node in it, and A quarter wave, and the path the current takes back is where the geometry that sets both couplings is computed rather than assumed. The delay that is not one number is the assumption this page inherits and does not test.

The gate

Three assertions about proportionality, not about values. The near end divided by kC+kLk_C + k_L is one constant across the whole slider to three parts in a thousand; the far end divided by ∣kC−kL∣|k_C - k_L| is one constant to the same tolerance; and the two constants are equal to five parts in a thousand. Together they say the two ends respond to the sum and the difference of the same pair.

The null is asserted as a null, below 10−1510^{-15} of the drive, rather than as “small”. A cancellation and a small number are different claims and only one of them survives being told that the coupling was increased.

Both slopes are measured — a decade of coupling per decade of frequency — because “both are derivatives” is the mechanism and a slope is what a mechanism looks like on a logarithmic axis.

And the model’s ceiling is on the axis. The figure stops at the frequency at which one of its own sections is one degree long, and the backward coefficient that a real line saturates at is printed as a number rather than drawn as a level, because drawing it would be drawing a feature this model cannot produce.

What makes the two couplings equal

The far-end cancellation is exact when the capacitive and inductive couplings are equal, and that is not a coincidence of the numbers chosen here — it is a statement about the medium. Two tracks in a homogeneous dielectric over a plane have Lm/L=Cm/CL_m/L = C_m/C, because both ratios are set by the same field geometry, so the cancellation is available whenever the two conductors share one dielectric and is not available when they do not. A microstrip has air above and laminate below, which is precisely the case where the two ratios differ and the far-end coefficient does not vanish; a stripline, buried between two planes, is the case where it does.

Which puts this essay’s result in the same family as two others in this field, and the family is worth naming because all three are about a symmetry rather than about a magnitude. Where the current comes back is what arranges the geometry: once the return has gathered beneath its track — half of the way by 283 kilohertz and nine tenths by 1.42 megahertz on that stack-up, as the corner that is three decades wide solves it — the two couplings are computed from one field, which is the condition this essay assumes. The millimetre that becomes common mode is the same symmetry broken deliberately by a length rather than by a dielectric — a millimetre and a half of skew converting a tenth of a differential signal into common mode by three gigahertz — and the inductor one mode cannot see is the component bought to remove what that conversion produced, presenting a millihenry to one mode and a microhenry to the other, a ratio of two thousand which is 2/(1−k)2/(1-k) and contains no inductance at all.

The three together are the field’s second argument, after impedance: a coupling is a difference between two paths, and what decides it is a symmetry that a layout either has or does not.

Which is why the near end is the one to design against. The near-end coefficient is a sum of the two couplings and so cannot cancel; the far-end coefficient is a difference and can. A designer who has achieved the symmetry has removed one of the two and left the other at full strength, and a designer who has not has both — so the useful figure of merit for a stack-up is the far-end coefficient, and the useful figure of merit for a spacing is the near-end one.

Part 1 on crosstalk

One argument about Crosstalk, and one of 2 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.

Where this goes next

The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres. Where the current comes back The return current under a track spreads out at low frequency and runs directly beneath it at high. Estimated from two paths chosen in advance, the change is a corner at 106 kilohertz for any track two hundred micrometres above a half-milliohm plane, with neither the length of the track nor its width in it. Solved, it is a band three decades wide rather than a corner, and the width enters it after all — but the length is still absent and the stack-up still sets the top, which is the part of the argument a designer needs. 20 inductor-capacitor sections, against the line they are meant to be. computed by solving, not by drawing by the trapezoidal rule over 2,600 steps. The LC 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. 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. The coupling coefficient, from two measurements that do not know it. computed by solving, not by drawing. Two windings in series, connected one way and then the other, each solved as a netlist and its inductance read out of the impedance. The two differ by four times the mutual inductance, so k comes out of the difference and the geometry never enters. The recovered value matches the one stamped into the coupling to 2.4e-15 at eight couplings from 0.1 to 0.99 — which is the second route the new element needed, since neither current law nor the energy balance can see a mutual inductance at all. Their sum stays at L₁ + L₂ throughout, which is the check that the two measurements are of one pair. One number from two measurements Neither of this site's two standing checks can see a mutual inductance. A coupling adds no current anywhere, so Kirchhoff's law is unmoved by it; a coupled pair dissipates nothing, so the energy balance is unmoved too. A coupling stamped into the wrong row would produce a well-formed solution to a different circuit and both checks would pass — so the field needed a third route, and the one it uses is the bench method: connect the windings in series one way, then the other, and the difference is four times the mutual inductance.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

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

CrosstalkHomogeneous dielectricLumped approximationModel rangeMutual capacitanceMutual inductanceTransmission line