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

Residues — where it appears

The coefficients of a response's partial-fraction expansion, one to a pole, which turn a factorised transfer function into an exact waveform in time. They give an exact waveform where a march gives an approximate one, and the difference between the two measures the march rather than the circuit.

Named by 25 essays across 5 fields — each of them below, with the objects they name alongside it.

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 How long a second-order step takes to arrive inside ±2%. computed by solving, not by drawing from the residue expansion at 260 damping ratios. The fastest is ζ = 0.780 at 3.60/ω₀; critical damping takes 5.83/ω₀, which is 62% longer. Between ζ = 0.775 and 0.780 the time falls by 33% in one step of the sweep, because which excursion is the last one outside the band changes there — the overshoot at the fastest damping is 1.99%, which is the band itself, and one step to the left it is larger. The faint curves are the other bands, each with its own step in a different place.

The cliff before the fastest settling

Settling time against damping is not a smooth curve with a minimum. It falls by a third in one step of a sweep of five thousandths, and the fastest damping sits on the edge of that step — so a design a hundredth of a damping ratio to the left of the optimum settles forty-eight per cent slower, with a waveform that looks no different.

transients · Damping A step on a series RLC at ζ = 0.079, and the two numbers read off H(s). computed by solving, not by drawing. A 50.3 kHz series RLC driven by a one-volt step, with the capacitor voltage and the inductor voltage drawn together. Two limits of the transfer function are two points of the waveform and neither needs the waveform: H(0) = 1.000000 is where the capacitor ends up, and H(∞) across the inductor is 1.0000, which is what it does at the first instant — the expansion gives 1.000000 for it at t = 0. Here the damping ratio is 0.0791, the response is inside ±2% after 7.6 cycles, and 100.0% of the last twenty-four cycles sit there. The poles are at a real part of -7.91e-2 of ω₀, which is the condition the final-value theorem actually has — not a property of H but of where sY(s) has its poles.

Two numbers without solving for the waveform

Where a step response starts and where it ends are two limits of the transfer function, and neither needs the waveform. Both are exact here — 1.000000000 volts at the end and the whole step at the first instant — and one of them is a lie waiting to happen: take the damping to zero and the final-value theorem still returns 1.000000 for a response that swings between 0 and 2 for ever. Its condition is not on the transfer function but on where the poles are, and the practical condition is narrower still: at five ohms the poles are safely in the left half-plane and sixty cycles is not enough time.

transients · Value theorems A 1.0% doublet: 0.078 dB in the magnitude, 36× the settling time. computed by solving, not by drawing. Above, the magnitude of a fast circuit followed by a pole and a zero that were meant to cancel and miss by 1.00%, against the same circuit with the cancellation exact: the worst disagreement anywhere up to the fast corner is 0.0777 dB. Below, the error left in the step response, in units of the tail's own amplitude of 0.909%. Settling to 0.10% takes 245.2 fast time constants against 6.9 with the cancellation exact, and the closed form τ·ln(A/B) gives 245.2 — a time that contains nothing of the fast circuit at all.

The cancellation that leaves a tail

A pole and a zero placed on top of each other disappear from the response. Miss by one per cent and the magnitude changes by 0.078 decibels, which no measurement would report as a fault, while the time to settle to a thousandth goes from 6.9 time constants to 245 — thirty-six times longer. The settling time has a closed form containing neither the fast circuit nor the doublet's separation as such, and its consequence is blunt: settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is.

transients · Doublet Three cliffs, not one, and the fastest damping is on the last of them. computed by solving, not by drawing. Settling time against damping for a third-order response — a complex pair at unit natural frequency and a real pole at 3 — with the second-order case behind it. Both are staircases: the settling time is set by the last excursion outside the band, so there is one step for each excursion that stops happening, and there are 3 of them between 0.3 and 0.98. They are at 0.378, 0.522, 0.773, with jumps of 1.24, 1.30, 1.42. The earlier essay found the last and largest of them and did not look below it. The fastest damping is 0.775, sitting on the edge of the last step, and a design a hundredth to the left of it settles 42 per cent slower.

