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

Transfer function — where it appears

The ratio of a network's output to its input as a function of complex frequency, which for a linear network is a rational function of s. Its poles are the network's natural frequencies and its zeros the frequencies at which the output vanishes, so the whole of a response follows from where they sit.

Named by 6 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

transients · Value theorems Which resistor the noise of a Chebyshev 5 actually comes from. computed by solving, not by drawing, one solve per resistor. Each bar is that resistor's share of the noise power at the output, found by splitting its node, putting a source of √(4kTR) in series with it and re-solving the whole network — so what is drawn is not how much noise each resistor makes but how much of it arrives. The largest contributor is F2R at 64.1 per cent, the smallest F0R at 9.0, and the shares add to 1.000000000000 because noise powers add. A resistor's share of the noise is not its share of the resistance: the largest departure between the two is 3.9 percentage points.

The resistor the noise comes from

Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.

noise · Johnson noise 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.

transients · Value theorems 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.

transients · Value theorems 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.

transients · Value theorems The dual bridged-T is the same function: at a ratio of 100 both floor at 0.0196 and both loops make a Q of 4.76. computed by solving, not by drawing. The transfer of an RC bridged-T — two 10 nF capacitors bridged by 100 kΩ with 1 kΩ from their junction to ground — and of its dual — two 10 kΩ resistors bridged by 1 nF with 100 nF from their junction to ground — over two decades about 1591.5 Hz. The two curves are one to 5.7e-8 of each other, which is the solver's own conditioning: the dual's capacitor ratio plays the resistor ratio's part exactly, the residue is 2/(2 + n) for both, and in a loop of gain 1000 both make a Q of 4.760. What differs is which components carry the ratio.

The dual that is the same function

Swap a bridged-T's resistors for capacitors and its capacitors for resistors and the transfer does not change: two 10 kΩ resistors bridged by 1 nF with 100 nF to ground are the RC network's 100 kΩ and 1 kΩ around two 10 nF capacitors, to 5.7×10⁻⁸, which is the solver's own conditioning. The residue is 2/(2 + n) for both and the loop's ceiling √n/2 for both. So the choice between them is not about the response; it is about which components carry the ratio. Near the ceiling the Q moves by half the ratio's error, and a ratio of two capacitors at ±5% spreads the Q by 10.0% where two resistors at ±1% spread it by 2.1% — and drifts 2.0% over a hundred kelvin where the resistors drift 0.25.

filters · Null depth

Named alongside it

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

Initial-value theoremPolesResiduesStep responseFinal value theoremPropagation delaySettling timeBridged-T networkComponent toleranceDesign tradeoffDualityEquivalent noise bandwidth

All concepts