Field

Where a signal becomes a number

A converter is an element in the netlist like any other: it has an anti-alias filter made of the same components as every filter here, a reference a resistor's Johnson noise sits on, and an aperture. What it also has is three boundaries of kinds the rest of the site does not carry — one with no gradient at all, one that stops being a floor and becomes distortion, and one whose variable is a duration.
A 9.0 kHz input sampled at 10 kHz arrives as 1.0 kHz. computed by solving, not by drawing. The dots are the samples. The input at 9.00 kHz is above half the 10 kHz rate, and every dot also lies on the 1.00 kHz curve drawn beside it — the two sample sequences differ by 9.3e-15, which is the arithmetic and not a small effect. Nothing is attenuated and nothing is distorted: the samples are the samples of a different signal, at full amplitude, and there is no measurement of them that could say which one was there.

The frequency a sample rate invents

Every other boundary in this field is a model getting gradually worse. This one has no gradient at all: below half the sample rate a set of samples has one sinusoid through it, above half the sample rate it has another, and the two sets of numbers are identical to three parts in ten thousand billion. Nothing is attenuated, nothing is distorted, and there is no measurement of the samples that could say which signal was there.

What the reconstruction returns, either side of 5.0 kHz. computed by solving, not by drawing. The Whittaker–Shannon sum is evaluated on the samples and compared with two things: the signal that was sampled, and the frequency the samples report. Below 5.00 kHz these are the same curve and the error is 4.88e-3 — the truncation of the sum at sixty-four samples either side, and nothing else. Above it they part: at 9.00 kHz the reconstruction is 1.59e-3 from the alias and 2.000 from the input. The small number is the interesting one. A reconstruction cannot be improved into the right answer, because it is already an exact answer to a different question.

An exact answer to a different question

The reconstruction that turns samples back into a signal is normally introduced as the thing that recovers what was there. Measured on both sides of half the sample rate it does something more interesting than failing: above the boundary it returns the alias to 1.6 parts in a thousand, which is the same accuracy it returns the input with below the boundary, and it is wrong about the input by twice the amplitude. Its error is not a degradation. It is exactness about something else.

The sample rate each anti-alias filter demands for 80 dB. computed by solving, not by drawing. A 20 kHz passband, and each filter must be 80 dB down by the frequency that folds back into it. The required rate follows, and it is a property of the filter rather than of the converter: Bessel 3.53× Nyquist, Butterworth 2.08×, Chebyshev 1.53×. The elliptic design at a selectivity of 0.8 is refused: its equiripple stopband has a floor at -74.1 dB, which is above the requirement at every frequency, so no sample rate satisfies it. At a selectivity of 0.5 the same order needs 1.34×. The floor is the selectivity's, not the order's.

What the filter in front costs

The filter that keeps a converter honest is normally chosen for its skirt. Measured against one requirement — eighty decibels down by the frequency that folds back into a 20 kHz band — the choice is not a decibel or two of skirt but a factor in the clock: Bessel demands 3.53 times Nyquist, Butterworth 2.08, Chebyshev 1.53. And one design is refused outright, because an elliptic stopband is a floor rather than a slope and no sample rate reaches past a floor.

Where a converter stops measuring the signal and starts measuring the resistor. computed by solving, not by drawing. The quantisation floor is q/√12 and halves with every bit; the Johnson floor of a 1 kΩ source in 100 kHz is 1.266 µV and does not move. They cross at 18.80 bits. Below that the converter is the limit; above it the resistor is, and a further bit buys a more precise measurement of thermal noise. A resolution quoted without a source impedance and a bandwidth is not a resolution — which is the same sentence the instruments field makes about a probe.

The floor a converter sets

A converter's resolution is quoted as a number of bits, which is a property of the converter. What it can actually resolve is a property of the circuit in front of it, and the two cross: measured against the Johnson noise of a 1 kΩ source in 100 kHz of bandwidth, the quantiser is the limit up to 18.80 bits and the resistor is the limit above it. Past that crossing every further bit buys a more precise measurement of thermal noise.

6.02 bits + 1.76 dB, measured — and the overload that is blamed on it. computed by solving, not by drawing. A coherent sinusoid is quantised and the ratio is read off the error sequence, never from the formula. One step inside full scale the two agree to 0.15 dB at every bit count from 8 to 16. Driven to exactly full scale the same converter measures 0.83 dB worse at 8 bits when driven to exactly full scale, and the reason is in the count beside it: 327 of 8192 samples hit the top code. A mid-tread converter's largest code is one step below full scale, so a sinusoid that reaches full scale is already overloading — half a step of overload, at the peak, where the error correlates with the signal. The loss halves with every bit for the same reason the clipped count does — 0.83, 0.47, 0.21, 0.06, -0.13 dB — and by sixteen bits it is inside the measurement's own scatter. The formula was never the approximation.

Six decibels a bit, and the half step blamed on it

Every converter data sheet quotes 6.02N + 1.76 dB and every bench measurement comes up short of it, which is usually explained by calling the formula an approximation. Measured one step inside full scale it agrees to 0.15 dB at every bit count from eight to sixteen. Driven to exactly full scale the same converter loses 0.83 dB at eight bits — because 327 of 8192 samples hit the top code, and a mid-tread converter's largest code is one step below full scale.

The amplitude at which a converter's floor becomes distortion. computed by solving, not by drawing. The quantisation error is transformed and the power in harmonics of the input is measured directly, at the bins those harmonics occupy. Undithered, the share rises from 0.29% at 230 levels to 76.5% at 1.3: the error has stopped being spread and has become a deterministic staircase locked to the signal. The sweep is drawn against the levels the input uses rather than against volts because it is then the same sweep for every converter — a twelve-bit part and a sixteen-bit part agree here to 0.0e+0, so what decides the character of the error is how many levels the signal crosses and not how many the converter owns. With 1 least significant bit of dither the share stays under 0.34% at every amplitude, for 3.03 dB of signal to noise. Adding noise to a converter's input improves what comes out of it, which is true and sounds like it should not be — and the amount is one whole step, not "some": a quarter of a step leaves 52.7% and a half leaves 25.6%, because a dither smaller than a step cannot make the quantiser cross one.

