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

Value theorems — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · transients
  2. 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.

    part 2 · transients
  3. A line cut into 8 LC sections passes 10% at 0.863 and 50% at 1.043 of its time of flight. computed by solving, not by drawing, marched at 8000 steps. A lossless line of 1 µH and 100 pF — a 100 Ω line with a 10 ns time of flight — built from 8 equal LC sections and driven through 100 Ω into 100 Ω, so the ideal line would deliver a clean half-volt step at exactly one time of flight (dashed). The ladder network passes 10% of its final value at 0.8630, 50% at 1.0428 and 90% at 1.1592 of the time of flight: a delay quoted at 10% is 18.0% of a flight time shorter than one quoted at 50%. It starts as the 16th power of time, for 16 poles and no zeros, and rings after the edge because a lumped ladder network cannot pass what lies above its sections' cut-off.

    The threshold a delay is quoted at

    A step through a network of r poles starts as the r-th power of time, so a delay measured at 10% moves with the network's order more than one measured at 50%. Cut a line into more sections and the order grows without limit, and what happens to the two delays depends on the line. A lossless LC line converges on its time of flight at every threshold: the spread between its 10% and 50% delays falls from 0.656 of a flight time with one section to 0.070 with thirty-two, as the −0.68 power of the section count, the −2/3 of an Airy front. An RC line converges too, but its 10% and 50% delays converge to 0.134 and 0.391 of R·C, a third of each other for good. On a propagating line a delay belongs to the path; on a diffusing one, two-thirds of it belongs to the threshold.

    part 3 · transients
  4. A parasitic of 0.01 ms changes the start only inside itself: t³ until 30.0 µs with a pole, t until 20.0 µs with a zero. 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 straight line of slope two (dashed); with one more pole of 0.01 ms (solid), slope three until the asymptotes cross at (r + 1)τp = 30.0 µs; and with a zero of the same time constant (dotted), slope one until r·τz = 20.0 µs. Well inside the parasitic's time constant the pole multiplies the bare step by t/((r + 1)τp) and the zero by r·τz/t, both checked on the expansions; outside it both curves rejoin the bare one, displaced. A parasitic a hundred times above the network's bandwidth decides the start's power law for its own few time constants and no longer.

    The parasitic that moves the start

    A pole far above a network's bandwidth raises its relative degree by one, so the initial-value theorem says its step now starts as tʳ⁺¹ instead of tʳ; a zero lowers it. Both are true and both expire. Added to a two-section step, a 10 µs parasitic pole holds the start at the cube of time until the asymptotes cross at (r + 1)τp = 30 µs and a 10 µs zero holds it at the first power until r·τz = 20 µs; thirty time constants later the local power is back within 0.07 of the bare step's. After that each does one thing only: it moves every threshold by its own time constant, the pole later and the zero earlier. Since the time is the same at every threshold, the fraction it costs is not — 9.1% of a 1% delay, 0.82% of a 50% one — and a pole and a zero together keep the start's power, scale it by τz/τp, and move every later threshold by τp − τz.

    part 4 · transients
  5. 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.

    part 5 · transients

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