Concept

Damping ratio — where it appears

The cosine of a pole pair's angle from the negative real axis, which decides overshoot and ringing while the distance from the origin decides speed. The settling time is not a smooth function of it: there is a cliff just before the fastest setting, and a margin measured at crossover cannot see it.

Named by 27 essays across 8 fields — each of them below, with the objects they name alongside it.

One loop, two measurements of the same margin. The loop gain crosses unity at 5.73 kHz with 34.9° of phase left. The closed-loop step overshoots by 35.2%, which the second-order relation says corresponds to 34.8°. They differ by 0.1°, and the difference is the third pole.

Two measurements of one margin

A phase margin is computed from the loop gain in the frequency domain, without ever looking at a step. An overshoot is measured from the closed-loop step response in the time domain, without ever looking at a Bode plot. Inverting the standard relation on the second returns 34.9° against the first's 34.9°, and the residue is the third pole.

feedback · Margin overshoot
Two poles at ζ = 0.3, recovered from the matrix. The poles are at -477.5 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows.

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

transients · Poles
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 pair at Q = 0.7071: the sketch is -3.01 dB out at the corner. computed by solving, not by drawing. The solved magnitude against the two straight lines that stand in for it. At a conjugate pair the error at the corner is 20 log Q = -3.010 dB, which is unbounded in both directions, and the worst error anywhere is 3.010 dB at 1.00× the corner. No damping brings it inside 0.770 dB: that is the minimax, at Q = 0.9152, where the corner error and an interior maximum are equal.

The straight lines, and where they are not the curve

Two straight lines through a corner is the most-used approximation in this subject and almost the only one with no number attached. It has one, and it is exact: at a single real pole the sketch is 3.0103 decibels high at the corner and nowhere worse, and its error is the same a factor above the corner as the same factor below — a symmetry the construction does not suggest. A pole pair has no such bound at all, and the best any two-slope sketch can do on one is 0.770 decibels, at a quality factor of 0.9152.

frequency · Asymptotic approximation
The inductance divides the current by 7 and leaves the capacitor 29% above the mains peak. computed by solving, not by drawing. The first conduction of a rectifier whose transformer has a leakage inductance as well as a winding resistance, marched from an empty capacitor at the worst instant of the mains. The peak falls from 36.4 A at 20 µH to 5.3 A at 5000, and — unlike the winding resistance measured on the first cycle alone, which limits the current and leaves ∫i²dt exactly where it was — the inductance takes the energy down with it, from 0.370 to 0.073 A²s. What it costs is the second curve: the inductor's current cannot stop at the instant the two voltages are equal, so the capacitor overshoots to 21.86 V at 1000 µH — 28.6% above the 17 V peak of its own supply — and the diodes will not let the charge back out. The overshoot has an interior maximum, because past it the mains reverses before the ring has finished.

The inductance that limits, and lifts

Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

power · Inrush
A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees.

The pair that is worse than either

A bulk capacitor and a ceramic are fitted together because each is good where the other is not, and between them is a frequency at which the pair presents six times the impedance either one does alone. The peak is a parallel resonance between one part's inductance and the other's capacitance, its height is one over the series resistance every data sheet asks to be minimised, and at it the two capacitors exchange 5.87 amps for every amp the load draws.

power · Decoupling
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 straight lines report 45.00° of margin, and the solved loop has 51.83°. computed by solving, not by drawing. The gain and phase of a loop made of an integrator and one pole, the corner at 1.00 kHz, with the integrator set so that the straight-line asymptotes cross unity at 1.00 kHz. The solved loop crosses at 786 Hz instead. Read at the lines' crossover, with the phase taken from the solve, the margin is 45.00°; at the loop's own crossover it is 51.83°. Closed, the loop is stable: the largest real part among its closed-loop poles is -5.00e-1 of the corner's angular frequency. The solved gain never rises above its asymptotes, so the lines cannot report more margin than the loop has.