Three cliffs, and where they are

Sweep a second-order step's damping and the settling time falls by a third in one step of a five-thousandth sweep, which is usually called the cliff. There are three of them between 0.3 and 0.98, one for each excursion that stops leaving the band, and adding a third pole moves all three left and makes all three shallower — so the classic 0.78 for fastest two per cent settling is a second-order number, and at a third pole one and a half times the natural frequency the answer is 0.745 and 0.78 is on the wrong side of the step.

transients · Damping The sensitivity of a pole against the room it has. computed by solving, not by drawing. A series R–L–C whose damping is walked from 0.3 to 0.999999, which slides its two poles together along a straight line and changes nothing else. The exact derivative of a pole with respect to the capacitor climbs from 0.5241 to 353.6 as the gap between them falls from 19078 to 28.28 radians a second. The fitted exponent over the closest four is -1.0000, and the product of the two is the natural frequency itself — 9999.6894 against 9999.6894, at every damping drawn and not merely in the limit, which a closed form gives and this computation never sees. The resistor's curve runs at 2ζ times the capacitor's — below it at 0.3 and at twice it by the time the poles have met — and the inductor's lies exactly under the capacitor's throughout.

The gap a derivative needs

The derivative of a pole is exact and has no step size in it, and beside the formula sits a sentence nobody had measured: it divides by a quantity that vanishes when two poles meet. Driven together, the sensitivity climbs as the reciprocal of the gap — fitted exponent −1.0000, the product a constant 1.00000 times the natural frequency — while the largest change it still describes falls as the gap *squared*. A one per cent capacitor is outside first order once the poles are 3194 radians a second apart, which is an ordinary critically damped design.

transients · Poles The thermal network's step response, and the sum of its own stages — 0.95 per cent apart at worst. computed by solving, not by drawing. A step of power into a three-stage thermal ladder, and the junction's rise divided by it. The solid curve is exact: the impedance is a continued fraction in s, its denominator has 3 real negative roots, and the partial-fraction expansion of Z(s)/s is a sum of that many ordinary exponentials — no march, no step size. The dashed curve is the sum every account of a thermal path writes, each stage's own resistance times 1 − exp(−t/RC) with its own local time constant, and it is an approximation because the stages load each other. What that costs is 0.950 per cent, once, at 12.9 ms — between the fastest stage's 2.4 ms and the next one's 200 ms, which is the only place two stages are moving together. It is one-sided: the sum never reads low.

Two ladders the terminals cannot tell apart

A thermal path drawn as a ladder and the same path drawn as a sum of exponentials are called different models of one object, and the difference between them has never been priced because pricing it needs an exact answer. Solved in closed form, the sum is 0.950 per cent high at worst and never low; the marched netlist is right to a part in 21,169; and the largest disagreement in the picture was 2.919 per cent that has nothing to do with heat at all, which reading the curve one sample differently removes.

transients · Reverse recovery 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 A step through r sections starts as (t/τ)^r: it reaches 1% at 10.1 µs, 105 µs, 243 µs, 380 µs, 508 µs for r = 1 to 5. Solved, and expanded two ways. The step response of buffered RC sections of time constants τ, τ/2, … τ/r, with τ = 1 ms, on logarithmic axes. The relative degree of the recovered transfer function is r, so the first r − 1 derivatives of the step are zero at the start and the r-th is lim s^r·H(s) = r!/τ^r, read off the network solved far above its poles and off the expansion of H about infinity; the step therefore starts as (t/τ)^r, a straight line of slope r. The expansion about infinity and the residue expansion agree to a part in a million where both are well conditioned. The output reaches 1% at 10.1 µs (r = 1), 105 µs (r = 2), 243 µs (r = 3), 380 µs (r = 4), 508 µs (r = 5), and half its final value at 693 µs, 1.23 ms, 1.58 ms, 1.84 ms, 2.04 ms. For these time constants the whole step is (1 − e^(−t/τ))^r, checked against both expansions, so the time to a fraction ε is −τ·ln(1 − ε^(1/r)).

