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

The thread: Measured, not tabulated

Butterworth is flat, Chebyshev is steep, Bessel has good delay — three adjectives and no quantities, in the most reproduced table in the subject. Every comparison on this site is taken off a solved network instead, which is how the third column, the one that decides whether a filter can pass an edge, turns out to have been missing.
The closed-loop poles at a gain of 3.05. The locus of the two poles as the amplifier's gain runs from 2.7 to 3.3. It crosses the imaginary axis at a gain of 3.000000 — bisected on the netlist, not quoted — and at 3.05 the real part is 2.500e+2 radians a second, which is an envelope multiplying by 1.17015 every cycle. The crosses are the closed form ω₀(k−3)/2 and they sit on the measured circles. Circuits that do a job, and the range they do it over

The gain that is exactly one

An oscillator is designed by making the loop gain one at the frequency where the phase is zero. The gain at which this circuit's poles reach the imaginary axis is 3.000000000000, bisected on the netlist — an equality, not a range. A gain three per cent high multiplies the envelope by 1.0987 every cycle and reaches the rails in 61 milliseconds; three per cent low divides it by the same factor. A one per cent resistor cannot hold the condition, and neither can any other component.

 A 1:1 transformer at k = 0.99, and the band it is a turns ratio over. computed by solving, not by drawing. Two 10 mH windings coupled at 0.99, driven from 50 Ω into 50 Ω, with 0.5 Ω of winding resistance and 100 pF across the secondary. The response is flat at 0.4901 — which is 98.02% of the 0.5000 an ideal transformer of this ratio would give, and that shortfall is the coupling itself: the flat part is k times the turns ratio, times what the two winding resistances leave of the loop, to four figures at every k on the slider — between 400 Hz and 81.3 kHz, which is 2.31 decades. Both edges are bisected on the solved network. Below the first, the magnetising inductance is a short across the source; above the second, the leakage inductance is in series with the load. The slider moves the coupling, and it moves the upper edge only. Two windings, and the band between them

The band a turns ratio holds over

Most circuit models are right below a number or above one. A transformer is wrong at both ends and right in the middle, and the flat part is not the turns ratio either — measured on the solve it is the turns ratio times the coupling, times what the two winding resistances leave of the whole loop — and the first version of that last factor was a coincidence that held for every coupling and broke at a different load.

 Started from a millivolt at a gain of 3.20. The output grows by 1.8804 a cycle — the factor the poles give — and then stops, at 858 mV of amplitude and 1.58 kHz. The lower panel is the envelope on a logarithmic axis, where the linear model's prediction is the straight line that keeps going. What ends it is the diode pair across the feedback resistor, and no direct-current analysis of this circuit returns that number. Circuits that do a job, and the range they do it over

The amplitude nothing linear predicts

A linear model's poles say the envelope grows by 1.88 a cycle and never say when it stops. Two diodes across the feedback resistor stop it, at 858 millivolts — a number no pole, no transfer function and no bias point contains. Getting it needs a netlist with a nonlinearity in it, marched forward in time, and that pairing is what the whole field is built on.

 One loop, two measurements of the same margin. The loop gain crosses unity at 5.73 kHz with 34.9° of phase left. The closed-loop step overshoots by 35.2%, which the second-order relation says corresponds to 34.8°. They differ by 0.1°, and the difference is the third pole. Feedback, and the margin

Two measurements of one margin

A phase margin is computed from the loop gain in the frequency domain, without ever looking at a step. An overshoot is measured from the closed-loop step response in the time domain, without ever looking at a Bode plot. Inverting the standard relation on the second returns 34.9° against the first's 34.9°, and the residue is the third pole.

 What each family costs, at order 5. Measured on the solved networks. The Chebyshev is 36 dB further down at three times the corner than the Bessel, and pays for it in delay: its group delay varies 49.0% across the passband against the Bessel's 0.06%. Filters, measured not tabulated

What a steep skirt costs

A filter's order buys attenuation at a known rate — twenty decibels per decade per pole, and no arrangement of components changes it. What varies between families is how quickly the slope is reached, and the currency it is paid for in is delay: the steepest of the three distorts delay eight hundred times more than the gentlest.

 What coupling buys: the upper edge only. computed by solving, not by drawing. Six couplings from 0.8 to 0.999, each transformer solved and both its edges bisected. The lower edge moves by 1.083× across the whole range — it is set by the magnetising inductance against the source and the reflected load, and the coupling barely enters it. The upper edge moves by 168×, from 4.84 kHz to 814 kHz, because it is set by the leakage — which is what the coupling is. Winding a better transformer widens the band at the top and does nothing at the bottom, where the answer is more inductance or a smaller load. Two windings, and the band between them

What coupling buys, and where it does not

Winding a transformer better is winding it more tightly coupled, and the coupling coefficient is the number a maker works on. Measured across six designs from k = 0.8 to k = 0.999, it moves the upper band edge by 168 times and the lower one by 1.083 — so every hour spent on the winding buys bandwidth at one end of the band and, to within eight per cent, nothing at all at the other.

 Group delay across the passband, at order 5. The Bessel filter's delay varies 0.1% below 0.8 of the corner; the Chebyshev's peaks near the band edge and is several times its low-frequency value. Every family here has the same half-power frequency, so this is a difference in behaviour rather than in scaling. Filters, measured not tabulated

Flat magnitude, unflat delay

A filter that passes every frequency in its band at the right amplitude and the wrong time has not passed the signal. Group delay is the measurement that says so, it is absent from the classical comparison, and it varies by fifty per cent across the passband of the two families everybody uses.

 The sample rate each anti-alias filter demands for 80 dB. computed by solving, not by drawing. A 20 kHz passband, and each filter must be 80 dB down by the frequency that folds back into it. The required rate follows, and it is a property of the filter rather than of the converter: Bessel 3.53× Nyquist, Butterworth 2.08×, Chebyshev 1.53×. The elliptic design at a selectivity of 0.8 is refused: its equiripple stopband has a floor at -74.1 dB, which is above the requirement at every frequency, so no sample rate satisfies it. At a selectivity of 0.5 the same order needs 1.34×. The floor is the selectivity's, not the order's. Where a signal becomes a number

What the filter in front costs

The filter that keeps a converter honest is normally chosen for its skirt. Measured against one requirement — eighty decibels down by the frequency that folds back into a 20 kHz band — the choice is not a decibel or two of skirt but a factor in the clock: Bessel demands 3.53 times Nyquist, Butterworth 2.08, Chebyshev 1.53. And one design is refused outright, because an elliptic stopband is a floor rather than a slope and no sample rate reaches past a floor.

 The limit cycle's spectrum at a gain of 3.20. Ten lines of the settled waveform. The third is 5.692% of the fundamental and the fifth 0.982%; the second and fourth are 6.8e-7, which is the arithmetic's floor and not a small residue. Two diodes facing opposite ways make a symmetric characteristic and a symmetric characteristic produces no even harmonic at all. Circuits that do a job, and the range they do it over

What the limiter charges for

The diodes that set the amplitude are the only nonlinear thing in the loop, so every harmonic in the output is theirs. Across the gain slider the amplitude rises by a factor of 1.51 and the distortion by 15.4 — an amplitude that goes as the 0.11 power of the excess gain and a distortion that goes as the 0.74 power. Two diodes facing opposite ways produce no even harmonic at all, at seven parts in ten million, which is the arithmetic's floor rather than a small residue.

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

Which picture sets the upper edge

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

 The equivalent of six elements, and the heat it does not account for. computed by solving, not by drawing at 73 loads. The network reduces to 8.56032 V behind 599.768 Ω, obtained twice and agreeing to twelve figures. Into every load on the axis the two put the same voltage to 2.2e-16 of it and deliver the same power to 7.8e-16 — there is no load that can tell them apart. Inside, at a load of 5.62 kΩ, the real network is dissipating 49.168 mW and the equivalent claims 1.1349 mW, a factor of 43.3. With the port open the equivalent says nothing at all is being burned and the network is burning 48.03 mW. The slider moves the resistor bridging the two sources, which joins two ideal sources and so cannot change anything the port can reach: the equivalent stays 8.56032 V behind 599.768 Ω to the last bit at every setting, while the heat inside moves by two and a half times. Networks, and how a solve is checked

Exact outside and wrong within

Six elements reduce to one source and one resistor that no load can distinguish from them: the same voltage into every load across six decades, to the last bit of a double. The reduction is wrong about the heat by a factor of forty-three, and with nothing connected it says the network is dissipating nothing while it burns 48 milliwatts.

 3 quarter-wave sections between 50 Ω and 200 Ω. computed by solving, not by drawing. |Γ| computed by cascading exact line impedances from the load back to the source, so every multiple reflection is in it. The binomial design is exact at the centre and holds |Γ| below 0.10 over 67.9% of the centre frequency, against 17.1% for a single section. The equal-ripple design, found by minimax search rather than from a table, covers 98.1% at the same worst reflection — 45% more — and its ripples come out level to 0.0e+0%, which is the check that the search converged. Lines, where a wire has a length

Several sections, and the band they buy

A quarter-wave transformer is exact at one frequency, and a 4:1 transformation holds |Γ| under 0.1 over 17.1% of it. Several sections whose reflections cancel over a band take that to 47.7, 67.9, 82.1 and 92.6% — which is filter design with the reflection as the shaped quantity. An equal-ripple design found by minimax search buys a further 35 to 46% at the same worst reflection, and its ripples come out level to four decimals.

 How long a second-order step takes to arrive inside ±2%. computed by solving, not by drawing from the residue expansion at 260 damping ratios. The fastest is ζ = 0.780 at 3.60/ω₀; critical damping takes 5.83/ω₀, which is 62% longer. Between ζ = 0.775 and 0.780 the time falls by 33% in one step of the sweep, because which excursion is the last one outside the band changes there — the overshoot at the fastest damping is 1.99%, which is the band itself, and one step to the left it is larger. The faint curves are the other bands, each with its own step in a different place. Before the steady state

The cliff before the fastest settling

Settling time against damping is not a smooth curve with a minimum. It falls by a third in one step of a sweep of five thousandths, and the fastest damping sits on the edge of that step — so a design a hundredth of a damping ratio to the left of the optimum settles forty-eight per cent slower, with a waveform that looks no different.

 Two reasons the frequency is not 1/2πRC. The measured oscillation frequency sits below the network's own zero-phase frequency, and by two separate amounts. The amplifier's share falls as 1/ρ — the product of shift and ratio is constant to 6.7% over two decades — and the limiter's share is flat at 0.7713%. They are equal at ρ = 578, and above that a faster amplifier moves the frequency by nothing that matters. Circuits that do a job, and the range they do it over

The frequency that is not the formula

The Wien network's zero-phase frequency is one over two pi RC to every digit the arithmetic has. The circuit does not run there. With a perfect amplifier it runs 0.771 per cent low, because the limiter's harmonics are part of the waveform whose period is being measured; with a real one it runs lower still, by an amount inversely proportional to the gain-bandwidth product. The two are equal at a ratio of 578, and above that a faster amplifier buys nothing.

 How much faster an instrument must be for 10% of inflation. computed by solving, not by drawing. The quadrature rule answers 2.182× and gives the same answer for every instrument, because it contains no instrument. Measured on the solved network, a one-pole front end needs 2.79×, two poles need 3.97×, three need 4.87× and four need 5.62× — between 28% and 215% more than the rule asks for. The rule errs optimistic at every pole count, which is the wrong direction. Measurement, which is a circuit on a circuit

The instrument's own rise time

Rise times add in quadrature, so ten per cent of inflation needs an instrument 2.18 times faster than the edge. That constant contains no instrument. Measured on the solved network it is 2.79 for a one-pole front end, 3.97 for two, 4.87 for three and 5.62 for four — the rule is optimistic at every pole count, which is the wrong direction for a rule of thumb to err in.

 What a 0.7 V constant costs, in the quantity it is used to predict. computed by solving, not by drawing by Newton's method on the exponential at 94 supplies through four resistors. The model is exact at 5.748 mA — the current at which the true drop is 0.7 V — and every curve crosses zero there, at four different supplies. Below it the model is low and above it high, and how much depends on the headroom rather than on the diode. Through the 87 Ω curve the drop is 49 mV out at 0.725 V and 147 mV out at 150.7 V — a factor of 3.0 — while the error in the current falls from -66% to 0.10%, a factor of 674, because the headroom underneath it has grown by 2017. On that curve the model is inside one per cent only above 1.12 V. Where the models stop

The one current a constant is right at

Seven-tenths of a volt is the true forward drop at 5.748 milliamperes and at no other current, and every circuit built on it crosses zero error there — four different resistors at four different supplies, all exact at the same current. What decides whether the model is any good is not the diode at all; it is how much of the supply the diode is taking.

 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.

 How much of the amplifier reaches the answer, at 35° of phase margin. computed by solving, not by drawing by multiplying the forward path's gain by 1 + 10⁻⁵ and solving again. At low frequency 0.0999% of a fractional change in the device reaches the closed-loop gain, which is one over the loop gain of 995.04. It changes sign at 258 Hz — above there a better amplifier gives a smaller gain — and reaches 1.61 at 8.08 kHz, which is worse than having no feedback, and settles at exactly 1 above crossover (5.73 kHz), where there is no loop gain left to spend. The second route, the real part of 1/(1 + T) from the cut loop, agrees to 9.6e-6. Feedback, and the margin

How much of the amplifier gets through

The reason to build an amplifier from a bad amplifier and two resistors is that a change in the device barely reaches the answer — a tenth of a per cent of it at low frequency, measured by making the device better and solving again. Near crossover the same fraction rises above one, so feedback makes the gain more device-dependent than no feedback at all, and below fifty degrees of margin it changes sign on the way.

 The coupling coefficient, from two measurements that do not know it. computed by solving, not by drawing. Two windings in series, connected one way and then the other, each solved as a netlist and its inductance read out of the impedance. The two differ by four times the mutual inductance, so k comes out of the difference and the geometry never enters. The recovered value matches the one stamped into the coupling to 2.4e-15 at eight couplings from 0.1 to 0.99 — which is the second route the new element needed, since neither current law nor the energy balance can see a mutual inductance at all. Their sum stays at L₁ + L₂ throughout, which is the check that the two measurements are of one pair. Two windings, and the band between them

One number from two measurements

Neither of the two standing checks on a solved circuit can see a mutual inductance. A coupling adds no current anywhere, so Kirchhoff's law is unmoved by it; a coupled pair dissipates nothing, so the energy balance is unmoved too. A coupling stamped into the wrong row would produce a well-formed solution to a different circuit and both checks would pass — so the field needed a third route, and the one it uses is the bench method: connect the windings in series one way, then the other, and the difference is four times the mutual inductance.

 6.02 bits + 1.76 dB, measured — and the overload that is blamed on it. computed by solving, not by drawing. A coherent sinusoid is quantised and the ratio is read off the error sequence, never from the formula. One step inside full scale the two agree to 0.15 dB at every bit count from 8 to 16. Driven to exactly full scale the same converter measures 0.83 dB worse at 8 bits when driven to exactly full scale, and the reason is in the count beside it: 327 of 8192 samples hit the top code. A mid-tread converter's largest code is one step below full scale, so a sinusoid that reaches full scale is already overloading — half a step of overload, at the peak, where the error correlates with the signal. The loss halves with every bit for the same reason the clipped count does — 0.83, 0.47, 0.21, 0.06, -0.13 dB — and by sixteen bits it is inside the measurement's own scatter. The formula was never the approximation. Where a signal becomes a number

Six decibels a bit, and the half step blamed on it

Every converter data sheet quotes 6.02N + 1.76 dB and every bench measurement comes up short of it, which is usually explained by calling the formula an approximation. Measured one step inside full scale it agrees to 0.15 dB at every bit count from eight to sixteen. Driven to exactly full scale the same converter loses 0.83 dB at eight bits — because 327 of 8192 samples hit the top code, and a mid-tread converter's largest code is one step below full scale.

 Averaging a white sequence, and averaging a pink one. computed by solving, not by drawing. Both sequences are the same seeded white stream, one of them put through the 1/f network. Averaged in non-overlapping blocks, the white one's spread falls as n to the -0.510 ± 0.006 across five seeds — the √N law — and the pink one's as n to the -0.087 ± 0.013, which is very nearly not at all. A thousand-sample average buys a factor of 36.9 on the first and 2.1 on the second. The floor, which bounds from below

The corner where averaging stops working

Average N samples and the noise falls by √N. That is a statement about independent samples, and flicker noise's samples are not independent — its correlation extends over every time scale, which is what a spectrum with no bottom means. Measured on the same seeded stream filtered and not: the white sequence falls as the −0.510 power of the block length and the pink one as the −0.087 power, so a thousand-sample average buys a factor of 36.9 on one and 2.1 on the other.

 1000 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 15.69 V with 1.331 V of ripple, against the 1.569 V the expression I/2fC gives — 15.1% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 28.8° of each half cycle and carries 2.10 A at the peak, which is 13.4 times the 157 mA the load draws. Circuits that do a job, and the range they do it over

The direct voltage that is a sawtooth

A rectifier and a reservoir capacitor make what everybody calls a direct voltage. Marched with the diodes in the netlist, a thousand microfarads across a hundred ohms gives 15.69 volts with 1.33 volts of ripple on it, against the 1.57 the textbook expression predicts. The expression is high by the fraction of the cycle the diode conducts for — measured at 0.94 to 0.96 of it across two sweeps — and it has no opinion at all about the quantity that actually sizes the transformer, which is a peak diode current 13.4 times the current the load draws.

 What a charge through 1 kΩ costs, against how long it is given. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. A step loses 1.00000 of ½CV² — 12.5 µJ here — and that number is the same through 10 Ω and through 100 kΩ, which is what the faint curves show. What the resistance decides is where the fall starts: the loss goes as 2τ/T once the ramp is long against τ = 1 ms, with a fitted exponent of -0.994 and no floor beneath it. The circles are marched; the line through them is a closed form that never sees a netlist. Before the steady state

The half that never arrives

Charging a capacitor from a step loses exactly as much energy as it stores, and the resistance it is lost in does not appear in the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms, to nine figures. Drive the same network with a ramp instead and the loss falls as two time constants over the ramp, with no floor beneath it at all.

 The load that takes the most power, and the load that wastes the least. computed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, which is the magnitude of the source impedance to six figures, and the efficiency there is 0.500000 — the source dissipates as much as the load receives. Ninety per cent efficiency needs a ratio of 9.7 and delivers 34% of the available power. Power, and the part that does no work

The load that takes the most

A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.

 The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres. Lines, where a wire has a length

Where the current comes back

The return current under a track spreads out at low frequency and runs directly beneath it at high. Estimated from two paths chosen in advance, the change is a corner at 106 kilohertz for any track two hundred micrometres above a half-milliohm plane, with neither the length of the track nor its width in it. Solved, it is a band three decades wide rather than a corner, and the width enters it after all — but the length is still absent and the stack-up still sets the top, which is the part of the argument a designer needs.

 A five-volt regulator's output impedance, with 1.00 Ω of series resistance. 0.430 mΩ at direct current, 1.95 Ω at 10.0 kHz — a factor of 4.52e+3 — and it has already doubled by 4.81 Hz. The upper curve is the same circuit with its loop opened, and the ratio between them is the loop gain. A regulator is a voltage source below a frequency and the datasheet's milliohms are the value at the bottom of it. Circuits that do a job, and the range they do it over

A source below a frequency

A five-volt regulator's output impedance is 0.43 milliohms, which is the number a datasheet quotes. It has doubled by 4.8 hertz, is ten times worse by 27, and reaches 1.95 ohms at ten kilohertz — four and a half thousand times its own specification, and higher there than the same circuit with its feedback loop cut. Two routes to that curve, sharing only the netlist, agree to a part in ten to the thirteenth.

 A sum that is exact, and the bandwidth estimate that is not. computed by solving, not by drawing, at 28 spreads of the three capacitor values in a resistor chain. The sum of the open-circuit time constants — each capacitor's own value times the resistance seen at its terminals with the other two removed — is 600.00 µs here, and it equals the ratio of the first two coefficients of the denominator to 2.0e-9 and the sum of the negated reciprocal poles to 2.0e-9. That much is a theorem. What is an estimate is the bandwidth: one over 2πΣτ gives 265.3 Hz against a measured 309.2 Hz, low by 14.2%. It is low at every spread on the axis — the estimate is never optimistic — and comes within ten per cent only once one of the three time constants is 7.48 times the others. Before the steady state

A sum that is exact, and the estimate that is not

Add each capacitor's value times the resistance seen at its own terminals with the others removed, and the total is the ratio of the first two coefficients of the denominator polynomial — a theorem, holding to a part in a billion at every spread tested. Divide one by two pi times it and you have a bandwidth estimate that is 14 per cent low with three equal capacitors and never once optimistic. Two settings of the slider have the same three time constants and bandwidths two per cent apart, which is why the sum can never be more than an estimate.

 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.

 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

The edges that are lengths

Almost every boundary in this collection is a frequency or an amplitude, and both of those are things a circuit designer chooses. A handful are lengths — the 0.60 millimetres a gap's field reaches into a window, the 200 microns between a track and its plane, the 10 centimetres at which Kirchhoff's laws are a degree out — and they behave differently in one way that matters: nobody chooses them at the schematic, they are set by whoever builds the thing, and they appear in no netlist at all.

 Two floors on one axis, and the 50.0 mV between them. computed by solving, not by drawing. The flat line is the Johnson current noise of 1 kΩ, √(4kT/R), which has no current in it. The rising line is shot noise, √(2qI), which has no resistance in it. They cross at 50 µA — and the direct voltage across the resistance there is 49.981 mV, which is 2kT/q and contains neither quantity. The slider moves the resistance over six decades; the crossing moves with it and the voltage at the crossing does not move at all, to the last bit of a double. The floor, which bounds from below

The floor a current sets

A resistor's noise contains no current and a current's noise contains no resistance, and the two are equal when the direct voltage across the thing carrying the current is 2kT/q — 50.0 millivolts at 290 kelvin, whatever the resistance and whatever the current. It is the only boundary in this collection whose axis is a direct voltage across an element. And a forward-biased junction, which has the same dynamic resistance as some resistor, produces exactly half its noise power at every current.

 What the summing junction of an inverting amplifier actually is, at 1.00 MHz of gain–bandwidth. computed by solving, not by drawing by driving a current into the node and reading the voltage. It is 100 mΩ at direct current, rises 1.000 decades per decade of frequency, and settles at 909.5 Ω — which is the 1 kΩ and 10 kΩ in parallel, with the amplifier contributing nothing. It passes one per cent of the input resistor at 995 Hz, a factor of 1,005 below the gain–bandwidth. The second route — the open-loop impedance over one plus the return ratio from the cut loop — agrees to 0.045%. Feedback, and the margin

The node that is at ground for a while

An inverting amplifier's summing junction is held at ground by the loop, so it is at ground exactly as well as the loop is strong. Driven with a current and measured, it is a tenth of an ohm at direct current, ten ohms at a kilohertz, and 909 ohms above a megahertz — which is the two feedback resistors in parallel, with the amplifier contributing nothing.

 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. Frequency, which is the same solve

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 coupled pair as a two-port, at k = 0.8. computed by solving, not by drawing. Each port driven in turn with the other open, four solves, and the four impedance parameters read out. The diagonal terms measure each winding's own inductance — 10.0000 mH and 40.0000 mH against 10 and 40 — and both transfer terms measure the mutual inductance, 16.0000 mH against k√(L₁L₂) = 16.0000. The two transfer terms agree to 2.83e-16, which is reciprocity — a property of the device rather than of the measurement, and the first thing a coupling stamped into the wrong row would break. Neither of this site's two standing checks can see it: a coupling adds no current and dissipates nothing. Two windings, and the band between them

Two ports from two one-ports

An inductor is a one-port: one impedance, one number. Two of them coupled is a two-port, and the four impedance parameters that describe it are recovered here by four solves — each port driven with the other open, the definition read literally. Two of the four come out equal to 2.8 × 10⁻¹⁶, which is reciprocity, and is the first property a coupling stamped into one row instead of two would break.

 Two meters, one current, and neither of them measuring the heat. computed by solving, not by drawing, at 35 conduction angles. The first curve is an average-responding meter: it rectifies, averages and multiplies by 1.1107, which is exactly right for a sinusoid — -7.8e-5% here — and exactly 11.07% high on a square wave, because the error is the ratio of two form factors and contains neither the amplitude nor the frequency. On a rectifier drawing its 100 W in sixty degrees of conduction it is -35.90% low. The second curve is a true-RMS meter that reaches 9 harmonics, which has no shape assumption in it and a bandwidth instead: it returns the root-sum-square of the lines it can see, and is one per cent low below every angle here of conduction. The crest factor at sixty degrees is 1.733, which is inside every instrument's rating — neither meter is failing because the peak is large. One is failing because the shape is not a sinusoid and the other because the spectrum is wider than it is. Power, and the part that does no work

What a meter multiplies by

An average-responding meter rectifies, averages and multiplies by 1.1107, which makes it exactly right for a sinusoid and wrong for everything else by the ratio of two form factors — 11.07 per cent high on a square wave and 35.9 per cent low on a rectifier drawing its current in sixty degrees. It is also exactly right at one other waveform, a 145.90 degree conduction angle, which is nobody's sinusoid. Beside it a true-RMS meter that reaches nine harmonics is two per cent low on a square wave and never within one per cent of anything narrower.

 The amplitude at which a converter's floor becomes distortion. computed by solving, not by drawing. The quantisation error is transformed and the power in harmonics of the input is measured directly, at the bins those harmonics occupy. Undithered, the share rises from 0.29% at 230 levels to 76.5% at 1.3: the error has stopped being spread and has become a deterministic staircase locked to the signal. The sweep is drawn against the levels the input uses rather than against volts because it is then the same sweep for every converter — a twelve-bit part and a sixteen-bit part agree here to 0.0e+0, so what decides the character of the error is how many levels the signal crosses and not how many the converter owns. With 1 least significant bit of dither the share stays under 0.34% at every amplitude, for 3.03 dB of signal to noise. Adding noise to a converter's input improves what comes out of it, which is true and sounds like it should not be — and the amount is one whole step, not "some": a quarter of a step leaves 52.7% and a half leaves 25.6%, because a dither smaller than a step cannot make the quantiser cross one. Where a signal becomes a number

When a floor stops being a floor

Quantisation error is treated as noise and behaves like noise while the input crosses many levels. As the amplitude falls it stops: measured at the bins its harmonics occupy, the share of the error's power sitting in harmonics of the input rises from 0.29% at 230 levels to 76% at 1.3. The floor has not moved. It has become a distortion product locked to the signal, and the fix is to add noise on purpose — one whole least significant bit, and no more.

 One bit, oversampled — and where the quantisation noise went. computed by solving, not by drawing. A first-order modulator is marched forward one sample at a time with a one-bit quantiser inside the loop, and the noise inside the band is read out of the transform of the error. It falls by 8.99 dB for every doubling of the oversampling ratio — measured 9.33, 10.12, 6.96, 9.54 — against 3.01 dB for plain oversampling, which is drawn beside it from the same starting point. The loop does not make less noise; it moves the noise out of the band, and at a ratio of 128 one bit is worth 9.18. Where a signal becomes a number

One bit, and where the noise went

Sampling faster spreads a fixed quantity of quantisation noise over a wider band, so the part inside the band of interest falls by 3.01 dB for every doubling — half a bit. Putting the quantiser inside a loop with an integrator does something different in kind: measured on a modulator marched forward one sample at a time, with its test tone inside the band the ratio is quoted over, the in-band noise falls by 8.99 dB per doubling. At an oversampling ratio of 128, one bit is worth 9.18 — and the octaves scatter by a decibel each, which is the loop telling the truth about what its error is made of.

 One diode curve, eight one-decade fits, and eight different ideality factors. computed by solving, not by drawing. A junction with two conduction mechanisms — recombination near n = 2 at low current, diffusion near n = 1 above it — and a series resistance, which is what a real diode is. Fitting ln(i) against v over each decade in turn returns an ideality factor for each, and they run from 1.227 to 1.984 without being monotonic: the factor rises through the recombination region, falls through the diffusion region, and rises again where the series resistance takes over. Two of the windows are straight to a few parts in a thousand, so the residual gives no warning. The bars are what each fit predicts for the forward voltage at 1 mA: the worst is out by -186 millivolts, which is a current 0.01 times the truth. Where the models stop

The constant that is a window

A diode's ideality factor is quoted as a number and defined as a derivative, which means it has a value at every current and no value anywhere. Eight one-decade fits to one curve return factors from 1.23 to 1.98, two of them straight to a few parts in a thousand — so the residual gives no warning at all. Asked for the forward voltage at a milliamp, the window containing it is right to a third of a millivolt and the worst is out by 186, which is a current a hundredth of the truth.

 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 gapped core: where the inductance goes, and where the energy is. computed by solving, not by drawing. Reluctance in series — the gap's lg/µ₀Ae and the core's le/µ₀µᵣAe — with the inductance N²/ℛ and the share of the stored energy in each proportional to its share of the reluctance. At two hundred microns on a µᵣ = 2000 core the inductance has fallen from 41.89 mH to 5.366, and 87.2% of the energy is in the gap — which is air. The share is lg/(lg + le/µᵣ), so it contains neither the turns nor the area and what decides it is µᵣ·lg against the path length; the slider shows the same gap holding 40% at µᵣ = 200 and 98% at 15,000. That is the reason a gap is a design parameter: it is the part of the magnetic circuit whose properties do not drift, do not saturate and do not depend on temperature. Two windings, and the band between them

The energy is in the gap

A ferrite core is chosen for its permeability and then deliberately cut, and the cut is not a compromise. Reluctance adds in series, so a two-hundred-micron gap in a µᵣ = 2000 core holds 87.0% of the stored energy while the ferrite holds thirteen — and the share is lg/(lg + le/µᵣ), a ratio of two lengths, containing neither the turns nor the area. The material chosen for its permeability holds almost none of what the component stores.

