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

The receiver that is a branch

A lattice diagram has two ends, and an interior receiver is not a point on a net — it is a short piece of track leading off it to a pin, open at the far end. Three lines of equal impedance meeting at a junction present each other with half the impedance, so a wave arriving is reflected by exactly minus a third and two thirds goes on: the far end receives two thirds of the swing and sits there for twice the stub's own delay, whatever the net is terminated with and wherever on it the branch is.
17 min read 8 figures Every model has an edgeThe constants decideTwo routes to a number

Assumes: The staircase in time · A ladder is not a line

The essay one rung down read the voltage at an interior point of a net and found the result that makes termination a real choice rather than a rule: a series-terminated net holds half its swing for two delays at a receiver a tenth of the way along, an unterminated one holds it for the mirror-image interval, and the two respectable schemes are worst at opposite ends of the same piece of track.

A point is not what an interior receiver is. A real one is a short piece of track leading off the main net to a pin, open at its far end, and that piece of track is a third transmission line. Where it meets the other two there is a junction, and no lattice diagram with two ends in it can express a junction.

This essay puts one in.

A stub holds the far end at two thirds for twice its own delaycomputed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.400 line delays — twice the stub's own delay of 0.41 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above.00.50010123456time, in line delaysvoltage at the far end, as a fraction of the swingtwo thirds of the swingrecovers at 1.40 delayswith the stub, without it, and at the junctionnet length30.0 cmline delay2.05 nsstub20.0% of the net…its delay0.41 nsjunction Γ−1/3 exactlyfirst arrival66.7% of the swingheld there for0.400 delays…which is2 × 0.200each round tripdivides the error by −3solved, then checked — a wave march against a lumped LC laddertwo thirds for 0.40 delays
Fig. 1 A series-terminated net with a branch halfway along it, marched as waves on a delay grid. The far end receives two thirds of the swing at one line delay instead of all of it, and stays there until the branch’s own round trip returns.

One line of new arithmetic

Three lines of characteristic impedance Z0Z_0 meeting at a point present each other with Z0/2Z_0/2, because from any one of them the other two are in parallel. So a wave arriving on any branch sees

Γ=Z0/2−Z0Z0/2+Z0=−13,τ=23\Gamma = \frac{Z_0/2 - Z_0}{Z_0/2 + Z_0} = -\frac{1}{3}, \qquad \tau = \frac{2}{3}

— a negative reflection back the way it came, and two thirds of the wave onward down each of the other two. In general, for nn branches of equal impedance meeting, the scattering matrix has Sii=2/n−1S_{ii} = 2/n - 1 on its diagonal and Sij=2/nS_{ij} = 2/n off it, which at n=2n = 2 reduces to a plain through connection and no reflection at all. That is the whole of what a junction is, and everything below follows from it.

The march is a wave-digital one: every length is an integer number of grid delays, so a wave moves one cell per step and nothing is interpolated. The three ends are the usual reflection coefficients — the source with its resistance, the far end with its load, the stub with its open circuit — and the middle is the three-line scattering above. It is exact for lossless commensurate lines and its only approximation is the grid: a stub whose length is not a multiple of a cell is rounded to one, and the figure says by how much.

Matching 50 Ω to 200 Ω with 51.7 mm of 100.0 Ω line. computed by solving, not by drawing at 261 frequencies. The reflection at the design frequency is 4.6e-17 — nothing, to the arithmetic — against 0.600 for the bare junction, which throws 36% of the power back. It stays under 0.1 from 0.914 to 1.086 of that frequency, a band of 17.1%.
Fig. 2 Another place in this field where two impedances meeting decide everything: a quarter-wave section between fifty ohms and two hundred, where the reflection at each end is the whole design.
A series-terminated net holds half a swing for 1.0 delays at 50% along it. computed by solving, not by drawing. What a receiver 50% of the way along one net sees under three terminations, from a lattice evaluated at that point rather than at the ends. The shaded strip is the interval in which a logic input has no defined answer — between 30% and 70% of the swing. The series-terminated net sits in it for 4.83 ns, which is 2(1 − x) delays exactly and is zero only at the far end; the unterminated net sits in it for 4.83 ns and then overshoots by 82%; the parallel-terminated net never does, and draws 60 mA down the line for as long as the level is held.
Fig. 3 The rung below this one: the same net read at an interior point rather than at a branch, where the interval a receiver spends undefined is 2(1−x) delays for a series termination and 2x for none.