The margin the straight lines report

A phase margin read off a sketch is read where the straight lines cross unity, and that is not where the loop does. For an integrator and one real pole the lines report 45.00 degrees on a loop that has 51.83 — short, and always short, because a real pole's response never rises above its asymptotes. A pair that peaks reverses the sign and removes the bound: at a quality factor of two the lines report 71.57 degrees on a loop that has no margin at all, and at five they report 82.41 on a loop that is unstable.

frequency · Asymptotic approximation
The fastest damping is a surface, and the band is worth 5 times the third pole. computed by solving, not by drawing. Each point is the last settling cliff, bisected — the damping at which the first overshoot's peak lands exactly on the band's edge, which is where the fastest settling is. Across the five bands the optimum moves by 0.231 of damping ratio; across a third pole from 1.5 times the natural frequency out to a second-order response it moves by 0.047. The two axes are worth 5.0 to one, and the expensive one is the specification rather than the parasitic. The classic 0.78 for fastest two per cent settling is the second-order curve's value at ±2%, 0.7797; at ±1% the same response wants 0.8261.

The best damping is not the one to build

The fastest settling damping is the right-hand limit at a discontinuity, so two thousandths below it costs 41 per cent and two thousandths above it costs 0.34 — a ratio of 120 in the penalty for the same error. With ±2 per cent on the damping ratio the nominal that minimises the worst case is 0.7927 rather than the optimum's 0.7734, and it guarantees 4.243/ωₙ against 5.943. The band moves the optimum by 0.231 of damping ratio and the third pole by 0.047, and 0.78 is exact at ±2% and 55 per cent slow at ±1%.

transients · Damping
The straight lines report 6.02 dB of gain margin on a loop that has none. computed by solving, not by drawing. The gain and phase of a loop made of an integrator and a pair at Q = 2, the corner at 1.00 kHz, the integrator set so that the straight-line asymptotes cross unity at 500 Hz. The loop's phase passes −180° at 1.00 kHz. There the lines put the loop gain at −6.02 dB and the solve at 0.00 dB, so the gain margin they report is 6.02 dB against 0.00 dB. The difference is 6.0206 dB, which is 20 log Q exactly with Q = 2, at any integrator gain. Closed, the loop is on the edge: the largest real part among its poles is -5.87e-17 of the corner's angular frequency.

The gain margin the straight lines get exactly wrong

A phase margin read off the straight lines is wrong by an amount that depends on where the loop crosses unity. A gain margin read off them is not: for an integrator and a pole pair the phase passes −180° at the pair's own frequency whatever the gain, and the lines are out there by exactly 20 log Q — 6.0206 dB for two real poles, 13.98 dB at a quality factor of five, at every integrator gain drawn. The stability condition for the loop turns out to be the same inequality: it is stable exactly when the gain margin the lines report exceeds that error.

frequency · Asymptotic approximation
The resistance that just stabilises it is 17 times smaller than the one that damps it. computed by solving, not by drawing. Two consequences of one base resistor, against how much of it there is, for a follower fed through 100 nH of wire with 47 pF on its emitter. The falling curve is the Q of the worst pole pair, rooted from the determinant so that no frequency grid is involved; the rising one is the output impedance the stage presents at low frequency. Bisecting on the SIGN of the pole's real part returns 69.7 Ω — at which the pair is stable with a Q of 1.7e+15, which is to say no damping and a peak whose height belongs to the arithmetic. A Q of one needs 1.15 kΩ, 16.6 times more, and that resistor takes the output impedance from 5.66 to 13.10 Ω — exactly R/(β+1) added, which is the bare follower's own expression with the base resistance in the place of the source resistance.

