The clock that is too fast

Three essays before this one have priced an anti-alias filter, and every number in all three is a lower bound on the sample rate. What the filter in front costs bisects the smallest rate at which a filter is down to a stated level by the frequency that folds back, and says so explicitly: the all-pole design falls at 48.2 decibels an octave, so every doubling of the clock is worth another 48 dB, and a faster clock is always an available answer. The floor every filter has qualified that — above a frequency a faster clock buys nothing, because the stopband has stopped falling — but it did not make a faster clock worse.

There is an arrangement in which it is. The three essays all take the signal to be at baseband, from direct current up to a passband edge, and a great many signals are not.

What the condition actually is #

The sampling theorem’s factor of two is not about a frequency; it is about a band, and the version everybody carries is the special case where the band starts at nought.

Sampling at $f_\mathrm{s}$ divides the frequency axis into Nyquist zones of width $f_\mathrm{s}/2$, and what a sampler does is fold every zone onto the first. A signal is recoverable exactly when it lies entirely inside one zone, because then nothing else folds on top of it. For a band from $f_L$ to $f_H$ that is: there is an integer $m$ with

$$\frac{m f_\mathrm{s}}{2} \le f_L \quad\text{and}\quad f_H \le \frac{(m+1) f_\mathrm{s}}{2}$$

For $m = 0$ the second condition is $f_\mathrm{s} \ge 2 f_H$, which is the familiar rule, and the first is vacuous. For $m \ge 1$ both bite, and rearranging gives a window rather than a bound:

$$\frac{2 f_H}{n} \le f_\mathrm{s} \le \frac{2 f_L}{n-1}, \qquad n = m+1$$

which is non-empty for $n$ up to $\lfloor f_H / B \rfloor$ with $B$ the band’s width. So a band away from direct current can be sampled at $\lfloor f_H/B \rfloor$ different rates rather than at one, and they are disjoint.

And the gaps between them — 200 to 240, 100 to 120, 66.67 to 80, 50 to 60, 40 to 48 megahertz — are rates at which the band straddles a zone boundary, so part of it folds onto the other part and no filter in front can separate them.

The clock that is too fast #

Read the table’s second and third rows together. A hundred megahertz works and a hundred and ten does not, and a hundred and twenty works again.

That is the sentence this essay exists for, because it makes that essay’s procedure fail. Raise the clock until the filter’s attenuation is enough is a procedure that terminates for a baseband signal and does not terminate here: raised out of the 80-to-100 window it enters a forbidden gap, and the signal is destroyed by being sampled faster.

What is destroyed is worth being precise about, because “aliasing” is doing a lot of work in that sentence. At a hundred and ten megahertz the zone boundaries are at multiples of fifty-five: 55, 110, 165. The band runs 100 to 120, so it straddles the boundary at 110 — the part below folds one way and the part above folds the other, and they land on top of each other reversed. The frequency a sample rate invents is what each half does individually, and the pair arriving at the same place is what makes it unrecoverable: two sinusoids, one set of samples, and no way to tell them apart.

Nothing in front of the sampler helps. A filter can remove what is outside the band and there is nothing outside the band — the whole signal is inside it, and it is folding onto itself. So this is a failure with no filter-shaped remedy at all, which is different in kind from every other failure on this sequence of essays.

Twice the width, not twice the top #

The slowest window’s lower edge is the useful number and it is forty megahertz — exactly twice the band’s twenty-megahertz width, and six times below the two hundred and forty that twice its top demands.

That is not a coincidence of this band. The last window is $n = \lfloor f_H/B \rfloor$, and as the band’s centre rises with its width fixed, its lower edge approaches $2B$ from above — so the absolute floor on a sample rate is twice the band’s width, whatever frequency the band sits at. Sweeping the centre:

Forty megahertz at every centre, and the number of available windows growing with the centre. So a band ten times higher in frequency needs no faster a clock and offers more choices of clock — which is the whole argument for the arrangement and is why a radio receiver samples at its intermediate frequency rather than after mixing to baseband.

The cost is stated in the table’s second column, and it is not a rate. The windows get narrower as the rate gets lower: eighty megahertz wide at the second zone, twenty at the third, and two at the fifth. A clock tolerance is a fraction of a window, so a design in the fifth zone needs its sample rate held to about four per cent and one in the second to about fifty. That is the exchange the arrangement actually makes — a slower clock for a tighter one — and it is nowhere in that essay’s arithmetic because a lower bound has no tolerance.

The two ways to be inside a zone, and why the low ones are narrow #

The window widths in the table fall away sharply — eighty megahertz, twenty, 6.67, two — and the pattern has a closed form worth having, because it is what a design’s clock tolerance is a fraction of.

