The ring a drive leaves behind

The arrow that goes past where it settles switched a resonator on from rest and found its voltage overshooting the settled amplitude. The solution from rest is two arrows: a fixed one, the phasor, turning at the drive’s frequency, and a second one turning at the circuit’s own and dying away. Where the two line up the voltage is larger than it will ever be again. That essay and the two after it — how long a sweep waits at each step and a sweep that never stops — were all about arriving at a steady state.

Every gated burst also ends, and the end is the other half of switching. When the drive stops, the fixed arrow is gone: there is no longer anything turning at the drive’s frequency. What is left is a single arrow turning at the circuit’s own frequency, whose length and angle are whatever the circuit’s state was at the instant of the switch. At resonance that is the settled amplitude and nothing more. Off resonance it depends on when the drive stopped, and the dependence has a simple and exact physical reading: the ring starts with whatever energy the circuit held at that instant, and a detuned resonator’s energy is not the same at every instant of its cycle.

The circuit, and two routes to its ring #

The circuit is the series one these essays have used throughout: ten millihenries, a microfarad, and a resistor setting the Q, its output the voltage across the capacitor. It is settled at a drive $\cos\omega t$, and at a chosen instant the source is replaced by a short. At that instant the capacitor holds $|V_c|\cos\varphi$, where $\varphi$ is the phase of its own voltage in its cycle, and carries a current $C\,dv/dt = -\omega C |V_c| \sin\varphi$. After the switch the circuit has no input and rings at its damped frequency $\omega_d$ inside $e^{-\alpha t}$, with exactly those two starting conditions.

The free solution with those conditions is written down in closed form, which is one route. The other is to march the same circuit — a capacitor’s voltage and an inductor’s current, the source shorted — forward by fourth-order steps from the same settled state. Three cycles after the switch the two agree to a millionth of the settled amplitude, at both switch-off instants the hero figure draws. Every number below comes from the closed form, with that agreement standing behind it.

Where in the cycle the drive stopped #

The closed form gives the ring’s starting size directly. With $v(0) = |V_c|\cos\varphi$ and $v'(0) = -\omega|V_c|\sin\varphi$, the free response’s envelope starts at

$$\frac{A_0}{|V_c|} = \sqrt{\cos^2\varphi + \left(\frac{\alpha\cos\varphi - \omega\sin\varphi}{\omega_d}\right)^2},$$

which at high Q is very nearly $\sqrt{\cos^2\varphi + (\omega/\omega_0)^2\sin^2\varphi}$.

Switched at the capacitor’s crest, $\varphi = 0$, the ring starts at the settled size whatever the drive’s frequency: all the state is in the capacitor’s voltage and none in the current, and the ring inherits the voltage directly. Switched as the capacitor’s voltage passes through zero, $\varphi = 90°$, all the state is in the current, and a current of $\omega C|V_c|$ through an inductor that rings at $\omega_0$ carries a voltage of $\omega/\omega_0$ times $|V_c|$. So the ring starts larger than the settled amplitude above resonance — 1.2048 at 1.2 times and 1.5034 at 1.5 times — and smaller below it, 0.7983 at 0.8 times. At resonance the two cases are the same size, and what remains is the small tilt the damping adds, 0.9759 to 1.0260 at Q 10.

The figure is the switch-on result run backwards. Switching a resonator on at the drive’s crest gave a second arrow $\omega_0/\omega_d$ times the settled one, and through zero $\omega/\omega_d$ times. Switching off leaves an arrow whose length runs between the same two kinds of ratio, and for the same reason: in each case the second arrow is whatever makes the voltage and current continuous across the switch.

The energy the ring starts with #

The closed form has a physical reading that makes it obvious rather than derived. A free resonator’s envelope is set by its energy, $\tfrac{1}{2}CA^2$, and at the instant of the switch that energy is whatever the circuit was holding.

At resonance a settled resonator’s energy is constant: it moves between the capacitor and the inductor and the total never changes, which is why the ring after a resonant drive starts at the same size whenever the drive stops. Off resonance it is not. The capacitor’s energy peaks at $\tfrac{1}{2}C|V_c|^2$ at its voltage’s crest, but the current is set by the drive’s frequency, $\omega C|V_c|$, and the inductor’s energy at the current’s peak is $\tfrac{1}{2}L\omega^2C^2|V_c|^2$, which is $(\omega/\omega_0)^2$ times the capacitor’s. At 1.2 times resonance that is 1.44: the circuit’s total energy breathes between one unit and 1.44, twice a cycle, and the source puts the difference in and takes it back out each time.

So the ring after a switch-off starts with the energy of that instant, and its starting envelope is the square root: $\sqrt{1.44} = 1.2$ at most. The figure holds that relation at every phase to within the damping’s order, a tenth at Q 10. Below resonance the inductor’s peak is the smaller one and the capacitor’s crest the largest the energy gets, which is why no switch-off instant below resonance leaves a ring larger than the settled amplitude.