What the cure at the base costs

Bisecting for the smallest base resistor that stops an emitter follower oscillating gives 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.

semiconductors · Emitter follower
From a 10 Ω driver with a 1 ns edge, a CMOS input of 1.5 pF allows only 0.86 mm of track — less than the bare open end's 1.80 — and one of 100 pF allows 10.9 mm. computed by solving, not by drawing, bisected on the march by characteristics: the longest 50 Ω track a 1 ns edge from a 10 Ω driver can cross into an open end with a capacitance across it and overshoot by less than one per cent, against the capacitance. With none it is 1.80 mm. 0.3 pF: 1.28 mm, 0.5 pF: 1.04 mm, 1 pF: 0.88 mm, 1.5 pF: 0.86 mm, 2 pF: 0.88 mm, 3 pF: 0.96 mm, 5 pF: 1.15 mm, 10 pF: 1.68 mm, 20 pF: 2.71 mm, 50 pF: 5.70 mm, 100 pF: 10.94 mm. The limit falls to its least, 0.86 mm, near 1.5 pF, and from there rises nearly in proportion to the capacitance (slope 0.87 between 20 and 100 pF), because the driver's resistance damps the ring the track's inductance makes with a heavy load. The slider is the driver's resistance.

The heavier input that allows a longer track

Put a capacitance across a CMOS input and the longest track a 1 ns edge from a 10 Ω driver may cross inside one per cent of overshoot does not shrink steadily. It falls from the bare open end's 1.80 mm to 0.86 mm at 1.5 pF, and then it rises — 2.71 mm at 20 pF, 10.9 at 100 — nearly in proportion to the capacitance. A heavy load turns the track into the inductor of a series RLC whose damping, (Rₛ/2)·√(C/L), the driver's resistance supplies, and at every heavy load the one-per-cent length sits at the same damping, about 0.8. So the limit scales as Rₛ²·C: a slower edge buys a 20 pF input only the 0.4 power of its length, and a stronger driver — 2 Ω instead of 10 — cuts the track it may drive from 2.71 mm to 0.20. Into a heavy input, a weak driver is the clean one.

limits · Lumped element
With a 0.30 µF coupling capacitor and a stiff driver the transformer's damping is 0.289: the pulse's tail crosses zero 8 times in sixty units of L/Rₗ. computed by solving, not by drawing, marched and checked against the closed form to 2.3e-4: a pulse two L/Rₗ long through a series coupling capacitor into a 1 mH magnetising inductance with 100 Ω reflected across it, from a stiff driver. The capacitor is 0.30 µF, x = CRₗ²/L = 3. The poles are those of 3.000e-10·s² + 1.000e-5·s + 1, a damping of 0.2887 and a natural frequency of 9.19 kHz. The top ends at 0.128 of the high-frequency gain, and after the pulse the tail crosses zero 8 times, with lobes above a thousandth of the pulse, in the sixty units drawn. The slider is the capacitor.

The ring a blocking capacitor buys

A series capacitor in front of a transformer stops its flux from walking: a 55%-duty drive that walks a bare core by 0.40 units of volt-time a period settles within a few dozen periods once the capacitor takes the drive's average. The price is a second-order high-pass with the magnetising inductance, and an inductance lets its poles go complex where two resistive coupling stages never could. Between equal resistances the damping cannot fall below 1/√2, and the poles are complex only for a capacitor between 0.17 and 5.8 times L/R². From a stiff driver the damping is √(L/C)/2R and has no floor: a capacitor large enough to make the droop small — the droop is 1/RC, the capacitor's and not the transformer's — makes the tail ring, from one zero crossing at a tenth of L/R² to thirty-five at a hundred.

instruments · AC coupling
At 20 steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9; the trapezoidal rule shows every Q, 1.6% high. computed by solving, not by drawing: a series resistor–inductor–capacitor resonance at 1 kHz, its quality factor set by the resistor, marched at 20 steps per undamped cycle with each rule; the quality factor the march shows — its decay per radian of its own ringing — against the circuit's, from 1 to 500. Lines are each rule's factor per step in closed form; dots are marched and measured from the record at Q = 3, 30, 300, agreeing to 0.3%. Backward Euler's shown Q flattens at 3.272 and BDF2's at 77.9, whatever the circuit's; the trapezoidal rule's sits 1.64% above the diagonal at every Q. Q 3: trapezoidal 3.05, BDF2 3.06, backward Euler 1.62; Q 30: trapezoidal 30.5, BDF2 22.5, backward Euler 2.97; Q 300: trapezoidal 305, BDF2 62.6, backward Euler 3.24.

