The order of the null, not the number of them

The filter an average is ended on a number that looks like a defect and is a theorem. A mean over a window rejects, exactly and completely, every frequency at which a whole number of cycles fits the window — which is why an integrating voltmeter integrates for a whole number of mains cycles. One per cent off that frequency it rejects by 39.9 decibels, and the window length does not enter: near the null at $k$ cycles the response is $\sin(\pi k \delta)/(\pi k(1 \pm \delta))$ and the $k$ cancels, so a five-cycle window and a fifty-cycle one both leave forty decibels at one per cent.

A mains supply wanders by rather more than one part in ten thousand over a day, and a reading taken through a shared return conductor has the supply’s current in it whether or not anything was meant to be measured at that frequency — the millivolts in the wire is where that arrives, and it is deterministic rather than random, so no amount of averaging touches it except through a null. So the infinite rejection an integrating converter advertises is, in practice, forty decibels — and a longer integration, which is the only knob the arrangement has, buys noise bandwidth and nothing whatever against the wander.

The obvious remedy is two means of slightly different lengths, putting a null either side of fifty hertz so that the trough between them covers the wander. This essay measures that against the thing it has to beat, which is not one mean.

The comparison that makes the answer honest #

Three arrangements, and fixing the right thing is most of the work.

A single mean over forty milliseconds. Its nulls are at 25, 50, 75 Hz and its noise bandwidth is 12.50 Hz.

An equal pair: two means of twenty milliseconds in cascade. Each has a null at 50 Hz, so the cascade has a null of order two there. Its noise bandwidth is $1/(3T)$ with $T$ the length of each, which is 16.67 Hz.

A split pair: two means of $20(1-s)$ and $20(1+s)$ milliseconds, whose nulls sit either side of 50 Hz at $50/(1-s)$ and $50/(1+s)$.

Every one of those has a total length of forty milliseconds, and that is the quantity held fixed rather than the noise bandwidth. The reason is settling. A mean over a window forgets everything older than the window completely, so it finishes in exactly one window; two means in cascade are a ramp convolved with a ramp, and their support is the sum of the two lengths. All three arrangements deliver a settled reading forty milliseconds after anything changes, to the last microsecond, so a reader comparing them is comparing rejection against noise and not against speed.

Fixing the noise bandwidth instead — which is the convention the neighbouring essays use, because equal noise is what a reading is designed to — would have made the comparison unreadable here. Two equal means with a null at 50 Hz must each be a whole number of mains cycles long, so their noise bandwidth is not a free parameter: it is $1/(3T)$ at whatever $T$ the mains period allows. The arrangement’s length is quantised and its noise bandwidth with it, which is why the length is the thing to hold and the noise is the thing to report.

Ten times the band, from the order of the zero #

The equal pair holds forty decibels from 45.3 to 55.8 Hz — 20.59 per cent of fifty hertz, against the single mean’s 2.00 — for 1.333 times the noise bandwidth and the same settling time.

A factor of ten, and it is not a coincidence of these numbers. A first-order null’s response rises linearly through it, so the offset at which the response reaches a level scales as that level’s amplitude, $10^{\mathrm{dB}/20}$. A second-order null’s rises quadratically, so its offset scales as the square root of that, $10^{\mathrm{dB}/40}$. The ratio of the two is $10^{-\mathrm{dB}/40}$:

So the order of the zero is worth more the deeper the rejection asked for, which is the opposite of how a rule of thumb would be quoted. It is the same exponent arithmetic what actually fills a null finds in a notch’s components: two errors that leave a resonance alone act second order and two that move it act first, and the two columns of its table are twenty decibels a decade apart for exactly this reason. At forty decibels it is a factor of ten; at eighty it is a hundred. A specification asking for sixty decibels over a per cent of frequency is unreachable with one mean of any length and comfortable with two.

This requirement also had to be written twice, and the first version is instructive about what a round number costs. It said the second-order null was “an order of magnitude wider”, which is true at forty decibels and at no other depth, and the measurement refused it at thirty — where the ratio is 6.3 and the band is forty per cent wide, which is far enough from a null for the linear-and-quadratic argument to have stopped applying at all. The law is asymptotic in the depth, and the figure’s slider starts at forty decibels for that reason rather than being given a looser tolerance.

What a third more noise bandwidth is, and what it is not #

The equal pair passes 16.67 Hz against the single mean’s 12.50, and the ratio $4/3$ is exact: $1/(3T)$ against $1/(2\cdot 2T)$.

In volts that is 1.155, a decibel and a quarter, and it is worth saying what it is not. It is not a longer measurement — the settling is identical. It is not a worse filter in any general sense — the cascade is nearer a brick wall than the single mean, as the figure above shows, because its noise bandwidth is 1.045 times its half-power point against the mean’s 1.129. What the cascade spends the extra bandwidth on is the top of its passband: it is a triangle rather than a rectangle in time, so it weights the middle of the window heavily and the ends lightly, and a weighting that throws away the ends of a record keeps less of the noise in it and less of the signal too. The same taper is what the ratio that does not walk to one is measuring when it finds a cascade’s noise bandwidth nearer a brick wall than a two-pole Butterworth’s.

