Where a signal becomes a number

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.

17 min read 5 figures Two routes to a number Measured, not tabulated The constants decide

Assumes: What the filter in front costs · The floor a resistor sets

What the filter in front costs prices an anti-alias filter in clock rate, and among the numbers it checks is that the demand falls with order for a fixed family — 5.50×, 2.82×, 2.08× and 1.76× for Butterworth at orders four, six, eight and ten. Read on its own that is an argument for the steepest filter anybody can afford, and the essay says so: the trade is legibly order against clock and not one against nothing.

It also names, in passing, the thing that ought to stop that argument. A steeper filter is more components, and every resistor in it is a noise source. The filter sits between the source and the converter, so whatever its own resistors produce arrives at the quantiser along with the signal.

Measured, that objection is much weaker than it sounds, and what it points at instead is a parameter nobody argues about.

Fig. 1 A Butterworth at a 20 kHz passband, at each order from two to ten and at three impedance levels, with the resolution its own resistors’ noise allows on the vertical axis and the sample rate it demands as the width of each bar. The slider is the capacitance a section.

Both currencies in familiar units

The comparison needs the two costs in units that can be read against each other, and both are available without inventing anything.

The clock is already a ratio. the sample rate is bisected the sample rate at which the filter is down to eighty decibels by the frequency that folds back, and reports it as a multiple of Nyquist — which is what that essay’s whole table is in.

The noise becomes a number of bits. Every resistor in the realised network contributes $4kTR$ , shaped by the response from its own node to the output, integrated over frequency; that gives a floor in volts, and the resolution at which a quantiser’s own floor equals it is $\log_2$ of the ratio. So a filter whose output noise is two microvolts in a two-volt span is a filter behind which no converter can do better than 18.1 bits, whatever its data sheet says.

order	clock demanded	filter’s own noise	resolution it allows
2	50.50×	2.00 µV	18.14 bits
4	5.50×	2.43 µV	17.86
6	2.82×	2.77 µV	17.68
8	2.08×	3.01 µV	17.56
10	1.76×	3.22 µV	17.45

Read the two right-hand columns against each other. Going from order two to order ten the clock demand falls by a factor of 28.7 and the resolution falls by 0.69 of a bit. Sixty per cent of a bit for a factor of nearly thirty in sample rate, which is thirty times the power in every digital stage downstream and thirty times the throughput in whatever consumes the result.

There is no design in which that is a close decision. The order is cheap in noise, and the objection that a steeper filter is a noisier one is arithmetically true and practically irrelevant.

Why the noise rises so slowly

The reason is worth having, because the intuition that fails here is a reasonable one.

A filter of order $n$ built as buffered second-order sections has about $n/2$ sections, each with its own resistor, so the number of noise sources grows linearly with the order. Uncorrelated noise sources add in power, so the total should grow as $\sqrt{n}$ — from order two to order ten that is $\sqrt5 = 2.24$ , which is 1.16 bits.

Measured it is 1.61 times, which is 0.69 bits. The shortfall is because a section’s own noise is filtered by every section after it, so only the last section contributes its whole noise and the first contributes almost none of it. That is the same arithmetic which section goes first is about — one filter realised twice from the same poles, with its sections in the two orders, giving the identical response to nine figures and noise differing by a factor because a section’s noise is filtered by everything after it and by nothing before it.

So adding a section to the front of a cascade adds a noise source whose contribution is attenuated by everything downstream, and adding one to the back adds a full one. A design that wanted to minimise the order’s noise cost would put its added sections first — and that is the same ordering the noise field’s own result prefers, for the same reason, so the two conclusions do not conflict.

The other half of the shortfall is that a higher-order filter is narrower in noise bandwidth at a given corner. The bandwidth noise sees measures the ratio: a one-pole response passes π/2 times the noise a brick wall at its corner does and a five-pole Butterworth passes 1.0065 times. So each added section brings a noise source and narrows the window every earlier source is seen through, and the two partly cancel.

Fig. 2 The same design at ten nanofarads a section: 19.80 bits at order two and 19.11 at ten, so the order’s cost is the same 0.69 of a bit and the whole curve has moved up by 1.66. The order’s cost does not depend on the impedance level and the level’s cost does not depend on the order, which is what makes the two separable.

The clock’s own cost, which is not linear either

The comparison above treats the clock as a currency without saying what it buys or what it costs, and the asymmetry between the two costs is worth making explicit because it is what makes the decision so one-sided.

A factor of 28.7 in sample rate is not a factor of 28.7 in anything a designer pays for once. It is:

Twenty-nine times the converter’s own power, near enough, since a converter’s dissipation is roughly proportional to its rate over a wide range. On a part that takes a hundred milliwatts that is two and a half watts, which is a different kind of product.

Twenty-nine times the data rate out of it, which is a bus, a memory bandwidth and a processing budget. An eighteen-bit converter at 40 kilosamples a second is 90 kilobytes a second and at 1.15 megasamples it is 2.6 megabytes, which is the difference between a serial link and a parallel one.