The breathing is the reactive power #

The energy figure has a name for what it draws. Averaged over a cycle the capacitor holds $\tfrac{1}{4}C|V_c|^2$ and the inductor $\tfrac{1}{4}C|V_c|^2(\omega/\omega_0)^2$, and the difference between the two, times twice the frequency, is the reactive power the source supplies: $\tfrac{1}{2}\omega C|V_c|^2\left((\omega/\omega_0)^2 - 1\right)$. At resonance it is zero and the stored energy is flat; above resonance it is positive, the circuit looks inductive, and the inductor’s peak exceeds the capacitor’s; below, the reverse. The current that does no work measured reactive power as the power that flows out to a reactance and back again each cycle, and the energy a unity power factor doubles found the energy stored in a corrected load rising rather than vanishing when the reactive power went to zero. The switch-off is where that flow stops being an accounting entry: the source that was putting energy in and taking it out twice a cycle is suddenly not there to take it out, and whatever was in at that instant stays in.

So the excess ring above resonance is the reactive power’s half-cycle of energy, caught at its peak. A designer who wants a gated resonator to ring out no larger than it was driven has two choices, and both come from this reading: drive it at or below its resonance, where the inductor’s peak is never the larger, or stop the drive at the instant the capacitor’s voltage is at its crest, when all the energy is in the capacitor and the inductor holds none.

A second route to the first crest #

The first crest after a switch at the capacitor’s zero can also be had without the search. The ring starts with no voltage and all of its state in the current, so it is $B\,e^{-\alpha t}\sin\omega_d t$ with $B$ the starting envelope, and its first crest comes close to a quarter of the ring’s period after the switch. In a quarter period the envelope falls by $e^{-\alpha\pi/2\omega_d}$, which at high Q is $e^{-\pi/4Q}$. At Q 10 that is 0.9245, and $1.2015 \times 0.9245 = 1.111$, against the 1.1120 the search finds — the small difference being that a damped sine’s crest comes a little before its quarter period. The same estimate at Q 50 gives 0.984 of the start, which is why the high-Q curve in the detuning figure runs so close to $\omega/\omega_0$.

It also says where the low-Q curve falls away from the high-Q one. At Q 3 a quarter period costs $e^{-\pi/12}$, about a quarter of the start, and at twice the resonant frequency the best switch-off instant moves off the capacitor’s zero toward a phase that trades a smaller start for an earlier crest. That is why the Q-3 curve’s largest crest, 1.606 at twice resonance, is more than the zero-instant estimate of $(\omega/\omega_d)\,e^{-\pi/12} \approx 1.56$ would give: the search finds a better instant, and the estimate only bounds the zero one.

The first crest, against the detuning #

A ring that starts large need not reach large: its envelope is falling from the first instant, and the first crest comes a quarter-cycle or so after the switch.

Below resonance the largest first crest is the settled amplitude at every Q, reached by stopping the drive at the capacitor’s crest. Above resonance it rises with the detuning and approaches $\omega/\omega_0$, the energy’s square root: at twice the resonant frequency, 1.969 at Q 50, against an ideal two. A lower Q starts the ring at the same size and loses part of it to damping before the first crest: 1.606 at Q 3, where a quarter of a cycle of decay at $\omega_0/2Q$ costs a noticeable fraction.

The figure answers the question the earlier essays left: the voltage on its way out can pass the settled value, and it does so only above resonance, by up to the ratio of the drive’s frequency to the circuit’s. A resonator driven below its frequency never rings out larger than it was driven. One driven above it can, by a factor that is not small: a circuit driven at twice its frequency, with a reasonable Q, rings out at nearly twice the voltage it was settled at.

The end of a burst #

A gated burst is a switch-on followed by a switch-off, and where it stops in the drive’s cycle is set by how many cycles it holds.

The two bursts differ by a quarter of a drive cycle and their tails by a fifth of the settled amplitude. Stopped after a whole number of cycles — with the capacitor near its crest, since above resonance it is nearly in antiphase with the drive — the tail never exceeds the settled level. Stopped a quarter-cycle later it reaches 1.207, a little more than the 1.112 a settled resonator would ring to, because at Q 10 eight cycles are not enough for the switch-on’s own second arrow to have died, and the two transients add.

That is the shape of a tone burst on a resonant transducer, a gated carrier into a tuned filter, a pulsed transmitter into its tank. The burst’s length is usually chosen in whole cycles of the drive for other reasons, and the tail it leaves is then a function of where the capacitor’s voltage happens to be at that count — which above resonance can differ from one burst length to the next by twenty per cent.