That is the same arithmetic as the previous essay’s windows, arriving from the other side. A Hann window is a weighting applied to a record before a transform and costs 1.5 bins of noise bandwidth against a rectangle’s one; two means in cascade are a triangle applied to a record before an average and cost 4/3 against a rectangle’s one. The bandwidth a bin is not prices the first and this prices the second, and both are the same statement: a taper buys something in the frequency domain and pays for it in area under the square.

Which is the cleanest way to state the whole trade. A decibel and a quarter of noise, no extra settling time, and a factor of ten in the band over which the mains is rejected. For any measurement where the mains is the interference that matters, that is not a close decision.

The split buys exactly root two, at a cliff #

Now the arrangement the previous essay proposed: two lengths straddling the frequency rather than equal to it.

It works, and it is worth a precise factor rather than an impression. Across every depth measured the widest split trough is $\sqrt2$ times the equal pair’s — 1.4539 at forty decibels, 1.4244 at fifty, 1.4173 at sixty, 1.4143 at seventy and 1.41429 at eighty, against $\sqrt2 = 1.41421$ — and the noise bandwidth is very slightly lower than the equal pair’s, 16.53 Hz against 16.67. So the split is free and it is worth forty-one per cent more band.

And it is unusable, which is the finding rather than a disappointment.

The second column of that table is the optimum and the cliff is immediately beside it. At forty decibels the widest trough is at a split of 10.117 per cent and there is no trough at all at 10.118: the peak between the two separated nulls, which is $\left[\sin^2(\pi s)/\pi^2(1-s^2)\right]^2$ and rises steadily with the split, has climbed through the target, and fifty hertz is no longer inside the band. One part in ten thousand of the split, which is two parts in ten million of a window length, separates the best trough available from none.

The tolerance is looser at greater depth and it is never loose. At eighty decibels the optimum is at a split of 1.000 per cent and the cliff at 1.000 per cent to four figures. A design asking for the $\sqrt2$ must hold two window lengths to a part in ten thousand of each other and know the mains frequency to the same precision, which is the quantity the whole arrangement was built because nobody knows.

So the answer to the question the essay before it left open is no, and the reason is worth carrying: what widens a null is the order of the zero, not the placement of two of them. The tenfold is free and robust; the $\sqrt2$ costs nothing and cannot be held. The design is the equal pair, and the correct number of window lengths to trim is zero.

A whole number of cycles of what #

The lengths in the tables above are not free, and the constraint is the reason the split is quantised in practice as well as fragile in principle.

A mean nulls a frequency only if a whole number of its cycles fits the window. So an equal pair nulling fifty hertz needs each of its two means to be $k/50$ seconds — twenty milliseconds, or forty, or sixty — and the pair’s noise bandwidth follows from the choice rather than being chosen: $1/(3T)$ at whichever $T$ is taken. There is no arrangement of two equal means with a null at fifty hertz and a noise bandwidth of, say, 14 Hz. The quantity is quantised in steps of a mains cycle.

That has three consequences worth separating.

Both supplies at once costs a hundred milliseconds. Five cycles of fifty hertz and six of sixty are both a hundred milliseconds, which is why an integrating converter offering “50/60 Hz rejection” has a hundred-millisecond setting and why its faster settings reject one supply and not the other. As an equal pair that becomes two means of fifty milliseconds — two and a half cycles of fifty hertz, which is not a whole number — so the pair must be two hundred-millisecond means, and the reading takes two hundred milliseconds rather than one. The order of the null is bought in units of the whole arrangement’s length, not in units of a cycle.

The split pair is not quantised at all, which is what makes it look attractive and is the same property that makes it fragile. Its two lengths are deliberately not whole numbers of cycles — that is the whole idea — so nothing snaps them into place and nothing holds them there. The equal pair’s lengths are held by an integer; the split pair’s are held by whatever generates the timing.

And the timebase is now in the answer. A window of twenty milliseconds means twenty milliseconds of the instrument’s own clock, and a clock one part in a thousand fast makes the window one part in a thousand short — which is the same offset as the supply being one part in a thousand high, and is indistinguishable from it in the reading. The forty decibels the single mean leaves at one per cent is therefore a budget shared between the supply’s wander and the instrument’s timebase, and the second of those is the one a design controls.

The depth and the width are different questions #

It is worth putting this beside the filters field’s own account of a null, because the two essays measure quantities that sound the same and are not.

What actually fills a null measures the depth of a notch and finds two mechanisms: a tolerance on the arm’s components moves the null rather than filling it, and loss in the arm fills it at twenty decibels per decade of series resistance, so that the depth is the arm’s quality factor in decibels and nothing else. Its conclusion is that a trimmer converts a tolerance problem into a loss problem and that loss usually binds.

Nothing in this essay has a depth problem. A mean’s null is exactly zero — it is a cancellation in arithmetic rather than an impedance going to infinity, so there is no arm, no quality factor and no loss to fill it. What it has instead is a width problem, which a notch built from components has too and which that essay states in one line: sixty decibels of rejection is available over one part in a thousand of frequency, and every twenty decibels of depth costs a factor of ten in width.

