Circuit Response
every model has an edge, and the figure carries it
Concept

Resonance — where it appears

The frequency at which a network's stored electric and magnetic energies are equal, so that its reactance vanishes. The frequency it happens at is not the frequency of the peak response, and the two separate as the damping rises.

Named by 19 essays across 4 fields — each of them below, with the objects they name alongside it.

A resonant circuit of Q = 8, and its measured bandwidth. The half-power points are 1.50 kHz and 1.69 kHz, a bandwidth of 198.9 Hz. The components predict f₀/Q = 198.9 Hz. They differ by 0.000%.

Resonance, and the bandwidth it sets exactly

The half-power bandwidth of a resonant circuit is f₀/Q — not approximately, but to every digit the arithmetic has, which is rare enough to be worth checking. What is not exact, and is drawn as though it were, is the idea that the band sits centred on the resonance. At a quality factor of one its middle is twelve per cent above.

frequency · Resonance Which mechanism sets the upper edge, against the load. computed by solving, not by drawing. Two candidate upper edges drawn against the measurement. The one every textbook names is a resonance between the leakage inductance and the winding capacitance; the one that actually binds at ordinary loads is the leakage in series with the load, a first-order corner at R/2πL. At 50 Ω they are 80.4 kHz and 1.13 MHz — a factor of 14 apart — and the measurement follows the first, to 19.5% at worst across nine loads. They swap at about 1500 Ω, above which the resonance is the binding one and the usual picture is right — which is why a transformer feeding a high impedance behaves as the textbooks say and one feeding fifty ohms does not.

Which picture sets the upper edge

Every account of a transformer's high-frequency limit names the same mechanism: the leakage inductance resonating with the winding capacitance. At fifty ohms that resonance is at 1.13 MHz and the measured edge is at 81.3 kHz — a factor of fourteen away — because what actually binds is the leakage in series with the load, a first-order corner with no resonance in it at all. The two swap at about 1500 Ω, and both accounts are current because both are sometimes right.

magnetics · Transformer The load at which a transformer becomes a resonant circuit, at k = 0.99. computed by solving, not by drawing. The peak output of the same 1:1 transformer against its load, as a multiple of what the turns ratio would give. Below about a kilohm the load damps the leakage resonance, the peak is the plateau, and the ratio is one: there is a band, and it is what the rest of this field measures. Above it the damping goes and the response peaks — 1.18× at 832 kHz into 1500 Ω, rising to 8.80× at the light end. A transformer with voltage gain is not a transformer behaving badly; it is a resonant circuit, and asking for "the band" of one returns the skirts of a resonance. So the measurement reports that the response is peaked, rather than returning two edge frequencies in the wrong order.

Where the band goes entirely

The three essays before this one measure a transformer's band, and all three assume there is one. Past a few hundred ohms of load there is not: the leakage that sets the upper edge is also what damps the resonance behind it, and a lightly loaded, well-coupled transformer peaks at 3.03 times its own turns ratio. A passive component with voltage gain is not a transformer behaving badly. It is a resonant circuit, and asking for its band returns the skirts of a resonance.

magnetics · Transformer A resonator's Q against its inductor's, with a capacitor of Q 1581. computed by solving, not by drawing. The dashed line is what the resonator's Q would be if the inductor were its only loss; the solid one is what it is with a capacitor of Q 1581 beside it. They part company where the inductor stops being the worst component. At the marked point the inductor's Q is 79.06, the capacitor's is 1581.1, the reciprocals predict 75.2923 and the solved network measures 75.2923 — 2.2e-7% apart, by two routes that share only the element values. The resonance stays at 1/2π√(LC) to a part in a million throughout.

The Q the components allow

Resonance and its bandwidth measured a half-power width of exactly f₀/Q at every Q tried, for a circuit whose only resistance was the one deliberately put there. Real components arrive with resistance of their own, and the consequence is a ceiling rather than a penalty. The reciprocals of the component quality factors add, so the total sits below the smallest of them: an inductor of 79 beside a capacitor of 1 581 gives a resonator of 75. The worst component decides and the best one cannot help.

frequency · Resonance 50 Ω + j100 Ω of line, and the load angle past which the far end rises. computed by solving, not by drawing at 71 load resistances and four load angles. A source of 50 Ω + j100 Ω feeding loads of the same resistance and different power factor: at unity power factor the voltage across the load climbs towards the source's and stops there, reaching 0.9675 of it at the largest load drawn. A lagging load leaves less. A leading one leaves more, and past a computable angle it leaves more than the source has: the condition is 2Rₗ(Rₛ + Xₛ·tanφ) + |Zₛ|² < 0, which for the largest load here is 28.35° of lead — bisected on the solve at 28.35° — tending to atan(Rₛ/Xₛ) = 26.57° as the load grows. So the edge is a property of the line and the load angle together, and "voltage regulation" quoted as a percentage carries neither.

