Circuit Response
every model has an edge, and the figure carries it
Concept

Termination — where it appears

The impedance a line or a two-port is closed with, which is part of its design rather than of the environment it is used in. A branch on the net breaks the property that makes a series termination work, since it no longer settles in one round trip.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

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.

lines · Reflections A 4.0:1 load reads 1.13:1 through twenty metres of cable. computed by solving, not by drawing. A 200 Ω load on a 50 Ω line — a standing-wave ratio of 4.00 at the load itself — measured from the other end of a length of cable that attenuates 0.500 dB per metre at 1 GHz. The reading falls as the length grows, exactly as |Γ| times ten to the minus twice the one-way loss over twenty, and reaches 1.5 at 9.5 m and 1.128 at twenty metres, which is 24.4 dB of return loss and would pass any acceptance test. The load is unchanged; the instrument is looking at it through 10.0 dB of attenuator. The length that hides it falls 3.16× per decade of frequency, which is the root of ten and is the attenuation's own law.

The mismatch that the cable hides

A lossless line carries a reflection back unchanged, so the standing-wave ratio at the instrument is the standing-wave ratio at the load. A real line does not, and the departure is exact: ten decibels of one-way loss improves any mismatch by twenty. A four-to-one load at the end of twenty metres of ordinary coaxial cable measures 1.13 at the near end, a return loss of 24 decibels, and passes an acceptance test the load could never pass. The boundary is a loss rather than a length, which makes it a frequency: 30.2 metres at 100 megahertz, 9.5 at a gigahertz, 3.0 at ten.

lines · Line loss 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.

lines · Termination An order-5 ladder driven from 2× the resistance it was designed between. computed by solving, not by drawing. A doubly-terminated Butterworth ladder is a two-port designed between two stated resistances, and the resistances are part of the design rather than the environment it happens to be used in. At match it loses 6.021 dB — exactly half the voltage — and its passband has no peak anywhere in it. Driving the same five reactances from 2× that resistance moves the shape by 2.345 dB and the insertion loss to 9.542 dB. The band inside which the shape is right to 0.5 dB runs 0.8857× to 1.1371× — a window of 25 per cent on a quantity usually written down as a round number. And the two sides are not alike: at twenty times the design resistance the departure has settled at 5.33 dB, while at a twentieth of it the passband is 15.51 dB out with 7.21 dB of peaking on it, so driving a ladder from too low an impedance is worse than driving it from too high a one.

The two resistors a ladder was designed between

A passive ladder filter is not a transfer function with some resistors attached; it is a two-port designed between two stated resistances, and the resistances are as much part of the design as the inductors. Driving an order-five Butterworth from anything outside 0.886 to 1.137 times its design resistance puts more than half a decibel of error on the passband — a tolerance tighter than the resistor is usually specified to — and the two sides of that window are not alike.

filters · Filter termination 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 0.400 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.

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.

lines · Termination 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.

The millimetre that becomes common mode

A pair carries two modes rather than two signals, and a length mismatch between its halves converts one into the other. A millimetre and a half of skew is ten picoseconds, a tenth of the signal is common mode by three gigahertz, and the differential signal has lost five thousand parts per million of itself getting there — so the error is not missing from the signal, which is why a pair can pass its own eye and fail an emissions test. The product in the answer is ωΔτ, which is what the instruments field's rejection corner is one over.

lines · Differential line 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

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.

lines · Termination The load alternates with parity between 1 and 1.9841, at every order. computed by solving, not by drawing. The last quotient of the continued fraction that turns a reflection polynomial into element values, which is the load resistance the design demands, drawn against order. A Butterworth returns exactly one at every order. A Chebyshev alternates: one at every odd order and 1.984056 at every even one, the same number each time, and it is the closed form (√(1+ε²)+ε)² to a part in 10¹⁵. The mechanism is in the last two rows of the panel. A lossless ladder is two resistances at direct current, so it must deliver the maximum available power there, which requires that zero be one of the frequencies the design reflects nothing at — and an even-order Chebyshev's reflection zeros are the roots of an even Chebyshev polynomial, none of which is zero. The synthesis stalls at order 9 for a Butterworth, which is why that series stops at eight.

The termination an even order cannot have

Every even-order Chebyshev ladder terminates in 1.984056 times its source resistance at half a decibel of ripple, the same number at orders two, four, six and eight, and it is (√(1+ε²)+ε)² to a part in 10¹⁵. Building one between equal terminations instead — which is what a table of g-values and a matched pair gives — turns 0.5000 dB of ripple into 1.8123 at order four, deletes one of the passband maxima outright, and moves the worst tolerance corner from the 2.000 power of the component tolerance to the 1.109 power: a factor of 95 at one per cent parts.

filters · Filter termination

Named alongside it

The objects these essays reach for when they reach for this one.

Reflection coefficientCharacteristic impedanceLattice diagramPropagation delayDesign tradeoffLogic thresholdComponent sensitivityDoubly terminated ladderInsertion lossMaximum power transferModel rangeTransmission line

All concepts