When a floor stops being a floor

Quantisation error is treated as noise and behaves like noise while the input crosses many levels. As the amplitude falls it stops: measured at the bins its harmonics occupy, the share of the error's power sitting in harmonics of the input rises from 0.29% at 230 levels to 76% at 1.3. The floor has not moved. It has become a distortion product locked to the signal, and the fix is to add noise on purpose — one whole least significant bit, and no more.

One bit, oversampled — and where the quantisation noise went. computed by solving, not by drawing. A first-order modulator is marched forward one sample at a time with a one-bit quantiser inside the loop, and the noise inside the band is read out of the transform of the error. It falls by 8.99 dB for every doubling of the oversampling ratio — measured 9.33, 10.12, 6.96, 9.54 — against 3.01 dB for plain oversampling, which is drawn beside it from the same starting point. The loop does not make less noise; it moves the noise out of the band, and at a ratio of 128 one bit is worth 9.18.

One bit, and where the noise went

Sampling faster spreads a fixed quantity of quantisation noise over a wider band, so the part inside the band of interest falls by 3.01 dB for every doubling — half a bit. Putting the quantiser inside a loop with an integrator does something different in kind: measured on a modulator marched forward one sample at a time, with its test tone inside the band the ratio is quoted over, the in-band noise falls by 8.99 dB per doubling. At an oversampling ratio of 128, one bit is worth 9.18 — and the octaves scatter by a decibel each, which is the loop telling the truth about what its error is made of.

The droop a zero-order hold imposes at 48 kHz. computed by solving, not by drawing. Holding each sample for a clock period convolves the output with a rectangle, so the spectrum is multiplied by a sinc: -0.143 dB down at a tenth of the sample rate, -0.912 at a quarter and -3.922 at half — which is exactly 20 log(2/π) and contains no design decision at all. The dots are the amplitude of the fundamental read out of the transform of the staircase itself, agreeing with the closed form to 0.008%. There is also half a sample of delay, 10.417 µs here, which is the reason a held reconstruction is not a droopy copy of the signal but a droopy copy that has moved.

The staircase on the way out

A converter does not emit impulses. It holds each sample for a whole clock period, which is a convolution with a rectangle and therefore a multiplication by a sinc — 0.14 dB down at a tenth of the sample rate, 0.91 at a quarter, and 3.92 at half, which is exactly 20 log(2/π). Nobody chose that droop and it is nearly eight times the half-decibel ripple of a Chebyshev passband. Measured on the transform of the staircase itself, it agrees with the closed form to 0.008%.

A 4 kHz Butterworth, mapped to 48 kHz two ways. computed by solving, not by drawing. The upper panel is the digital response on the unit circle; the lower one is where each analogue frequency lands. The bilinear transform compresses the axis as it approaches half the sample rate — -3.11% at a tenth of the rate and -15.2% at a quarter — so a design mapped straight through is -3.432 dB at its own corner instead of -3.010. Pre-warping puts that one frequency back exactly and no other: above it the pre-warped curve is the further of the two from the analogue prototype. It is a choice of where to be right, not a correction.

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.

A 8th-order Butterworth at 12 bits, as a cascade and as one polynomial. computed by solving, not by drawing. The open circles are the poles as designed, on the z-plane with the unit circle drawn. Filled marks are where they go once the coefficients are stored in 12 bits. A cascade of biquads keeps two coefficients per pole pair, so a rounding error moves that pair and nothing else: 8.43e-3, largest radius 0.97478. A direct form keeps one denominator whose coefficients are symmetric functions of every pole, so one rounding error moves all of them: 6.21e-1, largest radius 1.54393 — outside the unit circle, which is not an inaccurate filter but an unstable one. The two are the same filter until they are written down.

The same filter, rounded twice

The filters field measured two realisations of one analogue response and found the passband error growing as the 0.99 power of a component tolerance in a cascade and the 2.00 power in a ladder. The digital version of that argument comes out harder. At eighth order and sixteen bits, a cascade of biquads moves its poles by 6.6 × 10⁻⁴ and stays at a radius of 0.9748; the same filter written as one polynomial moves its poles to a radius of 1.4536, which is not an inaccurate filter but an unstable one.

What 10 ps of aperture jitter is worth, in bits. computed by solving, not by drawing. Samples are taken at instants displaced by a seeded Gaussian of 10 ps and the error is measured against the same sinusoid sampled exactly. The line is −20 log(2π f × jitter), which the measurement matches to 0.12 dB across three decades. The penalty is exactly twenty decibels a decade of input frequency, because the error is the signal's slope times the timing error and nothing else — so a converter holds 16 bits only up to 199 kHz and 12 bits up to 3.18 MHz. An aperture figure quoted without an input frequency states no resolution at all.

A picosecond, read as bits

Every other boundary in this collection has a frequency, an amplitude or a size on its axis. This one has a duration. A converter that samples at t + δ instead of t gets a value wrong by the slope times δ, so the damage is proportional to input frequency and to nothing else about the part: ten picoseconds holds sixteen bits up to 199 kHz and twelve bits up to 3.18 MHz, falling at exactly twenty decibels a decade. An aperture figure quoted without an input frequency states no resolution at all.

A second integrator is 5.8 more decibels an octave, and a third 5.4 more. computed by solving, not by drawing. Modulators of order 1, 2, 3 marched one sample at a time with a one-bit quantiser, their in-band noise read from the transform of the error with the test tone a quarter of the way up the band. The measured slopes are 8.17, 13.93, 19.33 decibels per doubling of the oversampling ratio, against the 9, 15, 21 the white-noise argument predicts and the 3.01 plain oversampling gives. Each order is short of its own prediction by about a decibel, in the same direction, which is the error in the band not being white. The straight lines are each order's prediction drawn through its own first point, so what is compared is a slope against a slope.

