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

The thread: Refused, not extrapolated

A network with no path to ground has no answer, and a solver that returns one is lying. A model asked to work outside its range declines and says why. Half the value of an assertion is what it rejects, so every refusal on this site was produced by running it rather than by describing it.
A diode fed from 5 V through 1.0 kΩ. computed by solving, not by drawing. The operating point is where the exponential meets the load line: 0.692544 V and 4.3075 mA, reached in 13 damped Newton steps from a cold start. The one-line Newton on Vs = v + R·i(v), which touches no matrix, gives 0.692544 V. The "drop" is not a constant: it moves 59.53 mV per decade of current, measured between two solved operating points. Devices, and the amplitude they stop being linear at

A bias point is a solution, not a choice

The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.

 Where four of these models stop being true. In order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is. Where the models stop

Every model has an edge

Four assumptions the circuits in this field run on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does.

 The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz. The floor, which bounds from below

The floor a resistor sets

A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.

 A network solved, and checked: a bridge, which no series-parallel reduction reaches. Node potentials from modified nodal analysis. The branch currents are then recomputed from each element's own law and summed at every node; the residual is 2.7e-16 of the largest current in the circuit, which is floating-point rounding and nothing else. Networks, and how a solve is checked

What a network answers, and how the answer is checked

A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number.

 A 10:1 divider with 12.8 pF across its top resistor. computed by solving, not by drawing. The divider's resistors set a ratio of 0.10000 and its capacitors 0.10000; the step starts at the second and relaxes to the first over 115 µs. The balance R₁C₁/R₂C₂ is 1.0000, and the edge lands 0.00% away from where it settles. At 12.8 pF the two ratios are the same number and the response is flat. Measurement, which is a circuit on a circuit

A divider with two ratios

Put capacitance in a resistive divider and it divides by resistance at direct current and by capacitance at high frequency, and those are two different numbers unless one equation holds. The adjustable trimmer on every oscilloscope probe exists for that single equation, and the square wave on the instrument's front panel is a display of which of the two ratios is currently winning.

 One capacitor of 77.3 µF, against every load it was not sized for. computed by solving, not by drawing. Sized from the 20 Ω load, the capacitor takes the power factor to 1.000000 there and leaves 0.0e+0 var of 1636 VA. At 178 Ω the same installation sits at 0.23 leading. The correction is exact at one point on this axis and nowhere else on it. Power, and the part that does no work

The capacitor that was right once

Cancelling a load's reactive power needs one division and no iteration, and the answer is exact. It is exact for the load it was computed from, at the frequency it was computed at, and the figure here is about what it does everywhere else — which includes making the installation worse than it was before anything was fitted.

 Linearising an exponential at 27 °C, and what it costs. The linear model understates the gain by 1% at 7.30 mV and by 10% at 22.8 mV. The thermal voltage at this temperature is 25.9 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong. Where the models stop

How small is small signal

Linearising an exponential replaces a curve by its tangent, which is exact at a point and progressively wrong away from it. The amplitude at which it is one per cent wrong is 7.3 millivolts at room temperature — 28 per cent of the thermal voltage, not a small fraction of it, and a good deal smaller than "small signal" suggests.

 A gain of 100 asked of an amplifier with 1.00 MHz of gain–bandwidth. The ideal amplifier — a nullor, so the two golden rules exactly — holds 100 at every frequency. The real one is 0.10% low at direct current, 1% low by 1.35 kHz, and 3 dB down at 10.0 kHz. Above 10.0 kHz there is no loop gain left and the ideal answer is not an approximation to anything. Feedback, and the margin

The ideal amplifier, and where it stops being one

An ideal operational amplifier's closed-loop gain is set by two resistors and nothing else — a horizontal line at every frequency. The real one is already a tenth of a per cent low at direct current, one per cent low by 1.35 kHz, and above 10 kHz has no loop gain left, at which point the ideal answer is not an approximation to anything.

 A 9 V source with 500 mΩ inside it. The ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A. Networks, and how a solve is checked

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

 Five steps, each divided by its own size, from an amplifier limited to 0.50 V/µs. A linear circuit would put these five curves exactly on top of each other. The 20.0 mV step is linear; everything above 79.6 mV is not, and the largest step takes 16.0 µs to travel a distance the linear model says takes 0.159 µs. Before the steady state

The step that is too big

A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

 A loop whose phase comes back, at a gain of 1.0e+7. computed by solving, not by drawing. Three identical poles take the phase to −270°, two lead sections bring it back to −158°, and their own poles take it down again — so ∠L = −180° at 189 Hz, 3.46 kHz and 23.5 kHz. The gain moves the magnitude curve and not those three frequencies, so it decides only which side of each the unity-gain point falls. At 1.00e+7 the crossover is 10.6 kHz, in the recovered band, and the loop is stable. Feedback, and the margin

Stable, and unstable with less gain

A three-pole loop with two lead sections is stable for gains between 1.81 × 10⁶ and 3.36 × 10⁷ and unstable on both sides of that window. Turning its gain *down* is what breaks it. Neither margin can see this, because a phase margin describes the loop at one frequency and a gain margin at one other, and this loop's phase crosses −180° at three: 189 Hz, 3.46 kHz and 23.5 kHz.

 Three balanced rectifier loads conducting 60°, and their neutral. computed by solving, not by drawing. The three phase currents are drawn faint and the neutral heavy. Balanced loads, identical in every respect, and the neutral carries 0.968 A against a line current of 0.559 A — a ratio of 1.7321, where √3 is 1.7321. The pulse trains are disjoint, so the neutral is their union and its mean square is three times one phase's. Rebuilding the same current from the multiples of three in one phase's spectrum gives 0.966 A, 0.13% away, by a route sharing only the waveform. Power, and the part that does no work

The neutral that carries more than a line

Three balanced loads draw currents summing to 5.3 × 10⁻¹⁵ amperes in the neutral. That is a theorem about sinusoids, and it uses nothing except that each current is a single frequency. A harmonic of order three is shifted by 360° between phases, which is no shift at all — so the third harmonics add, and for any conduction angle narrow enough that the pulse trains stay disjoint the neutral carries exactly √3 times a line current.

 Four families at order 5, and the 2 zeros in the stopband. computed by solving, not by drawing. The elliptic design is drawn at a selectivity of 0.8, so its stopband is asked to begin at 1.250 times the corner, and the degree equation returns 38.68 dB for it at 0.5 dB of ripple. From the stopband edge outward the elliptic response never rises above -38.7 dB; the steepest all-pole family of the same order is only -19.0 dB down there and does not reach -38.7 dB until 1.77 times the corner. That is what 2 transmission zeros buy. The price is in the same picture: the elliptic stopband has a floor and an all-pole one does not, so beyond 2.0×fc the all-pole response is the lower of the two and keeps going. Filters, measured not tabulated

The zeros that buy an order

Butterworth, Chebyshev and Bessel all fall because their denominator grows, so the only way to make one fall faster is another pole. An elliptic filter puts zeros in the stopband instead, and reaches forty decibels at twice the corner with two poles where the steepest all-pole family needs five. What it charges is a stopband that stops falling — measured here at −38.7 dB, and overtaken by an ordinary Chebyshev two corners out.

 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. Two windings, and the band between them

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.

 Where a switch is a switch: a band, and the 6.43 MHz at which it closes. computed by solving, not by drawing. A switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1.0% of being ideal only for loads between 49.5 Ω and 1.01 MΩ — 4.31 decades, and both edges are the same part. The upper edge is a frequency as well as a resistance, because the off-capacitance shunts the open switch: it falls a decade per decade above 318 Hz and meets the lower edge at 6.43 MHz, where the band closes and no load at all will do. Checked by scanning every load at 1.3 times that frequency and finding the best possible error to be 1.17%. Where the models stop

A band rather than an edge

Every other boundary in this collection is one-sided: a model is true below a frequency, or below an amplitude. A switch is a switch only for loads between 49.5 ohms and 1.01 megohms — bounded at both ends by the same part — and the upper end is a frequency as well as a resistance, so the band narrows as the frequency rises and shuts completely at 6.43 megahertz, above which no load resistance at all will do.

 A network the solver will answer, and should not be asked, into 0.01 Ω. computed by solving, not by drawing at 61 spreads. A hundredth of an ohm either side of a wire whose resistance is made smaller and smaller, into 0.01 Ω — an element written where the right answer is no element at all. The solution stays exact: 0.499999998422607 at the last spread before the refusal, against 0.500000000000000. What grows is the current-law residual, as the 0.96 power of the spread. The solve is refused at a spread of 2.5e+8, where the smallest pivot falls under 1e-12; the residual tolerance of 1e-7 would have been reached at about 3.3e+9. Two guards written for unrelated reasons, arriving within a decade of one another. Networks, and how a solve is checked

The answer that is perfect and absurd

A network with a wire written into it as a small resistance returns the right node voltage to fifteen figures, passes both verifications with a residual of two parts in ten to the sixteenth, and reports two hundred thousand amperes. The solver refuses it one decade further on, and by then it has been answering for eight decades.