The Q a rule can show

A resonance marched forward in time decays at the circuit's rate plus whatever the marching rule adds, and the rule's share has a ceiling built into it. At twenty steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9, whatever resistor the circuit has; the ceilings rise as N/2π and N³/4π³. The trapezoidal rule has no ceiling at all — it shows every Q 1.64 per cent high at twenty steps a cycle, a ratio fixed by the step alone. So the steps a cycle needed to read a resonance's Q to one per cent are about 26 for the trapezoidal rule at any Q, 99 for BDF2 at Q = 100 and 62,202 for backward Euler. Marched at Q = 50, the circuit has 0.533 V of ringing left after ten cycles, the trapezoidal rule 0.536 and BDF2 0.346.

transients · Step response
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
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
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 gain of ten round an amplifier with a second pole: its half-power bandwidth and its noise bandwidth, each over what the gain–bandwidth product predicts. computed by solving, not by drawing: the stage as a netlist round a two-pole amplifier of 1 MHz, the second pole placed at r times GBW/G. The half-power bandwidth agrees with the second-order closed form √r·g(√r/2) to two parts in a thousand at every placement: 1.000 at r = 0.5, 1.272 at r = 1, 1.414 at r = 2, 1.231 at r = 5, 1.110 at r = 10, 1.010 at r = 100 times GBW/G. It is 1.414 times the product's prediction at r = 2 and within ten per cent only beyond r = 10. The noise bandwidth is (π/2)·GBW/G to half a per cent at every r: the second pole moves the one and leaves the other where it was.

The bandwidth the product still gets right

Gain times bandwidth is a constant for an amplifier with one pole, and real amplifiers have a second. Put it at r times the closed-loop bandwidth the product predicts and the half-power bandwidth becomes √r·g(√r/2) times GBW/G — 41.4 per cent wide at r = 2, within ten per cent only beyond r = 10 — so a follower round a 1 MHz part with its second pole at 2.5 MHz runs 40 per cent wider than the rule says. The noise bandwidth does not move at all. A second-order loop's noise bandwidth is set by one coefficient the second pole cannot touch, and it stays (π/2)·GBW/G at every r and every gain, to half a per cent on the netlist. A third pole is what breaks it.

feedback · Gain-bandwidth
A 10 µs lead-inductance resonance holds the start at t⁴ until √12·τr = 34.6 µs, at a damping of 0.1 or any other. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a line of slope two (dashed), and with a buffered series resistance and lead inductance into a stray capacitance of natural time constant τr = 10 µs and damping 0.1 (solid). The relative degree rises by two, so the start is the bare step times t²/(12τr²) — the faint line, checked on the expansion to three per cent a hundredth of τr in — until the two meet at √12·τr = 34.6 µs. The damping does not enter that coefficient: it is the inductance and the capacitance, through τr² = LC, and nothing else. After it the curve rejoins the bare one, displaced by a delay that the damping does set.

The delay an inductance does not add

Lead inductance against a stray capacitance is a resonance, and far above a network's bandwidth it raises the relative degree by two: a two-section step starts as t⁴ instead of t² until √12·τr — 34.6 µs for a 10 µs resonance — at any damping, because the coefficient is LC and nothing else. After that the resonance moves a half-way threshold by R·C, its first moment, and the inductance is not in it. It moves every threshold by R·C only at ζ = 1/√2, where its second moment is a pure delay's, and below that its variance is negative: at ζ = 0.1 the 1% point moves by 1.48 R·C while the 10–90 rise time gets shorter. The ring that reaches the output is the network's own gain at the resonance, 2.0 × 10⁻⁴ at ω0τ = 100, and the quality factor decides only how long it lasts. Lightly damped, the start is a staircase that touches the bare step once a period and never overtakes it.