The loop that is worse at full scale

A second integrator in a one-bit loop takes the shaping law from nine decibels an octave to fourteen, and a third to nineteen. What it charges is not in decibels at all: past six tenths of full scale the ratio starts falling, and a fourth integrator's states run away at seven tenths. The best input to a second-order modulator is 0.6 of the reference it is measured against — an amplitude boundary of exactly the kind this collection is built for, on the one object in the field that has no continuous output to draw.

What a cascaded modulator is worth, against how well its two paths match. computed by solving, not by drawing. Two first-order loops marched sample by sample at an oversampling ratio of 64, with the first stage's quantisation error taken as the difference between what its comparator said and what was presented to it — nothing here reads a state a real converter could not. Perfectly matched, the cascade gives 72.5 decibels against the first stage's own 47.5: second-order shaping out of two first-order loops, neither of which can be unstable. The gain with which the first error reaches the second stage is then given an error, and the flat left-hand half of the curve is the arrangement working. It costs three decibels at 5.29 per cent, which is a capacitor ratio — achievable, and not free, and not something the digital side can measure or correct.

Two loops, and the mismatch between them

Single loops of second, third and fourth order, marched alone, each stop working above an amplitude that falls with the order — which is why nobody builds a fourth-order single loop. The standard answer is two first-order loops with the first one's error fed to the second and differentiated back out, giving second-order shaping out of parts that cannot be unstable. The cancellation is between an analogue path and a digital one, and it is worth 25 decibels until the two differ by five per cent.

Same 2 ps of jitter: a floor at -121 dB, or a line at -91. computed by solving, not by drawing. A 3.337 MHz sinusoid sampled at 10 MHz with two clocks of identical root-mean-square jitter. One clock's timing error is independent gaussians and the other's is a sinusoid at 130 kHz. The total error power is the same — -87.43 and -87.55 decibels below the signal, against the closed form's -87.55 — and the pictures are not. The random clock spreads it over 2048 bins, -120.5 dB each; the modulated one puts it into two lines at 3.337 ± 0.130 MHz, -91.5 dB down. A specification in picoseconds does not choose between them.

A floor, or a line

Two clocks with identical two-picosecond jitter sample the same sinusoid, and the total error power is the same to a tenth of a decibel — the closed form an earlier essay computed, right for both. One of them puts that error across two thousand bins at −120 dB each; the other puts it into two lines at −91.5. The gap is the processing gain, it grows with the record length because a density falls and a line does not, and nothing in a specification quoted in picoseconds root-mean-square distinguishes the two cases.

Fed nothing, an eighth-order cascade sits 88 least significant bits from zero. computed by solving, not by drawing. Where each section of a 12-bit rounded cascade settles with zero input and a seeded state. Nothing decays to zero: a constant state survives whenever rounding returns it to itself, which needs only |y·(1 + a₁ + a₂)| ≤ q/2, and that denominator is small precisely because the corner is far below the sample rate. Each section's own band is 30, 31, 32, 33 least significant bits and the settled offsets are -27, -50, -70, -88 — accumulating, because a section's dead-band output is the next section's input and the next section passes direct current. In volts the offset halves with every bit added; in least significant bits it does not move at all.

Zero in, and not zero out

Rounding a filter's coefficients and marching it in double precision leaves out the other half: the products are rounded too. Put that in and a twelve-bit eighth-order cascade fed nothing at all settles two per cent of full scale away from zero and stays there — and the offset does not shrink with the word length, it grows as the square of how far the corner sits below the sample rate, reaching eighteen per cent at a hundred and twenty-eight times.

Twenty-four orderings, and 11 of them are choices. computed by solving, not by drawing. Every ordering of the four sections of an eighth-order 0.5 dB Chebyshev, at 16 bits, drawn against the two things an ordering decides. The horizontal axis is the round-off floor the arrangement adds — 24.2 least significant bits at best and 99.8 at worst. The vertical axis is the largest value any section's output reaches, which is what decides whether a word overflows: 0.088 of full scale at best and 0.699 at worst, a range of 18.0 decibels. 13 of the twenty-four are beaten on both counts by another ordering and are simply mistakes; the 11 on the lower-left frontier are the actual choices, and no one of them is best.

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.

The direct form works at 28 bits and fails again at 29. computed by solving, not by drawing. The largest pole radius of an order-8 Butterworth at a 1 kHz corner and a 48 kHz clock, at every coefficient word length from 6 to 40 bits, for both realisations. Above the unit circle the filter is not inaccurate but unstable. The direct form is outside at 23 of the 35 word lengths drawn; the cascade at none of them, its worst radius being 1.000000. The failures are not an interval: 28 bits works and 29 bits does not, so the smallest word length that works is 1 below the largest that does not and a search for the crossing has no crossing to find. The direct form's curve is clipped at 1.24: a radius of 4.5 and a radius of 1.01 are the same verdict.

The word length that is not a threshold

The essay below this one measured an eighth-order direct form at five word lengths, found its poles outside the unit circle at every one, and left behind a phrase — the coefficient resolution required. Sixty word lengths later there is no such resolution. That filter is stable at 28 bits, unstable at 29 and stable again at 30, and the smallest word length a search would return is not one anybody can use. What survives is a law with a slope: 4.53 bits of coefficient for every pole added, rising to 47 bits at twelfth order, and none at all for a cascade.

Against its own shaping the error is 17.4× tonal, not 51×. computed by solving, not by drawing. The share of an order-1 loop's in-band error sitting in its five largest lines, from 0.02 to 0.9 of full scale, with both nulls drawn. The lower level is 1.953 per cent — five lines of a FLAT error over 256 — and it is what this measurement has always been quoted against. The upper level is 5.74 per cent, which is what five lines of the loop's OWN shaping hold with no tone anywhere in them. Measured against the first the error is 51 times tonal at 0.02 of full scale and 26 at 0.9; against the second, 17.4 and 8.9. The direction survives the correction and the size does not.

