A sweep that never stops

How long a sweep waits at each step measured a resonance the way a stepped instrument does: set a frequency, start from rest, wait, read, move on. It found that the resonance’s width comes out right to a per cent only after about 1.6Q cycles at each step, because every step begins with the circuit’s own transient — the arrow that goes past where it settles is that transient drawn as a second phasor — and the width’s error is the peak’s shortfall read twice.

It ended with the instrument that does not step. A continuous sweep through a resonance at a fixed rate has no dwell at all: the drive is always moving, the circuit is always catching up, and the phasor picture that makes a steady state a single arrow is never exactly true. That essay guessed the answer’s form — 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 — and asked for the number and for how far the apparent peak moves in the direction of the sweep.

The resonator and its chirp #

The circuit is the one those essays used: a series resonator of 10 mH and 1 µF, resonant at 1591.5 Hz, with its resistance chosen for a stated Q, read at the capacitor. The drive is a chirp — $\cos\bigl(2\pi(f_st + \tfrac12 rt^2)\bigr)$ — whose frequency rises linearly at $r$ hertz a second from ten bandwidths below resonance to ten above, with $B = f_0/Q$ the bandwidth.

The reading is the capacitor’s envelope, plotted against the frequency the drive has reached at that instant: exactly what a swept analyser draws. The envelope is taken at every step of the march as $\sqrt{v^2 + (i/C\omega)^2}$ with $\omega$ the drive’s instantaneous frequency — the amplitude a sinusoid at that frequency would have with that voltage and current — so it is smooth to the sample and needs no search for peaks.

The rate that has the bandwidth squared in it #

The earlier essay’s guess can be made exact by counting. A resonance of bandwidth $B$ has a time constant of $1/(\pi B)$ — it takes about that long for its amplitude to respond to a change, the pole’s distance from the imaginary axis that the circle a peak sits on draws — and a sweep at $r$ hertz a second moves $r/(\pi B)$ in that time. As a fraction of the bandwidth that is $r/(\pi B^2)$, and the only rate a single resonance can be swept at, measured in its own units, is

$$\eta = \frac{r}{B^2}$$

Equivalently, a sweep at $\eta = 1$ takes $1/B$ seconds to cross one bandwidth, which is $f_0/B = Q$ cycles of the resonance — the same Q cycles a step that the stepped sweep found the minimum useful dwell. The two instruments are measured on the same scale, and the table below is the continuous sweep’s version of the stepped one’s dwell curve.

At a Q of fifty the bandwidth is 31.8 Hz and $\eta = 1$ is 1,013 Hz a second: a sweep from ten bandwidths below to ten above, 636 Hz, takes 0.63 seconds. At that rate the peak reads 6.4 per cent low, lands 0.445 bandwidths above resonance and appears 26 per cent wide. To read within three-tenths of a per cent of the height takes $\eta = 0.1$ — six seconds for the same span — and still leaves the peak a twentieth of a bandwidth late.

One number, at any Q #

The claim that $\eta$ is the only rate is testable, and it is the reason for the second figure.

At $\eta = 1$ a Q-10 resonance reads 93.1 per cent of its height and a Q-200 one 93.6, beside Q 50’s 93.6: the same to under a per cent across a factor of twenty in Q, and the figure requires it to within two per cent at three rates. What differs at Q 10 is only the resonance’s asymmetry — a capacitor’s response carries a $1/\omega$, and at a Q of ten the bandwidth is a tenth of the frequency, so the two skirts are visibly unequal. Every narrow resonance, from a filter at a kilohertz to a crystal at ten megahertz, is swept with the same errors at the same $\eta$.

The practical content is the scaling. The rate a resonance tolerates goes as the square of its bandwidth, so halving the bandwidth quarters the rate and quadruples the time per bandwidth. A crystal with a bandwidth of ten hertz, swept to the three-tenths-of-a-per-cent accuracy of $\eta = 0.1$, must move at ten hertz a second — a second a bandwidth, and every bandwidth of span another second.

Why the peak arrives late #

The height and the width both have the stepped sweep’s explanation: the circuit has not reached the steady state for the frequency it is being driven at. The shift is new, and it is the continuous sweep’s own.

