Concept

Convergence order — where it appears

The power of the step, the sample count or the section count at which an approximation's error falls, fitted from measurements rather than quoted. An exponent measured this way is a stronger claim than any single error, because it says what halving the step or doubling the count will do.

Named by 24 essays across 10 fields — each of them below, with the objects they name alongside it.

A diode fed from 5 V through 1.0 kΩ. computed by solving, not by drawing. The operating point is where the exponential meets the load line: 0.692544 V and 4.3075 mA, reached in 13 damped Newton steps from a cold start. The one-line Newton on Vs = v + R·i(v), which touches no matrix, gives 0.692544 V. The "drop" is not a constant: it moves 59.53 mV per decade of current, measured between two solved operating points.

A bias point is a solution, not a choice

The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.

semiconductors · Operating point
One step response, computed twice: from the poles, and by walking the network forward. A damping ratio of 0.22, so the overshoot is 49.2%. The two curves are drawn on top of each other; the panel below is the difference between them, which is the trapezoidal rule's error at 500 steps and reaches 1.70e-3 V.

One step, computed twice

A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.

transients · Step response
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.

lines · Distributed vs lumped
What the reconstruction returns, either side of 5.0 kHz. computed by solving, not by drawing. The Whittaker–Shannon sum is evaluated on the samples and compared with two things: the signal that was sampled, and the frequency the samples report. Below 5.00 kHz these are the same curve and the error is 4.88e-3 — the truncation of the sum at sixty-four samples either side, and nothing else. Above it they part: at 9.00 kHz the reconstruction is 1.59e-3 from the alias and 2.000 from the input. The small number is the interesting one. A reconstruction cannot be improved into the right answer, because it is already an exact answer to a different question.

An exact answer to a different question

The reconstruction that turns samples back into a signal is normally introduced as the thing that recovers what was there. Measured on both sides of half the sample rate it does something more interesting than failing: above the boundary it returns the alias to 1.6 parts in a thousand, which is the same accuracy it returns the input with below the boundary, and it is wrong about the input by twice the amplitude. Its error is not a degradation. It is exactness about something else.

digital · Sample rate
How much faster an instrument must be for 10% of inflation. computed by solving, not by drawing. The quadrature rule answers 2.182× and gives the same answer for every instrument, because it contains no instrument. Measured on the solved network, a one-pole front end needs 2.79×, two poles need 3.97×, three need 4.87× and four need 5.62× — between 28% and 215% more than the rule asks for. The rule errs optimistic at every pole count, which is the wrong direction.

The instrument's own rise time

Rise times add in quadrature, so ten per cent of inflation needs an instrument 2.18 times faster than the edge. That constant contains no instrument. Measured on the solved network it is 2.79 for a one-pole front end, 3.97 for two, 4.87 for three and 5.62 for four — the rule is optimistic at every pole count, which is the wrong direction for a rule of thumb to err in.

instruments · Probe loading
Averaging a white sequence, and averaging a pink one. computed by solving, not by drawing. Both sequences are the same seeded white stream, one of them put through the 1/f network. Averaged in non-overlapping blocks, the white one's spread falls as n to the -0.510 ± 0.006 across five seeds — the √N law — and the pink one's as n to the -0.087 ± 0.013, which is very nearly not at all. A thousand-sample average buys a factor of 36.9 on the first and 2.1 on the second.

The corner where averaging stops working

Average N samples and the noise falls by √N. That is a statement about independent samples, and flicker noise's samples are not independent — its correlation extends over every time scale, which is what a spectrum with no bottom means. Measured on the same seeded stream filtered and not: the white sequence falls as the −0.510 power of the block length and the pink one as the −0.087 power, so a thousand-sample average buys a factor of 36.9 on one and 2.1 on the other.

noise · Device noise
One bit, oversampled — and where the quantisation noise went. computed by solving, not by drawing. A first-order modulator is marched forward one sample at a time with a one-bit quantiser inside the loop, and the noise inside the band is read out of the transform of the error. It falls by 8.99 dB for every doubling of the oversampling ratio — measured 9.33, 10.12, 6.96, 9.54 — against 3.01 dB for plain oversampling, which is drawn beside it from the same starting point. The loop does not make less noise; it moves the noise out of the band, and at a ratio of 128 one bit is worth 9.18.