A floor, or five tones

The field's sharpest statement about a one-bit loop is that three quarters of its in-band error sits in five lines, against the 1.953 per cent a white error would put in any five — a factor of thirty-eight. The comparison is to a white error, and a shaping loop exists to make its error anything but white. Measured against the loop's own transfer function, which the loop reproduces line by line to a part in five hundred when it is dithered, the same error is 17.4 times tonal at a fiftieth of full scale and 8.9 times at nine tenths. And 99.5 per cent of it at the quiet end is two harmonics of the input.

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.

The images a zero-order hold leaves, at 0.222 of the sample rate. computed by solving, not by drawing. A 10.67 kHz tone held at 48 kHz, with every line read out of a transform of the staircase itself. The sampled spectrum repeats at every multiple of the clock and the hold multiplies all of it by one sinc, so each image survives scaled by the sinc at its own frequency: the fundamental at -0.72 dB, the largest image (1fs−f, 37.3 kHz) at -11.60 dB, which is 10.88 dB of rejection. The sinc's nulls are exactly at the multiples of the clock and the two first-order images straddle the first of them without touching it — so the hold's rejection is 25.6 dB for a tone at 0.05 fs and 1.74 dB for one at 0.45, falling to nothing at half the clock. Measured and closed form agree to 0.026%.

The nulls are where nothing is

A zero-order hold multiplies the whole repeated spectrum by one sinc, so it attenuates every image at the image's own frequency and its nulls land exactly on the multiples of the clock. Nothing is ever at a null: the two first-order images straddle it, and the closer the signal comes to half the clock the closer they come to each other. The hold gives 25.6 dB of image rejection to a tone at a twentieth of the clock, 1.74 dB at 0.45 of it, and nothing at all at half — which is the frequency the band most needs it at.

What flatness costs, in the two places it can be bought. computed by solving, not by drawing. The hold's sinc across a band ending at 0.40 of the sample rate, and the same sinc with a one-zero one-pole shelf fitted to its reciprocal over that band. The droop to be removed is 2.420 dB. Corrected digitally the band comes flat exactly and the flat level sits 2.420 dB below what the uncorrected converter gave at direct current, because nothing may exceed full scale — the price is the disease. Corrected in analogue the band comes flat to 0.2308 dB and the shelf is still rising where the images are, so the worst image at 0.60 fs comes up by 3.799 dB, which is 1.57 times the droop it removed — and that boost is between 3.4 and 4.7 dB at every band on the slider, while the droop it cures runs from 0.58 to 3.75. Neither correction changes the signal-to-noise ratio, because the droop never cost any.

Flatness, and the two currencies it is bought in

The hold's droop takes the signal and everything arriving with it down together, so it costs no signal-to-noise ratio at all — a fact that is never stated and settles what correcting it can possibly be worth. Corrected digitally the price is headroom and is exactly the disease: 2.42 dB of flatness for 2.42 dB of output level. Corrected by an analogue shelf the price is image rejection and is nearly a constant: between 3.4 and 4.7 decibels whatever the band, so it is eight times the droop at a fifth of the clock and nine tenths of it at forty-nine hundredths.

What an oversampling ratio buys, and at two different rates. computed by solving, not by drawing. A 20 kHz band on a 48 kHz base clock, interpolated by ratios from 1 to 64. The hold's droop at the band edge falls with the SQUARE of the ratio — the fitted exponent over six doublings is -2.0113 — from 2.640 dB at the Nyquist rate to 0.0097 at sixteen times it. The nearest image moves out with the FIRST power, exponent 1.0222, from 1.40 times the band edge to 37.4. So one decision buys two things at rates differing by a factor of two in the exponent, and the third quantity — the poles a reconstruction filter needs for sixty decibels — collapses from 20.5 to 3.21 by a ratio of four alone.

One knob, and the two exponents it turns

Oversampling is quoted as buying one thing and buys two that improve at different rates. The hold's droop at the band edge falls with the SQUARE of the ratio — fitted exponent −2.011 over six doublings, from 2.640 dB at the Nyquist rate to 0.0097 at sixteen times it — while the nearest image moves out with the first power, exponent 1.022. The third quantity, the poles a reconstruction filter needs for sixty decibels, collapses from 20.5 to 3.21 by a ratio of four alone, because it is a logarithm of the second.

An all-pole filter's stopband floor: -120 dB, from 10 fF nobody drew. computed by solving, not by drawing. A butterworth 8 realised at a 20 kHz passband, drawn twice: from its poles, which fall at 48 decibels an octave for ever, and from the same netlist with 10 femtofarads of stray capacitance from its input to its last shunt node — a fraction of a picofarad between two tracks. The second stops falling at -120.0 dB at 136 kHz, because the stray bypasses every pole in front of it and what is left is its ratio to the shunt capacitance it divides against. The poles alone predict -293 dB a decade further out, which is 173 decibels lower. So an all-pole design has a stopband floor exactly as an elliptic one does — -140 dB at 1 fF, -130 dB at 3 fF, -120 dB at 10 fF, -110 dB at 30 fF, -100 dB at 100 fF, which is 20.0 decibels per decade of stray — and above that frequency a faster sample rate buys nothing at all, which is the refusal recorded for the elliptic row and not for this one.

The floor every filter has

The elliptic design was refused because its equiripple stopband is a floor and no sample rate reaches past a floor, while an all-pole design falls at six decibels an octave per pole for ever — so a faster clock is always an answer for one and never for the other. On paper. Ten femtofarads of stray capacitance from a filter's input to its output, which is a fraction of a picofarad between two tracks, puts the eighth-order Butterworth's stopband at −120 decibels from 136 kilohertz onwards where its poles predict −293. Every all-pole design has a floor, and it is a layout rather than a design.

