Generator

Two poles at ζ = 0.3, recovered from the matrix

The poles are at -477.5 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows.
Two poles at ζ = 0.3, recovered from the matrixThe poles are at -477.5 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows.the step this produces00.50011.5001234σζ = 0.3000ω₀ = 1592 Hzsolved, then checked — poles by rooting the determinantnatural frequency recovered to 6 digits

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 natural frequency recovered to 6 digits — 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 damping ratio 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 16 essays

which is the blast radius of changing it

One step, computed twice

A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.

Before the steady state

The gain that is exactly one

An oscillator is designed by making the loop gain one at the frequency where the phase is zero. The gain at which this circuit's poles reach the imaginary axis is 3.000000000000, bisected on the netlist — an equality, not a range. A gain three per cent high multiplies the envelope by 1.0987 every cycle and reaches the rails in 61 milliseconds; three per cent low divides it by the same factor. A one per cent resistor cannot hold the condition, and neither can any other component.

Circuits that do a job, and the range they do it over

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

Before the steady state

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.

Before the steady state

The frequency a device sets for itself

A common-emitter stage's bandwidth is decided by two picofarads between its collector and its base. The Miller approximation says how — lump it at the input, multiplied by one plus the gain — and predicts 643 kHz where the solved network gives 504 kHz. Twenty-two per cent optimistic, and it has no room at all for the second pole or for the zero in the right half-plane that the network also has.

Devices, and the amplitude they stop being linear at

The half that never arrives

Charging a capacitor from a step loses exactly as much energy as it stores, and the resistance it is lost in does not appear in the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms, to nine figures. Drive the same network with a ramp instead and the loss falls as two time constants over the ramp, with no floor beneath it at all.

Before the steady state

The cancellation that leaves a tail

A pole and a zero placed on top of each other disappear from the response. Miss by one per cent and the magnitude changes by 0.078 decibels, which no measurement would report as a fault, while the time to settle to a thousandth goes from 6.9 time constants to 245 — thirty-six times longer. The settling time has a closed form containing neither the fast circuit nor the doublet's separation as such, and its consequence is blunt: settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is.

Before the steady state

The corner that moved

The bilinear transform has to fit an infinite frequency axis onto a circle, so something must be compressed, and what is compressed is everything near half the sample rate. A 4 kHz Butterworth mapped to a 48 kHz clock is 3.432 dB down at its own corner instead of 3.010, and at a quarter of the sample rate the axis is 15.2% out. Pre-warping puts one frequency back exactly and no other — it is a choice of where to be right, not a correction.

Where a signal becomes a number

The inductance that limits, and lifts

Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

Power, and the part that does no work

The derivative of a root

The rung below turns one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.

Networks, and how a solve is checked

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.

Before the steady state

The gap a derivative needs

The derivative of a pole is exact and has no step size in it, and beside the formula sits a sentence nobody had measured: it divides by a quantity that vanishes when two poles meet. Driven together, the sensitivity climbs as the reciprocal of the gap — fitted exponent −1.0000, the product a constant 1.00000 times the natural frequency — while the largest change it still describes falls as the gap *squared*. A one per cent capacitor is outside first order once the poles are 3194 radians a second apart, which is an ordinary critically damped design.

Before the steady state

Which section goes first

A cascade of four biquads can be assembled in twenty-four orders, all of which realise exactly the same transfer function. They do not cost the same: the round-off floor runs from 24 to 100 least significant bits and the largest value any section reaches runs over eighteen decibels — and the two go opposite ways, so eleven of the twenty-four are genuine choices and thirteen are beaten on both counts by another arrangement.

Where a signal becomes a number

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%.

Before the steady state

The resistance that is below zero

An emitter follower's output resistance is 5.5 Ω at direct current and −21.9 Ω at 257 MHz, and the sign is not the transistor's: with an ideal source at the base there is no negative band at all, and a hundred nanohenries of wire between the source and the base produces one from 110 to 301 MHz. A capacitance resonating inside that band is a resonator with loss of the wrong sign, so 4.7 to 100 pF on the emitter oscillates while 1 pF and 470 pF do not — a band of load capacitance with quiet ground on both sides of it.

Devices, and the amplitude they stop being linear at

Where the estimate stops being a bound

The sum of open-circuit time constants is never optimistic on a network with real poles, and the claim is about the network rather than about the theorem. On a second-order section the ratio of the estimate to the truth is Q/√(k + √(k²+1)) with k = 1 − 1/2Q², which is exactly 1/√2 at the Butterworth quality factor — its worst point, 29.29 per cent low — and exactly 1 at a quality factor of √2. Above that the estimate is high, by 6.45 times at a Q of ten, and the crossing bisected on the solved response is 1.414213 against 1.414214.

Before the steady state

Every generator