Two coupling stages without a buffer

The second capacitor, and the tail that changes sign put two coupling stages in a row and found that a pulse through them no longer merely sags: its tail crosses zero and overshoots, because each capacitor puts a zero at the origin of the frequency plane and two zeros remove the waveform’s first moment as well as its area. Its two stages were joined by a unity-gain buffer so that neither loaded the other, and its open questions began with the arrangement that has no buffer. Two coupled stages sharing a node have poles at the roots of $1 + s(\tau_1 + \tau_2 + \tau_{12}) + s^2\tau_1\tau_2$, always further apart than the two time constants alone, and the crossing should move.

The unbuffered pair is not exotic. It is a coupling capacitor at an amplifier’s input followed by another at the next stage’s input, with nothing between them but the first stage’s bias resistor; it is a passive high-pass filter of second order, built cheaply; it is a probe’s coupling network feeding an instrument’s. This essay measures what the missing buffer costs, which turns out to depend sharply on what was held fixed when it was removed. A single stage’s corner, as the corner that says nothing about an edge found, is a poor guide to its pulse response, and a pair’s is a poorer one. The rejection two capacitors lose at their corner put two coupling networks side by side, one on each input of a differential instrument, and found their mismatch costing rejection at the corner; this essay puts two in series, with nothing between them, and finds the cost in the waveform instead.

The third time constant #

The first stage is the cascade essay’s: 1 µF into 1 MΩ, a one-second time constant. The second hangs directly on the first stage’s node, its resistor k times the first’s and its capacitor 1/k of it, so that each stage on its own still has a one-second time constant and the only thing that changes with k is how heavily the second loads the first.

Writing the node equations gives the pair’s transfer function directly:

$$H(s) = \frac{s^2\tau_1\tau_2}{1 + s\,(\tau_1 + \tau_2 + R_1C_2) + s^2\tau_1\tau_2}.$$

The numerator is the buffered pair’s — two zeros at the origin — and so is the coefficient of $s^2$. The difference is one term in the middle, $R_1C_2$: the second capacitor’s current has to come through the first stage’s resistor, and that resistor and that capacitor together are a third time constant that the buffer was there to remove. With equal parts, $R_1C_2$ is one second, the middle coefficient is three seconds rather than two, and the double pole of the buffered pair splits into two real poles at −0.382 and −2.618 per second, 6.85 apart.

The march agrees with the closed form to $4.5\times10^{-4}$ away from the edges. Through the buffered pair the half-second pulse’s top sagged to 0.303 and its tail crossed zero 2.27 seconds after the pulse began. Through the loaded pair the top sags to 0.175, the trailing edge undershoots to −0.825, and the tail crosses zero at 2.00 s and overshoots to +0.010 before settling.

The slider runs the impedance ratio from a third to ten. A second stage of ten times the first’s impedance barely loads it: the top sags to 0.287 and the crossing is at 2.24 s, close to the buffered pair’s. One of three times sags to 0.253 and crosses at 2.17 s. One of a third of the first’s impedance — a low-impedance second stage, its capacitor three times the first’s — sags to 0.044, nearly the whole of the pulse lost by its end, and crosses at 1.63 s. In every case the tail crosses: the two zeros at the origin do not depend on the loading, and two zeros remove two moments however the poles move.

Why the poles can only spread #

The split is not a feature of equal parts; it is guaranteed. The pair’s poles are real when the middle coefficient’s square exceeds four times the product of the time constants, and

$$(\tau_1 + \tau_2 + R_1C_2)^2 - 4\tau_1\tau_2 \;>\; (\tau_1 + \tau_2)^2 - 4\tau_1\tau_2 = (\tau_1 - \tau_2)^2 \;\ge\; 0.$$

The buffered pair sits exactly on the boundary when its two time constants are equal — a double pole — and the cross term pushes the loaded pair strictly inside the real region, whatever the parts. That is a special case of a general fact about networks made only of resistors and capacitors: their natural frequencies are all real and negative, so no RC network can ring. A coupled input built only of resistors and capacitors can sag, undershoot and overshoot once per pair of zeros at the origin, but it cannot oscillate about its baseline.

The general fact is also the boundary of this essay’s argument. The moment an inductance enters — a transformer’s magnetising inductance in place of one of the resistors, an inductive source — the guarantee is gone, the middle coefficient no longer has to be large, and the poles can be complex. Two zeros at the origin still force at least one crossing; complex poles then add as many more as the damping allows.

