The arrow that goes past where it settles

Three voltages that close on one drew a series resonator’s element voltages as arrows in the complex plane, found them closing on the source to a part in 10¹², and then put a condition on the whole picture: the arrows describe a circuit that has settled, and a transient takes about 1.47Q cycles to fall to a per cent. Before that, it said, the circuit is ringing at its own frequency while responding to the drive, and no arrow describes the sum.

That last sentence is true of the arrows it drew and not true of arrows in general. The transient can be drawn as an arrow too, in the right frame, and when it is, it shows something the settling count cannot: the circuit’s voltage does not approach its phasor. Away from resonance it overshoots it, by as much as twice, and at resonance — where the settled voltage is largest — it is the only frequency where it does not.

A fixed arrow and a second one #

In the frame that rotates with the drive, a settled sinusoid does not move: the phasor is a fixed point, and its arrow stays where it is. The transient is a sinusoid at the circuit’s own natural frequency, decaying at the rate its resistance sets. In the drive’s frame it is a second arrow, turning at the difference between the two frequencies and shrinking as it turns. The capacitor’s actual voltage is the tip of the sum.

At switch-on the circuit is at rest, so the sum is zero: the second arrow starts pointing straight back at the origin, as long as the first. Then it turns. If the drive is a little off resonance the second arrow turns slowly, at the detuning, and after half a turn it points the same way as the first. The tip is then far beyond the settled circle — as far as the two lengths add — and the second arrow has shrunk only by however much it decayed in half a beat.

That is the approximation in the caption, and its structure is visible in the drawing. Half a beat at a fractional detuning δ is π/(ω₀|δ|); the decay rate is ω₀/2Q; so the second arrow has fallen to exp(−π/2Q|δ|) of its starting length when it lines up, and the voltage reaches 1 + exp(−π/2Q|δ|) times its settled value. At a Q of 50 and five per cent below resonance that is 1.5335, and the exact solution reaches 1.5499 in the ninth cycle.

The two figures differ because the approximation makes two simplifications the figure does not. It takes the second arrow to start exactly as long as the first, when switching on at the crest it starts $\omega_0/\omega_d$ times as long; and it takes the alignment to happen at an instant when the voltage is at a crest, when the largest voltage is wherever the envelope’s peak and a crest of the drive coincide. The figure computes the exact solution of the switched circuit and, as a second route, marches the same netlist forward by the trapezoidal rule; the two agree to about a thousandth of the settled amplitude.

At resonance the second arrow does not turn #

Driven at resonance there is no detuning for the second arrow to turn at. It starts pointing back along the settled arrow and stays pointing that way, shrinking, so the tip moves straight out from the origin towards the settled point and approaches it from inside. At a Q of fifty it has come to within two parts in a hundred thousand of fifty volts by cycle 183, and it has never been above it.

Strictly, the second arrow does turn at resonance, because damping lowers the circuit’s own frequency below the undamped resonance the drive is set to. At a Q of fifty the natural frequency is √(1 − 1/4Q²) of it, 0.99995, so the second arrow turns once in about 8Q² drive cycles — twenty thousand. By the time it has turned half-way, ten thousand cycles in, it has decayed by a factor of e to the minus 628, which is to say it has not turned at all while it had any length. The approach from inside is not an approximation that happens to be good; it is the whole of the transient.

That makes a statement that sounds wrong until the arrows are drawn. The frequency at which the settled voltage is largest is the one frequency at which the settled voltage is never exceeded. Everywhere else on the axis the circuit overshoots its phasor, and the largest instantaneous voltage in the switched circuit can be at a frequency where the phasor is much smaller.

The resonance is also the frequency at which the approach is slowest: at resonance the voltage climbs to fifty times the drive and needs about 1.47Q cycles to be within a per cent of it, which is the count three voltages that close on one gave. Away from resonance the settled voltage is smaller and the first overshoot arrives within a few cycles, long before the settling is done.

