Field

Frequency, which is the same solve

Reactance, phase, resonance and the corner frequency are four readings of one number. The models that fail here are the straight-line sketch every engineer draws, which is three decibels wrong exactly where it is read, and the capacitor, which is a capacitor only below a frequency its own leads decide.

A single-pole low-pass with its corner at 995 Hz. Solved at 209 frequencies. The straight-line sketch, drawn faintly, is 3.01 dB wrong at the corner and within a tenth of a decibel only below 152 Hz. The phase is already −5.7° a decade before the corner and −84° a decade after it.

One solve, read four ways

Reactance, phase, the corner frequency and the roll-off are not four ideas. They are four readings of one complex number, obtained from the same matrix that answers direct-current questions — and the straight-line sketch every engineer draws of them is itself a model, three decibels wrong exactly where it is read.

Three element voltages closing on one source, at 1.59 kHz. Solved at 1.59 kHz, which is 1.00× the frequency at which the two reactances cancel. The three phasors add head to tail to the 1 V source exactly; their magnitudes sum to 5.26 V, which is not the same statement.

Three voltages that close on one, and the steady state they assume

Kirchhoff's voltage law drawn as a polygon in the complex plane. The three element voltages of a series circuit add head to tail to the source exactly — while their magnitudes add to five times it. And the whole picture is a statement about a settled circuit, which takes a computable number of cycles to arrive.

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.

A 100 nF capacitor, and what it is above 14.5 MHz. The dashed line is 1/(ωC), which is what the symbol means. The solid line is the same part with 30 mΩ of series resistance and 1.2 nH of series inductance, solved. They part company at 4.69 MHz and by a decade above resonance the part's impedance is 99.0× what its capacitance predicts.

The capacitor that is an inductor

A hundred-nanofarad capacitor follows 1/(2πfC) for four decades and then turns round and climbs. Above 14.5 MHz it is an inductor, and a decade past that its impedance is ninety-nine times what its capacitance predicts — all of it caused by about a nanohenry of lead and via that nobody chose and nobody drew.

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.

A 100 nF capacitor with 100 mΩ in series, written the other way round. computed by solving, not by drawing. At 100 kHz the series pair and the parallel pair are the same impedance to 8.7e-19 of itself — the arithmetic's floor, not a tolerance — with Rp = 2.533 kΩ against Rs = 0.100 Ω and Cp = 99.996 nF against Cs = 100 nF. Away from it they part company at a rate set by Q = 159.2: the substitution costs one per cent below 48.2 kHz and above 207 kHz, a band of 4.3 to one.

The same part written two ways

A capacitor's loss is quoted either as a resistance in series with it or as one across it, and the pair of expressions that converts between them is exact at one frequency and at no other. How wide the band is around that frequency is set entirely by the quality factor: an octave and a half at Q of eight, a thousand to one at Q of three thousand.

A 10 kΩ resistor, and the 19.3 MHz it is one below. computed by solving, not by drawing. The dashed line is R, which is what the symbol means. The solid line is the same part with 8.0 nH of lead inductance in series and 0.40 pF across the body, solved as a three-element network and checked against the closed form for the same three elements to 3.3e-16. It is ten per cent below its own value by 19.3 MHz, and which of the two parasitics does that depends on the resistance: the shunt capacitance wins above 91.02 Ω and the lead inductance below it. The slider is the resistance, and the departure frequency it moves is not monotonic — it rises a decade per decade of resistance, peaks near 91.02 Ω at 2.00 GHz, and falls a decade per decade after that.

The resistor that is only a resistor

A capacitor becomes an inductor above a frequency its leads decide, and an inductor becomes a capacitor. The third member of that family is the one nobody draws, and it is the only one whose edge is not monotonic in its own value: a ten-megohm resistor stops being one at 19 kilohertz, a ten-ohm resistor at 92 megahertz, and between them sits a resistance whose impedance is flat to fourth order — 91.02 ohms here, which is the square root of L over C divided by the root of one plus root two.

One magnitude curve, two phase curves, and the one of them the magnitude decides. computed by solving, not by drawing. A passive lead network — a resistor with a capacitor across it, over a second resistor, its zero at 1.00 kHz — and the same network followed by a first-order all-pass. The two magnitudes agree to the last bits of a double at every one of the 1601 frequencies sampled, and the phases differ by as much as 180°. Bode's gain–phase integral, fed the magnitudes alone with their phases discarded, returns 39.29° at the corner against a solved 39.29°, and tracks the minimum-phase curve to 0.02° across the band — while being wrong about the second network by the all-pass's own phase, which is what excess phase means. The edge here is the span of the sweep rather than a frequency of the circuit: ±3 decades of magnitude carries 99.92% of the integral's weight, and what is left out is the tail of a logarithm.