 A 4.0:1 load reads 1.13:1 through twenty metres of cable. computed by solving, not by drawing. A 200 Ω load on a 50 Ω line — a standing-wave ratio of 4.00 at the load itself — measured from the other end of a length of cable that attenuates 0.500 dB per metre at 1 GHz. The reading falls as the length grows, exactly as |Γ| times ten to the minus twice the one-way loss over twenty, and reaches 1.5 at 9.5 m and 1.128 at twenty metres, which is 24.4 dB of return loss and would pass any acceptance test. The load is unchanged; the instrument is looking at it through 10.0 dB of attenuator. The length that hides it falls 3.16× per decade of frequency, which is the root of ten and is the attenuation's own law. Lines, where a wire has a length

The mismatch that the cable hides

A lossless line carries a reflection back unchanged, so the standing-wave ratio at the instrument is the standing-wave ratio at the load. A real line does not, and the departure is exact: ten decibels of one-way loss improves any mismatch by twenty. A four-to-one load at the end of twenty metres of ordinary coaxial cable measures 1.13 at the near end, a return loss of 24 decibels, and passes an acceptance test the load could never pass. The boundary is a loss rather than a length, which makes it a frequency: 30.2 metres at 100 megahertz, 9.5 at a gigahertz, 3.0 at ten.

 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.

 What a mistuned arm leaves at 1300 Hz. computed by solving, not by drawing. Neither curve is the null's depth — the null is still bottomless, it has simply moved — but the depth at the frequency the notch was designed for, which is the number a filter is bought for. One per cent components leave -32.9 dB if both errors go the same way and -78.8 dB if they oppose, a factor of 197 from the same tolerance on the same two parts. The slopes are 20.0 and 40.0 decibels per decade: first order in the error on the product LC, second order in the error on the impedance level. Filters, measured not tabulated

What actually fills a null

A ten per cent error in the two components of a notch's arm leaves the null three hundred decibels deep — it moves it rather than filling it. What fills it is loss, at twenty decibels per decade of arm resistance exactly. And the depth at the frequency the notch was designed for splits into two orders depending on which way the two errors go: one per cent parts leave 79 decibels one way and 33 the other.

 The voltage a winding may carry, which is a volt-second limit read at a frequency. computed by solving, not by drawing. The dots are bisections on a marched flux — the voltage integrated sample by sample until the peak excursion reaches 0.35 T — and the line is N·Ae·Bsat·2πf. They agree to 0.001% over three decades, and the fitted slope is 1.000000: exactly proportional, because flux is the integral of voltage and nothing else. The quantity that belongs to the core is the 3.500 mWb-turn, which has no frequency in it. A transformer "rated for 50 Hz" is a transformer whose volt-second product was divided by 2π × 50 once. Two windings, and the band between them

A boundary in volt-seconds

A core saturates on the integral of the voltage applied to it, not on the current through it and not at a frequency. The quantity that belongs to the core is N·Ae·Bsat — 3.500 mWb-turn here — and it has no frequency in it at all. Everything a data sheet says about a transformer's frequency rating is that one number divided by 2πf once: measured on a marched flux, the voltage a winding may carry is proportional to frequency to a fitted exponent of 1.000000.

 The best shunt drops 7.75 mV, whatever the current is. computed by solving, not by drawing. Two errors on one axis, both from solved networks: the shunt's own drop, which lowers the current that was to be measured, and the amplifier's 5.0 µV of offset divided by the voltage the shunt develops. The first rises with the burden voltage and the second falls, so the worst case has an interior minimum at 7.7460 mV — the geometric mean of the offset and the 12 V supply — where the error is 0.1291%, being twice the root of the offset over the supply. Neither the shunt's resistance nor the current appears in either number: at 1 A the answer is 7.75 mΩ, and at a hundred times the current it is the same burden voltage across a hundredth of the resistance. What does depend on the current is the 7.7 mW the shunt then dissipates, and 40 K of self-heating at 50 ppm/K is 0.2000% on its own. The third curve is the shunt's own Johnson noise in a kilohertz of measurement bandwidth, as a fraction of the current: it is 4.5e-6% at the best burden and is the only line here that moves with the current at all, falling as one over its square root — so above about an ampere it leaves the bottom of these axes entirely and is drawn nowhere rather than flattened onto the floor. Measurement, which is a circuit on a circuit

The ammeter that is a resistor

Every direct measurement of a current is a measurement of a voltage across something the current was made to flow through, so the instrument has two errors pointing opposite ways: a larger shunt changes the current, a smaller one leaves less for the amplifier's offset to be compared with. Written in the burden voltage they are the burden over the supply and the amplifier's offset over the burden, and the best of them is at the geometric mean — 7.75 millivolts on a twelve-volt rail, with a worst-case error of 0.129 per cent. Neither number contains a resistance, and neither contains the current: ten milliamps and a hundred amps want the same 7.75 millivolts.

 Linearising an exponential at 125 °C, and what it costs. The linear model understates the gain by 1% at 9.69 mV and by 10% at 30.2 mV. The thermal voltage at this temperature is 34.3 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong. Where the models stop

The edges that move with the room

Every boundary in this collection is quoted at one temperature and most of them are functions of it. The small-signal edge is proportional to the thermal voltage, so it runs from 5.67 millivolts at −40 degrees to 9.69 at +125 — a factor of 1.71, the ratio of the absolute temperatures exactly. A realised Q is 1.54 per cent high at one end of that range and 2.58 at the other. The numbers are right; the condition attached to them was left off, and it is the same condition every time.

 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. Frequency, which is the same solve

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.

 25% compensation: 4.26% regulation, and a resonance at 25.0 Hz. computed by solving, not by drawing. The current a 230 V feeder draws, swept from two hertz to a hundred and fifty, with 25% of the line's reactance cancelled by a series capacitor. The regulation falls from 5.17% to 4.26%, which is what the capacitor was fitted for. What comes with it is a series resonance at 9.59 Hz with the load connected and 24.99 Hz with the far end shorted — the latter being exactly f₀√k = 25.00 Hz, a frequency the compensation fraction chooses on its own. The line's inductance does not appear in it and neither does the voltage. It is always below the fundamental, which is where the machines are. Power, and the part that does no work

The reactance cancelled, and the resonance it buys

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

 A series-terminated net holds half a swing for 1.0 delays at 50% along it. computed by solving, not by drawing. What a receiver 50% of the way along one net sees under three terminations, from a lattice evaluated at that point rather than at the ends. The shaded strip is the interval in which a logic input has no defined answer — between 30% and 70% of the swing. The series-terminated net sits in it for 4.83 ns, which is 2(1 − x) delays exactly and is zero only at the far end; the unterminated net sits in it for 4.83 ns and then overshoots by 82%; the parallel-terminated net never does, and draws 60 mA down the line for as long as the level is held. Lines, where a wire has a length

The resistor at the wrong end

A lattice diagram is read at the two ends of a line, and that is where the two respectable terminations look identical: a clean step, one delay late, at the receiver. Anyone standing halfway along a series-terminated net sees half the swing held for a full round trip, which for a logic input is not a level at all. The interval is 2(1−x) delays exactly, it is zero only at the far end, and the scheme that never has it draws sixty milliamperes for as long as the level is held.

 The droop a zero-order hold imposes at 48 kHz. computed by solving, not by drawing. Holding each sample for a clock period convolves the output with a rectangle, so the spectrum is multiplied by a sinc: -0.143 dB down at a tenth of the sample rate, -0.912 at a quarter and -3.922 at half — which is exactly 20 log(2/π) and contains no design decision at all. The dots are the amplitude of the fundamental read out of the transform of the staircase itself, agreeing with the closed form to 0.008%. There is also half a sample of delay, 10.417 µs here, which is the reason a held reconstruction is not a droopy copy of the signal but a droopy copy that has moved. Where a signal becomes a number

The staircase on the way out

A converter does not emit impulses. It holds each sample for a whole clock period, which is a convolution with a rectangle and therefore a multiplication by a sinc — 0.14 dB down at a tenth of the sample rate, 0.91 at a quarter, and 3.92 at half, which is exactly 20 log(2/π). Nobody chose that droop and it is nearly eight times the half-decibel ripple of a Chebyshev passband. Measured on the transform of the staircase itself, it agrees with the closed form to 0.008%.

 A divider of 4 equal 1.0% resistors, solved 3000 times. computed by solving, not by drawing. Every resistor drawn from its tolerance band and the divider solved, 3000 times. The worst case is ±1.000% — the part tolerance itself, and it does not improve when the divider is built from more parts — while the measured spread is 0.2944% and the worst of 3000 draws reached 82% of the bound. The root-sum-square, offered as though it were a standard deviation, is 1.70 of one here: for uniformly distributed parts it is √3 σ, a coverage of about 92%. Networks, and how a solve is checked

The tolerance that is not on any part

Four one per cent resistors in a divider give an answer whose worst case is one per cent, whose measured spread is 0.29 per cent, and whose root-sum-square bound — offered everywhere as though it were a standard deviation — is 1.70 of one. Adding parts does not move the worst case at all and shrinks the spread as one over their root, so the gap between the promise and the fact widens with every resistor. And the same arithmetic draws a boundary in tolerance rather than in frequency: an R–2R ladder is a twelve-bit converter only while its resistors are inside 0.14 per cent.

 An order-5 ladder driven from 2× the 1 kΩ it was designed between. computed by solving, not by drawing. A doubly-terminated Butterworth ladder is a two-port designed between two stated resistances, and the resistances are part of the design rather than the environment it happens to be used in. At match it loses 6.021 dB — exactly half the voltage — and its passband has no peak anywhere in it. Driving the same five reactances from 2× that resistance moves the shape by 2.345 dB and the insertion loss to 9.542 dB. The band inside which the shape is right to 0.5 dB runs 0.8857× to 1.1371× — a window of 25 per cent on a quantity usually written down as a round number. And the two sides are not alike: at twenty times the design resistance the departure has settled at 5.33 dB, while at a twentieth of it the passband is 15.51 dB out with 7.21 dB of peaking on it, so driving a ladder from too low an impedance is worse than driving it from too high a one. Filters, measured not tabulated

The two resistors a ladder was designed between

A passive ladder filter is not a transfer function with some resistors attached; it is a two-port designed between two stated resistances, and the resistances are as much part of the design as the inductors. Driving an order-five Butterworth from anything outside 0.886 to 1.137 times its design resistance puts more than half a decibel of error on the passband — a tolerance tighter than the resistor is usually specified to — and the two sides of that window are not alike.

 Rejection is not a property of the loop. The same regulator with the same loop gain and the same 46.5° of phase margin, drawn twice. With the amplifier's output referred to ground the rail reaches the output at -67.3 dB at the bottom and 5.79 dB at 10.0 kHz — where it is amplified. Referred to the rail instead, every point is 60.00 dB lower, which is 20 log(gm·ro) for the pass device and not a design choice. Circuits that do a job, and the range they do it over

What gets through from the rail

A regulator's job is to hold its output still while its input moves, and the measurement says it does that well at ten hertz, badly at a kilohertz, and not at all at ten kilohertz — where this one puts out 1.9 times what arrives. Then the same netlist with one node moved, the same loop gain and the same 46.5 degrees of margin, rejects 60.009 decibels better at every frequency in six decades. The sixty decibels is the pass device's own intrinsic gain and it is not a design choice.

 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 4 kHz Butterworth, mapped to 48 kHz two ways. computed by solving, not by drawing. The upper panel is the digital response on the unit circle; the lower one is where each analogue frequency lands. The bilinear transform compresses the axis as it approaches half the sample rate — -3.11% at a tenth of the rate and -15.2% at a quarter — so a design mapped straight through is -3.432 dB at its own corner instead of -3.010. Pre-warping puts that one frequency back exactly and no other: above it the pre-warped curve is the further of the two from the analogue prototype. It is a choice of where to be right, not a correction. Where a signal becomes a number

The corner that moved

The bilinear transform has to fit an infinite frequency axis onto a circle, so something must be compressed, and what is compressed is everything near half the sample rate. A 4 kHz Butterworth mapped to a 48 kHz clock is 3.432 dB down at its own corner instead of 3.010, and at a quarter of the sample rate the axis is 15.2% out. Pre-warping puts one frequency back exactly and no other — it is a choice of where to be right, not a correction.

 The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance. Power, and the part that does no work

The first cycle, which no steady state contains

Every number this field computes about a rectifier — the ripple, the crest factor, the conduction angle, the power factor — is read from the settled state, and the march that produces them starts from an empty capacitor and throws the first cycle away. That first conduction carries 32.4 amperes against a repetitive peak of 1.23, it is 26 times larger than anything the circuit ever does again, and how large it is depends on when somebody's hand closed the switch.

 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.

 A first stage of 100 buys 40.0 dB of rejection, and gives it back above 10.0 kHz. computed by solving, not by drawing. The rejection of a three-amplifier instrumentation amplifier against frequency, beside the one-amplifier difference stage it is built around. At low frequency the two differ by 39.99 dB against 20 log 100 = 40.00 dB, and the reason is that the input stage passes a common-mode voltage at exactly unity: the common-mode gain of the whole instrument is 1.998 mV/V, which is the difference stage's own. So the four resistors around the last amplifier decide the rejection and the two that set the gain do not — ten per cent between them moves it by less than a hundredth of a decibel. What ends it is bandwidth: above 10.0 kHz, which is the amplifier's gain–bandwidth divided by the gain that bought the rejection, the differential gain falls and the rejection falls with it at twenty decibels a decade. Measurement, which is a circuit on a circuit

The four resistors that decide, and the two that do not

A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.

 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. Frequency, which is the same solve

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.

 A threshold crossed once, in 20k samples of noise. With no hysteresis the comparator changes its mind 22.7 times on average and as many as 29, on a signal that crosses the threshold once. The vertical bars are the range over twelve seeds. The expected number of extra transitions falls below a tenth at 3.57 standard deviations — so the hysteresis a threshold needs is set by the noise under it and not by the signal over it. Circuits that do a job, and the range they do it over

Two thresholds because there is a floor

A comparator with one threshold, watching a slow signal cross it once, changes its mind 22.7 times on average and as many as 29 — because there is noise under the signal and no threshold is ever crossed once. Hysteresis fixes it, and how much is needed is a multiple of the noise's own standard deviation rather than a voltage: 3.57 of them here. The multiple grows with how long the threshold is watched, and slowly — a hundredfold longer record needs three times the hysteresis, not a hundred times.

 An exponential with 6× of degeneration: both edges move, and not together. computed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 36.6 mV and the gain falls one per cent short of its small-signal value at 85.0 mV. An emitter resistor dividing the gain by 6 moves the distortion edge by 35.4 times — the square of the factor, because the resistor both divides the drive reaching the junction and linearises what the junction does with it — while the gain edge moves by only 11.6 times. So the two edges close up: 2.32 times apart here against 7.06 bare, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known. Devices, and the amplitude they stop being linear at

What a resistor in the emitter buys

Degeneration is described as a trade: give up gain, get linearity. Measured on the transfer curve, the two sides of that trade are not the same size. Dividing the gain by six moves the amplitude at which distortion reaches one per cent by thirty-five times — the square of the factor — and the amplitude at which the gain is one per cent out by only twelve. So the two edges close up, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known.

 A difference quotient is best at a step of 1e-5, and is 1e+6 times worse at 10⁻¹¹. computed by solving, not by drawing. The worst disagreement between the adjoint network's derivatives and a central difference quotient of the same quantities, against the fractional step the quotient is taken with, on a 7-element LC ladder filter at 1000 Hz. The curve has a minimum because two errors pull opposite ways: the curvature the quotient neglects falls as the square of the step, and the digits its subtraction destroys rise as one over the step. The best it reaches is 1.4e-9, against the 2.3e-11 that the two-thirds power of the machine epsilon predicts. The exact route costs 2 solves against 15, and has neither error term. Networks, and how a solve is checked

Every derivative, and the one that is zero

How much does this response move if that capacitor is one per cent out? A difference quotient answers it one component at a time, in two solves each, and its best possible accuracy is four parts in a hundred million. Transposing the matrix and solving once more answers it for every component at once, exactly. Pointed at a claim the essays on ladder filters have made from the start and never tested directly — that a doubly terminated ladder filter's response is stationary in every element at its passband maxima — it returns two parts in ten billion, where the cascade realising the identical response returns 0.72.

 An order-6 cascade at 10 kΩ: a band 10× wide. computed by solving, not by drawing. A 10.0 kHz unity-gain Butterworth of order 6, 3 Sallen-Key sections in cascade, realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 4.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 316 Ω to 3.16 kΩ, with the least departure of 0.0763 dB at 1000 Ω; at this setting it is 0.182 dB at 12.9 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band. Filters, measured not tabulated

The band that closes with the order

One Sallen-Key section is inside a tenth of a decibel of its own design over three decades of impedance level, bounded below by fifty ohms of amplifier output resistance and above by two picofarads of stray. Give it three more sections and the band is one decade; give it four and there is no impedance level at all that meets a tenth of a decibel. Every section brings three more nodes each carrying their own stray and one more amplifier carrying its own output resistance, so the floor rises with the order until it crosses the tolerance — a boundary in the order rather than in the impedance.

 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. Frequency, which is the same solve

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.

 Below 910 kHz a trace is a diffusion, not a line — and its velocity goes as √f. computed by solving, not by drawing. The phase velocity of an ordinary FR-4 trace against frequency, computed from γ = √((R + jωL)(G + jωC)) with a series resistance that rises as √f above its skin-effect corner and a shunt conductance proportional to frequency. Above 910 kHz the velocity is 0.4767c and does not move, which is the number every other essay in this field uses. Below it the series resistance dominates the reactance, the line is a diffusion, and the velocity falls as the square root of frequency — measured at the 0.467 power. The characteristic impedance is not a constant down there either: 1508 Ω at a kilohertz against 50.0 Ω at ten gigahertz. Lines, where a wire has a length

The delay that is not one number

Nine essays in this field quote a delay: a length divided by a velocity, the same for every frequency, and the edge that comes out is the edge that went in. A real trace has a series resistance, and below the frequency where the reactance overtakes it — 910 kilohertz for ordinary copper — the line is a diffusion rather than a wave, with a velocity proportional to √f. What survives is that the arrival is still exactly linear in the length. What does not is the rise time, which grows as the square of it.

 The inductance divides the current by 7 and leaves the capacitor 29% above the mains peak. computed by solving, not by drawing. The first conduction of a rectifier whose transformer has a leakage inductance as well as a winding resistance, marched from an empty capacitor at the worst instant of the mains. The peak falls from 36.4 A at 20 µH to 5.3 A at 5000, and — unlike the winding resistance measured on the first cycle alone, which limits the current and leaves ∫i²dt exactly where it was — the inductance takes the energy down with it, from 0.370 to 0.073 A²s. What it costs is the second curve: the inductor's current cannot stop at the instant the two voltages are equal, so the capacitor overshoots to 21.86 V at 1000 µH — 28.6% above the 17 V peak of its own supply — and the diodes will not let the charge back out. The overshoot has an interior maximum, because past it the mains reverses before the ring has finished. Power, and the part that does no work

The inductance that limits, and lifts

Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

 The second path costs nothing at 12 pF and an order at 100 pF. computed by solving, not by drawing. Settling time to 0.01% of final value, marched on the closed loop, against the value of the second feedback path's capacitor — with the phase margin of the same circuit divided by ten drawn on the same axis so the two can be compared. The direct-current error the previous rung recorded as the isolation resistor's cost, 0.99% into 1 kΩ, falls to 1.20e-4% with the second path in. What the second path costs instead is a range: at 12 pF the circuit settles in 0.745 µs against the isolation resistor's 1.419 µs — faster than the thing it repairs — and at 100 pF it takes 9.18 µs, 12 times longer, at a phase margin of 47.9° that reports nothing whatever about it. What it is settling by there is one exponential of time constant 0.99 µs, which is the feedback network's own RC and contains no amplifier. Feedback, and the margin

The path that buys the error back

An isolation resistor restores a capacitively loaded amplifier's phase margin and costs it the thing feedback was for: the loop stops regulating the node the load is on, and a kilohm of load pulls the output down by a per cent. The standard repair is a second feedback path, and its cost is not an error or a margin — it is a range. At twelve picofarads it settles to a hundredth of a per cent in 0.745 microseconds, faster than the circuit it repairs; at a hundred it takes 9.18, and the phase margin there is better.

 A comparator's delay, at a divider ratio of 0.50. Each nanosecond of comparator delay adds 2.650 nanoseconds to the period — measured as a slope between two delays, both exact multiples of the marching step — against the 2.667 that 4/(1+β) gives and the 2 that counting the delay twice gives. A 50 ns comparator therefore holds the frequency to one per cent only below 75.0 kHz. Circuits that do a job, and the range they do it over

The period a delay lengthens

Feed a comparator's output back through a resistor to the capacitor on its own input and the two thresholds stop defending a decision and start setting a period. The closed form is two RC times the log of one plus beta over one minus beta, and a marched circuit recovers it as the step shortens. A comparator that responds fifty nanoseconds late does not add fifty nanoseconds to each half cycle: it adds 4/(1+beta) times the delay to the period, 2.65 here against the 2 that counting it twice gives, because during the delay the capacitor keeps going the way it was going.

 A 0.5 mm conductor's resistance against frequency, exact and asymptotic. computed by solving, not by drawing. The exact ratio is computed from the Kelvin functions by their series; the dashed curve is the asymptote everybody quotes, which treats the current as flowing in one skin depth of the rim and is drawn only where that annulus is inside the wire. At 17.4 kHz, where the skin depth equals the radius and the rule of thumb says the effect "starts", the asymptote says 1.0000 — no effect at all — and the exact answer is already 1.0208. The rule of thumb names a frequency the effect has passed, which is the same shape as the tenth-of-a-wavelength criterion marking a point at which the lumped model is already 30% wrong. Two decades above, the two agree to 0.00%, which is what makes it an asymptote rather than a formula. Two windings, and the band between them

The resistance that grows with frequency

The rule of thumb names the frequency at which the skin depth equals the conductor's radius as the point where the effect begins. Computed exactly from the Kelvin functions, the resistance is already 2.05% up there — and the rule's own asymptote says 1.0000, no effect at all. Two decades higher the two agree to 0.01%, which is what makes it an asymptote rather than a formula, and what makes the frequency it names the wrong one to design at.

 A 8th-order Butterworth at 12 bits, as a cascade and as one polynomial. computed by solving, not by drawing. The open circles are the poles as designed, on the z-plane with the unit circle drawn. Filled marks are where they go once the coefficients are stored in 12 bits. A cascade of biquads keeps two coefficients per pole pair, so a rounding error moves that pair and nothing else: 8.43e-3, largest radius 0.97478. A direct form keeps one denominator whose coefficients are symmetric functions of every pole, so one rounding error moves all of them: 6.21e-1, largest radius 1.54393 — outside the unit circle, which is not an inaccurate filter but an unstable one. The two are the same filter until they are written down. Where a signal becomes a number

The same filter, rounded twice

The filters field measured two realisations of one analogue response and found the passband error growing as the 0.99 power of a component tolerance in a cascade and the 2.00 power in a ladder. The digital version of that argument comes out harder. At eighth order and sixteen bits, a cascade of biquads moves its poles by 6.6 × 10⁻⁴ and stays at a radius of 0.9748; the same filter written as one polynomial moves its poles to a radius of 1.4536, which is not an inaccurate filter but an unstable one.

 Subtracting removes kT/C entirely and doubles the amplifier — worth 31× at a megahertz and a loss above 60 MHz. computed by solving, not by drawing. The noise on one sample of a switched-capacitor stage, and on the difference of two samples taken a settled interval apart, against clock frequency. The reset level is the same number in both samples and cancels exactly; the amplifier's own noise is two independent samples and its variance doubles, measured at 2.000 against the 2 the correlation predicts. At a megahertz that is 63.8 µV down to 11.53 — 31 times in power. The two curves cross at 60.2 MHz, which is where the amplifier's own noise equals kT/C, and above it the subtraction costs more than it removes. The floor, which bounds from below

The sample that is subtracted

Three earlier measurements found floors that no gain moves and no filter reaches, because both arrive as numbers already sampled. One of them can be subtracted: the reset level a capacitor holds is the same number in two consecutive samples and cancels exactly. What it costs is that the amplifier's own noise is not — two samples of it are independent, so its variance doubles. That is thirty times better at a megahertz, a loss above sixty, and the crossing is the one an earlier essay computed for a different question.

 Two of the three are one mechanism at 19.0°; the third arrives at ninety. computed by solving, not by drawing. The cosine between each pair of error waveforms at 30.0 kHz, against how much output the ideal model is asked for. Bandwidth and slewing sit at 0.9454 — 19.0 degrees — and close only slowly, reaching 0.8813 at 16 V. Clipping does not exist below 12.00 V, where its error is 4.0e-9 per cent of the signal and its direction is the direction of rounding; above it the mechanism is real — 18.1 per cent at the top of the sweep — and its cosine against both of the others stays under 0.0025. The bandwidth error is 0.615 per cent at every amplitude here, unchanged to 5.3e-15, because a linear stage's fractional error has no amplitude in it. Where the models stop

Where the mechanisms are one mechanism

An amplifier is said to run out of three separate things — bandwidth, slew rate and rails — and errors from separate mechanisms add in quadrature while errors from one mechanism add as magnitudes. Measured as waveforms rather than as numbers, two of the three sit 18.4349 degrees apart, which is exactly the angle between a sinusoid and its own cube, and the third sits at ninety: its cosine against both of the others stays under 0.0025 wherever it exists. So the arithmetic is neither of the two anybody reaches for, and a budget built the right way is within 2.9 per cent where quadrature is 17 per cent low and a straight sum 37 per cent high.

 A switched capacitor is a resistor below a ratio, not below a frequency. computed by solving, not by drawing. The exact response of a capacitor shuttled between the input and a holding capacitor at 1.00 MHz — a difference equation with one pole, evaluated on the unit circle — against the continuous R–C its equivalent resistance is supposed to make. The corner is 3.15 kHz against the model's 3.18 kHz, 0.99% out, and the discrepancy is set by the capacitor ratio alone: one per cent needs a ratio under 0.0201, which is a clock 315 times the corner. The second difference has no counterpart at all — the sampled response repeats at the clock, so the image rising on the right of this plot is signal at 997 kHz arriving as though it were at the corner. The third is settling: 1 pF charged through 1 kΩ gets 500.0 time constants a half period at this clock, and being short of full charge raises the equivalent resistance by 0.000%, which puts a ceiling on the clock at 108 MHz. Filters, measured not tabulated

A resistor made of a clock

A capacitor shuttled between two nodes at a megahertz behaves as a megohm, and a tenth of a picofarad shuttled at ten kilohertz behaves as a gigohm — which is how a filter with a one-hertz corner fits on a chip. What the equivalence costs is three conditions, and they bind on three different quantities: a capacitor ratio under 0.0201, a signal below half the clock, and a clock below the frequency at which the charge stops arriving.

 A 350 Ω bridge, with ideal leads. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 0.498% at 1.0% — so the tangent is worth one per cent only up to 2.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 4.040%. Circuits that do a job, and the range they do it over

The bridge that is linear near one point

The expression for a Wheatstone bridge's output is V delta over four, and the solve says it is V delta over two times two plus delta — low by half the fractional change, exactly, at every change tested. So the tangent is worth one per cent only up to a two per cent change and a tenth of a per cent only up to two tenths. Driving the bridge from a current source instead halves the departure at every point and doubles both edges, which is a change of one component and no change at all to the four resistors being measured.

 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. Frequency, which is the same solve

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

The inductor that is a capacitor

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

 The same three stages, in two orders. computed by solving, not by drawing by Friis's cascade. With the low-noise amplifier first the chain's noise figure is 3.31 dB; with the mixer first it is 12.01 dB. The gain is identical either way — 48.0 dB — so the 8.70 dB is bought with nothing but an ordering. Everything after the first stage contributes 26.0% of the total. The 2.0 dB of cable in front has gain below one, so it multiplies every later stage's contribution rather than dividing it. The floor, which bounds from below

The loss in front, counted twice

A decibel of cable before an amplifier costs a decibel of signal, which everybody expects, and a decibel of noise figure, which is a separate decibel arriving from a separate place. Measured across the slider it is exact: 1.31 dB of chain noise figure becomes 2.31 with one decibel of cable and 9.31 with eight, every time, while the 8.70 dB that stage ordering is worth does not move at all. A lossless reactive network in the same position costs nothing, because only the real part of an impedance is warm.

 A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees. Power, and the part that does no work

The pair that is worse than either

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

 Matched parts cost nothing; a 2 dB difference between them sets a 112 dB ceiling. computed by solving, not by drawing. The common-mode rejection of a three-amplifier instrumentation amplifier against the gain of its input stage, with each amplifier's own rejection in the netlist as an input-referred error of the common-mode voltage over the rejection. The architecture's own figure rises decibel for decibel with the gain, because the difference stage sees a larger differential signal beside the same common-mode one. The parts' contribution does not rise with anything, and the part of it that matters is not their rejection but the difference between their rejections: two amplifiers of 98 dB that are identical cost 0.000 dB, while 100 dB against 98 dB leaves a ceiling of 111.7 dB with no gain in it. The two mechanisms cross: below a gain of 1903 the four resistors decide everything, and above it more gain buys no more rejection at all — 111.9 dB at a gain of 100000, where the arrangement alone would have been worth 148. The one place the instrument beats its own floor is a gain of 1000, where the two errors cancel; that is a coincidence of signs and not something a design can hold. Measurement, which is a circuit on a circuit

The rejection the parts have

Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.

