Before the steady state

How many relaxations a settling curve holds

A settling curve read at eighty times, each good to ten parts in a million of the step, returns the amplitudes of three relaxations spread over two decades to 0.22 per cent and of eight to 46 — even when the fit is handed every time constant exactly. The condition number of the inversion grows by a factor of 7.7 for each extra relaxation packed into a decade, so each tenfold improvement in the reading buys a little more than one more relaxation a decade and no more. A settling measurement can count the dielectric's slowest relaxations; it cannot resolve a crowded spectrum, and how crowded is readable before any amplitude is.

Assumes: The cancellation that leaves a tail · The capacitor that remembers

The slowest one decides it ends with a measurement it could not yet make. Its ratio test — settle to one band and to a band ten times tighter, and divide the difference by $\ln 10$ — returns the slowest relaxation in a spread tail, eighteen per cent low at the bands anybody would use and to half a per cent at bands three decades deeper. The convergence of that number as the bands deepen, it noted, is itself a measurement of how crowded the spectrum is. And it named the next question: repeating it at several bands should return the whole spectrum — which would turn a settling measurement into a dielectric characterisation. Whether the inversion is well conditioned is the question.

It is not well conditioned, and how badly is a number with a clean dependence on one thing: how many relaxations there are per decade of time constant.

The most generous version of the question

The inversion here is given every advantage a real one would not have. A one-per-cent tail is divided equally among $N$ branches spread geometrically over two decades from 0.1 to 10 milliseconds — the arrangement the essay on the slowest doublet swept — and its error curve is read at eighty times, spaced logarithmically from a hundredth of a millisecond to beyond three hundred milliseconds. A settling measurement at many bands is the same information read the other way round: instead of the error at a chosen time, the time at a chosen error.

Every reading carries seeded Gaussian noise of a stated size, ten parts in a million of the step unless the figure says otherwise, which is roughly a seventeen-bit reading of a full-scale step. And the fit is given the time constants. It has only the amplitudes to find, which makes it linear: $e(t_j) = \sum_k a_k e^{-t_j/\tau_k}$ is a matrix times a vector, and least squares solves it. A real characterisation would not know the time constants either, and a fit for both is non-linear and strictly harder than this one. Whatever this version cannot do, no version can.

Fig. 1 Three relaxations over two decades, recovered from eighty readings each carrying ten parts in a million of noise, with the time constants given to the fit. Dots: one recovery, as a fraction of the true amplitude. Bars: the rms spread over two hundred seeded readings. The slider is how many relaxations share the two decades.

With three relaxations a decade apart, the recovery is excellent: the worst branch comes back with an rms error of 0.22 per cent. Drag to six and it is 5.4 per cent; to eight, 46 per cent — an amplitude that comes back, on a typical reading, anywhere between half and one and a half times what it is, from a curve read to ten parts in a million.

Nothing about the tail changed between three and eight except how the same one per cent is divided, and the curve being read is almost indistinguishable between the two. That is the whole difficulty in one sentence: eight exponentials two and a half decades wide sum to a curve that three exponentials can imitate to within the noise, and a fit cannot choose between two answers that the data do not.

The number that predicts it

Least squares turns noise on the readings into error on the amplitudes through the fit matrix, and the factor it multiplies by in the worst direction is the matrix’s condition number — the ratio of its largest singular value to its smallest. Here the matrix has a row per reading and a column per relaxation, each entry $e^{-t_j/\tau_k}$ , and its singular values are computed directly, by one-sided Jacobi rotations on the matrix itself.

That last detail is not pedantry. The textbook route forms the normal equations, whose condition number is the square of the matrix’s; at twelve relaxations over two decades the matrix’s is 8.1×10⁴, so the normal equations’ would be 6.5×10⁹, and a problem that is merely bad would have been made numerically meaningless before the measurement of it began. The matrix that is ill and the answer that is not draws that distinction for a circuit solve; here it decides whether the instrument measures the problem or itself.