One bit, and where the noise went

Sampling faster spreads a fixed quantity of quantisation noise over a wider band, so the part inside the band of interest falls by 3.01 dB for every doubling — half a bit. Putting the quantiser inside a loop with an integrator does something different in kind: measured on a modulator marched forward one sample at a time, with its test tone inside the band the ratio is quoted over, the in-band noise falls by 8.99 dB per doubling. At an oversampling ratio of 128, one bit is worth 9.18 — and the octaves scatter by a decibel each, which is the loop telling the truth about what its error is made of.

digital · Noise-shaping
A 0.5 mm conductor's resistance against frequency, exact and asymptotic. computed by solving, not by drawing. The exact ratio is computed from the Kelvin functions by their series; the dashed curve is the asymptote everybody quotes, which treats the current as flowing in one skin depth of the rim and is drawn only where that annulus is inside the wire. At 17.4 kHz, where the skin depth equals the radius and the rule of thumb says the effect "starts", the asymptote says 1.0000 — no effect at all — and the exact answer is already 1.0208. The rule of thumb names a frequency the effect has passed, which is the same shape as the tenth-of-a-wavelength criterion marking a point at which the lumped model is already 30% wrong. Two decades above, the two agree to 0.00%, which is what makes it an asymptote rather than a formula.

The resistance that grows with frequency

The rule of thumb names the frequency at which the skin depth equals the conductor's radius as the point where the effect begins. Computed exactly from the Kelvin functions, the resistance is already 2.05% up there — and the rule's own asymptote says 1.0000, no effect at all. Two decades higher the two agree to 0.01%, which is what makes it an asymptote rather than a formula, and what makes the frequency it names the wrong one to design at.

magnetics · Winding
Sixteen more digits move the boundary by sixteen decades and leave it exactly where it was. computed by solving, not by drawing. The earlier essay's bridge, walked towards balance and solved twice: once in double precision and once with a pair of doubles carrying about 31 decimal digits, against a closed form that cannot lose any. The 33 per cent error at an imbalance of 10⁻¹⁵ becomes 7.0e-18 — so that loss was the arithmetic's and not the network's, which is what the earlier essay could not say. Each arithmetic's error is its own round-off divided by the imbalance, drawn as the two straight lines, so the second boundary is the first one moved by exactly the extra digits. The condition number is 505 in both cases and at every point, which is the diagnostic being blind twice over.

The digits the arithmetic did not have

Bounding the arithmetic of double-precision circuit solving finds two boundaries that cannot be attributed to it: a bridge with no correct figures left at an imbalance of 10⁻¹⁵, and a filter synthesis that stalls at order 14. An ill-conditioned problem stays ill-conditioned however many digits are used, and a well-conditioned one computed badly gets better — so adding digits is the experiment that tells them apart. The bridge's loss is entirely the arithmetic's. The synthesis's is mostly the data's, and doubling the digits makes it worse.

networks · Conditioning
A bridge's reading turns over at 1.885 V, where the test level stops mattering. computed by solving, not by drawing. What a bridge reads is the fundamental of the charge waveform over the fundamental of the voltage, so it is a function of the bias AND of the amplitude, and a data sheet names one point of it: zero bias, one volt. The four curves are four test levels on one part. At zero bias they run from 9.999 µF down to 8.312 µF — the harder the drive the lower the reading, because the capacitance is at its maximum there and a sinusoid spends most of its time off the peak. At the rated 5 volts they run the other way, 2.000 µF up to 2.426 µF, because the curve is convex. Between them is one bias where the two effects cancel: at 1.8848 V a tenfold change of test level moves the reading by nothing at all, and that voltage is 0.6645 of the polarisation's own characteristic voltage.

The reading a data sheet does not take

A class II ceramic's temperature envelope leaves its working capacitance 29 points wide at one end and 79 across, because one printed number cannot pin a two-parameter model. One further bridge reading recovers almost all of it — and where the reading is taken decides everything. Turning the test level down to a fiftieth separates five parts a data sheet cannot tell apart by 17.76 per cent; moving the bias to half the rated voltage separates them by 176.02, and pins the working capacitance to ±0.512 per cent from a reading known to one.

frequency · Real capacitor
The fourth-order term explains 95% of what the third leaves out. computed by solving, not by drawing. Two closed forms for the amplitude at which a degenerated stage reaches 1% of second harmonic, each against the same measurement — a bisection on the harmonic content of a Newton-solved curve, which shares no arithmetic with either. The leading expression is 4Vₜ·t·D², derived by expanding the transfer curve and exact in the limit of small drive; its error grows as the 1.96 power of the drive it is evaluated at. At a degeneration of eleven that drive is 4.53 thermal voltages and the expression is 6.89% optimistic. Carrying the reversion one order further takes it to 0.362%.