 The ideal amplifier is good to 1% over a region, and its corner is 21% inside the specifications. computed by solving, not by drawing. The 1 per cent contour of the ideal-amplifier model for a non-inverting stage of gain 2 built from a 10 MHz part, drawn over frequency and output amplitude at once. Each point is bisected on a marched circuit: the error is the root-mean-square difference between the marched output and 2 times the input, which counts the gain that is low, the phase that is late and the peak that is flat. Three mechanisms bound the region — finite gain–bandwidth on the left, the input pair's slew rate on the diagonal, and the rails at 12.19 V along the top. The two dashed lines are the numbers a data sheet gives: a small-signal edge at 48.8 kHz with no amplitude in it, and a full-power bandwidth of slew rate over 2πV̂ with no gain–bandwidth in it. They cross at 10.60 V and 48.8 kHz; the measured contour passes 38.5 kHz at that amplitude, which is 0.790 of it. Where the models stop

The edge that is a region

Every boundary drawn so far in this subject is a number on one axis, and the figure that gathers four of them admits in its own caption that the fifth is an amplitude and cannot go there. Drawn on both axes at once, the ideal amplifier's one per cent boundary is a region with three sides and a corner — and the corner sits at 38.5 kilohertz where the two numbers a data sheet quotes cross at 48.8, because the two mechanisms are lags on the same waveform and add as magnitudes rather than in quadrature.

 A copy out by 1.3% for the reason everybody names, and 11% for the one nobody does. computed by solving, not by drawing at 60 output voltages, with the Early conductance iterated to self-consistency against the current that sets it. Two base currents are stolen from the reference, so the copy is β/(β+2) of it — 1.32% low at β = 150 — and that is exact at exactly one output voltage, 0.7043 V, which is 9.39 mV under the reference's own base-emitter voltage of 0.7137 V — a displacement that goes as 1/(β+2), so that the product of the two is 1.427 V at every β the slider offers. Everywhere else the Early effect is larger: the current rises at 1.21% per volt, so moving the output from one volt to ten changes it by 11.2%. One per cent holds over 0.810 V, which is the Early voltage over a hundred and contains neither the current nor any resistor. The slider is β: it moves the first error by fifty times and the second by nothing at all. Devices, and the amplitude they stop being linear at

The copy, and its two errors

Every account of a current mirror leads with the base currents: two are stolen from the reference, so the copy is beta over beta plus two, which is 1.32 per cent at beta of 150 and is what a third transistor is spent on. The Early effect is a footnote and is nine times larger over any useful swing — 11.2 per cent between one volt and ten. Moving beta from 20 to 1000 changes the first by a factor of forty-five and the second by nothing at all.

 A passive network, read from each end — and the two readings are one number. computed by solving, not by drawing. A five-element RLC ladder with a current injected at one port and the voltage read at the other, then the two exchanged. With no controlled source the two readings agree to 4.9e-14 of themselves over four decades, which is the arithmetic's noise rather than a physical difference — the network cannot tell which way round it is being used. A mutual inductance keeps that: a 1 mH and a 4 mH winding at k = 0.7 give 0.0e+0, because the coupling puts the same entry in both halves of the matrix. A transconductance does not, and the departure is proportional to it with a fitted exponent of 1.000 over four decades — so there is no small amount of gain that is harmless. It passes the arithmetic's own floor at 0.781 femtosiemens, and the smallest transistor in this collection is nine orders above that. Networks, and how a solve is checked

The reading that does not care which way round it is

Inject a current at one end of a network and read the voltage at the other; then swap the two. A passive network gives back the same number — not a similar one, but the same to five parts in ten to the fourteenth over four decades of frequency, and for a coupled pair of windings the same double. Put a transconductance of eight tenths of a femtosiemens anywhere in it and the two readings part company.

 A step on a series RLC at ζ = 0.079, and the two numbers read off H(s). computed by solving, not by drawing. A 50.3 kHz series RLC driven by a one-volt step, with the capacitor voltage and the inductor voltage drawn together. Two limits of the transfer function are two points of the waveform and neither needs the waveform: H(0) = 1.000000 is where the capacitor ends up, and H(∞) across the inductor is 1.0000, which is what it does at the first instant — the expansion gives 1.000000 for it at t = 0. Here the damping ratio is 0.0791, the response is inside ±2% after 7.6 cycles, and 100.0% of the last twenty-four cycles sit there. The poles are at a real part of -7.91e-2 of ω₀, which is the condition the final-value theorem actually has — not a property of H but of where sY(s) has its poles. Before the steady state

Two numbers without solving for the waveform

Where a step response starts and where it ends are two limits of the transfer function, and neither needs the waveform. Both are exact here — 1.000000000 volts at the end and the whole step at the first instant — and one of them is a lie waiting to happen: take the damping to zero and the final-value theorem still returns 1.000000 for a response that swings between 0 and 2 for ever. Its condition is not on the transfer function but on where the poles are, and the practical condition is narrower still: at five ohms the poles are safely in the left half-plane and sixty cycles is not enough time.

 The window a 10 µF capacitor leaves. Above 939 mΩ the loop holds 45° of margin; below it the regulator rings and then oscillates. The droop after a 100 mA step is smallest at 817 mΩ — 129 mV — and by 19.9 Ω it is 1.468 V, because at the first instant of a step the capacitor cannot move and the whole step falls across its series resistance. Two requirements, opposite directions, and the useful values are between them. Circuits that do a job, and the range they do it over

Two requirements pulling one capacitor

The output capacitor's series resistance is a stability requirement and a transient requirement at once, and they pull it in opposite directions. Below 939 milliohms this loop has less than 45 degrees of margin; above about 850 the droop after a load step starts to grow, because at the first instant of a step the capacitor cannot move and the whole step falls across that resistance. The droop is smallest 10 per cent inside the unstable region, which means the best transient this design can have is one it must not be built with.

 Five boundaries, one tolerance, and three exponents. computed by solving, not by drawing. Each of five model boundaries re-solved at forty-one tolerances from 0.1% to 30%, divided by its own value at 0.1% so that an amplitude in millivolts and four frequencies can share one axis — an exponent has no units. Fitted over the two decades to 10%: Kirchhoff's laws 1.000, the ideal amplifier 0.513, the small-signal model 0.497, the ideal capacitor 0.500, and the full-power bandwidth 0.000. A boundary set by a first-order departure moves in proportion to the tolerance, one set by a second-order departure moves as its square root, and a refusal does not move at all — so relaxing the tolerance from 0.1% to 10% buys a factor of 100 on the board and 10.0 on the capacitor. Where the models stop

A boundary is a model and a tolerance

Every edge in this collection is computed from a fraction of error nobody states, and the four on its opening axis use three different ones. Swept over two decades, each boundary moves as a power of that fraction — Kirchhoff's laws exactly as the first power, the amplifier and the capacitor and the small-signal model as its square root to within three per cent, and a full-power bandwidth not at all. The exponent identifies the mechanism, and it re-orders the axis twice: the board fails before the capacitor below 0.424 per cent, and the output before the amplifier above 21.7.

 A 1% imbalance, and the 64 cycles it survives. computed by solving, not by drawing. The upper panel is the peak flux density, marched cycle by cycle, under a square drive whose positive half is 1% larger in area than its negative half. It does not settle. It walks, by the same area every cycle, and reaches 0.35 T after 64 cycles — 1280 ms at 50 Hz — against a closed form of 63.7. The lower panel is the count against the imbalance, and it rises without bound and never becomes infinite. Halving the drive gives 128 cycles, which is exactly twice: reducing the amplitude buys time and not safety, and there is no amplitude at which this design is inside a limit. Two windings, and the band between them

The flux that walks

The previous essay's saturation limit is an amplitude, and an amplitude can be respected. This one cannot. A drive whose two half-cycles differ in volt-seconds by one per cent adds the same small area to the flux every cycle, so it reaches saturation after 64 cycles — and halving the drive gives 128, and a tenth of it gives 637. Reducing the amplitude buys time in exact proportion and removes nothing. There is no amplitude at which the design is inside a limit.

