How long a sweep waits at each step

The arrow that goes past where it settles switched a series resonator on from rest and watched its capacitor’s arrow circle the phasor instead of approaching it. Its last measurement was an instrument’s: a sweep that starts every frequency from rest and records the largest voltage it sees reads a resonance 1.20 times wider than its phasor says. It ended by asking how long such a sweep must wait at each step for that widening to go, and whether the 1.47Q cycles a transient takes to fall to a per cent is the right count or whether the skirts of the resonance, where the transient beats slowly, need longer.

The question has an answer with two halves, and neither is the one the question expected. An instrument that holds its largest reading never gets there, however long it waits. An instrument that reads after its wait gets there in about 1.6Q cycles — longer than the count, and not because of the skirts.

Two readings a sweep can take #

A stepped sweep sets a frequency, lets the circuit respond, reads a number, and moves on. What “reads a number” means decides everything, and there are two common answers.

It can hold its peak: record the largest voltage seen at that frequency since the step began. That is what an analogue peak detector does, and what a digitiser reporting the maximum of a record does.

It can read at the end of its wait: dwell for a stated time and record the amplitude over the last cycle of it. That is what a sweep with a settling delay does, and what a lock-in amplifier’s reading taken after its own time constant approximates.

The held reading cannot be improved by waiting, because what it holds has already happened. On the skirts, a switched-on resonator passes its settled voltage in its first few beats, and the holder keeps that excursion for as long as the step lasts. At a Q of ten it reads the resonance 20.2 per cent too wide at a dwell of one cycle or a million. Everything below is about the other reading.

A resonance after three waits #

Read at the end of a wait, a short dwell gives a resonance that is both too low and too wide, and the two are connected.

After half of Q cycles the reading at resonance has reached only 7.79 of its 10.01 volts, and the resonance it reports is twice as wide as the real one. After Q cycles the top is 4.7 per cent short and the width 19.6 per cent too large; after one and a half, one per cent short and two per cent too large.

The centre is the part of the resonance that arrives last. At resonance the second arrow does not turn: it starts pointing back along the settled arrow and shrinks, so the voltage creeps up from below at exp(−t/τ) and never gets ahead of itself. On the skirts the arrow turns and the reading swings past its settled value and back, so at any instant some skirt frequencies are already over and some under.

The width is measured at half the power of the top, and the top is the thing that is late. A top one per cent low puts the half-power level one per cent low, and the curve crosses a lower level further out. How much further follows from the resonance’s own slope: at its half-power frequencies |H| changes by a fraction Q per unit of fractional detuning, so a level one per cent lower moves each edge out by a hundredth of a Q and the width, which is one Q-th, grows by two per cent — the peak’s shortfall read twice.

The shortfall itself is nearly a closed form. At resonance the second arrow shrinks as $e^{-t/\tau}$ with $\tau = 2Q/\omega_0$, so after $D$ cycles it is $e^{-\pi D/Q}$ of its starting length. After Q cycles that is $e^{-\pi}$, 4.32 per cent, and the reading is 4.66 per cent low — a little more, because the second arrow starts $\omega_0/\omega_d$ times as long as the settled one and a reading over one cycle is taken slightly before the dwell’s end. The width’s 19.6 per cent is not 2 × 4.66, though, and the gap between the two is the linear argument running out: a peak nearly five per cent low moves the edges far enough down the curve that its slope there is no longer the half-power slope, and a longer lever gives a longer answer.

The count, measured #

That makes the width a harder reading than the peak, and the next figure measures by how much.

The peak is within one per cent from 1.50Q cycles, which is the settling count of three voltages that close on one — ln(100)/π = 1.466Q — read on a grid of a twentieth of Q. The count is right for the thing it was derived for.

The width is within one per cent only from 1.65Q. The two-to-one argument above says why: a width within a per cent needs a peak within half of one, which is ln(200)/π = 1.69Q cycles if the skirts contributed nothing. They contribute a little in the helpful direction, and the measured wait comes in just under that estimate.