The far end that rises

Voltage regulation is quoted as a percentage: how far the voltage at the end of a line falls when the load is applied. The percentage carries neither of the two things that decide it. Past a computable angle of leading load the voltage at the far end goes above the source's — 28.35 degrees for a line of fifty ohms and a hundred of reactance — and with no reactance in the line there is no such angle at all, because the rise is a partial resonance and needs both halves.

power · Voltage regulation 25% compensation: 4.26% regulation, and a resonance at 25.0 Hz. computed by solving, not by drawing. The current a 230 V feeder draws, swept from two hertz to a hundred and fifty, with 25% of the line's reactance cancelled by a series capacitor. The regulation falls from 5.17% to 4.26%, which is what the capacitor was fitted for. What comes with it is a series resonance at 9.59 Hz with the load connected and 24.99 Hz with the far end shorted — the latter being exactly f₀√k = 25.00 Hz, a frequency the compensation fraction chooses on its own. The line's inductance does not appear in it and neither does the voltage. It is always below the fundamental, which is where the machines are.

The reactance cancelled, and the resonance it buys

Putting a capacitor in series with a feeder cancels part of its reactance and the far end falls less: five per cent of regulation becomes four and a quarter at a quarter compensation and 2.7 per cent at seventy per cent. What comes with it is a series resonance that was not there before, at the line frequency times the root of the fraction cancelled — so a quarter compensation resonates at exactly half the line frequency and a ninth at exactly a third. The line's own inductance is not in that answer and neither is the voltage.

power · Series compensation A 10 mH inductor with 8 pF across it, and where ωL stops being its impedance. computed by solving, not by drawing. The dashed line is ωL, which is what an inductor is supposed to be; the solid one is the impedance of the same inductor with 8 pF of winding capacitance across it. They part company at 170 kHz, which is ten per cent, and the impedance peaks at 563 kHz and falls thereafter — above which the component is a capacitor. The ratio between the two is 3.317, and the slider shows it is the same ratio at every capacitance: the shape of the departure belongs to the resonance rather than to either part. This is the capacitor essay with the components exchanged, and it comes out with the same structure and a different number.

The inductor that is a capacitor

The frequency field's second essay measures where a capacitor stops being one, because its own leads are an inductance. This is the same measurement with the components exchanged, and it comes out with the same structure and a different number: the ten-per-cent departure from ωL sits at f₀/3.317, and it sits at f₀/3.317 at every winding capacitance and every inductance tried. The shape of the departure belongs to the resonance rather than to either part.

magnetics · Real inductor A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees.

The pair that is worse than either

A bulk capacitor and a ceramic are fitted together because each is good where the other is not, and between them is a frequency at which the pair presents six times the impedance either one does alone. The peak is a parallel resonance between one part's inductance and the other's capacitance, its height is one over the series resistance every data sheet asks to be minimised, and at it the two capacitors exchange 5.87 amps for every amp the load draws.

power · Decoupling A coil of 500 nF and a capacitor of 100 µH give the loop three features, not one. computed by solving, not by drawing. The current round the loop for a volt across it, with the coil carrying 500 nanofarads of its own capacitance and the capacitor 100 microhenries of its own inductance. Three features rather than one: the resonance the two nameplate values set, at 4.02 kHz; the coil's own self-resonance at 7.12 kHz, which is a parallel tank and so a null in a series loop, 5.67e+3 times below the peak beside it; and a second series resonance at 28.2 kHz that belongs to neither component. Above the null the coil is a capacitance, that capacitance is in series with the tuning capacitor, and the capacitor's own inductance resonates with the pair — which is why removing either parasitic removes this peak and neither alone can produce it.