 Fourteen decades of imbalance, fourteen digits gone, and a matrix in perfect health. computed by solving, not by drawing. A Wheatstone bridge walked towards balance, with the relative error of the solved output against a closed form that cannot lose digits. The condition number of the nodal matrix is 505.0 at every imbalance and the smallest pivot is 2.0e-3 of the matrix norm — orders above the 1e-12 at which this solver refuses to answer at all. Neither number moves, and the answer still loses one digit per decade of imbalance, reaching 33% at δ = 1e-15. The bound drawn over it is the round-off divided by the imbalance, which the measurement stays under at every point. The third curve is the same closed form written as ½ − 1/(2+δ) — algebraically identical, and it loses its digits at the same rate, which is where the loss lives: in the subtraction of two nearly equal numbers, not in the matrix. Networks, and how a solve is checked

The matrix that is ill, and the answer that is not

Every other essay here treats the solve as exact, and it is not. A bridge walked towards balance loses one digit per decade of imbalance and has none left at a part in 10¹⁵ — on a matrix whose condition number never moves and whose smallest pivot stays four orders above the threshold this solver refuses at. A feedback amplifier does the opposite: the solver declines to answer at a gain of 10⁹, one decade after it returned an answer that was exact to the last bit.

 Nine tenths of the heat is in the switch, and above 1.28 MHz there is no temperature at all. computed by solving, not by drawing. The junction temperature a recovering diode settles at, against switching frequency, with the carrier lifetime rising as the 1.8 power of absolute temperature so that the recovery gets worse as the junction gets hotter. Each event costs 15.43 µJ when cold, of which 90% is dissipated in whatever is pulling the current down rather than in the diode: while the junction is still conducting it holds almost no voltage, and the recovered charge is delivered through the switch at the full 100 V. That energy equals the diode's own conduction loss at 29.2 kHz. The fixed point T = Ta + Rth·P(T) is iterated from the ambient upward; the junction passes its rated 150 °C at 904 kHz, and above 1.28 MHz there is no temperature that satisfies it at all — the loop has gain and the solver reports the refusal rather than the last iterate. Before the steady state

The heat a recovery leaves behind

The essay below this one measured how much current a diode conducts backwards and for how long, and stopped there. Both numbers are multiplied by a voltage somewhere, and the surprise is where: while the junction is still conducting it holds almost nothing, so nine tenths of the energy is dissipated in the transistor pulling the current down and not in the diode. Repeat it a hundred thousand times a second and it is 1.5 watts, the lifetime rises with temperature, and above 1.28 megahertz the diode's own loop has no fixed point at all.

 Two loops on one heatsink give out at 135 kHz, and it is the switch that goes. computed by solving, not by drawing. The junction temperatures of the diode and the switch against switching frequency, with each device's own thermal resistance to a case they share. Each has a positive temperature loop and they are different loops — the diode's runs through its carrier lifetime and its recovery, the switch's through its on-resistance and its conduction — and the electrical coupling goes one way, since the charge the switch has to take at full supply is the diode's. The pair has no settled temperature above 135 kHz and the component that gives out is the switch, which has no exponential in it and is taking 84 per cent of the heat. The same two devices with the same total thermal resistance and no case in common survive to 485 kHz; the diode on its own to 1.28 MHz. Before the steady state

Two loops, and one heatsink

Nine tenths of a reverse recovery's energy is dissipated in the transistor and not in the diode, and the obvious calculation of the diode's junction temperature puts all of that energy in it anyway. Repaired, the diode alone survives to 1.28 megahertz instead of 128 kilohertz — a factor of exactly the ninety per cent. What replaces the number is the arrangement that exists: two devices with two different positive temperature loops on one piece of aluminium, giving out at 135 kilohertz, and it is the switch that goes.

 Sixteen more digits move the boundary by sixteen decades and leave it exactly where it was. computed by solving, not by drawing. The earlier essay's bridge, walked towards balance and solved twice: once in double precision and once with a pair of doubles carrying about 31 decimal digits, against a closed form that cannot lose any. The 33 per cent error at an imbalance of 10⁻¹⁵ becomes 7.0e-18 — so that loss was the arithmetic's and not the network's, which is what the earlier essay could not say. Each arithmetic's error is its own round-off divided by the imbalance, drawn as the two straight lines, so the second boundary is the first one moved by exactly the extra digits. The condition number is 505 in both cases and at every point, which is the diagnostic being blind twice over. Networks, and how a solve is checked

The digits the arithmetic did not have

Bounding the arithmetic of double-precision circuit solving finds two boundaries that cannot be attributed to it: a bridge with no correct figures left at an imbalance of 10⁻¹⁵, and a filter synthesis that stalls at order 14. An ill-conditioned problem stays ill-conditioned however many digits are used, and a well-conditioned one computed badly gets better — so adding digits is the experiment that tells them apart. The bridge's loss is entirely the arithmetic's. The synthesis's is mostly the data's, and doubling the digits makes it worse.

 A second integrator is 5.8 more decibels an octave, and a third 5.4 more. computed by solving, not by drawing. Modulators of order 1, 2, 3 marched one sample at a time with a one-bit quantiser, their in-band noise read from the transform of the error with the test tone a quarter of the way up the band. The measured slopes are 8.17, 13.93, 19.33 decibels per doubling of the oversampling ratio, against the 9, 15, 21 the white-noise argument predicts and the 3.01 plain oversampling gives. Each order is short of its own prediction by about a decibel, in the same direction, which is the error in the band not being white. The straight lines are each order's prediction drawn through its own first point, so what is compared is a slope against a slope. Where a signal becomes a number

The loop that is worse at full scale

A second integrator in a one-bit loop takes the shaping law from nine decibels an octave to fourteen, and a third to nineteen. What it charges is not in decibels at all: past six tenths of full scale the ratio starts falling, and a fourth integrator's states run away at seven tenths. The best input to a second-order modulator is 0.6 of the reference it is measured against — an amplitude boundary of exactly the kind this collection is built for, on the one object in the field that has no continuous output to draw.

 Where an amplifier's reading comes from, against the source it is reading. computed by solving, not by drawing. Three errors with three different dependences on the source, each measured by a solve with the other two set to zero. The offset voltage is flat — 50 microvolts wherever the source is. The bias current times the imbalance is linear in the source and is what balancing removes. The offset current times the source is linear too and is what balancing leaves. Unbalanced, the current overtakes the voltage at 1.77 kΩ; balanced, at 10.0 kΩ, which is the offset voltage divided by the OFFSET current and is the ratio of the two currents further along. Below about a kilohm, balancing makes the reading worse — the feedback network is already the larger resistance, and equalising means adding to the source. Measurement, which is a circuit on a circuit

The current the instrument draws

Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get.

 The floor rises as the 1.06 power of the order, and the arithmetic gives out first. computed by solving, not by drawing. The least departure a doubly terminated ladder can achieve at any impedance level, against order — the floor, which is what is left when the tolerance the band is drawn at is taken away. It rises from 0.0087 dB at order three to 0.0389 dB at order 13, as the 1.06 power of the order, so the order at which it reaches the 0.1 dB the band is drawn at is near 32. Drawn over it are the same networks with one imperfection removed at a time, and they do not add: the winding loss alone leaves a larger departure than the complete realisation does, because the track resistance is in series with the load and lifts the passband exactly where the winding loss droops it. The stray capacitance contributes nothing at all, since the level that minimises the departure is a few ohms. What ends the sweep is neither: the continued-fraction synthesis loses its leading coefficients to cancellation and stalls at order 9 for a Butterworth and 14 for a Chebyshev. Filters, measured not tabulated

The floor that outlives the arithmetic

A band of impedance levels is what a tolerance allows; a floor is what a structure has. Measured at six orders, a doubly terminated ladder's floor rises as the 1.06 power of the order and would not reach a tenth of a decibel until order thirty-two — an order nobody builds. What stops the sweep is neither the structure nor the parasitics: the continued fraction that turns a reflection polynomial into element values loses its leading coefficients to cancellation and stalls at order nine for a Butterworth and fourteen for a Chebyshev.

 A pair's third-order intercept is 7.8 dB above anything it can produce. computed by solving, not by drawing. Two equal tones through a differential pair, transformed coherently so every product lands in a bin of its own. The fundamental rises with slope 1.000 and the third-order product with slope 3.000, both fitted over the decade marked, and the dashed extensions are the extrapolation a specification quotes. They meet at a drive of 4.00 thermal voltages and an output of 2.00 — against a largest output of 0.8108, which is 8/π² and is what two equal tones give through a limiter. The intercept is 7.84 dB above it, which is π²/4 exactly. Devices, and the amplitude they stop being linear at

The point the device is never at

A device's linearity is specified by one number, and that number is a place on no curve. The third-order intercept is where two straight lines would cross if both went on being straight, and neither does. For a differential pair the crossing sits π²/4 — 7.84 decibels — above the largest output the device can produce at any drive whatever, and the arithmetic that says so contains no tail current and no temperature.

