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

Reflection coefficient — where it appears

The ratio of the wave returned from a discontinuity to the wave arriving at it, set by the two impedances either side of it. At a junction of three lines of equal impedance it is exactly minus a third, because from any one of them the other two are in parallel.

Named by 14 essays across 3 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 3 quarter-wave sections between 50 Ω and 200 Ω. computed by solving, not by drawing. |Γ| computed by cascading exact line impedances from the load back to the source, so every multiple reflection is in it. The binomial design is exact at the centre and holds |Γ| below 0.10 over 67.9% of the centre frequency, against 17.1% for a single section. The equal-ripple design, found by minimax search rather than from a table, covers 98.1% at the same worst reflection — 45% more — and its ripples come out level to 0.0e+0%, which is the check that the search converged.

Several sections, and the band they buy

A quarter-wave transformer is exact at one frequency, and a 4:1 transformation holds |Γ| under 0.1 over 17.1% of it. Several sections whose reflections cancel over a band take that to 47.7, 67.9, 82.1 and 92.6% — which is filter design with the reflection as the shaped quantity. An equal-ripple design found by minimax search buys a further 35 to 46% at the same worst reflection, and its ripples come out level to four decimals.

lines · Matching 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 5 sections, equal ripple, and the band that is 134% rather than 97. computed by solving, not by drawing. The repaired five-section equal-ripple design over the band it was designed for. The horizontal rule is the 0.1 the specification allows and the 4 interior peaks sit on it, level to 8.5e-4 per cent — which is the condition for a minimax solution and is now checked rather than assumed. The dots are the eighty-one frequencies the objective used to be evaluated at: the worst of them is 0.09999998 and the worst of the design over the whole band is 0.10013858, so an optimiser shown only the dots drove them down to the specification and left the true peaks 13.9 parts in ten thousand above it. That is nothing until something downstream is a threshold, and the band measurement was one: it reported 97.34 per cent for a design that holds 134.04.

The number that was wrong

The rung below printed 97.3 per cent of band for a five-section transformer where the answer is 134, said in its own text that the figure was wrong, and blamed a search that had converged to eight digits. The search was fine. The objective was the worst of a grid rather than the worst of a band, the band was then measured by bisecting a function that crosses its threshold five times, and the assertion guarding all of it passed — because 97.3 is still more than 92.6.

lines · Matching 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 Twenty metres buys 20 dB of apparent match and costs 10 dB of noise figure. computed by solving, not by drawing. The same cable and the same 200 Ω load as the reading, with the amplifier that is actually behind the instrument. The rising trace is the return loss the instrument reads, which is the load's own 4.44 dB plus twice the one-way loss. The lower pair is the chain's noise figure: a 2 dB amplifier with the cable in front of it, counted as a matched attenuator whose noise factor is its loss, and counted honestly from the available gain of a lossy line driven by a source that reflects 0.60. The first says the exchange rate is exactly two decibels of match per decibel of noise figure, at every length here. The second is higher everywhere — by (1−Γ²u²)/(1−Γ²), which is 0.028 dB at five centimetres and 1.938 dB, the load's own mismatch loss, once the cable is long enough to have absorbed the reflection. At twenty metres the instrument reads 24.4 dB and the chain costs 13.92 dB against the amplifier's own 2.

The cable that hides two things

A length of cable improves a return-loss reading by twice its loss and raises a noise figure by once it, so the rule of thumb is two decibels of apparent match per decibel of floor. Both halves are owned here and neither essay had the other. Put together they say what an acceptance limit costs: making a 4.0:1 load read 1.50:1 spends 6.53 decibels of noise figure, against a mismatch that was itself costing 1.938 — and the exchange rate is not two but 2(1−Γ²)/(1+Γ²), which is 0.94 where a pad is actually short.

lines · Line loss 1 pF across a 50 Ω line: a dip of 0.320 V and an area of 25 ps. computed by solving, not by drawing as a cascade of two-ports, with a raised-cosine edge of 59 ps sent into it. The incident edge is the faint curve; what comes back is the shaded dip and what goes on is the third. The dip reaches -0.3202 V and its area is 25 ps, which is Z₀C/2 to a part in ten thousand. Driven by an edge fifty times faster the same cascade returns the single exponential the closed form gives, to 9.3e-6 of a volt. The transmitted edge leaves at 81.1 ps, against 59 ps arriving.