That is the same law as this essay’s first table, for the same reason — a response passing linearly through zero has twenty decibels per decade of offset in it, whichever variable the offset is in — and the two essays are the two variables. There the offset comes from the components moving; here it comes from the signal moving. The remedy differs accordingly: a notch is trimmed onto the interference, and a mean cannot be, because the interference is what moves.

What is not measured here #

The reading’s own front end. Everything here is an average of a continuous signal. An instrument averages samples, and the window the square-root law has shows the noise bandwidths quoted above to be the right ones only when the filter in front of the sampler sits inside a band about a factor of three wide — so the decibel and a quarter this essay prices is the smaller of the two corrections a real chain carries.

A mains frequency that is known. The whole arrangement exists because the supply wanders, and it would be a different design if the period were measured. A window whose length tracks a measured period puts the null back on the interference and the residual becomes the measurement error of the period rather than the wander of the supply — which is a better arrangement than either of these and needs a frequency measurement this essay does not contain.

Harmonics. A mean’s nulls sit at every multiple of one over the window, so an arrangement nulling fifty hertz nulls a hundred and a hundred and fifty as well, and the trough widths at the harmonics are not the same as at the fundamental: near the null at $k$ cycles a fractional offset is $k$ times the absolute one. A supply one per cent off frequency is one per cent off at every harmonic, so the fractional widths above apply throughout — but an interference that is a fixed number of hertz away is not, and which of those a real disturbance is has not been established.

And the taper stops at two. Three means in cascade give a third-order null and, by the same argument, a width scaling as $10^{\mathrm{dB}/60}$ — a factor of a hundred over one mean at forty decibels rather than ten. The noise bandwidth of three equal means of a third the total is higher again, the settling is still the total length, and whether the sequence keeps paying has not been measured. There is a reason to think it stops: the cascade is converging on a Gaussian weighting, whose noise bandwidth keeps rising while its null order keeps rising too, and nothing here says which wins. The cascade’s own noise penalty is measurable at every order by the same quadrature the filter an average is uses, and nothing about the arrangement makes it hard — it simply has not been done.

Still open: the window that follows the supply, and the order that stops paying #

A length that tracks a measured period. Nulling the supply exactly requires knowing its period, and a converter that measures the period and sets its own integration time turns a forty-decibel rejection into one bounded by the period measurement. That measurement is itself an average with a noise bandwidth, so there is a loop: a longer period measurement is more precise and slower to follow a supply that is drifting. Pricing that loop would say what rejection is available against a supply whose frequency moves at a stated rate, which is the specification a real instrument is written to and which no figure here supplies.

Whether the third order pays. The width law is $10^{\mathrm{dB}/(20n)}$ for a null of order $n$, which promises a great deal, and the noise bandwidth of $n$ equal means of total length $L$ rises towards a limit as $n$ grows. Both sequences are computable in closed form and their ratio has a maximum somewhere. Finding it would turn “two means” from the arrangement this essay happens to have measured into the arrangement that is best, or say why it is not.

And the trough measured where it is used. Everything here is a magnitude response. What a reading actually carries after an imperfectly nulled mains is a residual sinusoid at the beat between the supply and the window, and its size in the reading is the response at the offset while its rate is the offset itself. A reading rejected by forty decibels wanders at half a hertz, and whether that reads as noise or as drift depends on how often the reading is taken — which is a sampling question, and the window the square-root law has is where this sequence of essays has the arithmetic for it.

What is checked #

Each arrangement’s noise bandwidth is held against its own closed form before anything is compared: $1/(2T)$ for one mean, $1/(3T)$ for the equal pair. The unequal cascade has no closed form and is integrated by quadrature on panels bounded by the sorted union of both windows’ nulls, so that no panel straddles either — and that integral is required to return the equal pair’s closed form as the split goes to zero, which is what says the panelling is right at all. The equal case itself is refused rather than integrated, because the mean of $\sin^4$ is three eighths where the mean of $\sin^2\!\cdot\sin^2$ is a quarter, and the tail term differs accordingly.

The tenfold is stated as the law and not as a factor. The ratio of the two widths against $10^{-\mathrm{dB}/40}$ to four per cent, at every depth the slider offers. Requiring “an order of magnitude” instead passed at forty decibels and was refused at thirty, which is how the law was found.

The root two is required to four per cent, at every depth, for the same reason: it is an exact factor and quoting it loosely would leave it looking like a coincidence of one window length.

The cliff is required by bisection on the condition that causes it — the response at fifty hertz rising above the target — and not on the width. The width has a sliver in it a few parts in ten thousand wide, because the peak between the two separated nulls sits at 49.983 Hz rather than at 50, so there is a range of splits where the response at fifty is still under the target and the peak beside it is not. A bisection on the width converges into that sliver and reports it as the optimum, which is what the first version did: it returned a trough of 17.3 per cent where the answer is 29.9.

And where there is no cliff, its absence is required. At twenty decibels the peak between the nulls never reaches so shallow a target within the sweep, so the figure has no collapse to find — and “none found” is an answer about the depth asked for rather than a failed search, which is the difference the check is written to keep.