transients · Value theorems
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
The bank's impedance peaks at 0.966 Ω at 7.12 MHz; a one-ampere load step moves the supply by 132.4 mV. computed by solving, not by drawing. The bulk-and-ceramic pair of the decoupling essays — 10 µF with 5 mΩ and 5 nH, 100 nF with 30 mΩ and 1 nH — standing on a regulator of 2 mΩ and 20 nH, and the voltage at the bank when the load's current steps by one ampere, from the impedance's own step response by residues (solid). The impedance this bank presents peaks at 0.966 Ω at 7.12 MHz (upper line), the parallel resonance of the bulk part's inductance with the ceramic. The step rings at that frequency and its largest swing is 132.4 mV, 31.3 ns after the step — about a quarter period — which is 14% of the peak and 59% of the resonance's characteristic impedance √(L/C) = 0.224 Ω (lower line).

The peak a load step does not see

A decoupling bank is specified by its impedance against frequency, and the number that fails a target is the anti-resonance peak — 0.966 Ω at 7.12 MHz for the bulk-and-ceramic pair standing on a regulator. A load that steps by one ampere moves that supply by 132 mV, 14% of what the peak suggests: a step rings the resonance once, and its first swing is set by the resonance's characteristic impedance √(L/C) = 0.224 Ω, not by its Q. The peak is the voltage per ampere of a load that sits at the resonance; switched at 7.12 MHz the same load builds over six cycles to ±611 mV, the peak times the current's fundamental. So damping, the cure for the peak, is a cure for that load only: sixteen times the ESR cuts the peak 8.9-fold and leaves a step's swing between 120 and 155 mV. What decides a step is its edge — 132 mV for anything faster than a few nanoseconds, 28 mV for a microsecond.

power · Decoupling
A 150 nH ground lead on a ten-to-one probe, at a 50 Ω source: the screen shows 49.4% of overshoot ringing at 121 MHz on an edge that has none. computed by solving, not by drawing, twice. A 1 V step from a 50 Ω source, watched through a ten-to-one probe with 11.5 pF at its tip, its ground returned through 150 nH of lead — about fifteen centimetres — and the scope reading the tip against the lead's end (solid, from residues; dots, a trapezoidal march, peak 1.4942). The same probe with its ground at the circuit's ground (dashed). The node itself steps cleanly. With the lead the tip capacitance and the lead inductance resonate at 1/(2π√(LC)) = 121 MHz, damped only by the source: ζ = (Rs/2)√(C/L) = 0.219, and the display overshoots by 49.4%.

The ring the ground lead adds

Every earlier measurement of a probe returned it to the circuit's ground directly. A real ten-to-one probe returns through fifteen centimetres of clip lead, about 150 nH, and its 11.5 pF tip and that lead are a series resonance at 121 MHz across the node being watched. The only damping it has is the source's own resistance, ζ = (Rs/2)·√(C/L), so the screen shows 49.4% of overshoot on a clean step from 50 Ω, 87.1% from 10 Ω, and none above 228 Ω — the stiffer the node, the worse the picture. Solved by residues and by a march of the netlist, the overshoot follows the second-order law at every source. A spring tip's 5 nH rings at 664 MHz and not at all from 50 Ω; a 150 Ω resistor in series with the tip brings a clip lead under 2% at the price of a 3.65 ns rise.

instruments · Probe loading
A Wien oscillator with a lamp for its lower gain arm, τ = 100 ms: the envelope rings 63.6 times in ten time constants at 63.6 Hz and settles at 1.000 V. computed by solving, not by drawing: three state equations marched through every cycle of the 1.59 kHz carrier. The peak at the network's node, over its settled value (solid), from a millivolt with a cold filament, against time in units of the filament's time constant τ = 100 ms. Cold, the 50 Ω filament asks for a gain of 11 and the output sits on the 12 V rail until the filament has warmed. It leaves the rail after 1.99 ms, falls through the operating point, overshoots and rings at 63.6 Hz with a damping ratio of 0.0225. The dashed bound is the linear loop's decay, exp(−(1 + m)t/2τ) with m = 0.8, and it contains neither the carrier's frequency nor the amplitude. The settled amplitude, 1.000 V at the node and 3.00 V at the output, is the one at which the filament is at 250 Ω.

