The delay an inductance does not add

The parasitic that moves the start added one stray element at a time to a two-section RC cascade and found that each did two things of very different size. A pole far above the bandwidth raised the relative degree by one and so changed the power of time the step starts with — but only for a few of its own time constants. After that it did one thing only: it moved every threshold later by its time constant. A zero moved every threshold earlier by its own. The start was a curiosity with an expiry date; the shift was the part a receiver would feel.

That essay left the commonest parasitic of all for later. A driver does not reach a stray capacitance through a resistance alone. It reaches it through a bond wire, a pin and a length of track, and those are inductance. A resistance and an inductance in series into a capacitance are a pair of poles, not one, and a pair of poles can be complex. So the question this essay asks is what a resonance does where a pole did something so simple. It adds two to the relative degree rather than one; it rings; and it has a damping ratio, which a single pole does not. Which of those reach the thresholds a receiver quotes?

The cascade is the one the earlier essays used: two buffered RC sections with time constants τ = 1 ms and 0.5 ms, so that the bare step is exactly $(1 - e^{-t/\tau})^2$. After it sits a buffered series resistance, an inductance and a shunt capacitance, with a natural time constant $\tau_r = \sqrt{LC}$ of 10 µs — a hundred times faster than the network — and a damping ratio $\zeta = (R/2)\sqrt{C/L}$ that each figure states.

Two more powers of time, and no damping in them #

Far above its own natural frequency the resonance’s transfer function falls as $1/(\tau_r^2 s^2)$, whatever its damping. That is the whole of what the initial-value theorem needs. The bare cascade’s step starts as $t^2/\tau^2$; through the resonance it starts as that times $t^2/(12\tau_r^2)$, because each extra power of $1/s$ integrates once and the second integration brings the factor of twelve from $4!/2!$. The two asymptotes meet at $\sqrt{12}\,\tau_r$, which is 34.6 µs here.

What is worth noticing is what is missing from that coefficient. The resistance is not in it, so neither is the damping. Drag the damping from 0.05 to 2 and the left-hand end of the curve barely moves: its coefficient is fixed by the product $LC$ and by the bare cascade’s own coefficient, the damping enters only in the next term of the expansion, which grows as $\zeta t/\tau_r$, and the ratio holds to three per cent a hundredth of $\tau_r$ in at every setting. The theorem reads a network at infinite frequency, and at infinite frequency a series resistance is negligible beside an inductance’s reactance.

So the first part of the answer repeats the single pole’s lesson with the power doubled. A resonance a hundred times above the bandwidth decides how the step starts for a few of its own time constants. After $\sqrt{12}\,\tau_r$ the curve rejoins the bare one, displaced — and the displacement, unlike the start, depends on the damping.

The delay is R·C, and the inductance is not in it #

A threshold reached long after the resonance has finished acting sees it as a delay, and the delay is the resonance’s first moment: the centre of area of its impulse response. For any stable transfer function normalised to one at direct current, that moment is the first coefficient of the denominator. Here the denominator is $1 + RCs + LCs^2$, so the first moment is $RC$ — which is also $2\zeta\tau_r$.

Read in components, that sentence says something a designer might not expect. The inductance adds no delay at first order. A series resistance driving a stray capacitance delays a half-way crossing by $RC$, and putting lead inductance in series with that resistance leaves the half-way delay where it was. The delay that is a centre of mass showed that a filter’s group delay at zero frequency is exactly where its impulse response balances, and the balance point of this one has no $L$ in it.

The same thing comes out of the sum of time constants. A sum that is exact showed that the first denominator coefficient is the sum of each capacitor’s value times the resistance it sees with every other capacitor removed. An inductor enters that sum as its inductance divided by the resistance it sees with the capacitors opened, and with the stray capacitance opened this inductor drives a buffer’s input — an open circuit. The resistance it sees is infinite and its term is zero. The theorem and the moment are two routes to one statement.

The damping at which a resonance is only a delay #

The previous essay found that a single pole moves every threshold reached after it by exactly its time constant — at 1% and at 50% alike, to six per cent — because a single pole’s impulse response is all on one side and its higher moments are those of a smoothed delay. A resonance has no such guarantee, and the next figure measures it. Each curve is the delay the resonance adds at a threshold, divided by $RC$, at four dampings.