 Two currents called saturation: one doubles every 4.49 K, the other every 8.98 K. computed by solving, not by drawing. The two current scales of one model junction against temperature, on a logarithmic axis. The saturation current of the exponential law goes as the square of the intrinsic carrier density — a cube of the temperature and the whole band gap in a Boltzmann factor — and doubles every 4.489 K at 300 K. The generation current a reverse-biased junction actually conducts goes as the density itself, with half the band gap, and doubles every 8.978 K: exactly twice as long, at every temperature. The dashed line is "doubles every ten kelvin" drawn through the generation scale, which is the current the rule belongs to. At 300 K this junction's two scales are 10 fA and 2 nA, which are its own parameters and not a property of silicon, and they become equal only at 616.8 K, or 343.6 °C. Where the models stop

Two currents with one name

A junction's saturation current is two currents with one name. The one in the forward law doubles every 4.49 kelvin; the one a reverse-biased junction actually conducts is generated in its depletion region, doubles every 8.98, and on this model junction is 3.06 × 10⁵ times larger at a volt of reverse bias. Doubles every ten kelvin is the second current's rule, and applied to the first it turns the forward drop's −1.81 millivolts per kelvin into +0.39.

 The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range. Feedback, and the margin

What the second path costs at the floor

The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

 At the order a cascade runs out, a ladder still has 4.2 decades of level. computed by solving, not by drawing. How many decades of impedance level each structure can be built at while staying inside 0.1 dB of its own design, against order. Both carry two picofarads of stray at every node. The cascade also carries fifty ohms of amplifier output resistance, a fixed resistance, which binds it from below; its floor rises 8.5× over the four orders drawn, to 0.114 dB at order eight. The ladder's inductors carry a fixed resistance per henry instead — a fixed quality factor, which scales with the design and bounds nothing — so what limits it from below is a few milliohms of track, and at order nine it still has 4.20 decades with a floor of 0.0337 dB, 3.3× its own floor at order three against the cascade's 8.5×. Filters, measured not tabulated

The band that does not close

A cascade of active sections can be built at three decades of impedance level at second order, one at sixth, and none at all at eighth — the band shuts by half a decade per order because two fixed quantities bind it from opposite ends. A doubly terminated ladder has only one of those quantities, because an inductor's loss is a fixed quality factor rather than a fixed resistance and therefore scales with the design. At order nine it still has four decades.

 Alternating-current resistance against foil thickness, 4 layers. computed by solving, not by drawing. The falling dashed curve is the direct-current resistance, which is what more copper buys. The solid curve is the alternating-current resistance at 100 kHz for a portion of 4 layers, and it turns over: past ξ = 0.663 skin depths, thicker foil has MORE resistance, not less. The minimum sits at 1.3368 times the direct-current resistance of the same foil, which is four thirds and is the same number for every layer count above one. The resistance per turn there is 2.016 against √m = 2.000, which is the law the layer count obeys. Two windings, and the band between them

The copper that makes it worse

The rung below measured one conductor pushing its own current to its rim, and there is nothing to optimise in it: thicker wire is always less resistance. Stack the conductors and the quantity changes character. Each layer sits in the field of the ones below it, the loss that field drives has no upper bound in the thickness, and the product turns over — so a portion of four layers has a best foil thickness, and above it more copper is more resistance. The best thickness is the fourth root of three over the square root of the layer count, in skin depths, and the penalty at it is four thirds for every layer count above one.

 95 dB of instrument, 290 Hz corner — and the corner belongs to the source. computed by solving, not by drawing. The common-mode rejection of the same three-amplifier instrument measured earlier measured, with 1 kΩ of imbalance between the two source resistances and 10 pF at each input. The instrument's own curve is drawn beside it. Below 290 Hz the two agree; above it the measurement falls at twenty decibels a decade while the instrument does not, reaching 84.0 dB at a kilohertz against the instrument's 95.0. What converts common mode into differential is the difference of the two input time constants — 10.0 ns here — and once it is differential no rejection repairs it. Measurement, which is a circuit on a circuit

The corner the instrument has no part in

Three earlier essays measured a three-amplifier instrumentation amplifier's rejection at direct current and found 95 dB, of which the resistors' matching decides one part and the amplifiers' own mismatch another. Connect it to a source with a kilohm of imbalance and ten picofarads at each input and the rejection has a corner at 290 Hz and falls twenty decibels a decade after it — reaching 84 dB at a kilohertz on an instrument that is still doing 95. What converts common mode to differential is the difference of two time constants, and the cure is a capacitor on the quiet input.

 Twenty nanoseconds of skew is 0.28% of duty, not 0.40. computed by solving, not by drawing. The duty cycle of a relaxation oscillator whose comparator takes longer to go one way than the other, marched, against the expression that lengthens each half cycle by its own delay. The two part company immediately and by a constant factor of about 1.42: during a delay the capacitor keeps charging past the threshold it already crossed, so the next half cycle starts further out and takes longer, and the two halves partly cancel. At 200 ns of skew on a 2585 ns period the duty is 52.805 per cent where the expression says 54.004. The open circles are the same quantity in closed form — the overshoot is V(1 − (1 − β)e^(−d/τ)) and the next half starts from it — which the march reproduces to parts in ten thousand. Circuits that do a job, and the range they do it over

The delay that is two delays

Modelling one comparator delay applied to both transitions makes the two half cycles equal by construction, and the fix is a change of one line. Made, the duty cycle moves by 0.28 per cent for twenty nanoseconds of skew rather than the 0.40 the obvious expression gives, more hysteresis improves the duty cycle and worsens the volt-seconds at the same time, and ten nanoseconds of skew saturates a hundred-turn core in 231 cycles.

 The square law is within 1% over a factor of 1.06 in overdrive. computed by solving, not by drawing. One field-effect device drawn against the two models it is between: the square law, which is zero below threshold and rises as the square of the overdrive, and the weak-inversion exponential, which rises at 77.4 mV per decade. The device is neither and approaches both. The two errors point opposite ways — the subthreshold current lifts it above the square law below, and velocity saturation holds it below above — so the square law is exact at 156.5 mV and the shaded band is where it is inside 1%: 152.0 mV to 161.7 mV, a factor of 1.06. The slider is the velocity-saturation voltage, which is the channel length times a critical field, so what it moves is the band's width and not its position. Devices, and the amplitude they stop being linear at

The exponent that is a square

Every device in this collection so far has been an exponential, and the whole of its arithmetic — 59.5 millivolts a decade, a distortion edge at 1.03 millivolts, 3,333 parts per million a kelvin — comes out of that one law. A field-effect device obeys a different one, and the interesting part is that it obeys both: an exponential below threshold and a square above it. The square law is within one per cent over a factor of 1.06 in overdrive at half a micron, and the two errors that bound it point opposite ways.

 One part, two corners: 3.98 kHz to the common mode and 7.86 MHz to the signal. computed by solving, not by drawing. Two windings on one core with a coupling of 0.999, driven twice from the same netlist — once with the two conductors in opposition, which is the signal, and once with them in parallel, which is everything the cable picked up. The mode that goes the same way round both windings meets (1+k)L and is down three decibels by 3.98 kHz; the mode that goes opposite ways meets the leakage, (1−k)L, and is untouched until 7.86 MHz. The ratio is 1975, which is 2/(1−k) and contains no inductance at all. Neither number is computed here: both modes are driven and the answer is read. Lines, where a wire has a length

The inductor one mode cannot see

Two windings on one core present a millihenry to a current that goes the same way round both and a microhenry to one that goes opposite ways, so the same component has a corner at 3.98 kHz and another at 7.86 MHz — a ratio of two thousand, which is 2/(1−k) and contains no inductance at all. It is bought to remove the conversion the previous essay measured, and five picofarads across each winding turn it over at 1.59 MHz and leave it worth 0.02 decibels by ten gigahertz.

 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.

 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. Frequency, which is the same solve

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.

 Which resistor the noise of a Chebyshev 5 actually comes from. computed by solving, not by drawing, one solve per resistor. Each bar is that resistor's share of the noise power at the output, found by splitting its node, putting a source of √(4kTR) in series with it and re-solving the whole network — so what is drawn is not how much noise each resistor makes but how much of it arrives. The largest contributor is F2R at 64.1 per cent, the smallest F0R at 9.0, and the shares add to 1.000000000000 because noise powers add. A resistor's share of the noise is not its share of the resistance: the largest departure between the two is 3.9 percentage points. The floor, which bounds from below

The resistor the noise comes from

Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.

 The load sees 10.0 Ω, 1.2e-3 Ω or 1.2e-3 Ω at direct current, and the peak is lowest for the arrangement with both paths. computed by solving, not by drawing. The impedance at the load node of all three arrangements, measured by grounding the input and driving a unit current into the load. Feedback from the amplifier leaves the load looking at the isolation resistor — 10.0 Ω, with no loop gain in it at all. Feedback from the load gives 1.2e-3 Ω, and the two-path arrangement has the same, which is what its direct-current path is for. All three resonate with the load capacitance near 3.2 MHz, and the two-path arrangement's peak is the lowest — 27.6 Ω against 37.4 and 59.2. What it gives up is between: above the 159 kHz handover it has let go of the load node. Feedback, and the margin

What the load sees looking back

Four rungs of this argument have measured what the amplifier does to the signal — the margin, the settling, the error, the noise. None has asked the question from the other end. A load that draws its own current sees an impedance looking back, and with the feedback taken from the amplifier that impedance is the isolation resistor, with no loop gain in it whatever: ten ohms, and a load step leaves an error that never goes away. The two-path arrangement recovers to a thousandth of it and charges for that in a quantity none of the four rungs below measured.

 Where one pole goes when each component is 5% high. computed by solving, not by drawing. A series R–L–C, its poles recovered by rooting the determinant, and the derivative of the upper one with respect to each element taken exactly from the two null vectors at the pole. The dashed lines are the first-order prediction for a 5 per cent change; the filled circles are where the root actually goes when the element is changed and the determinant re-rooted. The three directions are the argument: the resistance moves the pole along a circle of constant radius, because the natural frequency does not contain it — its normalised sensitivity has a real part of 3.1e-16. The inductance and the capacitance each carry exactly −½ of the radius, and imaginary parts that are exact negatives. At 5 per cent the prediction is out by 0.122 per cent of the pole's own magnitude. Networks, and how a solve is checked

The derivative of a root

An earlier essay turned one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder filter stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder filter's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.

 Three nanohenries of copper move the peak to 5.63 MHz and raise it to 1.29 Ω. computed by solving, not by drawing. The same two capacitors, with and without the inductance of the way to them: one nanohenry of mounting loop per part and two nanohenries of plane between the bank and the load. The dashed curve is the bank with no inductance in the way, peaking at 1.187 Ω at 6.52 MHz; the solid one is what the load sees, peaking at 1.293 Ω at 5.63 MHz. The peak moves down because the branch that is inductive at that frequency got more inductive, and it rises for the same reason. Above about twenty megahertz the two part company entirely: the bank is still falling toward its parts' own resistances and the load is rising on two nanohenries that no capacitor is across. Power, and the part that does no work

The capacitor that is not where the load is

A decoupling bank is usually drawn as two capacitors joined to a load through nothing, which is a stated simplification. Put three nanohenries of ordinary copper in — one of mounting loop per part and two of plane between the bank and the load — and the anti-resonance moves down to 5.63 megahertz and up to 1.29 ohms, a probe touching the ceramic reads a twelfth of what the load sees at 16.7 megahertz and three and a half times too much at 8.35, and the twentieth capacitor is worse than the second.

 Every node of a Chebyshev 5, and the one that clips first. computed by solving, not by drawing. The largest signal each node of the realised network ever carries, over the whole frequency sweep, relative to the input. The output reaches 1.000 times the input and F0b reaches 4.537, so on a ±15 V supply the input can be driven to 3.31 V rather than 15.00 before something clips — and the thing that clips is not the output. With the floor at 249.7 nV that is 142.44 dB of range against the 155.57 dB an instrument on the output would report, a difference of 13.14 dB that no measurement at the output can see. The floor, which bounds from below

The ceiling is not at the output

A fifth-order Chebyshev's noisiest node is its output and its largest signal is not. F0b carries 4.537 times the input, so on ±15 V the range is 142.44 decibels and not the 155.57 an instrument on the output reports. Across fifteen realised filters the output reading spans 3.97 dB and the range they actually have spans 22.34 — and one family of the three has no internal peaking at all, at any order.

 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.

 One part, three saturation currents: 1.90 A, 2.40 A, 2.69 A. computed by solving, not by drawing. B(H) is μ₀H plus a saturating magnetisation, written as a flux linkage, and the inductance drawn here is dλ/di — the slope of that flux, which is what a small signal on a direct current actually meets. It is 18.92 µH at no current, 16.73 µH at two amps and 10.43 µH at three, and it never reaches zero: the vacuum is still there, so the part falls to its air-core 0.05 µH and stays. The three marks are the ten, twenty and thirty per cent drops different manufacturers print as the saturation current — 1.896, 2.395, 2.694 amps, a spread of 42 per cent on one part. Two windings, and the band between them

The inductance the current decides

The rung below bounded the flux and the boundary is exact: the volt-seconds decide the flux swing whatever the material does, and a cycle whose current ripple runs from 519 mA to 75 A has the same flux excursion to better than two per cent. What saturation breaks is the relationship between that flux and the current — so a ripple the design expression puts at 514 mA is 1,022 mA at 2.56 A of load, the peak reaches 3.57 A where the part is at a fifth of its nameplate inductance, and the same part has three saturation currents depending on which per cent it was quoted at.

 At 300 K one junction holds a logarithm to ±1% over 2.4 decades as a diode and 8.5 at its collector. computed by solving, not by drawing. The voltage of one model junction against the logarithm of the current it carries, as a percentage error of that current from a straight line fitted over the widest range that stays within ±1%. Taken as a diode — both mechanisms and 0.6 Ω of series resistance — the range is 2.40 decades, from 50.1 nA to 12.6 µA, and its slope is an ideality of 1.982. Taken at the collector, where the recombination current is supplied from the base and 1.604 Ω remains, it is 8.50 decades, from the axis's own end at 1 pA to 316 µA, at an ideality of 1.0001. Nothing arrives beside the collector current, so its lower end on this axis is the axis. Where the models stop

The logarithm is in the collector

A diode is the textbook logarithm, and a real junction holds one to within one per cent over only 2.40 decades — from 50 nanoamps to 12.6 microamps, at an ideality of 1.98 — because two mechanisms and a series resistance share its terminals. The same junction read at a transistor's collector, with its recombination current supplied from the base, holds 8.50 decades at an ideality of 1.0001. How far the logarithm reaches is decided by which terminal the current is taken from, and at the bottom of the range by a leakage current a millivolt is enough to switch on.

 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.

 At a stationary point a tolerance has a mean, not a spread. computed by solving, not by drawing. Six hundred ladder filters with every reactance drawn independently from ±1 per cent, measured at the ripple peak and at a frequency between the peaks. Away from the peak the distribution is centred on nominal and 325 of 600 are above it. At the peak none is: the whole distribution lies below, with a mean of -0.0030 per cent and a worst case of -0.0140. A yield calculation that assumes a symmetric spread at a frequency that has a stationary point is wrong in both directions at once — it allows parts above a limit that cannot exist, and it misses that the whole batch has moved. Networks, and how a solve is checked

The tolerance that can only take away

At a doubly terminated ladder filter's ripple peak the first derivative of the magnitude with respect to every reactance is zero to ten digits, which an earlier essay measured and which says nothing about how much the response moves. This says how: every second derivative is negative, so of six hundred ladder filters built from one per cent components not one is above nominal, the mean has shifted rather than the spread having grown, and doubling the tolerance quadruples the damage instead of doubling it.

 The growth per cycle, and the form that is a fifth low at the top of the range. computed by solving, not by drawing. The factor the envelope is multiplied by each cycle, against the gain. The solid curve is exp(π(k−3)/√(1 − ((k−3)/2)²)), which is what the characteristic equation gives and what the netlist's own poles return to twelve digits; the dashed one is exp(π(k−3)), which drops the denominator. The circles are the marched envelope, fitted over the cycles that are still small — 80 of them at k = 3.01 and 4 at k = 3.2, and none at all above that. The two expressions differ by 3.9e-7 at k = 3.01 and by 20.462% at k = 3.8, so the approximation fails exactly where nothing is left to check it against. Circuits that do a job, and the range they do it over

Two exponentials, and where they meet

An oscillator's envelope grows by exp(π(k−3)/√(1 − ((k−3)/2)²)) a cycle, and the form usually quoted drops the denominator — exact to four parts in ten million at a hundredth above three, and 20.462 per cent low at 3.8, which is precisely where too few small cycles are left to measure it. Where the growth stops is the diode's own exponential: 108.5 millivolts of amplitude for every decade of saturation current, proportional to the ideality to four parts in a thousand. Above 60.121 nanoamperes the limiter is already conducting at zero signal and there is no oscillation at all.

 What a cascaded modulator is worth, against how well its two paths match. computed by solving, not by drawing. Two first-order loops marched sample by sample at an oversampling ratio of 64, with the first stage's quantisation error taken as the difference between what its comparator said and what was presented to it — nothing here reads a state a real converter could not. Perfectly matched, the cascade gives 72.5 decibels against the first stage's own 47.5: second-order shaping out of two first-order loops, neither of which can be unstable. The gain with which the first error reaches the second stage is then given an error, and the flat left-hand half of the curve is the arrangement working. It costs three decibels at 5.29 per cent, which is a capacitor ratio — achievable, and not free, and not something the digital side can measure or correct. Where a signal becomes a number

Two loops, and the mismatch between them

Single loops of second, third and fourth order, marched alone, each stop working above an amplitude that falls with the order — which is why nobody builds a fourth-order single loop. The standard answer is two first-order loops with the first one's error fed to the second and differentiated back out, giving second-order shaping out of parts that cannot be unstable. The cancellation is between an analogue path and a digital one, and it is worth 25 decibels until the two differ by five per cent.

 A photodiode's own capacitance sets the bandwidth, as its -0.50 power. computed by solving, not by drawing. The bandwidth of a 1.0 MΩ transimpedance stage against the capacitance of the diode driving it, with the feedback capacitor at each point bisected to give exactly forty-five degrees of phase margin on the solved loop. The classical expression √(GBW/2π·rf·cd) is drawn over it: the right shape, and conservative by about a fifth at every capacitance. The bandwidth falls as the -0.497 power of the capacitance — a square-root law, so a diode of four times the area costs half the bandwidth rather than three quarters of it. At 30 pF the compensation is 0.598 pF against the expression's 0.725, and the bandwidth 295 kHz against 220. Feedback, and the margin

Where the trouble is at the input

Every other arrangement in this field has its difficulty at the output — a capacitive load, an isolation resistor, a load that draws current. A photodiode amplifier has it at the input, and the capacitance causing it is not a parasitic: it is the diode's junction, which is the price of its area, and area is what a photodiode is bought for. The feedback resistor's own noise is 127 nV/√Hz against the amplifier's 4, and the amplifier is still ninety-five per cent of the noise at a large diode.

 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.

 Same 2 ps of jitter: a floor at -121 dB, or a line at -91. computed by solving, not by drawing. A 3.337 MHz sinusoid sampled at 10 MHz with two clocks of identical root-mean-square jitter. One clock's timing error is independent gaussians and the other's is a sinusoid at 130 kHz. The total error power is the same — -87.43 and -87.55 decibels below the signal, against the closed form's -87.55 — and the pictures are not. The random clock spreads it over 2048 bins, -120.5 dB each; the modulated one puts it into two lines at 3.337 ± 0.130 MHz, -91.5 dB down. A specification in picoseconds does not choose between them. Where a signal becomes a number

A floor, or a line

Two clocks with identical two-picosecond jitter sample the same sinusoid, and the total error power is the same to a tenth of a decibel — the closed form an earlier essay computed, right for both. One of them puts that error across two thousand bins at −120 dB each; the other puts it into two lines at −91.5. The gap is the processing gain, it grows with the record length because a density falls and a line does not, and nothing in a specification quoted in picoseconds root-mean-square distinguishes the two cases.

 Terminated at both ends: no interval anywhere, at 1.65 V of 3.3. computed by solving, not by drawing. The same net as the three-way comparison, with a fourth trace: a series resistor at the driver AND a parallel one at the receiver. Both reflection coefficients are zero, so the wave that arrives at a receiver 50% of the way along is already the final value and there is no second arrival to wait for — the departure after the first edge is 0.0e+0 per cent, which is the arithmetic's floor. What it costs is the level: 1.650 V of 3.3, exactly half, because two equal resistances divide the supply and nothing reflects to double it back. The series scheme in the same place is undefined for 1.00 delays and the unterminated one overshoots by 82 per cent. The hold current is 33.0 mA against the parallel scheme's 60, because the path to ground now has two resistances in it. Lines, where a wire has a length

Terminated at both ends

A series resistor at the driver and a parallel one at the receiver cost exactly half the swing — 1.650 volts of 3.300 — and no reflection ever gives it back, because there is no reflection. What the half buys is measured rather than assumed: a driver thirty per cent off its assumed impedance rings a series-terminated net by 16.3 per cent and a doubly terminated one by nothing at all, and an interior receiver on a series-terminated net sits in the undefined band for every far-end resistance above 125 ohms, which is Z₀/(1−2b) and contains no length, no driver and no frequency.

 A major hysteresis loop at 9.0 A/m of coercivity, and the anhysteretic curve it closes onto. computed by solving, not by drawing. The B–H loop of a core driven sinusoidally to ±400 A/m, marched through a superposition of twenty-four play operators and drawn over the single-valued magnetisation curve with no hysteresis in it. The loop encloses 12.481 joules per cubic metre per cycle, which is the core loss and which no single-valued model can produce, because a curve has no area. The coercivity is 8.96 amperes per metre and the remanence 22.3 millitesla; both are read off the marched descending branch rather than handed in. The slider takes the threshold spread down to zero, where the two branches become one, the area falls to 9.8e-15 J/m³, and the object is exactly the core the field already had. Two windings, and the band between them

The area a curve cannot have

Every magnetic model in this collection is a single-valued B(H), and a single-valued B(H) cannot dissipate: ∮H dB around a curve is zero, so the transformers and inductors here have all run cold. Giving the material a second branch costs one object — a play operator, which lags the field by a threshold and is otherwise nothing — and a superposition of twenty-four of them produces a loop with 12.481 joules per cubic metre in it, a coercivity of 8.958 amperes per metre and a remanence of 22.3 millitesla, none of which was handed in. At zero threshold the whole thing collapses onto the curve the field already had, to fifteen figures.

 A clean board is 14.0 mV of error, and a guard makes it 1.0 nV. computed by solving, not by drawing. Two boards differing by one wire, solved with the leakage present and again with it removed, so the number plotted is the leakage's own contribution and nothing else. Unguarded, a teraohm across the laminate from a 15 V rail through a 1 GΩ source is 13.99 millivolts — and a humid morning takes that resistance down two decades, which is the left-hand end of this axis. Guarded, the ring is held at the input's own potential by the amplifier, so what is across the leakage is the amplifier's own error and the result is 1.00 nanovolts. The guarded line is flat in the rail and proportional to the signal: the offset has become a gain error of 1.00 parts per billion, and the reading is low by it rather than high: the ring sits a little below the input, so the last of the leakage pulls the input down. Measurement, which is a circuit on a circuit

The current that does not reach the input

A teraohm across a board from a fifteen-volt rail is fourteen millivolts of error through a gigohm source, and a humid morning takes that resistance down two decades. A ring held at the input's own potential leaves a nanovolt — proportional to the signal rather than to the rail, so an offset has become a gain error of one part in a billion — and the same wire multiplies the input resistance by the loop gain, which makes it 10¹⁸ Ω at direct current and 10¹² Ω at a megahertz.

 The temperature through a 20 mm core that makes its own heat. computed by solving, not by drawing. Conduction with volumetric generation, solved on forty-one cells with a surface film at each face and the loss density evaluated at each cell's own temperature — so the middle of the core makes less heat than its faces do, which a closed form for uniform generation cannot express. The peak is 115.05 °C and the surface 112.45: a gradient of 2.59 kelvin, which is 2.9 per cent of the 90.0 kelvin rise. That share is Bi/(Bi + 2) — 3.0 per cent at a Biot number of 0.063 — so it is decided by how well the surface is cooled and not by how much heat is made. Power, and the part that does no work

The degrees a thermocouple cannot see

Every thermal answer in this collection has been one temperature, and a core makes its heat in its volume and loses it from a surface, so it has two. Solved as a conduction problem, a twenty-millimetre core in still air is 2.59 kelvin hotter in the middle than on the outside — 2.9 per cent of a ninety-kelvin rise, which is why the lumped answer has been good enough. Cool the same core on a plate and the gradient does not shrink; it grows to 3.37 kelvin and becomes 78 per cent of what is left.

 A tenth of a per cent of lead is 999 µε of strain that is not there. computed by solving, not by drawing. A 350 Ω quarter bridge at zero load, against the resistance in each lead. With both leads in the changing arm the output is 4.99500 mV at 350 mΩ — an apparent fractional change of 0.1998%, which at a gauge factor of 2 is 999 microstrain. With one lead in that arm and one in the arm beside it the output is zero to the last bit, at every lead resistance drawn. The lower panel is what the second arrangement costs: the sensitivity falls as 1/(1 + Rlead/R), which is 0.100% at the same lead and is a calibration constant rather than a drift. Circuits that do a job, and the range they do it over

The leads that are in the bridge

A strain gauge on the end of two long wires cannot be told from a strain gauge under load: both leads in the changing arm is bit for bit the same netlist as a quarter bridge whose fractional change is larger by twice the lead over the gauge, which at 350 milliohms on a 350 ohm gauge is 999 microstrain that is not there. Moving one of those leads into the arm beside it leaves the output at exactly zero for every lead resistance drawn, and turns a twenty-kelvin drift of 76.8 microstrain into 0.077. That factor is two over the strain being read — 999 at a thousand microstrain and 9990 at two hundred — and it costs 0.100 per cent of sensitivity.

 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. Frequency, which is the same solve

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 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.

 Five tolerances, and the response moves in two directions. computed by solving, not by drawing. The eigenvalues of the relative second-derivative matrix of a fifth-order 0.5 dB Chebyshev ladder's magnitude at its lower ripple peak, over its five reactances. Two are of order one — -0.9473 and -0.8051, both negative — and the other three are 3.9e-9, which is zero at the precision the arithmetic has. So the quadratic form is negative semi-definite of rank two, and there is a three-dimensional subspace of component errors that the peak cannot see. The open circles are the same matrix computed by four re-solves per pair, sharing no adjoint arithmetic with the filled ones: they agree to parts in ten thousand on the two that are there and place the three zeros about two decades higher, which is the price of differencing a difference. Networks, and how a solve is checked

The three tolerances that do nothing

An earlier essay computed every second derivative of a ladder filter's magnitude at a ripple peak, found them all negative, and built six hundred ladder filters to argue that no combination of tolerances could raise the response. The whole matrix says so outright — and says something six hundred samples could not have found, because a sample of a five-dimensional box never lands on a three-dimensional subspace: two of the five eigenvalues are of order one and the other three are nine decades down.

 The fourth-order term explains 95% of what the third leaves out. computed by solving, not by drawing. Two closed forms for the amplitude at which a degenerated stage reaches 1% of second harmonic, each against the same measurement — a bisection on the harmonic content of a Newton-solved curve, which shares no arithmetic with either. The leading expression is 4Vₜ·t·D², derived by expanding the transfer curve and exact in the limit of small drive; its error grows as the 1.96 power of the drive it is evaluated at. At a degeneration of eleven that drive is 4.53 thermal voltages and the expression is 6.89% optimistic. Carrying the reversion one order further takes it to 0.362%. Devices, and the amplitude they stop being linear at

What the fourth order says about the third

A closed form derived by neglecting the fourth-order term is a claim with an error, and the error is a quantity the fourth-order term can be asked about. Carried one order further, the degenerated exponential's reversion predicts 7.33 per cent where the leading expression was measured to be 6.89 per cent optimistic — and takes the residue to 0.362 per cent. The correction has its own edge, in the same quantity, and it is measured too.

 Three cliffs, not one, and the fastest damping is on the last of them. computed by solving, not by drawing. Settling time against damping for a third-order response — a complex pair at unit natural frequency and a real pole at 3 — with the second-order case behind it. Both are staircases: the settling time is set by the last excursion outside the band, so there is one step for each excursion that stops happening, and there are 3 of them between 0.3 and 0.98. They are at 0.378, 0.522, 0.773, with jumps of 1.24, 1.30, 1.42. The earlier essay found the last and largest of them and did not look below it. The fastest damping is 0.775, sitting on the edge of the last step, and a design a hundredth to the left of it settles 42 per cent slower. Before the steady state

Three cliffs, and where they are

Sweep a second-order step's damping and the settling time falls by a third in one step of a five-thousandth sweep, which is usually called the cliff. There are three of them between 0.3 and 0.98, one for each excursion that stops leaving the band, and adding a third pole moves all three left and makes all three shallower — so the classic 0.78 for fastest two per cent settling is a second-order number, and at a third pole one and a half times the natural frequency the answer is 0.745 and 0.78 is on the wrong side of the step.