The start a step takes from infinity

The initial-value theorem reads where a step starts off H at infinite frequency. Apply it again to s·H, s²·H and on, and it reads how the step starts: the first r − 1 derivatives are zero for a network r degrees more poles than zeros, and the r-th is the ratio of the leading coefficients. So a step through r sections begins as a power of time — for sections of τ, τ/2, … τ/r, exactly (t/τ) to the r — and reaches one per cent at 10.1 µs through one section, 105 µs through two and 508 µs through five. Put a zero anywhere, even a thousand times above every pole, and the step starts linearly instead, with a slope of twice the zero's time constant over τ² that is the larger term for the first two of them.

transients · Value theorems 8 doublets over 2 decades settle 31.6 times slower than one. A tail of 1.0 per cent split into 8 doublets whose time constants are spread over 2 decades from 0.100 ms, each carrying 0.125 per cent. Settling to 0.10 per cent takes 7.278 ms against 0.2303 for the same total tail at one time constant — a factor of 31.6 — and it is between τₘₐₓ·ln((A/N)/B) = 2.231 and τₘₐₓ·ln(A/B) = 23.03 ms, which says the SLOWEST doublet decides it whatever its share. The spread is what costs and the number is not: four decades of spread multiply the settling by 2.04e+3 while sixteen times the count changes it by 0.386. And the trimming result, halving the tail for τ·ln 2, now buys 3.86 ms — the slowest time constant's ln 2 rather than the fastest's.

The slowest one decides it

One doublet's settling is its own time constant times the log of the tail over the band, so halving the mismatch buys ln 2 of it and no more. Split the same one-per-cent tail into eight doublets spread over four decades and the settling goes from 0.23 milliseconds to 470 — a factor of two thousand — while sixteen times as many doublets over the same spread changes it by less than two. It is the spread that costs and not the number, because the slowest doublet decides the settling however small its own share, and its share enters only through a logarithm.

transients · Doublet Four networks of one magnitude: the first and last are fixed at every instant, and the two between swap at 12.64 µs. computed by solving, not by drawing. A low-pass with poles at 1, 10 and 100 kHz and zeros at 3.00 kHz and 30.0 kHz, and the three networks that share its magnitude with one or both zeros moved into the right half-plane by an all-pass. The magnitudes agree to 6.7e-16 and each impulse response carries 9261.008 of energy. The fraction delivered by each instant: the all-left network is never behind any other and the all-right network never ahead of any other. The two between are not ordered. Half the energy arrives at 2.13, 2.80, 10.1, 11.2 µs in the order listed, so mirroring the higher zero is the later; nine tenths arrives at 50.2, 156, 60.9, 167 µs, so mirroring the lower zero is the later. The curves cross at 12.64 µs.

The two mirrors that change places

A low-pass with two zeros has four networks that share its magnitude, one for each choice of side for each zero. The one with both zeros on the left delivers its energy first at every instant, and the one with both on the right delivers last. The two in between are not ranked at all. With zeros at 3 and 30 kHz, mirroring the higher zero delays half the energy more and mirroring the lower zero delays nine tenths more. They swap at 12.64 µs, and that instant depends only on the product of the two zeros: it is where the all-pole response is rising at a rate of √(ω₁ω₂). And the network that delivers last does not start the wrong way at all.

frequency · Minimum-phase Driven through 1kΩ, the absorbing capacitor's tail is 2.6e-5 of the step, not 1%. computed by solving, not by drawing. The error at a capacitor's terminal after a step applied through 1k ohms, where the capacitor is 1 nF with eight absorption branches spread over two decades from 0.1 ms, together holding one per cent of its capacitance. The poles are found by bisection and the residues in closed form, and a trapezoidal march of the same netlist agrees. The tail the branches leave is 2.58e-5 of the step — the source resistance over the branch resistances in parallel — and settling to 1e-5 takes 173µs against 11.5µs with no branches; the model that weights each branch by its share of the capacitance says 48.4ms.

The tail the source decides

A capacitor with one per cent of dielectric absorption does not leave a one-per-cent tail in the circuit it is in. Driven through a kilohm, the eight slow branches that model the absorption leave 2.58×10⁻⁵ of the step at its terminals — the source resistance over the branch resistances in parallel — and settling to 10⁻⁴ takes 9.49 µs against 9.21 with no absorption at all. The share of the capacitance is not the amplitude; it is a ceiling the circuit reaches only when the source's own time constant falls inside the spread of the branches, and there the cost is 2.2 times the settling at 10⁻⁴ and 16 times at 10⁻⁵.

transients · Doublet The dual bridged-T is the same function: at a ratio of 100 both floor at 0.0196 and both loops make a Q of 4.76. computed by solving, not by drawing. The transfer of an RC bridged-T — two 10 nF capacitors bridged by 100 kΩ with 1 kΩ from their junction to ground — and of its dual — two 10 kΩ resistors bridged by 1 nF with 100 nF from their junction to ground — over two decades about 1591.5 Hz. The two curves are one to 5.7e-8 of each other, which is the solver's own conditioning: the dual's capacitor ratio plays the resistor ratio's part exactly, the residue is 2/(2 + n) for both, and in a loop of gain 1000 both make a Q of 4.760. What differs is which components carry the ratio.