A resonator responds to the drive with a delay of about its time constant, $1/(\pi B)$. In that delay the drive has moved on by $r/(\pi B)$, so the amplitude the circuit shows at any instant belongs to a frequency a little behind the one the drive has reached. The reading is the steady curve shifted towards where the drive came from — and plotted against where the drive has got to, the peak appears beyond resonance in the direction of travel. The one-delay estimate is $\eta/\pi$ bandwidths, 0.32 at $\eta = 1$, against the 0.445 measured. It is short because a delay is not the only thing happening: the envelope is still growing when the drive crosses resonance, so its maximum is reached later than a pure delay would put it. The simple model gives the direction and the size; the march gives the number.

The ringing is the same delay seen after the peak. Once the drive has passed resonance, the circuit still holds energy at its own frequency and releases it at a frequency the drive has already left, and the two beat. The envelope’s oscillation on the far side is that beat, and the faster the sweep the larger it is, as the hero figure’s η = 3 curve shows.

The ringing, which is the steady state catching up #

At three times the rate the reading has stopped looking like a resonance curve at all. The rising side is a slower, lower version of the steady curve; the falling side is a decaying oscillation. The oscillation is the phasor picture made visible: three voltages that close drew a steady state as one arrow, and the earlier essays drew a switch-on as a second arrow turning at the circuit’s own frequency and shrinking. A sweep lays down a second arrow continuously, at every frequency it passes, and after resonance those arrows add to a free oscillation at $f_0$ that the drive, now above $f_0$, beats against. The beat’s period is the difference frequency, which grows as the drive moves away, so the ripples crowd together and shrink as the free oscillation decays.

That makes the far side of a fast sweep informative rather than wrong. The decay of its ripples is the resonance’s own time constant, and their spacing is set by how far the drive has moved; a swept measurement that records the ripples can read the Q from the ring-down rather than from the width, which is how a resonance too narrow to sweep slowly is sometimes measured.

Q, read from a sweep #

A resonance is usually swept to find its Q, as the centre frequency over the width, and the width is the error this rate law bites hardest. At $\eta = 1$ the width reads 1.263 bandwidths, so the Q reads 79 per cent of its value: a component that is being graded by its Q is graded low by a fifth. At $\eta = 0.1$ the width is 0.7 per cent wide and the Q 0.7 per cent low.

The Q the components allow is the ceiling a resonator’s parts put on its Q, and a measurement of that ceiling swept at the wrong rate reports a lower one; since both the components and the sweep lower the reading, there is nothing in the reading to say which did it. The only defence is the one the table gives: know $\eta$, and keep it below a tenth when the width matters.

The peak’s height is the other half of a Q measurement for a resonator driven through a known source, and it is less sensitive: at $\eta = 0.3$ the height is 98.7 per cent while the width is already 4.2 per cent wide. Read through the height the Q is three times less sensitive to the rate than read through the width, which is the opposite of the stepped sweep, where the width’s error was the peak’s read twice.

What a swept analyser’s rule is saying #

Swept instruments carry a rule for the fastest sweep a resolution filter allows, and the rule is always of the form this essay found: the sweep time must exceed a constant times the span divided by the square of the bandwidth. That is $\eta$ below a constant, stated backwards — span over time is the rate, and dividing by the bandwidth squared makes it the same dimensionless number. The constant differs between instruments because their filters differ in shape; the bandwidth noise sees is a reminder that a filter’s bandwidth is itself a choice of definition, and a rule written for a Gaussian filter’s bandwidth does not transfer to a single resonance’s without re-deriving the constant.

What the resonance adds to the rule is the reason for it. The square appears because the bandwidth enters twice: once as the width of frequency the sweep has to resolve, and once, inverted, as the time the circuit needs to respond. A filter that is half as wide has to be crossed in twice the time to cover the same fraction of itself, and it takes twice as long to respond — so the sweep must be four times slower, and a span at a quarter of the rate.

Two sweeps, and the resonance between them #

Because the shift is always in the direction of travel, sweeping the other way puts it on the other side.

