Concept

Seeded generator — where it appears

A pseudo-random sequence produced from a stated seed, so that a figure whose content is a sample is the same figure every time it is built. It is what allows a claim about a sample to be asserted at all, since a figure whose content changes between builds cannot carry an assertion.

Named by 8 essays across 5 fields — each of them below, with the objects they name alongside it.

The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz.

The floor a resistor sets

A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.

noise · Johnson noise
A divider of 4 equal 1.0% resistors, solved 3000 times. computed by solving, not by drawing. Every resistor drawn from its tolerance band and the divider solved, 3000 times. The worst case is ±1.000% — the part tolerance itself, and it does not improve when the divider is built from more parts — while the measured spread is 0.2944% and the worst of 3000 draws reached 82% of the bound. The root-sum-square, offered as though it were a standard deviation, is 1.70 of one here: for uniformly distributed parts it is √3 σ, a coverage of about 92%.

The tolerance that is not on any part

Four one per cent resistors in a divider give an answer whose worst case is one per cent, whose measured spread is 0.29 per cent, and whose root-sum-square bound — offered everywhere as though it were a standard deviation — is 1.70 of one. Adding parts does not move the worst case at all and shrinks the spread as one over their root, so the gap between the promise and the fact widens with every resistor. And the same arithmetic draws a boundary in tolerance rather than in frequency: an R–2R ladder is a twelve-bit converter only while its resistors are inside 0.14 per cent.

networks · Component tolerance
A threshold crossed once, in 20k samples of noise. With no hysteresis the comparator changes its mind 22.7 times on average and as many as 29, on a signal that crosses the threshold once. The vertical bars are the range over twelve seeds. The expected number of extra transitions falls below a tenth at 3.57 standard deviations — so the hysteresis a threshold needs is set by the noise under it and not by the signal over it.

Two thresholds because there is a floor

A comparator with one threshold, watching a slow signal cross it once, changes its mind 22.7 times on average and as many as 29 — because there is noise under the signal and no threshold is ever crossed once. Hysteresis fixes it, and how much is needed is a multiple of the noise's own standard deviation rather than a voltage: 3.57 of them here. The multiple grows with how long the threshold is watched, and slowly — a hundredfold longer record needs three times the hysteresis, not a hundred times.

applied · Hysteresis
The period's spread is 1.20 times what counting two crossings gives. computed by solving, not by drawing. 19999 periods of a relaxation oscillator with 5 mV rms of noise on its thresholds, computed from the exact flip instants rather than marched, at β = 0.5. The measured standard deviation is 3.4 ns and the closed form — three partial derivatives of the period with respect to the three draws it depends on — gives 3.4 ns. The estimate that counts two threshold crossings and divides the noise by the slope at each gives 2.83 ns, which is 17 per cent low. The curve is the closed form's Gaussian, drawn on the measured histogram rather than fitted to it.

The decision taken where the ramp is slowest

A relaxation oscillator decides at its thresholds, and a threshold is the one place on a charging exponential where the slope is smallest. Noise there costs 3.399 units of period against the 2.828 that counting two crossings gives, because a draw moves the crossing it is armed for and the level the next ramp starts from. Consecutive periods share that draw, so they are positively correlated and the jitter accumulates at 3.771 per root period rather than at 3.399. And at a fixed frequency there is a best hysteresis: β = 0.648, where β·ln((1+β)/(1−β)) = 1.

applied · Hysteresis
A whole number of steps: 1, 2, 3 and 4 agree to 7% and a step and a half is 1.9× worse. computed by solving, not by drawing. The upper panel is the share of the quantisation error sitting in harmonics of the input against the amount of dither added — the upper curve the largest share anywhere on the amplitude sweep, with the spread across the five tones each point averages drawn as a bar, and the lower curve that sweep's mean. Undithered the worst is 76.5 per cent. It falls steeply up to one whole step (4.70 per cent at three quarters, 0.34 at one) and then stops improving — but only AT whole steps. The sweep means at 1, 2, 3, 4 steps are 0.283, 0.297, 0.297, 0.301 per cent, flat to 7 per cent; at 1.5, 2.5, 3.5 they are 0.526, 0.345, 0.311, each above both whole steps beside it. The lower panel is what each costs in signal-to-noise ratio, with 10·log₁₀(1 + L²) drawn through it — the measurement is that curve to 0.118 dB everywhere, so the price is known in advance and only the benefit has to be measured. The choice is a corner and a comb: nothing here is minimised, something stops improving, and between the places where it has stopped it is worse again.