The phase the magnitude already knows

For one class of network the phase is not an independent measurement: it is fixed everywhere by the magnitude, through an integral Bode wrote down in 1945. Fed nothing but the magnitudes of a solved lead network, that integral returns 39.289 degrees at the corner against a solved 39.289, and tracks the whole curve to two hundredths of a degree. Cascade an all-pass and the magnitudes agree to the last bits of a double while the phases part by 180 degrees — so the recovery is exact for one of the two and cannot see the other at all.

A pair at Q = 0.7071: the sketch is -3.01 dB out at the corner. computed by solving, not by drawing. The solved magnitude against the two straight lines that stand in for it. At a conjugate pair the error at the corner is 20 log Q = -3.010 dB, which is unbounded in both directions, and the worst error anywhere is 3.010 dB at 1.00× the corner. No damping brings it inside 0.770 dB: that is the minimax, at Q = 0.9152, where the corner error and an interior maximum are equal.

The straight lines, and where they are not the curve

Two straight lines through a corner is the most-used approximation in this subject and almost the only one with no number attached. It has one, and it is exact: at a single real pole the sketch is 3.0103 decibels high at the corner and nowhere worse, and its error is the same a factor above the corner as the same factor below — a symmetry the construction does not suggest. A pole pair has no such bound at all, and the best any two-slope sketch can do on one is 0.770 decibels, at a quality factor of 0.9152.

Three capacitances, all correct: 2.000 µF, 5.814 µF and 2.105 µF at 5 V. computed by solving, not by drawing. The charge is C∞·v + Q_s·tanh(v/V_k) — a linear backbone and a polarisation that saturates — with both parameters pinned by the capacitance at zero volts and at the rated voltage, so there is no third degree of freedom to tune the answer with. The three curves are three questions. The small-signal value is the slope at the bias, which is what a ripple sees. The charge-average is the total charge moved from zero divided by the voltage, which is what a reservoir or a hold capacitor obeys. What a bridge reads is neither: it is the fundamental of the charge waveform under a one-volt test, which is a measurement condition. At 5 V they are 2.000 µF, 5.814 µF and 2.105 µF — a factor of 2.91 between the extremes, and every one of them is the capacitance.

The capacitance that is not one number

A ten-microfarad ceramic at its rated five volts is 2.000 µF as a slope, 5.814 µF as a charge average and 2.105 µF as a bridge reads it — three answers to three different questions, all correct, all called the capacitance. At zero bias the standard one-volt test alone reads 5.2 per cent low. Marched in a circuit the same part distorts as the square of the drive with no bias and in proportion to it with a bias, because the bias is what puts a second harmonic there.

A ±15% envelope on what a bridge reads permits 29 points of working capacitance. computed by solving, not by drawing. The charge-average capacitance between zero and the rated voltage — the number a reservoir or a hold capacitor obeys — against how the data sheet's stated temperature change is divided between the model's two parameters. Every point honours the envelope exactly: the measured value at zero bias with a one-volt test signal is 15 per cent from nominal at each point on each curve, by construction. The working capacitance is not. At the cold end it is anywhere from -42.9 to -13.9 per cent — 29.0 points of ambiguity at a temperature where the measured value is pinned exactly — and at the hot end from 13.9 to 36.2. Across the whole envelope that is 79.0 points against the 30 the specification bounds, a factor of 2.63. The left-hand end of the upper curve is missing because it is impossible: with the amplitude fixed, no characteristic voltage makes the measured value exceed the zero-bias capacitance.

The coefficient that is about one reading

A class II ceramic's temperature coefficient is a third printed number, and it is a coefficient of the one capacitance a data sheet reports: what a bridge sees at zero bias with a one-volt test. The model behind the part has two parameters, one number does not determine two, and every way of dividing a ±15 per cent envelope between them honours the envelope exactly while putting the working capacitance anywhere across twenty-nine points — and two parts a bridge cannot tell apart differ by 1.80 at the voltage they are used at.

A bridge's reading turns over at 1.885 V, where the test level stops mattering. computed by solving, not by drawing. What a bridge reads is the fundamental of the charge waveform over the fundamental of the voltage, so it is a function of the bias AND of the amplitude, and a data sheet names one point of it: zero bias, one volt. The four curves are four test levels on one part. At zero bias they run from 9.999 µF down to 8.312 µF — the harder the drive the lower the reading, because the capacitance is at its maximum there and a sinusoid spends most of its time off the peak. At the rated 5 volts they run the other way, 2.000 µF up to 2.426 µF, because the curve is convex. Between them is one bias where the two effects cancel: at 1.8848 V a tenfold change of test level moves the reading by nothing at all, and that voltage is 0.6645 of the polarisation's own characteristic voltage.