Two thirds, for twice the stub’s delay

With a source resistance equal to the line impedance — a series termination, which is what puts half the swing on the line by construction — the far end of a branched net does the following.

At one line delay the wave arrives, having lost a third of itself at the junction, and the open far end doubles what reaches it: the receiver sits at two thirds of the swing. It stays there for exactly the stub’s own round trip, 2Tstub2T_\mathrm{stub}, which is the time the reflected third takes to reach the open end of the branch and come back. Then the branch’s return arrives at the junction, two thirds of it goes onward, and the far end steps up.

Not to the full swing. To eleven ninths of it, because the same doubling that helped the first arrival helps this one too, and the deficit has changed sign. The next round trip brings it back to within a ninth from below, then a twenty-seventh from above, and so on: each round trip of the branch divides what is left of the error by three and turns it over.

Every number in that paragraph is exact and none of them contains the length of the net, the position of the branch or the termination at either end. The junction does not know about any of those. Which is why the result is worth having as a rule: a stub costs a deficit of a third and an interval of twice its own delay, and the only quantity a designer controls is the second.

A stub holds the far end at two thirds for twice its own delay. computed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.100 line delays — twice the stub's own delay of 0.10 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above.
Fig. 4 A branch a twentieth of the net long. The same third is lost at the junction and the same staircase follows; only the interval has shortened, to a tenth of a line delay.
A stub holds the far end at two thirds for twice its own delay. computed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.200 line delays — twice the stub's own delay of 0.21 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above.
Fig. 5 A stub a tenth of the net’s length — 0.21 ns. The two-thirds plateau is held for 0.20 delays, which is twice the stub’s own delay: the wave reaches the branch, splits, reflects off the open receiver and comes back, and until it does the driver sees two lines in parallel.

A real edge is not instantaneous, and that changes the shape

A step with zero rise time loses a third of the swing whatever the stub is, and only the duration moves. That is a true statement about a signal nobody has.

Give the edge a rise time and the branch’s round trip is compared against it. While the edge is still rising, the far end is receiving two thirds of a ramp and one third of the same ramp delayed by 2Tstub2T_\mathrm{stub}, and the difference between a ramp and its own delayed copy is a constant: one third of the fraction of the edge that the round trip occupies. So

dip≈13⋅2Tstubtr\text{dip} \approx \frac{1}{3}\cdot\frac{2T_\mathrm{stub}}{t_r}

until the round trip is as long as the edge, whereupon it saturates at a third — the far end has time to reach the two-thirds level and sit at it.

Measured against that expression, over the whole range where the ratio is below one, the march agrees to 12.5 per cent at the shortest branch drawn and better than two per cent at the longest. What is left is the march’s own grid rather than the expression: halving the cell halves the departure, which is the signature of a half-cell offset and not of a missing term.

The dip is a third of the round trip's share of the edge, until it is a third. computed by solving, not by drawing. The dip a branch puts in the far end of a net, measured against the same net without the branch, against how much of the signal's own rise time the branch's round trip takes up. With an instantaneous edge the dip is a third of the swing for any stub at all; with a real edge the stub's round trip is compared against it, and the measured dip follows one third of that ratio to within 12.5% — of which what is left is the march's own grid, since halving the cell takes it to 6.2%. Above a ratio of one it saturates at 33.6% of the swing. The design rule follows and is a length rather than a ratio: on this net a rise time of 1.03 ns allows a stub of 2.25 cm before the dip reaches a tenth of the swing.
Fig. 6 The dip against how much of the edge the branch’s round trip takes up, measured against the same net without the branch rather than against the settled level — which is the only comparison that does not charge the branch with the edge’s own rise time.
A stub holds the far end at two thirds for twice its own delay. computed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.042 line delays — twice the stub's own delay of 0.04 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above.
Fig. 7 Two per cent, 0.04 ns: the plateau lasts 0.04 delays. A real edge is not instantaneous, and that changes the shape — once the stub’s round trip is shorter than the edge’s rise time, the plateau is no longer a plateau but a notch in the middle of the transition, and the receiver never sees two thirds of anything.

