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

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.
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.-10120204060time (cycles of the 50.3 kHz natural frequency)output (volts), for a 1 V step inH(0) = 1.000000 — where the capacitor endsacross the inductor — H(∞) at the first instantdamping ratio0.0791H(0), solved1.000000000 Vmarched, at the end1.00000 VH(∞) across the inductor1.000000 Vexpansion, first instant1.000000 Varrives within ±2% at7.6 cycleslast 24 cycles inside ±2%100.0%their mean1.00000 Vmarch against expansion3.8e-4 Vworst pole, real part-7.91e-2 ω₀60 cycles runs out below6.50 Ωsolved, then checked — two limits against a marchinside ±2% after 8 cycles

Drawn above at its default parameters, which is almost never how an essay calls it. A placement states the numbers that essay is arguing about, so the figure a reader meets is about that argument rather than about the generator — 98% of the placements on this site pass one, and the phase that raised that number from 12% found eight captions describing a figure the page was not showing.

At those defaults the edge it states is inside ±2% after 8 cycles — the right-hand slot of the caption strip, which on this site is never used for anything else, and which is read back out of the drawing above rather than out of the code that wrote it. It belongs to Before the steady state, which is to say a change to it is a change to lib/figures/transients.js. It takes a slider on series resistance (ohms) — 0 removes it with 7 settings, and every one of them has passed the same assertions as the frame above — a figure whose circuit stops doing what its caption says at any setting stops the build.

Called by 2 essays

which is the blast radius of changing it

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.

Before the steady state

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.

Before the steady state

Every generator