The dual that is the same function

Swap a bridged-T's resistors for capacitors and its capacitors for resistors and the transfer does not change: two 10 kΩ resistors bridged by 1 nF with 100 nF to ground are the RC network's 100 kΩ and 1 kΩ around two 10 nF capacitors, to 5.7×10⁻⁸, which is the solver's own conditioning. The residue is 2/(2 + n) for both and the loop's ceiling √n/2 for both. So the choice between them is not about the response; it is about which components carry the ratio. Near the ceiling the Q moves by half the ratio's error, and a ratio of two capacitors at ±5% spreads the Q by 10.0% where two resistors at ±1% spread it by 2.1% — and drifts 2.0% over a hundred kelvin where the resistors drift 0.25.

filters · Null depth 3 relaxations over two decades come back within 0.22% from readings good to 1e-5. computed by solving, not by drawing. A one-per-cent tail divided equally among 3 branches spread over two decades from 0.1 ms, read at 80 times with 1e-5 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 8.29e+0, and the worst branch comes back with an rms error of 0.224 per cent.

How many relaxations a settling curve holds

A settling curve read at eighty times, each good to ten parts in a million of the step, returns the amplitudes of three relaxations spread over two decades to 0.22 per cent and of eight to 46 — even when the fit is handed every time constant exactly. The condition number of the inversion grows by a factor of 7.7 for each extra relaxation packed into a decade, so each tenfold improvement in the reading buys a little more than one more relaxation a decade and no more. A settling measurement can count the dielectric's slowest relaxations; it cannot resolve a crowded spectrum, and how crowded is readable before any amplitude is.

transients · Doublet Returning the shunt resistor to k times the output divides the residue by up to 30 and never takes it through zero. computed by solving, not by drawing. The residue of an RC bridged-T with R₁/R₂ = 100 at its null frequency, with its shunt resistor returned to k times its own output rather than to ground, against 1 − k: solved (dots) and 2/(2 + n(1 − k)) (line). An inverting return raises the effective ratio to n(1 − k) — at k = −15 it is 1600 — so the residue falls to 0.00125. But the null's slope falls with it, and the loop's ceiling, the slope over twice the residue, is 4.994 against 4.902 earthed: it approaches √n/2 = 5.000 of the resistors' own ratio and never passes it. At no k below one does the residue reach zero, because the trim is a ratio and the residue is positive for every ratio.

The residue that falls with its slope

A bridged-T's residue at its null is 2/(2 + n), and a loop around it stops at a Q near √n/2. Returning its shunt resistor to −15 times the output instead of to ground divides the residue by sixteen, to 0.00125 — and the loop's ceiling stays at 5.00, because the slope through the null falls with the residue. It has to: at a transmission zero the output is zero, so a trim that acts through the output leaves the zeros where they were, and the ceiling of a loop with unlimited gain is the zeros' own Q. What the trim does change is the gain the ceiling needs — nine-tenths of it at A = 9/ρ, 449 earthed and 14,400 at k = −31. No return gain below one takes the residue through zero, and a non-inverting one lowers the ceiling.

filters · Null depth Weighted towards the fast end as τ^(−1), the tail falls as a power of time and each decade of band costs 9.9× the time. computed by solving, not by drawing. A one-per-cent tail spread densely over five decades of time constant from 0.1 ms to 10 s, with each branch's amplitude weighted as τ to the power −m, for m = −0.5, 0, 0.5 and 1. The drawn weighting is m = 1: it settles to 10⁻³, 10⁻⁴, 10⁻⁵ and 10⁻⁶ in 972µs, 9.71ms, 96.3ms, 891ms. With m above zero the fast end carries more and the error between the ends is a power of time; at or below zero the slow end decides.

