The mean that makes a stopband

The order of the null, not the number of them put two means of twenty milliseconds in cascade and found the mains null widened tenfold: forty decibels held over 20.6 per cent of fifty hertz rather than the single mean’s 2.0, for a third more noise bandwidth and the same settling time. The factor was not a coincidence. A first-order null rises linearly through a depth and a second-order one quadratically, so the band held scales as $10^{\mathrm{dB}/(20n)}$ for a null of order $n$, and the ratio between the two grows the deeper the rejection asked for.

It closed by asking whether a third order pays. The width law promises a great deal; the noise bandwidth of more means rises; somewhere the two should balance. The measurement here finds something sharper than a balance. Three means do widen the band again, and four do something the width law does not describe at all: they stop holding a band and start holding everything above a frequency. Past a certain count the cascade of means is not a better null. It is a low-pass filter with a stopband, and the count at which that happens is a single division.

The cascade, and the response held to samples #

Each mean here is one mains period long, twenty milliseconds, so each has a null at fifty hertz and at every multiple of it. A mean of length $T$ has the response $\sin(\pi fT)/(\pi fT)$, and $n$ of them in cascade have that raised to the $n$-th power. The essay before this one held a fixed total length of forty milliseconds and compared one long mean against two short ones. That comparison cannot be extended to three: three means each nulling fifty hertz must each be a whole number of mains periods, so they are at least sixty milliseconds together, and at sixty milliseconds the only other arrangement that nulls the mains is a single mean of three periods. So the means are held at one period each, and the cascade’s length — its settling time — grows by twenty milliseconds with each mean added.

The closed form is checked against the thing it describes. A running mean of two hundred samples at ten kilohertz is twenty milliseconds, and $n$ of them applied in turn to a sampled sine give an output whose amplitude, read by correlation over a whole number of the sine’s periods, should be the closed form’s response. At two means and 47 hertz, three and 70, four and 120, the two agree to two parts in a thousand, which is the sampling’s own resolution of a twenty-millisecond window.

One identity #

The band a cascade holds below a depth $D$ is the set of frequencies where $|H|^n \le 10^{D/20}$. Taking the $n$-th root, that is exactly where $|H| \le 10^{D/(20n)}$ — the band a single mean holds at a depth of $D/n$.

Every arrangement from one to five means at depths from twenty to a hundred decibels lies on the single mean’s curve, to a part in a million. Two means at forty decibels hold 10.29 hertz, which is what one mean holds at twenty; three at sixty hold the same; four at eighty the same. The tenfold the earlier essay found is this identity at one point, and the width law $10^{\mathrm{dB}/(20n)}$ is its asymptote near the null, where a single mean’s response rises linearly.

The identity does more than reproduce the width law. It says the whole question of how many means to use is a question about one mean’s response at a fraction of the depth. And one mean’s response has a feature the width law ignores: it does not keep falling away from the null. Between the null at fifty hertz and the next at a hundred it rises to a sidelobe, $-13.26$ decibels, the largest value $|\sin x/x|$ reaches beyond its main lobe. A band defined as the region below a depth ends where the response climbs back through it, and for depths shallower than the sidelobe it does not climb back at all.

The count at which a null becomes a stopband #

So the cascade’s behaviour changes character when $D/n$ crosses 13.26 decibels. Below that, deeper than the sidelobe, it holds a band around fifty hertz, bounded on both sides. Above it, every frequency past the main lobe is below the depth: the response’s sidelobes, all at or below the first, have dropped out of the picture, and what is left is a low-pass filter whose stopband begins where the main lobe crosses $D/n$.

The count is the depth divided by 13.26, rounded up. Forty decibels needs four means: three are two tenths of a decibel short, their sidelobe at $-39.78$ decibels poking just above the line, which is why the hero figure’s three-mean curve holds a band from 40.7 to 69.5 hertz and then lets the region round 75 hertz through by a whisker. Sixty decibels needs five; a hundred needs eight.

The stopband’s edge barely moves with the depth. It sits where one mean’s main lobe crosses $D/n$, and since $n$ is chosen to bring $D/n$ just inside the sidelobe, $D/n$ is always between 10 and 13 decibels, and the edge always between 37 and 41 hertz. A cascade of means rejecting a hundred decibels from 40 hertz up is the same shape as one rejecting forty: more means, the same corner.