At 50% every curve sits within five per cent of one: half-way, the delay is $RC$ at any damping, which is the first moment doing its work. Away from 50% the curves separate. At ζ = 0.1 the 1% threshold moves by 1.48 $RC$, and at ζ = 1.5 by 0.90. Only one damping draws a flat line: at $\zeta = 1/\sqrt{2}$ every threshold the step reaches after ten $\tau_r$ moves by $RC$ to within a tenth of a per cent.

That damping has a name, and the reason it is the one is worth stating. Expand the resonance’s transfer function about zero frequency: $1/(1 + as + bs^2) = 1 - as + (a^2 - b)s^2 + \dots$ with $a = RC$ and $b = LC$. A pure delay of $a$ expands as $1 - as + a^2s^2/2 + \dots$. The two agree to second order when $b = a^2/2$, which is $\tau_r^2 = 2\zeta^2\tau_r^2$, which is $\zeta = 1/\sqrt{2}$.

Why it is the Butterworth damping #

That condition is also the one that makes the magnitude response maximally flat. For any real impulse response the squared magnitude at low frequency is $|H(j\omega)|^2 = 1 - \sigma^2\omega^2 + \dots$, where $\sigma^2$ is the impulse response’s second central moment, its variance. A response flat to second order has zero variance, and a response with zero variance passes a slow waveform as a pure delay to second order. The second-order Butterworth response is the first of those statements and the flat curve above is the second.

It is not the Bessel damping, and the difference is worth a sentence. The line from Bessel to Butterworth traced the family between the two, and a second-order Bessel response has $b = a^2/3$ — flatter group delay, which is a statement about the phase, and flat magnitude, unflat delay showed how far the two properties part company inside a passband. A threshold crossing on a slow edge reads the impulse response’s moments, and the second moment belongs to the magnitude. So the damping that makes a parasitic resonance cost exactly its $RC$ at every threshold is the one that keeps its magnitude flattest, which is 0.707 rather than the Bessel 0.866.

A variance below zero #

Below that damping $\sigma^2 = a^2 - 2b = \tau_r^2(4\zeta^2 - 2)$ is negative. No impulse response that stays above zero can have a negative variance; one that rings below zero can, because its negative lobes count against its spread. That one quantity predicts the whole pattern of the shift figure, through a two-term expansion of the output about the bare step: $y(t) \approx y_b(t - a) + \tfrac{1}{2}\sigma^2 y_b''(t)$. Solve for the crossing and the shift is $a - \tfrac{1}{2}\sigma^2\,y_b''/y_b'$ — the first moment, corrected by the variance times the bare step’s own curvature over its slope at that threshold.

The dots are bisected crossings and the lines are that prediction. At 50% the bare step is near its inflection, its bend is small, and the shift is the straight line $2\zeta\tau_r$: 0.094 $\tau_r$ at ζ = 0.05 against the first moment’s 0.1. At 1% the bare step is still curving upward hard — it reaches 1% at 105 µs, where its bend is about one over 125 µs — and a negative variance turns into extra delay. As the damping falls the first moment shrinks towards nothing and the variance term does not, so the shift at 1% stops falling and turns back up: 0.20 $\tau_r$ at ζ = 0.05, twice the first moment. The prediction holds against the bisection at 50% and at 1% for all fourteen dampings, to within three hundredths of $\tau_r$ plus three per cent.

At 0.1% the step is reached at 32.1 µs, three $\tau_r$ in, inside the start where it is still rising as $t^4$. The prediction is drawn and departs, which is what the earlier essays on the start would say it must: a two-moment expansion describes a step that has forgotten its start, and a threshold inside the start has not.

What the inductance does to a low threshold #

Written in components, the two-moment shift is $RC + (LC - \tfrac{1}{2}R^2C^2)\,y_b''/y_b'$. That form says where the inductance’s delay has gone. It has not vanished; it has moved from the first moment, where every threshold would feel it, into the second, where it is weighted by how sharply the bare step is bending at the threshold in question. Early in a step that bend is roughly one over the time since the edge, so an inductance adds about $LC/t$ at a threshold reached at $t$.