The window for zone $n$ runs from $2f_H/n$ to $2f_L/(n-1)$, so its width is

$$\frac{2f_L}{n-1} - \frac{2f_H}{n} = \frac{2\left(n f_L - (n-1) f_H\right)}{n(n-1)}$$

and the numerator shrinks as $n$ rises while the denominator grows as $n^2$. The last non-empty window, at $n = \lfloor f_H/B \rfloor$, has a width that is a small fraction of the rate itself.

As a fractional tolerance on the clock — which is the form a design needs — the window at zone $n$ is about $B/f_H$ divided by $n$, near enough, so:

The last row is the honest one and it is why the “absolute floor is twice the width” result is a limit rather than a design. Forty megahertz is allowed and forty point one is not, so a clock at the floor has no tolerance whatever — and the useful slowest rate is the one whose window is wide enough for the oscillator that generates it, which for a crystal at a hundred parts per million is every row in that table and for a resistor-and-capacitor oscillator at five per cent is the first three.

Which makes the arrangement’s real design variable the zone number rather than the rate. Each zone down halves the clock and halves its tolerance, twice over — once from the numerator and once from the denominator — so the exchange is a good one for two or three zones and stops being one after that. The table above is a table of that exchange and it is the quantity that essay’s lower bound has no room for at all, because a bound is satisfied or not and a window has a middle.

What the filter becomes #

The filter in front is still necessary and it is a different filter, which changes that essay’s table rather than cancelling it.

For a baseband signal the requirement is a lowpass whose stopband starts at the frequency that folds back, and the earlier measurement prices that in clock rate. Here the requirement is a bandpass whose stopband must suppress the adjacent Nyquist zones — the zones above and below the one the signal is in, because their contents fold onto the signal exactly as a baseband signal’s out-of-band content does.

Three consequences, and the second is the one that makes the arrangement practical at all.

The transition width is the gap between zones, which is $f_\mathrm{s}/2 - B$. At the third zone with a twenty-megahertz band and an eighty-megahertz clock that is twenty megahertz of transition on each side, which is an undemanding filter — one part in five of fractional width, against a baseband anti-alias filter’s requirement of going from passband to stopband in the space between $f_\mathrm{pass}$ and $f_\mathrm{s} - f_\mathrm{pass}$.

The filter’s own centre frequency has to be right, and a bandpass filter’s centre is a component tolerance where a lowpass filter’s corner is too. So the arrangement trades a steep lowpass for a moderately steep bandpass whose placement matters, which is a real exchange and is why the filters are usually crystal or ceramic rather than built from resistors and capacitors.

And the noise bandwidth is the band’s, not the clock’s. A baseband converter behind a filter at half the sample rate admits noise over that whole width. A bandpass arrangement admits it over $B$ only, so the folded noise from every other zone is removed by the same filter that removes the signals there — and the noise arithmetic is the band’s width rather than the sample rate’s, which the window the square-root law has is the general statement of.

The arrangement seen the other way round #

Everything above treats the zone number as something to be chosen and the band as given. Turning it round gives the same arithmetic as a statement about what a converter can do, and it is the form in which the arrangement is usually met.

A converter with a stated sample rate $f_\mathrm{s}$ can digitise any band of width up to $f_\mathrm{s}/2$ that lies inside one of its zones — so a forty-megasample part is not limited to signals below twenty megahertz, it is limited to bands twenty megahertz wide and can take them from anywhere its own input bandwidth reaches. A part with a 250-megahertz input bandwidth and a forty-megahertz clock has twelve available bands, each twenty megahertz wide, at 0–20, 20–40, 40–60 and so on.

Two consequences, and the first is the one that surprises.

The signal’s frequency is not a converter specification. The rate and the input bandwidth are, and they constrain different things: the rate constrains the band’s width and the bandwidth constrains where the band may sit. A part’s data sheet quotes both and the arrangement uses them independently, where a baseband design uses only their ratio.

And the band must not straddle a zone edge, which at a fixed clock is a constraint on the signal rather than on the clock. A twenty-megahertz band at a forty-megahertz clock has to sit exactly in a zone — 40 to 60, or 60 to 80 — and a band from 50 to 70 is unsamplable at that rate however wide its guard bands are. That is the same refusal the floor every filter has reports for a stopband: a condition no amount of the usual remedy satisfies. That is the same fact as the forbidden gaps, with the roles of the two variables exchanged, and it is the reason intermediate frequencies in receivers are chosen as they are: the frequency plan and the sample rate are one decision.

The exchange between the two views is worth stating because it says which is the free variable in a real design. A receiver’s intermediate frequency is chosen for its filters, its image rejection and what crystals exist — so it is fixed early and by other people. The sample rate is then chosen from this essay’s windows, which is why the arrangement is normally met as “pick a rate from the allowed set” rather than as “pick a zone”. An exact answer to a different question is the reconstruction half of the same story, where the arrangement’s images have to be removed on the way out and the filter’s requirement has the arrow reversed.