 One component of a gyrator moves the response by 0.50; the passive ladder's inductor moves it by 3e-8. computed by solving, not by drawing. The magnitude sensitivity at the passband maximum of a gyrator ladder, against the resistance scale the gyrators are built at, for a single component and for the combination of both halves that actually changes the synthesised inductance. The passive ladder realising the identical response is at 3.2e-8. A single component sits at about a half whatever the resistance scale is, and the combination falls as one over it — the 0.97 power over 3 decades. So the stationarity has not been destroyed, it has become a statement about a combination of components rather than about a component, and independent parts do not come in combinations. Filters, measured not tabulated

One inductor, and ten components

The rung below built an inductor out of an amplifier and ended with a boundary that is not a frequency: one end of it is soldered to ground. A ladder's series inductors are floating, so making one takes four amplifiers rather than two — and the four-amplifier version is exactly floating, to five figures, and reproduces the ladder's response to a hundredth of a decibel. It does not reproduce its stationarity. Each of a gyrator's ten components moves the response by exactly a half, where the inductor it replaced moved it by ten to the minus eight.

 Which limit binds is a property of the load, and they change places near 22 nF. computed by solving, not by drawing. Each limit measured on its own, as the departure of its march from the linear one, at a load step of half the output stage's rating. The input pair's departure falls with load capacitance — a bigger reservoir holds the node while the loop responds, which is the sixth rung's own result — and the output stage's does not fall nearly as fast, because what it has to supply is the charge the capacitor wants. Below about 22 nanofarads the thermal voltage decides the answer and above it the output stage does, and nothing about the amplifier changed. Feedback, and the margin

The current above which there is no impedance

The sixth rung found the impedance leaving at 10.6 mA, where the input pair's own tanh takes over and the slew rate is twice the thermal voltage times the gain-bandwidth in radians, with no design choice in it. A real output stage has a second limit that is nothing but design choice, and the two do not bind at the same load: at 0.47 nF the input pair's departure is 19.4 per cent against the output stage's 4.2, at 22 nF it is 0.9 against 2.3, and above the output stage's rating the excursion does not come back at all — 2,254 Ω for a quantity that was 37.

 The crest factor is 13.4 and the winding is sized by 3.01. computed by solving, not by drawing. Three ratios of the same settled march, against the reservoir. The crest factor — the peak diode current over the load's direct current — runs 6.43 to 24.25. The form factor, which is the root-mean-square current over the same direct current and is what a winding heats by, runs 2.114 to 4.077. Its square is the copper loss against a winding carrying the direct current alone, and that runs 4.47 to 16.62. The first ratio is 3.04 times the second at 220 µF and 5.95 times at 4700, so quoting one of them tells a reader nothing about the other. Circuits that do a job, and the range they do it over

The current that sizes the transformer

A reservoir's crest factor is 13.374 at a thousand microfarads and the winding is not sized by it. The root-mean-square of the same marched current is 3.0069 times the load's direct current, so the copper dissipates 9.0417 times what it would carrying the direct current alone — and the two ratios diverge, from 3.04 apart at 220 microfarads to 5.95 apart at 4700. The expression for the mean output is wrong in three places whose signs differ, and at 313.9 microfarads they cancel to six microvolts while the ripple expression inside it is still 32.3 per cent high.

 The most sensitive direction at one ripple peak is not the one at the next. computed by solving, not by drawing. The eigenvector of the largest eigenvalue of the relative curvature matrix, at each of the 2 ripple peaks of a 5th-order 0.5 dB Chebyshev ladder, drawn as its components over the reactances in order along the ladder filter. Every one is symmetric under the filter's own reversal and no two of them are the same direction: the largest overlap between any pair is 0.0603, which is 86.5 degrees apart. The eigenvalue that belongs to them grows from -0.9473 at 554.9 Hz to -12.005 at 897.9 Hz, a factor of 12.7, so the peak nearest the band edge is both the most sensitive place in the passband and sensitive to a different combination of parts. Each direction is confirmed by a second computation of the whole matrix — four re-solves per pair, sharing no adjoint arithmetic — which returns the same direction to 7.6e-7 radians. Networks, and how a solve is checked

The direction a response is most sensitive to

An earlier essay diagonalised a ladder filter's curvature at one ripple peak and read only the eigenvalues: rank two of five, three directions of nothing. The eigenvectors say what the two directions are, and doing it at every peak rather than one changes the conclusion. The most sensitive combination at 554.9 Hz and the one at 897.9 Hz are 86.5 degrees apart, the curvature that belongs to them grows from −0.556 to −75.4 across a ninth-order passband, and exactly one combination survives the whole band at every order — which turns out to be the ripple depth.

 The Steinmetz exponent is a local slope, and how far it moves is a property of the material. computed by solving, not by drawing. Loss per cycle against peak flux density over three decades, marched on a play-operator core, with the local exponent d ln W / d ln B drawn across the top of the same frame. It is not a constant anywhere: 2.797 at 5.5 millitesla, heading for the three that Rayleigh's law gives, and 1.462 near saturation where the material has run out of magnetisation to give — a range of 1.420. How wide that range is is itself a property of the material: over the same amplitudes a soft core's exponent moves by 1.73 and a hard one's by 0.21. A single power law fitted across the whole range returns β = 2.518 and misses by 72.2 per cent; the same law fitted over the quarter of it from 9.7 to 24 millitesla returns 2.743 and misses by 0.97. Below 0.58 millitesla this discretisation has no loss at all, which is the finite operator count showing and not the material; the sweep starts above it. Two windings, and the band between them

The exponent nobody put in

A catalogue prints core loss as a coefficient times the frequency raised to one power and the flux to another, and the two exponents look like material constants. Neither is. On a loop built from play operators the frequency exponent is exactly one — a theorem, not a fit, because a rate-independent locus has the same area however fast it is traced — and the flux exponent is a local slope that runs from 2.94 at half a millitesla to 1.46 near saturation, so five windows on one measured curve give β from 1.58 to 2.84 and predictions three times apart at a hundred and fifty millitesla.

 The bank reaches 1.500 mΩ and the load sees 141.5 mΩ at that same frequency. computed by solving, not by drawing. One bulk part and 20 ceramics, with a nanohenry of mounting loop each and two nanohenries of plane between the bank and the load, solved once per frequency and read at both nodes. The dashed curve is the bank's own node — what a probe on the parts measures. The solid one is the load. The bank's least impedance is 1.500 mΩ at 11.3 MHz, and at that frequency the load sees 141.5 mΩ, which is 94.3 times more, against 141.5 mΩ of plane reactance at that frequency. Whatever the parts do, the load's reading cannot fall below the reactance of the copper in front of them, and the parts reach their best by moving up the frequency axis into it. Power, and the part that does no work

The floor and the ceiling move apart

A decoupling bank is judged by two numbers — the lowest impedance it reaches and the highest frequency at which it still meets its target — and with no copper between the parts and the load both improve together as capacitors are added, 5.000 milliohms down to 1.500 and 19.8 megahertz up to 162. Three nanohenries of ordinary board separate them. The bank's own floor still falls 3.33 times while the load's falls 1.49, and the ceiling read at the parts climbs to 82.4 megahertz while the load's peaks at 8.06 and falls to 5.05. At twenty parts the two nodes disagree by a factor of 94 about the same solve.

 Every order buys less range than the one before, and above 9 Vₜ the sixth is worse than the fourth. computed by solving, not by drawing. The error of the same expression truncated at three orders, against the drive it is evaluated at, with the measurement it is chasing being a Newton-solved transfer curve that knows about no series at all. Each truncation's error grows as its own order in the drive — fitted at 2.00, 4.00, 5.88 against 2, 4 and 6 — so each buys a further range at a stated accuracy: inside 1% the leading expression is good to 1.12 thermal voltages, the fourth order to 3.88 and the sixth to 7.03, factors of 3.46 and 1.81. Beyond all of them the series stops helping: at 8.9 thermal voltages, where the second harmonic is 11.4%, the sixth-order expression is exactly as wrong as the fourth and is worse above it. What a designer does there is bisect the curve. Devices, and the amplitude they stop being linear at

The order that stops helping

Three essays in this field have derived expressions for the amplitude at which a degenerated stage's distortion reaches a target, each one order longer than the last, and each one nearer the measurement. This is where that stops. The error of an expression truncated at order m grows as the m-th power of the drive — 2.00, 4.00 and 5.88 measured — so every added order buys a range that ends sooner than the last one bought, and above 8.9 thermal voltages the six-term expression is further from the device than the four-term one.

 Fed nothing, an eighth-order cascade sits 88 least significant bits from zero. computed by solving, not by drawing. Where each section of a 12-bit rounded cascade settles with zero input and a seeded state. Nothing decays to zero: a constant state survives whenever rounding returns it to itself, which needs only |y·(1 + a₁ + a₂)| ≤ q/2, and that denominator is small precisely because the corner is far below the sample rate. Each section's own band is 30, 31, 32, 33 least significant bits and the settled offsets are -27, -50, -70, -88 — accumulating, because a section's dead-band output is the next section's input and the next section passes direct current. In volts the offset halves with every bit added; in least significant bits it does not move at all. Where a signal becomes a number

Zero in, and not zero out

Rounding a filter's coefficients and marching it in double precision leaves out the other half: the products are rounded too. Put that in and a twelve-bit eighth-order cascade fed nothing at all settles two per cent of full scale away from zero and stays there — and the offset does not shrink with the word length, it grows as the square of how far the corner sits below the sample rate, reaching eighteen per cent at a hundred and twenty-eight times.

 Twenty metres buys 20 dB of apparent match and costs 10 dB of noise figure. computed by solving, not by drawing. The same cable and the same 200 Ω load as the reading, with the amplifier that is actually behind the instrument. The rising trace is the return loss the instrument reads, which is the load's own 4.44 dB plus twice the one-way loss. The lower pair is the chain's noise figure: a 2 dB amplifier with the cable in front of it, counted as a matched attenuator whose noise factor is its loss, and counted honestly from the available gain of a lossy line driven by a source that reflects 0.60. The first says the exchange rate is exactly two decibels of match per decibel of noise figure, at every length here. The second is higher everywhere — by (1−Γ²u²)/(1−Γ²), which is 0.028 dB at five centimetres and 1.938 dB, the load's own mismatch loss, once the cable is long enough to have absorbed the reflection. At twenty metres the instrument reads 24.4 dB and the chain costs 13.92 dB against the amplifier's own 2. Lines, where a wire has a length

The cable that hides two things

A length of cable improves a return-loss reading by twice its loss and raises a noise figure by once it, so the rule of thumb is two decibels of apparent match per decibel of floor. Both halves are owned here and neither essay had the other. Put together they say what an acceptance limit costs: making a 4.0:1 load read 1.50:1 spends 6.53 decibels of noise figure, against a mismatch that was itself costing 1.938 — and the exchange rate is not two but 2(1−Γ²)/(1+Γ²), which is 0.94 where a pad is actually short.

 A cascode multiplies rₒ by β, not by gₘrₒ — and the two are 21× apart. computed by solving, not by drawing. The output resistance of a cascode stage, measured by driving the output node with a current source and reading the voltage, against the current gain of the upper device. The plain stage's is 80 kΩ — rₒ and nothing else. The cascode's is 11.5 MΩ at β = 150, which is βrₒ to within a tenth and is 21 times below the gₘrₒ² every reference gives. The reason is in the netlist rather than in the algebra: the upper device's base draws current, so its rπ sits from the lower device's collector to signal ground and shunts the node the feedback works through. What the arrangement buys therefore scales with β and stops when β does, and the curve is the two expressions drawn against the measurement. Devices, and the amplitude they stop being linear at

The device that never sees the swing

A second transistor standing between the first and the load does two things that every reference gives one expression each for, and one of the two expressions has no ceiling in it. The output resistance is not gₘrₒ² — that is 248 megohms here, and the measurement is 11.5 — it is βrₒ, because the upper device's base draws current and shunts the very node the feedback works through. The bandwidth really is fourteen times better, and what it costs is two volts of a five-volt supply.

 A 4.0:1 flux slope ratio and a sinusoid enclose the same loop to 0.00%. computed by solving, not by drawing. A core driven in flux rather than in field — the way a winding drives it, by integrating a rectangular voltage — around a triangle of ±100 millitesla at a duty cycle of 0.2, whose two slopes differ by 4.00 to one. The loop it traces encloses 2.1006 joules per cubic metre, against 2.1007 for a symmetric triangle and 2.1006 for a sinusoid of the same peak: the same number to 0.001 per cent. That is not an approximation, it is a theorem about the model — a rate-independent locus depends on where the flux went and not on how fast — and it is the prediction that real cores disagree with by tens of per cent. The disagreement is the measurement of what the model has left out. Two windings, and the band between them

The duty cycle that costs nothing

A converter drives its core with a rectangular voltage, so the flux is a triangle whose two slopes differ by nineteen to one at a five per cent duty. The play-operator core charges exactly the same for all of them — 2.1006 joules per cubic metre at every duty and for a sinusoid of the same peak, to three parts in ten thousand — because a rate-independent locus depends on where the flux went and not on how fast. Real cores charge tens of per cent more, and the standard correction hides its entire waveform dependence in α − 1, which is the one term a rate-independent model has none of.

 A staircase costs one Nth, computed rather than quoted. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. The charge is broken into N equal risers, each held for 16 time constants so that it completes. The measured losses are 1.00000, 0.500001, 0.250000, 0.125000, 0.0625000, 0.0312500 of ½CV² — which is 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 times 1/N, so the law is exact to four parts in a million at the worst rather than approximately true. The fitted exponent is -1.00000 and the energy account closes to 2.17e-6 at the worst. Before the steady state

The half a switch keeps

Charging a capacitor from a step loses half the delivered energy whatever the resistance, and a ramp takes the loss down as 2τ/T with no floor. A staircase of N settled risers costs one Nth of the step, exact to four parts in a million, and the law ends at a dwell of 5.272 time constants. A switch is the other half of the same product and buys nothing at all: with the supply held at five volts and the channel conductance ramped over a thousand time constants, the loss is 1.00000000 of ½CV².

 One channel's load step reaches another through the supply, and the compensation decides by 4497×. computed by solving, not by drawing. Two identical amplifiers on one rail — sharing no signal node — with an ampere of load step pulled from the first and the second's output read. With the wiring left out the coupling is exactly zero, which is what the seven rungs below this one computed. With 30 nanohenries and fifty milliohms of rail and 10 microfarads of decoupling it is not: 3.84 microvolts per ampere at 271 kHz if the compensation capacitor returns to ground, and 17.29 millivolts per ampere at 2.33 MHz if it returns to the rail. That is a factor of 4497 decided by a modelling choice, which is why both are drawn. The channel that caused the step is unaffected: its own loop corrects the disturbance along with everything else, and the crosstalk is entirely a problem for the channel that did not. Feedback, and the margin

The rail the load moves

Seven rungs of this ladder end by saying the same thing: the supply is an ideal voltage source, so a load step is drawn from a node that cannot be disturbed. Giving the rail an impedance turns out to change the disturbing channel's own output impedance by three parts in ten million — its loop corrects the supply along with everything else — and to open a path to a second amplifier that shares nothing with it but a wire. How large that path is is a modelling choice: 3.84 microvolts per ampere with the compensation capacitor returned to ground, 17.3 millivolts with it returned to the rail, a factor of four and a half thousand.

 The ripple 1000 µF leaves, through the regulator: 60.62 mV rather than 21.11 mV. One settled cycle of the reservoir's output — 1.331 V peak to peak across 1000 µF — split into 1000 Fourier lines, each passed through the solved regulator's rail-to-output response, and summed. The output is 60.62 mV peak to peak. The ripple times the rejection at 100 Hz, -36.00 dB, gives 21.11 mV, which is 2.872 times too little: the rejection worsens at twenty decibels a decade, so each harmonic of the sawtooth arrives at nearly the size of the first. The 100 Hz line carries 23.4% of the output's mean square and the lines above 1 kHz 10.4%; the first three arrive at 7.731 mV at 100 Hz, 7.200 mV at 200 Hz, 6.388 mV at 300 Hz. The dashed curve is the 100 Hz line alone. Circuits that do a job, and the range they do it over

The ripple that arrives as a comb

The rejection essay multiplied two numbers: 1.331 volts of reservoir ripple and the regulator's 36.0 decibels of rejection at a hundred hertz, for 21 millivolts at the output. The ripple is a sawtooth, a comb of lines at every multiple of a hundred hertz, and the rejection worsens at twenty decibels a decade — so the second line arrives at nearly the size of the first, and the third too. Summed with their phases, the output is 60.62 millivolts, 2.87 times the estimate, and the hundred-hertz line carries only 23.4 per cent of it. A reservoir twenty-one times larger cuts the rail's ripple 14.8 times and the regulated ripple 5.95.

 Twenty-four orderings, and 11 of them are choices. computed by solving, not by drawing. Every ordering of the four sections of an eighth-order 0.5 dB Chebyshev, at 16 bits, drawn against the two things an ordering decides. The horizontal axis is the round-off floor the arrangement adds — 24.2 least significant bits at best and 99.8 at worst. The vertical axis is the largest value any section's output reaches, which is what decides whether a word overflows: 0.088 of full scale at best and 0.699 at worst, a range of 18.0 decibels. 13 of the twenty-four are beaten on both counts by another ordering and are simply mistakes; the 11 on the lower-left frontier are the actual choices, and no one of them is best. Where a signal becomes a number

Which section goes first

A cascade of four biquads can be assembled in twenty-four orders, all of which realise exactly the same transfer function. They do not cost the same: the round-off floor runs from 24 to 100 least significant bits and the largest value any section reaches runs over eighteen decibels — and the two go opposite ways, so eleven of the twenty-four are genuine choices and thirteen are beaten on both counts by another arrangement.

 What the accuracy costs: dynamic range against the resistance scale. computed by solving, not by drawing. The earlier essay found the active realisation's response converging on the passive one's as the resistance scale rises. This is the price. The floor rises as the square root of the scale — fitted at 0.500 — because the resistors are the noise. The largest internal swing rises as the scale itself — fitted at 1.012 — because each gyrator forces the inductor's own current through its own resistors, so an amplifier inside it carries that current times R. Dynamic range on a ±15 V supply therefore falls as the three-halves power: 78 dB at 100 kΩ and 18 dB at 10 MΩ. The passive ladder realising the same response has 141 dB, and its worst internal node carries 1.10 times the input. Filters, measured not tabulated

Eight amplifiers, and what they add

The rung below realised a Chebyshev ladder out of floating gyrators and found the response converging on the passive one's as the resistance scale rises — twenty-two decibels out at ten kilohms, a twentieth of a decibel at ten megohms. It closed by naming two quantities it had not measured. They are the same quantity: the resistors that buy the accuracy are the noise, and the amplifiers inside the gyrators carry the inductor's own current through them, so the floor rises as the square root of the scale and the ceiling falls as the scale.

 The reading is a count of decades: 1.0288 parts per thousand of them. computed by solving, not by drawing. 17 marched tests, three families of absolute time — a tenth of a second, one second and ten seconds of short — plotted against the number of decades between the short and the reading. The families lie on one another, which is the finding: the answer is not a property of the part alone and not a property of either duration, it is a count of the decades of relaxation time the test leaves in. The line is a least-squares fit through the origin at 1.0288e-3 per decade; the model's own capacitance per decade of relaxation time is α = 1.0343e-3, which nothing in the fit was told — the fit sits 0.53 per cent under it, because the charge that comes back is shared with the slow branches it came off. The worst residual is 4.80 per cent, at the narrowest ratio drawn, and 1.23 per cent over the 8 tests that are two decades wide and read before the slowest relaxation the model has; the 3 read after it fall away to 4.49 per cent, which is where the law ends. Families a hundred times apart in absolute time differ by at most 0.84 per cent, which is the whole of the collapse. Before the steady state

Ten seconds, and fifteen minutes

A data sheet's dielectric absorption is quoted as a property of the part. It is not: the same modelled capacitor reads 0.4050 per cent with a tenth-of-a-second short and 0.0305 per cent with a thousand-second one, and 0.0047 against 0.2948 depending on when the reading is taken. Seventeen marched tests collapse onto one line — the recovery is 1.0288 parts per thousand for every decade between the two durations — and four dielectrics the specified test declares identical read a factor of 3.31 apart one decade away from it.

 The return under a track gathers from 26.3 kHz to 1.42 MHz, not at one frequency. computed by solving, not by drawing, on a cross-section of a 50 mm plane cut into 120 strips, each with its resistance and its partial inductance to every other strip and to the track. The solid curve is the share of the return current inside one track-height of the point beneath the track; the second is the share inside ten heights. At direct current the return spreads evenly — 0.8 per cent within one height — and far above the band it is the image-current distribution, 48.7 per cent, which the closed form gives as 48.7. Between them it gathers across three decades: a tenth of the way by 26.3 kHz, half by 283 kHz, nine tenths by 1.42 MHz, shaded. The resistance of the path equals its reactance at 1.59 kHz, where the return has not yet moved. The single corner estimated from a path three track-widths wide and a parallel-plate inductance is 106 kHz, 28 per cent of the way through the band. Lines, where a wire has a length

The corner that is three decades wide

Where the current comes back put the change in a return current's path at 106 kilohertz, from a low-frequency path assumed three track-widths wide and an inductance taken from a parallel-plate formula. Solved across a plane cut into a hundred and twenty strips, the loop's resistance equals its reactance at 1.59 kilohertz, where the current has not moved at all, and the return then gathers beneath the track over three decades — half of the way by 283 kilohertz, nine tenths by 1.42 megahertz. The single corner is a point about a quarter of the way through a band.

 Two of the four errors are divided by the gain; the best the instrument gets is 110.1 dB, at ×776. computed by solving, not by drawing. The four mechanisms that limit a three-amplifier instrumentation amplifier's common-mode rejection, each measured alone against the gain of its input stage and then all four together, at 50.0 Hz with 1 kΩ of imbalance between the source resistances and 10 pF at each input. The difference stage's four resistors and the difference amplifier's own rejection are injected after the gain, so their common-mode gain is a constant — 1.998 mV/V and 0.0100 mV/V — and the rejection they allow rises decibel for decibel with the gain. The input pair's mismatch and the source's time-constant gap are injected before it, so they are amplified by exactly the gain the signal is and the rejection they allow is flat. The four add as complex numbers: the two largest are real and of opposite sign, they cancel at a gain of 776, and what is left there is the source's 3.142 µV/V, which is purely imaginary because it is ωΔτ. The instrument's best is 110.07 dB against the source's own 110.06, and above that gain more of it buys nothing. Measurement, which is a circuit on a circuit

The errors that arrive before the gain

Four earlier measurements each found a different owner of one instrument's common-mode rejection and each measured it alone. Solved together, the four are complex numbers that add — to 1.25 parts in ten thousand — and two of them carry a factor of the first stage's gain while two do not. The two that do cancel the two that do not at a gain of 776, and what is left there is 110.066 decibels, which is exactly the number the cable sets. Better-matched amplifiers move that gain from 93 to 3392 and do not move the ceiling by a hundredth of a decibel.

 The two sequences a neutral current says nothing about. computed by solving, not by drawing at 61 imbalances. Three 20 Ω loads on a 230 V, 50 Hz star supply, one of them raised by a fraction of itself, with the neutral in place. The zero-sequence current is the one the neutral carries three times and is the only one a neutral ammeter reads; the negative sequence is a balanced set of three phasors rotating the other way. At 30.0 per cent imbalance it is 0.8846 A against 10.6154 A of positive sequence, 8.333 per cent, against 8.333 per cent from x/(3 + 2x). Two per cent arrives at 6.250 per cent imbalance, bisected on the network. Power, and the part that does no work

The half the neutral does not carry

A star load unbalanced in one phase produces two things, not one. The neutral carries three times the zero-sequence current, which is the half a neutral ammeter reads; the other half is a negative-sequence set of exactly the same size, rotating backwards, that the neutral never sees. With 0.5 Ω of line in front of 20 Ω loads, losing a phase entirely puts 50.00 per cent negative sequence in the current and 0.8265 per cent in the voltage a switchboard meter reads.

 A cascoded mirror is 90× the output resistance, and 43% of it goes back into the reference. computed by solving, not by drawing. The output resistance of a two-transistor mirror and of the same mirror with a cascode on each branch, measured by moving the output a little either side of its operating point and reading the current, against the current gain of every device. The plain mirror sits at rₒ = 89 kΩ and does not move. The cascoded one reaches 7.39 MΩ at β = 150 and rises with β until β stops being the smaller of the two quantities, where it saturates on gₘrₒ² = 268 MΩ. The third curve replaces the diode-connected upper device with a held voltage at the same potential and recovers 1.76 times the resistance, which is the upper device's base current being charged a second time — to the reference branch, where it moves the mirror's own bias. Devices, and the amplitude they stop being linear at

The source that holds to the supply

Putting a second transistor on each branch of a current mirror is always described as buying output resistance and costing headroom, and both halves of that are measured here rather than repeated. The resistance goes from 82 kΩ to 7.39 MΩ, the floor rises by 0.71 volts — and the range over which the current is actually what it was set to goes from 1.70 volts to 9.09, because a plain mirror's current never stops climbing. Forty-three per cent of the resistance that should be there is missing, and it is in the reference branch.

 The resistors own the floor between 2.91k Ω and 85.9k Ω, and the part owns it outside. computed by solving, not by drawing. The noise at the load of the two-path compensation against the impedance of its own feedback network, with the resistors scaled together and the compensation capacitor taken down in proportion so that Rf·Cf — the handover between the two feedback paths — does not move. Four contributions are integrated over 10 Hz to 100 MHz: the amplifier's 4 nV/√Hz, fitted as the 0.005 power of the impedance and so flat; the two resistors' √(4kTR), the 0.501 power; and the amplifier's 0.60 pA/√Hz flowing in the feedback resistor, the 0.997 power. Two different powers of one quantity cross twice. The resistors carry more than half the power only between 2.91k Ω and 85.9k Ω; outside that window, in both directions, the part does. The part's share is least at 15.8k Ω, which is not eₙ/iₙ — it is that ratio multiplied by the noise gain of 2.000 and again by 1.187, the square root of the ratio of the bandwidths the two generators actually see; there it carries 26.26 per cent. The model stops where the amplifier's output current does: at 100 Ω the feedback resistor alone draws 10 mA a volt. Feedback, and the margin

The window the resistors own

Eight essays have priced one compensated stage, and the fourth of them left two of the amplifier's own generators named and uncounted. With the current generator put in the netlist the floor at ten kilohms goes from 28.88 microvolts to 30.13, and the resistors carry more than half the noise power only between 2.91 kΩ and 85.9 kΩ — outside that window, in both directions, the part does. The flicker corner turns out to be worth 1.00009 in this stage's own band, and 3.474 one band away.

 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.

 Against its own shaping the error is 17.4× tonal, not 51×. computed by solving, not by drawing. The share of an order-1 loop's in-band error sitting in its five largest lines, from 0.02 to 0.9 of full scale, with both nulls drawn. The lower level is 1.953 per cent — five lines of a FLAT error over 256 — and it is what this measurement has always been quoted against. The upper level is 5.74 per cent, which is what five lines of the loop's OWN shaping hold with no tone anywhere in them. Measured against the first the error is 51 times tonal at 0.02 of full scale and 26 at 0.9; against the second, 17.4 and 8.9. The direction survives the correction and the size does not. Where a signal becomes a number

A floor, or five tones

The field's sharpest statement about a one-bit loop is that three quarters of its in-band error sits in five lines, against the 1.953 per cent a white error would put in any five — a factor of thirty-eight. The comparison is to a white error, and a shaping loop exists to make its error anything but white. Measured against the loop's own transfer function, which the loop reproduces line by line to a part in five hundred when it is dithered, the same error is 17.4 times tonal at a fiftieth of full scale and 8.9 times at nine tenths. And 99.5 per cent of it at the quiet end is two harmonics of the input.

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

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.

 A true-RMS reading of a sine: ripple a second filter removes, and a bias it cannot. An explicit converter — square, average through a one-pole of τ = 100 ms, take the root — in steady state on a sine of unit root-mean-square value, integrated exactly over a period at 91 frequencies and by a fourth-order march of its own equation at 6, which agree to 1.9e-8. The upper curve is half the ripple on the reading and the lower one the amount by which its mean is low. The reading is low at every frequency, because the square root is concave; it is 1% low below 1.86 Hz, while the ripple is inside ±1% only above 39.8 Hz. The dashed curve is the small-ripple form, an eighth of the averaged square's ripple power, which the bias approaches as the ripple shrinks. Power, and the part that does no work

The average a square root pulls low

A true-RMS converter squares, averages and takes the root, and the root of a quantity that ripples averages below the root of its mean. With a hundred-millisecond averager a sine is read one per cent low below 1.86 hertz, where the ripple is still ±20 per cent — and a second filter that steadies the display takes the ripple away and leaves the reading exactly as low as it was. A square wave is read exactly at any averaging time; a rectifier current conducting for twenty degrees needs 2.41 times the averaging a sine does. The implicit converter is the explicit one at half the time constant, and a reading falls 1.38 times slower than it rises.

