Every draw shared, and nothing in common

A relaxation oscillator that drives a transformer can fail in two ways during a burst. Its timing can wander further than the system it clocks will tolerate, and its two half cycles can disagree in volt-seconds for long enough that the core runs out of flux. The count a core survives measured the second as a first-passage problem and closed with the observation that both failures are driven by the same thing: every threshold the comparator decides at carries one noise draw, and that draw lands in the timing and in the imbalance alike. A burst that fails one way, it reasoned, should be more likely than average to fail the other, and a specification that budgets the two separately would then be counting part of its margin twice.

The reasoning is natural, and it is wrong in an exact way. The two errors are made of the same draws and they are uncorrelated — not approximately, and not on average over some range of settings, but identically, as an algebraic consequence of how a draw enters a half cycle. What they do share is confined to the periods on either side of each other, with opposite signs, and over any whole number of periods those two terms cancel as well. Failing both ways is then exactly as likely as the product of failing each way, which is the arithmetic a designer who never thought about it would have used anyway.

Three draws, weighed twice #

The oscillator is the one this family of essays has used throughout: a comparator with hysteresis, a resistor from its output back to a capacitor on its input, and thresholds at ±β of the rails. Each threshold is armed with a fresh normal draw added to it and holds that draw until it is crossed — noise whose correlation time is long against a crossing and short against a half cycle. The decision taken where the ramp is slowest worked out what one draw does. It moves the crossing it was armed for, by the draw divided by the ramp’s slope at the threshold, and it leaves the capacitor at the wrong level, so the next half cycle starts from somewhere else and runs short or long by the draw divided by the slope at the other end of the ramp.

Write those two sensitivities as B for the first and A for the second. With a time constant RC and rails at ±V,

$$A = \frac{RC}{V(1+\beta)}, \qquad B = \frac{RC}{V(1-\beta)},$$

and at β = 0.5 with RC = 1 µs and V = 5 V they are 0.133 and 0.400 nanoseconds a millivolt. Every half cycle is then B times its own draw plus A times the draw that ended the half before it.

A period is a high half followed by a low half, so it contains three draws: the one that started the high half, with weight A; the one between the halves, which ends the first and starts the second, with weight A + B; and the one that ends the low half, with weight B. The high half less the low half — the quantity a core integrates, since the drive is +V for one and −V for the other — contains the same three draws. The first enters with A, as before. The middle one ends the high half and starts the low, so it enters with B from the first and −A from the second: B − A. The last enters only the low half, so it enters with −B.

The covariance of two weighted sums of independent draws is the sum of the products of the weights:

$$\operatorname{cov}(P, D) = \sigma^2\left[A\cdot A + (A+B)(B-A) + B\cdot(-B)\right] = \sigma^2\left[A^2 + B^2 - A^2 - B^2\right] = 0.$$

Nothing about the oscillator’s values survives into that line. It is zero at every β, every RC, every rail and every noise level, because the middle draw’s contribution, $B^2 - A^2$, is exactly paid back by the outer two. The hero figure is the check: forty thousand seeded periods, each plotted at its period error against its own duty error, and the cloud is round. The measured correlation is 0.0080, which is inside the ±0.0100 that forty thousand samples resolve. And because the draws are normal, a pair of weighted sums that are uncorrelated are independent outright, not merely uncorrelated in their linear part: the cloud is round at every radius, not only on average.

The same draw, read in two directions #

The identity looks like an accident of algebra and is better read as a consequence of the drive’s symmetry. A draw shared by the two halves of a period makes the first half long and the second half short, or the other way round, according to where it falls. In the period, the halves are added, and the shared draw counts with the sum of its two effects. In the duty error, the halves are subtracted, and it counts with their difference. The two outer draws each belong to one half only, so they count the same way in both readings, up to a sign.

A sum and a difference of the same two numbers are uncorrelated whenever the two numbers have equal variance — that is the whole content of the familiar fact that $X + Y$ and $X - Y$ are independent for independent normal $X$ and $Y$ of equal spread. The two halves of a period do have equal variance here, $\sigma^2(A^2 + B^2)$ each, because the oscillator is symmetric: the high half is the low half reflected through zero. They are correlated with each other, by the shared draw, but that correlation enters a sum and a difference with opposite sign and the equal variances do the rest.

That also says when the identity would fail. Give the two halves different noise — a comparator whose upper threshold is noisier than its lower, or a drive whose rails are unequal so the two ramps have different slopes — and the variances of the halves differ, and so does the covariance of their sum and difference, by exactly the difference of those variances. A symmetric oscillator with an asymmetric comparator is the case that would couple the two failures, and it is not the case the walk the core sees or its successors model.

Uncorrelated with itself, correlated with its neighbours #

A period and its own duty error share nothing linear. A period and the next period’s duty error do share something, because the draw that ends one period starts the next.

