Lines, where a wire has a length

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.

19 min read 7 figures The constants decide Two routes to a number Every model has an edge

Assumes: The far end that cancels · Where the current comes back

The far end that cancels solved two coupled ladders and found that the near end of the quiet track responds to the sum of the capacitive and inductive couplings and the far end to their difference, with the same constant in front of both. When the two couplings are equal the far end is $3.5\times10^{-19}$ of the drive against a near end of $1.06\times10^{-3}$ , and that essay was careful to say what kind of statement that is: a cancellation rather than a small number, and one that is exact in the model rather than physical.

It is exact in the model at exactly one point. Every real stack-up is somewhere else, and the question that decides whether the null is worth anything is not how deep it is but how far the two couplings may differ before it comes back.

That is a question about the ratio axis, and the ratio was the slider. Sitting on the null says nothing about its shape.

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. — Fig. 1 The same twelve coupled sections read at 10.0 MHz with the ratio swept across the null rather than sat at one setting, both ends divided by the near end at a ratio of one. The lower trace is a straight-sided V through zero — 4.98×10⁻³ at a ratio one per cent off, 4.76×10⁻² at ten per cent — and the upper trace is the near end, which changes by 10.5 per cent over the same ±10 per cent while the far end moves through 13.9 decades.

The bottom, which is where the previous reading was taken

Before sweeping across it, it is worth being clear about what the reading at the bottom is and is not, because it is the calibration this essay is quoted against.

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 frequency picture at a ratio of exactly one, which is where the rung below stopped. The near end rises a decade per decade to 1.06×10⁻³ of the drive at 10.1 MHz; the far end is 7.52×10⁻¹⁹ and is below the axis entirely, so the figure prints it rather than drawing it. The model’s own ceiling is on the axis: a section is one degree long at 50.0 MHz.

That number is not a physical prediction and was never offered as one. It is what a subtraction of two equal quantities leaves in double-precision arithmetic, and its only content is that the two mechanisms are the same size, computed independently, to the last bits available. A lossy pair, a real dielectric, a connector, a via — each of them puts a floor under it that is fifteen decades higher.

So the depth is a verification and the width is the result. The two are different quantities and the rest of this essay is about the second, which needs the ratio to be an axis rather than a setting.

What the shape is, exactly

The two proportionalities from the rung below give it in one line. Writing $k_C = C_m/C_t$ and $k_L = L_m/L_t$ and $r = k_L/k_C$ , the near end goes as $k_C + k_L$ and the far end as $|k_C - k_L|$ with the same constant, so their ratio is

\frac{\text{far}}{\text{near}} = \frac{|k_C - k_L|}{k_C + k_L} = \frac{|1-r|}{1+r}

Measured on the solved netlist at fifty-seven settings of the ratio from a half to two, that expression is right to two parts in a thousand at every one of them. It is asserted as an expression rather than as a table, which is the difference between a shape and a set of readings.

Three things follow from it and only the first is obvious.

It is first order in the departure, with a coefficient of exactly one half. Near the null, $|1-r|/(1+r) \approx |1-r|/2$ , so one per cent of ratio error gives half a per cent of the near end at the far end. Measured: 0.0498 per cent of the near end at a thousandth off, 0.498 at a hundredth, 4.76 at a tenth.

So the null is a V and not a bowl. A resonant minimum has a quadratic bottom and therefore a band inside which nothing much changes; a cancellation between two first-order terms does not. There is no tolerance band to sit in, and asking for a stated suppression is asking for a stated ratio tolerance — twice it, near the null.

And the far end can be reversed, not merely reduced. The expression carries an absolute value because the quantity underneath changes sign at $r = 1$ . At a ratio of 0.99 and a ratio of 1.01 the far end is $5.29\times10^{-6}$ of the drive in both cases and the two are opposite in sign, which is something a magnitude specification cannot express and which matters when two coupled regions of a board are in series.

The two sides of the null are not mirror images in every quantity, and the difference is small enough to be worth stating exactly. The far end is symmetric to three figures because it goes as $|1-r|$ , and one per cent of departure is one per cent either way. The near end is not, because it goes as $1+r$ , and $2.01/1.99$ is a per cent: $1.062\times10^{-3}$ of the drive at a ratio of 1.01 against $1.051\times10^{-3}$ at 0.99. So the ratio between the two ends is a per cent worse on the low side — $5.03\times10^{-3}$ at 0.99 against $4.98\times10^{-3}$ at 1.01 — which is the $(1+r)$ in the denominator and is the only asymmetry the expression has.