 The fastest damping is a surface, and the band is worth 5 times the third pole. computed by solving, not by drawing. Each point is the last settling cliff, bisected — the damping at which the first overshoot's peak lands exactly on the band's edge, which is where the fastest settling is. Across the five bands the optimum moves by 0.231 of damping ratio; across a third pole from 1.5 times the natural frequency out to a second-order response it moves by 0.047. The two axes are worth 5.0 to one, and the expensive one is the specification rather than the parasitic. The classic 0.78 for fastest two per cent settling is the second-order curve's value at ±2%, 0.7797; at ±1% the same response wants 0.8261. Before the steady state

The best damping is not the one to build

The fastest settling damping is the right-hand limit at a discontinuity, so two thousandths below it costs 41 per cent and two thousandths above it costs 0.34 — a ratio of 120 in the penalty for the same error. With ±2 per cent on the damping ratio the nominal that minimises the worst case is 0.7927 rather than the optimum's 0.7734, and it guarantees 4.243/ωₙ against 5.943. The band moves the optimum by 0.231 of damping ratio and the third pole by 0.047, and 0.78 is exact at ±2% and 55 per cent slow at ±1%.

 The eddy term is f² below the skin-depth frequency and f^1.5 above it, both exactly. computed by solving, not by drawing. Eddy-current loss in a 0.35 millimetre lamination held at a mean flux of 1 tesla, against frequency, with the classical uniform-flux expression drawn beside it and the local exponent across the top. Below the frequency at which the sheet is two skin depths thick — 465 Hz here — the two agree and the exponent is 2.0000. Above it the flux is confined to a layer whose thickness falls as one over the square root of the frequency, and the exponent is 1.5000: three halves, exactly. It does not arrive there monotonically — it undershoots to 1.4847 at ξ = 3.28 and comes back up, which is the bounded cosine term the asymptotic statement drops. At 1.00 kHz the classical term is already 1.74 times the truth. The solve and the closed form agree to 7.5e-3 per cent across six decades. Two windings, and the band between them

The current inside the iron

Every core-loss law has an eddy term of the form d²f²B²/6ρ, and it is derived by assuming the flux is uniform across the lamination — an assumption that is a frequency and that the expression does not carry. Solved instead as a diffusion, the exponent is exactly 2 below the frequency at which the sheet is two skin depths thick, exactly 1.5 above it, and it undershoots to 1.485 on the way. For a 0.35 mm sheet the crossing is 465 hertz, so at a kilohertz the classical term is 1.74 times the truth and at ten kilohertz it is forty-seven times.

 The period's spread is 1.20 times what counting two crossings gives. computed by solving, not by drawing. 19999 periods of a relaxation oscillator with 5 mV rms of noise on its thresholds, computed from the exact flip instants rather than marched, at β = 0.5. The measured standard deviation is 3.4 ns and the closed form — three partial derivatives of the period with respect to the three draws it depends on — gives 3.4 ns. The estimate that counts two threshold crossings and divides the noise by the slope at each gives 2.83 ns, which is 17 per cent low. The curve is the closed form's Gaussian, drawn on the measured histogram rather than fitted to it. Circuits that do a job, and the range they do it over

The decision taken where the ramp is slowest

A relaxation oscillator decides at its thresholds, and a threshold is the one place on a charging exponential where the slope is smallest. Noise there costs 3.399 units of period against the 2.828 that counting two crossings gives, because a draw moves the crossing it is armed for and the level the next ramp starts from. Consecutive periods share that draw, so they are positively correlated and the jitter accumulates at 3.771 per root period rather than at 3.399. And at a fixed frequency there is a best hysteresis: β = 0.648, where β·ln((1+β)/(1−β)) = 1.

 300 mirrors built to one design, with 2% device mismatch. computed by solving, not by drawing. Every pair in the population is a full Newton solve of the same netlist with two saturation currents drawn from a normal distribution, the Early conductances iterated to self-consistency for each. The mean is 3.836 per cent, which is the systematic error the rung below computed with identical devices (3.937 per cent) — the mismatch does not move it. The spread about it is 1.955 per cent, which is the device mismatch arriving with nothing dividing it, and the worst pair of the 300 is 8.43 per cent out. A design whose specification is the mean has specified the one mirror nobody has. Devices, and the amplitude they stop being linear at

The error that is a distribution

The rung below solved a mirror and separated two errors — one that falls with beta and one that does not. Neither is what limits a real mirror. Two transistors on the same die differ, a fractional difference in saturation current is a fractional difference in collector current with nothing dividing it, and the honest object is a spread rather than a number: mean 3.84 per cent, standard deviation 1.96, worst of three hundred 8.43. Degeneration divides it by one plus gm·R and stops at the resistors' own tolerance, and where it stops is a voltage — a hundred millivolts, containing nothing but the ratio of two tolerances.

 One capacitor moves three quantities, and the expression's own answer peaks by 1.18 dB. computed by solving, not by drawing. The bandwidth, the peaking and the total output noise of a 1.0 MΩ transimpedance stage against its feedback capacitor, swept from 0.30 to 4.20 times what the classical expression asks for. Over that factor of fourteen the bandwidth falls from 334 to 55.4 kHz, the total noise from 230 to 65 µV, and the peaking from 10.0 dB to nothing. The expression's own answer sits at 1.18 dB of peaking, 0.82× is where the loop reaches forty-five degrees, and √2× is where the response is flat — so the choice usually quoted as maximally flat is neither of the two conditions it is quoted for. The faint families are the same three quantities at 3 pF and 300 pF of diode: normalised this way they are one curve, so the diode sets the scale and the multiple sets the shape. Feedback, and the margin

The factor the expression leaves out

The classical compensation for a photodiode amplifier is quoted both as the forty-five degree choice and as the maximally flat one, and it is neither: it leaves 1.18 dB of peaking and 52.4° of margin. Flat is at exactly √2 times it — fitted at 1.4186 against 1.4142, at every detector from 3 pF to 1 nF. And the third quantity the capacitor is supposed to trade, the noise, does not move at all inside the signal band: three compensations spanning a factor of fourteen give 8.37 against 8.36 µV in a 4.36 kHz measurement and 230 against 65 µV over the whole plane.

 The load that takes the most power is 3.67 Ω, and the open-circuit voltage over the short-circuit current is 3.97 Ω. computed by solving, not by drawing. The terminal characteristic of a photocurrent with a junction across it, with the power along it drawn on the same axes and scaled to fill them. The most power, 80.99 W, is delivered at 17.244 V and 4.696 A, which is a load of 3.67 Ω. The incremental resistance of the source there — the negative of the characteristic's own slope — is 3.67 Ω, the same number to four figures. A straight line drawn between the two end points has a resistance of 3.97 Ω and would promise 24.81 W. The ratio of the true maximum to that promise is 3.264, and its reciprocal is the fill factor, 0.8160. Networks, and how a solve is checked

The load a curve recommends

The maximum power theorem says to match the load to the source's internal resistance, and for a straight-line source the internal resistance, the slope and the open-circuit voltage over the short-circuit current are one number. On a photovoltaic panel they are three: the slope at the maximum is 3.6717 ohms, the load there is 3.6717, and the open-circuit voltage over the short-circuit current is 3.9709 — and a quarter of the open-circuit voltage times the short-circuit current, which is what a straight line would deliver, is 24.81 watts against the 80.99 that is there. The theorem survives with the incremental resistance in place of the internal one, and it survives only where the characteristic has a slope: on a supply whose current limit folds back, the most power is delivered at a corner.

 An amplifier a hundred times the corner moves the unity-gain section's Q by 2.0%. computed by solving, not by drawing. A unity-gain Sallen–Key section designed for Q = 2 at 1.00 kHz, built with a one-pole amplifier, and its poles recovered by rooting the determinant. The section as drawn has two poles; as built it has three, and the pair is not where it was put. At a gain-bandwidth of a hundred times the corner — which is the rule of thumb — the quality factor is 1.97 per cent high and the pole frequency is 1.97 per cent low. The two are the same number with opposite signs over the window where the amplifier is well clear of the section and its own pole is still resolvable, and the number is the designed quality factor, 2, divided by the ratio: 2.00 per cent at 100 times the corner. Filters, measured not tabulated

The Q the amplifier decides

A second-order section's quality factor is set by a capacitance ratio and its pole frequency by a product of four passive values, and neither expression contains the amplifier. Build it with one that has a gain-bandwidth a hundred times the corner — the usual rule — and the Q comes out 1.97 per cent high while the pole comes out 1.97 per cent low, the same number in both directions, and the number is the designed Q divided by the ratio. A fifth-order half-decibel Chebyshev built that way has 2.1 decibels of ripple.

 A whole number of steps: 1, 2, 3 and 4 agree to 7% and a step and a half is 1.9× worse. computed by solving, not by drawing. The upper panel is the share of the quantisation error sitting in harmonics of the input against the amount of dither added — the upper curve the largest share anywhere on the amplitude sweep, with the spread across the five tones each point averages drawn as a bar, and the lower curve that sweep's mean. Undithered the worst is 76.5 per cent. It falls steeply up to one whole step (4.70 per cent at three quarters, 0.34 at one) and then stops improving — but only AT whole steps. The sweep means at 1, 2, 3, 4 steps are 0.283, 0.297, 0.297, 0.301 per cent, flat to 7 per cent; at 1.5, 2.5, 3.5 they are 0.526, 0.345, 0.311, each above both whole steps beside it. The lower panel is what each costs in signal-to-noise ratio, with 10·log₁₀(1 + L²) drawn through it — the measurement is that curve to 0.118 dB everywhere, so the price is known in advance and only the benefit has to be measured. The choice is a corner and a comb: nothing here is minimised, something stops improving, and between the places where it has stopped it is worse again. Where a signal becomes a number

The dither that is a decision

One whole least significant bit is quoted everywhere as the dither, which makes a decision look like a constant. Swept, the axis is a corner and a comb. An eighth of a step leaves 64.5 per cent of the error locked to the signal and one whole step leaves 0.34; above that the sweep mean is 0.283, 0.297, 0.297 and 0.301 per cent at one, two, three and four steps and 0.526 at a step and a half, which fails at exactly the small amplitudes dither exists for. The price is 10·log₁₀(1 + L²) to 0.118 of a decibel, and four steps cost 12.41 for nothing.

 Three averagers passing the same noise, and three different half-power points. Integrated by eight-point quadrature on every lobe, with the tail past the last lobe in closed form. A mean over 20 ms has the response sin(πfT)/(πfT), and the area under its square is 25.000000 Hz against the 25 Hz of 1/(2T) — the brick wall drawn shaded. A mean over the window passes half its power at 22.147 Hz, so the noise bandwidth is 1.12880 times that frequency; A one-pole averager passes half its power at 15.915 Hz, so the noise bandwidth is 1.57080 times that frequency; Two means in cascade pass half their power at 23.919 Hz, so the noise bandwidth is 1.04521 times that frequency. Every curve drawn encloses the same area; they differ in where they spend it. The floor, which bounds from below

The filter an average is

A mean taken over a window is a filter, and the area under its squared response is exactly one over twice the window — 25 hertz of noise bandwidth for twenty milliseconds, passing half its power at 22.15. Built to the same noise, a one-pole averager passes half its power at 15.92 hertz and takes 2.33 times as long to settle to one per cent, and two means in cascade pass half at 23.92 and take 1.25 times as long. Between its nulls a mean rejects the mains no better than the one-pole does, and one per cent off a null it rejects it by forty decibels however many cycles the window holds.

 A capacitor across the upper divider resistor removes the output capacitor's resistance floor. computed by solving, not by drawing. Two series resistances against the capacitance across the upper divider resistor: the smallest the loop tolerates at 45° of margin (lower curve), and the one that gives the smallest droop after a 100 mA load step (upper). With no capacitor they are 939 mΩ and 885 mΩ — the second BELOW the first, which is the conflict this design has: the best transient is one the loop refuses. The floor falls as the capacitor grows and between 500 and 836.5 pF it leaves the sweep altogether, so every series resistance down to a milliohm is stable. Past about 5000 pF the floor climbs back and overtakes the optimum again. The shaded band is where the design a transient wants is one the loop allows. Circuits that do a job, and the range they do it over

The floor a second capacitor removes

Two requirements pulling one capacitor found a regulator whose best transient is one it must not be built with: below 939 milliohms of output-capacitor series resistance the loop has under 45 degrees of margin, and the droop is smallest at 885. The capacitor across the upper divider resistor, added for the reference's sake, dissolves that conflict. At 836.5 picofarads the 45-degree floor leaves the sweep entirely — every series resistance down to a milliohm is stable — and the droop falls 40 per cent at the same time. The band of capacitances that do it runs from 100 picofarads to 5 nanofarads, and above it the conflict returns.

 The "offset" a switch leaves moves by 1.863 mV across a volt of signal. computed by solving, not by drawing. The charge a switch leaves on a 1 pF hold capacitor, as the input it was sampling changes. It is W·L·C_ox·(V_gs − V_th) with the clock's high level for the gate, so it falls as the input rises for two reasons at once: less gate drive, and a threshold that rises with the source potential through the body effect. The dashed line is the part that does not know what the signal is — the clock coupling through the gate overlap capacitance, -1.649 mV — and it is the only part of this that is honestly an offset. The curve stops where the switch does: an n-channel device passes nothing within a threshold of its own gate, and the model refuses rather than returning a zero. Filters, measured not tabulated

The offset that knows the signal

The charge a switch leaves on the capacitor it was sampling is the gate oxide capacitance times the channel area times the overdrive, and both terms in that overdrive depend on the input — one directly and one through the body effect. So the 5.67 mV a data sheet would call an offset moves by 1.86 mV across a volt of signal, which is a gain error of 0.186 per cent and only 3.4 µV of anything else. Opening the summing-node switch first divides the gain error by the amplifier's own gain, exactly, and what is left goes from twenty-nine times the sampled-noise floor to nineteen times below it.

 The guard leaves a negative resistance, and it reaches −1.59 kΩ. computed by solving, not by drawing. The magnitude of the conductance a source sees looking into the input, guarded and not, with 100 pF of cable and a 1.00 MHz amplifier. The unguarded input's conductance is positive everywhere — a capacitance to ground and a leakage to a rail are both losses. The guarded one is negative above 0.0404 Hz, and its magnitude rises as the square of frequency: −15.9 MΩ at 10 kHz, −161 kΩ at 100 kHz, −3.18 kΩ at a megahertz. Above the amplifier's gain-bandwidth product it flattens at ωₜ·C, which is −1.59 kΩ. That is the same input the guard raises to 10¹⁸ Ω at direct current, and nothing about the leakage the guard was installed for appears in it: the negative resistance is a product of the amplifier's bandwidth and the cable it is driving. Measurement, which is a circuit on a circuit

The sign of what the guard gives back

A guard ring is sold on two numbers and they are both about magnitudes: a teraohm of leakage multiplied to 10¹⁸ ohms, and a hundred picofarads of cable bootstrapped out of the way. The guard is also driving that capacitance with a copy of the input that lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with the voltage across it. What the guarded input presents is a negative conductance rising as the square of frequency — −15.9 megohms at ten kilohertz, −3.18 kilohms at a megahertz, flattening at the gain-bandwidth product times the capacitance — and a faster amplifier makes it worse.

 Over one decade a constant is out by ±29.76 mV and a drop plus a resistance by ±8.002 mV. computed by solving, not by drawing. The forward drop of a pure exponential junction over the decade from 1e-3 to 1e-2 amperes, with the best constant drop and the best drop-plus-resistance drawn across it. Both are fitted minimax — the model whose WORST error over the window is smallest, which is what a design has to tolerate — rather than by least squares. The constant is 684.6 mV and is out by ±29.76 mV; the line is 656.2 mV plus 6.61 Ω and is out by ±8.002 mV, which is 3.72 times better. The best constant needs no search: it is the midpoint of the window's highest and lowest voltage, and its error is half their difference. Where the models stop

The straight line between two models

Between a constant seven-tenths of a volt and an exponential sits the model a designer actually reaches for: a drop plus a resistance. Fitted so that its worst error over a decade of current is as small as it can be, it is out by ±8.00 millivolts where the best constant is out by ±29.76 — and both numbers are the same over every decade, because a decade of a logarithm is the same shape wherever it is taken. On a real two-mechanism junction the line does best in the top decade, ±8.71 against a constant's ±58.53, because a series resistance is exactly the term the line has and the constant has none. And the resistance the fit returns is the part's plus 701 milliohms that is not there.

 A track needs about three heights of copper beside it, and it is the resistance that says so. computed by solving, not by drawing at 100 MHz, each point a strip solve of its own on a 50 mm plane of the same area, moved sideways. The horizontal axis is where the track's centre sits relative to the plane's edge, in units of the track's height above it; negative is a track hanging past the edge with no copper beneath it. With the centre directly over the edge the loop's inductance is 1.161 times its centred value and its resistance 2.90 times, because the return has to crowd into the last few hundred micrometres of copper. Three heights in, the inductance is 1.006 times and the resistance 1.09; ten heights in, both are within 0.7 per cent. Three heights past the edge the inductance is 1.82 times. At direct current every point on this axis is exactly one, because the copper has been moved and not removed. Lines, where a wire has a length

Where the plane runs out

The corner that is three decades wide solved a return current over a plane that extends well past the track on both sides. Where it does not, the two costs arrive at opposite ends of the band: at direct current a plane that ends under the track costs 27 per cent of inductance and not one part in a million of resistance, and above the band it costs 16 per cent of inductance and 199 per cent of resistance. Three track-heights of copper beside the track removes almost all of both, and the number three has no millimetres in it — sixteen times the whole cross-section gives the same ratios to a part in a billion.

 The lower corner, estimated from short-circuit time constants. computed by solving, not by drawing, on three coupling capacitors and three shunt resistors at 28 spreads of the capacitor values. Each capacitor's short-circuit time constant is its own value times the resistance between its terminals with the other two shorted; the sum of the RECIPROCALS is 60000 s⁻¹ here, and it equals the ratio of the denominator's two highest coefficients to 1.4e-12 and the negated sum of the poles to 1.4e-12. That much is the same theorem as the other end. What is an estimate is the corner: 9549 Hz against a measured 8192 Hz, high by 16.6%. It is high at every spread drawn — the error reverses direction with the construction, so both ends of a band are estimated inwards. Before the steady state

Shorted instead of opened, and the error changes sign

The same construction with the other capacitors shorted rather than removed sums the reciprocals of the products, and that sum is the ratio of the denominator's two HIGHEST coefficients — the negated sum of the poles, exact to a part in 10¹². Divided by 2π it estimates the lower corner of a band, and it is 16.6 per cent HIGH with three coupling capacitors and never once low. Two settings of the slider give the same three time constants in a different order, the same sum, and corners two per cent apart.

 1 pF across a 50 Ω line: a dip of 0.320 V and an area of 25 ps. computed by solving, not by drawing as a cascade of two-ports, with a raised-cosine edge of 59 ps sent into it. The incident edge is the faint curve; what comes back is the shaded dip and what goes on is the third. The dip reaches -0.3202 V and its area is 25 ps, which is Z₀C/2 to a part in ten thousand. Driven by an edge fifty times faster the same cascade returns the single exponential the closed form gives, to 9.3e-6 of a volt. The transmitted edge leaves at 81.1 ps, against 59 ps arriving. Lines, where a wire has a length

The dip whose area is fixed

A picofarad across a 50 Ω line makes a dip in what comes back. Its depth is 0.833 volts to a six-picosecond edge and 0.0196 volts to a nanosecond one, forty-two times less; its area is 25 picoseconds to both, to six parts in a hundred thousand, because the area is Z₀C/2 and contains nothing about the edge. Two half-picofarad discontinuities too close to tell apart read as exactly one picofarad, and so do two far enough apart to be separate — the area is additive where the depth is not. And what a reflectometer calls the capacitance of an impedance step is the step's real excess capacitance times 1 + Z/Z₀.

 The load that may be complex, and what a resistor alone gives up. computed by solving, not by drawing, with the best load searched over BOTH of its parts on the solved network rather than substituted. Against a source of 50 Ω + j100 Ω the search returns 50.00 − j100.0 Ω — the conjugate — which delivers the available 500.00 mW at exactly 50.00% efficiency, and does so at every source reactance. A load that may only be a resistance takes 2/(1 + √(1 + x²)) of that, matching the solved answer to 4.4e-16 at ten reactances — and at x = 2 that fraction is exactly the golden ratio less one, 0.618034. The resistor-only load is the MORE efficient of the two at every non-zero reactance, rising towards one while the conjugate match sits at a half for ever, so the familiar "maximum power at fifty per cent" belongs to the conjugate and not to the load. Power, and the part that does no work

The load that may be complex

Freed of the constraint that it be a resistance, the best load is the source's conjugate — found here by a two-dimensional search on the solved network rather than assumed — and it takes the available power at exactly fifty per cent efficiency whatever the source's reactance. A load that may only be a resistance takes 2/(1 + √(1+x²)) of that, and at a source reactance of twice its resistance that is exactly the golden ratio less one, 0.618034. The resistor-only load is also the MORE efficient of the two, rising towards one while the conjugate sits at a half for ever.

 Two large-signal limits, each alone and then both, at 20 mA and half of it. computed by solving, not by drawing. Four marches of one netlist at each load: neither limit, the input pair's tanh alone, the output stage's 20 mA alone, and both, driven by a 10 mA step. The three curves are each limit's departure from the linear march and the departure with both present; the faint line is the two singles added. Both lies on the sum and a little above it — 1.112 times it at 0.47 nF and 1.022 at 47 nF — so the limits are present together rather than taking turns. Where the two singles cross, near 10 nanofarads, the pair costs 1.88 times what the worse of them costs alone. Feedback, and the margin

The load that neither limit owns

Nine rungs of this argument asked which of an amplifier's two large-signal limits binds, and drew the load capacitance where the answer changes hands. Both are present at every load: the excursion with both in the netlist is the two departures added and between 2 and 12 per cent more, never the larger of them. So the crossing is not a handover but a maximum — at 12 nanofarads the pair costs 2.084 times what the worse of them costs alone, against 1.35 at 2.2 nanofarads and 1.07 at 47 — and the same peak sits on the resistance axis at 20 ohms and the gain-bandwidth axis at 50 megahertz.

 The images a zero-order hold leaves, at 0.222 of the sample rate. computed by solving, not by drawing. A 10.67 kHz tone held at 48 kHz, with every line read out of a transform of the staircase itself. The sampled spectrum repeats at every multiple of the clock and the hold multiplies all of it by one sinc, so each image survives scaled by the sinc at its own frequency: the fundamental at -0.72 dB, the largest image (1fs−f, 37.3 kHz) at -11.60 dB, which is 10.88 dB of rejection. The sinc's nulls are exactly at the multiples of the clock and the two first-order images straddle the first of them without touching it — so the hold's rejection is 25.6 dB for a tone at 0.05 fs and 1.74 dB for one at 0.45, falling to nothing at half the clock. Measured and closed form agree to 0.026%. Where a signal becomes a number

The nulls are where nothing is

A zero-order hold multiplies the whole repeated spectrum by one sinc, so it attenuates every image at the image's own frequency and its nulls land exactly on the multiples of the clock. Nothing is ever at a null: the two first-order images straddle it, and the closer the signal comes to half the clock the closer they come to each other. The hold gives 25.6 dB of image rejection to a tone at a twentieth of the clock, 1.74 dB at 0.45 of it, and nothing at all at half — which is the frequency the band most needs it at.

 The corner a 2 nH shunt has, against the current it is sized for. computed by solving, not by drawing. A shunt held at the best burden voltage of 7.75 mV has R = u⁄I, so its own 2 nH of series inductance puts a corner at u⁄(2πLI) — 616 kHz at an ampere and 6.16 kHz at a hundred, for the same piece of metal. The optimum that contains no current at all therefore hands the bandwidth a current dependence: the corner falls in exact proportion. At 1 A the shunt is 7.75 mΩ with a time constant of 258.2 ns, so a 10 ns edge is read 2.58e+3% high and a 1 ns edge 259.20 times too large. A resistor and a 258.2 pF capacitor across it — the value found by search on the solved response, agreeing with L/(R·Rc) to 7.9e-5% — flatten the reading to 7.8e-5% across six decades, and a fifth too much makes it ten times worse. The dots are the corner bisected on the solved impedance rather than taken from R/2πL. Measurement, which is a circuit on a circuit

The optimum that hands back a bandwidth

The best burden voltage across a shunt is 7.75 mV and contains neither the current nor the resistance, which is what made it worth having. A shunt has two nanohenries whatever it is made of, so holding the burden fixed fixes the resistance at u*/I — and the corner R/2πL then falls in exact proportion to the current: 6.16 MHz at a tenth of an ampere, 616 kHz at one, 6.16 kHz at a hundred. A resistor and a 258.2 pF capacitor across it, found by search on the solved response, flatten the reading to 8×10⁻⁵ per cent across six decades.

 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.

 A half-decibel filter that meets one decibel at 25 °C and does not at 89. computed by solving, not by drawing. The passband ripple of a fifth-order 0.5 dB Chebyshev built from three Sallen–Key sections, against temperature, with a part whose gain-bandwidth at 300 K is 300 kHz against a 1 kHz corner. Not one passive component has a temperature coefficient in this model. The ripple is 0.824 dB at −40 °C and 1.049 at +125, and it crosses a one-decibel specification at 88.6 °C — a boundary in temperature, which every other edge in this collection is not. The dashed line is the same filter with a tail current proportional to absolute temperature: 0.916 dB at both ends, and no crossing anywhere. Filters, measured not tabulated

The ripple that is a temperature

An amplifier a hundred times the corner leaves a section's quality factor two per cent high. That two per cent has a temperature in it: with a tail current a resistor sets, the transconductance falls as one over absolute temperature, and a fifth-order half-decibel design whose passives have no temperature coefficient at all goes from 0.82 decibels of ripple at minus forty to 1.05 at a hundred and twenty-five — crossing a one-decibel specification at 89 °C. With the other bias it does not move at all.

 Two leads nobody counts cost 1995 parts per million, and two more leads remove all of it. computed by solving, not by drawing. The error in the reported strain against the resistance in each of the two excitation leads, at 1000 µε on a quarter bridge of 350 Ω excited at 10 V. With four wires the instrument takes the excitation to be the supply's voltage, so every reading is scaled by R/(R + 2r): 1995 parts per million at 0.35 Ω, 54029 at 10. Two more leads brought back from the bridge's own terminals, carrying only the instrument's input current, leave 3.5e-4 parts per million. Exciting with a current instead of a voltage does the same thing with no extra leads at all, because the lead resistance is in series with a source that does not care. Circuits that do a job, and the range they do it over

The two leads nobody counts

The leads that are in the bridge moved a lead out of the changing arm and turned a 999-microstrain error into nothing. The two leads carrying the excitation are still there, and they scale every reading: 0.35 ohms each on a 350-ohm bridge is 1,995 parts per million, exactly −2r/(R + 2r), the same on a full bridge as on a quarter one, and drifting 0.153 microstrain over twenty kelvin — twice what the three-wire fix left behind. Two more wires brought back from the bridge's own terminals leave 0.00035 parts per million. So does exciting the bridge with a current, which needs no extra wires at all.

 What flatness costs, in the two places it can be bought. computed by solving, not by drawing. The hold's sinc across a band ending at 0.40 of the sample rate, and the same sinc with a one-zero one-pole shelf fitted to its reciprocal over that band. The droop to be removed is 2.420 dB. Corrected digitally the band comes flat exactly and the flat level sits 2.420 dB below what the uncorrected converter gave at direct current, because nothing may exceed full scale — the price is the disease. Corrected in analogue the band comes flat to 0.2308 dB and the shelf is still rising where the images are, so the worst image at 0.60 fs comes up by 3.799 dB, which is 1.57 times the droop it removed — and that boost is between 3.4 and 4.7 dB at every band on the slider, while the droop it cures runs from 0.58 to 3.75. Neither correction changes the signal-to-noise ratio, because the droop never cost any. Where a signal becomes a number

Flatness, and the two currencies it is bought in

The hold's droop takes the signal and everything arriving with it down together, so it costs no signal-to-noise ratio at all — a fact that is never stated and settles what correcting it can possibly be worth. Corrected digitally the price is headroom and is exactly the disease: 2.42 dB of flatness for 2.42 dB of output level. Corrected by an analogue shelf the price is image rejection and is nearly a constant: between 3.4 and 4.7 decibels whatever the band, so it is eight times the droop at a fifth of the clock and nine tenths of it at forty-nine hundredths.

 Four threshold densities, all matched to the same measured coercivity. computed by solving, not by drawing. Every hysteresis figure below this one used a uniform density of play-operator thresholds, and the reason given was that a uniform density makes the small-signal loss law come out cubic — Rayleigh's law, which is what soft ferrites do. That is a real constraint and it is a weak one. These four are all bisected to a measured coercivity of 9 amperes per metre on a deeply driven loop, and they need threshold spreads from 30.1 to 76.0 amperes per metre to get there. The slider moves the coercivity they are all matched to. Two windings, and the band between them

The distribution the bench cannot see

Six rungs of this ladder assumed a uniform density of operator thresholds, and the reason given was that a uniform density makes the small-signal law come out cubic. That reason is weaker than it looks: every density finite at the origin does the same. Matched on a measured coercivity, four distributions whose spreads differ by 2.5 times give the same remanence to 0.6 per cent, the same loop area to 1.0, and the same branch-slope ratio to 0.54 — and small-signal exponents from 2.81 to 3.81. The major loop cannot see the shape and the small signal cannot see anything else.

