Circuit Response
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Series

Anti-alias — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · digital
  2. 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.

    part 2 · digital
  3. 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.

    part 3 · digital
  4. 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.

    part 4 · digital

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