Two parasitics, and the resonance neither of them has

A resonator's quality factor is supposed to sit below the worst of its components, because reciprocals add. Give the capacitor half a millihenry of its own inductance and the loop measures 92.21 against a coil that allows 64.55 and a reciprocal sum that predicts 62.47 — the ceiling passed exactly where ESL/L crosses ESR/DCR, at 5.00 per cent. Add the coil's own capacitance beside it and the loop grows a second resonance at 28.2 kHz that neither part has alone, taller than the first by 40.4, and the half-power level is then crossed four times.

frequency · Resonance Switched on at 0.95× resonance, a Q 50 capacitor reaches 1.550 times its settled voltage. computed by solving, not by drawing. The capacitor voltage of the series circuit, switched on from rest at the crest of the drive, drawn as the tip of its arrow in the frame that turns with the drive, so that the settled state is a fixed point — the arrow from the centre, 10.067 V long, with the circle of that radius around the centre. The path is the settled arrow plus a second one turning at the circuit's own frequency and shrinking, drawn until the second is a hundredth of its first length. The voltage reaches 15.603 V in cycle 9, 1.5499 times the settled amplitude, and the tip's farthest point is 1.5499 times it. The approximation 1 + exp(−π/2Q|δ|) gives 1.5335. Marched in time by the trapezoidal rule, the network agrees with the exact solution to 1.2e-3 of the settled amplitude over its first 11 cycles.

The arrow that goes past where it settles

A phasor is where a driven resonator ends up, and counting the cycles it takes to get there says nothing about the path. Switched on from rest a little away from resonance, the capacitor's arrow circles its settled tip instead of approaching it: a Q of 50 driven at 0.8 of resonance reaches 1.854 times its settled voltage, and a Q of 200 at 1.25 switched on through zero reaches 2.182. At resonance exactly, where the settled voltage is largest, the arrow never passes its mark. So a resonance read by the largest voltage after switch-on is 1.20 times wider than its phasor says.

frequency · Phasors A Q 10 resonance read by a stepped sweep is within 1% of its width after 1.65Q cycles a step. computed by solving, not by drawing. A stepped sweep switches the series circuit on from rest at each of 201 frequencies, waits a stated number of cycles of resonance, and reads the largest capacitor voltage in the last cycle of the wait; the resonance's half-power width is then bisected on those readings and compared with the settled width, 0.1003 of the resonant frequency. After Q cycles the reading is 19.6% too wide and its peak 4.66% low. The width stays within 1% from 1.65Q cycles on and the peak from 1.50Q, against the ln(100)/π = 1.466Q cycles the second arrow takes to fall to a per cent. A reading that holds its largest value instead is 20.2% too wide at any dwell, since the overshoot it holds has already happened.

How long a sweep waits at each step

A resonance measured by a stepped sweep that starts each frequency from rest and reads the last cycle of its wait comes out 19.6 per cent too wide after Q cycles a step, and within one per cent of its width only from 1.65Q cycles at a Q of ten and 1.60Q at fifty. That is longer than the 1.47Q the transient takes to fall to a per cent, and the reason is the centre rather than the skirts: the width's error is the peak's shortfall read twice, while at the half-power frequencies the transient swings through its settled value and partly cancels itself. A reading that holds its peak instead is twenty per cent wide however long it waits.

frequency · Phasors A pole pair at Q = 2: the response peaks where the circle centred on the pair's real part and through both poles meets the axis, 0.9354 of ω₀, at 2.0656. computed by solving, not by drawing. The poles of a series RLC low-pass (10 mH, 100 nF, and the resistance for Q = 2), solved from the netlist, at −0.2500 ± j0.9682 in units of ω₀ = 1/√(LC) = 5.03 kHz: on the circle of radius ω₀, and 1/(2Q) of it from the imaginary axis. The magnitude at a point jω is ω₀² over the product of that point's two distances to the poles, to 9e-16 of the solve at 241 frequencies. The circle centred at −σ with radius ωd passes through both poles; it meets the axis at 0.935414 of ω₀, which is where the product of the two distances is least and the response peaks — 2.065591, which is ω₀²/(2σωd), found by a search on the solve at 0.935414.

The circle a peak sits on

A pole pair's response at a frequency is ω₀² over the product of two distances — from the point on the imaginary axis to each pole — to a part in 10¹⁵ of the nodal solve. That turns a resonance into geometry. The peak is where a circle centred on the pair's real part and through both poles meets the axis: at Q = 2, 0.9354 of ω₀ and a height of 2.0656, found the same by a search. The quality factor is an angle, Q = 1/(2 cos θ), and a peak exists exactly when the pair is closer to the imaginary axis than to the real one — past forty-five degrees. The half-power width is where the short leg has grown by √2, twice the pair's distance from the axis. And the step, marched, rings at the pole's height and decays at its distance: a projection, not a length.

frequency · Frequency response Swept at η = 1, a resonance reads 93.6% of its height, 0.45 bandwidths late and 1.26 bandwidths wide. computed by solving, not by drawing. The envelope of a series resonator's capacitor voltage, Q 50 at 1591.5 Hz, while its drive sweeps linearly upwards through resonance, plotted against the frequency the drive has reached, for sweep rates η = r/B² of 0.1, 0.3, 1 and 3; dashed, the steady-state response the sweep is chasing. At η = 1 — 1013 Hz a second, Q cycles for every bandwidth crossed when η is one — the reading peaks at 93.6 per cent of the true height, 0.445 bandwidths above resonance, and is 1.263 bandwidths wide. Swept at η = 0.03 it is the steady curve to 0.2 per cent.