The near end is not measuring the same thing

The upper trace of the sweep is worth as much as the lower one, and it is the part that says the two ends are not two versions of one number.

Across a ratio moving by a tenth either way the near end changes by 10.5 per cent — from $9.95\times10^{-4}$ to $1.10\times10^{-3}$ — which is what a sum of two things does when one of them moves by a fifth. Over the same interval the far end falls through 13.9 decades and comes back. One of those is a quantity being adjusted and the other is a quantity being cancelled, and no measurement of the first says anything about the second.

That is the practical consequence and it inverts a common procedure. A near-end reading is easy to take, is large, and is insensitive; it is a good measurement of the coupling’s size. A far-end reading is small, is hard to take, and moves thirteen decades for a ten per cent change in something nobody specified; it is a measurement of the coupling’s symmetry and of nothing else. Treating the second as a harder version of the first is the mistake, and it is the same one every model has an edge collects: two numbers that share a mechanism and answer different questions.

The second route, which solves nothing

A statement about a cancellation deserves a route that does not go through the thing being cancelled, because a subtraction is exactly where a solve loses its digits.

There is one, and it is in the section values rather than in the solve. An even-mode wave — both conductors at the same potential — meets a series inductance raised by the mutual one and the shunt capacitance alone, because a capacitor between two equal potentials carries no charge. An odd-mode wave meets the series inductance lowered by the mutual one and the shunt capacitance plus twice the mutual one, because the capacitor between them sees the whole difference. So

\tau_{\text{even}} = N\sqrt{L(1+k_L)(C-C_m)}, \qquad \tau_{\text{odd}} = N\sqrt{L(1-k_L)(C+C_m)}

and the two are equal exactly when $k_L = k_C$ . The far end is a reading of their difference, and predicting it from the two delays alone gives

\frac{\text{far}}{\text{near}} = \frac{\Delta\tau/\tau}{2k_C + \Delta\tau/\tau}

which agrees with the solved netlist to 0.3 per cent across three and a half decades of skew. The two routes share the section values and nothing else: one factors a coupled network of a hundred and fifty elements and reads two node voltages, and the other multiplies four numbers and takes a square root. That is a genuinely independent check of the kind one step, computed twice is about, and it is the reason the $10^{-19}$ at the bottom of the null can be quoted as a cancellation rather than as a residue.

The 0.3 per cent that is left is not noise. It is the difference between a twelve-section ladder’s transfer and an exact modal decomposition, and it is the same finite-section error a ladder is not a line measures directly.

What the tolerance is, in something a board is built to

A ratio of two coupling coefficients is not a manufacturing quantity. A delay is, and so is a permittivity, and the mode delays turn one into the other with no modelling in between.

A delay is an effective permittivity: $\tau = \ell\sqrt{\varepsilon}/c$ , so a fractional difference in delay is half a fractional difference in permittivity, exactly, with a second-order residual of $6.3\times10^{-4}$ at the widest setting drawn. Two modes with different delays are two modes seeing different effective permittivities, and that is a statement about where the field is rather than about a coupling coefficient.

Holding the far end a hundredth below the near one needs the two modes within Δε = 0.008 of four. computed by solving, not by drawing. The far-end null redrawn on the axis a board is built to. An even-mode wave meets the shunt capacitance alone and an odd-mode wave meets it plus twice the mutual one, so the two modes have different delays whenever the two couplings differ — and a delay is an effective permittivity. The rising trace is the far end from the solved netlist; the trace on top of it is the same quantity predicted from the two mode delays alone, agreeing to 0.3 per cent over three and a half decades without solving anything. The upper trace is the near end, which reads the sum of the couplings and moves by only 50 per cent across the whole range. Keeping the far end a hundredth below the near one asks the two modes to agree to Δε = 0.0081 out of 3.99, which is 0.20 per cent and 0.67 picoseconds of mode skew over a hundred millimetres. — Fig. 5 The null on the axis a stack-up is built to. The rising trace is the far end from the solved netlist, the trace on top of it is the same quantity predicted from the two mode delays alone, and the upper one is the near end, which moves by half over a range in which the far end moves by three and a half decades. Holding the far end a hundredth below the near one needs the two modes within Δε = 0.0081 of 3.99, which is 0.20 per cent and 0.675 picoseconds of mode skew over a hundred millimetres.