The rule, in millimetres

The expression above inverts into the number a designer needs, and it is a length rather than a ratio.

To keep the dip under a tenth of the swing, the branch’s round trip must be under three tenths of the rise time. On FR-4, where a signal travels about 15 centimetres per nanosecond, that is

ℓstub<0.15 tr2×0.3 metres per nanosecond\ell_\mathrm{stub} < 0.15\,\frac{t_r}{2}\times 0.3 \ \text{metres per nanosecond}

which for a one-nanosecond edge is 2.2 centimetres, and for a hundred-picosecond edge is 2.2 millimetres. The second of those is shorter than the pin field of a connector, which is the practical content of the whole essay: at modern edge rates a branch is not a wiring detail, it is a component, and “daisy-chain the parts and drop a short stub to each” stops being a layout style and becomes a timing error.

The rule as usually stated — keep stubs under a sixth of the rise length — is the same number with the constant rounded and the factor of two for the round trip already folded in. It is pleasant to be able to say that the constant is a third, that it comes from three lines meeting, and that the ratio it multiplies is a time over a time.

Where the branch is does not matter, and that is worth proving

The claim that the deficit and its duration do not depend on the branch’s position is the kind of statement that is easy to assert and easy to be wrong about, so it is worth saying exactly why it holds and where it stops.

The junction’s scattering does not contain a position: it is three impedances meeting, and the wave that arrives there has no memory of how far it travelled. So the fraction transmitted is two thirds wherever the branch is. What the position changes is the time at which everything happens — the first arrival at the far end is still one line delay, since the wave crosses the whole net whether or not it lost amplitude halfway, and the deficit therefore appears at the same instant regardless.

What the position does change is what happens at the source. The third that is reflected at the junction travels back a distance that depends on the position, arrives at the source at 2x2x delays, and is absorbed there if the source is series terminated. With any other source termination it is reflected and comes back, and from then on the position matters to everything.

So the clean statement has a condition on it: with a series termination at the source, the far end’s deficit and its duration are independent of where the branch is; without one, they are not. The slider on the figure moves the branch’s length rather than its position for that reason, and the position-independence is the claim the figure’s own assertions carry.

A stub holds the far end at two thirds for twice its own delay. computed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.300 line delays — twice the stub's own delay of 0.31 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above.
Fig. 8 The same branch a fifth of the way along instead of halfway. The first arrival, the deficit and the interval are identical; what has moved is the small return at the source, which a series termination absorbs.

What this says about a bus

The arrangement this essay describes is not a corner case. It is what a memory bus is: one driver, one main net, and a branch to every device on it.

Two consequences follow directly from the numbers above and are worth stating in the form a designer meets them.

Branches do not average out — they multiply. Two branches at different points each take a third of what reaches them, so a receiver beyond both sees four ninths of the swing at its first arrival rather than two thirds. Three branches leave eight twenty-sevenths, which is under a third of the swing, and no termination at either end of the main net changes that.

A branch that is not being read is still a branch. The junction reflects whether or not anybody is listening at the end of the stub; an unpowered device, a test point, an unpopulated footprint with a via in it, all present the same open circuit. The only stub that costs nothing is one that is short compared with the edge, which brings the whole argument back to the rule in millimetres above.

The scheme that avoids it is the one every fast bus has converged on: no branches at all. A point-to-point link with a single receiver has two ends and one line, the lattice diagram of the essay below applies exactly, and a series or parallel termination does what it was designed to do. That is a topology chosen to make an arithmetic problem disappear, and the arithmetic is the minus one third above.

What a branch does to a series termination

There is a second finding here and it is about the scheme rather than about the branch.

A series termination works because it settles in one round trip: half the swing goes out, the open far end doubles it, the returning wave is absorbed at the source because the source resistance matches the line, and nothing is left. That property is what makes it the cheapest termination there is — it draws no static current at all — and it is exactly what the junction destroys.