The order is worth 0.68 bits and a factor of 29 in clock. computed by solving, not by drawing. A butterworth design at a 20 kHz passband, at each order from two to ten and at three impedance levels, with the sample rate it demands for eighty decibels and the resolution its own resistors' noise allows. The clock demand falls steeply — 50.50×, 11.27×, 5.50×, 3.65× and on down to 1.76×, a factor of 28.8 — and the bar widths carry it. The noise rises: 2.00 µV at order two against 3.22 µV at ten, which is 0.68 of a bit. So the order is cheap in the currency everybody worries about and expensive in nothing. What is expensive is the impedance level: every resistance goes as one over the capacitance and a noise voltage as the square root of a resistance, so a decade of capacitance is 1.66 bits — the same at every order, to 0.000 of a bit — and the three curves are ten, one and a tenth of a nanofarad a section. The ceiling on resolution is set by a choice nobody prices, and the order, which everybody argues about, moves it by two thirds of a bit.

Two thirds of a bit for a factor of twenty-eight

An anti-alias filter costed in clock rate gets steeply cheaper with order — fifty times Nyquist at the second and 1.76 at the tenth — and the obvious objection is noise, since the filter is in the signal path and every section adds resistors. Measured, the objection barely holds: from order two to order ten the clock demand falls by a factor of 28.7 and the resolution the filter's own noise allows falls by 0.69 of a bit. What does cost resolution is the impedance level, at five thirds of a bit per decade of capacitance, and nobody argues about that at all.

Sampling 100–120 MHz: 6 allowed windows and 5 forbidden gaps. A band 20 MHz wide centred at 110 MHz, and the sample rates at which it and all its images land without overlapping. Twice the top of the band is 240 MHz and every rate above that works, which is the familiar answer. Below it there are 5 more windows, the slowest at 40.00 MHz — 1.00 times twice the band's WIDTH, which is the quantity that bounds a sample rate when the signal is not at baseband. Between the windows are 5 forbidden gaps, shaded: at 219.1 MHz the band's images overlap it, while 200.00 MHz works and 240.0 works. So a clock raised out of a working window has been made worse by being made faster, and "a faster clock is always an available answer" is a statement about baseband and not about sampling.

The clock that is too fast

Every sample rate in the requirement is a lower bound: the filter must be down by the frequency that folds back, and a faster clock is always at least as good. That holds for a signal at baseband. A band from 100 to 120 megahertz can be sampled at 240 and above, or at 120 to 200, or 80 to 100, or 60 to 66.7, or 48 to 50, or 40 — six disjoint windows with five forbidden gaps between them, so 110 megahertz fails while 100 works and 120 works. The bound is twice the band's width, 40 megahertz, and it is six times below twice its top.

Joining the samples with straight lines squares the hold's sinc: 5.28 dB of droop and 5.8 dB of image rejection, both doubled. computed by solving, not by drawing. A 20 kHz tone on a 48 kHz clock, reconstructed two ways: held for a whole clock period (dashed envelope, open stems) and joined sample to sample by straight lines (solid envelope, filled stems). Every line is read from a transform of that waveform itself. The held one comes out 2.640 dB down and its nearest image 2.92 dB below it; the interpolated one 5.280 dB down with the image 5.85 dB below. A straight line between samples is a triangle two clocks wide, which is a rectangle convolved with itself, so its transform is the sinc squared: every decibel doubles and the nulls stay where they were.

The pole a straight line is worth

A converter that joins its samples with straight lines instead of holding each one is convolving with a triangle rather than a rectangle, and its transform is the hold's sinc squared. Every decibel doubles: 5.28 dB of droop at a 20 kHz band edge on a 48 kHz clock instead of 2.64, and 5.85 dB of image rejection instead of 2.92. Both exponents of the oversampling ratio are unchanged. The second sinc is worth exactly one pole of reconstruction filter at every ratio, because a sinc's rejection of the nearest image is twenty times the log of that image's distance from the band edge, identically. A single analogue pole allowed the same droop gives a fraction of that rejection. The price is a whole clock of delay rather than half.

4 rectangles: the hold's sinc to the power 4, 10.56 dB of droop and the first image 11.7 dB down. computed by solving, not by drawing. A 20 kHz tone on a 48 kHz clock, reconstructed by a pulse of 4 rectangles convolved together — the hold, the straight line, the quadratic and the cubic B-spline for one to four — with the envelopes of all four drawn and the lines of the 4-rectangle waveform read from its own transform (stems). They agree with the sinc to the 4th power to 0.000 per cent. Droop 10.560 dB; the nearest image 11.69 dB below the signal, 4 times the hold's 2.92. The nulls stay on the multiples of the clock; each rectangle deepens everything else.

Four rectangles and no filter

A converter's output pulse can be any number of one-clock rectangles convolved together: one is the hold, two the straight line, four the cubic B-spline. Measured on its own waveform, n rectangles give the hold's sinc to the n-th power to a few parts in ten million, so each one rejects the nearest image by exactly what one analogue pole could at best — and each saves exactly one pole of reconstruction filter until none is left. At the Nyquist rate that is three poles of nineteen and a half, bought with 10.56 dB of droop. From about 2.8 times oversampling the cubic pulse needs no analogue filter at all for 60 dB, for 0.62 dB of droop at four times and two clocks of delay. The cost is that from three rectangles on the output no longer passes through its samples.

With the crossover at 10% of the clock the straight line overshoots 13.7%, the hold with its pole 32.8%. computed by solving, not by drawing, marched at 160 steps a clock. A unit step through a loop whose plant is an integrator crossing over at 10% of the sample rate, reconstructed three ways: held (dotted), joined by straight lines (solid), and held then passed through the analogue pole that attenuates the nearest image, at 90% of the clock, exactly as much as the straight line's second sinc does (dashed) — a pole at 9.90% of the clock, at the crossover itself. Overshoot: 0.0%, 13.7% and 32.8%. The straight line buys the same image rejection for half a clock of delay; the pole buys it for forty-five degrees.

