Concept

Baseline shift — where it appears

The drift of a waveform's resting level when a coupling capacitor passes its changes but blocks its average. A pulse train's baseline settles where the areas above and below it balance, so the level a threshold sees depends on the duty cycle and not only on the signal.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

A 100 ms pulse through a 0.159 Hz corner, 9.52% shorter by the end of it. computed by solving, not by drawing. Marched. The dashed line is the pulse that was sent. The solid line is what a 1 MΩ input with 1 µF in front of it receives: the top decays as exp(−t/RC) for the whole 100 ms, ending 9.515% down, and the trailing edge undershoots by exactly the same amount. The closed form for the same two components gives 9.516%. The input's specification is a corner at 0.159 Hz; a top flat to one per cent needs a pulse shorter than 10.1 ms, which is a rate 625.2 times the corner — a constant with no component in it, and the reason a low-frequency specification says nothing useful about an edge.

The corner that says nothing about an edge

An AC-coupled input is sold on a low-frequency corner, and a corner is a statement about steady sinusoids. What an instrument is usually shown is a pulse, and for a pulse the number is a sag: a hundred-millisecond pulse through a 0.159 hertz corner comes out 9.5 per cent shorter than it went in. A one per cent flat top needs a pulse rate 625 times the corner, which is a constant with no component in it — and above fifty per cent duty an AC-coupled pulse never reaches half its own height at all.

instruments · AC coupling

A second coupling stage with the same time constant: the top sags to 0.303 instead of 0.607, and the tail crosses zero at 2.27 s and overshoots to +0.025. computed by solving, not by drawing, marched and checked against the closed form to 3.0e-4: a half-second pulse through a 1 µF, 1 MΩ coupling stage (dashed), and through that stage followed by a buffered second one with a time constant of 1 s (solid). One stage sags to 0.6065 by the pulse's end and its undershoot decays monotonically to zero. Two stages sag to 0.3033, undershoot to −0.6966, cross zero 2.271 s after the pulse began and overshoot positive — the second capacitor removes the first moment of the waveform as well as its area, so the tail has to change sign.

The second capacitor, and the tail that changes sign

One coupling stage turns a pulse into a sagging top and an undershoot that returns to zero from below. Add a second stage with the same one-second time constant and a half-second pulse sags to 0.303 of itself instead of 0.607, undershoots to −0.697, crosses zero 2.27 s after it began and overshoots positive before settling. The step response becomes (1 − t/τ)·exp(−t/τ), through zero at exactly one time constant and down to −e⁻² at two. The reason is arithmetic: one capacitor removes the waveform's area, two remove its area and its first moment, and a pulse with a one-signed tail cannot have both. Built to the same stated corner, two stages still sag 1.29 times as fast, and a 1% flat top needs a pulse rate 806 times the corner rather than 625.

instruments · AC coupling

Without the buffer, a second stage of 1 times the first's impedance: the top sags to 0.175 instead of 0.303, and the tail crosses at 2.00 s and overshoots to +0.010. computed by solving, not by drawing, marched and checked against the closed form to 4.5e-4: a half-second pulse through two one-second coupling stages, buffered (dashed) and with the second hung directly on the first (solid), its resistor 1 times the first's and its capacitor 1/1 of it, so that each stage alone keeps its one-second time constant. Buffered, the top sags to 0.3033 and the tail crosses zero 2.27 s after the pulse began. Unbuffered, the denominator gains the cross term R1·C2 and its poles move to −0.382 and −2.618 per second; the top sags to 0.1751, undershoots to −0.8245, crosses zero at 2.002 s and overshoots to +0.0103. The slider is the second stage's impedance relative to the first's.

Two coupling stages without a buffer

Hang a second coupling stage directly on the first, with equal parts, and the pair's denominator gains a third time constant — the first resistor carrying the second capacitor's current. The poles spread from a double pole to −0.382 and −2.618 per time constant, a half-second pulse's top sags to 0.175 instead of the buffered pair's 0.303, and the sag rate is exactly 2 + 1/k per time constant for a second stage k times the first's impedance — the sum of the two short-circuit time constants. The tail still crosses zero, because the two zeros at the origin are untouched, and a step's lowest point is still at exactly twice its zero crossing. Then a reversal: specified by the overall corner rather than by the parts, the loaded pair sags only 1.12 times as fast as one stage and undershoots 7.6% rather than 13.5% — the buffer the cascade essay assumed makes the specified input worse.

instruments · AC coupling

A 3.3 V edge from a 10 Ω driver down a 50 Ω line of 1 ns, received through 50 Ω in series with four capacitors, beside a plain 50 Ω and an open end. computed by solving, not by drawing: the line marched by its characteristics. The open end rings up to 5.50 V; the plain resistor holds 2.750 V, the launched level, and never reaches the swing. With the capacitor the far end arrives at the same 2.75 V and then climbs as the capacitor charges and stops terminating: with a time constant of half a round trip (20.0 pF) it overshoots to 4.48 V; with one, 3.83 V; with three (120 pF) it rises without overshoot to 3.30 V by 24 ns; with ten, 3.13 V by 24 ns, still on its way.

The termination that runs out

A 50 Ω resistor at the end of a line terminates every edge and draws current for as long as the line is held high: 75.6 mW on an alternating pattern from a 3.3 V, 10 Ω driver. Put a capacitor in series with it and the standing current is gone, and so, eventually, is the termination: the far end reflects nothing as an edge arrives and the whole of it once the capacitor has charged, by one less the exponential of −t/2Z₀C exactly. Under two round trips of time constant the receiver overshoots — 54.8 per cent at a quarter of one — and above it the receiver waits on the capacitor to reach the swing. On a pseudo-random pattern the levels are tightest at three round trips, and on a fast pattern the capacitor holds the line's mean and the resistor still burns half of what a plain one does.

lines · Termination

Named alongside it

The objects these essays reach for when they reach for this one.

AC coupling Charging exponential Corner frequency Duty cycle Step response Characteristic impedance Design tradeoff Loading Model range Poles Reflection coefficient Short-circuit time constants

All concepts