And twenty-nine times as demanding on the sampling instant. A converter’s aperture jitter costs a number of bits that depends on the input frequency rather than on the sample rate, so this one does not scale — but the clock generating a 1.15-megahertz sample rate to the same absolute jitter as a 40-kilohertz one is a harder clock, and a picosecond, read as bits is where that arithmetic is done.

Against all of that, two thirds of a bit. And two thirds of a bit is recoverable for nothing by moving the impedance level up a third of a decade, which costs a larger capacitor.

So the honest summary of the trade is not that the order is cheap. It is that the two costs are not in the same units and one of them is recoverable. The noise the order adds can be bought back with a component value; the clock rate the order saves cannot be bought back with anything. A design that treats them as comparable — a decibel here against a decibel there — has missed that asymmetry, and it is the asymmetry rather than the sizes that settles the question.

What is expensive

The impedance level is set by one component value and it costs more than the whole range of order does.

A section’s resistance is $\sqrt{L/C}/Q$ with $L = 1/(\omega_0^2 C)$ , so $R \propto 1/C$ : halving the capacitance doubles every resistance in the filter. Johnson noise is $\sqrt{4kTR}$ , so the noise voltage goes as $1/\sqrt C$ — and a decade of capacitance is a factor of 3.16 in noise, which is 1.66 bits.

capacitance a section	resolution at order 2	at order 10
10 nF	19.80 bits	19.11 bits
3 nF	18.93	18.25
1 nF	18.14	17.45
0.3 nF	17.27	16.58
0.1 nF	16.48	15.79

Five thirds of a bit a decade, and the same at every order to within four hundredths of a bit — which is what says the two parameters are separable. The impedance level is worth 2.4 times as many bits as the entire range of order, and it is chosen for reasons that have nothing to do with noise: what capacitors are available, how large the inductors get, what the amplifier can drive.

Which inverts the design conversation. The order is argued about because it is in the specification, and the capacitance is picked because something has to be picked. The first is worth two thirds of a bit across its whole range and the second is worth a bit and two thirds per decade, and there are two or three decades of it available.

The reason the level is not simply made as large as possible is worth one paragraph, because it is not free either. Ten nanofarads at 20 kHz needs 6.3 millihenries of inductance, which is a wound component with a quality factor and a saturation current and a self-resonance; the q the components allow is the ceiling that puts on a section’s realised quality factor, and it binds well before the noise does. In an active realisation the same choice becomes a resistance the amplifier has to drive, and the band that does not close is where the two ends of the available range are measured — fifty ohms of amplifier output resistance from one side and two picofarads of stray from the other, shutting the band by half a decade per order.

So the level has a range rather than a direction, and this essay’s contribution is that noise is a reason to sit at the top of that range rather than in the middle of it, which is not how it is usually chosen.

Fig. 3 A tenth of a nanofarad a section: 16.48 bits at order two and 15.79 at ten. Two decades of capacitance below the previous figure and three and a third bits of resolution gone, with the clock demand and the response identical at every order. A sixteen-bit converter behind this filter is measuring the filter.

Fig. 4 Three tenths of a nanofarad a section: 17.27 bits at order two and 16.58 at ten. A third of a decade of capacitance below the default and half a bit gone, with the clock demand and the response identical at every order — which is the asymmetry this essay is about, on one axis.

Reading the two costs against a converter

Neither currency is a cost until something is being paid for, and the thing being paid for is a converter’s resolution.

The floor a converter sets establishes the arithmetic: a quantiser’s floor is $q/\sqrt{12}$ and falls by a factor of two per bit, and the resistance in front of it has a thermal floor that does not move at all — they cross at 18.8 bits for a kilohm in a hundred kilohertz, and past the crossing every further bit is a more precise measurement of Johnson noise.

This essay’s filter is that resistance, made specific. At one nanofarad a section it allows 17.5 to 18.1 bits, so:

A twelve-bit converter is unaffected. Its own floor is 74 decibels below full scale, which is far above the filter’s, and the filter’s order and impedance level are free choices as far as noise goes.

A sixteen-bit converter is nearly unaffected. Its floor is 17.6 microvolts against the filter’s 2 to 3.2, so the filter adds about 1.5 per cent in power — a hundredth of a bit. This is the ordinary case and it is the reason the objection this essay started from is not usually noticed.

An eighteen-bit converter is limited by the filter at every order, at one nanofarad a section. Its floor is 4.4 microvolts and the filter’s is 2.0 to 3.2, so the two are comparable and the filter costs between a third and two thirds of a bit. Moving to ten nanofarads a section recovers all of it.

And a twenty-bit converter cannot be used at all behind this filter at this impedance level: the filter’s own noise is two to three times the quantiser’s floor, so eighteen of the twenty bits are measuring the filter’s resistors. At ten nanofarads a section it allows 19.1 to 19.8 bits, which is most of the way there.

So the rule the two tables give together is short. Choose the order on the clock, the impedance level on the converter’s bits, and expect the level to bind first. That essay’s three currencies for choosing a family — clock rate, delay flatness, noise bandwidth — are untouched by any of this, because every number here is a Butterworth and the family enters only through the clock column.