 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. Frequency, which is the same solve

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 direct-current error a bigger feedback resistor does not fix. computed by solving, not by drawing, at direct current with the amplifier in the netlist. A 1.0 nA photocurrent into 100 MΩ gives -0.100 V. The bias current flows in the same resistor, so its contribution is the ratio of the two currents — 0.1000% at 25 °C, and the same percentage at 1 MΩ and at 1 GΩ, which is checked by changing the resistor rather than by reading an expression. The amplifier's 100 µV of offset behaves the other way: it is multiplied by one plus the resistor over the diode's own leakage, so it grows with the resistor and with temperature. Both terms double every ten kelvin, for two different reasons, and at 85 °C they are 6.40% and 0.740% against 0.100% and 0.110% at 25. Feedback, and the margin

The error a bigger resistor cannot help

Every other quantity in a photodiode amplifier improves with a larger feedback resistor: the signal grows as R and the resistor's own noise as √R, so the ratio goes as √R. The bias current does not behave like either — it flows in the same resistor the signal does, so its contribution is the ratio of the two currents with the resistance cancelled, 0.1 per cent at 25 °C and the same 0.1 per cent at a megohm and at a gigohm. The offset behaves the other way, growing as one plus the resistor over the diode's own leakage, and at 85 °C the two are 6.40 and 0.74 per cent against 0.10 and 0.11 at room temperature.

 The floor a biased resistor is not standing on. computed by solving, not by drawing. A 100 kΩ resistor with 10 V across it. Its Johnson noise is 40.02 nV/√Hz and does not depend on the voltage; its excess noise is a 1/f density proportional to that voltage, 0.1 µV per volt per decade here, and the two are equal at 271 Hz. Below that the resistor is dominated by a generator that is a property of how it was made rather than of its resistance, and the crossover moves with the SQUARE of the voltage. Splitting the same total resistance into eight parts in series leaves the Johnson noise exactly where it was and divides the excess by √8 = 2.828, taking the total at 1 Hz from 660.2 to 236.4 nV/√Hz. With no voltage across it the second generator is absent rather than small. The floor, which bounds from below

The floor that is only a floor while nothing flows

Johnson noise depends on a resistance and a temperature and on nothing else, which is what makes it a floor. A real resistor has a second generator that depends on how it was made and on the voltage across it: 0.1 µV per volt per decade for a metal film, rising as 1/√f. On 100 kΩ with 10 V across it the two are equal at 271 Hz, and the crossover moves with the SQUARE of the voltage — 0.678 Hz at half a volt, 2.44 kHz at thirty. Splitting the same total resistance into eight parts in series divides the excess by exactly √8 and leaves the Johnson noise where it was.

 A permittivity quoted as one number falls 0.645 across five decades. computed by solving, not by drawing. The real part of the relative permittivity against frequency for FR-4 (ε′ = 4.4 at 1 GHz, tanδ = 0.02), from a continuum of relaxations spread uniformly in log-frequency — the arrangement that makes the loss tangent flat. The dashed line is the single number a datasheet quotes. FR-4: 4.787 at 1 MHz and 4.142 at 100 GHz, a fall of 14.7 per cent. Nothing here is fitted: the slope is what a flat loss tangent forces. Lines, where a wire has a length

The permittivity a loss forbids

A datasheet quotes a relative permittivity and a loss tangent as two independent numbers, and they are not two numbers. A material that dissipates has a permittivity that falls logarithmically with frequency at a rate its own loss fixes — 0.129 of permittivity a decade for FR-4, so the 4.4 quoted at a gigahertz is 4.79 at a megahertz and 4.14 at a hundred. Three hundred millimetres of track loses 62.9 picoseconds of delay between 100 MHz and 10 GHz, which a constant permittivity puts at 1.3; and the constant-permittivity model smears an edge backwards, taking 180 picoseconds to reach half height and 133 more to reach nine tenths.

 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.

 The series resistance that makes the ripple smaller. computed by solving, not by drawing, marched with the diodes in the netlist. A 1000 µF reservoir with 30 mΩ of its own series resistance. The resistance adds a step of ESR times the diode's peak current to the output and at the same time limits that peak current, and the two nearly cancel: the ripple has an interior minimum of 1.330 V at 17.0 mΩ, BELOW the 1.331 V a perfect capacitor gives, and rises to 1.711 V at an ohm. What the resistance buys monotonically is the peak current: the crest factor falls from 13.37 to 6.550 at an ohm, which more than halves the current that sizes the transformer, for 380 mV of mean output and a root-mean-square diode current that falls from 0.4717 A to 0.3484. The textbook ripple expression says 1.568 V here and moves by 2.4% across the whole axis, because it has no term for a series resistance at all. Circuits that do a job, and the range they do it over

The resistance that lowers the ripple

Two earlier essays here marched a reservoir with a perfect capacitor. A real one has tens of milliohms of its own, and the obvious expectation — that the resistive step it adds makes the ripple worse — is wrong in an interesting direction: the resistance also limits the charging current, and the ripple has an interior minimum of 1.3303 V at 17.0 mΩ, below the 1.3312 V a perfect capacitor gives. What the resistance buys monotonically is the peak current, which falls from 13.37 times the load's to 6.55 at an ohm, for 380 mV of mean output.

 A diode thermometer measuring its own sense current, at 600 K/W. computed by solving, not by drawing. The same fixed point as the core and the thermistor, on a junction: the dissipation is I·V and V falls with temperature, so the loop gain is negative and the equation has one root at every current. What it costs is two errors. The junction sits above ambient by 0.29 millikelvin at a microamp and 4.26 kelvin at ten milliamperes, which a calibration removes; and a kelvin of ambient produces less than a kelvin of junction, by 1/(1 − R_th·dP/dT), which it does not. After a calibration at 25 degrees the reading at 85 is out by -583 millikelvin at ten milliamperes and -0.09 at a microamp. The coefficient itself moves too — -2.403 against -1.613 millivolts a kelvin — so a quoted tempco carries a sense current as well as a junction. Devices, and the amplitude they stop being linear at

The sensor inside its own answer

A junction driven from a current source cannot run away, because its forward voltage falls with temperature and its loop gain is therefore negative. What that costs is a thermometer that is warmer than what it is measuring by 4.26 kelvin at ten milliamperes, and — the part a calibration cannot remove — under-reports every change in ambient by 9,584 parts per million, because the sense current's own dissipation falls as the reading rises. Calibrated at 25 degrees, it is out by 583 millikelvin at 85.

 Three dividers in a row, and where the error actually is. computed by solving, not by drawing. Three two-resistor dividers cascaded with nothing between them. The product of their ratios is 0.1250 and the solved output is 0.076923, 38.5% low at a staircase of ×1. Decomposed stage by stage with the rest of the chain in place — and the product of those three is the answer exactly — they are 0.3846, 0.4000, 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is at the front, which is the opposite of where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion, fitted exponent -0.993, so a decade a stage is within 4.88% and two decades within 0.50% — which is why a chain that has to be right is built as a staircase and not out of one value repeated. Networks, and how a solve is checked

The stage that is wrong is the far one

Three identical ten-kilohm dividers in a row give 0.076923 rather than the product of their ratios, 0.125 — 38.5 per cent low. Decomposed stage by stage with the rest of the chain in place, and the product of those three is the answer exactly, they are 0.3846, 0.4000 and 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is entirely at the front, which is the opposite end from where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion — fitted exponent −0.993.

 What an oversampling ratio buys, and at two different rates. computed by solving, not by drawing. A 20 kHz band on a 48 kHz base clock, interpolated by ratios from 1 to 64. The hold's droop at the band edge falls with the SQUARE of the ratio — the fitted exponent over six doublings is -2.0113 — from 2.640 dB at the Nyquist rate to 0.0097 at sixteen times it. The nearest image moves out with the FIRST power, exponent 1.0222, from 1.40 times the band edge to 37.4. So one decision buys two things at rates differing by a factor of two in the exponent, and the third quantity — the poles a reconstruction filter needs for sixty decibels — collapses from 20.5 to 3.21 by a ratio of four alone. Where a signal becomes a number

One knob, and the two exponents it turns

Oversampling is quoted as buying one thing and buys two that improve at different rates. The hold's droop at the band edge falls with the SQUARE of the ratio — fitted exponent −2.011 over six doublings, from 2.640 dB at the Nyquist rate to 0.0097 at sixteen times it — while the nearest image moves out with the first power, exponent 1.022. The third quantity, the poles a reconstruction filter needs for sixty decibels, collapses from 20.5 to 3.21 by a ratio of four alone, because it is a logarithm of the second.

 The winding window solved in two dimensions, copper filling 100% of it. computed by solving, not by drawing. The grey frame is iron of infinite permeability, which in this formulation is a Neumann boundary — flux enters it at right angles and pays nothing. The thin curves are flux lines, which are contours of the vector potential, so equal spacing is equal flux. The copper is shaded by its own share of the loss. At 100 per cent fill the solved ratio is 16.280 against Dowell's 16.382, and the difference is entirely the flux that curls round the ends of the foils — which the one-dimensional model has no way to hold. Two windings, and the band between them

The assumption that is a geometry

Every alternating-resistance number this collection has computed for a winding rests on one sentence — the field is parallel to the layers everywhere — and the sentence has never been tested, because testing it needs a field. Solved as one, a portion of foils that fills its window returns Dowell's expression to 0.155 per cent; the same copper filling a quarter of it returns 9.00 against the expression's 16.38, and dissipates 0.528 watts a metre against 0.232. The ratio falls by 45 per cent and the loss more than doubles.

 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 band closes over a stage's own bias at 7.93 microns. computed by solving, not by drawing. The overdrive over which the square law is within 1.0 per cent, against the channel length that sets it — the velocity-saturation voltage is Ec·L, so the axis is a size and not a bias. The shaded region is the band; the curve through it is the overdrive at which the square law is exact, which exists at every length because the subthreshold and velocity-saturation errors have opposite signs. Both edges move: the lower one from 79.6 mV to 252.9 mV and the upper from 81.2 mV to 1200.0 mV, so the band is a factor of 1.021 at 0.050 µm and 4.7 at 30 µm. The fourth curve is the overdrive a common-source stage with a fixed gate voltage and a fixed source resistor solves to, which barely moves at all; it leaves the band at 7.929 microns and is outside it for every shorter device. What the square law would have said about that stage is the last two rows: 220 per cent too much current on the 0.050 µm device and 53 per cent too much efficiency, against 0.15 and 0.66 per cent at 30 µm. Devices, and the amplitude they stop being linear at

The length that is a voltage

The square law's band is closed from above by a parameter that is not a bias, a current or a temperature: it is the channel length, wearing a voltage's units. Swept, the band goes from a factor of 4.74 on a thirty-micron device to 1.021 at fifty nanometres — and the overdrive a stage actually biases itself to barely moves at all, so the two cross at 7.93 microns and every shorter device is biased outside the band. The band was also measured in the wrong quantity: the square law is exact in the current somewhere at every length, and its error in the transconductance is never below 42.5 per cent at fifty nanometres and reaches one per cent only above 6.502 microns.

 The same 100 mΩ in the winding instead of the capacitor: 1.317 V of ripple against 1.335 V, at the same crest factor of 11.15. computed by solving, not by drawing, marched with the diodes in the netlist: a 1000 µF reservoir behind a centre-tapped rectifier, with a series resistance from 1 mΩ to 1 Ω placed either in the capacitor or in each half-winding. The crest factor is the same in both places to two parts in a thousand at every resistance — 13.34, 13.26, 13.03, 12.49, 11.15, 9.069, 6.548 — because both limit the charging current alike. The ripple is not: in the capacitor it has a minimum and rises to 1.711 V at an ohm; in the winding it falls throughout, to 1.172 V, against 1.331 V with no resistance. At an ohm the winding costs 534 mV of mean output and the capacitor 380 mV. The diode's root-mean-square current at an ohm is 0.3449 A with the resistance in the winding and 0.3484 A with it in the capacitor. Circuits that do a job, and the range they do it over

The resistance that belongs in the winding

A reservoir capacitor's series resistance lowers the ripple to a minimum of 1.3303 V at 17 mΩ and raises it past that. The same resistance moved into the transformer's winding limits the peak current by the same amount — the crest factor agrees to two parts in a thousand at every value from a milliohm to an ohm — and the minimum is gone: the ripple falls throughout, to 1.172 V at an ohm against 1.711 V in the capacitor. The two resistances each carry a current the other does not, and that one asymmetry decides where a deliberate one should go.

 An order buys 38.7 dB at 1.25× the corner and 56.6 at 1.67×. computed by solving, not by drawing. The degree equation for an order-5 elliptic filter, swept over the two quantities a designer sets. The horizontal axis is where the stopband is required to begin; the five curves are passband ripples from 0.01 to 3 decibels. Nothing on this page is a choice: pick a ripple and a transition width and the attenuation is decided. At a half decibel of ripple, order 5 gives 38.68 dB with the stopband beginning at 1.25 times the corner and 66.09 dB with it beginning at twice — a factor of two in transition width for 27.4 decibels. The vertical spacing between the curves is the ripple's own term and is the same at every transition width: relaxing from a half decibel to three buys 9.12 dB wherever it is spent. Filters, measured not tabulated

The selectivity that is not free

The filter trade is normally drawn with two quantities in it. It has three, and an order fixes a relation between all of them: at order five and half a decibel of ripple, a stopband asked to begin at twice the corner is worth 66.1 decibels and one asked to begin at 1.25 times is worth 38.7. The exchange is exact addition in decibels — relaxing the ripple from a half to three buys 9.12 dB wherever it is spent — and there is a fourth price nobody writes down: settling to a tenth of a per cent goes from 9.04 milliseconds to 23.23 while the overshoot does not move.

 A via's area changes sign at 44.7 Ω, and two impedances give its 0.5 pF and 1 nH back. computed by solving, not by drawing, as a cascade of two-ports: a via of 0.25 pF, 1 nH and 0.25 pF, met by an edge of 59 ps from reference lines of 20 to 150 Ω. The area under the reflection is −Z₀C/2 + L/2Z₀ at every impedance: a bump below 44.7 Ω, where the inductance's term is the larger, nothing at it, and a dip above. From 50 Ω the area is 2.5 ps of dip, which a single-capacitance reading calls 0.100 pF. From 50 and 75 Ω together the two areas give 0.5000 pF and 1.0000 nH. An error of 50 fs on each area moves them by up to 0.8% and 1.5%. Lines, where a wire has a length

The via two lines can weigh

The area under a reflection is a property of the discontinuity rather than of the edge, and for a via it is one number made of two: −Z₀C/2 from its pads and +L/2Z₀ from its barrel, with opposite signs. A via of half a picofarad and a nanohenry, seen from fifty ohms, leaves 2.5 picoseconds of dip — which a reading that assumes a capacitor calls 0.100 pF, a fifth of what is there. From fifty and seventy-five ohms together the two areas give 0.5000 pF and 1.0000 nH back, and fifty femtoseconds of error on each costs 0.8 per cent of the capacitance and 1.5 of the inductance. A second line at fifty-five ohms costs five times as much.

 An on-resistance 10% highest at mid-range: 74.16 ppm uncalibrated, 13.89 ppm once the straight line is removed. computed by solving, not by drawing, at direct current, at 41 levels across the range. A 0.5 Ω, 100 MΩ switch whose on-resistance moves by 10%, highest at mid-range, against the load. Uncalibrated, the worse of its closed error at the worst level and its open leak is least at 7.42 kΩ, 74.16 ppm — 13.72 bits, the lone switch's floor at its largest on-resistance, against 70.71 ppm for a constant one. With the gain and offset calibrated away, what is left of the closed error is the curvature, and against the leak it is least at 1.39 kΩ: 13.89 ppm, 16.14 bits. Where the models stop

The resistance that bends the signal

A switch of half an ohm and a hundred megohms has a floor of 70.71 parts per million, 13.79 bits, because its on-resistance and its off-resistance cannot both be small beside one load. Most of that floor is a gain error, and a gain error calibrates away. Give the on-resistance a realistic ten per cent of movement across the signal range and the uncalibrated floor slips to 74.16 parts per million, while the part no calibration can touch — the curvature — balances the leak at 13.89 parts per million, 16.14 bits, into 1.39 kilohms. The floor was never set by the on-resistance. It is set by how much the on-resistance moves, as its square root.

 The fringing field, and the turns standing in it. computed by solving, not by drawing. The slot on the left is the gap, cut through the centre leg to the core's own symmetry plane where the potential is zero. Flux crossing it does not stay in the slot: it bulges into the window and crosses the copper at right angles to the layers, which is the one direction Dowell's expression and every one-dimensional winding model assumes has no field in it. The turns are shaded by their own loss. The worst is turn 4, level with the gap, at 32.5 times its direct-current dissipation; the best is 1.39 times. Same wire, same current, same winding, and a spread of 23.4 between them. Two windings, and the band between them

The turns nearest the gap

A gapped inductor's flux does not turn a corner into the iron on its way out of the gap; it bulges into the window and crosses the copper at right angles to the layers. Four tenths of a millimetre from a one-millimetre gap, the worst turn of an eight-turn winding dissipates 37.5 times its direct-current loss and the winding as a whole 14.1 times. Move the same winding three millimetres further out and those become 2.9 and 2.4 — and the distance that governs it is 0.60 millimetres, which is not the gap length and does not scale with it.

 Two 1000 µF parts of 100 mΩ each against one 2000 µF part of 100 mΩ: 709 mV against 728 mV of ripple, and a crest factor of 14.41 against 12.61. computed by solving, not by drawing, marched with each capacitor behind its own series resistance. Two identical 500 µF, 60 mΩ parts give 1.330303 V of ripple and one 1000 µF, 30 mΩ part 1.330303 V — the same network. Two 1000 µF parts of a stated resistance each, against one 2000 µF part of the same resistance: at 10 mΩ, 704 mV against 704 mV of ripple and a crest factor of 17.40 against 16.93; at 30 mΩ, 704 mV against 705 mV of ripple and a crest factor of 16.51 against 15.48; at 100 mΩ, 709 mV against 728 mV of ripple and a crest factor of 14.41 against 12.61; at 300 mΩ, 758 mV against 867 mV of ripple and a crest factor of 11.45 against 9.490; at 1000 mΩ, 1.013 V against 1.334 V of ripple and a crest factor of 8.170 against 6.620. The pair's ripple is lower only where one part's resistance is past the ripple's minimum, and its peak current is higher at every resistance. Circuits that do a job, and the range they do it over

Two capacitors that are one

Two identical reservoir capacitors in parallel are not a new circuit to be marched: 500 µF at 60 mΩ twice gives 1.330303 V of ripple and so does 1000 µF at 30 mΩ once, to the sixth decimal. So a pair against one part of the same capacitance is one curve read at two resistances, and the pair always sits at the lower one — which lowers the ripple only past the curve's minimum, 709 mV against 728 at 100 mΩ a part, and raises the peak current at every value, a crest factor of 14.41 against 12.61. Mismatch the pair and the current still divides by capacitance, so a part with twice the resistance carries 2% less current and 1.93 times the heat.

 The bandwidth a windowed density is divided by, against the one bin an analyser assumes. Computed from the weights, from the coefficients in closed form, and out of the bins by Parseval — three routes agreeing to a part in 10¹². A record of N samples weighted by a window and transformed gives bins whose noise bandwidth is N·Σw²/(Σw)² in units of the bin spacing: 1 exactly for an unwindowed record, 1.5000 for Hann — three halves, exactly, because (a₀² + ½a₁²)/a₀² is (0.25 + 0.125)/0.25 with no record length in it — 1.3628 for Hamming, 1.7268 for Blackman and 2.0044 for Blackman–Harris. A density divided by one bin instead is wrong by the figure beside each bar. The bandwidth is not ordered with the sidelobe: Hamming's first sidelobe is 11.2 dB below Hann's and it costs less bandwidth as well, so the received sequence of windows fails at its second entry. What Hann buys is the far skirt instead, and that is a different measurement. The floor, which bounds from below

The bandwidth a bin is not

A density read off a transform is divided by a bandwidth, and the bandwidth belongs to the window rather than to the bin spacing — exactly three halves of a bin for a Hann window, so a density divided by one bin is 1.761 decibels high in power. Measured three ways that share only the weights. And the received sequence of windows fails at its second entry: Hamming's first sidelobe is 11.2 decibels below Hann's and it costs less bandwidth as well, with Hann collecting its debt at the fourth sidelobe and every one after it.

 A summing junction with 8 inputs on it, and the ceiling that fell. computed by solving, not by drawing by driving a current into the node and reading the voltage, with the input source zeroed, at each of 6 channel counts. With one input the node's ceiling is 909.5 Ω — the input and feedback resistors in parallel — and with 32 it is 31.15 Ω, a factor of 29.19 stiffer for 32 times the inputs — less than the count, because the feedback resistor is in the parallel combination too. Only the ceiling moves: the direct-current value and the whole rise are the same to 5.4 per cent at every count, because the open-loop impedance and the loop gain both scale as 1/N and cancel. So what the channel count does is bring the ceiling down to meet the rise earlier, and the per-pair leakage saturates rather than reaching unity — -0.8 dB at one input against -30.1 dB at 32. The total into every other channel together does not saturate: it is (N − 1)·Z/Rin, and with Z ≈ Rin/N that tends to unity — 0.966 at 32 inputs, 0.476 at two. A summing junction above its loop hands over essentially all of one input's signal to the others collectively, and the count decides only how it is divided. Feedback, and the margin

The node that does not care how many

Put thirty-two inputs on a summing junction instead of one and its ceiling falls from 909 ohms to 31.2 — the input resistors in parallel with the feedback resistor, twenty-nine times stiffer. Every other part of the curve is unchanged to five per cent, because the open-loop impedance and the loop gain both scale as one over the count and cancel. So the stiffer node buys nothing: the per-channel leakage saturates instead of reaching unity, and the leakage into every other channel together tends to exactly one, whatever the count.

 A switching load's comb in a 10 mV reading: every tooth the same height. computed by solving, not by drawing. The harmonics of a 50-per-cent trapezoid of 100 mA at 100 kHz with a 10 ns edge, each multiplied by the shared conductor's impedance at its own frequency. Below 79.6 kHz the conductor is 5.00 mΩ of resistance and the teeth fall as 1/n. Above it the conductor rises as n while the harmonics still fall as 1/n, so the two cancel and every tooth from 239 kHz to 10.6 MHz is 0.828 mV — flat to 5.4 per cent across 52 of them. The comb stops at 31.8 MHz, which is the edge's own corner, so the total of 12.6 mV is decided by how many teeth there are: fitted over five edge rates it grows as the edge time to the power -0.497, against the −½ that a quadrature sum of equal teeth gives. A faster edge raises no harmonic and adds teeth. What a slow measurement keeps is a different term entirely: the zeroth harmonic, which is the 100 mA average through 5.00 mΩ of resistance — 500.0 µV, with no inductance in it and no edge, against the comb's 12.6 mV. Measurement, which is a circuit on a circuit

Every tooth the same height

A shared return conductor's impedance rises a decade per decade above 79.6 kilohertz and a switching load's harmonics fall a decade per decade, so the two cancel exactly: every harmonic of a hundred-milliamp square wave puts 0.828 millivolts into a ten-millivolt reading, from the conductor's own corner up to the edge's. The comb is flat rather than falling, so the total is decided by how many teeth there are — and since they add in quadrature it grows as the square root of the edge rate, 39.7 millivolts at a nanosecond against 4.03 at a hundred.

 One rms value, and a peak that moves 41 per cent with a phase. A 10 V fundamental and a 3.33 V 3th harmonic, summed, with the harmonic's relative phase swept. The two are orthogonal over a period, so the root-mean-square value is the quadrature sum 7.45356 V at every phase — to 6.2e-15, which is the one case superposition allows and is a theorem rather than an approximation. The peak is not a sum of anything: it runs from 9.428 V to 13.33 V, a factor of 1.414, against a bound of 13.33 V that it reaches exactly at 180° — where this harmonic's own crest lands on the fundamental's, which for n ≡ 3 (mod 4) is that end of the sweep and not the other. At the far end it flattens the crest to 9.428 V, and there the peak has no closed form at all. So one signal has one heating and a headroom requirement that varies by 41 per cent, decided by a phase that no magnitude spectrum carries. Networks, and how a solve is checked

The peak that only has a bound

Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.

 The resistance that just stabilises it is 17 times smaller than the one that damps it. computed by solving, not by drawing. Two consequences of one base resistor, against how much of it there is, for a follower fed through 100 nH of wire with 47 pF on its emitter. The falling curve is the Q of the worst pole pair, rooted from the determinant so that no frequency grid is involved; the rising one is the output impedance the stage presents at low frequency. Bisecting on the SIGN of the pole's real part returns 69.7 Ω — at which the pair is stable with a Q of 1.7e+15, which is to say no damping and a peak whose height belongs to the arithmetic. A Q of one needs 1.15 kΩ, 16.6 times more, and that resistor takes the output impedance from 5.66 to 13.10 Ω — exactly R/(β+1) added, which is the bare follower's own expression with the base resistance in the place of the source resistance. Devices, and the amplitude they stop being linear at

What the cure at the base costs

Bisecting for the smallest base resistor that stops an emitter follower oscillating gives 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.

 A true-RMS converter reads noise low by 1/(16Bτ): 0.600% at Bτ = 10 and 672 ppm at 100, where a sine at fτ = 100 is 0.0396 ppm. Seeded, and measured. An explicit true-RMS converter — square, one-pole average of time constant τ, root — reading Gaussian noise of unit power, against the product of the noise's bandwidth and τ, from 1 to 100. Each point is 2²¹ samples at eight times the bandwidth, with the reading compared against the record's own root-mean-square and its standard error from batch means. Bτ = 1: 4.550% ± 150 ppm (band from zero), 4.081% (band of the same width about 3B). Bτ = 2: 2.598% ± 114 ppm (band from zero), 2.411% (band of the same width about 3B). Bτ = 5: 1.149% ± 75.4 ppm (band from zero), 1.098% (band of the same width about 3B). Bτ = 10: 0.600% ± 53.1 ppm (band from zero), 0.576% (band of the same width about 3B). Bτ = 20: 0.309% ± 39.1 ppm (band from zero), 0.295% (band of the same width about 3B). Bτ = 50: 0.127% ± 26.3 ppm (band from zero), 0.119% (band of the same width about 3B). Bτ = 100: 672 ppm ± 19 ppm (band from zero), 623 ppm (band of the same width about 3B). The dashed line is 1/(16Bτ), an eighth of the averaged square's variance, which the readings approach above Bτ = 10 and fall short of below it. A sine read by the same converter at the same product of frequency and τ is low by 3.96 ppm at 10 and 0.0396 ppm at 100 — the square of the noise's rate rather than its first power. Power, and the part that does no work

The noise a true-RMS meter reads low

A true-RMS converter reads a sine low by an amount that falls as the square of its frequency, and for anything but a slow sine that amount vanishes: 3.96 parts per million at ten times the averager's corner. Noise is not a sine. Its square fluctuates at every frequency down to zero, and the averager passes a share of that set by its own bandwidth against the noise's, so the reading is low by 1/(16Bτ) — the first power, not the second. Measured on seeded noise at Bτ = 10 it is 0.600 per cent low against a predicted 0.625; at 100, 672 parts per million. One reading scatters by eighteen times that, so no single reading shows the bias and the mean of a few hundred is nothing but bias.

 Degeneration removes the second-order product 9 times less well than the third. computed by solving, not by drawing. Both intermodulation products of a degenerated stage at 5 mV a tone, against the degeneration factor. The second-order product falls as D⁻² — the straight reference is exactly that law, anchored at D = 1 — and the fitted exponent is -2.000. The third-order product falls faster, as D⁴/|3 − 2D|, which is one more power of D at large factors, and it collapses altogether at D = 1.5 where its coefficient changes sign. So the ratio between the two goes from 20.7 at D = 1 to 183 at D = 16: the more linear the stage is made, the more completely its distortion is the product a third-order account never measures. Devices, and the amplitude they stop being linear at

The product that is not the third

Every distortion result in this field is odd-order, and every two-tone calculation behind those results computed the second-order product and threw it away. On a bare exponential it is the drive over twice the thermal voltage — 1.934 × 10⁻² of the fundamental at a millivolt, against 1.870 × 10⁻⁴ for the third-order product, a ratio of 4Vₜ/a and a hundred and three to one. A differential pair puts it at 6.2 × 10⁻¹⁶. And degeneration removes it as D⁻² where it removes the third order as D⁴/|3 − 2D|, so a stage linearised until its third-order product is negligible is a stage whose distortion is almost entirely the one nobody measured.

 The load alternates with parity between 1 and 1.9841, at every order. computed by solving, not by drawing. The last quotient of the continued fraction that turns a reflection polynomial into element values, which is the load resistance the design demands, drawn against order. A Butterworth returns exactly one at every order. A Chebyshev alternates: one at every odd order and 1.984056 at every even one, the same number each time, and it is the closed form (√(1+ε²)+ε)² to a part in 10¹⁵. The mechanism is in the last two rows of the panel. A lossless ladder filter is two resistances at direct current, so it must deliver the maximum available power there, which requires that zero be one of the frequencies the design reflects nothing at — and an even-order Chebyshev's reflection zeros are the roots of an even Chebyshev polynomial, none of which is zero. The synthesis stalls at order 9 for a Butterworth, which is why that series stops at eight. Filters, measured not tabulated

The termination an even order cannot have

Every even-order Chebyshev ladder terminates in 1.984056 times its source resistance at half a decibel of ripple, the same number at orders two, four, six and eight, and it is (√(1+ε²)+ε)² to a part in 10¹⁵. Building one between equal terminations instead — which is what a table of g-values and a matched pair gives — turns 0.5000 dB of ripple into 1.8123 at order four, deletes one of the passband maxima outright, and moves the worst tolerance corner from the 2.000 power of the component tolerance to the 1.109 power: a factor of 95 at one per cent parts.