Each extra relaxation a decade multiplies the fit's condition number by 7.7. computed by solving, not by drawing. The condition number of the least-squares fit that recovers a spread tail's amplitudes from 80 readings of its error curve, with the time constants given, for two to twelve branches spread over two decades. It rises from 3.3 to 8.09e+4, and its logarithm is a straight line in branches a decade with a slope of 0.884 — a factor of 7.7 of noise amplification for each relaxation added to a decade. — Fig. 2 The condition number of the fit that recovers the amplitudes from eighty readings, with the time constants given, for two to twelve relaxations spread over two decades, against relaxations a decade. Its logarithm is a straight line with a slope of 0.884: every extra relaxation packed into a decade multiplies the noise amplification by 7.7.

relaxations in two decades	a decade	neighbours apart by	condition number
2	0.5	100×	3.29
3	1.0	10×	8.29
4	1.5	4.64×	23.1
5	2.0	3.16×	65.1
6	2.5	2.51×	182
8	3.5	1.93×	1,410
12	5.5	1.52×	8.09×10⁴

The growth is geometric in the density, not in the count. The same eight relaxations spread over eight decades instead of two would be one a decade, and would recover like the three here. It is the ratio between neighbouring time constants that matters: two exponentials whose rates differ by a factor of ten look different over the decade in which one of them decays, and two whose rates differ by a factor of 1.9 look alike over every decade.

What each tenfold of reading buys

Because the rms error of each amplitude is the reading’s noise multiplied by a factor that grows with the condition number and falls as the square root of the number of readings, the error scales linearly with the noise, exactly — the same relaxations at a tenth of the noise come back ten times closer. So the table of errors is one table, shifted:

relaxations in two decades	reading to 10⁻⁴	to 10⁻⁵	to 10⁻⁶
2	0.86%	0.086%	0.0086%
3	2.2%	0.22%	0.022%
4	6.5%	0.65%	0.065%
5	18%	1.8%	0.18%
6	54%	5.4%	0.54%
8	460%	46%	4.6%

Read along a row at a fixed accuracy — say ten per cent on every amplitude, which is enough to call a relaxation measured — and the resolvable density climbs by a step each time the reading improves tenfold: four relaxations in the two decades from a reading to a part in ten thousand, six from a part in a hundred thousand, eight from a part in a million. With the condition growing as $7.7^{\,n}$ in relaxations a decade, each decade of signal-to-noise buys $1/\log_{10} 7.7 = 1.13$ relaxations a decade, and nothing else buys any.

6 relaxations over two decades come back within 5.4% from readings good to 1e-5. computed by solving, not by drawing. A one-per-cent tail divided equally among 6 branches spread over two decades from 0.1 ms, read at 80 times with 1e-5 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 1.82e+2, and the worst branch comes back with an rms error of 5.40 per cent. — Fig. 3 Six relaxations over two decades — neighbours 2.5 times apart — from the same readings. The rms error is 5.4 per cent on the worst branch, and on this reading the errors mostly alternate in sign between neighbours: a fit that overstates one relaxation tends to understate the one beside it, because the pair’s sum is what the curve actually constrains.

The alternation in that figure is the structure of the ill-conditioning made visible. The smallest singular value belongs to a combination of columns in which neighbouring relaxations enter with opposite signs, since two neighbouring exponentials are nearly the same function and their difference is nearly nothing. Noise along that combination moves the recovered amplitudes up and down in alternation while barely changing the curve they sum to. The sum of each neighbouring pair is well determined; the split within it is not.

That is a useful thing to know about what a measurement can return, because it says what to ask for instead of the full spectrum. The amplitude in a decade — the sum of whichever relaxations fall in it — is recoverable when the individual ones are not.

Why more readings do not rescue it

The obvious remedy for noise is to take more readings and average, and against this problem it is weak. The error on each amplitude falls as the square root of the number of readings, other things equal, so a hundred times as many readings buys a factor of ten — the same as one decade of signal-to-noise, and so the same 1.13 relaxations a decade. Ten thousand readings where eighty were taken buys about two more relaxations in the two decades.

And the readings are not free in the way that averaging assumes. The slow end of the curve has a ten-millisecond time constant and has to be followed past three hundred milliseconds, so every repeat of the reading is a repeat of the whole slow step; the corner where averaging stops working is what happens to a long average when the instrument’s own drift has a slow component of its own, and a dielectric measurement’s slowest relaxations sit exactly where a bench’s drift does.

So the resolving power of a settling measurement is set by the reading’s accuracy and almost nothing else, and the accuracy of a practical reading is a few parts in a hundred thousand of a step. That puts the resolvable density near two relaxations a decade.