What the fourth order says about the third

A closed form derived by neglecting the fourth-order term is a claim with an error, and the error is a quantity the fourth-order term can be asked about. Carried one order further, the degenerated exponential's reversion predicts 7.33 per cent where the leading expression was measured to be 6.89 per cent optimistic — and takes the residue to 0.362 per cent. The correction has its own edge, in the same quantity, and it is measured too.

semiconductors · Emitter degeneration
Every order buys less range than the one before, and above 9 Vₜ the sixth is worse than the fourth. computed by solving, not by drawing. The error of the same expression truncated at three orders, against the drive it is evaluated at, with the measurement it is chasing being a Newton-solved transfer curve that knows about no series at all. Each truncation's error grows as its own order in the drive — fitted at 2.00, 4.00, 5.88 against 2, 4 and 6 — so each buys a further range at a stated accuracy: inside 1% the leading expression is good to 1.12 thermal voltages, the fourth order to 3.88 and the sixth to 7.03, factors of 3.46 and 1.81. Beyond all of them the series stops helping: at 8.9 thermal voltages, where the second harmonic is 11.4%, the sixth-order expression is exactly as wrong as the fourth and is worse above it. What a designer does there is bisect the curve.

The order that stops helping

Three essays in this field have derived expressions for the amplitude at which a degenerated stage's distortion reaches a target, each one order longer than the last, and each one nearer the measurement. This is where that stops. The error of an expression truncated at order m grows as the m-th power of the drive — 2.00, 4.00 and 5.88 measured — so every added order buys a range that ends sooner than the last one bought, and above 8.9 thermal voltages the six-term expression is further from the device than the four-term one.

semiconductors · Emitter degeneration
Stepped at 20 of its time constant, a 1 µs pole rings between 1.818 and 0.331 V, and needs 23 steps to settle. Marched with the trapezoidal rule at a step of 20.0 µs. A 1 µs pole (1 kΩ, 1 nF) drives, through a unity buffer, a 1 ms pole (1 kΩ, 1 µF). The fast node's exact response reaches its final volt within a few microseconds; the march's first values are 1.8182, 0.3306, 1.5477, 0.5519, 1.3666 V. Its distance from its final volt is multiplied by (1 − h/2τ)/(1 + h/2τ) = −0.8182 every step, measured and checked against that form, so it changes sign every step and takes 23 steps to fall below 1% — 460 µs. The slow node it drives is 1.23e-5 V from exact at 1 ms, because a 1 ms pole averages an alternation at half the stepping rate to nothing.

The ringing that belongs to the rule

The trapezoidal rule is stable for every stable circuit and every step size, and it is not damping. March a one-microsecond pole with twenty-microsecond steps and its node reads 1.818, 0.331, 1.548, 0.552 volts — an oscillation at half the stepping rate, its distance from the final volt multiplied by exactly −0.8182 every step, taking twenty-three steps to fall below one per cent. The slow node that pole drives is right to 1.2 × 10⁻⁵ V at a millisecond. One backward-Euler step at the discontinuity cuts the first swing from 0.818 V to 0.048 and two to 0.0023, because backward Euler multiplies the same error by 1/(1 + h/τ) and the trapezoidal rule by (1 − h/2τ)/(1 + h/2τ), which approaches −1.

transients · Step response
At 20 steps a cycle, ten cycles of an undamped LC: the trapezoidal rule keeps the amplitude and falls 29.2° behind; backward Euler keeps 0.0082% of it. Marched, both rules, against 1 − cos ωt for a 1 kHz inductor–capacitor pair stepped with no resistance at all. At 20 steps a cycle the trapezoidal march's amplitude stays at 1.00000 a cycle and its frequency is slow: it loses 2.918° a cycle, measured from the march's own recurrence, against 2π − 2N·atan(π/N) = 2.918°, so after ten cycles it is 29.2° behind. Backward Euler keeps 0.3901 of its amplitude a cycle, against (1 + (2π/N)²)^(−N/2) = 0.3901, so 0.0082% is left after ten, and it loses 11.19° a cycle. No resistance is in the circuit; every loss is the rule's.

