The branches a square root needs

The time constant one trimmer cannot reach gave a compensated divider a second time constant and found that one more branch in its upper arm, a scaled copy of the late one, makes the whole arm flat to rounding. It then noticed what its lumped kilohm and 15 pF were standing in for. A resistive cable is a distributed line, and its contribution to the lower arm goes as $1/\sqrt{s}$ over a band rather than as a single exponential, so a finite number of copy branches can only approximate it over a band. It stated the question and left it: how many branches hold an edge within one per cent from a nanosecond to a millisecond, which amounts to asking how well a ladder network of sections fits a square root?

The usual way to answer a question like that is by a numerical fit, since a square root over a stated band has no finite network. But the object being fitted, an RC line open at its far end, is not an abstract square root, and it happens to have an exact decomposition with a simple form. That form decides almost everything before any fit is made, and it makes the question easier than it first looks, though the answer depends on what is being asked of the branches.

A square root with an exact expansion #

Keep the earlier essay’s divider: nine megohms and a trimmer over one megohm, 100 pF the divider reaches directly. Replace the lumped late branch with the same 15 pF spread along 2 kΩ of resistive line, open at its far end. That total capacitance leaves the calibrator’s reading and the trimmer’s setting as they were. The line’s input admittance is

$$Y(s) = \frac{\gamma}{R}\tanh\gamma, \qquad \gamma = \sqrt{sRC}.$$

Tanh has an old and exact partial fraction expansion over its poles, and substituting gives

$$Y(s) = \sum_{k=1}^{\infty}\frac{sC_k}{1 + sR_kC_k}, \qquad C_k = \frac{8C}{(2k-1)^2\pi^2}, \qquad R_k = \frac{R}{2}.$$

Every term is a resistor and capacitor in series. So a line is, exactly, a parallel set of RC branches, each behind the same resistance $R/2$, here 1 kΩ, with capacitances falling as the reciprocal of an odd square. That is a stronger statement than any fit, and it gives the time constants at once: 12.2 ns, 1.35 ns, 486 ps, 248 ps and so on, going as $1/(2k - 1)^2$.

A netlist of eighty sections matches the expansion within 2 per cent over the three decades it is compared on. The lump the earlier essay used is close to the first term: 15 pF behind a kilohm against 12.2 pF behind a kilohm, so that essay was already looking mostly at the first branch, with 19 per cent of the capacitance unaccounted for. The difference between them is above a few tens of megahertz, where the lump’s admittance flattens at a millisiemens and the line’s keeps rising as the square root of frequency. Ever more of a line’s capacitance is reached through ever less of its resistance as the frequency rises, which a ladder is not a line discusses for an LC structure. In these branches, that fact has a form: the higher terms all have the same resistance, and their capacitance is falling.

Two different results therefore have to be kept apart. The uncopied capacitance, which is what a low-frequency argument would call the size of the problem, falls slowly: the tail of $\sum 1/(2k-1)^2$ goes as $1/4N$ beyond the $N$th term, so 18.9 per cent remains after one branch, 9.89 after two, 4.99 after four and 2.48 after eight. A problem whose size halves only when its cost doubles sounds hopeless. It would be, if the size of the remainder were what an edge sees.

An upper arm built one piece at a time #

A scaled copy of any lower-arm branch makes that branch’s contribution vanish from the ratio. The first essay on this showed it for one branch, and the algebra is the same for each: an upper-arm branch of $C_k$ times $R_2/R_1$ in series with $R_k$ times $R_1/R_2$ has exactly the admittance of the corresponding lower branch, divided by nine. So copying the first $N$ branches, and reducing the plain trimmer by the capacitance they took over so the slow balance is unchanged, leaves a divider whose fast departure is due only to the terms that were not copied.

The output has no convenient closed form, so it is computed by inverting the exact $H(s)$ numerically, with Talbot’s method applied to the ramp response and the edge obtained as a difference of two ramps. It takes a fraction of a second, and that is important here since the worst departure over an edge’s aftermath needs a search over many times. The method’s only input is the exact sum, kept to 400 terms, of which only a few dozen are needed.