8 relaxations over two decades come back within 4.6% from readings good to 1e-6. computed by solving, not by drawing. A one-per-cent tail divided equally among 8 branches spread over two decades from 0.1 ms, read at 80 times with 1e-6 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 1.41e+3, and the worst branch comes back with an rms error of 4.59 per cent. — Fig. 4 Eight relaxations over two decades from readings a tenfold better — a part in a million of the step. The worst branch comes back to 4.6 per cent, and the eight errors of this one reading alternate in sign from the fastest branch to the slowest without a single exception. A reading that good buys two more relaxations in the two decades than the reading drawn above it, and it would cost a twenty-bit measurement of a settling step.

A practical reading, and what it resolves

The table’s left column is the one worth dwelling on, because it is the reading a bench actually gets. A settling measurement read to a part in ten thousand of the step is roughly what a fourteen-bit converter delivers after the step has been scaled to fill it — the floor a converter sets is the arithmetic — and an oscilloscope with a clamped input and a careful offset does no better.

4 relaxations over two decades come back within 6.5% from readings good to 1e-4. computed by solving, not by drawing. A one-per-cent tail divided equally among 4 branches spread over two decades from 0.1 ms, read at 80 times with 1e-4 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 2.31e+1, and the worst branch comes back with an rms error of 6.49 per cent. — Fig. 5 Four relaxations over two decades, neighbours 4.6 times apart, from readings good to a part in ten thousand. The worst amplitude comes back with an rms error of 6.5 per cent. At this reading two relaxations a decade is the limit of what the curve holds; a fifth relaxation in the same two decades takes the worst error to eighteen per cent.

At that reading the curve holds four relaxations in two decades to a few per cent, and a fifth is already a guess. That is a coarse instrument for a spectrum and a precise one for the question settling actually asks, because the slowest relaxation, which dominates every settling time the essay on the slowest doublet measured, is recovered in all of these cases to better than the worst branch’s figure: it is the relaxation with the most curve to itself, since nothing slower is present to share the late readings with it. At eight relaxations from readings to 10⁻⁵ the middle branch comes back to 46 per cent and the slowest to 4.8; at four from 10⁻⁴, the slowest to 2.9 per cent. The two ends of a spectrum each have a side with no neighbour, and the ill-conditioning lives in the middle.

The spread of recovered amplitudes over two hundred seeded readings is also a distribution in its own right, and it has the shape the error that is a distribution describes for a mirror’s mismatch: centred on the truth, symmetric, and with a width set by how the noise projects onto the fit. It is not a bias, and averaging many independent characterisations of one part would converge on the true amplitudes — slowly, at the square root of their number, which is the same arithmetic as more readings and buys the same 1.13 relaxations a decade per hundredfold.

The same limit in a circuit solve

The problem has an exact counterpart in the circuit solves of nodal analysis, and the counterpart explains why the growth is geometric rather than something milder.

Where the matrix is worst, and where the answer is not measures a nodal matrix whose condition number swings over many decades with frequency while the answer it returns barely changes. The resolution there is that the condition number bounds the worst direction and the answer does not lie along it. Here the answer lies squarely along it: the split between neighbouring relaxations is exactly the combination the smallest singular value belongs to, so the pessimistic bound is the realised error rather than a ceiling nobody reaches.

And the digits the arithmetic did not have is the reason the fit had to avoid the normal equations: a condition number is a count of digits lost, and squaring it loses twice as many. At eight relaxations the fit loses about three of a double’s sixteen digits and the reading loses the rest — a reading to 10⁻⁵ against amplitudes of 10⁻³ has two digits to spare before the fit begins, and a condition of 1,410 spends them.

That last sentence is the whole problem restated in digits, and it gives a rule of thumb that needs no table: the digits of accuracy in the reading, minus the digits of amplitude below the step, minus the logarithm of the condition number, is how many digits the recovered amplitudes have. When the answer is below one digit the relaxation has not been measured.

What the ratio test was measuring

The ratio test from the essay on the slowest doublet fits into this picture as its most conservative case. It asks for one number — the slowest time constant — from the deepest part of the curve, where only one relaxation is left, and a single exponential is the best-conditioned fit there is. That is why it converges, and why it needed bands three decades deeper than the practical pair to converge well: at shallow bands two or three slow relaxations are still contributing and the one-exponential model is simply wrong, not badly conditioned.