 An inrush limiter's steady state, and how little of it is still a limiter. computed by solving, not by drawing. A negative-temperature-coefficient thermistor in series with a supply, at 1 ampere of load current. The falling curve is what it dissipates at a temperature — I²R with R following the two-point β fit a catalogue prints — and the rising line is what its mounting removes. They cross once, at 83.1 degrees, and the loop gain there is -1.374: negative, so the part is stable at every current rather than below a boundary. What is left of its cold 10 ohms at that temperature is 1.937 — 19.4 per cent. The slider moves the load current, and more current leaves less resistance. Power, and the part that does no work

The protection that is gone by the second time

An inrush thermistor is ten ohms cold and holds the first cycle down; then the load current warms it and it settles at 83 degrees and 1.94 ohms, which is 19 per cent of what was bought. That is the design working. It is also a part that takes 198 seconds to recover half its cold resistance, against a reservoir capacitor that empties in tens of milliseconds — so a mains dip in that window hands the rectifier an unlimited inrush into an empty capacitor, which is the exact event the part is on the bill of materials for.

 5 sections, equal ripple, and the band that is 134% rather than 97. computed by solving, not by drawing. The repaired five-section equal-ripple design over the band it was designed for. The horizontal rule is the 0.1 the specification allows and the 4 interior peaks sit on it, level to 8.5e-4 per cent — which is the condition for a minimax solution and is now checked rather than assumed. The dots are the eighty-one frequencies the objective used to be evaluated at: the worst of them is 0.09999998 and the worst of the design over the whole band is 0.10013858, so an optimiser shown only the dots drove them down to the specification and left the true peaks 13.9 parts in ten thousand above it. That is nothing until something downstream is a threshold, and the band measurement was one: it reported 97.34 per cent for a design that holds 134.04. Lines, where a wire has a length

The number that was wrong

An earlier figure printed 97.3 per cent of band for a five-section transformer where the answer is 134, said in its own text that the figure was wrong, and blamed a search that had converged to eight digits. The search was fine. The objective was the worst of a grid rather than the worst of a band, the band was then measured by bisecting a function that crosses its threshold five times, and the check guarding all of it passed — because 97.3 is still more than 92.6.

 The noise bandwidth of four families at 8 orders, and the one that has none. computed by solving, not by drawing. Every one of the 32 entries is an integral of the realised network's own squared magnitude, divided by that network's own measured −3 dB point. At order one the four families are the same filter and return 1.5706, which is π/2 — the calibration the rest of the table is quoted against. Only Butterworth then does what the ratio is usually said to do: it falls at every order, to 1.0065 at 8. Bessel is least at order 5 (1.0385) and rises to 1.0441; Chebyshev alternates with parity, 0.9637 at five against 1.0686 at six; and an even-order elliptic has no noise bandwidth at all, because its stopband comes back up to a constant — its magnitude at the top of the range moves by 0.00 decades per decade of frequency, so the integral grows with whatever limit it is stopped at. The floor, which bounds from below

The ratio that does not walk to one

A single pole passes π/2 times as much noise as a brick wall at its corner, and every account of it says the ratio falls towards one as the skirt steepens. Over thirty-two realised filters only Butterworth does that. Bessel is least at order five, 1.0385, and rises again; Chebyshev alternates with parity and the two branches separate only above 0.1968 decibels of ripple; and an even-order elliptic has no noise bandwidth at all, its integral returning 380 or 38,005 depending on where it was stopped.

 The sensitivity of a pole against the room it has. computed by solving, not by drawing. A series R–L–C whose damping is walked from 0.3 to 0.999999, which slides its two poles together along a straight line and changes nothing else. The exact derivative of a pole with respect to the capacitor climbs from 0.5241 to 353.6 as the gap between them falls from 19078 to 28.28 radians a second. The fitted exponent over the closest four is -1.0000, and the product of the two is the natural frequency itself — 9999.6894 against 9999.6894, at every damping drawn and not merely in the limit, which a closed form gives and this computation never sees. The resistor's curve runs at 2ζ times the capacitor's — below it at 0.3 and at twice it by the time the poles have met — and the inductor's lies exactly under the capacitor's throughout. Before the steady state

The gap a derivative needs

The derivative of a pole is exact and has no step size in it, and beside the formula sits a sentence nobody had measured: it divides by a quantity that vanishes when two poles meet. Driven together, the sensitivity climbs as the reciprocal of the gap — fitted exponent −1.0000, the product a constant 1.00000 times the natural frequency — while the largest change it still describes falls as the gap *squared*. A one per cent capacitor is outside first order once the poles are 3194 radians a second apart, which is an ordinary critically damped design.

 What one temperature costs the loop gain of a part that has a gradient. computed by solving, not by drawing. The thermal loop gain of a 30 mm core, solved as a body with its own internal temperature profile and again as a single lump at that profile's mean, against the Biot number. Both are negative, so the core is a stabilising feedback either way — but the body's loop is the more negative of the two at every point, by 3.0 per cent at a Biot number of 0.108 and 38 per cent at 10.8. A lumped calculation therefore understates how stable a wound part is, and the amount it understates by is not a property of the material but of how well the surface is cooled relative to how well the inside conducts. Below a Biot number of about a tenth it is worth under two per cent and the lump is the right model; at the cooled end the part has 7 kelvin inside it and half the feedback is invisible to a single temperature. Power, and the part that does no work

The loop gain one temperature understates

Every thermal loop gain this collection has computed was computed at a single temperature, because a lumped fixed point has only one — and the essay that measured the gradient inside a core recorded, without measuring it, that this makes each of those numbers a lower bound. It is a lower bound by three per cent where a ferrite usually sits and by thirty-eight per cent at the well-cooled end, always in the direction that makes the part safer than the calculation said. The obvious candidate for what decides it is refused: three geometries at one Biot number are 3.3 times apart.

 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. Frequency, which is the same solve

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.

 Four wires against a 10 MΩ voltmeter. computed by solving, not by drawing at 81 resistances, twice each, with a voltmeter of 10 MΩ and 50 mΩ in every lead. The four-wire error is not zero: it is the voltmeter's own divider, −(R + 2R_lead)/(R + 2R_lead + R_m), which grows with the resistance being measured rather than shrinking. The two-wire error is that same quantity plus the leads, so it passes through zero at 1000 Ω — where the reading is right to 1.8e-12 while the four-wire reading is 0.0100% low — and above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance. Measurement, which is a circuit on a circuit

The voltmeter four wires do not remove

A four-terminal measurement is described everywhere as removing the leads from the answer. It moves them. What is left is the voltmeter's own input resistance, and it grows with the resistance being measured rather than shrinking: with a ten-megohm voltmeter and fifty milliohms of lead, the four-wire reading is 0.0100 per cent low at a kilohm, where the two-wire reading is exactly right — 1.8 × 10⁻¹² — because its lead error and its loading error cancel. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

 Inside the passband the matrix is worst at the edge, and at no ripple peak. computed by solving, not by drawing. The condition number of a 5th-order Chebyshev filter's nodal matrix at every frequency from a hundredth of its cutoff to ten times it. It is 4.004e+3 at direct current, rises to 1.5868e+7 at 964.0 Hz — located by golden section rather than read off the sweep — falls to 4.970e+6 at 1520 Hz and rises again through the stopband. The maximum is above every ripple peak (the highest is at 897.9 Hz) and within 3.6 per cent of the half-power frequency: the worst-conditioned place in the passband is where the filter stops passing and starts blocking, which is a place the response curve has no feature at. It is a local maximum: above the band κ climbs again, reaching 2.702e+7 at ten times the cutoff, because the susceptances in the matrix grow with the frequency and κ counts them. Nothing about any of these numbers says whether a digit is actually lost anywhere. Networks, and how a solve is checked

Where the matrix is worst, and where the answer is not

When a solve stops being exact has been answered with two networks at direct current. Swept along the frequency axis, a resistive chain's nodal matrix has a condition number of 4.00×10³ at direct current and 1.59×10⁷ at 964 Hz — a maximum at the band edge, at no ripple peak and at no feature the response has. It predicts nothing. The solution vector is right to three units of round-off everywhere, the response taken out of it loses four decades into the stopband, and the same filter written at 400 kΩ instead of 10 Ω has a condition number 1.4×10⁹ times larger and returns the same twelve digits.