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. — Fig. 5 The crossing this essay’s numbers are read against: a quantiser’s floor falling by a factor of two per bit against the thermal floor of a kilohm in a hundred kilohertz, which does not move — meeting at 18.8 bits. The filter measured here is that resistance made specific, and at one nanofarad a section it allows 17.5 to 18.1 bits, which is just below the crossing.

What this does not measure

The amplifiers’ own noise. The realisation uses ideal buffers, so the only noise sources are the resistors. A real section’s amplifier contributes a voltage noise in series with its input and a current noise across it, and for a filter at a low impedance level the amplifier’s voltage noise dominates the resistors’ — which flattens the bottom of the table above and means the impedance level stops paying below some capacitance. Where that is depends on the part, and the bowl, and the bottom of it is where the same crossing is measured for a source resistance rather than for a filter.

The source’s own noise. Everything in front of the filter is silent here. A real source has a resistance of its own whose noise the filter passes, and if that dominates then neither the order nor the level matters at all — which is the most common real situation and the reason this whole measurement is about a ceiling rather than about a floor.

And the order is a Butterworth’s order. A Chebyshev of the same order has higher-Q sections, and a section’s noise gain at its own resonance rises with its quality factor, so its noise is higher for the same number of resistors. The clock demand is lower too. Both move in the direction that makes the trade closer, and neither is measured.

The order that is actually chosen, and for what

Nothing above explains why anybody builds a second-order anti-alias filter, and plenty of designs do — so it is worth saying what the other terms in the decision are, because this essay’s two are not the only two and the conclusion changes when a third is added.

Component count and cost. Ten poles is five sections, five amplifiers, ten resistors and ten capacitors, against two poles’ one of each. On a design where the converter is a small part of a large system that is the whole argument, and it is not an argument about signals at all.

Group delay, which is that essay’s own second currency. A steeper filter is a longer delay and a less flat one. The essay before it puts the numbers beside each other: Chebyshev’s group delay varies 49 per cent over the band against Bessel’s 0.06, and the delay itself grows with order in every family. For anything in a control loop the delay is the binding constraint and the clock is not.

And realisability, which is the constraint that actually stops the argument. The band that does not close measures how much impedance level is available at each order and finds three decades at second order, one at sixth, and none at all at eighth — because fifty ohms of amplifier output resistance binds from one end and two picofarads of stray from the other, and every added section brings three more nodes and one more amplifier. So an eighth-order active cascade at a single impedance level does not exist, whatever its noise or its clock demand says.

That last one interacts with this essay’s finding in a way worth spelling out. The impedance level is the parameter this essay identifies as expensive in bits, and it is also the parameter that runs out with order. A design at order two can sit anywhere in three decades of it and should sit at the top, buying five bits over the bottom; a design at order six has one decade and 1.66 bits of choice; a design at order eight has none, so its resolution is whatever the one available level gives.

So the order does have a noise cost after all, and it is not the resistors’ — it is the loss of the freedom to choose the impedance level. Two thirds of a bit from the sections themselves, and up to three or four more from no longer being able to sit where the noise is lowest. That is a real cost, it is several times the one this essay measured, and it arrives through a constraint in a different field.

Still open: the amplifier that ends the impedance argument, and the family’s own noise

Where the amplifier takes over. The impedance level pays 1.66 bits a decade only while the resistors dominate. An amplifier with a few nanovolts per root hertz of its own stops that at some capacitance, and below it the level costs nothing and buys nothing — so the table above has a flat bottom that is not drawn. Locating it needs one number from a data sheet and turns this essay’s rule into a bounded one.

The family’s noise, beside its clock. The essay before it ranks five families by clock demand and this essay measures one of them for noise. Doing all five would say whether the ranking survives — Chebyshev demands the lowest clock of the all-pole families and has the highest-Q sections, so it may be the noisiest, and the two currencies would then rank the families oppositely. That is exactly the shape of result the essay before it reports for clock against delay flatness, and a third such conflict is worth knowing about.

And the section ordering, priced here. A section’s noise is filtered by everything after it, so a cascade’s noise depends on the order of its sections as well as on how many there are. The noise field measures the factor; what this essay needs is the best ordering’s noise rather than the realisation’s default, because that is the number a design can actually achieve and it moves the whole table up by an amount nobody has quoted.

What is checked

The clock demand is required to fall at every step of order, which is that essay’s own result and is the trade this figure is against, and required to fall by more than a factor of ten across the range so that the comparison has something in it.

The noise is required to rise with order and to cost less than one bit across the whole range. Both halves: the rise is the objection this essay is about and the size of it is the answer.

The impedance level’s cost is stated as an exponent and as a constant. Five thirds of a bit per decade of capacitance — which is a noise voltage going as the square root of a resistance going as one over the capacitance — and the same to four hundredths of a bit at every one of nine orders, which is what says the two parameters separate.

And the level is required to cost more than the order does, by more than a factor of two, because that comparison is the essay’s conclusion and the figure would otherwise be two curves with no claim between them.