The draw that ends period n enters it with weight B. It also starts period n + 1’s high half, so it enters that period’s duty error with weight A. The covariance between a period and the following duty error is therefore $\sigma^2 AB$, and as a correlation, dividing by the two spreads, it is 0.1572 at β = 0.5. The measurement over forty thousand periods gives 0.155. Run the other way, the draw that starts period n ended period n − 1’s low half, where it carried weight −B in that period’s duty error, and in period n it carries A: a covariance of $-\sigma^2 AB$ and a correlation of −0.157, measured as −0.154. Two periods apart the two quantities have no draw in common at all, and the measured correlations at ±2, ±3 and ±4 are all under 0.011 in size.

So the statement “they share every draw” is true and incomplete. Each period’s error shares two draws with its own duty error, in a way that cancels, and one draw each with the duty errors of its two neighbours, in a way that does not. A long period predicts that the next period’s high half will run long against its low half, by about a sixth of a standard deviation for every standard deviation of period error, and that the previous one’s ran short. This is a real coupling, visible on a time-interval analyser that records both quantities cycle by cycle, and it is the only one.

Why a burst forgets the neighbours #

A burst-level failure is not about one period. A timing failure is the accumulated period error leaving a band, and a saturation failure is the accumulated high-minus-low time leaving the core’s headroom; both are walks, sums over every period so far. Summing changes which covariances matter, and it is here that the lag terms disappear.

Over N periods the summed period contains every draw: the first with A, the last with B and each of the 2N − 1 in between with A + B. The summed duty error contains the same draws: the first with A, the last with −B, and those in between alternately with A − B and B − A, because each interior draw is the middle of one period or the boundary between two. The covariance of the two sums is again a sum of products:

$$\sigma^2\Big[A^2 - B^2 + (N-1)(A+B)(A-B) + N(A+B)(B-A)\Big] = \sigma^2(A^2 - B^2)\big[1 + (N-1) - N\big] = 0.$$

Exactly zero after every whole number of periods, with the end draws included. In terms of the lags, every period’s +AB with its successor is matched by its successor’s −AB with it, and in a whole burst each pairing occurs the same number of times.

A burst is not read at one instant, though. A first passage asks whether the walk has ever left its band, so the timing walk at one count and the imbalance walk at another both matter. Their covariance is no longer exactly zero, because a truncation can cut a pair of lags in half and leave one term without its partner. But one term is all it can leave. The covariance of the timing summed to m periods with the imbalance summed to n is at most $\sigma^2 AB$ in size whatever m and n are, while the two variances grow as the counts — $2m\sigma^2(A + B)^2$ and $2n\sigma^2(B - A)^2$ for long walks. Measured across fifteen hundred bursts and computed exactly from the weight vectors, the largest correlation anywhere in the figure is 0.021, at the shortest timing walk; by ten periods it is 0.006 and by fifty 0.003, falling as the walks lengthen.

That is the sense in which the two walks share draws and not paths. Each step of each walk is built from the same noise, but the part of a step one walk keeps is the part the other throws away, and what is left over from that bookkeeping is a single draw’s worth at the edges of the window. The window the square-root law has found the same structure in a different place: whether an average gains as the square root of its count depends entirely on the covariances between the things averaged, and a covariance that is bounded while the variance grows is a covariance the long run cannot see.

Two first passages, one set of draws #

The decisive test is the one the question was actually about: run bursts, judge each against a timing limit and a headroom limit, and count how often both fail.

Four thousand bursts of two hundred periods were run at β = 0.5 and 5 mV of noise, each recording the largest excursion of both walks. Each limit was set at five percentiles of its own walk’s peak — so that the separate failure rates run from a half down to one in fifty — and every combination of the two was counted. If a burst that failed one way were more likely to fail the other, the dots would sit above the diagonal, increasingly so where the product is small, because a correlation shows most in the joint tail. They do not. The departures from the product are 0.69 standard errors rms over the twenty-five pairs, which is what sampling alone produces.

The practical reading is reassuring and slightly deflating. A designer who allots a burst a one per cent chance of a timing excursion and a one per cent chance of saturating the core has a failure rate of 1.99 per cent from either cause, and the overlap they might have worried about double-counting is one part in ten thousand. Budgeting the two separately is right. Budgeting them as though they were one event — tightening one limit because the other is already being watched — would be the error, because watching one tells nothing about the other.

What hysteresis does to the one correlation left #

The one non-zero coupling, between a period and its neighbours’ duty errors, depends on the hysteresis through A and B, and it vanishes at one end of the range.

With very little hysteresis the thresholds sit near zero and the ramp is equally steep at both ends, so A and B are nearly equal. The period’s weights then approach A, 2A, A and the duty error’s A, 0, −A — the middle draw drops out of the duty error completely — and the lag-one correlation becomes $A^2/(\sqrt{6}A\cdot\sqrt{2}A) = 1/\sqrt{12}$, 0.289. With wide hysteresis the thresholds approach the rails, the ramp arrives at each threshold almost flat, B grows without bound against A, and the shared draw’s weight in the neighbour’s duty error, A, becomes negligible beside everything else: 0.079 at β = 0.73, 0.013 at 0.95. The seeded checks at three settings agree with the closed form inside their resolution, and the same-period correlation is under 0.007 in size at all three.