The phase the rule loses

An inductor and a capacitor with no resistance ring for ever, and two ways of marching them disagree about how. The trapezoidal rule keeps the amplitude exactly — its factor per step has a magnitude of one — and loses phase instead: 2π − 2N·atan(π/N) a cycle, 2.918° at twenty steps a cycle, so ten cycles later it is 29.2° behind the circuit. Backward Euler keeps 0.3901 of the amplitude a cycle at the same step, and after ten cycles 0.0082 per cent of the ringing is left, in a circuit that has no loss. The two errors fall at different rates: the trapezoidal rule's phase as the square of the steps a cycle, backward Euler's amplitude as the first power. A hundred cycles to within one per cent needs 182 steps a cycle of one and 196,404 of the other.

transients · Step response
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
BDF2 loses amplitude as N⁻³ and phase as N⁻², four times the trapezoidal rule's: a hundred cycles to 1% needs 363 steps a cycle, against 182 and 196,404. Each rule's error a cycle on an undamped inductor–capacitor pair, against the steps a cycle, measured from BDF2 marches and drawn with the trapezoidal rule's and backward Euler's closed forms. BDF2 loses amplitude a cycle as the −2.91 power of N and phase as the −1.95 power, its phase loss 3.988 times the trapezoidal rule's at 128 steps a cycle. To keep 99% of the amplitude over a hundred cycles it needs 157 steps a cycle, and to stay within 1% of a cycle in phase 363; the trapezoidal rule needs 182 for the phase and loses no amplitude, and backward Euler needs 196,404 for the amplitude.

The rule between the two

The trapezoidal rule keeps an undamped ringing's amplitude and loses its phase, and it preserves a stiff pole's error with its sign alternating. Backward Euler damps the stiff error at once and damps the ringing too, keeping 0.0082 per cent after ten cycles at twenty steps a cycle. The second-order backward difference sits between them in a precise sense. It damps the stiff error by 1/√(3 + 2h/τ), 0.152 a step at twenty time constants, and settles in three steps where the trapezoidal rule needs twenty-three. It loses a ringing's amplitude only as the cube of the steps a cycle, and its phase as the square with a constant four times the trapezoidal rule's. A hundred cycles to one per cent costs 363 steps a cycle, against 182 and 196,404.

transients · Step response
Three bias points below a coupling of 56.6 kΩ, one above — and the middle one is where Newton from nought lands. computed by solving, not by drawing. Two transistors, each with 1 kΩ to a 5 V supply and each biased from the other's collector through a coupling resistor into a 10 kΩ base resistor, solved by Newton on the netlist from six starting guesses and de-duplicated. Below a coupling of 56.6 kΩ there are three operating points: two latched, with one collector at 0.061 V and the other at 4.615 V at 10 kΩ, and a symmetric one at 1.702 V whose Jacobian determinant is negative, so the circuit cannot rest there. Above it the three merge into one symmetric point, and the circuit is an amplifier rather than a memory.

The circuit with three bias points

Two transistors, each biased from the other's collector, have three operating points: one collector at 0.061 V and the other at 4.615, the same the other way round, and both at 1.702. All three solve the circuit to a part in 10⁹, and Newton started from nought — where a simulator starts — finds the symmetric one, which is the one the circuit can never sit at: its Jacobian determinant is negative, so any positive capacitance at the nodes makes it run. Weaken the coupling past 56.6 kΩ and the three merge into one, the loop gain around the symmetric point having fallen through one. Which answer a solver returns depends on where it begins, and from 1,681 starting guesses at 10 kΩ, 437 return the one that is not a state.

semiconductors · Operating point
The best foil thickness for 4 layers, for three currents with the same fundamental. computed by solving, not by drawing. The loss of a portion of 4 layers against foil thickness, with the loss weighted by the current in each harmonic rather than computed for one frequency. A sinusoid wants 0.6631 skin depths and lands at 1.3368 times the direct-current resistance — four thirds, the single-frequency constant, reproduced. A triangular ripple wants 0.6432, which is the same answer to within 3.0 per cent, so a winding carrying one needs none of this. A square current of the same fundamental wants 0.3838 — thinner by a factor of 1.728 — and lands at 1.8313, which is not four thirds and is not any constant the geometry knows. Building to the sinusoid's answer costs 16.8 per cent more loss.