With a branch on the net, the wave that returns to the source is two thirds of what left. The remaining third is still bouncing between the branch’s open end and the junction, and each of its returns launches a fresh wave down the main line in both directions. The net no longer settles in one round trip; it settles geometrically, at a factor of three per branch round trip, and the number of branch round trips inside one line delay is Td/2TstubT_d/2T_\mathrm{stub}.

So a branch converts a series-terminated net from a scheme that is exactly right after one round trip into one that is approximately right after several. For a short branch that is a distinction without a difference, since three round trips of a two-millimetre stub is under a hundred picoseconds. For a branch a fifth of the net long it is three quarters of a line delay of residual ringing, and the designer who chose a series termination to avoid ringing has bought some anyway.

The second route, and how far it gets

The march above is a wave picture and it would be a poor essay in this collection if that were the only picture available.

The same three lines can be built as three lumped ladders of series inductance and shunt capacitance meeting at one node, and marched with the trapezoidal rule. Nothing in that netlist knows what a characteristic impedance is, what a delay is, or that a scattering matrix exists. Thirty sections reproduce the staircase to within ten per cent of the swing, worst at the first arrival where a lumped ladder’s own rise time smears the step — which is the error this field has already measured and fitted, falling as the section count to a power near a half.

That is a weaker agreement than this collection usually reports, and the reason is not a disagreement about physics: it is that the lumped route converges slowly and the wave route is exact, so the difference between them is entirely the ladder’s own discretisation. The comparison is kept because the shape is reproduced — two thirds, then a step at the right instant, then a decaying alternation — by a computation that could not have been told about any of it.

What the branch is measured against

A stub is a third node on a two-node model, and everything on this page is a departure from that model. The staircase in time is the two-node picture, where the far end sits at exactly zero until one delay has elapsed. A ladder is not a line is what happens when the same wave is approximated by lumps, and it converges far too slowly to be used here. The resistor at the wrong end is the termination decision this branch complicates — a series-terminated net has an undefined stretch whose length depends on where the receiver stands. Where the current comes back is where the geometry that sets the branch’s impedance is computed, and The delay that is not one number is the assumption the lattice inherits.

What is checked

The junction’s reflection coefficient is asserted to be exactly −1/3-1/3, which is arithmetic rather than a measurement and is checked so that a future change to the scattering matrix cannot pass silently. The first arrival is asserted to be two thirds of the swing, and the interval it is held for to be twice the stub’s delay to within two per cent — with the condition under which that is the stub’s alone, that its round trip is shorter than the net’s own delay, asserted beside it rather than assumed.

The recovery is asserted to be a third of the previous error with the sign changed. The dip expression is asserted against the march over the whole range where the round trip is under an edge, and what remains is asserted to be the grid rather than the expression, by halving the cell and requiring the departure to halve. And the saturation is asserted to be exactly a third of the swing for an edge much shorter than the round trip.

What is not built: a branch with a load on the end of it rather than an open circuit, which is what a terminated receiver at the end of a stub would be, and a net with two branches, where the junctions interact. Both are the same machinery with a different reflection coefficient and neither is here.

Why the two-thirds is the harder number

The interval measured on this page and the one the resistor at the wrong end measures look alike and behave quite differently, and the difference decides which of the two a designer can do anything about.

That one is 2(1−x)2(1-x) delays of half swing, and it is a property of the termination scheme: it goes to zero at the far end of the net, so putting every receiver where the line ends removes it entirely, and the scheme that never has it draws sixty milliamperes for as long as the level is held. It is a trade with two named options and a price in current.

This one is two thirds of the swing held for twice the stub’s own delay, and it is a property of the junction: three lines of equal impedance meeting present each other with half the impedance, so the minus a third and the two thirds follow from the topology and hold whatever the net is terminated with and wherever on it the branch is. There is no scheme that removes it. The only quantity that moves it is the stub’s length, and a length is not a schematic parameter — the edges that are lengths is the collection’s name for exactly that class: boundaries nobody chooses at the schematic, set by whoever builds the thing, appearing in no netlist at all.