The reading a data sheet does not take

A class II ceramic's temperature envelope leaves its working capacitance 29 points wide at one end and 79 across, because one printed number cannot pin a two-parameter model. One further bridge reading recovers almost all of it — and where the reading is taken decides everything. Turning the test level down to a fiftieth separates five parts a data sheet cannot tell apart by 17.76 per cent; moving the bias to half the rated voltage separates them by 176.02, and pins the working capacitance to ±0.512 per cent from a reading known to one.

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.

40.0 dB of gain change and 180.0 degree-decades of phase, peaking at 109.8°. computed by solving, not by drawing. The gain and the phase of two leads of 10:1, together, each section buffered from the next. The gain changes by 40.000 dB from one end of the sweep to the other. The area under the phase, integrated on the solve over seven decades beyond the outermost corner, is 180.000 degree-decades, against 180.000 for ninety per twenty decibels of gain change. The peak is 109.806° at 3.16 kHz, below the 131.93° no arrangement of real leads sharing that rise can pass.

The phase a decibel buys

The area under a minimum-phase network's phase, counted in degrees across decades of frequency, is fixed by how far its gain moves from one end of the spectrum to the other: ninety degree-decades for every twenty decibels, however the network is arranged. One 100:1 lead peaks at 78.579 degrees and two 10:1 leads stacked together at 109.806, and both enclose 180.000. The peak is a design decision and the area is a bill — and a network that gives its gain back gives its phase back with it.

The straight lines report 45.00° of margin, and the solved loop has 51.83°. computed by solving, not by drawing. The gain and phase of a loop made of an integrator and one pole, the corner at 1.00 kHz, with the integrator set so that the straight-line asymptotes cross unity at 1.00 kHz. The solved loop crosses at 786 Hz instead. Read at the lines' crossover, with the phase taken from the solve, the margin is 45.00°; at the loop's own crossover it is 51.83°. Closed, the loop is stable: the largest real part among its closed-loop poles is -5.00e-1 of the corner's angular frequency. The solved gain never rises above its asymptotes, so the lines cannot report more margin than the loop has.

The margin the straight lines report

A phase margin read off a sketch is read where the straight lines cross unity, and that is not where the loop does. For an integrator and one real pole the lines report 45.00 degrees on a loop that has 51.83 — short, and always short, because a real pole's response never rises above its asymptotes. A pair that peaks reverses the sign and removes the bound: at a quality factor of two the lines report 71.57 degrees on a loop that has no margin at all, and at five they report 82.41 on a loop that is unstable.

Flat in angle at √(L/C), 141.42 Ω, and flat in size at 91.018 Ω. computed by solving, not by drawing, at 127 resistances on the closed form the network was checked against. The upper curve is the frequency at which the part's size is 1% away from R, the lower one the frequency at which its angle reaches 1°. Both are V-shaped and their points are in different places: the angle's first-order term vanishes at √(L/C) = 141.42 Ω, bisected on the measured slope to ten figures, and the size's second-order term at 91.018 Ω. The widest 1° band is 1.13 GHz, at 150.69 Ω; the widest 1% band is 1.66 GHz, at 95.806 Ω. At 91.018 Ω, flattest in size, the angle reaches 1° by 53.9 MHz. The flat line is 139 MHz, where a 6 mm body is one degree long: it binds the angle's edge from 118.72 Ω to 168.88 Ω and the size's from 43.947 Ω to 359.53 Ω.

The resistor that is right in size and wrong in angle

A resistor's impedance departs from its value in size as the square of frequency and in angle as the first power, so the angle always leaves first: at ten milliohms and at a megohm alike, where the angle has reached a degree the size is still only 152 parts per million out. One time constant, L/R − RC, sets that degree — 4.00 nanoseconds and 695 kilohertz at ten kilohms, 800 nanoseconds and 3.47 kilohertz for a ten-milliohm shunt. The resistance flattest in angle is exactly √(L/C), 141.42 ohms, the value the size question rejected; at the 91.02 ohms flattest in size the angle reaches a degree by 53.9 megahertz, and no resistance is flat in both.

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.

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.