A sweep that never stops

Sweep a drive continuously through a resonance and read the envelope against the frequency the drive has reached, and the reading depends on one number: the rate in hertz a second divided by the square of the bandwidth. At that rate equal to one — Q cycles for every bandwidth crossed, the dwell a stepped sweep needed — a Q-50 resonance reads 93.6 per cent of its height, 0.445 bandwidths late and 1.263 bandwidths wide, and at Q 10 or 200 the same within a per cent. A tenth of that rate gets within 0.3 per cent of the height and costs ten times the time. The peak always lands beyond resonance in the direction of travel, so an upward and a downward sweep together find it to 0.0075 bandwidths where either alone is off by nearly half.

frequency · Phasors One series resonator of Q = 1, read five ways. computed by solving, not by drawing; every reading from the solved network, none from its formula. The half-power bandwidth, the phase slope at resonance and the energy ratio driven at resonance give 1.0000, 1.0000 and 1.0000 — Q exactly. The ring-down's logarithmic decrement gives 0.8660, and the largest capacitor-voltage magnification 1.1547.

Five readings of one Q

A resonator's Q can be read from its bandwidth, from the slope of its phase, from the energy it stores against the energy it loses, from how fast its ringing dies and from how high its response peaks. On one series resonator at Q = 1 the first three give 1.0000 and the last two 0.8660 and 1.1547. Bandwidth, phase slope and energy are Q exactly at every Q, because they are one equation read three ways; the ring-down reads Q√(1 − 1/4Q²) because a damped cycle is longer than a natural one, and the peak reads the reciprocal because the peak is not at resonance. Each is within 1% of Q only above 3.5.

frequency · Resonance A 1 kHz tone burst through a band-pass of Q = 10, centred on its resonance. computed by solving, not by drawing, marched twice — sine and cosine carriers — so the output's envelope is the modulus of the two, exactly. The input envelope (pale) peaks at 240 ms; the output's (dark) at 243.171 ms, 3.171 ms late. The group delay at resonance is 2Q/ω₀ = 3.183 ms. The phase there is zero, so the carrier itself is not delayed at all.

The envelope that leaves early

A band-pass's group delay at its centre is the delay of an envelope, not of a carrier. Marched twice — on a sine and a cosine carrier, so the envelope is their modulus exactly — a 1 kHz burst through a band-pass of Q = 10 comes out 3.171 ms late against 2Q/ω₀ = 3.183, while the carrier's zero crossings come out within 6.4 µs of where they went in. Through a notch, whose group delay at its centre is −3.810 ms, the envelope peaks 3.746 ms before the input's. Neither is a paradox, and both hold only for bursts long against the inverse bandwidth: 60.6 per cent wrong at 0.6 of it, 0.7 per cent at 20.

filters · Group delay The Q a 10 mH coil with 1.2 Ω of winding and 8 pF across it shows, against frequency. computed by solving, not by drawing, the coil's impedance at each frequency and its reactance over its resistance. The straight ωL/R line (pale) rises without limit; the coil's Q follows it at first, peaks at 11340 at 325 kHz — 0.5773 of the 563 kHz self-resonance, which is 1/√3 — and falls to zero at the resonance. The peak is 0.3849 of what the line reaches at resonance, 2/(3√3).

The Q that peaks before resonance

A coil's quality factor is quoted as ωL/R and drawn rising with frequency, and its own winding capacitance turns it over: Q peaks at the self-resonance over √3 and falls to zero at the resonance. Eighteen coils from 1 mH with 300 pF to 100 mH with 1 pF all peak at 0.5774 of their resonance, at 2/(3√3) of the line's value there. Skin effect moves the peak down toward the resonance over √5. And the falling Q is a property of the coil read alone: in a tank tuned at 0.8 of its self-resonance, the same coil gives the full ωL/R, 23,570 against the 8,485 a meter reads off the bare part.

magnetics · Real inductor A resonator of Q 10 driven at 1.2 times its frequency, the drive switched off at the capacitor's crest and at its zero. computed by solving, not by drawing: the series circuit's exact free response from its settled state, held to a fourth-order march of the same circuit. Switched off at the capacitor's crest the ring starts at 1.0013 of the settled amplitude and never passes it; switched off at its zero it starts at 1.2015 — ω/ωd — and its first crest reaches 1.1120. The settled arrow turning at the drive's frequency is gone at the switch; what is left turns at the circuit's own, and its length is set by where in its cycle the settled arrow was.