The first branch is most of the value. On a 1 ns edge, one trimmer alone is 11.2 per cent out. Copying one branch takes that to 1.07, a tenfold improvement from a single resistor and capacitor. Two branches give 0.252, three 0.0822, four 0.0353 and five 0.0192. These fall much faster than the remainder’s capacitance does: from one branch to four the departure falls by a factor of thirty and the remainder by four. On an edge of 200 ps, which sees further down the series, the numbers fall less steeply: 1.77 per cent, 0.66, 0.31 and 0.163 for one to four.

The reason the departure falls so much faster is the edge itself. A branch whose time constant is shorter than the edge’s rise has finished charging by the end of the edge, so nothing of it is left to see afterwards. On a 1 ns edge the third branch, at 486 ps, has largely finished, and so has every one after it. What remains uncopied has 5 per cent of the capacitance and has no time to show it.

So the remainder’s slow $1/4N$ is real but it is not the measure of error. Almost all of the unmatched capacitance sits in fast terms, and a rise time filters away exactly those. The error budget is decided by the branches slower than the edge, and there are only about $\frac{1}{2}\sqrt{\tau_1/t_r}$ of them, 1.7 at a nanosecond and 5.5 at 100 ps.

How many for a given edge #

Turning that around gives the question the earlier essay asked.

Within 1 per cent, one branch is enough for an edge of 2 ns or slower and two serve all the way down to 100 ps. That is a far better answer than the earlier essay expected when it asked about a nanosecond to a millisecond. It suggested a question about how many sections fit a square root, and it would have supposed tens of them. Two branches are enough for this line, at 1 per cent, because the error from every branch below the second is smaller than the edge can show and smaller than the tolerance.

Within 0.1 per cent the count rises with the edge: two on a 10 ns edge, three on 1 ns and seven on 100 ps. So a hundred-fold faster edge asks for about three and a half times as many branches, and the tightening of the band costs about as much as the speed does. That pattern resembles what the sections a wavelength needs found for an LC ladder, whose section count is set by the fastest frequency of interest, but here the relation is weaker than a proportion, because the branches’ time constants fall as an inverse square.

There is also a statement here about instruments. A probe for a 100 MHz oscilloscope, with edges no faster than about 3.5 ns, has essentially one adjustment for this line’s fast end, and at 1 per cent that is adequate. A probe for a gigahertz oscilloscope, with 350 ps edges, has to do better than that at the same 1 per cent, and it ships with two or more adjustments. Real probes have several fast adjustment points, usually labelled as low, middle and high frequency, and three is about what this argument predicts for a good high-speed one.

Tolerance ends the series first #

The arithmetic above assumes exact copies, and each branch here is the scaled copy of a term with a stated capacitance and time constant. A real upper arm is built from a resistor and a capacitor, each with a tolerance, or a trimmer at the factory.

The first branch is most of the capacitance, and a fractional error in its time constant leaves an error in the hook in proportion. With tolerances on the upper arm’s resistors, the error from the first branch quickly dominates the small contribution a third or fourth branch was going to remove. At ±10 per cent nothing further comes of a branch beyond the second: 0.369 per cent with two and 0.392 with three, the worst of twenty-four seeded sets either way. At ±2 per cent the series gets to three, 0.0947, and then stops near 0.08 per cent. That is the correct way to frame the budget: a branch is worth adding only while what it would remove is larger than the tolerance on the branches already built. It suggests the sequence in which real probes are set up, and why it starts from the slowest adjustment and works upward: the largest and slowest branch has to be right first, since its error is larger than everything that comes after it.

That also explains why a real probe’s compensation is done by trimmers and not fixed parts. A trimmer on the first branch, set at the factory with a suitable fast step, puts its effect below the tolerance of the second and third, and each is adjusted in turn. It is a very physical version of what a calibration at the carrier removes, in which a measured quantity removes an error the design could not know beforehand.

What the expansion says about measuring a line #

The same exact sum reads backwards, as a statement about what a settling curve on a resistive line can reveal. How many relaxations a settling curve holds found that a measured curve supports only a few independent exponentials, whatever the physics behind it. A line’s expansion is a clean test of that: its later terms’ time constants are spaced by ratios of 1.8 and less, so a curve that fits two exponentials well would report two time constants and a capacitance share that are not any of these terms. The first term’s 81 per cent and 12.2 ns are distinct and should be recoverable. The rest merge. What a designer gets from a step on a real cable, in other words, is the first branch and an amount of remainder, and that happens to be what a probe’s adjustments need. The delay that is a centre of mass offers another measure that needs no fit at all: the mean time of a distributed response is a sum over the whole line, here half of $RC$, and it is correct whatever the number of branches.