At the 1% threshold here that is $(10\ \mu\mathrm{s})^2/125\ \mu\mathrm{s}$, about 0.8 µs, which beside an $RC$ of 2 µs at ζ = 0.1 is the difference between 1.0 and about 1.4 $RC$ — and the bisection says 1.48. At 50% the bare step’s bend has turned negative and the same term removes a little delay instead: six thousandths of $\tau_r$. Lower still, the effect grows faster than the prediction. At ζ = 0.02 and a threshold of one part in ten thousand a separate bisection finds the crossing moved by 23 times $RC$ — nearly a whole $\tau_r$, where the first moment says a twenty-fifth of one.

So the rule of thumb a designer can carry is short. Lead inductance costs nothing at a half-way threshold and costs $LC$ divided by the elapsed time at an early one. The threshold a delay is quoted at already found that the low thresholds are where a step’s shape lives; this is the same finding with a resonance as the shape.

How much of the ring gets through #

A lightly damped resonance rings, and a reader looking at the shift figure might suspect the ring of causing the odd behaviour at low thresholds. It is worth measuring how large the ring in the output actually is. The step’s residue at the resonance’s own pole gives the amplitude directly: twice its magnitude is the size of the oscillating term.

The residue factors, and the factorisation is the result. At a resonance pole $p$ the residue of $H(s)/s$ is the bare cascade’s gain $H_b(p)$ times the resonance’s own residue, and the second has magnitude $1/(2\sqrt{1 - \zeta^2})$ because $|p| = \omega_0$ and the two poles are $2\omega_0\sqrt{1 - \zeta^2}$ apart. So the ring’s amplitude is $|H_b(p)|/\sqrt{1 - \zeta^2}$, and that agrees with the solved network’s residue to a part in a million at all twenty-one settings. Far above the bandwidth $|H_b(p)|$ is the cascade’s gain at the resonant frequency, $2/(\omega_0\tau)^2$: 2.0 × 10⁻⁴ at $\omega_0\tau = 100$.

The quality factor sets how long the ring lasts and almost nothing about how large it is. At ζ = 0.05 the amplitude is 2.01 × 10⁻⁴ and the ring lasts $\tau_r/\zeta$ = 200 µs; at ζ = 0.707 it is 2.89 × 10⁻⁴, larger by the $\sqrt{2}$ in the denominator, and gone in 14.1 µs. The network in front of the resonance has already filtered the edge that would excite it, and what is left to ring with is the network’s own gain at the ringing frequency. A resonance that rings hard on a fast edge rings at two parts in ten thousand behind a slow one.

That is also why ringing cannot explain a shift of 1.48 $RC$ at 1%. A term of 2 × 10⁻⁴ cannot move a crossing at 10⁻² by a measurable amount; the variance can.

A staircase that never goes back #

The ring is small against the final value but not against the start, where the bare step itself is only $t^2/\tau^2$. The last figure draws the first eight $\tau_r$ on linear axes, scaled by $(\tau_r/\tau)^2$ so that the bare step reads almost exactly $(t/\tau_r)^2$.

Lightly damped, the output follows $(t/\tau_r)^2 - 2(1 - \cos\omega_0 t)$: a parabola less a term that swings between zero and four. At ζ = 0.02 the measured curve is 3.94 behind at half a period and back within 0.34 at a whole one. It is a staircase whose treads flatten as the ramp’s slope outgrows the ring’s.

Two things about it are guaranteed rather than observed, and both are checked at every damping drawn. The step never turns back: the slope of that expression is $2(\omega_0 t - \sin\omega_0 t)$ in the same units, and $x - \sin x$ is never negative. And it never rises above the bare step, because $1 - \cos\omega_0 t$ is never negative either. A lossless resonance driven by a ramp can only lag it. It touches the ramp once a period and falls behind in between, which is why a delay measured at a low threshold under light damping depends on where in the period the threshold happens to fall — the twenty-three-fold shift at 10⁻⁴ is a threshold that landed on a tread.

The rise time a resonance shortens #

The instrument’s own rise time tested the rule that rise times add in quadrature and found it optimistic for real poles. Behind that rule is the addition of variances under convolution: the spreads of two impulse responses add, and a rise time is roughly proportional to the square root of a spread. The rule assumes every variance is positive.