Half a clock against a pole

Inside a digitally controlled loop a converter's delay is phase at the crossover, and a straight-line output pays one more half clock than a hold: 180 degrees times the crossover over the clock. The analogue pole that rejects the nearest image as much as the straight line's second sinc does turns out to sit at the crossover itself — 0.990 of it when the loop crosses over at a tenth of its clock, sinc(r) of it in general — so it costs forty-five degrees however slow the loop is. By the phase estimate the half clock is the cheaper filter below 26% of the clock. Marched as a sampled loop the answer is 17.6%, and at a tenth of the clock a step overshoots 13.7% through the straight line against 32.8% through the hold and its pole, for the same image rejection measured on both waveforms.

A band from zero to its top frequency takes 4.77 dB less from 10 ps of jitter than a tone at the top: it is the rms frequency that counts, not the highest. computed by solving, not by drawing. Forty-eight equal tones at random phases filling a band from 0 to 1 of its top frequency, sampled 16,384 times at instants displaced by a seeded 10 ps Gaussian, at seven top frequencies from 100 kHz to 100 MHz. The measured signal-to-error ratio (dots) is 88.64 dB at 1 MHz, against −20 log(2π × rms frequency × jitter) = 88.76 dB from the band's power-weighted rms frequency, 580 kHz — and against 84.04 dB for a tone at 1 MHz itself. The gap is 4.77 dB at every frequency, 10 log(3/(1 + r + r²)) for a flat band from r to 1 of the top: 4.77 dB (a factor of √3 in frequency) for a band from zero, nothing for a narrow one. At 1 MHz the band holds 14.43 bits where the tone holds 13.67.

The frequency a band moves at

The jitter rule is written for a sinusoid: ten picoseconds holds a tone at 1 MHz to 84.04 dB. A signal that fills the band up to 1 MHz holds 88.64 dB against the same clock, because each sample's error is its own slope times its own timing error, and a band's slope is set by its rms frequency — the top frequency over √3 for a flat band, 4.77 dB better. Pink content up to 1 MHz is 11 dB better than the rule and content falling as 1/f² is 29 dB better. So the same ten picoseconds hold sixteen bits to 199 kHz for a tone, 344 kHz for a flat band and 734 kHz for a pink one: an aperture figure needs a signal as well as a frequency.

The same 0.1 ps of period jitter: 107.0 dB at any record length if each edge is independent, 80.3 dB over 4096 samples if the edges walk. computed by solving, not by drawing, averaged over 8 seeded clocks at each of 5 record lengths. A 10.01 MHz tone sampled at 160 MHz by a clock whose every period has a spread of 0.1 ps, in two ways: each edge independently displaced (white), and each edge displaced from where the last one landed (a free-running oscillator's walk). The sampled tone is fitted by least squares for amplitude, delay and a small frequency offset, and what is left is the error. White jitter gives 107.2, 107.0, 107.0, 107.1, 107.0 dB — the closed form 107.0 dB, at every length. The walk gives 92.6, 86.4, 80.3, 74.5, 68.2 dB for 256, 1024, 4096, 16384, 65536 samples, falling 10.1 dB a decade, as the period spread times √(n/15): a jitter with no value until the length of the look is named.

A jitter that grows with the look

Two clocks with the same 0.1 ps spread in every period sample the same 10 MHz tone. If each edge is displaced independently, the error is 107.0 dB down at any record length. If each edge is displaced from where the last one landed — a free-running oscillator — the edges walk, and the error after fitting the best delay and frequency is 92.6 dB down over 256 samples, 80.3 over 4,096 and 68.2 over 65,536: ten decibels a decade of record, as the period spread times √(n/15). The one oscillator honestly has jitter figures from 0.10 to 4.7 ps depending on the length named. A locking loop is what gives the jitter a value: with a time constant of a hundred periods it stops growing at 87.0 dB.

At 60 dB of rejection an eighth-order filter keeps 3 of the brick wall's 6 windows; a fourth-order one keeps 2. computed by solving, not by drawing. The sample rates at which a band from 100 to 120 MHz can be sampled without anything the filter in front passes at more than −60 dB folding onto the band, for a band with hard edges and for Butterworth bandpass filters of order 12, 8, 6 and 4 on the band's own edges; each row is labelled with the span, in megahertz, where the filter is still above −60 dB. A real filter moves each window's edges inward by half its transition band. The window from 120 to 200 MHz narrows to 124.4–193.2, 127.9–188.4, 132.8–182.4, 149.7–166.9 MHz as the order falls, and the lower windows close altogether. Every window is checked by folding every frequency the filter passes.

The windows a real filter leaves

A band from 100 to 120 MHz can be sampled at any clock in six disjoint windows — for a band with hard edges. Put a real bandpass filter in front and ask that whatever folds onto the band be 60 dB down, and every window's edges move inward by half the filter's transition band: an eighth-order Butterworth keeps three of the six, a fourth-order one two, and the window from 120 to 200 MHz shrinks from ±25 per cent of clock tolerance to ±19.1 per cent at order eight and ±5.4 per cent at order four. The windows depend on the rejection only through the rejection per order: 40 dB at order four, 60 at order six and 80 at order eight all leave the same ±15.7 per cent. So a filter's order is bought in clock tolerance, at one order per ten decibels.

With a 2 ps reference and a 0.1 ps-a-period oscillator the best loop has a time constant of 20.0 periods, where the clock's error is 0.453 ps — 90.9 dB on a 10 MHz tone. computed by solving, not by drawing, with seeded locked clocks as dots. A free-running oscillator whose period has a spread of 0.1 ps, locked by a first-order loop to a reference whose timing error, as the phase detector sees it each period, is 2 ps rms and white. The loop passes the reference's error below its corner and the oscillator's walk above it, so the stationary error is σₒ²/(1 − a²) + σᵣ²(1 − a)/(1 + a) with a = exp(−1/τ). The oscillator's part rises as √(τ/2)·σₒ and the reference's falls as σᵣ/√(2τ); they are equal, and their sum least, at τ = 20.00 periods, the ratio σᵣ/σₒ, where the clock's error is 0.4528 ps — √(σᵣ·σₒ). Seeded clocks of 65,536 periods: τ = 2, 1.000 ps; τ = 5, 0.656 ps; τ = 20, 0.452 ps; τ = 80, 0.649 ps; τ = 300, 1.187 ps; τ = 1000, 2.105 ps.