The numbers are severe and they are the essay’s point.

For the far end at a tenth of the near end, the two modes must agree to $\Delta\varepsilon = 0.089$ out of 3.99 — 2.2 per cent, and 7.42 picoseconds of skew over a hundred millimetres. That is loose.

For a hundredth, $\Delta\varepsilon = 0.0081$ , which is 0.20 per cent and 0.675 picoseconds.

For a thousandth, $\Delta\varepsilon = 0.00080$ , which is 0.020 per cent and 67 femtoseconds.

A microstrip is nowhere near any of these. The ratio of 1.4 the rung below drew as a rough microstrip is a mode-permittivity difference of 0.16 — four per cent — and a mode skew of 13.4 picoseconds, and it puts the far end at a sixth of the near one. Between a stripline, where the field is in one material and $\Delta\varepsilon$ is zero by construction, and a microstrip, where a solder mask, a coverlay, a resin-starved region or an air gap under a coverlay each moves it by more than a per cent, there is no intermediate stack-up that lands inside a tolerance of two parts in a thousand by accident. The null is available and it is not adjustable. It is had by burying the pair, and not by getting close.

What a hundredth of the near end is worth

A suppression stated as a fraction of the near end is a ratio, and a ratio is not a specification. The near end here is $1.06\times10^{-3}$ of the drive, so a far end held a hundredth below it is $1.06\times10^{-5}$ — eleven microvolts on a one-volt edge, which is under any threshold that matters and is comfortably below the noise a board has anyway.

That sounds like an easy target and it is the reason the tolerance above is worth stating in the form it is. The suppression asked for is modest; the symmetry it demands is not, because the two are connected by a factor that has the coupling strength in it. At a five per cent capacitive coupling a hundredfold suppression needs the ratio within two per cent; at a two per cent coupling it needs the same two per cent of ratio, and the far end that results is smaller in absolute terms because both ends are. The absolute far-end level and the ratio tolerance move independently, and a specification written on one says nothing about the other.

There is a second reason not to read the fraction as the answer, and it belongs to the pair’s length. The rung below records the asymmetry: near-end crosstalk is a sum of contributions arriving over a full round trip, so a longer pair spreads it rather than raising it, while far-end crosstalk is a difference of contributions that all arrive together, so a longer pair raises it in proportion. Doubling the coupled length therefore doubles the far end and leaves the near end where it was — which doubles the ratio this essay measures without any coupling changing at all. The two per cent of ratio tolerance is quoted for one length, and it halves for twice that length.

What separation does not do

The expression $|1-r|/(1+r)$ contains no coupling strength at all, and that is worth stating as a result rather than noticing as an absence.

Moving two tracks apart reduces $k_C$ and $k_L$ together and leaves their ratio where it was, so the far end falls in exact proportion to the near end and the shape of the null is untouched. Both repairs are real and they are not substitutes:

Spacing buys an attenuation. It works on both ends at once, it has diminishing returns, and it cannot produce a zero because it cannot make the difference of two shrinking numbers vanish.

Symmetry buys the null. It works on one end only, it is exact rather than asymptotic, and it is had by changing where the field is — which costs a layer and is decided by whoever builds the board rather than by whoever draws the schematic, putting it with the edges that are lengths.

A designer who has doubled the spacing and measured the far end falling by the expected factor has learnt nothing about the symmetry, because the ratio did not move. The measurement that says whether the null is available is the far end divided by the near end, and that ratio is invariant to the one repair most often reached for.

What is checked

Four assertions, and not one of them is a value at a setting.

That the far end divided by the near end is $|1-r|/(1+r)$ , to two parts in a thousand, at fifty-seven ratios spanning a factor of four — an expression in the ratio rather than a reading at one of them, which is what makes the V a shape rather than three points.

That at a ratio of one the far end is absent rather than small: below $10^{-14}$ of the near end, asserted as a bound on a cancellation rather than as a number, because a cancellation and a small number are different claims and only one of them survives being told the coupling was increased.