Further off, and higher #

Twenty per cent below resonance the second arrow turns fast — half a beat is two drive cycles — and decays hardly at all before it lines up. The voltage reaches 1.8543 times its settled amplitude in the third cycle, and the approximation is right to three parts in ten thousand, because a fast beat is the case in which lining up the arrows at an instant is a good description of what happens.

The figure distinguishes two quantities that are easy to conflate. The tip’s farthest distance from the origin is 1.8565 times the settled amplitude; the largest voltage the capacitor actually reaches is 1.8543 times. The first is the envelope, and the second is the envelope sampled by the drive’s own phase: the voltage is the arrow’s projection at each instant, and the arrow is not quite pointing along the axis of measurement when its length is greatest.

Switched on through zero, above resonance, the overshoot passes twice the settled voltage: 2.1820 at a Q of 200. The approximation cannot say that, because it takes the second arrow to start exactly as long as the first and so tops out below two. Switching on through zero starts the second arrow $\omega/\omega_d$ times the settled one’s length, and above resonance that ratio is larger than one — here 1.25 — so when the two line up the sum exceeds twice the settled length before any decay has been subtracted.

The energy stored in the capacitor goes as the square of its voltage, so at that peak the capacitor holds 4.76 times its settled peak energy. A part rated for the voltage a phasor predicts, in a resonant circuit that is switched on off its resonance, is rated for less than half of what it will see.

The circuit whose three voltages close #

The circuit three voltages that close on one drew has a quality factor of 2.13, and at that Q the second arrow has decayed to almost nothing before it lines up with the first. Its largest overshoot on the whole sweep is thirteen per cent at the crest and twenty through zero, which is why its arrows could be drawn as a settled picture after three cycles without anything visibly wrong. The effect is a property of Q, and that circuit sits at a Q at which it is small.

The numbers make the reason plain. At a quality factor of 2.13 the second arrow keeps exp(−π/Q) of its length each of the circuit’s own cycles, 0.229, so it loses more than three quarters of itself every cycle. Thirty per cent above resonance half a beat is 1.67 of those cycles, by which time the arrow is 8.6 per cent of its starting length: that is the approximation’s 1.086 at 1.3 times resonance. The exact 1.1310 is larger because the arrow starts longer than the settled one and the alignment does not fall on a crest, and at a low Q neither of those is a small correction.

The approximation is at its worst here, 4.40 per cent out at 1.24 times resonance. That is expected: it is built on the second arrow surviving half a beat with little decay, and at a Q of two there is no half beat to speak of.

At a Q of fifty the ratio climbs from exactly one at resonance to nearly two within ten per cent of it, and the through-zero curve passes two above resonance. The approximation is symmetric in the detuning and the exact curve has no reason to be, since the settled amplitude is not symmetric about resonance and the second arrow’s starting length depends on which side of it the drive is. The approximation, which is symmetric in the detuning, misses by up to 3.15 per cent — at 0.72 times resonance, where the beat is fastest and its assumption about when the arrows align is least accurate.

At two hundred the curve is a narrow notch at resonance inside a plateau near two. The approximation’s worst error has fallen to 2.55 per cent, and it will keep falling as the Q rises, but it is not converging to the exact curve at the edges of the sweep, where the ratio above resonance through zero is 2.26 and no approximation that tops out at two can reach it. It is a good estimate of the effect’s size and a poor model of its shape, which is the distinction where the behaviour is written down draws between a pole’s location and the waveform it produces.

A resonance read by its peaks #

A resonance is measured by sweeping a drive and reading an amplitude, and the phasor reading of that measurement assumes each frequency is read after the circuit has settled. An instrument that steps its frequency, starts each step from rest and records the largest reading at each step is measuring something else.

The two curves agree at the top, because at resonance there is no overshoot. They part on the skirts, where every frequency overshoots its settled value, and so the peak-read curve is broader: at a Q of ten its half-power width is 0.1205 of the resonant frequency against the phasor’s 0.1003. The phasor’s width is f₀/Q to the digits resonance, and the bandwidth it sets exactly found; the peak-read width is twenty per cent more.