The dither that is a decision

One whole least significant bit is quoted everywhere as the dither, which makes a decision look like a constant. Swept, the axis is a corner and a comb. An eighth of a step leaves 64.5 per cent of the error locked to the signal and one whole step leaves 0.34; above that the sweep mean is 0.283, 0.297, 0.297 and 0.301 per cent at one, two, three and four steps and 0.526 at a step and a half, which fails at exactly the small amplitudes dither exists for. The price is 10·log₁₀(1 + L²) to 0.118 of a decibel, and four steps cost 12.41 for nothing.

digital · Quantisation
Three averagers passing the same noise, and three different half-power points. Integrated by eight-point quadrature on every lobe, with the tail past the last lobe in closed form. A mean over 20 ms has the response sin(πfT)/(πfT), and the area under its square is 25.000000 Hz against the 25 Hz of 1/(2T) — the brick wall drawn shaded. A mean over the window passes half its power at 22.147 Hz, so the noise bandwidth is 1.12880 times that frequency; A one-pole averager passes half its power at 15.915 Hz, so the noise bandwidth is 1.57080 times that frequency; Two means in cascade pass half their power at 23.919 Hz, so the noise bandwidth is 1.04521 times that frequency. Every curve drawn encloses the same area; they differ in where they spend it.

The filter an average is

A mean taken over a window is a filter, and the area under its squared response is exactly one over twice the window — 25 hertz of noise bandwidth for twenty milliseconds, passing half its power at 22.15. Built to the same noise, a one-pole averager passes half its power at 15.92 hertz and takes 2.33 times as long to settle to one per cent, and two means in cascade pass half at 23.92 and take 1.25 times as long. Between its nulls a mean rejects the mains no better than the one-pole does, and one per cent off a null it rejects it by forty decibels however many cycles the window holds.

noise · Noise bandwidth
A true-RMS converter reads noise low by 1/(16Bτ): 0.600% at Bτ = 10 and 672 ppm at 100, where a sine at fτ = 100 is 0.0396 ppm. Seeded, and measured. An explicit true-RMS converter — square, one-pole average of time constant τ, root — reading Gaussian noise of unit power, against the product of the noise's bandwidth and τ, from 1 to 100. Each point is 2²¹ samples at eight times the bandwidth, with the reading compared against the record's own root-mean-square and its standard error from batch means. Bτ = 1: 4.550% ± 150 ppm (band from zero), 4.081% (band of the same width about 3B). Bτ = 2: 2.598% ± 114 ppm (band from zero), 2.411% (band of the same width about 3B). Bτ = 5: 1.149% ± 75.4 ppm (band from zero), 1.098% (band of the same width about 3B). Bτ = 10: 0.600% ± 53.1 ppm (band from zero), 0.576% (band of the same width about 3B). Bτ = 20: 0.309% ± 39.1 ppm (band from zero), 0.295% (band of the same width about 3B). Bτ = 50: 0.127% ± 26.3 ppm (band from zero), 0.119% (band of the same width about 3B). Bτ = 100: 672 ppm ± 19 ppm (band from zero), 623 ppm (band of the same width about 3B). The dashed line is 1/(16Bτ), an eighth of the averaged square's variance, which the readings approach above Bτ = 10 and fall short of below it. A sine read by the same converter at the same product of frequency and τ is low by 3.96 ppm at 10 and 0.0396 ppm at 100 — the square of the noise's rate rather than its first power.

The noise a true-RMS meter reads low

A true-RMS converter reads a sine low by an amount that falls as the square of its frequency, and for anything but a slow sine that amount vanishes: 3.96 parts per million at ten times the averager's corner. Noise is not a sine. Its square fluctuates at every frequency down to zero, and the averager passes a share of that set by its own bandwidth against the noise's, so the reading is low by 1/(16Bτ) — the first power, not the second. Measured on seeded noise at Bτ = 10 it is 0.600 per cent low against a predicted 0.625; at 100, 672 parts per million. One reading scatters by eighteen times that, so no single reading shows the bias and the mean of a few hundred is nothing but bias.

power · RMS and average
At β = 0.5, the duty error scatters by 2.49 ns a period against the period's 3.39 ns — and consecutive duty errors are anticorrelated, −0.217. Seeded: forty thousand periods of the event map with 5 mV of threshold noise, β = 0.5. The period scatters by 3.39 ns against σ√(A² + (A+B)² + B²) = 3.4 ns; the high half less the low half scatters by 2.49 ns against σ√(A² + (B−A)² + B²) = 2.49 ns, with A = RC/V(1+β) and B = RC/V(1−β) — 0.667 and 2.000 in units of RC/V. The draw both halves share enters the period with A + B and the difference with B − A. Consecutive periods correlate by 0.107 (closed form 0.115); consecutive duty errors by −0.217 (closed form −0.214).

The walk the core sees

Threshold noise in a relaxation oscillator walks its timing at 3.77 nanoseconds per root period at β = 0.5, because the draw two half cycles share adds. A transformer driven by the same square wave sees the difference of the halves instead, where the shared draw subtracts, and that walks at 1.89 — exactly β times the timing, 18.3 ns against 35.9 after a hundred periods over six hundred seeded runs. The per-period duty error does not vanish with the hysteresis and the walk does, consecutive duty errors are anticorrelated where consecutive periods are not, a comparator skew outruns the walk after 2(σB/d)² periods, and at a fixed frequency the core wants less hysteresis than the clock does.

applied · Hysteresis

Named alongside it

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

Design tradeoffMonte carloComparatorHysteresisModel rangeAveragingEquivalent noise bandwidthNoise floorPeriod jitterRelaxation oscillatorSettling timeSpectral density

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