An expansion that says where to stop #

The neat thing about having an exact series is that it answers a question an ordinary fit cannot. A fit over a stated band produces a set of branches that are optimal for the band, and the fitting does not reveal how much further a different band would need. The exact expansion is not optimal: a fit with two branches, chosen to minimise worst departure on a 1 ns edge, could beat the two exact first terms by moving some capacitance from the first branch to make up for what the rest are not doing, though by how much is not computed here. So the numbers above are an upper bound on what two branches can do, and that is the direction a designer would go on to take in practice. But they provide a clear standard for what a branch is: the first branch here is a real part of the line and not a parameter in a fit, and when a probe is adjusted against a real cable, it is being adjusted to a real term of this sum.

Flat delay bought with more delay is the case of a filter with the same kind of structure, where an all-pass section is added for each order of correction and each costs real delay. Here the cost of each branch is a resistor and capacitor that have to be set, and the marginal return falls off about as fast.

A source, once again #

The figures here are all for an ideal source, and the source the hook hides behind showed what a source’s resistance does to one branch: it hides the hook near where the two time constants are equal. For a line there are several time constants, and the source’s lag covers them from the fast end up. At 50 Ω the lag is 0.6 ns, and all the terms past the second are concealed before any edge is involved. At a kilohm the lag is 11.5 ns and even the first term has merged with it. So a soft node demands less of the probe’s fast adjustment, and a stiff one demands all of it. The branch count above is the worst case and belongs to the stiffest node a probe is put on. A probe used only on nodes of a few hundred ohms, or a probe for slow edges, is being asked for very little.

What this leaves out #

The cable here is a line of pure resistance and capacitance. A real one has inductance too, and above some frequency it is a transmission line with a resistive centre conductor, which is deliberately lossy in order to damp its own reflections: that is how a high-impedance probe’s cable is made, and the shape of the reflection then matters as much as anything above. The far end here is open, whereas a real cable’s far end is the instrument’s input with its own capacitance. That moves every pole of the tanh, though not the form of the sum, and it would need the corresponding expansion. The line’s resistance is uniform and a cable’s may not be, and dielectric absorption in its insulation adds another distributed relaxation on much longer time scales, of the kind the capacitor that remembers measures.

The numbers to keep #

A resistive line is exactly a set of RC branches, all behind $R/2$, with capacitances as $8C/((2k-1)^2\pi^2)$: 81.1, 9.01 and 3.24 per cent for the first three. On a 1 ns edge a divider with one trimmer is 11.2 per cent out, with one copied branch 1.07, two 0.252 and four 0.0353. The uncopied capacitance falls only as $1/4N$, but a branch faster than the edge is averaged away, so only about $\frac{1}{2}\sqrt{\tau_1/t_r}$ of them count. Two branches hold 1 per cent down to a 100 ps edge. Tolerance ends the series first, since the first branch’s error soon exceeds everything the later ones would remove.

Still open: a fitted pair of branches, a far end that is an instrument, and a line that drifts #

A fitted pair. Two exact terms are not the best two branches for a stated edge. A fit that moves capacitance from the first to make up for the rest could beat them by some definite factor, and that factor says how much an adjustment procedure gains over simply copying the cable. It also says what the trimmers in a real probe are settling on, which is not any term of this sum.

A far end that is an instrument. With a capacitance at the far end of the line, the poles of the tanh move to the roots of a transcendental equation, and the branches no longer have equal resistance. Whether the first branch still holds most of the capacitance, and what a heavy input does to the count, is a small extension of the same expansion.

A late branch that moves with temperature. Resistance wire has a coefficient of hundreds of parts per million a kelvin and the first term has most of the capacitance, so the first copied branch’s time constant drifts against a fixed copy. How far a warm probe’s error sits above the tolerance’s is a question with a number in it, and whether the second and third branches are worth their adjustment at temperature extremes is the same budget at a different corner.