The second capacitor, and the tail that changes sign

The corner that says nothing about an edge measured one coupling capacitor with pulses and found two effects a corner frequency does not report: a sag across the top of every pulse, and a baseline that moves until the waveform’s own average sits at zero. Both came from one exponential. The top of a pulse decays as $e^{-t/RC}$, the trailing edge undershoots by what the top lost, and the undershoot climbs back to zero from below, never crossing it.

A signal rarely meets only one coupling capacitor. A probe’s input is AC-coupled, the amplifier behind it may block DC between its stages, a transformer in the chain is a coupling element of its own, and a recording system may high-pass its data again to remove drift. The rejection two capacitors lose at their corner put two coupling networks side by side and found them converting common mode at their corner; this essay puts two in a row. The second stage does not simply add a second sag to the first. It changes the shape of everything after the pulse, and the change has a cause that can be stated in one line.

One stage and two, marched #

The first stage is the earlier essay’s: 1 µF into 1 MΩ, a time constant of one second and a corner at 0.159 Hz. The second is another of the same, driven from the first through a unity-gain buffer so that neither loads the other, which is how two coupled stages in an amplifier are arranged. A half-second pulse goes in, and the output of each is marched on the netlist and compared with the closed form, which it matches to eight parts in ten thousand away from the pulse’s two edges.

Through one stage the top sags to 0.607 of the pulse by its end, which is $e^{-0.5}$; the trailing edge drops by the full pulse height to −0.393, and the undershoot decays to zero from below. Through two, the top sags to 0.303, which is $e^{-0.5}(1 - 0.5)$: twice the single stage’s loss over the first moments of the pulse, and 1.77 times it by the end. The trailing edge drops to −0.697. And the tail does something the single stage never does — it climbs back through zero 2.27 s after the pulse began and overshoots to +0.025 before settling.

The slider changes the second stage’s time constant. A second stage three times slower than the first changes less — a top of 0.487, a crossing at 3.56 s and an overshoot of +0.011 — and one ten times slower barely changes the pulse at all, 0.568 at the top and +0.001 of overshoot at 5.38 s. A second stage faster than the first dominates it: at 0.3 s the top has fallen to 0.010 by the pulse’s end and the tail crosses at 1.33 s; at 0.1 s the pulse is differentiated into a spike at each edge, the top ends at −0.060, and the trailing spike reaches −1.06, more than the pulse’s own height. In every case with a second stage, the tail crosses zero.

Why the tail must cross #

The earlier essay’s single stage removed the pulse’s average: a capacitor passes no direct current, so the output has zero area. The area lost from the top is exactly the area of the undershoot, and the tail need never cross zero to arrange that — it can approach from below for ever.

A second capacitor removes one more thing. Each coupling stage puts a zero of its transfer function at the origin of the frequency plane, and a zero at the origin removes a moment of the waveform: one zero takes the area, $\int y\,dt$, and two take the area and the first moment, $\int t\,y\,dt$, as well.

Integrated over thirty seconds of march, the single stage’s output has no area — to six parts in a hundred trillion — and a first moment of −0.500 $\text{V·s}^2$, which is minus the pulse’s area times the time constant: the undershoot is centred later than the pulse, by one time constant. The two-stage output has neither area nor first moment, to parts in $10^{11}$.

That second condition is what forces the crossing. A pulse followed by a tail that stays below zero has a positive part early and a negative part late, so its first moment is negative — the balance point of the negative area lies after that of the positive. To make the first moment zero, some positive area must lie later still, beyond the negative lobe, and that is the overshoot. The tail cannot be one-signed, whatever the two time constants are, because it has two conditions to meet and one sign change is the least that meets both. A third capacitor would remove the second moment too, and the tail would have to cross at least twice.

The step through two equal stages #

The step response of the pair makes the same point in its simplest form.

For two equal stages of time constant $\tau$ the step response is $e^{-t/\tau}(1 - t/\tau)$. It starts at one and falls twice as fast as a single stage’s; it crosses zero at exactly one time constant; it reaches its deepest undershoot at two, where it is $-e^{-2}$, −13.5 per cent of the step; and it returns to zero from below over the next several time constants. The march finds the crossing at 1.000 s and the minimum at 2.00 s and −0.1353, and matches the closed form to one part in a thousand.