The dip whose area is fixed

A picofarad across a 50 Ω line makes a dip in what comes back. Its depth is 0.833 volts to a six-picosecond edge and 0.0196 volts to a nanosecond one, forty-two times less; its area is 25 picoseconds to both, to six parts in a hundred thousand, because the area is Z₀C/2 and contains nothing about the edge. Two half-picofarad discontinuities too close to tell apart read as exactly one picofarad, and so do two far enough apart to be separate — the area is additive where the depth is not. And what a reflectometer calls the capacitance of an impedance step is the step's real excess capacitance times 1 + Z/Z₀.

lines · Reflections An L-section from 50 Ω to 1 kΩ: Q 4.359, fixed by the two resistances, and a band of 4.73%. computed by solving, not by drawing. A series inductor and a shunt capacitor matching 50 Ω to 1 kΩ at 1.00 MHz, their values from the series–parallel conversion: the load with the capacitor across it is 50 Ω in series with a reactance of 217.9 Ω at the design frequency, and the inductor cancels the reactance. The reflection there is 3.6e-16. The section's Q is √(20 − 1) = 4.3589 and no choice of parts changes it. |Γ| stays under a tenth from 976 kHz to 1.02 MHz, 4.73% of the design frequency, against 0.2/Q = 4.59%; and under half the power from 727 kHz to 1.21 MHz, 48.53%, against 2/Q = 45.88%.

The match with no knob

An L-section — a series inductor and a shunt capacitor — is the series–parallel conversion used on purpose: a load with a capacitor across it is, at one frequency, the source's resistance in series with a reactance an inductor cancels. Matching fifty ohms to a kilohm that way reflects 3.6 × 10⁻¹⁶ at its design frequency and has a Q of √19 = 4.359 that no choice of parts can change, so it holds its reflection under a tenth over 4.73 per cent of band whatever it is built from. The band is 0.2/Q to within three per cent, it depends on nothing but the ratio, and only splitting the match widens it: 14.98 per cent in two sections, 30.34 in three — and 30.83 in four.

frequency · Series parallel A via's area changes sign at 44.7 Ω, and two impedances give its 0.5 pF and 1 nH back. computed by solving, not by drawing, as a cascade of two-ports: a via of 0.25 pF, 1 nH and 0.25 pF, met by an edge of 59 ps from reference lines of 20 to 150 Ω. The area under the reflection is −Z₀C/2 + L/2Z₀ at every impedance: a bump below 44.7 Ω, where the inductance's term is the larger, nothing at it, and a dip above. From 50 Ω the area is 2.5 ps of dip, which a single-capacitance reading calls 0.100 pF. From 50 and 75 Ω together the two areas give 0.5000 pF and 1.0000 nH. An error of 50 fs on each area moves them by up to 0.8% and 1.5%.

The via two lines can weigh

The area under a reflection is a property of the discontinuity rather than of the edge, and for a via it is one number made of two: −Z₀C/2 from its pads and +L/2Z₀ from its barrel, with opposite signs. A via of half a picofarad and a nanohenry, seen from fifty ohms, leaves 2.5 picoseconds of dip — which a reading that assumes a capacitor calls 0.100 pF, a fifth of what is there. From fifty and seventy-five ohms together the two areas give 0.5000 pF and 1.0000 nH back, and fifty femtoseconds of error on each costs 0.8 per cent of the capacitance and 1.5 of the inductance. A second line at fifty-five ohms costs five times as much.

lines · Reflections At 44.7 Ω the via's area is zero; its reflection is a doublet falling as the 1.99 power of the edge, not the first. computed by solving, not by drawing, as a cascade. A via of 0.25 pF, 1 nH and 0.25 pF on a line of √(L/C) = 44.72 Ω, met by an edge of 59 ps, against the same capacitance alone on the same line. The via's reflection is a doublet — a dip and a bump of equal area — whose largest excursion is 24.4 mV against 166 mV for the capacitance alone. Over edges from 5.9 ps to 295 ps the capacitance's reflection falls as the −0.97 power of the edge and the via's as the −1.99 power: 24.4 mV at 59 ps, 994 µV at 295 ps. The via delays the edge going past it by 22.4 ps, against √(LC) = 22.4 ps.

The via that is a piece of line

A via of half a picofarad of pad and a nanohenry of barrel puts no area under its reflection from a line of √(L/C) = 44.72 ohms, from any edge. It has not vanished. It delays the edge going past it by 22.4 picoseconds, which is √(LC) exactly, and it still reflects: a doublet whose largest excursion falls as the −1.99 power of the edge where a lone capacitance's falls as the −0.97 — 24.4 millivolts for a 59-picosecond edge against 166 for the pads alone, and 994 microvolts for a 295-picosecond one against 35. The balanced via is a short piece of line, and the reflection it leaves is the reflection of its length rather than of its size.

lines · Reflections 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.

TerminationTransmission lineCharacteristic impedanceImpedance matchingModel rangePropagation delayDesign tradeoffLattice diagramChebyshevInsertion lossLogic thresholdLumped-element

All concepts