 At β = 0.5, the duty error scatters by 2.49 ns a period against the period's 3.39 ns — and consecutive duty errors are anticorrelated, −0.217. Seeded: forty thousand periods of the event map with 5 mV of threshold noise, β = 0.5. The period scatters by 3.39 ns against σ√(A² + (A+B)² + B²) = 3.4 ns; the high half less the low half scatters by 2.49 ns against σ√(A² + (B−A)² + B²) = 2.49 ns, with A = RC/V(1+β) and B = RC/V(1−β) — 0.667 and 2.000 in units of RC/V. The draw both halves share enters the period with A + B and the difference with B − A. Consecutive periods correlate by 0.107 (closed form 0.115); consecutive duty errors by −0.217 (closed form −0.214). Circuits that do a job, and the range they do it over

The walk the core sees

Threshold noise in a relaxation oscillator walks its timing at 3.77 nanoseconds per root period at β = 0.5, because the draw two half cycles share adds. A transformer driven by the same square wave sees the difference of the halves instead, where the shared draw subtracts, and that walks at 1.89 — exactly β times the timing, 18.3 ns against 35.9 after a hundred periods over six hundred seeded runs. The per-period duty error does not vanish with the hysteresis and the walk does, consecutive duty errors are anticorrelated where consecutive periods are not, a comparator skew outruns the walk after 2(σB/d)² periods, and at a fixed frequency the core wants less hysteresis than the clock does.

 A porosity of 0.50, with the field the substitution smooths away. computed by solving, not by drawing. The flux lines between the conductors are the whole difference. The porosity substitution replaces this layer with a foil of the same direct-current resistance spread over the full breadth, in which the field is parallel to the layers by construction; here it is not, and it crowds between the turns. The solved ratio is 5.816 against the substitution's 6.212, 6.4 per cent apart. The copper is shaded by its own loss, which is what says the turns inside a layer are not alike either. Two windings, and the band between them

The wire that is not a foil

Almost no winding is made of foil, and the closed form for a winding's alternating-current resistance is about foils. The bridge between them is a substitution — squeeze the layer's conductors together, spread the result back across the breadth, divide the conductivity by the porosity — and it replaces a two-dimensional geometry with a one-dimensional one. Solved as a field it is exact where it must be, at a porosity of one, and 7.2 per cent high at a porosity of 0.40. It errs on the safe side, which is the half of the answer nobody could have assumed.

 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. Frequency, which is the same solve

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°.

 Where "the noise is the density over twice the window" is true for a mean of 600 samples. The exact sampled autocorrelation of a one-pole front end, summed in closed form and checked against seeded samples through the same filter. The solid curve is the reading's noise power divided by what the rule gives. It crosses one at 0.1355 of half the sample rate and nowhere else, so the rule is right at one front-end bandwidth rather than over a range: it is within one per cent from 0.0712 to 0.205 of half the sample rate, a window a factor of 2.88 wide. Above it the front end admits noise from past half the sample rate and the sampler folds it in, the ratio tending to (x/2)·coth(x/2) with x = 1/(τ·fₛ), which has no N in it — so the sample rate must be 9.7 times the front end's noise bandwidth for one per cent, not twice it. Below it the front end has already removed noise the rule charges for, and the faint curve — the √N law's own error, which is a claim about the samples rather than about a density — shows why: consecutive samples are correlated by ρ = e^(−1/(τ·fₛ)) and a mean of 600 of them is a mean of fewer things. The floor, which bounds from below

The window the square-root law has

Averaging N readings divides the noise power by N only if the readings are independent, and samples of a filtered noise are independent only over a band. Measured on the exact sampled autocorrelation, the rule that a reading's noise is the density over twice the window is right at one front-end bandwidth and wrong on both sides — a factor of three wide at six hundred samples, asking the sample rate to exceed the front end's noise bandwidth by about ten rather than by two. Above it the folded noise arrives at full amplitude; below it the front end has removed noise the rule is still charging for.

 A summing junction with a second pole above it: 3.1 dB of peak, and inductive. computed by solving, not by drawing on a two-pole amplifier — a gain block, two sections and a buffer, so the feedback network loads the output exactly as it does with one pole. The faint curve is the one-pole case, which rises to 909.1 Ω and stops. With a second pole at 100 kHz the junction reaches 1304 Ω at 112 kHz — 3.13 dB above the two resistors in parallel — because the division is by |1 + T| and near crossover that is smaller than one. A third of a decade below the peak the node's impedance leads its current by 84.4°, which is an inductance: the node is no longer a resistance that a capacitance loads, it is a reactance that a capacitance resonates with. The slider puts capacitance on the node and moves the peak. Feedback, and the margin

The node that is an inductance

A summing junction with a one-pole amplifier round it rises to the two feedback resistors in parallel and stops there, and that ceiling is the whole of what a virtual earth can be. Give the amplifier a second pole and the node overshoots it — 1074 ohms against 909 with the second pole at 300 kilohertz, 1868 with it at thirty — and below the peak the impedance leads its current by more than twenty degrees, which is an inductance. So a stray capacitance on the node is no longer a load on a resistance; it is a resonance, and the peak grows with it rather than being damped by it.

 Two sensors on one return: 511 µV in each reading, 100 nV in the difference. computed by solving, not by drawing. Two sensors referenced to the same local node, each reaching it through 2 mm of its own copper and each drawing 1 mA through that copper, with a 100 mA load returning through the 10 mm they all share. The interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it: 511.0 µV in each reading against 100.0 nV in the difference at a kilohertz, which is 74.2 dB. No resistor tolerance enters it — with the two stubs equal the rejection is exact to the solver, 166 decibels. What is left is the 10 per cent MISMATCH between the stubs times the sensors' own current, which the closed form puts at 100.0 nV, and it is proportional to the mismatch: 10.0 nV at 1%, 30.0 nV at 3%, 100 nV at 10%, 300 nV at 30%. Measurement, which is a circuit on a circuit

One voltage added to two readings

A difference amplifier across a shared return rejects somebody else's current by 54 decibels, and the number is four resistors rather than the amplifier. Two sensors referenced to the same node do better by a different mechanism entirely: the interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it — 511 microvolts in each reading against two and a half picovolts in the difference, which is 166 decibels and is exact rather than good. What is left is not a tolerance but a mismatch between the two sensors' own return paths, ten nanovolts per per cent of it.

 The order is worth 0.68 bits and a factor of 29 in clock. computed by solving, not by drawing. A butterworth design at a 20 kHz passband, at each order from two to ten and at three impedance levels, with the sample rate it demands for eighty decibels and the resolution its own resistors' noise allows. The clock demand falls steeply — 50.50×, 11.27×, 5.50×, 3.65× and on down to 1.76×, a factor of 28.8 — and the bar widths carry it. The noise rises: 2.00 µV at order two against 3.22 µV at ten, which is 0.68 of a bit. So the order is cheap in the currency everybody worries about and expensive in nothing. What is expensive is the impedance level: every resistance goes as one over the capacitance and a noise voltage as the square root of a resistance, so a decade of capacitance is 1.66 bits — the same at every order, to 0.000 of a bit — and the three curves are ten, one and a tenth of a nanofarad a section. The ceiling on resolution is set by a choice nobody prices, and the order, which everybody argues about, moves it by two thirds of a bit. Where a signal becomes a number

Two thirds of a bit for a factor of twenty-eight

An anti-alias filter costed in clock rate gets steeply cheaper with order — fifty times Nyquist at the second and 1.76 at the tenth — and the obvious objection is noise, since the filter is in the signal path and every section adds resistors. Measured, the objection barely holds: from order two to order ten the clock demand falls by a factor of 28.7 and the resolution the filter's own noise allows falls by 0.69 of a bit. What does cost resolution is the impedance level, at five thirds of a bit per decade of capacitance, and nobody argues about that at all.

 How much larger a gap is than its own length says. computed by solving, not by drawing. A magnetic circuit prices a gap as g/(µ₀A) and everybody knows that is low, because the flux bulges out of the sides. The usual repair is to add one gap length to each dimension of the gap's area, which is the dashed line. The measurement is the solid one: at a 0.3 mm gap in a 6 mm leg the true correction is 1.100 and the rule offers 1.050, so the rule supplies 50 per cent of a correction worth 10 per cent of the inductance; at 1.7 mm it supplies 85 per cent. The rule is not wrong so much as it is a rule whose accuracy depends on the thing it is correcting. Two windings, and the band between them

The gap that is bigger than it is

A magnetic circuit prices a gap as g/µ₀A and everybody knows that is low, because the flux bulges out of the sides. The usual repair — add one gap length to each dimension of the gap's area — supplies half the correction at a 0.3 mm gap and 85 per cent at 1.7 mm, on a correction worth 10 per cent of the inductance at the first and 33 at the second. It is not a rule that is right or wrong; it is a rule whose accuracy is a function of the very thing it is correcting.

 Three mismatches, and only one of them reaches the output. computed by solving, not by drawing. Each of the three quantities that can differ between the two transistors is given a spread of its own, one at a time, and 200 pairs are solved at each. The saturation currents produce a spread that follows them exactly — exponent 0.998, so 1.910 per cent of copy error for two per cent of mismatch. The current gains produce a line of slope 2.003, which is second order rather than first, and land at 2.20e-4 per cent for the same two. There is no third line because there is no third component: with no emitter resistors there is nothing for a resistor tolerance to be a tolerance of, and matching a mirror is a statement about emitter area and about nothing else. Devices, and the amplitude they stop being linear at

The mismatch that cancels itself

A current mirror's copy error is spread by three things the two transistors can differ in, and the population that measures it has always drawn all three at once. Turned on one at a time, a two per cent spread of saturation currents gives 1.910 per cent of copy error and a two per cent spread of current gains gives 0.00022 — because the gains enter only as a sum of reciprocals, which has no first derivative where they are equal. Then the standard cure un-cancels it, by a factor of 86.

 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.

 An integrator's summing junction is flat at 15.7 Ω, and worst at direct current. computed by solving, not by drawing. A current is driven into the node and the voltage read. The dashed curve is the same amplifier with a 10 kΩ resistor as its feedback element: a tenth of an ohm at direct current, rising a decade per decade, 909.5 Ω at the top. With a 10 nF capacitor instead, the loop gain has no frequency in it — the amplifier's gain falls as 1/f while the feedback factor rises as f — so it is flat at 62.83 and the node is flat with it, at Rin/(1 + 2π·GBW·C·Rin) = 15.666 Ω, measured 15.668 Ω and holding to 0.50 per cent from 100 Hz to 5.31 kHz — between the amplifier's own pole and the capacitor's corner, which is where a loop gain with no frequency in it lives. Both ends are the other way round from the resistive case: at direct current the capacitor is an open circuit, there is no loop at all, and the node is 843.0 Ω — a factor of 8388 worse than the resistor's 100 mΩ. Above the amplifier's crossover the capacitor is a short and the node is 47.6 Ω against 909.5. The two cross at 1.56 kHz, which is the frequency above which an integrator is the better virtual earth of the two. Feedback, and the margin

The shelf a capacitor makes

An inverting amplifier's summing node rises a decade per decade to the two resistors in parallel. Replace the feedback resistor with a capacitor and it does neither: an integrator's loop gain has no frequency in it, so the node is flat at Rin/(1 + 2π·GBW·C·Rin) — 15.67 ohms over four decades, to half a per cent. Both ends invert. At direct current the capacitor is an open circuit, there is no loop at all, and the node is 843 ohms against the resistor's tenth of one; above the amplifier's crossover the capacitor is a short and the node is nineteen times better.

 A star ground is a 0.50 mm daisy chain, not a point. computed by solving, not by drawing. 8 circuits of 100 mA each meeting at one pad, which reaches the plane through a via of 0.25 mΩ and 0.35 nH. The via is shared by every one of them, so the star's residual is 7 other currents through it — 175.0 µV at 1.00 kHz against the 1.800e+4 µV a daisy chain of 10 mm hops would give at its far end, a factor of 103. Measured as a length of the same track, the via's resistance is worth 0.500 mm and its inductance 0.350 mm — two different lengths, so a via has a corner of its own at 114 kHz where a track's is at 79.6 kHz wherever it is cut. The faint curves are other circuit counts. The two grow differently: fitted over counts from 2 to 32 the star's error goes as N^1.22 and the chain's as N^1.87, because a chain's hop carries every circuit beyond it — so the star's advantage grows with the count rather than shrinking. And a star has no position in it: a chain's far end reads 4.5 times its near end, which is a diagnostic no real fault should have. Measurement, which is a circuit on a circuit

A star is half a millimetre long

The repair for a shared return is a star ground, and it is described as making the shared length zero. It does not: every return meets at a pad, the pad reaches the plane through a via, and a via of a quarter-milliohm and 0.35 nanohenries is worth half a millimetre of one-ounce track by its resistance and 0.35 by its inductance — two different lengths, so a via has a corner of its own at 114 kilohertz where a track's is at 79.6. What that buys is a change of law rather than a factor: a daisy chain's error grows as the square of the circuit count and a star's as the first power, so the advantage runs from sixty times at two circuits to three hundred and forty at thirty-two.

 Adding powers is wrong by the number of sources, when they are coherent. computed by solving, not by drawing, at five source counts. 8 ten-volt sources reach one hundred-ohm load through a hundred ohms each. With every source in phase the load takes 790.1 mW against the 98.77 mW that adding their separate powers gives — exactly 8 times, at every count, to a part in 10¹², because all 28 cross terms are positive and equal. With seeded random phases it takes 0.9865 times the sum over 64 runs, against a standard error of 0.125 — the cross terms average to nothing, which is the reason noise powers add and a root-sum-square is the right arithmetic for uncorrelated interference. The factor between the two limits is the source count itself, so it grows without bound: adding powers is not wrong by a bounded amount, it is wrong by how many things are being added. Networks, and how a solve is checked

The cross terms that outnumber the sources

With two sources the power in a load has one cross term, and adding the separate powers is wrong by a factor of two. With thirty-two there are four hundred and ninety-six of them, and the factor is thirty-two: 940 milliwatts delivered against the 29.4 that adding gives. With the same thirty-two sources at seeded random phases it is 26.9 milliwatts — the sum itself, inside the standard error of sixty-four runs, because every cross term averages to nothing. One identity, two limits, and the two are a phased array's gain and a noise sum's.

 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.

 A cable's carrier limit is its conductors' matching, not its capacitance. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε, excited at a carrier frequency, with 10 metres of cable at 100 pF a metre on each output terminal. Each terminal sees the two arms meeting there in parallel against that capacitance, so each lags by arctan(2πf·(r/2)·C) with a corner at 909 kHz — and with the two conductors matched the lag CANCELS in the difference: -0.117° where each terminal has -5.71°. A mismatch does not cancel, and it arrives multiplied by the ratio of the excitation to the signal, which at 1000 µε is 2000: at 1 per cent the phase is 40.3° at a hundred kilohertz. So the excitation a 10-metre cable allows for 200 parts per million falls from 205 kHz matched to 1.82 kHz at 1 per cent. And it depends on the strain being measured — 183 Hz at 100 µε, 547 Hz at 300 µε, 1.82 kHz at 1000 µε, 5.45 kHz at 3000 µε — because the amplification is the reciprocal of the signal. Circuits that do a job, and the range they do it over

The carrier a cable allows

A carrier-frequency amplifier removes an instrument's offset drift by exciting the bridge with alternating current, and the cable that was a resistance at direct current becomes a capacitance. Both output terminals see the same arms and the same capacitance, so the lag cancels in the difference: ten metres matched leaves 0.15 degrees at a hundred kilohertz where each terminal has 6.3. What does not cancel is the mismatch between the two conductors, arriving multiplied by the ratio of the excitation to the signal — two thousand at a thousand microstrain — so one per cent of it takes the allowed carrier from 205 kilohertz to 1.8.

 Four readings of one distance: |H| is |p| over the length of this arrow. The complex plane of a single-pole low-pass with its corner at 995 Hz. The pole sits at −995 Hz on the real axis and the faint circles are contours of constant magnitude — circles centred on the pole, because |H| = |p|/|s − p| and a constant magnitude is a constant distance. The imaginary axis crosses them, and where it crosses is the frequency response: 1.00 corners up the distance is 1.4142 times the pole's own, so the magnitude is 0.70711 and the phase is minus the 45.00° the separation subtends. Checked against the nodal solve at sixty frequencies over five decades, the two agree to 6.7e-16. So the four readings are one geometric fact: the corner is where the distance is √2 times its smallest, the roll-off is the distance becoming ω, the flat passband is the distance not yet changing, and the phase is the angle. Frequency, which is the same solve

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.

 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.

 8 doublets over 2 decades settle 31.6 times slower than one. A tail of 1.0 per cent split into 8 doublets whose time constants are spread over 2 decades from 0.100 ms, each carrying 0.125 per cent. Settling to 0.10 per cent takes 7.278 ms against 0.2303 for the same total tail at one time constant — a factor of 31.6 — and it is between τₘₐₓ·ln((A/N)/B) = 2.231 and τₘₐₓ·ln(A/B) = 23.03 ms, which says the SLOWEST doublet decides it whatever its share. The spread is what costs and the number is not: four decades of spread multiply the settling by 2.04e+3 while sixteen times the count changes it by 0.386. And the trimming result, halving the tail for τ·ln 2, now buys 3.86 ms — the slowest time constant's ln 2 rather than the fastest's. Before the steady state

The slowest one decides it

One doublet's settling is its own time constant times the log of the tail over the band, so halving the mismatch buys ln 2 of it and no more. Split the same one-per-cent tail into eight doublets spread over four decades and the settling goes from 0.23 milliseconds to 470 — a factor of two thousand — while sixteen times as many doublets over the same spread changes it by less than two. It is the spread that costs and not the number, because the slowest doublet decides the settling however small its own share, and its share enters only through a logarithm.

 A twin-T's null: everything fills it, and at one order. computed by solving, not by drawing. A twin-T of 10 kΩ and 10 nF nulls at 1591.5 Hz, where a low-pass T of two resistors and a high-pass T of two capacitors deliver equal and opposite contributions. With exact components the null is -Infinity dB, which is the arithmetic's floor. A tolerance fills it at 20.0 decibels per decade — first order, where an LC notch splits into twenty and forty depending on which way two errors go, because there a product can be left alone and here there is no product. Loss fills it at 20.0 decibels a decade too, and barely moves it — 1591.55 to 1595.52 Hz across four decades of series resistance — because a resistance of r in a capacitor branch is a fractional error of r/R and nothing else: one ohm gives -88.5 dB against a hundredth of a per cent of tolerance's -92.0. So a notch's two mechanisms are not exchanged here — they are one mechanism, because a twin-T's null is stationary in nothing and an LC notch's frequency is stationary in the product of its two components. Filters, measured not tabulated

The null that is stationary in nothing

An LC notch has two mechanisms and they do different things: a tolerance moves its null rather than filling it, because the resonance depends on a product two errors can leave alone, and loss fills it at twenty decibels per decade of arm resistance. A twin-T has no resonance and no arm. Measured, everything fills it at twenty decibels a decade and nothing moves it — one ohm in series with its capacitors gives 88 decibels where a hundredth of a per cent of tolerance gives 92, because a series resistance in a capacitor branch is a fractional error of r over R and nothing else.

 Where a transformer's leakage inductance actually is. computed by solving, not by drawing. Both windings carry the same ampere-turns in opposite directions, which is the short-circuit condition a leakage measurement is made under, so the flux drawn here is the flux that fails to link the two — the leakage field, and nothing else. It is largest in the insulation between the portions, where the magnetomotive force is at its full value and there is no copper to be in. The energy in this window is 2.058 microjoules per metre, which is 4.116 microhenries per metre referred to the primary against a closed form of 5.213. Two windings, and the band between them

The inductance that is a shape

Leakage inductance is the one transformer parameter that belongs to the geometry rather than to the material: twice the magnetic energy in the window under equal and opposite ampere-turns, divided by the square of the current. Solved as a field it is 3.086 microhenries a metre against a closed form's 3.128 when the copper fills the window, and 4.608 against 6.255 when it fills half of it. Interleaving is worth 3.11 times and not the four it is quoted as, and the missing 0.89 is the insulation nobody puts in the formula.

 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. Frequency, which is the same solve

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.

 4 branches read with 10 Ω ammeters: one at a time every reading is low, all at once they add up and the small branches are high. computed by solving, not by drawing. A 1 A source into 4 parallel branches of 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ, each read with an ammeter of 10 Ω. With one ammeter moved from branch to branch, each reading is low — by −0.989%, −0.902%, −0.099%, −0.010% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 990.275 mA rather than 1000.000. With an ammeter in every branch at once the readings add to 1000.000 mA and are wrong by −0.892%, +7.939%, +8.910%, +9.008%: the meter in the largest branch pushes current out of it, and the small branches take it. Networks, and how a solve is checked

The readings that add up and are wrong

Four branches of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ share a 1 A source. Read them one at a time with a single 10 Ω ammeter moved from branch to branch, and every reading is within one per cent of the truth, but the four add to 990.275 mA. Put a 10 Ω ammeter in every branch at once and the four readings add to 1000.000 mA exactly, satisfying Kirchhoff's current law to the last digit. Yet three of them are 7.9 to 9.0 per cent high. The check a reader would run on the readings is passed by the wrong set and failed by the right one.

 A 10 ps stub under a balanced via, and the pad it has to come out of. computed by solving, not by drawing, as a cascade. The reflection from a via of 0.5 pF of pad and 1 nH of barrel on its balancing line of 44.72 Ω, met by an edge of 59 ps, four ways: with no stub, whose largest excursion is 24.4 mV; with a 10 ps open stub below its bottom pad, 90.5 mV and an area of 5 ps; with the stub's 0.224 pF taken out of both pads, 56.1 mV; and taken out of the bottom pad alone, 24.6 mV. Both trims balance the area again. Only the second restores the doublet. Lines, where a wire has a length

The pad the stub comes out of

A via used to a middle layer leaves barrel hanging below the connection: an open stub. Under a via balanced on its √(L/C) line, a 10 ps stub is 0.224 pF of extra capacitance, 5 ps of area, and a reflection of 90.5 mV where the balanced via gave 24.4. Taking that capacitance out of the pads balances the area again whichever pad it comes from. Only one restores the doublet: out of both pads the via still reflects 56.1 mV, out of the bottom pad the stub hangs from, 24.6. The trim works for every stub whose capacitance fits inside that pad, delays up to Z·C/2 = 11.2 ps, whose quarter wave is above 22.4 GHz. A longer stub has nothing left to come out of. At 40 ps it is a notch in the transmission at 6.25 GHz.

 BDF2 loses amplitude as N⁻³ and phase as N⁻², four times the trapezoidal rule's: a hundred cycles to 1% needs 363 steps a cycle, against 182 and 196,404. Each rule's error a cycle on an undamped inductor–capacitor pair, against the steps a cycle, measured from BDF2 marches and drawn with the trapezoidal rule's and backward Euler's closed forms. BDF2 loses amplitude a cycle as the −2.91 power of N and phase as the −1.95 power, its phase loss 3.988 times the trapezoidal rule's at 128 steps a cycle. To keep 99% of the amplitude over a hundred cycles it needs 157 steps a cycle, and to stay within 1% of a cycle in phase 363; the trapezoidal rule needs 182 for the phase and loses no amplitude, and backward Euler needs 196,404 for the amplitude. Before the steady state

The rule between the two

The trapezoidal rule keeps an undamped ringing's amplitude and loses its phase, and it preserves a stiff pole's error with its sign alternating. Backward Euler damps the stiff error at once and damps the ringing too, keeping 0.0082 per cent after ten cycles at twenty steps a cycle. The second-order backward difference sits between them in a precise sense. It damps the stiff error by 1/√(3 + 2h/τ), 0.152 a step at twenty time constants, and settles in three steps where the trapezoidal rule needs twenty-three. It loses a ringing's amplitude only as the cube of the steps a cycle, and its phase as the square with a constant four times the trapezoidal rule's. A hundred cycles to one per cent costs 363 steps a cycle, against 182 and 196,404.

 Losing capacitance to bias raises this loop's margin; the corner that binds is the ceramic at full value. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5.00 mΩ ceramic output capacitor, against the capacitance that capacitor keeps at the working voltage, from the full 10 µF down to 1.5 µF, for five capacitors across the upper divider resistor. With none, the margin climbs from 18.0° at full value to 43.6° at 15% kept, so the lost capacitance is stabilising rather than the reverse. 630 pF holds at least 49.7° across the whole range. 836.5 pF, the value that removed the resistance floor at the nominal capacitance, gives 48.8° at full value and 44.3° at 15%: the capacitor that has lost the most is the one it fails on. Circuits that do a job, and the range they do it over

The capacitance the bias takes away

A ceramic output capacitor has a few milliohms of series resistance, which a capacitor across the upper divider resistor made legal, and a capacitance that falls with the voltage across it — a ten-microfarad part at five volts is two microfarads to a small signal. Swept from the full ten down to that fifth, the regulator's margin does not collapse. It rises: with no divider capacitor from 18.0° to 38.5°, because the output pole climbs away from the amplifier's. The corner that binds is the capacitor at full value. One divider capacitor, 423 to 808 pF, holds 45° at every capacitance between. The 836.5 pF that removed the resistance floor sits just above that band, and the lost capacitance is paid for in droop: 103 mV at full value, 202 mV at a fifth.

 6 straight lines over six decades of a diode: ±8.002 mV everywhere, and they meet at every join. computed by solving, not by drawing. A pure exponential junction from a nanoamp to a milliamp, modelled by 6 drop-plus-resistance segments each spanning a current ratio of 10.00, each fitted minimax on its own window. The model's error swings between +8.002 mV and −8.002 mV in every segment, which is the closed form for that ratio, because a span of a logarithm is the same shape wherever it is taken. The dots are the joins: at each one the two neighbouring segments are off by the same amount in the same direction, so they meet, and the chain is continuous although no segment was asked to be. Where the models stop

How many straight lines a diode is

One drop-plus-resistance line fitted minimax over a decade of a diode's current is out by ±8.00 millivolts, and it is the same over every decade. Chain such segments over six decades, a nanoamp to a milliamp, and the error of each is a closed form in its current ratio alone. Minimax segments meet at their joins without being asked to, so the chain is a continuous piecewise-linear model for free. The error falls as the inverse square of the segment count: one segment is out by 132 mV, six by 8.00, eighteen by 0.94. A millivolt takes 18 segments and a tenth of one takes 56. A real junction needs more, 23 and 72, because recombination gives it an ideality near two over most of the range and each segment narrows by the square root of the local ideality.

 Joining the samples with straight lines squares the hold's sinc: 5.28 dB of droop and 5.8 dB of image rejection, both doubled. computed by solving, not by drawing. A 20 kHz tone on a 48 kHz clock, reconstructed two ways: held for a whole clock period (dashed envelope, open stems) and joined sample to sample by straight lines (solid envelope, filled stems). Every line is read from a transform of that waveform itself. The held one comes out 2.640 dB down and its nearest image 2.92 dB below it; the interpolated one 5.280 dB down with the image 5.85 dB below. A straight line between samples is a triangle two clocks wide, which is a rectangle convolved with itself, so its transform is the sinc squared: every decibel doubles and the nulls stay where they were. Where a signal becomes a number

The pole a straight line is worth

A converter that joins its samples with straight lines instead of holding each one is convolving with a triangle rather than a rectangle, and its transform is the hold's sinc squared. Every decibel doubles: 5.28 dB of droop at a 20 kHz band edge on a 48 kHz clock instead of 2.64, and 5.85 dB of image rejection instead of 2.92. Both exponents of the oversampling ratio are unchanged. The second sinc is worth exactly one pole of reconstruction filter at every ratio, because a sinc's rejection of the nearest image is twenty times the log of that image's distance from the band edge, identically. A single analogue pole allowed the same droop gives a fraction of that rejection. The price is a whole clock of delay rather than half.

 Inside a loop of gain 1000, a twin-T error of +0.10% raises the peak's Q from 250 to 333, and the same error the other way lowers it to 200. computed by solving, not by drawing. An amplifier of gain 1000 with a 10 kΩ, 10 nF twin-T as its feedback network, whose closed-loop gain A/(1 + A·T) peaks at the null, 1591.5 Hz. With exact components the peak is 1000 and its quality factor 250.2, a quarter of one more than the gain. With the twin-T's shunt elements −0.10% out the other way the peak falls to 800 and the Q to 200.2; +0.10% raises them to 1333 and 333.5, and +0.30% to 3973 and 994. At +0.40% the loop oscillates. Filters, measured not tabulated

The sign the null leaves behind

Put a twin-T in the feedback path of an amplifier of gain A and its null becomes a peak with a quality factor of (A + 1)/4. A filled null is not a small phasor of any phase. It is a real number, minus a quarter of the error in the shunt elements, and once the loop multiplies it by A its sign decides everything. At a gain of a thousand a 0.1 per cent error lowers the Q from 250 to 200 one way and raises it to 334 the other, and at 0.4 per cent the second direction oscillates. Every gain falls on one curve, the exact-null Q divided by 1 − A·t/4. So a Q is set to ten per cent only by a null deeper than 20 log(44·Q): one per cent parts allow a Q of 9.6.