 The same core at the same current has two inductances, 1.80 times apart. computed by solving, not by drawing. The small-signal inductance of a sixty-turn winding on a core walked down from 400 amperes per metre, measured by pushing the excitation up and by pushing it down at each bias. They are never the same: 14.899 millihenries against 8.297 at zero bias, a factor of 1.796, and up to 1.796 across the sweep. The mechanism is that an operator held inside its own backlash contributes nothing to dB/dH, so a reversal is measured by whichever operators are still moving, and that is a different set in each direction. The single-valued curve the field already had is drawn above both, which is what it is: an optimistic reading of an object with two answers. Two windings, and the band between them

Two inductances at one current

A data sheet prints one L(i) curve and there are two. An operator sitting inside its own backlash contributes nothing to dB/dH, so a small excitation sees only the operators still moving — every one at the tip of a loop, none just after a reversal — and the same core at zero bias measures 14.90 millihenries pushed downward and 8.30 pushed upward, a factor of 1.80. The same split decides what a converter's ripple costs: held at the flux swing volt-seconds actually fix, a twenty-millitesla ripple costs twelve times more at a hundred and seventy-five amperes per metre of bias than at none, and above a hundred and eighty-three the question has no answer at all.

 The direct form works at 28 bits and fails again at 29. computed by solving, not by drawing. The largest pole radius of an order-8 Butterworth at a 1 kHz corner and a 48 kHz clock, at every coefficient word length from 6 to 40 bits, for both realisations. Above the unit circle the filter is not inaccurate but unstable. The direct form is outside at 23 of the 35 word lengths drawn; the cascade at none of them, its worst radius being 1.000000. The failures are not an interval: 28 bits works and 29 bits does not, so the smallest word length that works is 1 below the largest that does not and a search for the crossing has no crossing to find. The direct form's curve is clipped at 1.24: a radius of 4.5 and a radius of 1.01 are the same verdict. Where a signal becomes a number

The word length that is not a threshold

The essay below this one measured an eighth-order direct form at five word lengths, found its poles outside the unit circle at every one, and left behind a phrase — the coefficient resolution required. Sixty word lengths later there is no such resolution. That filter is stable at 28 bits, unstable at 29 and stable again at 30, and the smallest word length a search would return is not one anybody can use. What survives is a law with a slope: 4.53 bits of coefficient for every pole added, rising to 47 bits at twelfth order, and none at all for a cascade.

 A core loss of 100 kΩ gives the resonance a floor at −66 dB, and by 10.0 GHz the choke is worth 0.02 dB. computed by solving, not by drawing. The common mode at the load, in decibels against the same circuit with no choke in it, with 5 pF across each winding and a core loss of 100 kΩ across each, drawn over the lossless curve. At the winding's resonance, 1.59 MHz, the inductance and the capacitance cancel and what is left is the loss, so the deepest point is 66.02 dB — 20·log(1 + (Rp/2)/(Zs + Zl)), with no inductance and no capacitance in it. Above the resonance the part is a capacitor across the path it was fitted to block, and by 10.0 GHz it attenuates 0.02 dB. Lines, where a wire has a length

The depth a resonance does not have

The inductor one mode cannot see printed its choke's best attenuation as 102 decibels at 1.59 megahertz. Swept with 1,999, 2,000 and 2,001 samples the same lossless part reports 110.0, 96.3 and 103.6 decibels, because a resonance with nothing to dissipate has no bottom and a sweep reports how near its nearest sample fell. Give each winding a core loss and the depth is 20·log(1 + (Rp/2)/(Zs + Zl)) — 66.02 decibels at a hundred kilohms, twenty more per decade of loss, with no inductance and no capacitance in it, and half of it belonging to the circuit the choke sits in.

 The winding current a hysteretic core asks for, and the current the same circuit's single-valued core asks for. computed by solving, not by drawing. One netlist — a 22.6 volt peak sinusoid at 5.00 kHz through 4 ohms into a sixty-turn winding — marched twice, once with the core as a superposition of play operators and once with the single-valued saturating curve every earlier figure in this field used. The two currents differ in shape and not only in size: the hysteretic one leads the flux by an angle that is not ninety degrees, which is the whole of the core loss, and its peak is 59.45 milliamperes against 50.12. The energy the source delivers over a complete cycle is 17.19 microjoules for the loop and 2.5e-5 for the curve, which is zero to the resolution of the march. The slider takes the flux the drive demands from a twentieth of the material's saturation to most of it. Two windings, and the band between them

The core the solver has to remember

A saturating inductor's state is one number, because the current is a function of the flux linkage. A hysteretic core's is not: half an amp is one flux on the way up and a different flux on the way down, and there is no function of the current that returns the flux. So the march's state vector gains twenty-four more numbers, one per play operator, and the Newton loop is forbidden to touch them. The circuit and the bench then agree on the loss to three parts in ten million — one integral taken at two wires, the other inside the material.

 Loop gain of a two-pole amplifier closed for a gain of 100. Unity loop gain at 5.73 kHz, where 35.0° of phase remains before −180°. The phase never reaches −180° at any frequency, so there is no gain margin to quote: 2 poles contribute at most 180° and the last of it arrives only at infinity. Feedback, and the margin

The loop that never crosses

Every loop this field draws carries two margins, and one of them is not always a number. Take the third pole out of the standard loop and its crossover moves by 2.45 parts per million and its phase margin by 0.164 degrees — the third pole's own arctangent there, to five decimal places — while its gain margin goes from 46.06 decibels to no number at all. The phase reaches −180° only where the magnitude has already reached −62 decibels, and the two instruments that are supposed to notice report 3.19 × 10⁻⁶ either way.

 One gain takes the equivalent resistance from 5 kΩ through infinity to negative. computed by solving, not by drawing. A 5 V source drives a node through 10 kΩ; the node also reaches 10 kΩ whose far end is held at A times the node's own voltage. The Thévenin resistance looking into that node is r1 in parallel with r2/(1 − A), which the solve returns to a part in a billion without being told: 5 kΩ at no gain, 10 kΩ at unity where r2 takes no current at all, and unbounded at A = 2.00 where the two conductances cancel. Above that it is negative. The open-circuit voltage follows it, because the short-circuit current is 500.0 µA at every gain — a short across the controlling node leaves the dependent source nothing to be controlled by — so the open-circuit voltage is simply the short-circuit current times whatever the resistance is, and reaches 150.0 volts from a five-volt source inside the range drawn. Networks, and how a solve is checked

The resistor that is not made of the resistors

Exact outside and wrong within reduced six elements to one source and one resistor and found the resistor two ways that agreed to the last bit. Put a dependent source in the network and one of those routes stops working, because setting the sources dead kills the independent ones and leaves the dependent one where it is. On a bootstrap of two ten-kilohm resistors the Thévenin resistance runs from five kilohms through infinity to minus ten, the open-circuit voltage of a five-volt source reaches 225, and above one gain the equivalent's resistor is negative — which the netlist refuses to stamp, correctly, because a negative resistance is a controlled source and not a resistor.

 The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance. Two windings, and the band between them

The loss that depends on what it causes

Every thermal figure in this collection has had the power handed in. A ferrite's has no business being: its saturation flux falls with temperature, its permeability rises, and both move the loss. Closing that loop makes the temperature a fixed point rather than a product — and the fixed point has a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists.

 The circuit does not care which node is called zero, and the matrix does. computed by solving, not by drawing. The condition number of the nodal matrix for a 12-section chain, against which of its nodes was taken as the reference. The network, its elements and its physics are identical in every case — only a label has moved — and every branch voltage and branch current comes back the same to 1.1e-13. The condition number runs from 5.25e+4 at "n6" to 1.72e+5 at "n12", a factor of 3.28, which is 0.52 decimal digits of the arithmetic's own margin. Networks, and how a solve is checked

The node that is not in the circuit

Nodal analysis needs a node to call zero and no circuit contains one. Moving it changes every node voltage by the same amount and no branch voltage or branch current at all — to a part in ten to the fourteenth on a well-behaved chain. What it does change is the matrix: the condition number of a twelve-section chain moves by a factor of 3.3 with the reference, and on a star whose resistances span nine decades by 8.0. Measured against the same matrix solved in twice the precision, the worst reference costs four decimal digits of the answer, and it is the reference the condition number named before the error was looked at.

 One output in saturation takes 6.5% off another that is at five volts. computed by solving, not by drawing. A three-transistor mirror: a reference, an output taken down into saturation, and a third output held at five volts throughout. The upper curve is the saturating output's own loss and the lower one is the sibling's. At 50 mV the saturating output is 27.3 per cent down and the sibling, which is nowhere near saturation, is 6.54 per cent down. The reference current moves by -0.0188 per cent, which is nothing: a base current is a hundred and fiftieth of a collector current and the reference is set by a resistor from the supply. What does move is the base-emitter voltage every output shares — by -1.75 millivolts — and every output is an exponential of it. Devices, and the amplitude they stop being linear at

The refusal, and what it was protecting

A current mirror's model has declined to answer below two hundred millivolts of collector-emitter voltage since it was written, because it has no base-collector junction and would return a forward-active current for a saturated transistor. Put the junction in and the refusal turns out to have been placed where the model it protects is still right to seven parts in ten thousand — and the mirror's real failure is somewhere else entirely: a saturated output takes six and a half per cent off an output that is sitting at five volts.