The two shifts are nearly equal and opposite — 0.445 up, 0.460 down, three per cent apart for the same capacitor asymmetry as before — so the mean of the two peak frequencies finds the resonance to 0.0075 bandwidths, a sixtieth of the error either sweep makes alone. An instrument that sweeps both ways and averages can locate a resonance at a rate at which one sweep is off by half a bandwidth, and it pays for it with one extra sweep rather than with the tenfold slower rate a single sweep would need for the same accuracy.

What the pair does not recover is the height or the width. Both sweeps read the peak low by the same amount and wide by the same amount, since neither error has a direction; averaging them changes nothing. So the two-way trick is for finding where a resonance is, and a slow sweep is still the only way to find how sharp.

The dwell, compared #

The stepped sweep and the continuous one now sit on one scale, and the comparison is worth making in the unit both share.

The stepped sweep waited for the transient to die at each step: its width came right to a per cent at 1.6Q cycles a step. The continuous sweep at $\eta = 1$ spends exactly Q cycles on each bandwidth, and its width is 26 per cent wide; to get the width to within a per cent it needs a rate a little above $\eta = 0.1$, about eight Q cycles a bandwidth. For the same accuracy the continuous sweep is several times slower per bandwidth — but it spends no time on steps at all, where a stepped sweep needs several steps per bandwidth to draw the shape, each costing 1.6Q cycles. Which is faster over a whole measurement depends on how finely the stepped one steps, and that is a design choice the stepped essay’s arithmetic now meets this one’s.

The deeper difference is in what goes wrong. A stepped sweep that waits too briefly reads the peak low at the right frequency. A continuous one that moves too quickly reads it low at the wrong frequency. The first error is symmetric and the second is not, which is why the second has a cure — sweep both ways — and the first has only patience.

How the sweeps were computed #

Each sweep is a Runge–Kutta march of the series resonator’s two state variables at sixty-four steps a cycle of its resonance, driven by the chirp in closed form, from ten bandwidths below resonance to ten above (or as many as fit above half the resonant frequency, for a low Q). The envelope is read at every step from the capacitor’s voltage and current at the drive’s instantaneous frequency. The peak is the largest envelope; the width is the distance between the half-power crossings either side of it, interpolated between samples.

Checked two ways. Swept at $\eta = 0.03$ the reading is required to be the steady-state response — the peak within 0.2 per cent of Q, the width within half a per cent of one bandwidth — which is the check that the envelope and the march measure the resonance and not themselves. And a short sweep of a Q-10 resonator at $\eta = 3$ is marched a second time, as the netlist under the trapezoidal rule with the same chirp as its drive, and the two marches are required to end at the same capacitor voltage to two thousandths of the resonance’s own gain.

A linear sweep past one resonance #

The sweep is linear in frequency. Analysers often sweep logarithmically, which makes the rate proportional to frequency; at the resonance the only rate that matters is the local one, so the table applies with $r$ read at the peak.

The detector is ideal. A real swept analyser follows its resonance with a filter and a detector of its own, each with its own time constant, and the rate limit of the whole instrument is set by whichever of those is narrowest; the filter an average is is the detector’s share of that chain written as a bandwidth. The resonance being measured is one more filter in that chain, and it is the one this essay isolates.

And the resonance is one pole pair. Two resonances within a few bandwidths of each other are swept into each other’s ringing, and the beat of one lands on the skirt of the next; resonance and its bandwidth is the single case, and the double one is a different measurement.

Still open: switching off, the sweep that keeps its state, and the optimum rate with an averaging detector #

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 none of these essays has measured.

A stepped sweep that keeps its state. Carrying the circuit’s state from one frequency to the next makes each step’s transient the difference between two neighbouring steady states rather than the whole of one, and a stepped sweep with small steps that never resets is a continuous sweep in disguise. Where the stepped sweep’s dwell law turns into this essay’s rate law, as the steps shrink, is the measurement that would join them.

The detector’s own time constant. A real instrument reads the envelope through a filter, which adds its own delay and its own smoothing. With the resonance and the detector both lagging, the peak’s shift is the sum of two delays, and the best detector for a given sweep rate is the one whose lag the two-way average cancels along with the resonance’s.