Where the model stops #

The band has hard edges. Everything above requires the signal to be exactly nothing outside $f_L$ and $f_H$, and the frequency a sample rate invents is emphatic that no filter does that: a signal with nothing above a frequency is a signal that never started. So the real condition is a level rather than a boundary, and the windows have soft edges whose width depends on how far down the filter’s skirt is at the zone boundary. A window quoted as 80 to 100 megahertz is 82 to 98 at some stated rejection.

The sampler’s own bandwidth has to reach. Sampling a 120-megahertz signal at 80 megasamples a second needs a track-and-hold whose input bandwidth covers 120 megahertz, which is a converter specification independent of its rate and is the reason not every part can do this. It is also the specification that a picosecond, read as bits prices: aperture jitter costs a number of bits set by the input frequency and not by the sample rate, so a band at 210 megahertz sampled at forty pays the jitter of 210 megahertz.

And the zones are exact only for an exact clock. A sample rate off by a fraction moves every zone boundary by that fraction times $m$, so the band’s margin inside its zone shrinks with the zone number — which is the tolerance the table’s window widths are really about, and it is the arrangement’s one genuinely awkward requirement.

What this does to the three currencies #

The essay before it closes by naming three currencies a filter family is chosen in — clock rate, delay variation and noise bandwidth — and observing that they do not rank the families the same way, which is why it draws all three. This arrangement adds nothing to that list and changes what two of the three mean.

Clock rate stops being a scalar. It was a lower bound and it is now a set, so a family’s demand is no longer a single number to be compared with another family’s. What the filter’s skirt decides here is how far inside its zone the band sits — which is a margin rather than a rate — and a steeper filter buys margin rather than a slower clock. So that essay’s factor of 2.3 between Bessel and Chebyshev, which was a factor in the sample rate, becomes a factor in the guard band instead, and whether that matters depends on how much room the frequency plan already left.

Noise bandwidth changes by more than the arrangement does. A baseband converter admits noise over half its sample rate; this one admits it over the band’s width, which at the third zone is a quarter of that. So the arrangement is worth six decibels of noise before anything else is decided, and that is larger than the whole spread between the filter families’ own noise bandwidths — which the ratio that does not walk to one measures at 0.96 to 1.57 across thirty-two realised designs, a range of four decibels.

Delay variation is unaffected, and it is worth saying so because it is the one currency that transfers directly. A bandpass filter’s group delay varies across its passband exactly as a lowpass filter’s does, the band is the same width, and nothing about sampling it at a lower rate changes what the filter did to it.

Which leaves the arrangement’s own currency, and this essay has measured it: clock tolerance. It is not on that essay’s list because a lower bound has no tolerance, and it is the quantity that decides how far down the zones a design can usefully go. A fourth currency, ranking nothing about the families and everything about the oscillator.

Still open: the soft edges, and the jitter the zone number multiplies #

The windows at a stated rejection. The boundaries above are computed from a band with hard edges. Doing them at, say, sixty decibels of rejection — with a real bandpass response in front — would give windows narrower than these by an amount that is itself a function of the filter’s order, which converts this essay’s clean arithmetic into that essay’s currency: a filter order priced in clock tolerance rather than in clock rate.

The jitter that the zone number multiplies. A clock’s own jitter moves the sampling instants, and the phase error it produces on a signal at $f_\mathrm{in}$ is $2\pi f_\mathrm{in}$ times it — so sampling a 210-megahertz band at forty megahertz pays five times the jitter penalty of sampling a forty-megahertz band at the same rate. That is the arrangement’s real cost in resolution and it is nowhere in the window arithmetic, which is about placement only.

And the zone the signal lands in, spectrally reversed. Odd zones fold the band through with its spectrum intact and even ones reverse it, so half the windows in the table above deliver a mirrored signal. That is free to undo digitally and it is not free to forget, and which windows do it is a parity that the figure reports as a zone number and does not draw.

What is checked #

Every window is tested by placing the band rather than by evaluating the inequality. For a rate inside each window and a rate inside each gap, the figure asks directly whether there is an integer $m$ with the band inside the $m$-th zone — and requires the first set to pass and the second to fail.

That check found its own first version wrong. Recovering $m$ from $\lfloor 2f_H/f_\mathrm{s} \rfloor$ is correct at a window’s lower edge and off by one everywhere inside it, so six windows were refused that had just admitted — and the failure looked like the windows being wrong rather than the test. $\lfloor 2f_L/f_\mathrm{s} \rfloor$ is the right index, and the difference between the two is the whole content of a window having two edges.

The windows are required to be disjoint, with at least two gaps between them, because a set of overlapping windows would be a lower bound wearing a disguise and the essay’s claim is that they are not.

And the specific failure is required: that there is a working rate immediately below a gap, so a clock raised out of a window has been made worse by being made faster. That is the sentence the title is about and it is required rather than illustrated.