The loop a reference allows

A sampling clock is a free-running oscillator held by a loop to a reference, and the reference has an error of its own. With 0.1 ps of period spread in the oscillator and 2 ps of white error in the reference, a first-order loop of 20 periods leaves the clock 0.453 ps from ideal — 90.9 dB on a 10 MHz tone — and that is the best any first-order loop can do: the oscillator's share rises as the square root of the loop's time constant, the reference's falls as one over it, and they are equal at σᵣ/σₒ. The clock's error there is the geometric mean of the two, so a reference a hundred times worse costs ten. On a spectrum the rule reads as one instruction: put the loop's corner where the free oscillator's skirt meets the reference's floor. And it holds only for long captures. Below about seven loop time constants the free oscillator beats every loop.

With a 100 kHz flicker corner the 0.1 ps oscillator's ratio stops falling ten decibels a decade of capture and starts falling twenty, past 2,277 samples — 1.42 periods of the corner. computed by solving, not by drawing, from the exact expected residual of a summed process, with seeded clocks as dots (24 at each length to 16,384). A free-running oscillator with a 0.1 ps period spread and a flicker of its frequency whose timing spectrum meets the walk's at 100 kHz, sampling a 10.01 MHz tone at 160 MHz. Signal-to-timing-error after removing each capture's best delay and frequency: the walk alone falls ten decibels a decade, 85.7 dB at 1024, 67.6 dB at 65536; with the flicker, 84.1 and 52.9 dB. The flicker's share equals the walk's at 2,277 samples, 14.2 µs, which is 1.42 periods of the corner frequency; beyond it the ratio falls 19.5 dB a decade. Removing a quadratic as well buys 3.80 dB at 65,536 samples and changes no slope.

A frequency that wanders

Below its flicker corner an oscillator's phase noise falls thirty decibels a decade instead of twenty, because the oscillator's frequency itself wanders. A clock with 0.1 ps of period spread and a 100 kHz corner samples a 10 MHz tone at 84.1 dB over 1,024 samples and 52.9 over 65,536 — 14.7 dB worse than the walk alone — because past 2,277 samples its error grows as the capture rather than its square root. That crossing is 1.42 periods of the corner frequency for every corner. Fitting a parabola as well as a line buys 3.88 dB at every length and changes no slope; each further term buys a fixed fraction. And a first-order locking loop cannot stop it: the locked clock's variance grows with the logarithm of the record, 1.92 dB lost between 4,096 and 262,144 samples, until a loop with an integrator holds it at 87.5.

A sinusoid sampled 8 times a cycle, at the phase that puts its peak between two samples. computed by solving, not by drawing: the samples of a unit sinusoid at 8 a cycle, synchronous, at the sampling phase that hides its peak worst. 8 samples a cycle is 4 times the Nyquist rate and holds the waveform completely; the largest sample is still 0.9239, 7.61% below the peak, which is 1 − cos(π/8). Read straight off the samples the peak needs 22.2 samples a cycle for 1% and 70.2 for 0.1%.

The rate each reading needs

Twice the frequency is the rate at which a set of samples holds a sinusoid completely, and it is not the rate at which any particular property can be read off them. Eight samples a cycle — four times the Nyquist rate — hold the waveform, and their largest is still 7.61% below the peak; the largest sample needs 22.2 a cycle for 1% and 70.2 for 0.1%, because its error falls only as the square of the rate. A parabola through the three largest gets 1% at 8. A zero crossing found by a straight line between two samples is a hundredth of a cycle right at 5 a cycle, its error falling as the cube, because a sine is straight at its zero. The rms of whole cycles is exact from three a cycle — and at exactly two it reads anything from zero to √2. A slow rate with a long record of a steady signal does what a fast rate does, once the samples have slipped a whole interval past it: 42 cycles at 4.0236 a cycle.

The quantisation error's power as a function of where the input sits, for four dithers. computed by solving, not by drawing; the error's mean square given the input, integrated over each dither's density through the converter's own quantiser, and checked against seeded draws to 6 × 10⁻³. Undithered it is the square of a sawtooth, x² up to half a step and (1 − x)² beyond. One uniform step makes it x(1 − x): zero at a code, a quarter of a step² between codes — the error's power is carried by the signal. Two steps (triangular) hold it at ¼ everywhere, and one step subtracted after the converter at 1/12 everywhere. Averaged over the input the prices are 3.01, 4.77 and 0.00 dB against the undithered 1/12.

One step of dither per moment

A whole step of uniform dither makes a quantiser's error average to zero wherever the input sits, and leaves its power at x(1 − x) steps²: zero at a code and a quarter between codes, so the noise rises and falls with the signal. Two steps — a triangular dither — hold the power at a quarter everywhere and leave the third moment swinging by 0.096 steps³; three steps fix that too. Each added step fixes one more moment of the error and costs a twelfth of a step² of power: 3.01 dB for one, 4.77 for two. Subtracted after the converter, one step fixes the error's whole distribution for 0.00 dB.

A second-order loop at a ratio of 64, with a quantiser of one to five bits: the ratio in the band against the input amplitude. computed by solving, not by drawing: the loop whose noise transfer is exactly (1 − z⁻¹)², marched with a sine inside the band, the error measured through a squared Hann window. At half scale, 1 bit 69.4 dB, 2 bits 82.5 dB, 3 bits 90.3 dB, 4 bits 96.8 dB, 5 bits 102.0 dB. The first extra bit adds 13.1 dB because a one-bit quantiser's error is larger than a step's uniform error; after that each adds six or seven. The amplitude each does best at moves up with the bits — 0.6 for 1, 0.6 for 2, 0.8 for 3, 0.85 for 4, 0.95 for 5 — because a fine quantiser is never driven past its last level by the shaped error.