The optimum a spectrum moves

The best foil thickness for a winding is derived for one sinusoid and quoted as a property of the geometry: a minimum at four thirds of the direct-current resistance, whatever the layer count. Weight the loss by the current in each harmonic instead and a square current of the same fundamental wants foil 1.728 times thinner and lands at 1.83, and a narrow pulse wants it 3.68 times thinner. Four thirds is a property of the current. The constant that replaces it for an ideal square edge is exactly two, and a real winding sits between them at a place its edge rate decides.

magnetics · Winding
A sinusoid sampled 8 times a cycle, at the phase that puts its peak between two samples. computed by solving, not by drawing: the samples of a unit sinusoid at 8 a cycle, synchronous, at the sampling phase that hides its peak worst. 8 samples a cycle is 4 times the Nyquist rate and holds the waveform completely; the largest sample is still 0.9239, 7.61% below the peak, which is 1 − cos(π/8). Read straight off the samples the peak needs 22.2 samples a cycle for 1% and 70.2 for 0.1%.

The rate each reading needs

Twice the frequency is the rate at which a set of samples holds a sinusoid completely, and it is not the rate at which any particular property can be read off them. Eight samples a cycle — four times the Nyquist rate — hold the waveform, and their largest is still 7.61% below the peak; the largest sample needs 22.2 a cycle for 1% and 70.2 for 0.1%, because its error falls only as the square of the rate. A parabola through the three largest gets 1% at 8. A zero crossing found by a straight line between two samples is a hundredth of a cycle right at 5 a cycle, its error falling as the cube, because a sine is straight at its zero. The rms of whole cycles is exact from three a cycle — and at exactly two it reads anything from zero to √2. A slow rate with a long record of a steady signal does what a fast rate does, once the samples have slipped a whole interval past it: 42 cycles at 4.0236 a cycle.

digital · Sample rate
Three trackers finding a panel's maximum power from open circuit, against the number of readings each takes. computed by solving, not by drawing: the single-diode panel of 81.0 W at 17.244 V, every tracker starting at 0.995 of the 19.85 V open-circuit voltage. Perturb-and-observe with 0.2 V steps walks down and then oscillates about the maximum, within two parts in a thousand of it from reading 39. Newton on I + V·dI/dV, two readings a step for the current and its slope, is within a part in a million from reading 10. The fractional open-circuit rule needs one reading and lands exactly, because its fraction, 0.8686, was taken at this light.

How many readings a maximum takes

A panel's maximum power sits where I + V·dI/dV is zero, and a tracker finds it by reading the panel. From open circuit, perturb-and-observe in 0.2 V steps takes 39 readings to come within two parts in a thousand and then oscillates about the maximum forever, giving up 7|P''|δ²/24 of it; Newton's method on the condition itself takes 10 readings to come within a part in a million, the digits doubling at each step. Newton converges only from a window: above 15.82 V on a panel whose maximum is at 17.24, because below it the condition is nearly the short-circuit current and nearly flat, and one step leaves the characteristic. And the one-reading rule — a fixed fraction of the open-circuit voltage — loses nothing at the light it was set at and 17 per cent at a fiftieth of it.

networks · Source model
Three waveforms at 16 samples a period, sampled at the worst instant: the largest sample of a sine is 1.92% short, of a triangle 12.5%, of a clipped sine 0. computed by solving, not by drawing: three waveforms with the same period and the same peak, sampled 16 times a period at the phase that puts the peak midway between two samples (dots). The sine leaves its peak as the square of the time, the triangle in proportion to it, and the sine clipped at nine-tenths of its height does not leave it at all for as long as it stays flat. The largest sample therefore falls short by 1.92% for the sine, 12.5% for the triangle, 0 for the clipped sine, at one rate, for three waveforms with the same period.