Its convergence as the bands deepen — eighteen per cent low, then three, then half a per cent — is the tail peeling off relaxations one at a time. The first band pair sees a blend of the three slowest, the next sees a blend of two, the last sees one. A measurement that recorded the ratio test at every band pair would be recording, in order, how many relaxations lie within each decade of the slowest, which is exactly the “amplitude in a decade” that the fit above can recover when the individual amplitudes are lost.

So the two instruments agree about what is measurable: the slowest relaxation, cleanly; the weight in each decade below it, roughly; and the individual relaxations only when they are a decade apart or the reading is very good.

What a characterisation can therefore claim

A settling measurement is a real instrument for a dielectric, and this puts a boundary around what it can report.

It can report the slowest relaxation. That is the quantity the slowest one decides it found governs settling, it is recoverable to a per cent, and no data sheet prints it. This is the useful output and it is fully available.

It can report the weight per decade. The sum over a decade is a well-conditioned combination, so a histogram of absorption by decade of time constant is measurable at the densities real dielectrics have.

It cannot report a line spectrum finer than about two a decade from a reading anybody would make, and any characterisation that claims to — a fitted model with eight or ten discrete relaxations in two decades, each with an amplitude to three figures — is reporting the fit’s regularisation, not the part. That includes the discrete model this collection uses, and it is why the models of dielectric absorption in the capacitor that remembers are stated as models: their branches stand for weight in a decade, and nothing measured would distinguish them from a different set with the same weights per decade.

That last point cuts back to the settling problem itself, and in a useful direction. The tail the source decides found that a driven capacitor’s tail depends on the branch resistances, not on the split of amplitude between neighbours; and the settling time depends on the slowest relaxation and on the logarithm of the weight near it. Both depend on the well-conditioned quantities. The spectrum a measurement cannot resolve is the part of the spectrum settling does not care about.

How the recovery was set up

The fit is linear least squares, solved through the singular value decomposition of the eighty-by- $N$ matrix of exponentials, computed by one-sided Jacobi rotations on the matrix itself so that its condition number is measured and not squared.

With no noise the fit returns the amplitudes it was given to rounding the condition number allows — a part in a million multiplied by the condition number is the tolerance, and at eight relaxations the worst branch is far inside it. That is the check that the solver is not the thing being measured.

The noise is seeded — a fixed random sequence per relaxation count — so every figure on this page is the same figure on every reading, and the two-hundred-run spread is reproducible to the last digit.

The condition number is required to increase with every relaxation added to the same two decades, and its logarithm to lie within 0.35 of a straight line in relaxations a decade. The slope is 0.884; the straight line is the finding, and a curve would have meant the growth depended on something other than the density.

Given time constants, equal amplitudes and white noise

The time constants are given to the fit. Estimating them too makes the problem non-linear and strictly harder, and the literature on fitting sums of exponentials is largely about how much harder. Every limit here is a best case.

The amplitudes are equal. A spectrum weighted towards one end recovers the heavy end better and the light end worse, because the noise is the same size against a smaller signal.

And the noise is white and independent between readings. A bench’s noise at the slow end of a dielectric measurement is neither, and correlated noise is worse than this for the same rms, since it can imitate a slow exponential outright.

Still open: the weight per decade, the unknown time constants, and a band that is not a reading

The weight in each decade as the target. The well-conditioned combination is the sum over a decade. Fitting that directly — a histogram of absorption by decade of time constant, with its own error bars — should need a far less accurate reading than any line spectrum, and the accuracy it needs is the number that would say whether a practical settling measurement can characterise a capacitor at all.

Fitting the time constants as well. With the time constants free, the fit can move a relaxation to where the data are weakest, and the question becomes how far apart two relaxations must be before a fit places them as two. That is a resolution in the sense a spectroscopist uses the word, and it should be a ratio of time constants depending only on the reading’s accuracy.

Settling times instead of readings. A real instrument measures the time to reach a band, not the error at a time, and its error is a timing error whose size depends on the slope of the curve at the crossing. Near the slow end the curve is flat and a small voltage error is a large time error, which weights the information differently from the readings drawn here — and might change which relaxations are recoverable.