So hysteresis, which two thresholds because there is a floor introduced to stop a comparator chattering and the walk the core sees found scaling the volt-second walk to β times the timing walk, also decides how visible the coupling between the two is from one cycle to the next. It decides nothing about whether the two failures are independent over a burst, which they are at every β.

Where the independence comes from, and where it would go #

It is worth being exact about which assumptions the result rests on, since each is an assumption a real oscillator can break.

The draws are independent from threshold to threshold. A comparator’s input noise with a correlation time longer than a half cycle carries part of each draw into the next, and then the weights in both sums pick up cross-terms between consecutive draws. The walk the core sees named a wideband floor as the opposite case — noise whose correlation time is short against the crossing itself — and found it would weaken the cancellation in the duty error. Either departure changes the per-period covariance from zero to something proportional to the draws’ own correlation, and would have to be measured.

The two halves are mirror images. As above, the cancellation is a sum and a difference of two quantities of equal variance. A rail asymmetry or a threshold asymmetry gives the halves different slopes at their thresholds, hence different A and B in each half, and the per-period covariance becomes the difference of the halves’ two variances. A skew of the kind the delay that is two delays measured — a comparator that is slower to switch one way than the other — does not do this, because it shifts the halves by constant amounts and adds no variance; it ramps the imbalance walk, as the count a core survives found, and leaves its covariance with the timing walk untouched.

The core is ideal. With winding resistance the imbalance walk leaks back towards zero, as the spread a winding resistance holds measured, and the accumulated imbalance becomes an exponentially weighted sum rather than a plain one. Its weights on the draws are then no longer equal from period to period, so the term-for-term cancellation of the lags no longer holds exactly; but the covariance it leaves is still a sum of single-draw terms, AB each, weighted by a geometric series, and so bounded by about AB times the leak’s memory in periods — against a variance that grows with the same memory. The independence survives, approximately, for the same reason it survives at unequal counts.

Reading a margin budget with this in hand #

A burst-mode drive is usually specified with a timing tolerance from the system it clocks and a headroom from the magnetic design, and the two are written by different people. The worry this essay started from is that the two specifications interact through the oscillator’s noise, and that the second person is relying on margin the first already spent. For a symmetric oscillator with threshold noise they do not interact. The rates multiply, the union is their sum less a negligible product, and each limit can be set from its own walk’s distribution — the inverse Gaussian of the count a core survives for the core, and the same law with drift zero and spread $\sigma\sqrt{2}(A + B)$ per root period for the timing.

There is a corollary that cuts the other way. Because the two walks are independent, observing that the timing has behaved well during a burst says nothing about the core. A system that infers its magnetic state from its clock — resetting a transformer only when the timing has drifted, say — is inferring from an uncorrelated signal. The one thing timing does predict is one period ahead: a long period makes the next duty error positive with correlation 0.157 at β = 0.5, and that is too weak and too short-lived to steer anything by.

The arithmetic, collected #

At β = 0.5, RC = 1 µs and 5 mV of threshold noise, A and B are 0.133 and 0.400 nanoseconds a millivolt. A period scatters by 3.40 ns and its high-minus-low time by 2.49 ns, and the two are uncorrelated: $\sigma^2[A^2 + (A + B)(B - A) - B^2]$ is zero for any A and B, measured as 0.008 over forty thousand periods. A period correlates with the next duty error by +AB/(σPσD) = 0.157 and with the previous one by −0.157, $1/\sqrt{12}$ with no hysteresis and 0.013 at β = 0.95. Summed over a burst the two lags cancel and the walks are uncorrelated at every whole count, within 0.021 at any pair of counts. Four thousand bursts fail both ways as often as the product of the separate rates, to 0.69 standard errors.

Still open: an asymmetric comparator, a noise with memory, and a reset that reads the clock #

An asymmetric comparator. If the upper threshold’s noise is larger than the lower’s, the halves’ variances differ, and the per-period covariance is their difference: $(B^2 - A^2)(\sigma_+^2 - \sigma_-^2)$, with each half weighing its own threshold’s draw by B and the other’s by A. How large an asymmetry is needed before the joint failure rate departs measurably from the product, for a burst of stated length, is the number that would say whether a real comparator’s data sheet ever needs reading on this point.

A noise with memory. A draw that carries part of itself into the next threshold makes consecutive draws correlated, and then the sum and the difference no longer separate. The covariance per period picks up terms in that correlation; whether it then grows with the burst or stays bounded is the comparison that would say which kind of comparator noise couples the two failures.

A reset that reads the clock. Some burst-mode drives reset their core between bursts by a fixed recipe and some by a measurement. With the two walks independent, a reset triggered by the timing walk carries no information about the flux; a reset triggered by the integrated drive carries all of it. Measuring the residual flux each strategy leaves, as a distribution, is the renewal question the count essay left beside this one. The walk that stops found the point a constant imbalance settles to once resistance is in the loop; a reset is the same question asked at the end of every burst.