The tail with no slowest relaxation

Spread a one-per-cent tail densely over five decades of time constant and weight each relaxation as τ to the power −m. With m = 1 the tail falls as one over time and settles to 10⁻⁴ in 9.71 ms and to 10⁻⁶ in 891 ms — each tenfold tighter band costing ten times the time, not the fixed τ·ln 10 one relaxation charges. With even weighting it takes 14.4 s to reach 10⁻⁴; weighted towards the slow end, 26.5 s. The exponent decides whether any relaxation is the slowest one that matters: above zero none is, the settling is a power of the band with slope −1/m, and the spread's far end is never reached.

transients · Doublet A 10:1 divider into 100 pF and 15 pF more behind 1.0 kΩ, trimmed for the calibrator, on a 1 ns edge. computed by solving, not by drawing from the residues of its four poles, on a time axis from 0.1 ns to 1 ms. The trimmer is set to 12.8 pF, where a 1 kHz square wave reads flat: the upper arm then matches the lower arm's total of 115 pF. On a 1 ns edge the output overshoots to 1.128 times where it settles, because for the first few nanoseconds the late 15 pF is not yet part of the divider, and falls back over that branch's own time constant, 15 ns, within 1% only by 35.5 ns. With a second branch in the upper arm, 9.0 kΩ and 1.7 pF — a scaled copy of the late branch — the same edge is flat to 7.0e-11 (dashed).

The time constant one trimmer cannot reach

A ten-to-one divider whose lower arm is 100 pF of instrument input and 15 pF more reached through a kilohm reads flat on a 1 kHz calibrator with its trimmer at 12.8 pF — the same setting as a lower arm of 115 pF lumped — and overshoots 12.8% on a 1 ns edge, settling within 1% only after 35.5 ns. No setting of the one trimmer is flat: the best judged on a fast edge is 6.1%, at 11.9 pF, and a trimmer set by eye so the fast edge lands flat leaves the calibrator's square wave 9.9% low with a 114 µs slope on its top. Flat at every time needs the upper arm to be a scaled copy of the lower one — a second branch of 9 kΩ and 1.7 pF — and then a 1 ns edge is flat to 10⁻¹⁰, with a window of −18 to +24% on that resistor and ±6% on its capacitor.

instruments · Compensation A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV².

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

transients · Switching energy A chebyshev low-pass of order 5: where its impulse response balances, and where its step crosses half. computed by solving, not by drawing, the step built from the network's poles and the impulse response its derivative. The impulse response (dark, scaled to its peak) balances at 0.7090 ms; the group delay at a thousandth of the corner is 0.7090 ms — the same number by a different route. The step (pale) crosses half its final value at 0.7238 ms, 2.1% after it.

The delay that is a centre of mass

A filter's group delay at zero frequency is not an approximation to when its step arrives; it is exactly where its impulse response balances. Across three families and orders one to seven the two agree to 5.7 parts in a million by routes that share only the netlist. The step's half-way point is a different thing: ln 2 of the delay for one pole, approaching it from below for a Bessel filter, and passing it — by up to 16.9 per cent — for Butterworth and Chebyshev filters whose impulse responses swing negative. Elmore's RC-tree delay is the same centre of mass, and a uniform ladder network crosses half at 0.757 of it.

filters · Group delay One series resonator of Q = 1, read five ways. computed by solving, not by drawing; every reading from the solved network, none from its formula. The half-power bandwidth, the phase slope at resonance and the energy ratio driven at resonance give 1.0000, 1.0000 and 1.0000 — Q exactly. The ring-down's logarithmic decrement gives 0.8660, and the largest capacitor-voltage magnification 1.1547.

Five readings of one Q

A resonator's Q can be read from its bandwidth, from the slope of its phase, from the energy it stores against the energy it loses, from how fast its ringing dies and from how high its response peaks. On one series resonator at Q = 1 the first three give 1.0000 and the last two 0.8660 and 1.1547. Bandwidth, phase slope and energy are Q exactly at every Q, because they are one equation read three ways; the ring-down reads Q√(1 − 1/4Q²) because a damped cycle is longer than a natural one, and the peak reads the reciprocal because the peak is not at resonance. Each is within 1% of Q only above 3.5.

frequency · Resonance A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it.

