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

Damping — the series

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

    part 1 · transients
  2. 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 rung below 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

    The rung below sweeps a second-order step's damping, finds the settling time falling by a third in one step of a five-thousandth sweep, and calls it 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.

    part 2 · transients
  3. 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%.

    part 3 · transients

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