A step held on a doubly coupled input is therefore not merely forgotten. It is remembered upside down first, at a seventh of its size, and the upside-down memory peaks two time constants after the step — when a single stage would still be holding the same 13.5 per cent the right way up. On a display the effect looks like a baseline that overshoots after every large change of level, and it is often mistaken for a fault in whatever stage is suspected of ringing, when nothing in the circuit rings at all.

Twice the sag, and 1.29 times at the same corner #

Over the top of a short pulse the sag is what the earlier essay made its central number, and two stages double it.

For a pulse much shorter than the time constant, a single stage loses a fraction $w/\tau$ of its top over a width $w$, and two equal stages lose $2w/\tau$ — each takes its own share. A top flat to one per cent needs a pulse shorter than 10.05 ms through one stage and 5.02 ms through two, so the earlier essay’s constant — a pulse rate 625 times the corner — becomes 1,252 times each stage’s corner.

That comparison is unfair to the pair, because two stages of the same time constant have a higher overall corner than one: the pair’s response is $(\omega\tau)^2/(1 + \omega^2\tau^2)$, which is $1/\sqrt2$ at $\omega\tau = \sqrt{1 + \sqrt2}$, 1.554 times each stage’s corner. An instrument is specified by its overall corner, so the fair comparison scales the pair’s time constants up by 1.554 until its corner matches the single stage’s. Then the pair sags at $2/1.554 =$ 1.287 times the single stage’s rate. A top flat to one per cent needs a pulse shorter than 7.80 ms, a rate 806 times the stated corner instead of 625.

So a corner specification understates the sag of a two-stage input by 29 per cent even when both corners are read correctly, and understates the undershoot and ignores the overshoot entirely. The constant that relates a corner to a flat top depends on how many capacitors the corner is made of, which is information a single frequency cannot carry.

Three stages, and n #

The moments argument does not stop at two. A chain of $n$ equal coupling stages puts $n$ zeros at the origin and removes the first $n$ moments of whatever passes through it, so its step response must change sign at least $n - 1$ times. The closed form bears that out: the step response of $n$ equal first-order high-passes is $e^{-t/\tau}L_{n-1}(t/\tau)$, with $L_{n-1}$ the Laguerre polynomial of that degree. For one stage that is the plain exponential; for two, $e^{-t/\tau}(1 - t/\tau)$, the curve drawn above; for three, $e^{-t/\tau}(1 - 2t/\tau + t^2/2\tau^2)$, which crosses zero at $2 - \sqrt2 = 0.586$ and $2 + \sqrt2 = 3.414$ time constants — once downwards, early, and once back up, late.

Each added stage therefore adds one more lobe to the tail, each smaller and later than the last, and each crossing is a place a threshold near the baseline can be triggered. The chain’s overall corner rises too, roughly as the square root of the count — 1.55 times each stage’s corner for two, 1.96 for three — so a designer adding stages at a fixed per-stage corner is also moving the input’s specified corner up, while the tail’s lobes, which follow each stage’s time constant, stay where they were.

Chains of that length are ordinary. An audio path with a coupling capacitor at each stage, a data-acquisition front end with a blocking capacitor at the input and another after a gain stage, a probe with its own coupling feeding an oscilloscope that couples again: each is a cascade, and the pulse response a user sees is this one, with the lobes that the count of capacitors sets. The step on a doubly coupled input overshooting its baseline is the commonest symptom; two thresholds because there is a floor is what a comparator does when a slowly moving level wanders back across its threshold, and the second lobe is exactly such a wander.

The lobes also make a diagnostic. A long step applied to an unknown chain and watched for long enough shows its coupling stages as sign changes in the tail: none for one stage, one for two, two for three, each crossing later than the last. Counting them identifies how many capacitors stand between the input and the display, which is otherwise a question for the schematic. An oscilloscope fed through an AC-coupled active probe is the commonest two-stage chain in a laboratory, and the instrument’s own rise time is the fast end of the same bookkeeping: an instrument has a response at both ends of its band, and at the slow end the count of capacitors decides the shape as surely as the count of poles decides it at the fast end.

A pulse train whose pattern changes #

The earlier essay’s most practical finding was the baseline shift: a coupled pulse train settles with its own average at zero, so a change of duty cycle moves the whole waveform, and above fifty per cent duty the top never reaches half the amplitude. Two stages keep the settled answer and change the way to it.