That is the answer to whether the third mean pays, and it is better than yes or no. At forty decibels the third mean nearly triples the band held — 28.8 hertz against 10.3 — and stops just short of the change of kind that the fourth delivers. At shallower depths the change of kind comes with fewer means: thirty decibels is a stopband with three.

What each mean costs #

More means of the same length pass less noise, not more: 25.00 hertz for one, 16.67 for two, 13.75 for three, 11.98 for four. Each additional mean narrows the cascade, which is why a cascade of four is a better filter in every sense than one mean of twenty milliseconds. The closed forms — a half, a third, eleven fortieths, 151 over 630 — are the classical integrals of $(\sin x/x)^{2n}$, and the quadrature returns each to a part in ten thousand.

The fair comparison is not with one short mean but with one long one, because the cascade takes $n$ mains periods to settle and a single mean of $n$ periods settles in the same time. That mean passes $1/(2nT)$: 12.50 hertz for two periods, 8.33 for three, 6.25 for four. Against it the cascade’s noise bandwidth is 1.333, 1.650 and 1.917 times larger, and the factor does not level off. For many means the cascade’s response tends to a Gaussian, $e^{-n(\pi fT)^2/3}$, and its noise bandwidth to $\sqrt{3/(4\pi n)}/T$ — sixteen means are 6.08 hertz, within three per cent of it — so the excess over the single long mean grows as $\sqrt{3n/\pi}$ without limit.

The earlier essay’s guess that the noise cost “rises towards a limit” is therefore wrong in the form it was put, and the correction is worth having. Measured against a single mean of equal settling time, every added mean costs more noise volts than the last, in proportion to the square root of the count. The cascade spends that noise on rejection the long mean cannot buy at any length, because a long mean’s nulls stay first-order however long it is: its forty-decibel band at fifty hertz is two per cent wide at a mains period and two per cent wide at a hundred.

A stopband makes the instrument indifferent to its mains #

A null is a property of one frequency, and an instrument built round it is built for one supply. One twenty-millisecond mean nulls fifty hertz completely and passes sixty at −16.1 decibels, so an integrating meter designed for a fifty-hertz grid and carried to a sixty-hertz one loses almost all its mains rejection — which is why such meters have a switch, or integrate over a hundred milliseconds, five periods of one supply and six of the other. A stopband removes the question. Each mean contributes its −16.1 decibels at sixty hertz, so two means reject it by 32.3, three by 48.4 and four by 64.6: the four-mean cascade that holds forty decibels above 36.9 hertz holds sixty-four at the other grid’s frequency without being told it exists.

The same is true of the mains’ harmonics, which a single mean nulls only at exact multiples of the fundamental and passes at the few hertz of wander each one carries. The millivolts in the wire found the supply’s current reaching a reading through a shared return, and a rectifier’s current is rich in harmonics; a reading that rejects the fundamental and lets the third harmonic’s wander through has moved the problem rather than solved it. Above the stopband’s edge every harmonic is below the depth whatever it is doing.

What a null cannot reach, a stopband does not need to: what actually fills a null found the depth of a notch set by its components’ errors, the floor every null in practice sits on. A cascade of digital means has no components to err, and a stopband reached by four of them has no narrow feature whose depth an error could fill.

The count a specification asks for #

Put together, the figures turn a specification into a count. A reading has to hold a stated depth over a stated band around the mains frequency, and each mean it uses costs a mains period of latency.

A mains supply that wanders by a per cent needs two means for anything up to seventy decibels and three beyond. A five per cent band, the kind a supply in a weak grid or a generator set presents, needs three means at sixty decibels and four at eighty. Ten per cent — a reading made beside a variable-speed drive whose interference sits near but not at the mains — needs four at sixty and six at a hundred. In every case the cost is the count times twenty milliseconds of latency and a noise bandwidth from the figure above, and in no case is a single longer mean an alternative, because its band does not widen.

The same limit as a chain #

The Gaussian that the cascade’s response tends to is not a coincidence of means. A mean’s impulse response is a rectangle, and $n$ rectangles convolved tend to a Gaussian by the central limit theorem, exactly as the delay a chain adds found $n$ exponential lags tending to a normal edge. Both cascades acquire a delay that adds — here, half a mains period a mean — and a spread that adds in quadrature, $T/\sqrt{12}$ a mean. A cascade of means is the noise-filtering form of a chain of stages, and its Gaussian limit is why its noise bandwidth falls as $1/\sqrt n$ rather than as $1/n$: the spread, not the length, sets the bandwidth.