The ring a drive leaves behind

Stopping a resonator's drive removes the arrow that turned at the drive's frequency and leaves the one that turns at the circuit's own, and how long that arrow is depends on the instant the drive stopped. Its length is the energy the circuit held at that instant, and off resonance the energy is not constant through a cycle: it breathes between the capacitor's peak and the inductor's, which differ by (ω/ω₀)². So above resonance a drive stopped as the capacitor's voltage passes through zero leaves a ring 1.2015 times the settled amplitude at 1.2 times resonance, and its first crest passes the settled value; below resonance no instant does.

frequency · Phasors The bank's impedance peaks at 0.966 Ω at 7.12 MHz; a one-ampere load step moves the supply by 132.4 mV. computed by solving, not by drawing. The bulk-and-ceramic pair of the decoupling essays — 10 µF with 5 mΩ and 5 nH, 100 nF with 30 mΩ and 1 nH — standing on a regulator of 2 mΩ and 20 nH, and the voltage at the bank when the load's current steps by one ampere, from the impedance's own step response by residues (solid). The impedance this bank presents peaks at 0.966 Ω at 7.12 MHz (upper line), the parallel resonance of the bulk part's inductance with the ceramic. The step rings at that frequency and its largest swing is 132.4 mV, 31.3 ns after the step — about a quarter period — which is 14% of the peak and 59% of the resonance's characteristic impedance √(L/C) = 0.224 Ω (lower line).

The peak a load step does not see

A decoupling bank is specified by its impedance against frequency, and the number that fails a target is the anti-resonance peak — 0.966 Ω at 7.12 MHz for the bulk-and-ceramic pair standing on a regulator. A load that steps by one ampere moves that supply by 132 mV, 14% of what the peak suggests: a step rings the resonance once, and its first swing is set by the resonance's characteristic impedance √(L/C) = 0.224 Ω, not by its Q. The peak is the voltage per ampere of a load that sits at the resonance; switched at 7.12 MHz the same load builds over six cycles to ±611 mV, the peak times the current's fundamental. So damping, the cure for the peak, is a cure for that load only: sixteen times the ESR cuts the peak 8.9-fold and leaves a step's swing between 120 and 155 mV. What decides a step is its edge — 132 mV for anything faster than a few nanoseconds, 28 mV for a microsecond.

power · Decoupling A tank with its loss in its coil, Q = 3: the voltage is largest at 0.9972 of 1/√(LC) and in phase with the current at 0.9428, and neither is where the poles sit. computed by solving, not by drawing, by closed form and by a netlist of L, Rs and C. The impedance of a coil of inductance L and series resistance Rs across a capacitor, against frequency, for Q = √(L/C)/Rs = 3. The poles are the series circuit's, at 1/√(LC) with a damping of Rs/2L. The coil's own zero at −Rs/L moves the peak to ωₘ = 0.9972·ω₀ and the zero-phase point to ωᵣ = 0.9428·ω₀, and the ring is at 0.9860·ω₀. Four frequencies, which a series circuit's voltage across its resistor would have as one.

The reading a coil leaves nearly exact

On a series resonator three readings of Q agreed exactly and two did not. A parallel tank whose loss is in its coil has no exact readings at all, and the order is not the one its schematic suggests. Its poles are the series circuit's, but the coil's own zero puts the peak, the zero-phase point and the ring at three different frequencies, and all of them below 1/√(LC). At Q = 3 the phase slope reads 16.2 per cent low, the stored energy 5.72 low and the peak height 5.42 high, while the bandwidth reads only 0.307 low: the loss that a parallel resistance stands in for tilts across the band, and the tilt is nearly antisymmetric about the centre, so the reading that spans it cancels it. The instrument's own source then decides what is being read.

frequency · Resonance

Named alongside it

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

Quality factorHalf-power bandwidthEquivalent series resistanceMeasurement conditionModel rangeSelf-resonanceThe quality factorSteady stateDamping ratioInterwinding capacitancePhasorSelectivity

All concepts