 What a high-side shunt's optimum is made of, at 100 dB of rejection. computed by solving, not by drawing. The same two errors as a low-side shunt, with the amplifier now standing at the rail rather than at the return. 100 dB of rejection turns 12 V of common mode into 120.0 µV of equivalent input error, which is 24 times the amplifier's own 5.0 µV of offset. The optimum keeps its form — the geometric mean of an input error and the supply, golden-sectioned on the solved worst case rather than substituted — and changes its value: 38.73 mV of burden and 0.6440% of error, against 7.75 mV and 0.129% low-side. A shunt sized by the low-side answer reads 1.678% wrong. With the common-mode term removed the optimum returns to the low-side value exactly, which is what says the term is the whole of the difference. Measurement, which is a circuit on a circuit

The rail that is an input error

The best burden voltage across a shunt is the geometric mean of the amplifier's offset and the supply, and moving the shunt to the high side does not change that form — it changes what the offset is. A hundred-decibel amplifier on a twelve-volt rail turns the rail into 120 µV of equivalent input error, twenty-four times its own five, so the optimum moves from 7.75 mV and 0.129 per cent to 38.73 mV and 0.644. The two are equal only at 128 dB, and with the common-mode term removed the optimum returns to the low-side value exactly.

 What a pair does to the three boundaries: removes two, moves one. computed by solving, not by drawing, with every harmonic taken from a transform of the waveform rather than from a series. A differential pair's transfer is an odd function, so the mean and every even harmonic are zero — -3.2e-17 and 9.1e-17 at a drive of two thermal voltages, which is absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit of a tight criterion and by 1.42610 at the one per cent quoted here — 10.42 mV against 7.304 mV at 27 °C, against √2 = 1.41421 and 1.41433 at a criterion a hundred times tighter. So a pair reached for as headroom has bought forty per cent of it and a pair reached for as linearity has bought something else entirely. Twenty millivolts of imbalance brings the even orders straight back, which is what says the cancellation belongs to the symmetry rather than to the topology. Where the models stop

Two boundaries removed, and one moved

A differential pair's transfer is odd, so its mean and every even harmonic are zero — −3.2×10⁻¹⁷ and 9.1×10⁻¹⁷ at a drive of two thermal voltages, absent rather than small. Two of the three one-per-cent boundaries therefore do not exist for a pair at all. The third, the loss of gain, moves by √2 in the limit and by 1.42610 at the one per cent usually quoted: 10.42 mV against 7.304. So a pair reached for as headroom has bought forty per cent of it, and twenty millivolts of imbalance brings the even orders straight back.

 At Q = 2 and a hundred times the corner, one arrangement is 1.97% out and the other 2.83%. computed by solving, not by drawing. The same second-order section designed twice — a follower with a capacitance ratio, and equal passives with the Q supplied by a closed-loop gain of 2.5000 — built with the same one-pole amplifier and swept over its gain-bandwidth, with both pole pairs recovered by rooting the determinant. The unity-gain arrangement's error is Q over the ratio: 2.00 per cent at a hundred times the corner. The equal-component arrangement's is K²/2 over the ratio, 3.13 per cent, because its amplifier is a gain-of-K stage and therefore has K times less bandwidth to spend. Below about thirty times the corner both laws fail, and the second one changes sign. Filters, measured not tabulated

Where the Q comes from

Two second-order sections with no component value in common have the same transfer function to 1.8×10⁻¹⁰ of a decibel, and are not the same circuit. Against a slow amplifier the follower's Q error is Q over the gain-bandwidth ratio and the gain stage's is K²/2 over it — so the second is worse below Q = 3.08 and better above it. Against component tolerance the follower's worst element carries ½ and the gain stage's carries 2Q − ½, nineteen times as much at Q = 5. Neither oscillates at any gain-bandwidth at all.

 The current divider, and the resistance that is not in the branch. computed by solving, not by drawing. A current source into two parallel branches, the metered one 10.0 kΩ and the other 1.00 kΩ. The metered branch takes 0.090909 of the current, which is the OTHER branch's resistance over the sum; writing the subscripts the way a voltage divider writes them gives 0.90909, a different number at every ratio but one. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with the roles exchanged: one per cent of error at 111.1 Ω, which is that resistance over ninety-nine to 1.7e-12%. And the headline of the loaded divider holds in the dual too — two current dividers of identical ratio read 0.04762 and 0.09090 into one hundred-ohm meter — while a perfect ammeter reads them identically. Networks, and how a solve is checked

The branch the other resistance decides

A voltage divider's output is set by the resistance the output is taken across; a current divider's is set by the resistance the current does not go through. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with one word changed: one per cent at a meter resistance of R/99 where R is what the meter looks back into — and removing an ideal current source means OPENING it, so that R is the two branches in series, 11 kΩ here, not the 909 Ω of their parallel combination — a factor of twelve in the same construction on the same network.

 The shunt's resistance as a function of what it is measuring. computed by solving, not by drawing, as a fixed point: the shunt dissipates I²R, its temperature rises by 20 K per watt, and at 50 ppm/K its resistance rises with its temperature — so the resistance the reading is divided by depends on the reading. Iterated to convergence it agrees with the closed form R₀/(1 − αθI²R₀) to 2.2e-16. Along the burden-voltage optimum, where R = u⁄I, the dissipation is I·u rather than I²R, so the temperature rise is 155 mK per ampere and the error is the FIRST power of the current — fitted exponent 1.0007 over five decades. That is the only one of the shunt's errors with the current in it, and it puts a term in I² into the reading, which is a curvature no single-current calibration removes. The upper curve is a shunt of fixed resistance, where the error is quadratic. The fixed point stops existing at 129 kA and never at a current a shunt will see. Measurement, which is a circuit on a circuit

The resistance that depends on the reading

Three of a shunt's errors are free of the current being measured, which is the whole content of the burden-voltage optimum. The fourth is not: the shunt dissipates, warms, and its resistance rises — so the divisor the reading uses is a function of the reading. Solved as a fixed point it agrees with R₀/(1 − αθI²R₀) to 2×10⁻¹⁶, and along the optimum, where the dissipation is I·u* rather than I²R, the error is the FIRST power of the current: 7.75 ppm at an ampere, 775 at a hundred, fitted exponent 1.0007.

 The net noise power between two resistors at two temperatures. computed by solving, not by drawing. A 1 kΩ resistor at 400 K joined to a second one at 290 K, whose resistance runs over six decades. Each delivers 4kTR₁R₂/(R₁+R₂)² per hertz to the other; the net is 1.5187 zW/Hz at the match, which is exactly kΔT and contains no resistance — a gigohm pair at the same two temperatures exchanges the same. Away from the match the exchange falls, to 0.0596 zW/Hz at a ratio of a hundred, so the maximum-power argument applies to noise as it does to a signal. At one temperature the net is zero at every ratio, to 7.5e-37 W/Hz, which is the statement a wrong noise model would break. The floor, which bounds from below

Which way the noise goes

Two warm resistors joined together each drive the other, and the net flow is 4kΔT·R₁R₂/(R₁+R₂)² per hertz. At the match that is kΔT exactly — 1.5187 zeptowatts per hertz between 400 K and 290 K — and a kilohm pair and a gigohm pair at the same two temperatures exchange the same, which is why noise is quoted as a temperature. At one temperature the net is zero at every ratio to a part in 10³⁷, and that zero is the second law rather than a tolerance.

 A regulator holding 10 V on a load that asks for 101% of the nose power collapses it in 256 s; at 110%, in 79 s. Marched with a fourth-order rule. A 10 V source behind 1 Ω feeds a load resistance through an ideal ratio n, and a regulator raises n at 0.05 per volt-second of error to hold the load at 10 V. The load's resistance is chosen so that at 10 V it takes the stated fraction of the most the line can deliver, 25 W. At 90% the regulator settles at n = 1.5195, below the nose ratio √(Rₗ/R) = 2.1082. At 99% the regulator settles at n = 1.8182, below the nose ratio √(Rₗ/R) = 2.0101. At 101% the voltage climbs to 9.950 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 101.7 s, and falls below half the setpoint at 256.3 s while the regulator keeps raising the ratio. At 110% the voltage climbs to 9.535 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 23.5 s, and falls below half the setpoint at 78.6 s while the regulator keeps raising the ratio. The regulator's gain, the slope of the load's voltage against the ratio, is positive below √(Rₗ/R) and negative above it. Power, and the part that does no work

The regulator that pushes past the nose

A regulator that raises a ratio whenever its load's voltage is low is a stabiliser only while raising the ratio raises the voltage, and on a line that stops being true at exactly the nose: the load's voltage, n·V₀ times the load resistance over n²R plus that resistance, peaks at a turns ratio of the square root of the load over the line and falls beyond it. Ask the load for 90 per cent of the nose power and the regulator settles at n = 1.519 — unless it starts above n = 2.925, where the same setpoint is met on the wrong side of the peak, and then it collapses the voltage. Ask for 101 per cent and the voltage climbs to 9.950 volts, the most the line allows, and is below half its setpoint 256 seconds later. Near the nose the collapse takes a time that grows as the inverse square root of the excess: 2,521 seconds at a hundredth of a per cent.