The filter an average is found a single mean’s noise bandwidth exactly one over twice its window, and the bandwidth noise sees the factor of π/2 by which a single pole exceeds a brick wall at its corner. A cascade of four means sits between: its response is nearly Gaussian, its sidelobes are below fifty decibels, and its noise bandwidth is close to the half-power width — the shape a designer would draw if asked for a low-pass filter with a clean stopband and no ringing, made from nothing but sums.

The window and the cascade are the same choice #

A cascade of means is also a choice of window, and the bandwidth a bin is not met the same trade from the transform side. Two means in cascade weight their samples as a triangle; three as a piecewise parabola; four as a piecewise cubic, already hard to tell from a Gaussian. Each weighting has lower sidelobes than the last — 13.26 decibels a mean, so 26.5, 39.8 and 53.0 for two to four — and a wider main lobe, which is the noise bandwidth’s price. An integrating converter that cascades its integrations is choosing a window whose sidelobes fall by 13.26 decibels a stage, and the mains frequency is simply the place where every one of those windows happens to have a null.

Against the filters a designer would otherwise build #

The usual alternative to cascaded means is an analogue low-pass in front of the converter, and the comparison is instructive because the two buy their stopbands differently. The ratio that does not walk to one found the realised filter families’ noise bandwidths approaching their half-power widths only for some families and some orders; a Butterworth or a Chebyshev pays for a steep skirt in ringing, and its stopband attenuation grows with its order at twenty decibels a decade per pole. A cascade of four means reaches forty decibels from 36.9 hertz with no ringing at all, because each mean’s impulse response is positive and so is any convolution of them: the step never overshoots.

What the means give up is the shape of the passband. A twenty-millisecond mean is already half a decibel down at nine hertz, and four of them in cascade are 2.3 decibels down at ten, so a signal meant to be read up to ten hertz is read with a gain that depends on its frequency. That is no cost to an instrument that reads a slowly varying quantity once a reading, and it is a disqualification for one meant to follow a waveform. The cascade is a filter for readings, and the stopband it buys is free only because the passband it gives up was never wanted.

What the measurement leaves out #

Every mean is exactly one mains period. A converter whose period is set by its own clock nulls fifty hertz only to the clock’s accuracy, and the null of order $n$ is then $n$ nulls a hair apart, which is the split pair of the earlier essay arriving by accident. Close to the null that changes little, since the cascade’s band at any depth is set far from the null’s exact position; the stopband, once reached, does not care where the nulls are at all.

The means are ideal. A digital mean is exact, and a cascade of digital means is what most integrating instruments do after their converter. An analogue integration is a mean only to the accuracy of its integrator, and the integrator that leaks the zero away is the measurement of what a leak does to a null of this kind.

The number to keep #

A cascade of n means holds at a depth D exactly the band one mean holds at D/n. Once D/n is shallower than one mean’s first sidelobe, 13.26 dB, the null becomes a stopband: forty decibels needs four twenty-millisecond means and holds everything above 36.9 Hz, and the stopband’s edge stays between 37 and 41 Hz at every depth. Each mean costs a mains period of settling, and against a single mean of that length the cascade’s noise bandwidth is larger by $\sqrt{3n/\pi}$ — 1.333, 1.650 and 1.917 for two to four, with no limit.

Still open: the window that follows the supply, unequal means, and the residue a reading carries #

A length that tracks the supply. A cascade whose means follow a measured mains period holds its null wherever the supply goes, at the cost of a period measurement that is itself an average. With a stopband in hand the tracking matters less: whether four fixed means outperform two tracking ones against a supply that wanders by a stated amount is the comparison that would decide which an instrument should build.

Unequal means. Means of slightly different lengths put their nulls at slightly different frequencies and their sidelobes at different places, so a cascade of unequal means may reach a stopband with fewer stages than the 13.26-decibel rule allows. Whether three means of chosen lengths can hold forty decibels everywhere above a frequency, which three equal means miss by two tenths of a decibel, is a question with a definite answer.

The residue a reading carries. Everything here is a magnitude response. A reading taken through a cascade at a supply a per cent off carries a residual sinusoid at the beat between the supply and the window, and whether that reads as noise or as drift depends on how often readings are taken. The count of means sets its size; the reading rate decides what it looks like.