Switch the train from twenty per cent duty to eighty and its mean rises by 0.6 of its height. Through a single stage the per-period mean of the output jumps by that amount, 0.594 as marched, and decays back to its old level without ever crossing it: the settled waveform sits lower, as the baseline shift requires, and it gets there from above. Through two stages the jump is the same and the return is not. The mean slides back through its old level and undershoots to −0.081 — the change times $-e^{-2}$, exactly the step response of the pair applied to the train’s average — two seconds after the switch, before settling over the following several.

That is the fault the earlier essay described — a comparator whose threshold was set on the settled waveform misbehaving when the pattern changes — made worse. With one stage, a threshold set just above the settled bottom of the 80 per cent train is crossed once, on the way down. With two, the waveform’s baseline sweeps below its final level and back, and a threshold near that level is crossed twice: one extra decision per change of pattern, arriving two time constants late, in a system where nothing appears to have changed at that moment.

Two routes, and what they share #

Every trace here is marched on the netlist, and each is checked against a closed form computed from the time constants alone: the difference of two exponentials, $(\tau_1 e^{-t/\tau_2} - \tau_2 e^{-t/\tau_1})/(\tau_1 - \tau_2)$, for unequal stages, and $e^{-t/\tau}(1 - t/\tau)$ for equal ones, with a pulse as the difference of two steps. They share the element values and nothing else.

The march itself needed a choice that is worth recording, because it is a small instance of the subject. The second stage’s capacitor hangs between two nodes whose voltages are both set algebraically — the buffer’s output and the second resistor’s top — and the trapezoidal rule that one step, computed twice checks the other marches in these essays against answered each edge of the pulse with a sample-to-sample alternation of the full pulse height that never decayed: 0, 2, 0, 2. The rule is stable and not damping, so a mode it cannot resolve is flipped rather than suppressed — the behaviour the ringing that belongs to the rule measured on a fast pole, here in its limiting form, since a capacitor between two algebraically fixed nodes is a pole of zero time constant. The second-order backward difference damps it in a step, as the rule between the two found it damping a stiff error, at the cost of one step of first-order error at each edge, which the comparisons skip; the pulse-train figure, with a thousand edges, shows that cost as a bias in the per-period mean that halves when the step halves, and is run at a step small enough to leave it below 0.003.

What it leaves out #

Loading between stages. Every pair here is buffered. Two coupling networks connected directly load each other — the second’s resistor sits across the first’s — and the pair’s poles are no longer the two time constants but the roots of a quadratic that splits them apart. The moments argument survives unchanged, since two capacitors in the signal path still put two zeros at the origin, and so does the crossing.

A transformer as one of the stages. A transformer couples through its magnetising inductance, which is a first-order high-pass of its own with a corner set by the inductance and the source resistance — the low edge of the band a turns ratio holds over. In a chain with a coupling capacitor it is a second stage, and the droop of a transformer-coupled pulse is this essay’s two-stage sag.

The first moment’s size. The two-stage output has zero first moment, and what it has instead is a second moment, which sets how large the overshoot must be. That is a quantity with its own closed form, and it is what decides whether a threshold near the baseline is crossed at all.

Still open: loaded stages, the transformer, and the digital high-pass #

Unbuffered stages. Two coupling networks sharing a node have poles at the roots of $1 + s(\tau_1 + \tau_2 + \tau_{12}) + s^2\tau_1\tau_2$, with $\tau_{12}$ the first capacitor into the second resistor, and they are always further apart than the two time constants alone. The tail still crosses; how its crossing time and overshoot move with the loading is a march on the unbuffered netlist.

A transformer with a coupling capacitor. The magnetising inductance and a series capacitor make a second-order high-pass whose two roots can be complex, and then the tail does not merely cross zero once — it rings. Where a transformer-coupled pulse’s droop turns into an oscillation is a boundary in inductance and capacitance that no one-stage argument sees.

The same arithmetic in software. A recording system that high-passes its samples to remove drift adds a digital coupling stage, and a recursive one has the same zero at the origin as a capacitor. Whether a second-order digital drift filter, chosen for steeper rejection, puts the same crossing into every event it records is a question about moments that holds for any linear high-pass, analogue or digital.