The lamp that has to be slow

A small lamp in a Wien oscillator's gain network adjusts the gain to whatever each unit needs, so the margin a limiter spends is no longer spent, and the price is an amplitude loop of its own. The envelope is an integrator and the filament is a lag, which makes a second-order loop that rings at the square root of 2ω₀m/τ and decays at (1 + m)/2τ. The filament's ripple at twice the carrier is the only distortion, 0.201/ω₀τ. So the slowness that lowers the distortion lengthens the settling in exact proportion: distortion times settling is 0.163 cycles at any time constant, and 0.01 per cent takes a second at 1.59 kHz and 82 seconds at 20 Hz.

applied · Oscillator
A stage with its second pole at 3 times the bandwidth the product predicts settles within 0.1% in 5.06τ, against 6.91τ for the one-pole stage the product describes. computed by solving, not by drawing: the step of a stage whose amplifier has a second pole at r = 3 times GBW/G (solid), against the one-pole stage of the same product (dashed). Time is in units of τ = G/(2π·GBW). Settled within 0.1% at 5.055τ, against ln(1000) = 6.908τ, with the netlist of a gain-of-ten stage agreeing at 5.057τ. The stage with the second pole is usually faster, whatever it does on the way.

The gain that stops making a stage faster

The gain–bandwidth product says a stage of gain G settles in time proportional to G, and a stage with a second pole does something else. Below r = 4 the closed loop's poles are a complex pair whose real part is half the sum of the open loop's, π times the second pole's frequency, with neither the gain nor the product in it: a 1 MHz amplifier with a 0.5 MHz second pole takes 4.10 µs to settle within 0.1 per cent at unity gain, 4.04 at two and 4.22 at five, where the product says 1.10 to 5.50. Above the knee at G = 4·GBW/f₂ the product returns. In between lies the fastest stage, r = 3.32 for a 0.1 per cent band, which settles in 3.44 time constants against the one-pole stage's 6.91: half the time, and not at the r that gives the widest band.

feedback · Gain-bandwidth
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
Two loops with exactly 45° of margin and the same crossover: with 100% of the extra lag a delay, the step overshoots 28.7%, against 23.3% for two poles. computed by solving, not by drawing: the closed-loop step of a loop K·exp(−sD)/(s(1 + s/p)) with its crossover at one, its phase margin exactly 45°, and 100% of the 45° beyond the integrator supplied by the delay (solid), against the same margin from a lag alone (dashed). Delay 0.785, lag's pole absent. Overshoot 28.7% against 23.3%; settled within 2% at 7.03 against 7.06; gain margin 6.0 dB against no limit. Marched with a history buffer, and at δ = 1 checked against the exact series of the method of steps.

The margin a delay spends

A phase margin can be made of a lag or of a delay, and the number is the same either way. Forty-five degrees from a lag overshoots 23.3 per cent, and forty-five degrees from a delay overshoots 28.7; at sixty degrees the sign reverses, 8.77 against 5.57, and the two cross near 54 degrees. The overshoot is the smaller difference. A two-pole loop's gain margin has no limit at all, and five per cent of its extra lag as delay brings that to 25.5 dB; with all of it delay the gain margin is 90°/(90° − margin), which is 6.0 dB at 45°, and nothing else is in the expression. Across twenty-one such loops the margin alone misses the overshoot by 13.1 points. The closed-loop peak, which already reads the whole band, misses by 1.2.

feedback · Margin overshoot

Named alongside it

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

OvershootModel rangeDesign tradeoffResiduesSettling timePhase marginQuality factorLoop gainGain marginNatural frequencyPolesSecond-order approximation

All concepts