$Two networks of one magnitude deliver the same energy, and the minimum-phase one delivers half of it 5.61 times sooner. computed by solving, not by drawing. A low-pass with poles at 1.00 kHz and 10.0 kHz and a zero at 3.00 kHz, and the same network with an all-pass behind it that moves the zero into the right half-plane. Their magnitudes agree at every frequency sampled to 4.4e-16. The energy of each impulse response, from the residues in closed form, is 6029.319 for both, and the integral of |H|² over frequency gives 6029.305. What differs is when it arrives: the minimum-phase network has delivered half its energy by 11.27 µs and its mirror by 63.23 µs; by 20 µs the fractions are 0.658 and 0.352, by 100 µs 0.918 and 0.676; and at no instant has the mirror delivered more.$

The energy that arrives first

Two networks with the same magnitude at every frequency have the same impulse-response energy, and Parseval's theorem says so before either is solved. They do not deliver it on the same schedule. A low-pass with poles at one and ten kilohertz and a zero at three delivers half its energy by 11.27 microseconds; the same network with its zero mirrored into the right half-plane, which changes no magnitude anywhere, takes 63.23, and at no instant has it delivered more. Its step response starts the wrong way, to −0.170 of the final value, before it turns round. Minimum phase is minimum delay, and the delay is in the energy rather than in any one number a frequency plot shows.

An L-section from 50 Ω to 1 kΩ: Q 4.359, fixed by the two resistances, and a band of 4.73%. computed by solving, not by drawing. A series inductor and a shunt capacitor matching 50 Ω to 1 kΩ at 1.00 MHz, their values from the series–parallel conversion: the load with the capacitor across it is 50 Ω in series with a reactance of 217.9 Ω at the design frequency, and the inductor cancels the reactance. The reflection there is 3.6e-16. The section's Q is √(20 − 1) = 4.3589 and no choice of parts changes it. |Γ| stays under a tenth from 976 kHz to 1.02 MHz, 4.73% of the design frequency, against 0.2/Q = 4.59%; and under half the power from 727 kHz to 1.21 MHz, 48.53%, against 2/Q = 45.88%.

The match with no knob

An L-section — a series inductor and a shunt capacitor — is the series–parallel conversion used on purpose: a load with a capacitor across it is, at one frequency, the source's resistance in series with a reactance an inductor cancels. Matching fifty ohms to a kilohm that way reflects 3.6 × 10⁻¹⁶ at its design frequency and has a Q of √19 = 4.359 that no choice of parts can change, so it holds its reflection under a tenth over 4.73 per cent of band whatever it is built from. The band is 0.2/Q to within three per cent, it depends on nothing but the ratio, and only splitting the match widens it: 14.98 per cent in two sections, 30.34 in three — and 30.83 in four.

What an L-section costs when its parts have a quality factor of 100. computed by solving, not by drawing. The same L-section as the ideal one, with each component given a series resistance of its own reactance over 100, and the efficiency read off a solve rather than from an expression. The section circulates Qₛ times the load's current through its own parts, and Qₛ is √(ratio − 1) with nothing left to choose, so the loss is TWICE Qₛ/100 — once in the inductor and once in the capacitor — to 4.40% wherever it is small. The consequence is that a match starts to cost something at a ratio nobody would call demanding: one per cent at a ratio of 1.253 — which is fifty ohms to 62.7, and is 1 + (Q/200)² to 0.25%. Splitting the match buys efficiency only above a ratio of 10.0: at a ratio of three one section loses 2.76% against 3.35% in two, and at a hundred 16.6% against 11.2%. The best number of sections for efficiency is 2 at a ratio of twenty and 4 at a ratio of a thousand — which is not the answer the band gives, where the band keeps improving with every section.

The efficiency a fixed Q costs

An L-section's quality factor is √(ratio − 1) with nothing left to choose, and the same fixed Q that decides its band decides what it dissipates. The circulating current is Q times the load's and it goes through both components, so the loss is twice Qₛ over the components' own Q — measured to 0.03 per cent. With parts of Q 100 that is one per cent at a resistance ratio of 1.253, which is fifty ohms to sixty-three. And splitting the match buys efficiency only above a ratio of 10.02: below it a second section adds two more lossy parts for less than it takes off anybody's Q.

The straight lines report 6.02 dB of gain margin on a loop that has none. computed by solving, not by drawing. The gain and phase of a loop made of an integrator and a pair at Q = 2, the corner at 1.00 kHz, the integrator set so that the straight-line asymptotes cross unity at 500 Hz. The loop's phase passes −180° at 1.00 kHz. There the lines put the loop gain at −6.02 dB and the solve at 0.00 dB, so the gain margin they report is 6.02 dB against 0.00 dB. The difference is 6.0206 dB, which is 20 log Q exactly with Q = 2, at any integrator gain. Closed, the loop is on the edge: the largest real part among its poles is -5.87e-17 of the corner's angular frequency.