A resonance below $1/\sqrt2$ offers a negative one, and the same bisections show the consequence. At ζ = 0.1 the 10% crossing moves by 1.052 $RC$ and the 90% crossing by 0.954 $RC$, so the 10–90 rise time is shorter than the bare cascade’s, by about 0.2 µs on 2.59 ms. The amount is negligible here and the sign is the point: a lightly damped parasitic steepens a slow edge’s middle while it delays the edge’s foot. A rule of thumb that adds parasitics’ rise times in quadrature has the sign of this one wrong.

What a designer should take #

Three statements, each sized by the figures. A half-way delay costs the parasitic’s $RC$, and lead inductance is not in it — budget the driver’s resistance against the stray capacitance and leave the inductance out. A low threshold costs $RC$ plus about $LC$ over the time the bare step takes to reach it, so a receiver that switches early on a slow edge sees lead inductance and one that switches at half-way does not. And the damping that makes the first statement true at every threshold is the Butterworth damping; a series resistance chosen to put a parasitic resonance at ζ = 0.707 makes its cost the same number wherever the threshold sits. Unlike the settling time in the best damping is not the one to build, this optimum has no cliff beside it: the shift at every threshold is a smooth function of the damping, so a resistor two per cent high moves the 1% delay by about two tenths of a per cent of $RC$ and no more.

What a designer should not take is the ring as the measure of harm. Behind a slow network the ring’s size is that network’s gain at the resonant frequency, fixed by $LC$ and the network, and adding resistance to quench it shortens it without making it smaller.

How the numbers were obtained #

Each network is assembled as a netlist — two resistor-capacitor sections, each behind an ideal buffer, then a buffered series resistance, inductance and shunt capacitance — and its transfer function is recovered from the modified nodal matrix. The step is evaluated from the expansion about infinite frequency inside three-tenths of the fastest pole’s time constant, and from the residues after. Thresholds are bisected in logarithmic time over a hundred iterations. The bare step is checked against its closed form $(1 - e^{-t/\tau})^2$ to a part in a million, and its curvature over slope for the two-moment prediction is taken from that closed form, so the prediction and the measurement share only the netlist’s element values. The ring amplitudes are residues of the solved network, checked against the cascade’s closed-form gain evaluated at the resonance’s pole.

What it leaves out #

The resonance is buffered, so it neither loads the cascade nor is loaded by it. A real stray capacitance hangs on the node it slows, and a real lead inductance carries the cascade’s own current; the loaded case is the one the parasitic that moves the start also left open. The resistance here is linear and frequency-independent, where at the frequencies of a fast resonance skin effect raises it and so raises the damping. And the step is ideal: a real driver’s edge has its own rise time, which filters the resonance further and moves every number here towards the heavily damped column.

The number to keep #

$RC + LC/t$. A parasitic resonance delays a threshold by its resistance-capacitance product, which is its first moment and has no inductance in it, plus a variance term that at a threshold reached a time $t$ into the step is about $LC/t$. At half-way the second term is negligible and a lead inductance costs nothing; at 1% of a 1 ms step it is 0.8 µs against an $RC$ of 2 µs. The two agree at every threshold only when $\zeta = 1/\sqrt{2}$.

Still open: the loaded resonance, the driven edge, and two resonances in a row #

The loaded resonance. A stray capacitance that loads the node it sits on moves the cascade’s own poles as well as adding two. The exact sum of time constants still holds for the first moment, but the variance now has cross terms between the resonance and the section it loads, and whether the Butterworth damping still flattens the shift is a question the same bisections can answer.

The driven edge. A driver with a finite rise time is a low-pass in front of the resonance. If the edge’s rise time is long against $\tau_r$ the ring vanishes and only the moments remain; somewhere between, the ring’s amplitude crosses from the network’s gain to the edge’s own, and where that happens sets when a fast driver needs a damping resistor.

Two resonances in series. A bond wire and a package pin are two resonances. Their first moments add, as every sum of time constants does, and their variances add too — so two resonances each damped below $1/\sqrt2$ make a variance more negative than either. Whether the rise time keeps shortening, and where the delay a chain adds meets the delay a chain of resonances subtracts from the rise, is the next measurement the expansion can make.