Where the step crosses, and where it bottoms out #

The step response shows the effect most cleanly. For any pair of real poles $p_1$ and $p_2$ and two zeros at the origin, a unit step produces

$$y(t) = \frac{p_1e^{p_1t} - p_2e^{p_2t}}{p_1 - p_2},$$

which starts at one, falls through zero where $p_1e^{p_1t} = p_2e^{p_2t}$ — at $t^* = \ln(p_2/p_1)/(p_1 - p_2)$ — and reaches its lowest point where its derivative vanishes, $p_1^2e^{p_1t} = p_2^2e^{p_2t}$, which is at exactly $2t^*$. The buffered pair’s crossing at one time constant and minimum at two were one case of this.

The march finds the buffered pair crossing at 1.000 s and lowest at 2.000 s, at −0.1353, as before. At equal impedance the loaded pair crosses at 0.861 s and is lowest at 1.722 s, at −0.0756; at a third of the impedance, 0.662 s, 1.324 s and −0.0293. The minimum is at twice the crossing in every case, to the march’s resolution, which is the kind of statement worth having in hand at a bench: a doubly coupled input’s undershoot after a step bottoms out at twice the time it took to cross zero, whatever the parts, so a measurement of either predicts the other.

The undershoot is the striking change. Loading the first stage makes the step’s inverted memory shallower — 7.6 per cent of the step at equal impedances instead of 13.5 — because the spread poles make the pair behave more like one fast stage followed by one slow one, and the slow one alone would not undershoot at all. That observation is the start of the reversal below.

The two features of a step — where it crosses and how steeply it starts — are also a measurement. The initial slope is minus the sum of the two poles and the crossing time is $\ln(p_2/p_1)/(p_1 - p_2)$, so a single captured step gives two equations in the two poles, and both can be recovered without knowing a single component value. For the equal-impedance pair, a slope of three per second and a crossing at 0.861 seconds admit only the poles at −0.382 and −2.618. A bench that can capture one slow step through an unknown doubly coupled input can therefore say whether its stages are buffered — a double pole, crossing at exactly the reciprocal of half the initial slope — or loaded, and by how much, which is otherwise a question for the schematic and the schematic is not always available.

The sag is the short-circuit time constants #

Over the top of a short pulse the sag rate is the step’s initial slope, and for a second-order high-pass it is the ratio of the denominator’s middle coefficient to its last: $(\tau_1 + \tau_2 + R_1C_2)/\tau_1\tau_2$, which with equal time constants is $(2 + 1/k)$ per second. A buffered pair sags at two per second, twice a single stage; an unbuffered pair of equal impedances at three — half again as much.

The same number comes out of the netlist by a different route. Short each capacitor in turn, find the resistance the other one sees, and form its time constant: the first capacitor, with the second shorted, sees R1 in parallel with R2; the second, with the first shorted, sees R2 alone. The sum of the reciprocals of those two short-circuit time constants is $2 + 1/k$ per second, to $10^{-16}$. It is the high-pass counterpart of the construction in a sum that is exact, where a low-pass network’s open-circuit time constants added up exactly to its first coefficient: here the short-circuit time constants’ reciprocals add up exactly to the sag rate. And the march, reading the slope of a step’s first millisecond, gives 11.91, 5.32, 3.00, 2.33 and 2.10 per second at impedance ratios of 0.1, 0.3, 1, 3 and 10, against 12, 5.33, 3, 2.33 and 2.1.

The identity makes the sag computable without the transfer function, for any number of coupling capacitors in any arrangement: short all of them but one, find what the one sees, and sum the reciprocals. A designer who adds a coupling capacitor to an existing path can read its cost to the pulse’s flatness off one resistance.

A worked case shows the size of the effect with unequal parts. An amplifier’s input coupling of 1 µF into a 100 kΩ bias resistor, followed directly by the next stage’s input coupling of 1 µF into 47 kΩ, has time constants of 0.1 and 0.047 seconds and a cross term $R_1C_2$ of 0.1 seconds. The sum of its short-circuit time constants’ reciprocals is 10 + 21.3 + 21.3 = 52.6 per second; with a buffer between the stages it would be 31.3. A 1 ms pulse through the unbuffered pair loses 5.3 per cent of its top against 3.1 per cent buffered, and the difference is entirely the 100 kΩ resistor carrying the second capacitor’s current — a resistor the designer chose for the first stage’s bias and never thought of as part of the second stage’s coupling.

The reversal: the same corner, less sag #

All of that compares the loaded pair with the buffered pair at the same parts. An instrument is not specified by its parts; it is specified by its low-frequency corner, and a designer choosing between the two arrangements scales the parts to meet it. The cascade essay made the same correction for its buffered pair: two stages of one time constant have a higher overall corner than one, 1.554 times each stage’s, so meeting a single stage’s corner means slowing both, and then the pair sags 1.287 times as fast as the single stage.