The gain margin the straight lines get exactly wrong

A phase margin read off the straight lines is wrong by an amount that depends on where the loop crosses unity. A gain margin read off them is not: for an integrator and a pole pair the phase passes −180° at the pair's own frequency whatever the gain, and the lines are out there by exactly 20 log Q — 6.0206 dB for two real poles, 13.98 dB at a quality factor of five, at every integrator gain drawn. The stability condition for the loop turns out to be the same inequality: it is stable exactly when the gain margin the lines report exceeds that error.

An order-8 Butterworth: the whole sketch is 3.0103 dB out at the corner, and its 4 sections −5.85 to +8.17 dB. computed by solving, not by drawing. The error of the straight-line sketch against the solved response — for each buffered section of an order-8 Butterworth lowpass at 1.00 kHz, and for the whole cascade. At the corner the sections are out by −5.852 dB (Q = 0.5098), −4.418 dB (Q = 0.6013), −0.915 dB (Q = 0.9000), +8.175 dB (Q = 2.5629), which add to −3.0103 dB: the whole filter's error, the same 10 log 2 as a single pole. The whole sketch is never further out than that anywhere; the section with the highest quality factor is +8.343 dB out at 1.04 kHz. The quality factors multiply to 1/√2.

The corner error a filter hides in its sections

The straight lines of an eighth-order Butterworth filter are 3.0103 dB out at the corner and nowhere worse — the same as one pole, at every order. The lines of the four sections it is built from are out by −5.85, −4.42, −0.92 and +8.17 dB there, which must add to the whole because the sections multiply. The whole sketch never gets worse with order and the worst section's error grows as 20 log(n/π). And the section the sketch misrepresents most is the one whose frequency error moves the filter most: 0.312 dB for one per cent, against the 0.173 dB any section's stopband shift gives.

Lead zero at 500 Hz: the lines over-report by at most 3.12° with the loop's phase, and short by up to 20.6°. computed by solving, not by drawing. The error in the phase margin read off the straight lines of an integrator, two poles at 1.00 kHz and a lead section with its zero at 500 Hz and its pole 10 times higher, against where the lines cross unity, measured in decades from the zero. Gaps are placements where the lines are flat at unity or the loop crosses more than once. Read with the loop's own phase the reading is over by at most +3.12°, with the lines crossing −0.01 decades from the zero on a loop with 73.4° of margin, and never over on any loop with 60° or less; it is short by as much as 20.65°. Read with the phase off its straight lines too, on loops with 60° or less, it is never over.

The zero that lifts the lines

A phase margin read off the straight lines of an all-pole loop can only be short, and the proof takes three steps. A lead section breaks two of them: a zero's response lies above its lines, and between a zero and its pole the phase rises. The two breaks pull opposite ways, and measured across every placement of the crossing, in every sweep drawn, they leave an over-report of at most 6.58°, on a loop with 87.5°; on loops with sixty degrees or less it never exceeds 1.52°. The phase sketch is another matter: with the lines crossing at a lead zero above the plant's corner, it reports 17.91° on a loop with 1.17°.

Four readings of one distance

Reactance, phase, the corner frequency and the roll-off are four readings of one complex number, and the number has a geometry. A single pole's magnitude is the pole's distance from the origin divided by its distance from the point on the imaginary axis, and the phase is minus the angle that separation subtends — agreeing with the nodal solve to a part in 10¹² over five decades. So the corner is where the distance is root two times its smallest, the roll-off is the distance becoming omega, the flat passband is the distance not yet changing, and the phase is the angle. One distance, four readings.

The gain recovered from the phase alone, right in shape, with its level supplied by the gain at direct current. computed by solving, not by drawing. The lead, 1.00 kHz to 10.0 kHz, solved at 1601 frequencies over sixteen decades, and its gain rebuilt by Bode's companion integral from the phase alone over ±3 decades of 1.00 kHz. The integral returns a difference of log-magnitudes and nothing else, so the curve it gives has no level: drawn as if the direct-current gain were one, it is the dashed curve. Supplying the direct-current gain, −20.000 dB, places it. The placed curve lies within 0.055 dB of the solved gain across the drawn band, worst at 794 kHz.