The difference between 1.50 and 1.65 is a tenth of the wait, which sounds small until the wait is counted in time. A dwell of $1.65Q$ cycles of a resonance at $f_0$ is $1.65Q/f_0$ seconds, and $f_0/Q$ is the resonance’s bandwidth in hertz, so the wait at each step is 1.65 divided by the bandwidth, whatever the frequency. A resonance ten hertz wide needs 165 milliseconds a step, and a sweep of two hundred steps across it takes half a minute. Reading its peak alone would save three seconds of that; reading its width at the peak’s wait would report it two per cent wide.

A wait that is a constant over a bandwidth is a family resemblance rather than a coincidence. The instrument’s own rise time is 0.35 over a single pole’s bandwidth, and the filter an average is settles to one per cent in 0.99 over twice its noise bandwidth. Each is a statement that a system which passes a band of width B cannot finish responding in much less than 1/B, and each has its own constant because each asks a different question of the response. The stepped sweep’s constant is large because its question is stringent twice over — a per cent of a width, which is two per cent of a height — and because the transient it is waiting for starts as large as the answer.

Put in hertz, the constant turns into times that decide how a measurement is planned. A resonance ten hertz wide needs 165 milliseconds a step, whatever frequency it sits at. A quartz crystal at ten megahertz with a quality factor of fifty thousand is two hundred hertz wide and needs about eight milliseconds a step, which is why a crystal can be swept in a second. A mechanical resonator at a kilohertz with a quality factor of ten thousand is a tenth of a hertz wide and needs 16.5 seconds at every step — two hundred steps across it is most of an hour, and a sweep that allows the peak’s count instead saves five minutes of that and reports the resonator’s quality factor two per cent low. The count measured here was measured at quality factors of ten and fifty; that it holds at fifty thousand is the dimensionless argument above rather than a measurement, and it is stated as one.

At a Q of fifty the numbers barely move when the wait is counted in multiples of Q: the peak from 1.50Q, the width from 1.60Q, 17.0 per cent wide after Q cycles against 19.6 at ten. The small drift is the resonance becoming more symmetric as Q rises — a capacitor’s voltage peaks slightly below resonance, by an amount that shrinks as $1/Q^2$ — and the dwell has the same shape at both. So the count is a property of reading a second-order resonance’s width by a stepped sweep from rest, and the time it costs is set by the bandwidth alone.

Two estimates of Q that stop agreeing #

A sweep over a resonance is usually taken for one number, its quality factor, and the three readings above give two ways of getting it: from the width, as the resonant frequency over the half-power bandwidth, and from the height, since a series resonator’s capacitor reaches Q times its drive. On the settled curve the two are 9.970 and 10.013, three parts in a thousand apart because a capacitor’s voltage peaks a little below resonance and its curve is not quite symmetric.

On the short dwells they part company. After 1.5Q cycles the width gives 9.766 and the height 9.916. After Q cycles the width gives 8.340 and the height 9.546, fourteen per cent apart. After half of Q cycles the width gives 4.885 and the height 7.792. Both are low, both are wrong, and they are wrong by different amounts, because the height is short by the centre’s deficit once and the width by roughly twice that.

That makes the disagreement a test a sweep can run on itself. Two estimates of one quality factor that differ by more than the settled curve’s own asymmetry are a measurement that did not wait, and the direction of the difference says which way the error lies: a width-derived Q below the height-derived one is a resonance read too wide. The Q the components allow and resonance, and the bandwidth it sets exactly both read a quality factor from a solved curve, where the two estimates agree to the resonance’s asymmetry; a measured curve offers the same check for nothing.

Why the skirts do not set the wait #

The question the arrow that goes past where it settles left was whether the skirts, where the transient’s beat is slow, need longer than the centre. They do not, and the reason is a coincidence that is not one.

A series resonance’s half-power frequencies sit at a fractional detuning of 1/2Q from resonance. The transient decays at a rate of $\omega_0/2Q$, and in the frame of the drive it turns at the detuning, which at those frequencies is also $\omega_0/2Q$. At the half-power frequencies, then, the second arrow turns one radian for every time constant it decays — 1.0528 of that at a Q of ten, where the capacitor’s asymmetric peak moves the half-power edge a little further out, and closer to one as Q rises.