A quality factor computed from that width as the resonant frequency over the bandwidth would be reported as 8.30 for a circuit whose Q is ten.

At a Q of fifty the same comparison gives 0.0240 against 0.0200, 1.198 times as wide. So across a factor of five in Q the peak-read resonance is about 1.2 times wider than the phasor one, which is measured at two quality factors and not shown here to be a law. What it does show is the sign and size of an error in a measurement that looks like the definition: a quality factor inferred from a peak-read sweep is about a sixth too low, and the error does not come from the instrument’s bandwidth or its noise but from reading a transient as a steady state.

Where a phasor is overshot #

The effect belongs to any resonance switched on away from its own frequency, and it appears in circuits nobody would call resonators.

A tuned stage gated in bursts. A resonant load driven in bursts — a transducer, a resonant tank in a converter, the matching network in front of an antenna — starts every burst from rest. If the burst is at the resonance the voltage creeps up and never passes its settled value; if it is a little off, as it will be once a component drifts, the first few cycles overshoot by up to about a factor of two and the stored energy by up to 4.76, and a part rated from the phasor is under-rated for the transient it meets most often.

A rectifier’s first cycle. The first cycle no steady state contains found a reservoir capacitor’s first charging current twenty-six times anything the settled circuit ever draws again, decided by when a switch closed. That circuit is nonlinear and its excess is an inrush rather than a beat, but the lesson is the one drawn here: a quantity computed from the settled state is not a bound on the transient, and the instant of switch-on is a parameter.

A loop that never settles. The gain that is exactly one has a second arrow that grows rather than shrinks — a loop gain three per cent high multiplies the envelope by 1.0987 a cycle — so there is no fixed point for the tip to circle and nothing to be overshot. The spiral drawn here is the stable half of that essay’s picture.

A measured quality factor. Every quality factor the Q the components allow and two parasitics, and the resonance neither has measure is a steady-state quantity read from a solved response. The second of those found a loop with two resonances whose half-power level is crossed four times. Switched on, such a loop carries two second arrows turning at two detunings, and how far past its settled voltage it goes has not been measured here.

What the arrows here assume #

The drive is an ideal sinusoidal source, switched on at an instant, into a circuit holding no energy. A real switch has its own transient and a real source its own impedance, and either would change the second arrow’s starting length and angle. The circuit is linear; the amplitude nothing linear predicts is the case where the component that sets the steady state is itself nonlinear, and there no phasor exists to be overshot.

The frame that turns with the drive is a drawing device and nothing more — the solution is the time-domain solution of the switched network, and the arrows are that solution plotted in the frame in which its settled part stands still. One solve, read four ways introduced the phasor as the number the frequency-domain solve returns; this figure is what the time-domain solve looks like when it is asked to agree with that number, and the answer is that it agrees eventually and not on the way.

Still open: switch-off, the dwell, and a second frequency #

Switching off. A settled resonator whose drive stops loses its fixed arrow and keeps the second one, which then decays from the settled length towards zero while turning at the circuit’s own frequency. Whether a circuit switched off can reach a voltage beyond its settled one — and whether switching off at a different phase of the drive changes what the circuit’s energy does on its way out — is a separate question from switch-on, and the same arrows would answer it.

How long a stepped sweep must dwell. The peak-read widening disappears if each step waits for the second arrow to decay before reading. Measuring the dwell, in cycles, at which a peak-read width is within a per cent of f₀/Q would put a number on a measurement condition — and would tell whether 1.47Q is the right count for it or whether the skirts, where the beat is slow, need longer.

A drive that changes frequency. A resonator driven at one frequency and switched to another starts from a non-zero settled arrow rather than from rest, so the second arrow’s starting length is the distance between two fixed points. That is how frequency-shift keying looks to a resonant filter, and the overshoot at each change would be larger or smaller than switch-on’s according to where the two fixed points sit on the resonance curve.