 Zeroing the dependent source too: an answer that does not depend on the gain. computed by solving, not by drawing. Two independent sources reach a node through resistors; a voltage-controlled source of gain −10 is driven from that node and fed back to it through 10 kΩ. Solved whole, the output is -25.0000 V. Solved by superposition with the dependent source LEFT IN each of the two solves it is -25.0000 V — the same number to 1.1e-16, so superposition is exact here as it is everywhere else in a linear network. Zeroing the dependent source along with each independent one gives nought volts at the output and 4.06250 V at the node, against the node's true 2.50000 V — 62.5 per cent out — because a controlled voltage source set to zero is a short circuit and the three solves are then of three different circuits. What makes it unmistakable is that the wrong answer is 4.06250 V at EVERY gain from one to three hundred: the route has deleted the only element that knows what the gain is, while the true answer moves by more than a factor across the same range. Networks, and how a solve is checked

The source that must not be zeroed

Superposition says to zero all the sources but one, and the care usually taken over that instruction is that zeroing is not deleting — a voltage source with no voltage is a short circuit and it stays in the netlist. A dependent source needs the opposite care: it must not be zeroed at all, because it is an element whose law involves another node rather than a source the theorem is about. Zeroed, it is a short circuit, and the answer that comes back is 4.0625 volts at every gain from one to three hundred — while the true answer moves from 3.82 volts to 0.206.

 What widens a null: the order of the zero, not the placement of two. Integrated by quadrature on the union of both windows' nulls, at a fixed total length of 40 ms so that every arrangement settles in the same time. One mean of 40 ms holds -40 dB over 2.002 per cent of 50 Hz and passes 12.50 Hz of noise. Two means of 20 ms in cascade — a second-order null in the same place — hold it over 20.59 per cent — a factor of 10.3, which is 10^(−dB/40) because a first-order null rises linearly through the target and a second-order one quadratically — for 1.333 times the noise bandwidth. Splitting the two lengths puts two first-order nulls either side instead, and the widest trough is at a split of 10.12 per cent — 17.98 and 22.02 ms — holding 29.93 per cent for 16.53 Hz — which is exactly √2 times the equal pair's band, at every depth, for slightly LESS noise. It is also a cliff: a split of 10.118 per cent — 0.01 per cent further — leaves no band at 50 Hz at all, because the peak between the two nulls has climbed through the target. The floor, which bounds from below

The order of the null, not the number of them

A mean rejects the mains exactly and rejects it one per cent off frequency by forty decibels, whatever the window length. Two means in cascade of the same total length settle in the same time, cost a third more noise bandwidth and hold that forty decibels over ten times the band — the factor being ten raised to minus the depth in decibels over forty, because a first-order null rises linearly through a target and a second-order one quadratically. Splitting the two lengths to straddle the frequency buys a further factor of exactly root two, for no noise at all, and puts the design two parts in ten thousand from a cliff.

 Sampling 100–120 MHz: 6 allowed windows and 5 forbidden gaps. A band 20 MHz wide centred at 110 MHz, and the sample rates at which it and all its images land without overlapping. Twice the top of the band is 240 MHz and every rate above that works, which is the familiar answer. Below it there are 5 more windows, the slowest at 40.00 MHz — 1.00 times twice the band's WIDTH, which is the quantity that bounds a sample rate when the signal is not at baseband. Between the windows are 5 forbidden gaps, shaded: at 219.1 MHz the band's images overlap it, while 200.00 MHz works and 240.0 works. So a clock raised out of a working window has been made worse by being made faster, and "a faster clock is always an available answer" is a statement about baseband and not about sampling. Where a signal becomes a number

The clock that is too fast

Every sample rate in the requirement is a lower bound: the filter must be down by the frequency that folds back, and a faster clock is always at least as good. That holds for a signal at baseband. A band from 100 to 120 megahertz can be sampled at 240 and above, or at 120 to 200, or 80 to 100, or 60 to 66.7, or 48 to 50, or 40 — six disjoint windows with five forbidden gaps between them, so 110 megahertz fails while 100 works and 120 works. The bound is twice the band's width, 40 megahertz, and it is six times below twice its top.

 10.0 cm of track leaves 40 per cent on a 1 ns edge. The far end of 10.0 cm of 50 Ω track, driven from 10 Ω into 200 Ω, marched as the exact staircase of reflections it is and driven by a ramp rather than by a step. The staircase converges on the resistive divider the circuit was going to be all along, to a part in 10⁹, which is the number the lumped model returns immediately and with no staircase. What is left on the edge is an overshoot: 40.0 per cent at 0.1 ns, which is the first wave's own reflection coefficient and contains no rise time at all, falling to 0.399 per cent at 100 ns where the staircase happens inside the edge. It reaches ten per cent at an edge of 4.24 ns and one per cent at 40.0 — which is 28.6 times the 1.40 ns round trip. Nothing in the measurement is a repetition rate: the boundary is a transit time, a mismatch and a rise time, and the clock the signal repeats at does not enter. Where the models stop

The boundary that is a rise time

Kirchhoff's own frequency is 3.97 megahertz for ten centimetres of track, and a one-kilohertz clock is four thousand times below it — which is the reading that concludes wrongly. Marched as the exact staircase of reflections it is, the same track leaves forty per cent of overshoot on a one-nanosecond edge and 0.40 per cent on a hundred-nanosecond one, reaching one per cent at 40.0 nanoseconds. The boundary is a rise time, it is twenty-nine times the conductor's round trip, and the repetition rate is not in it anywhere.

 The temperature a part cannot come back from, and how long it takes to leave. computed by solving, not by drawing. The same fixed-point equation as the rung below, marched in time with a thermal capacitance rather than solved for its steady states: C dT/dt = P(T) − (T − T_a)/R_th, stepped adaptively on the temperature change because dT/dt goes through zero at each fixed point. Every trajectory starting below 191.1 °C returns to 88.8, however far above the operating point it began; every one starting above it leaves the material's range entirely, the closest in 0.2 minutes. The two nearest starts are 3.0 kelvin apart. The ignition temperature is a boundary in the STARTING CONDITION, and no steady-state analysis contains one. Two windings, and the band between them

The boundary that is a starting point

A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes.

 With 1 mV of offset the pair's second harmonic returns in proportion to the drive, and the third harmonic reaches one per cent first. computed by solving, not by drawing, every harmonic read from a solved waveform. A differential pair at 27 °C with 1 mV of input offset: its second harmonic (solid) and third (dashed) over its fundamental, against the peak drive, beside a single transistor's second harmonic (dotted). The third is the balanced pair's own, u²/12, and reaches one per cent at 17.9 mV whatever the offset. The second is (u/2)·tanh(Vos/2Vₜ) at small drive, the single transistor's second harmonic times tanh(Vos/2Vₜ) = 1.93e-2, and never reaches one per cent: it peaks at 0.74% as the pair begins to saturate and falls again. Where the models stop

The offset that brings the even orders back

A balanced pair has no second harmonic. An offset gives it one, and the closed form is exact: the second harmonic over the fundamental is (u/2)·tanh(Vos/2Vₜ), with u the drive over 2Vₜ — a single transistor's second harmonic times the fraction of the tail the offset steers aside, 1.93% of it at a millivolt, not the twentieth usually estimated. It is first order in the offset up to 9.0 mV. It does not grow with the drive for ever: it peaks as the pair saturates and falls again, so below 1.36 mV of offset it never reaches one per cent at any drive, and above 3.14 mV it reaches one per cent before the pair's own third harmonic does, at 17.9 mV. Read the other way, a matching specification is a distortion specification: −60 dB of second harmonic at 10 mV of drive allows 0.535 mV of offset.