The peak that is a corner

The largest of a set of samples falls short of a sine wave's peak as the square of the sample interval, and that law belongs to the sine. Measured over every sampling phase, a triangle's corner falls short as the first power, 12.5 per cent at sixteen samples a period against the sine's 1.92, a peak whose curvature vanishes as the fourth power, and a sine clipped at nine-tenths of its height reads exactly above seven samples a period. The order in which a waveform leaves its peak is the whole of the law, and it also fixes how much better the average reading is than the worst. A real corner is round at some scale, so the law bends where the samples resolve its top harmonic, and a parabola through three samples helps a round peak, barely touches a corner and makes a flat one worse.

digital · Sample rate
Eight stages of damping 0.8 against eight one-pole stages of the same bandwidth: settled within 1% at 18.75τ against 16.00τ, with 2.02% of overshoot. computed by solving, not by drawing: the step of eight identical second-order stages of damping ζ = 0.8, each with its half-power bandwidth at ω = 1, marched (solid), against eight one-pole stages with the same bandwidth from the closed Erlang step (dashed). Within 1% at 18.75τ against 16.00τ, the overshoot being 2.02% and each stage's delay 1.393 against one. At ζ = 1 the march agrees with the Erlang step of order sixteen.

The stage that overshoots a little

A stage with a little overshoot settles sooner than a one-pole stage of the same bandwidth, 3.53 time constants against 4.61 within one per cent, and a chain of them should do better still, since a stage at the right damping adds almost no spread. It does not. The saving is 17 per cent at two stages, 11 at four and 2 at eight, and at sixteen the chain is 5 per cent slower than the one-pole chain, because a stage's extra delay adds as the number of stages and the spread it removes adds only as the root. The best damping stays near 0.83 at every length, just short of letting the overshoot out of the band. In frequency the same stages are five times flatter at their best, at a different damping, and so the two edges of a chain ask for two different stages.

limits · Model edges
A Gaussian edge read at 3 samples per rise, at the worst sampling phase: straight chords between the samples say the rise is 8.78% from its true value. computed by solving, not by drawing: a Gaussian edge (solid), its samples at 3 per 10–90% rise at the sampling phase that reads worst (dots), and the straight lines an instrument's linear interpolation draws through them (dashed). The chord crosses 10 and 90 per cent earlier and later than the edge does, because a Gaussian edge curves in opposite senses at the two thresholds. The reading is 1.0878 rises against 1.

The samples a rise time needs

A rise time is read from two thresholds, and an instrument that draws straight chords between its samples is wrong at each by an amount that goes as the square of the interval. For a Gaussian edge at the worst sampling phase that is 5.02 per cent at four samples per rise and 0.818 at ten; the rule of thumb of three to five is a five per cent rule, and one per cent takes nine. The reading is not monotonic in the rate: above 20.9 per rise a first-order edge reads exactly at whole numbers and 0.046 per cent at 24.5, because its two errors subtract at one phase, and a Gaussian edge reads worse there because its two add. Reconstructing before reading needs 1.87 per rise for one per cent, and at four times the edge's 3 dB bandwidth the reading is still 11.7 per cent out.

digital · Sample rate
A line ten rises long stood in for by 20 LC sections, matched at both ends: the edge overshoots 17.5% and takes 1.66 rises to climb, where the line delivers it unchanged. computed by solving, not by drawing: an edge of 10–90% rise tr at the far end of a chain of 20 LC sections with a total delay of 10 tr, matched at both ends, marched (solid), against the same edge delayed by exactly 10 tr, which is what a line delivers (dashed). Overshoot 17.5%, rise 1.66 tr. With no reflections at either end, everything here is the chain's own dispersion: a section has a group delay that rises with frequency, so the fast end of the edge arrives late and rings behind it.

The sections an edge needs

A chain of LC sections rings on an ideal step however many sections it has, because a step has energy all the way to the chain's cut-off. An edge has a top frequency, and for a matched chain carrying one the section count has a clean answer: for one per cent of overshoot, 6 sections on a line one rise long, 183 on ten rises and 5,760 on a hundred, 5.8 times the length in rises to the three-halves power, within a few per cent from end to end. So sections per rise is not a constant. A section's phase error goes as the cube of frequency and the inverse square of the count, and it accumulates along the line. A 1 ns edge on a metre of cable asks for 62 sections, a 100 ps edge for 2,048, and the delay is right long before the edge is.

lines · Distributed vs lumped

Named alongside it

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

Model rangeNumerical errorSample rateDesign tradeoffNyquist rateReconstructionStep responseTrapezoidal ruleVerificationAliasingArtificial delay lineLumped approximation

All concepts