A step in frequency is a switch-off and a switch-on #

A stepped sweep that carries its state from one frequency to the next, rather than resetting, makes every step a switch-off of the old drive and a switch-on of the new one at the same instant. How long a sweep waits at each step started each step from rest, where the whole of the new settled state had to be built from nothing. Carrying the state changes what has to be built: the old drive’s arrow is removed and the new one’s added, and the second arrow that does the transient work is their difference, with the energy breathing figure deciding how much of the old state is in the inductor and how much in the capacitor at the instant of the step.

That makes the step’s own instant a design variable a sweep usually ignores. Stepping when the capacitor’s voltage is at its crest hands the new drive a state whose energy is all in the capacitor, which is the state with the least excess on the far side of resonance. Stepping at its zero hands over the inductor’s peak. With a resonance’s bandwidth set by its Q, as resonance, and the bandwidth it sets exactly measured, neighbouring steps inside the bandwidth differ little and the instant hardly matters; steps that cross from one side of the resonance to the other hand over an energy that differs by the ratio of the two frequencies squared.

Reading a Q from the tail #

A ring-down is the standard way to measure a resonator’s Q: count the cycles for the envelope to fall by a known factor. Five readings of one Q found the ring-down reading $Q\sqrt{1 - 1/4Q^2}$, close to the true Q above three or so. What the switch-off figures add is that the ring-down’s starting size is not the settled amplitude unless the drive was at resonance or stopped at the capacitor’s crest. A measurement that takes the settled amplitude as the ring’s first value and the first crest after the switch as its second will mistake the starting excess for a slower decay, or the starting deficit for a faster one.

At 1.2 times resonance the excess is up to 20 per cent, which read as one cycle’s decay is a Q in error by a large factor. The cure is the one the closed form suggests: read the decay between crests after the switch, never between the settled amplitude and the first crest, or drive at resonance, where the energy does not breathe and the start is the settled size whatever the instant.

The arrows, once more #

Three voltages that close on one drew the series circuit’s phasors as a closed polygon and noted that it describes a settled circuit. The switch-off is the most direct demonstration of what “settled” means in that picture. At the instant the drive stops the polygon is still exact — every element’s voltage is still the phasor’s real part — and one instant later it is meaningless, because there is no longer a frequency for the arrows to turn at. What survives is the state: one voltage and one current, two numbers, from which the whole future follows. The phasor was always a description of that state that happened to repeat; the switch-off shows which part of it was the circuit’s and which was the source’s.

And it closes the account the switch-on began. Turning a drive on adds an arrow of the circuit’s own frequency, of a length the switch-on instant decides; turning it off removes the drive’s arrow and leaves one of the circuit’s own, of a length the switch-off instant decides. In both cases the arrow that does the transient work turns at $\omega_d$, and in both cases its length is between one and $\omega/\omega_0$ of the settled amplitude.

What the closed form assumes #

The switch is perfect: the source becomes a short in no time, and nothing is left behind of it. A real gate has a resistance that changes over a finite time, and a switch that opens rather than shorts leaves the series circuit with no path at all, in which case the inductor’s current has nowhere to go and the circuit is a different one. The tail of a burst is drawn with the source shorted throughout the ring, which is what a driver with a low output impedance does when its input stops.

The resonator is linear, and its Q does not depend on its amplitude. A ring that starts at twice the settled voltage is a ring at twice the stress, and a capacitor or core whose loss rises with voltage would lose more in its first cycles than the closed form allows.

The number to keep #

A resonator whose drive stops rings with the energy it held at that instant. Off resonance that energy breathes between the capacitor’s peak and the inductor’s, in the ratio $(\omega/\omega_0)^2$, so the ring starts between one and $\omega/\omega_0$ of the settled amplitude according to the phase of the switch. Below resonance no instant leaves it larger than it was driven; above, the first crest approaches $\omega/\omega_0$ — 1.386 at 1.5 times resonance and 1.856 at twice, at Q 10.

Still open: the switch that opens, the burst that repeats, and the detector’s lag #

A switch that opens instead of shorting. A driver that goes to a high impedance leaves the series circuit open, and the inductor’s current must then go somewhere — into a stray capacitance, through a clamp. The energy the circuit held is the same; where it goes, and what voltage the open switch sees while it goes there, is a different and more dangerous question.

A burst train. Repeated bursts with gaps shorter than the ring’s decay start each burst from the previous tail, so the switch-on is not from rest. Whether the train settles to a periodic state whose tails are larger or smaller than one burst’s, and how that depends on the gap, is the next measurement for a pulsed system.

A ring read through a detector. An envelope detector after the resonator adds its own lag, and the tail it reports starts later and lower than the circuit’s. Whether the starting excess above resonance survives a detector of a stated time constant would say whether a burst’s tail can be measured at all without the phase of the switch.