The level the phase cannot know

Bode's gain–phase relation runs both ways. Given only the phase of a lead network, its companion integral returns the gain's shape across six decades to 0.055 dB. It cannot return the gain's level, because multiplying a network by a constant changes no phase anywhere. The level is one number, −20.000 dB here, and it has to come from outside the phase: one gain reading at any frequency supplies it. The same integral fed an all-pass reads a flat gain as a double pole, 6.021 dB down at the all-pass's frequency and 40.086 dB down a decade above it.

$Four networks of one magnitude: the first and last are fixed at every instant, and the two between swap at 12.64 µs. computed by solving, not by drawing. A low-pass with poles at 1, 10 and 100 kHz and zeros at 3.00 kHz and 30.0 kHz, and the three networks that share its magnitude with one or both zeros moved into the right half-plane by an all-pass. The magnitudes agree to 6.7e-16 and each impulse response carries 9261.008 of energy. The fraction delivered by each instant: the all-left network is never behind any other and the all-right network never ahead of any other. The two between are not ordered. Half the energy arrives at 2.13, 2.80, 10.1, 11.2 µs in the order listed, so mirroring the higher zero is the later; nine tenths arrives at 50.2, 156, 60.9, 167 µs, so mirroring the lower zero is the later. The curves cross at 12.64 µs.$

The two mirrors that change places

A low-pass with two zeros has four networks that share its magnitude, one for each choice of side for each zero. The one with both zeros on the left delivers its energy first at every instant, and the one with both on the right delivers last. The two in between are not ranked at all. With zeros at 3 and 30 kHz, mirroring the higher zero delays half the energy more and mirroring the lower zero delays nine tenths more. They swap at 12.64 µs, and that instant depends only on the product of the two zeros: it is where the all-pole response is rising at a rate of √(ω₁ω₂). And the network that delivers last does not start the wrong way at all.

With 100% of its capacitance to its mounting, a 141.4 Ω helix is 1.37° out at 100 MHz, not 2.7 × 10⁻³°. computed by solving, not by drawing. The angle of the impedance of a 141.4 Ω resistor — √(L/C) for 8 nH of lead and 0.4 pF — against frequency: the three-element part with all 0.4 pF across its film (dashed), and a helix of 32 turns with the same totals (solid), 0% of the capacitance between adjacent turns and 100% from the turns to the part's mounting, with one end grounded. At √(L/C) the three-element part has no first-order angle and rises as the cube of frequency. The helix is not flat there: its capacitance to the mounting counts only a third, so its own flat resistance is √(3L/C) = 244.9 Ω, and at 141.4 Ω its angle is first order in frequency — 1.37° at 100 MHz against 2.7 × 10⁻³°.

The resistor that is a helix

A spiral-cut film resistor is a helix, and its capacitance is distributed along the cut rather than lumped across the ends. Where it goes decides whether the three-element model survives. Capacitance between adjacent turns bridges each turn's own piece of film, and a helix of any number of turns built that way is the three-element part exactly — to 8×10⁻¹³ at 64 turns. Capacitance from the turns to what the part is mounted over is different: its current returns through only part of the film, it weighs (4N² − 1)/(12N²) of its value, a third in the limit, and the resistance with the flattest angle moves from √(L/C) = 141.4 Ω to √(3L/C) = 245 Ω. At 141.4 Ω such a part is 1.37° out at 100 MHz where the lumped model says 0.0027°. The third is a first-order rule; above about a gigahertz the helix is a line.

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.

$One per cent more C moves the Q = 5 pair straight down, parallel to the axis, and the response by up to 3.20% at 1.093 ω₀. computed by solving, not by drawing. The upper pole of the Q = 5 series RLC low-pass before and after a 1% increase in C, solved from both netlists (upper panel: the displacement, magnified to a fixed length, beside the tangent to the circle |s| = ω₀ and the direction parallel to the axis), and the fractional change in the response's magnitude against frequency (lower panel): solved as the ratio of the two responses (solid) and predicted as the displacement projected onto the two legs (dashed, on top of it), agreeing to 3.9% of the largest change. The pole moves 0.00 × 10⁻⁴ ω₀ towards the axis and 49.88 × 10⁻⁴ ω₀ down. The largest change in the response is −3.198% at 1.093 ω₀: beside the peak, where the short leg is at forty-five degrees.$

A tolerance is an arrow

Change one part of a series RLC low-pass by one per cent and its pole pair moves by a few ten-thousandths of ω₀ — and the response moves by the displacement projected onto the two distances that make it, to within four per cent of the solve. The direction is set by the part. A resistor moves the pair round its circle, away from the axis, and is felt at the peak: 0.99 per cent at Q = 5. A capacitor moves it straight down, parallel to the axis, and is felt beside the peak: +1.78 and −3.20 per cent at the half-power points and almost nothing at the peak itself. Because the short leg is ω₀/2Q long, the capacitor's effect grows by half a per cent for every unit of Q, to 10.1 per cent at Q = 20. Two hundred builds from 1% parts spread by ±2.5 per cent at the band edge.