That is fast enough to matter and not fast enough to hurt. The reading at the skirt swings through its settled value about once every six time constants, so its error changes sign, and what reaches the width from the two edges is a mixture of over and under that partly cancels. The centre’s error has one sign throughout. Both shrink at the same rate, exp(−t/τ), because both are the same decay; the centre’s simply never helps.

The spiral is the same fact drawn the other way. At resonance the tip runs straight out along its arrow; well away from resonance it circles many times before it has shrunk; at half power it makes about one loop per e-fold, which is the curve that settles fastest in the sense that matters to a reading — it spends as long above its mark as below it.

How the numbers were obtained #

Every reading comes from the closed form of the switched series circuit, the same solution whose arrows the arrow that goes past where it settles drew and whose march by the trapezoidal rule it checked. At each of 201 frequencies across ±2.5 resonance widths the largest $|v_C|$ in the last drive cycle of the dwell is found from 97 samples of that cycle; the resonance’s top is the largest of those readings, and each crossing of the top’s half-power level is bisected on the readings themselves. The width is compared with the settled width found by the same rule on the settled amplitude, and the dwell from which an error stays inside one per cent is read on a grid of 0.05Q from 0.5Q to 5Q. The held-peak reading is searched over the whole transient, as it was where the widening was first measured.

What it does not say #

It switches every step on at the crest of the drive. Switched on as the drive rises through zero instead, the second arrow starts with a different length and angle, and at a Q of ten the reading after Q cycles is 20.0 per cent too wide with its peak 4.41 per cent low, where the crest gave 19.6 and 4.66. The width is still within one per cent from 1.65Q cycles and the peak from 1.55Q rather than 1.50Q, and a held peak is 19.7 per cent wide rather than 20.2. The phase of switch-on moves the transient’s details and not the count, which is what a count that belongs to the decay rather than to the start should do.

It starts every step from rest. A real stepped sweep changes frequency without discharging the circuit, so each step starts from the previous frequency’s nearly settled state and the second arrow begins as the distance between two neighbouring phasors rather than as a whole one. For fine steps that is a much shorter arrow and a much shorter wait, and the count here is an upper bound on it rather than a measurement of it.

It reads the largest value over one cycle. A synchronous detector that multiplies by the drive and averages reads a different combination of the two arrows, and over one cycle of that averaging the second arrow’s contribution is filtered as well as decayed — the filter an average is is the arithmetic of that filtering — which would shorten the wait again.

And it measures one resonance with nothing near it. Two parasitics, and the resonance neither has finds a loop whose half-power level is crossed four times; a stepped sweep over that loop reads two resonances whose second arrows turn at two detunings, and no count measured on a single resonance transfers to it.

Still open: a sweep that keeps its state, a sweep that never stops, and switching off #

A step that starts where the last one ended. Carrying the circuit’s state from one frequency to the next makes the second arrow at each step the difference between two phasors a step apart. Near resonance that difference is largest, because the phasor turns fastest there, so a sweep that keeps its state should need its longest wait at the centre and almost none on the skirts — the opposite distribution to a sweep from rest. Measuring the wait against the step size would say how much a real instrument saves by not resetting.

A sweep that never stops. A continuous sweep through a resonance at a fixed rate in hertz per second has no dwell at all, and its error is set by how far the frequency moves in one time constant compared with the bandwidth — a rate with the bandwidth squared in it. The same closed form, driven by a chirp instead of a stepped sine, would put a number on the rate at which a swept resonance is read one per cent wide, and on how far its apparent peak moves in the direction of the sweep.

Switching off. A settled resonator whose drive stops keeps its second arrow and loses its fixed one, and whether the voltage on its way out can pass the settled value depends on the phase at which the drive stopped. That is the decay a gated burst meets at the end of every burst, and it is the half of switching this page has not measured.