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

transients · Switching energy Three loops with exactly 45° of phase margin, and their step responses. computed by solving, not by drawing; three lags spaced by a ratio r, the amplifier's gain bisected in each for a 45.000° margin, the step from the closed loop's residues. Lags a hundredfold apart need a loop gain of 141 and overshoot 23.6%. Spaced by three they need 4.7 and overshoot 34.5%; coincident, 2.83 and 41.0%. The same margin, and overshoots 1.73 times apart.

One margin, and the gain under it

A phase margin of 45° is quoted as an overshoot of 23 per cent, and on a loop with plenty of gain below its crossover it is: three lags a hundredfold apart, closed at 45°, overshoot 23.6%. Bring the same three lags together and the loop reaches 45° at a loop gain of 2.83 instead of 141, and overshoots 41.0% — the same margin, 1.73 times the ringing. The closed-loop peak follows the overshoot where the margin does not: predicted from the peak the worst error over 27 loops is 3.1 points, from the margin 22.2. The gain margin moves with it, from 37.2 dB to 9.0.

feedback · Margin overshoot A zero at 1ωₙ moves the fastest-settling damping from 0.782 to 0.873 and the fastest settling from 3.62 to 2.38/ωₙ. computed by solving, not by drawing. The 2% settling time of a second-order step against its damping ratio, with no zero (grey), a zero at 2ωₙ (dashed) and at 1ωₙ (solid), each located by residues on the recovered polynomial and refined by bisection; the step by residues equals the plain step plus (1/b) times its derivative to 2e-16. zero none: fastest at ζ = 0.7825, 3.623/ωₙ; zero 2ωₙ: fastest at ζ = 0.7975, 3.028/ωₙ; zero 1ωₙ: fastest at ζ = 0.8725, 2.376/ωₙ. The zero adds the step's own derivative, which adds overshoot, so more damping is needed to keep the last excursion inside the band — and it speeds the rise, so the best settling is faster than with no zero at all. The cliffs in the curve are still there, moved.

The damping a zero asks for

The fastest-settling damping of 0.78 belongs to a step with no zero in it. Put a zero at the natural frequency and the step becomes the plain one plus its own derivative: the best damping moves to 0.873 and the best settling falls from 3.62/ωₙ to 2.38/ωₙ. A zero five times further out moves nothing, and a zero at three times moves the optimum by only 0.005, yet keeping 0.78 there costs 52 per cent, because the optimum sits on a cliff. A zero in the right half-plane moves the best damping by 0.015 and costs 20 per cent in settling that no damping can buy back.

transients · Damping A ten-to-one divider with 15 pF of its arm behind a kilohm, stepped through 1 kΩ: one trimmer overshoots 0.00936% where an ideal source shows 13.2%, and the second branch that was exact no longer settles first. computed by solving, not by drawing: the step of the divider of the earlier essay, 9 MΩ and a trimmer over 1 MΩ with 100 pF direct and 15 pF behind a kilohm, through a source resistance of 1 kΩ, by residues. One trimmer at its calibrator setting (solid) overshoots 0.00936%, and the second branch that made the arm a scaled copy (dashed) rises without overshoot on the source's own lag. An ideal source (dotted) shows a 13.2% hook with one trimmer and nothing with two. A source resistance charges the input capacitance, 11.5 pF, so its lag and the hook act on one edge, and when they are comparable the hook's excess goes into shortening the lag.

The source the hook hides behind

A compensated divider with a late branch in its lower arm overshoots 13.2 per cent on an ideal step, and every measurement of that hook has been made from a source with no resistance. Through a real source the input capacitance charges first, and the hook and the source's lag act on the same edge. The overshoot is 10.5 per cent from 50 Ω, 4.13 from 300 Ω and gone by a kilohm, where the source's lag equals the hook's. The excess has not gone away: it is in the edge, which the screen shows 15 per cent faster than the source and the probe's capacitance allow. And the second branch that made the divider exact only makes its ratio exact, so from a kilohm the exact divider settles in 65.6 ns and the one with the hook in 44.

instruments · Compensation The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections.

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

instruments · Compensation

Named alongside it

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

Settling timeModel rangeStep responseDamping ratioPolesOvershootTransient responseClosed formTime constant matchingCompensated dividerDielectric absorptionQuality factor

All concepts