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.

$With the capacitor tracking the inductor, one per cent of drift moves the Q = 5 response by at most 1.00%, at the peak — against −2.73% beside it with the capacitor held. computed by solving, not by drawing, from two netlists. The Q = 5 series RLC low-pass with its inductor raised by 1% and its capacitor changed by (1.01)^(−t) − 1, so that at t = 1 their product is unchanged; here t = 1. The fractional change in the response against frequency (solid), against the inductor alone with the capacitor held (dashed). The upper pole moves 9.90 × 10⁻⁴ ω₀ towards the axis and 0.99 × 10⁻⁴ ω₀ up it, and ω₀ by 0.00%. The largest change is 1.00% at 1.000 ω₀; with the capacitor held it is −2.73% at 1.115 ω₀. The slider is the tracking.$

Two parts that drift together

A one per cent change in the inductor of a Q = 5 resonant low-pass moves its pole pair mostly downwards, and the response by 2.73% beside the peak. Let the capacitor fall by the same fraction as the inductor rises and their product is fixed, so ω₀ is: the pair can only go round its circle, sideways near the top, and the response changes by 1.00% at the peak and nowhere else — the change a resistor makes. The difference grows with Q: at Q = 20 one per cent of drift moves the band edge 10.2% with the capacitor held, 20.7% with it drifting the same way, and 1.00% with it tracking. Over −40 to 85 °C an inductor of +150 ppm/K moves a Q = 20 band edge 9.1% with a zero-coefficient capacitor and 0.57% with a −150 ppm/K one. Correlation between two parts' errors is itself a tolerance.

An L-section whose coil has a Q of 50 and whose capacitor has a Q of 1000. computed by solving, not by drawing at 89 ratios, the loss read as one minus power into the load over power into the section. It follows Qₛ(1/Qcoil + 1/Qcap) (dashed) to 3% wherever it is under 2%: at a ratio of 20 it is 8.39% against 9.15%. Two parts of Q 95.2 each — the harmonic mean of 50 and 1000 — lose 8.40% there (faint), and within 0.4% of the unequal pair at every ratio: the section hardly cares which part is lossy, only what the two reciprocals sum to.

The part that decides the match

An L-section's loss is not twice its Q over the components' Q unless the two components are equally good. Given a coil of Q 50 and a capacitor of Q 1000, a 50 Ω to 1 kΩ match loses 8.39%, which is what two parts of Q 95.2 — the harmonic mean — would lose: the reciprocals of the two Qs add, each weighted by the loaded Q, and 95% of the heat is in the coil. Both parts carry the same circulating energy, so they share the loss in the inverse ratio of their Qs, which makes the coil the one to spend on: Q 50 to 100 halves the loss, and past the capacitor's own Q a better coil buys almost nothing. The high-pass section, with each part in the other arm, loses the same to 0.01%; per hundred watts delivered the coil runs at 4.36 W; and the ratio above which splitting the match pays, 9 to 10, does not move with the parts at all.

A pi network matching 50 Ω to 1 kΩ at a loaded Q of 8, coil Q 100, capacitor Q 1000. computed by solving, not by drawing at 10 design Qs from √19 = 4.36 — the least a pi network can have, at which it is the L-section and loses 4.58% — upwards. At a Q of 8 it loses 9.47% (solid) and its band for |Γ| under 0.1 is 2.3% of the frequency (dashed, scaled to the left end), against 4.7% for the L-section. At 30 it loses 28.73% for a band of 0.6%. The chosen Q is paid for in both currencies at once.

The Q a pi network makes a designer choose

An L-section has no choice of Q; a pi network adds a third part and can be built at any loaded Q above √(ratio − 1). Matching 50 Ω to 1 kΩ with coils of Q 100 and capacitors of Q 1000, it is the L-section at its least Q of 4.36 — 4.58% lost, a band of 4.7% — and at a Q of 8 it loses 9.47% for a band of 2.3%; at 30, 28.7% for 0.6%. What the extra loss buys is rejection: the second harmonic 16.0 dB down at the least Q, 29.8 at Q 10, 39.1 at 30. A tee network, with two coils where the pi has one, loses within 0.8% of the pi at every Q, because the loss follows the energy the Q makes the network circulate rather than the number of coils that carry it. And the freedom runs one way only: no pi or tee is less selective than the L-section; a lower Q needs more sections.