The levels the loop cannot shape

Give a second-order noise-shaping loop a quantiser of several bits instead of one and every bit beyond the second buys six or seven decibels, the best input amplitude climbs from 0.6 of full scale to 0.95, and loops of third and fourth order that will not run on one bit run on three or four. What it charges is outside the loop's reach. The levels the quantiser feeds back enter where the input does, so their errors are passed rather than shaped: a three-bit loop loses its lead over a one-bit one when each level is out by 0.027 per cent, its error stops falling with oversampling, and the error arrives as harmonics. One bit is immune because two levels always lie on a line.

A second-order four-bit loop whose integrators are built round amplifiers of finite gain: the ratio in the band against oversampling. computed by solving, not by drawing: each integrator's pole at 1 − 1/A, half-scale input, half a step of shaped dither, three seeds averaged, squared Hann window. With perfect integrators the ratio rises 15.1 dB an octave, to 138.9 dB at 512. With A = 1000: 138.5 dB at 512, 0.4 below; with A = 100: 122.3 dB at 512, 16.6 below; with A = 30: 102.9 dB at 512, 35.9 below; with A = 10: 85.0 dB at 512, 53.9 below. Each curve leaves the ideal one near a ratio of about πA; for A = 10 the last octave gains 2.7 dB against the ideal's 15.1 — toward the three an octave a floor that is not shaped improves at.

The integrator that leaks the zero away

A noise-shaping loop puts its error's zero at zero frequency because its integrators never forget. An integrator built round an amplifier of gain A forgets a part in A every sample, and the zero moves off zero frequency to a floor. The cost is a function of one number, the oversampling ratio over π times the gain: for two leaky integrators it is 10 log(1 + (10/3)x² + 5x⁴), which the marched loops follow to within two decibels from gains of 10 to 300, and keeping it under a decibel needs a gain of 1.2 times the oversampling ratio. A first-order loop pays a second way: inputs within exactly 1/(2A − 1) of zero leave its output where it was.

Fed nothing, a 12-bit eighth-order cascade sits 88 least significant bits from zero; with its rounding error fed back, never more than 1. computed by solving, not by drawing. The same cascade with its corner at 1 kHz and its clock 48 times higher, each section's state seeded at eight least significant bits and every product rounded to 12 bits, marched with no input. Plainly rounded, the sections settle at 27, 50, 70, 88 least significant bits — a constant state survives whenever rounding returns it to itself, and each section's offset is the next one's input. With each section's last two rounding errors added back with weights −2 and +1, the error reaches the output through (1 − z⁻¹)²/A(z), a constant state y needs (1 + a₁ + a₂)·y = 0, and the only one is zero: the largest excursion over the last two thousand samples is 1, 1, 1, 1 least significant bit, single isolated steps rather than a settled offset.

The rounding a section keeps

A fixed-point biquad that remembers its last two rounding errors and adds them back with weights −2 and +1 before it rounds again sends its error to the output through (1 − z⁻¹)²/A(z) instead of 1/A(z). For a low-pass clocked far above its corner that double zero lands exactly on the recursion's largest gain. Fed nothing, an eighth-order 12-bit cascade at 48 times its corner sat 88 least significant bits from zero; with the error kept, a constant state needs (1 + a₁ + a₂)·y = 0, its only solution is zero, and the output never strays more than one bit at any clock rate from 8× to 128×. The round-off floor at 128× falls 89 times, 6.5 bits, for two additions and a shift per section — and the white-noise model, which the plain cascade broke by a factor of ten at eight bits, holds to 2.3% from six bits to twenty.

Three waveforms at 16 samples a period, sampled at the worst instant: the largest sample of a sine is 1.92% short, of a triangle 12.5%, of a clipped sine 0. computed by solving, not by drawing: three waveforms with the same period and the same peak, sampled 16 times a period at the phase that puts the peak midway between two samples (dots). The sine leaves its peak as the square of the time, the triangle in proportion to it, and the sine clipped at nine-tenths of its height does not leave it at all for as long as it stays flat. The largest sample therefore falls short by 1.92% for the sine, 12.5% for the triangle, 0 for the clipped sine, at one rate, for three waveforms with the same period.

The peak that is a corner

The largest of a set of samples falls short of a sine wave's peak as the square of the sample interval, and that law belongs to the sine. Measured over every sampling phase, a triangle's corner falls short as the first power, 12.5 per cent at sixteen samples a period against the sine's 1.92, a peak whose curvature vanishes as the fourth power, and a sine clipped at nine-tenths of its height reads exactly above seven samples a period. The order in which a waveform leaves its peak is the whole of the law, and it also fixes how much better the average reading is than the worst. A real corner is round at some scale, so the law bends where the samples resolve its top harmonic, and a parabola through three samples helps a round peak, barely touches a corner and makes a flat one worse.

A Gaussian edge read at 3 samples per rise, at the worst sampling phase: straight chords between the samples say the rise is 8.78% from its true value. computed by solving, not by drawing: a Gaussian edge (solid), its samples at 3 per 10–90% rise at the sampling phase that reads worst (dots), and the straight lines an instrument's linear interpolation draws through them (dashed). The chord crosses 10 and 90 per cent earlier and later than the edge does, because a Gaussian edge curves in opposite senses at the two thresholds. The reading is 1.0878 rises against 1.

The samples a rise time needs

A rise time is read from two thresholds, and an instrument that draws straight chords between its samples is wrong at each by an amount that goes as the square of the interval. For a Gaussian edge at the worst sampling phase that is 5.02 per cent at four samples per rise and 0.818 at ten; the rule of thumb of three to five is a five per cent rule, and one per cent takes nine. The reading is not monotonic in the rate: above 20.9 per rise a first-order edge reads exactly at whole numbers and 0.046 per cent at 24.5, because its two errors subtract at one phase, and a Gaussian edge reads worse there because its two add. Reconstructing before reading needs 1.87 per rise for one per cent, and at four times the edge's 3 dB bandwidth the reading is still 11.7 per cent out.

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