Do the same for the loaded pair. At equal impedances its overall corner is 2.672 times each stage’s — higher than the buffered pair’s, because the loading adds to the middle coefficient — so it must be slowed further to meet the specification. And its sag, scaled that way, is 1.123 times the single stage’s, not 1.287.

The factor falls further as the loading grows: 1.262 at an impedance ratio of ten, 1.211 at three, 1.123 at one, 1.036 at a third and 1.007 at a tenth. The reason is the pole spread. A buffered pair has its two poles together and both contribute to the corner and to the sag; a loaded pair has one pole far below the other — 6.85 to one at equal impedances, 26.4 at a third — and the faster of the two then sets the corner almost alone, with the slower one adding little. In the limit the pair behaves like a single stage with a very slow second one behind it, and sags like a single stage.

So the arrangement the cascade essay treated as the clean one, buffered, is the worse one at a fixed specification. A designer who inserts a buffer between two coupling networks to stop them interacting, and then scales them to the same overall corner, has made the pulse’s top sag 15 per cent faster than if the buffer had been left out, and has nearly doubled the step’s undershoot. The interaction the buffer removed was helping.

What the loading costs, after all #

None of this makes loading free. At fixed parts it sags the top half again as fast, and an instrument whose parts are fixed — an existing front end with a second network added — pays that. At a fixed corner the loaded pair needs larger time constants than the buffered one, which means larger capacitors, and a coupling capacitor large enough for a low corner at a high impedance is often the largest and least stable part on the board. And the slow pole that the loading creates is slow: at equal impedances it is at 0.382 per second, so the pair’s recovery from a large step, though shallower, is longer, its undershoot taking about 2.6 seconds per time constant to fade rather than one.

What the measurements say is that the choice between buffered and loaded is a choice between two shapes of the same constraint, not between a clean arrangement and a compromised one. Buffered, the two poles share the corner and the pulse sees both at once — a faster sag and a deeper, quicker undershoot. Loaded, one pole carries the corner and the other lingers — a slower sag, a shallower undershoot, and a long, slow return. The two zeros at the origin are common to both, and whatever else changes, the tail crosses. Which shape suits an instrument depends on what it is asked to show: a pulse amplitude read at the end of the pulse favours the loaded pair’s slower sag, while a baseline that must be back within a tolerance soon after a large event favours the buffered pair’s quicker recovery.

How the numbers were obtained #

Both pairs are netlists marched by the second-order backward difference with twenty thousand steps over six seconds, the step size chosen as in the cascade essay because a capacitor between two algebraically set nodes makes the trapezoidal rule ring at every edge. Each march is compared with the closed form for the same pole pair, the pulse as the difference of two steps, skipping two and a half steps either side of each edge where the method is recovering from the discontinuity. The short-circuit time constants are read from the netlist by replacing one capacitor with a unit test current, the other with an ideal short, and solving for the voltage. The overall corners are bisected on the pair’s magnitude.

What it leaves out #

A source with resistance. The pulse source here is ideal. A source resistance adds to the first stage’s and changes its time constant; it also adds a further cross term, because the source resistance carries both capacitors’ currents, and a third short-circuit time constant to the sum.

More than two stages. Three loaded stages have a cubic denominator whose poles spread further still, and the cascade essay found that a third zero at the origin forces a second crossing of the tail. Whether the loaded triple’s undershoot and overshoot shrink in proportion, as the pair’s did, is the natural next measurement.

Still open: the transformer, the digital high-pass, and the probe in front #

A transformer with a coupling capacitor. A series capacitor feeding a transformer’s magnetising inductance is a second-order high-pass whose poles, unlike this pair’s, can be complex. Then the tail does not merely cross zero once — it rings, as the ring a blocking capacitor buys measures — and where a transformer-coupled pulse’s droop turns into an oscillation is a boundary in inductance and capacitance that no pair of resistive stages can reach.

The same arithmetic in software. A recording system that removes drift with a recursive digital high-pass adds a zero at the origin of its own. Whether a second-order digital drift filter, chosen for steeper rejection, puts the same zero crossing into every event it records — and whether its designer can choose the equivalent of the loaded pair’s spread poles to make the undershoot shallow — is the question the loaded pair raises for a filter that has no parts to load.

A probe’s coupling in front of an instrument’s. A probe with its own coupling network, plugged into an instrument with another, is an unbuffered pair whose impedance ratio is set by two manufacturers who never met. The short-circuit sum says what the combination’s sag is from two resistances that are on both data sheets; whether the combination’s corner is ever stated is another matter.