 Three bias points below a coupling of 56.6 kΩ, one above — and the middle one is where Newton from nought lands. computed by solving, not by drawing. Two transistors, each with 1 kΩ to a 5 V supply and each biased from the other's collector through a coupling resistor into a 10 kΩ base resistor, solved by Newton on the netlist from six starting guesses and de-duplicated. Below a coupling of 56.6 kΩ there are three operating points: two latched, with one collector at 0.061 V and the other at 4.615 V at 10 kΩ, and a symmetric one at 1.702 V whose Jacobian determinant is negative, so the circuit cannot rest there. Above it the three merge into one symmetric point, and the circuit is an amplifier rather than a memory. Devices, and the amplitude they stop being linear at

The circuit with three bias points

Two transistors, each biased from the other's collector, have three operating points: one collector at 0.061 V and the other at 4.615, the same the other way round, and both at 1.702. All three solve the circuit to a part in 10⁹, and Newton started from nought — where a simulator starts — finds the symmetric one, which is the one the circuit can never sit at: its Jacobian determinant is negative, so any positive capacitance at the nodes makes it run. Weaken the coupling past 56.6 kΩ and the three merge into one, the loop gain around the symmetric point having fallen through one. Which answer a solver returns depends on where it begins, and from 1,681 starting guesses at 10 kΩ, 437 return the one that is not a state.

 Returning the shunt resistor to k times the output divides the residue by up to 30 and never takes it through zero. computed by solving, not by drawing. The residue of an RC bridged-T with R₁/R₂ = 100 at its null frequency, with its shunt resistor returned to k times its own output rather than to ground, against 1 − k: solved (dots) and 2/(2 + n(1 − k)) (line). An inverting return raises the effective ratio to n(1 − k) — at k = −15 it is 1600 — so the residue falls to 0.00125. But the null's slope falls with it, and the loop's ceiling, the slope over twice the residue, is 4.994 against 4.902 earthed: it approaches √n/2 = 5.000 of the resistors' own ratio and never passes it. At no k below one does the residue reach zero, because the trim is a ratio and the residue is positive for every ratio. Filters, measured not tabulated

The residue that falls with its slope

A bridged-T's residue at its null is 2/(2 + n), and a loop around it stops at a Q near √n/2. Returning its shunt resistor to −15 times the output instead of to ground divides the residue by sixteen, to 0.00125 — and the loop's ceiling stays at 5.00, because the slope through the null falls with the residue. It has to: at a transmission zero the output is zero, so a trim that acts through the output leaves the zeros where they were, and the ceiling of a loop with unlimited gain is the zeros' own Q. What the trim does change is the gain the ceiling needs — nine-tenths of it at A = 9/ρ, 449 earthed and 14,400 at k = −31. No return gain below one takes the residue through zero, and a non-inverting one lowers the ceiling.

 The bootstrap with 1 nF at its node: one natural frequency, against the gain. computed by solving, not by drawing: the natural frequency of the network found as the root of its determinant at each of 31 gains. It is −G/C to a part in a million at every gain, where G is the Thévenin conductance the reduction gives: −200 thousand per second at A = 0, zero at A = 2 where the resistance passes through infinity, and +100 thousand at A = 3 — a growth rather than a decay. The gain at which the pole crosses into the right half-plane is the gain at which the resistance changes sign, and the capacitor's size moves the pole's speed and nothing else. Networks, and how a solve is checked

The pole that crosses with the resistance

A bootstrap's Thévenin resistance passes through infinity and turns negative at a gain of two. Put one nanofarad at its node and the network has one natural frequency, −G/C to a part in a million: −200 thousand per second with no gain, zero at A = 2, +100 thousand at A = 3. The gain at which the pole crosses is the gain at which the resistance changes sign — for this network. Loaded by a resistor it crosses at one plus R₂ times the sum of the conductances to ground, so a negative resistance is stable across a smaller load and not a larger one; a lag in the source slows the growth and does not move the crossing; a tank on the node turns the runaway into a 30.5 kHz oscillation once its coil is below about 1.5 Ω; and with an inverting source the resistance is positive at every gain and the node still oscillates beyond A = −16.

 The bootstrap reduced both ways: Thévenin's resistance diverges at A = 2, Norton's conductance crosses zero there. computed by solving, not by drawing at 121 gains, R₁ = R₂ = 10 kΩ. The Norton current is Vs/R₁, 0.500 mA, at every gain to a part in a billion (dashed, at 1), because shorting the node also shorts the bootstrap's input. The Norton conductance times R₁ is a straight line, 2 at A = 0 through zero at A = 2 to −1 at A = 3 (solid). The Thévenin resistance over R₁ is its reciprocal (faint, clipped at ±3): 0.5, 1 at A = 1, infinite at A = 2, −1 at A = 3. One network, two coordinate systems, and only one of them is singular exactly where the network is interesting. Networks, and how a solve is checked

The reduction that stays finite

The bootstrap's Thévenin pair diverges at A = 2: its resistance and its open-circuit voltage both pass through infinity and the solve that finds them refuses. Its Norton pair does not. The short-circuit current is Vs/R₁, 0.5 mA, at every gain to a part in a billion, and the conductance is a straight line through zero. A bench that drives a current and reads a volt loses the network for every gain from 1.992 to 2.008; a bench that drives a volt and reads a current sees it to a nanosiemens throughout. A 0.1% error in the gain moves the conductance by a constant 0.1 µS and the resistance by an amount that diverges at the pole — and four hundred bootstraps built from 0.1% parts to sit on it have conductances within ±0.3 µS of zero and resistances that are a median 10.6 MΩ, 196 of them negative.

 A 1 MHz sine of 1 V into an emitter follower with 1 nF on its emitter and a 2 mA sink: the input falls at up to 6.28 V/µs, the output can fall at only 2.0, and the transistor switches off. Marched, the base driven by a stiff source and the emitter current Is·exp((vᵢₙ − vₒᵤₜ)/Vₜ). The input (dashed) and the output (solid) over the last of four cycles. The input's steepest fall is 2πfA = 6.28 V/µs; the load can be discharged no faster than the sink allows, I/C = 2.00 V/µs. On every falling stretch the base–emitter junction reverse-biases, the emitter current drops below a hundredth of the sink's (dots), and the output falls as a straight line of slope −2.000 V/µs until the input comes back up to meet it. On the rising half the transistor pushes whatever current the capacitor needs and the output follows a Vbe below the input. Devices, and the amplitude they stop being linear at

The follower that lets go on the way down

An emitter follower can push current into its load without limit and pull it out only through whatever sinks the emitter. Marched as a large-signal circuit — a 2 mA sink, 1 nF on the emitter — a 1 V, 1 MHz sine asks the output to fall at up to 6.28 V/µs, the sink allows 2.00, and on every falling stretch the junction reverse-biases, the emitter current drops below a hundredth of its quiescent value for 57% of the cycle, and the output falls as a straight line of exactly −I/C. The boundary is a slope, 2πfA ≤ I/C: 2.87 V at 100 kHz, 0.309 V at 1 MHz, creeping above the line only as the period approaches the junction's own 12.9 ns. Past it the follower rectifies — its output's average rises 415 mV at 1 V of drive and 1.30 V at 2 V — and with a resistor in place of the sink the fall slows as the output descends, from 2.41 V/µs to 1.99. None of it is visible to the small-signal model the earlier essays used.

 One step of refinement in the same double precision returns the stopband's lost digits: 1.1e-12 at 10 kHz before, under 2.4 units of round-off after. computed by solving, not by drawing. The 5th-order Chebyshev ladder solved at every frequency in double precision (solid), and again after one step of iterative refinement — the residual b − A·x formed in double, the correction solved with the same elimination and added (dashed) — each against the same network eliminated in a pair of doubles. Plain, the response's error tracks the round-off divided by the response (dotted), from 1.1e-16 at 10 Hz to 1.1e-12 at 10 kHz, where the output is 6.78e-7 of the input. One step brings every frequency to within 2.4 units of round-off of the truth. No more digits were used: the loss belonged to the elimination's errors in the small component, and the residual, computed against the original matrix, sees them. Networks, and how a solve is checked

The step that returns the digits

A fifth-order ladder filter solved in double precision loses four decades in its stopband — 1.1 × 10⁻¹² of relative error at ten times the cutoff — on a solution vector that is right to round-off. One step of iterative refinement, in the same double precision, returns all of it: form the residual b − A·x, solve for a correction with the same elimination, add it, and every frequency is within 2.4 units of round-off of a 31-digit solve. The reason is that the plain solve is stable by the norm and not by the row — its componentwise backward error is 5.3 × 10⁻¹³ in the stopband rows — and one step makes it componentwise stable. The bridge walked to balance cannot be helped at all. Its node voltages are already right to the last bit; its digits are lost storing two numbers whose difference is the answer, and again storing its conductances, which moves the network itself by round-off over δ before anything is solved.

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