Which is the practical reading of the pair. Choose the termination and the receiver positions together, because the first has a scheme-shaped answer and the second does not; then shorten the stubs, because that is the only remaining variable and it is one the layout owns rather than the design.

The minus a third is worth one more sentence, because it is the same arithmetic as the essay this one stands on. The staircase in time is a sequence of reflections at two ends of one line, each of which is a divider between the line’s impedance and whatever it meets; a junction of three lines is the same divider with two lines in parallel on the far side, which is where the half and therefore the third come from. Nothing new is being computed — what is new is that the discontinuity is in the middle of the net rather than at either end, so it is a place a lattice diagram has no row for.

Part 2 on termination

One argument about Termination, and one of 3 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

A series-terminated net holds half a swing for 1.0 delays at 50% along it. computed by solving, not by drawing. What a receiver 50% of the way along one net sees under three terminations, from a lattice evaluated at that point rather than at the ends. The shaded strip is the interval in which a logic input has no defined answer — between 30% and 70% of the swing. The series-terminated net sits in it for 4.83 ns, which is 2(1 − x) delays exactly and is zero only at the far end; the unterminated net sits in it for 4.83 ns and then overshoots by 82%; the parallel-terminated net never does, and draws 60 mA down the line for as long as the level is held. The resistor at the wrong end Part 1 — A lattice diagram is read at the two ends of a line, and that is where the two respectable terminations look identical: a clean step, one delay late, at the receiver. Anyone standing halfway along a series-terminated net sees half the swing held for a full round trip, which for a logic input is not a level at all. The interval is 2(1−x) delays exactly, it is zero only at the far end, and the scheme that never has it draws sixty milliamperes for as long as the level is held. Terminated at both ends: no interval anywhere, at 1.65 V of 3.3. computed by solving, not by drawing. The same net as the three-way comparison, with a fourth trace: a series resistor at the driver AND a parallel one at the receiver. Both reflection coefficients are zero, so the wave that arrives at a receiver 50% of the way along is already the final value and there is no second arrival to wait for — the departure after the first edge is 0.0e+0 per cent, which is the arithmetic's floor. What it costs is the level: 1.650 V of 3.3, exactly half, because two equal resistances divide the supply and nothing reflects to double it back. The series scheme in the same place is undefined for 1.00 delays and the unterminated one overshoots by 82 per cent. The hold current is 33.0 mA against the parallel scheme's 60, because the path to ground now has two resistances in it. Terminated at both ends Part 3 — A series resistor at the driver and a parallel one at the receiver cost exactly half the swing — 1.650 volts of 3.300 — and no reflection ever gives it back, because there is no reflection. What the half buys is measured rather than asserted: a driver thirty per cent off its assumed impedance rings a series-terminated net by 16.3 per cent and a doubly terminated one by nothing at all, and an interior receiver on a series-terminated net sits in the undefined band for every far-end resistance above 125 ohms, which is Z₀/(1−2b) and contains no length, no driver and no frequency.

Where this goes next

A series-terminated net holds half a swing for 1.0 delays at 50% along it. computed by solving, not by drawing. What a receiver 50% of the way along one net sees under three terminations, from a lattice evaluated at that point rather than at the ends. The shaded strip is the interval in which a logic input has no defined answer — between 30% and 70% of the swing. The series-terminated net sits in it for 4.83 ns, which is 2(1 − x) delays exactly and is zero only at the far end; the unterminated net sits in it for 4.83 ns and then overshoots by 82%; the parallel-terminated net never does, and draws 60 mA down the line for as long as the level is held. The resistor at the wrong end A lattice diagram is read at the two ends of a line, and that is where the two respectable terminations look identical: a clean step, one delay late, at the receiver. Anyone standing halfway along a series-terminated net sees half the swing held for a full round trip, which for a logic input is not a level at all. The interval is 2(1−x) delays exactly, it is zero only at the far end, and the scheme that never has it draws sixty milliamperes for as long as the level is held. 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. 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 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.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down, the 8 sharing most with it of 10.

The objects named here

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

Characteristic impedanceDesign tradeoffLattice diagramLogic thresholdPropagation delayReflection coefficientTermination