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

Transmission line — where it appears

A pair of conductors long enough that a signal's travel time along them matters, described by waves rather than by one impedance. The lumped ladder that stands in for one converges slowly — about the section count to the power of a half — which is measured here rather than asserted.

Named by 11 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 10.0 cm of track, solved as a lumped circuit and as a line. The two agree to 0.030% at 3.97 MHz, where the track is one degree long, and to 30.1% at 143 MHz, where it is a tenth of a wavelength. Above that the lumped model is not approximately right; it is describing a different object.

Kirchhoff's own frequency

The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board.

limits · Lumped-element 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.

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.

lines · Crosstalk 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 The same three stages, in two orders. computed by solving, not by drawing by Friis's cascade. With the low-noise amplifier first the chain's noise figure is 3.31 dB; with the mixer first it is 12.01 dB. The gain is identical either way — 48.0 dB — so the 8.70 dB is bought with nothing but an ordering. Everything after the first stage contributes 26.0% of the total. The 2.0 dB of cable in front has gain below one, so it multiplies every later stage's contribution rather than dividing it.

The loss in front, counted twice

A decibel of cable before an amplifier costs a decibel of signal, which everybody expects, and a decibel of noise figure, which is a separate decibel arriving from a separate place. Measured across the slider it is exact: 1.31 dB of chain noise figure becomes 2.31 with one decibel of cable and 9.31 with eight, every time, while the 8.70 dB that stage ordering is worth does not move at all. A lossless reactive network in the same position costs nothing, because only the real part of an impedance is warm.

noise · Johnson noise 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

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.

lines · Crosstalk 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 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 20 sections imitating a 1 m line: its delay is 1% long at 318 MHz and its group delay at 185 MHz, and it passes nothing above 1.32 GHz. computed by solving, not by drawing, as a chain of 20 series inductors and shunt capacitors carrying the inductance and capacitance of a metre of 50 Ω line of delay 4.83 ns, terminated in 50 Ω at both ends, beside the Bloch phase of an endless chain, 2·arcsin of ω over the cutoff, a section. The chain's cutoff is the cutoff 2/√(LₛCₛ), 1.32 GHz. Its phase delay is too long by arcsin(x)/x − 1 with x = f over that cutoff, 1% at 318 MHz (x = 0.2417); its group delay by 1/√(1 − x²) − 1, 1% at 185 MHz (x = 0.1404). Well below cutoff the solved chain's delay follows the closed form; nearer it the fifty-ohm terminations, which are not the LC ladder's own impedance there, add a ripple. Ten sections per wavelength is 414 MHz for this chain.

The sections a wavelength needs

A ladder of inductors and capacitors is a line only below its own cutoff, 2/√(LC) of one section, and in the frequency domain how far below can be written down exactly: its delay is too long by arcsin(x)/x − 1 and its group delay by 1/√(1 − x²) − 1, where x is π over the number of sections per wavelength. One per cent of delay needs 13.0 sections per wavelength; one per cent of group delay, 22.4. The rule of ten per wavelength is 1.72 per cent slow in phase and 5.33 in group delay. Twenty sections imitating a metre of cable are a line to a per cent of group delay up to 185 megahertz and pass nothing at all above 1.32 gigahertz.

lines · Distributed vs lumped

Named alongside it

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

Reflection coefficientModel rangeLumped-elementCharacteristic impedanceImpedance matchingInsertion lossParasiticsPropagation delayRise timeAvailable powerCrosstalkDesign tradeoff

All concepts