That across a ratio moving a tenth either way the near end moves by under an eighth and the far end by more than twelve decades, which is the sum and the difference behaving as a sum and a difference, asserted as the pair rather than as either one.

And that the far end predicted from the two mode delays — a route that solves nothing, reads no node, and never forms the difference the solve is losing digits to — agrees with the solved netlist to better than one per cent over three and a half decades of skew, with the delay-to-permittivity relation itself checked to a part in a thousand and to a part in a billion once its exact second-order factor is put back.

What this does not claim

That the ratio is computed from a geometry. It is not, and the rung below said so: $k_C$ and $k_L$ are inputs here, and getting either from a stack-up is a field problem this collection does not own. What is owned is the map from the ratio to a mode-permittivity difference, which is arithmetic, and the map from that to a tolerance, which is the sweep above.

That the null survives loss. It does not exactly. A lossy pair has a small phase difference between the two mechanisms and the null becomes a minimum, in the same way and for the same reason that the millimetre that becomes common mode finds a closed-form null at 95.3 gigahertz filled to −28.9 decibels by an amplitude imbalance the closed form has no term for. The V drawn here is a lossless V, and the floor a real board puts under it is set by the same asymmetry that fills that one.

That twelve sections is a line. A section is one degree long at 50.0 megahertz and the sweep is read at 10.0, which is inside the model by a factor of five. The backward coefficient a real pair saturates at is 0.02500 here and would need several hundred sections to reach — the same ceiling the delay that is not one number is about from the other side.

What 1.5 mm of length mismatch does to a differential pair. computed by solving, not by drawing. Two lines of the same impedance and different lengths, driven differentially. The solid rising curve is what arrives as common mode; the dashed one beside it is sin(ωΔτ/2), which is what a lossless pair gives and is the same curve until the null. The flat curve at the top is the differential signal, and it is the point: at 3.04 GHz a tenth of the launched amplitude is common mode and the differential has lost 5011 parts per million of itself. The conversion is first order in the skew and the loss is second order, so the error is not missing from the signal — which is why a pair can pass its own eye and fail an emissions test. At 95.3 GHz the closed form has a null and the real pair does not: the longer conductor is also the lossier one, and an amplitude imbalance has no null in it. — Fig. 7 The same symmetry broken by a length rather than by a dielectric. A millimetre and a half of skew between two conductors of a differential pair converts a tenth of the launched amplitude into common mode by 3.04 GHz while the differential signal has lost 5011 parts per million of itself — first order one way and second order the other, which is why the error is not missing from the eye.

Where the same shape appears

A cancellation between two first-order terms is a recurring object in this collection, and its tolerance always has the same form.

The inductor one mode cannot see is a component whose whole function is a difference: one mode meets $(1+k)L$ and the other meets $(1-k)L$ , and the ratio of its two corners is $2/(1-k)$ , which contains no inductance. A one per cent winding mismatch there converts at a level first order in the mismatch, exactly as a one per cent ratio error does here.

One number from two measurements recovers a coupling coefficient from a sum and a difference of two series inductances, which is the same construction read as an instrument rather than as a defect — and the rung below pointed at it for the same reason.

And a quarter wave, and the path the current takes back is where the geometry that sets both couplings is computed rather than assumed, which is the layer underneath all of this.

What the three have in common is that the useful figure is a ratio of two readings rather than either reading, and that the quantity in the denominator is the one that is easy to measure and irrelevant to the answer.

The number worth carrying

Two parts in a thousand of permittivity, for a hundredfold suppression.

The chain is short: a hundredth of the near end at the far end needs the ratio within two per cent, two per cent of ratio is 0.20 per cent of mode-permittivity difference at a five per cent coupling, and 0.20 per cent of 3.99 is 0.0081 — or 0.675 picoseconds of skew between the two modes over a hundred millimetres of pair. Every step of that is first order and none of it has a flat region in it.

The habit is about which measurement is being asked to carry the claim. A null measured at its bottom reports the arithmetic’s floor and says nothing; a null measured across reports its own tolerance, and the tolerance is the only part a board can be held to. Sitting on a cancellation is not a measurement of it — the edge that is a region is the general version, and this is the sharpest instance of it in the field, because the depth at the bottom is $10^{-16}$ and the width is two parts in a thousand.