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.

A zero on the right at 1.00 kHz: the straight lines report 46.47° of margin, and the loop has 18.19°. computed by solving, not by drawing, the closed form held to the netlist — an inverting differentiator added back to its input makes the zero — at three frequencies and to the closed loop's poles. An integrator with a zero in the right half-plane at 1.00 kHz, the integrator set so that the straight lines cross unity at 950 Hz. The magnitude, and so the lines, are those of a zero on the left; the phase falls by the zero's angle where the left zero's would rise (dashed). The loop crosses at 3.04 kHz. Read at the lines' crossing with the loop's own phase the margin is 46.47°, and 133.53° with the phase the magnitude implies; at the loop's own crossing it is 18.19°.

The zero the lines put on the wrong side

A zero in the right half-plane has exactly the magnitude of its mirror on the left, so it has exactly the same straight lines, and nothing drawn from a magnitude can say which side it is on. The phase can: it falls where the mirror's rises. An integrator and such a zero, read off its lines, reports 90° − atan x of margin on a loop that has acos x, where x is where the lines cross as a fraction of the zero — so the lines say 45° of a loop with none. Behind one real pole the over-report reaches 37.6° on a loop of 2.1° when the zero sits below the pole, and vanishes when it sits above it: the reading is unsafe exactly when the zero comes first.

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.

$A complex pair of zeros at 30.0 kHz, damping 0.5, and its mirror: one magnitude, and half the energy 10.6 µs later — 4ζ/ω₀ is 10.6. computed by solving, not by drawing. A low-pass with poles at 1, 10 and 100 kHz and a pair of zeros at 30.0 kHz with damping ζ = 0.5, against the same network with the pair mirrored into the right half-plane; the fraction of the impulse response's energy delivered by each instant, from the residues, with the mirrored network also marched from its polynomials (dots, within 6e-7). Both carry the same energy. The left-half-plane network is ahead at every instant; half its energy arrives at 75.4 µs against 86.0, a delay of 10.62 µs, where the mirror's all-pass delays a low frequency by 4ζ/ω₀ = 10.61 µs.$

The zeros that have no side

A complex pair of zeros can be mirrored into the right half-plane without changing a magnitude anywhere, and what the mirror adds is a second-order all-pass whose delay is 4ζ/ω₀ at low frequency. Above the band that is the whole story: with the pair at 30 kHz on a 1 kHz low-pass, half the energy arrives 0.21 µs later at ζ = 0.01, 2.12 µs at 0.1 and 21.2 µs at 1, linear in the damping — so the zeros of a notch, which sit on the axis, have no side worth choosing. Inside the band the all-pass's delay peaks at a millisecond exactly where the notch has removed the energy, and the energy's centroid moves by the delay averaged over |H|², which for fifteen pairs from 2 to 100 kHz matches a quadrature over frequency to five parts in a hundred thousand. And unlike one mirrored zero, a mirrored pair never starts a step the wrong way: it is only late.

A megohm part with one picofarad to its mounting, heavier at one end: 0.433 pF with that end driven and 0.233 pF turned round, 1.56° against 0.84° at 10 kHz. computed by solving, not by drawing: the angle of the impedance of a 1 MΩ resistor as 200 sections, with 1 pF in total from its film to its mounting tapered as c(x) ∝ 1 + a(1 − 2x), a = 0.6, read at the terminal with the far end grounded. Driving the end nearer the mounting (solid) it presents 0.433 pF and 1.56° at 10 kHz; turned round (dashed), 0.233 pF and 0.84°. The first-order weight is ∫c(x)(1 − x)²dx, which is C(1/3 + a/6) and C(1/3 − a/6). The part's total capacitance has not changed, nor has its resistance, and neither of the readings is that total.

The resistor turned round

A resistor's stray capacitance is usually drawn as one capacitor across its ends, and a stray to the part's mounting is not one number. Read at the terminal with the far end grounded it is weighted by the square of the distance from ground, so a uniform stray counts as a third of itself, and turned round a part with a tapered stray reads differently: 0.433 pF one way and 0.233 pF the other from one picofarad in all. In the current through the part the weight is x(1 − x), a sixth, and the current reaching the far end is 3 dB down at ωRC = 9.18 rather than at one. Two readings of one part separate the total from the taper, and a distributed stray's phase never reaches a capacitor's: it stops at 45 degrees.

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.

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