 1 pA in 10.0 kHz: 312 carriers a window, and a Gaussian of the same variance that looks right and is not in its tails. computed by solving, not by drawing. A direct current of 1 pA counted in windows of 1/(2B) for a noise bandwidth of 10.0 kHz: the mean count is N = I/(2qB) = 312.1, and the count is Poisson, drawn as bars, against a Gaussian of the same mean and variance, drawn as a curve. The variance is N exactly, so 2qI is right; the signal-to-noise ratio is √N = 17.67 and the skewness 1/√N = 0.0566. The count exceeds its mean by five standard deviations with probability 7.84 × 10⁻⁷, against the Gaussian's 2.87 × 10⁻⁷. A seeded arrival train counted in 20,000 windows gives a mean of 312.200 and a variance of 314.461. The floor, which bounds from below

The count under the density

Shot noise's density, 2qI, is exact at every current, because it is the variance of a count of independent carriers, and that count is Poisson at any size. Reading that variance as a Gaussian floor is not exact. A picoampere in a 10 kHz noise bandwidth is 312 carriers a window, and it passes five standard deviations above its mean 2.74 times as often as a Gaussian of the same variance says. At 10 femtoamperes it is 359 times as often, and 4.4 per cent of the windows are empty. A single pole reads a third more skewness than a count does, (4/3)/√N against 1/√N. The density is right and the shape is wrong, and the shape is what a threshold uses.

 At 50° of margin a capacitor across the resistor carries 85.2 µV and a tee 347 µV: the tee is dominated everywhere. computed by solving, not by drawing. A photodiode amplifier of 500 kΩ transimpedance on 33 pF of detector, behind a 10 MHz amplifier with 4 nV/√Hz of voltage noise, compensated either by a capacitor across a single feedback resistor, swept from none to 4 pF, or by a tee of the same transimpedance and factor 3 to 64. Each design's phase margin is drawn against its output noise integrated to its own bandwidth. At every margin a tee reaches, the capacitor reaches it with less noise: at 50.4° the capacitor's design (0.97 pF) carries 85.2 µV in 396 kHz and the tee (factor 32) 347 µV in 401 kHz, 4.1 times more. The ratio runs from 1.29 to 7.47 across the margins compared. Feedback, and the margin

The compensation that costs nothing

A photodiode amplifier can be compensated with a capacitor across its feedback resistor, or with a feedback tee, whose raised noise gain also raises the phase margin. At one transimpedance, 500 kΩ on 33 pF of detector, each design's margin is drawn against its output noise over its own bandwidth. The capacitor dominates at every margin the tee can reach: at 50° it carries 85 µV where the tee carries 347, in the same 400 kHz. The two keep the same band to within 1.3 per cent at equal margin, so the tee's noise buys no bandwidth. The conjecture that the answer would turn on which noise dominates does not survive either. With the amplifier's voltage noise cut tenfold, the tee is still 5.6 times worse.

 Held one drop and 300 mV up, the cascode's floor falls from 0.832 V to 0.403 V. computed by solving, not by drawing. The output current of two cascode mirrors against their output voltage near the bottom of the supply, each divided by its own value at 5 V, on a model whose transistors saturate. The stacked mirror holds its upper device's base two drops up and stays within one per cent down to 0.832 V; the wide-swing mirror holds it one drop and 300 mV up, which leaves the lower device 0.302 V of collector-emitter voltage, and stays within one per cent down to 0.403 V. Its output resistance at 5 V is 13.2 MΩ against 7.39 MΩ, and its copy ratio 0.9758 against 0.9747. Devices, and the amplitude they stop being linear at

The drop the lower device did not need

A cascode mirror's floor is a whole base-emitter drop plus the upper device's own minimum, because the stack that biases the upper base holds it two drops above ground. The lower device needs only a few hundred millivolts. Hold the upper base one drop and 300 mV up instead, and on a model whose transistors saturate the one-per-cent floor falls from 0.832 V to 0.403 V. The output resistance does not fall with it: it rises from 7.39 to 13.2 MΩ, because the upper device's base current no longer comes out of the reference. The cost has moved into the margin itself. Below about 150 mV the lower device saturates, and at 50 mV the resistance is 0.51 MΩ and the copy ratio 0.7665.

 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.

 At the same rise time a lossy edge draws 2.9–4.0 times more from a balanced via than a clean one, and the square law survives. computed by solving, not by drawing. The largest excursion of the reflection from a balanced via and from a 0.5 pF capacitive via against the 10–90 rise time of the edge that reaches them: behind 5 to 50 cm of lossy line, with a 20 ps edge sent in, and from clean raised-cosine edges of the same rise times on a lossless line. Behind the loss the balanced via's reflection falls as the −1.97 power of the rise time and the capacitive one's as the −0.82. Against clean edges of the same rise time the lossy edge draws 2.93 to 3.97 times as much from the balanced via and 1.07 to 1.54 times as much from the capacitive one. Lines, where a wire has a length

The edge the loss delivers

A balanced via's reflection falls as the square of the edge's rise time, so the loss in front of it should hide it. A 20 ps edge sent down 20 cm of lossy line arrives with a 10–90 rise time of 194 ps, and the via's reflection falls to 9.1 mV. But a clean edge with that same 194 ps rise time draws only 2.29 mV from the same via. The lossy edge excites it 3.97 times as much, because loss makes an edge steep in the middle and slow at both ends, 3.07 times steeper at its centre than a raised cosine of the same 10–90 time. Across 10 to 50 cm the square law survives, an exponent of −1.97. It survives shifted: the 10–90 time overstates, by a factor of three or four, how slow the edge is to anything that reads its curvature.

 Matched at 1 kHz, the conjugate load wins only from 730 Hz to 1.84 kHz; the resistor-only load is a low-pass, not a broadband answer. computed by solving, not by drawing. A source of 50 Ω and an inductance of 100 Ω at 1 kHz, with the conjugate load for 1 kHz — 50 Ω and a capacitor of −j100 Ω — and the resistor-only load for 1 kHz, 111.8 Ω, both held fixed while the frequency moves. The conjugate match takes all the available power at 1 kHz and stays within a decibel of it from 777 Hz to 1.29 kHz, a band of 50.9% of the centre, 1.0176/x. The resistor-only load takes 0.6180 at 1 kHz, 0.8509 at 100 Hz, and less at every frequency above: a low-pass whose peak is at direct current. The conjugate load is the better of the two only between 730 Hz and 1.84 kHz. Power, and the part that does no work

The load for a source that will not hold still

A conjugate match is exact at one frequency. Held fixed while the frequency moves, it stays within a decibel of the available power over 1.0176/x of the centre: 51 per cent of it when the source's reactance is twice its resistance, 12.7 per cent at eight times. It beats the resistor-only load only between 0.73 and 1.84 of the centre frequency. The resistor-only load is not a broadband answer, as it is usually called. It is a low-pass whose best frequency is direct current. And for a source whose reactance could be anywhere within ±Δx of a centre, the fixed load that guarantees the most power cancels the centre exactly and takes a resistance of √(1 + Δx²) times the source's. That guarantees 2/(1 + √(1 + Δx²)): the resistor-only rule, applied to the uncertainty instead of to the reactance.

 4 rectangles: the hold's sinc to the power 4, 10.56 dB of droop and the first image 11.7 dB down. computed by solving, not by drawing. A 20 kHz tone on a 48 kHz clock, reconstructed by a pulse of 4 rectangles convolved together — the hold, the straight line, the quadratic and the cubic B-spline for one to four — with the envelopes of all four drawn and the lines of the 4-rectangle waveform read from its own transform (stems). They agree with the sinc to the 4th power to 0.000 per cent. Droop 10.560 dB; the nearest image 11.69 dB below the signal, 4 times the hold's 2.92. The nulls stay on the multiples of the clock; each rectangle deepens everything else. Where a signal becomes a number

Four rectangles and no filter

A converter's output pulse can be any number of one-clock rectangles convolved together: one is the hold, two the straight line, four the cubic B-spline. Measured on its own waveform, n rectangles give the hold's sinc to the n-th power to a few parts in ten million, so each one rejects the nearest image by exactly what one analogue pole could at best — and each saves exactly one pole of reconstruction filter until none is left. At the Nyquist rate that is three poles of nineteen and a half, bought with 10.56 dB of droop. From about 2.8 times oversampling the cubic pulse needs no analogue filter at all for 60 dB, for 0.62 dB of droop at four times and two clocks of delay. The cost is that from three rectangles on the output no longer passes through its samples.

 A line cut into 8 LC sections passes 10% at 0.863 and 50% at 1.043 of its time of flight. computed by solving, not by drawing, marched at 8000 steps. A lossless line of 1 µH and 100 pF — a 100 Ω line with a 10 ns time of flight — built from 8 equal LC sections and driven through 100 Ω into 100 Ω, so the ideal line would deliver a clean half-volt step at exactly one time of flight (dashed). The ladder network passes 10% of its final value at 0.8630, 50% at 1.0428 and 90% at 1.1592 of the time of flight: a delay quoted at 10% is 18.0% of a flight time shorter than one quoted at 50%. It starts as the 16th power of time, for 16 poles and no zeros, and rings after the edge because a lumped ladder network cannot pass what lies above its sections' cut-off. Before the steady state

The threshold a delay is quoted at

A step through a network of r poles starts as the r-th power of time, so a delay measured at 10% moves with the network's order more than one measured at 50%. Cut a line into more sections and the order grows without limit, and what happens to the two delays depends on the line. A lossless LC line converges on its time of flight at every threshold: the spread between its 10% and 50% delays falls from 0.656 of a flight time with one section to 0.070 with thirty-two, as the −0.68 power of the section count, the −2/3 of an Airy front. An RC line converges too, but its 10% and 50% delays converge to 0.134 and 0.391 of R·C, a third of each other for good. On a propagating line a delay belongs to the path; on a diffusing one, two-thirds of it belongs to the threshold.

 The one number the window does not move. computed by solving, not by drawing. Loss against foil thickness, at three window fills, with each curve's minimum located by a parabola through its three lowest points rather than read off the grid. The optimum sits at 0.654 skin depths at full fill, 0.708 at forty per cent, against the closed form's 0.663 — a drift of 8.3 per cent while the ratio the same winding carries moves by eighty. The alternating-current resistance at the optimum is 1.340, 1.351, 1.406, against four thirds. What did move is the loss it costs: 0.0555 watts a metre at full fill and 0.1383 at forty per cent, for the same current in the same number of layers. Two windings, and the band between them

The optimum that does not move

A foil winding has a best thickness — past it, more copper is more resistance — and at that thickness the alternating-current resistance is four thirds of the direct-current resistance, whatever the layer count. Both of those are one-dimensional results, and this ladder has spent three rungs finding that the one-dimensional picture is 82 per cent wrong about the resistance ratio. Solved as a field, the optimum drifts by 8.3 per cent between a full window and a quarter-full one, and four thirds becomes 1.34, 1.35, 1.41. The trade barely moves while everything it is made of moves a great deal.

 A 0.4 mm slot between the tracks leaves all of the resistance shared at DC and raises the in-band share from 4.03% to 4.98%. computed by solving, not by drawing, across the plane in strips. Two tracks 200 µm above a 50 mm plane and 2 mm apart, one driven and the other listened to: the share of the driven loop's resistance that the quiet loop's return also carries, against frequency, over an uncut plane (dashed) and with a slot 0.4 mm wide cut along the plane midway between the tracks (solid). The slot's two halves are one conductor, joined where the run ends, so at direct current the return still spreads across both and the whole of the plane's resistance is shared, 1.0000 of it. Above the band each return has gathered under its own track, and the share is 4.98% with the slot against 4.03% without: the slot is where the copper nearest the other track used to be. Lines, where a wire has a length

The slot that does not separate

Cutting a slot in the plane between two tracks is the common response to coupling between their returns, and in a cross-section it looks as if it separates them. It does not, if the plane is one ground: the halves stay joined where the run ends, so at direct current both returns still spread across the whole plane and share all of its resistance — 1.0000 with a 0.4 mm slot — and above the band the slot makes the sharing worse, 4.98% against 4.03%, rising to 15.7% as the slot widens to 1.6 mm, because it takes away the copper between the tracks and pushes each return towards the other. Only a split into two grounds removes the shared resistance at DC, at the price of each loop's resistance multiplied by 2.10, and it leaves the in-band coupling 29% lower, not gone: most of two tracks' coupling at 100 MHz is their own magnetic field, which the plane's copper does not carry.

 With each source 0.8 of the one before, 32 coherent sources cost 8.99 times the sum of their powers, and never more than 9.00. computed by solving, not by drawing, at five counts and five amplitude ratios. Sources of 10 V, 10q V, 10q² V … each reach one 100 Ω load through 100 Ω; with all in phase the load's power over the sum of each source's power alone is (Σ|a|)²/Σa² to ten figures. Equal sources (q = 1) give the count itself, 2, 4, 8, 16, 32. Falling amplitudes give less, and a limit: for q = 0.8, 1.98, 3.77, 6.41, 8.51, 8.99, tending to (1 + q)/(1 − q) = 9; for q = 0.5, 3.000 against 3. The coherent penalty counts sources by their participation, not their number. Networks, and how a solve is checked

How many sources the cross terms count

N equal coherent sources on one load deliver N times the sum of their separate powers. Sources of unequal size deliver (Σ|a|)²/Σa² times it — solved on the network to ten figures — which is the number of equal sources that would carry the same penalty: a participation ratio. Amplitudes falling by 0.8 a source cap it at (1 + q)/(1 − q) = 9 however many there are; one large source and thirty-one at a tenth of its amplitude give 12.8, not 32, and for small sources the excess is 2(N − 1)ε — their amplitudes, not their powers, so thirty-one sources carrying a hundredth of the large one's power each turn 1.31 times its power into 16.8 times it when they line up. Random amplitudes take a fixed fraction off the count, (mean)²/(mean square): 0.75 for uniform ones. And with random phases none of it matters: the load takes the sum.

 A 6 dB pad at 290 K turns a cold source into 217.2 K and a 290 K one into 290 K. computed by solving, not by drawing, every resistor's Johnson noise carried to the pad's open output. A matched 50 Ω T-pad of 6 dB (arms 16.61 Ω, shunt 66.93 Ω) at 290 K, fed from a 50 Ω source at the temperature on the horizontal axis. The output's noise temperature (dots) is Tₛ/L + T·(1 − 1/L), with T the pad's own temperature, to nine figures (line): a slope of 1/L = 0.2512 and an intercept of 217.2 K — the pad's own noise, (1 − 1/L)·290 K, seen at its output. The dotted diagonal is no pad at all. The two cross at 290 K, where the source is at the pad's temperature and the output is too. The floor, which bounds from below

The noise a pad adds is its temperature

A matched resistive pad of loss L at physical temperature T presents a noise temperature of Tₛ/L + T(1 − 1/L) at its output, for a source at Tₛ,, and summing its four resistors' Johnson noise to the open output confirms it to nine figures. Referred to its input its own noise is (L − 1)·T — 288.6 K at 3 dB, 28,710 K at 20 dB — so its noise figure in decibels is its loss in decibels at 290 K, which Friis states from the definition and the resistors reach by themselves. At its output a pad can never add more than its own temperature: cooled to 4 K, a 20 dB pad adds 3.96 K there, and still 396 K referred to its input. Most of a heavy pad's noise is made in its last resistor — 82.6% at 20 dB. And set between a 400 K source and a 290 K load, a pad stops the net flow of noise power at a temperature that depends on its loss, or, below 1.40 dB, at none.

 With the period 1.3% long, eight periods read a sine 0.599% wrong through a rectangle, 79.4 ppm through a triangle, 22.8 ppm through a Hann window. Integrated over each weighted window, the worst over every starting phase. The error in a sine's root-mean-square summed over N assumed periods 1.3% too long, against N, for four weights across the window. The rectangle is the plain sum: 0.640% at two periods and flat in N, δ/2 as the earlier measurement found. The triangle is flat in N too but 75 times smaller at eight periods. The Hann window falls with N, as the −2.67 power: 414 ppm at two periods, 0.293 ppm at thirty-two. The four-term Blackman–Harris is below 4.85 ppm at every count from two. A weight that falls to zero at the window's ends counts the extra sliver of period at almost nothing. Power, and the part that does no work

The window that forgives the period

A converter that sums a waveform's square over whole periods reads its RMS exactly, and one that sums over periods 1.3% too long reads a sine 0.6% wrong — δ/2, for any number of periods — because the extra sliver of period counts at full weight. Weight the window down to zero at its ends and the sliver barely counts. Over eight periods a triangular weight reads the same sine 79 ppm wrong and a Hann weight 23 ppm; and they are different kinds of cure. The triangle is the only one that changes the error's order, from first to second in δ, so it wins when the period is nearly right; the Hann weight keeps first order but falls with the number of periods, about as its inverse square, so it wins over long windows. Both cost nothing with the period exact — from two periods up. Over one, a Blackman–Harris weight reads even a sine 10% wrong. The price is noise: a Hann-weighted reading of noise scatters √1.5 as widely.

 3 gaps of 0.40 mm, where one of 1.20 would do. computed by solving, not by drawing. The same total gap, the same turns, the same core, and very nearly the same inductance — cut into 3 instead of one. Each gap now drops one part in 3 of the magnetomotive force, so the field it throws into the window is that much weaker; and because the field's own energy goes as its square, the loss it causes falls faster than the field does. The winding dissipates 3.00 times its direct-current loss here against 6.52 with a single gap, and its worst turn 3.7 against 13.3. Two windings, and the band between them

The gap that is three gaps

Reluctances in series add, so one gap of 1.2 millimetres and three of 0.4 are the same magnetic circuit: the same inductance, the same saturation current, the same energy in the air. They are not the same field. Solved in two dimensions, the winding beside the single gap dissipates 6.52 times its direct-current loss and its worst turn 13.3; beside three gaps those are 3.00 and 3.7. And there is a best number of gaps rather than a monotone gain — past three, spreading them along the leg brings each one close to a different part of the winding.

 A band from zero to its top frequency takes 4.77 dB less from 10 ps of jitter than a tone at the top: it is the rms frequency that counts, not the highest. computed by solving, not by drawing. Forty-eight equal tones at random phases filling a band from 0 to 1 of its top frequency, sampled 16,384 times at instants displaced by a seeded 10 ps Gaussian, at seven top frequencies from 100 kHz to 100 MHz. The measured signal-to-error ratio (dots) is 88.64 dB at 1 MHz, against −20 log(2π × rms frequency × jitter) = 88.76 dB from the band's power-weighted rms frequency, 580 kHz — and against 84.04 dB for a tone at 1 MHz itself. The gap is 4.77 dB at every frequency, 10 log(3/(1 + r + r²)) for a flat band from r to 1 of the top: 4.77 dB (a factor of √3 in frequency) for a band from zero, nothing for a narrow one. At 1 MHz the band holds 14.43 bits where the tone holds 13.67. Where a signal becomes a number

The frequency a band moves at

The jitter rule is written for a sinusoid: ten picoseconds holds a tone at 1 MHz to 84.04 dB. A signal that fills the band up to 1 MHz holds 88.64 dB against the same clock, because each sample's error is its own slope times its own timing error, and a band's slope is set by its rms frequency — the top frequency over √3 for a flat band, 4.77 dB better. Pink content up to 1 MHz is 11 dB better than the rule and content falling as 1/f² is 29 dB better. So the same ten picoseconds hold sixteen bits to 199 kHz for a tone, 344 kHz for a flat band and 734 kHz for a pink one: an aperture figure needs a signal as well as a frequency.

 Four 1% parts spread evenly across their band: the divider's error has a spread of 0.289%, and 8.6% of builds are beyond ±0.5%. computed by solving, not by drawing: a divider of four equal 10 kΩ resistors tapped at the middle, built 6000 times from seeded parts uniform across ±1%, each copy's output error taken from the ratio in closed form and the first four hundred also solved on the netlist, agreeing to 7e-16. The inset is the parts' own distribution; the histogram is the divider's error. Its standard deviation is 0.2894%, half the parts' 0.5780%, because each part moves the output by a quarter of its own error and four quarters add in quadrature to a half. 8.60% of builds fall beyond ±0.5%, which is half the worst case the printed tolerance allows. Networks, and how a solve is checked

The shape a tolerance does not print

A resistor marked 1% promises only that its value lies within one per cent. How the values fill that band is not printed, and it decides almost everything a design does with the part. Four populations that all honour ±1% — spread evenly, Gaussian with the band at three standard deviations, screened from a wider population, and sorted with the best parts removed — give a four-resistor divider spreads of 0.289, 0.168, 0.267 and 0.378 per cent, a factor of 2.26 under one label. Specify the divider at ±0.5 per cent and the yield is 91.4, 99.8, 93.7 and 85.3 per cent. The shape's signature fades as one over the number of parts, so it bites hardest in the circuits with fewest.

 An unregulated supply's output against its load: 3.38 Ω of slope with perfect parts, 3.64 with 0.3 Ω in the winding, 3.32 with it in the capacitor. computed by solving, not by drawing, marched with the diodes in the netlist: a 17 V peak centre-tapped rectifier into 1000 µF, loaded from 200 Ω down to 25 Ω. The mean output with perfect parts falls from 15.99 V at 80 mA to 14.33 V at 573 mA, an average slope of 3.380 Ω against the textbook's 1/(4fC) = 5 Ω, drawn dashed without its diode drop. With 0.3 Ω in each half-winding the slope is 3.642 Ω; with the same 0.3 Ω in the capacitor it is 3.317 Ω. The slider is the resistance. Circuits that do a job, and the range they do it over

The resistance a supply's regulation counts

An unregulated supply's output falls as its load rises, and the slope is its output resistance: 4.17 Ω at 80 mA and 2.80 at 575 for a thousand microfarads, below the textbook's 1/(4fC) = 5 Ω and falling, because the diodes conduct for longer as the load grows. A winding resistance adds to that slope at about its own value, 0.94 to 1.18 of it; a capacitor's adds −0.14 to −0.25. The guess that each should count by the share of the cycle it carries current in — a sixth and five sixths — is wrong in both places, and the right weighting is the mean current each carries: all of the load's through the winding, none through the capacitor.

 A 0.5 dB Chebyshev ladder of order five passes what it does not reflect: the two powers sum to one to 2e-15. computed by solving, not by drawing: a fifth-order Chebyshev ladder of 0.5 dB ripple between 1 kΩ terminations, solved at 481 frequencies. The transmitted power is the transducer gain, read from the load voltage; the reflected power is |ρ|², read from the impedance the filter presents to its source. Their sum is one to 2.0e-15 everywhere. The reflection reaches 0.1087 at each ripple trough, against ε²/(1 + ε²) = 0.1087 with ε² = 10 to the power ripple/10, less one, a return loss of 9.64 dB, and falls to zero at direct current and at 0.555 and 0.898 of the corner, where the transmission is at its maxima. The ripple band ends at 0.944 of the corner. The slider is the ripple. Filters, measured not tabulated

What a filter does not pass, it sends back

A passive LC ladder built from lossless reactances has nowhere to put power except the load and the source. So the power it transmits and the power it reflects sum to what the source makes available — to 2 × 10⁻¹⁵ across a fifth-order Chebyshev, read from two different ports — and every passband specification is a return-loss specification in disguise: half a decibel of ripple is 9.64 dB of return loss at every order, a hundredth of a decibel is 26.4. The same identity predicts the filter's response from any source resistance out of one input impedance, and bounds the damage a wrong source can do by composing two reflections.

 The winding window solved electrostatically, in two portions. computed by solving, not by drawing. The same cross-section the loss solve reads, read with ∇·(ε∇φ) = 0 instead. Two things are the opposite way round from the magnetic problem and both are the whole difference. The iron is now a Dirichlet boundary rather than a Neumann one — an earthed core is an equipotential, so the field meets it at right angles instead of running along it — and a conductor carries a prescribed potential rather than a prescribed current. The thin curves are equipotentials, which are contours of φ, so equal spacing is equal potential step and crowded curves are a strong field. The copper is shaded by the potential each foil sits at, which rises along the winding rather than being one number. Winding to winding this window is 926.9 picofarads a metre, and 89 per cent of the energy is inside insulation that occupies a fraction of the window. Two windings, and the band between them

The other half of the same window

The two-dimensional solve that settled what a winding's alternating-current resistance really is computed one of the window's two parameters and never mentioned the other. Read with Laplace instead of the vector potential, the same cross-section returns 926.9 picofarads a metre — and 89 per cent of that energy sits inside films that occupy 14.1 per cent of the window. The instrument agrees with a layered slab to three parts in ten thousand billion and converges on a real winding at order 1.34, and the reason for the shortfall is not the arithmetic but the corner of a conductor.

 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. Frequency, which is the same solve

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.

 A 1 ns edge from 10 Ω into 200 Ω on 50 Ω track stays inside one per cent of overshoot only up to 2.5 mm of track. computed by solving, not by drawing, the staircase of reflections marched against a ramp: the far end's overshoot against the track's length for a 1 ns edge, 50 Ω track on FR-4 driven from 10 Ω into 200 Ω. 2 mm: 0.79%; 5 mm: 1.98%; 10 mm: 3.98%; 20 mm: 8.07%; 50 mm: 22.93%; 100 mm: 40.00%; 200 mm: 40.00%; 500 mm: 40.00%. The longest track inside one per cent, bisected, is 2.51 mm — a round trip of 35 ps, the edge being 28.5 times it — and the rise-time boundary of that length is 0.994 ns, this edge given back. Where the models stop

How long a track may be

The lumped-element boundary read the way a layout is designed: given the edge, how long may the track be? For 50 Ω track driven from 10 Ω into 200 Ω, a 1 ns edge stays inside one per cent of overshoot for 2.51 mm, and every nanosecond of edge buys 2.52 mm more — proportional to within half a per cent from 0.1 ns to 30. The mismatch decides the rest. From 45 Ω the same edge may cross 23.5 mm; from 50 Ω, or anything above it into this load, no length overshoots at all; and into an open end, a CMOS input, 1.80 mm. The common rule — a track is short while its round trip is under the edge — allows 49 per cent of overshoot from a 5 Ω driver, and a tenth of its length still allows 3.3.

 The same 0.1 ps of period jitter: 107.0 dB at any record length if each edge is independent, 80.3 dB over 4096 samples if the edges walk. computed by solving, not by drawing, averaged over 8 seeded clocks at each of 5 record lengths. A 10.01 MHz tone sampled at 160 MHz by a clock whose every period has a spread of 0.1 ps, in two ways: each edge independently displaced (white), and each edge displaced from where the last one landed (a free-running oscillator's walk). The sampled tone is fitted by least squares for amplitude, delay and a small frequency offset, and what is left is the error. White jitter gives 107.2, 107.0, 107.0, 107.1, 107.0 dB — the closed form 107.0 dB, at every length. The walk gives 92.6, 86.4, 80.3, 74.5, 68.2 dB for 256, 1024, 4096, 16384, 65536 samples, falling 10.1 dB a decade, as the period spread times √(n/15): a jitter with no value until the length of the look is named. Where a signal becomes a number

A jitter that grows with the look

Two clocks with the same 0.1 ps spread in every period sample the same 10 MHz tone. If each edge is displaced independently, the error is 107.0 dB down at any record length. If each edge is displaced from where the last one landed — a free-running oscillator — the edges walk, and the error after fitting the best delay and frequency is 92.6 dB down over 256 samples, 80.3 over 4,096 and 68.2 over 65,536: ten decibels a decade of record, as the period spread times √(n/15). The one oscillator honestly has jitter figures from 0.10 to 4.7 ps depending on the length named. A locking loop is what gives the jitter a value: with a time constant of a hundred periods it stops growing at 87.0 dB.

 On a linear frequency axis the loop's desensitisation is an area, and the area it gives back above crossover is the same: 6.74 and 6.74 kilohertz-nepers at 34.9° of margin. computed by solving, not by drawing. The logarithm of the device sensitivity |1/(1 + T)|, from the cut loop at 3601 frequencies between 0.01 Hz and 10 GHz, drawn in decibels against linear frequency to 60 kHz. It is −60.0 dB at low frequency, crosses zero at 4.47 kHz (the crossover is 5.73 kHz), peaks at +5.70 dB at 6.87 kHz, and returns towards zero as the loop gain goes. The area below zero is 6.74 kHz·Np (58.6 dB·kHz); the area above, integrated to 10 GHz, is 6.74, of which 6.11 lies inside the drawing. The net is −1.48 × 10⁻⁸ Hz·Np. Bode's sensitivity integral says it must be zero for a loop that falls two poles or more and is stable when cut. Feedback, and the margin

The area that feedback only moves

A feedback amplifier divides its device's variation by the loop gain, sixty decibels of it at low frequency, and pays for it with a sensitivity peak near crossover. The payment is exact. Integrated over linear frequency, the logarithm of the sensitivity has as much area above zero as below — 6.74 kilohertz-nepers each side at 34.9° of margin, net zero to eight parts in a billion — at every margin and every gain, because the loop falls by more than one pole. Margin decides only the shape of the repayment: a 14-decibel spike at 11.7°, a one-decibel hump spread past half a megahertz at 80.4°. Only a loop that falls as a single pole keeps any of it, (π/2) times its crossover, and no amplifier is one.

 Parts sharing 90% of their error: the divider's ratio spreads 0.091% and the chain's total resistance 0.559% — one from the same four parts, cancelling, and one adding. computed by solving, not by drawing, over 20,000 seeded builds: four 1% resistors whose errors share a fraction ρ = 0.9 of their variance — as parts from one reel, one wafer or one temperature do — built into a divider tapped at the middle. The ratio's error (solid bars) has a standard deviation of 0.0912%, half the parts' 0.577% times √(1 − ρ); the total resistance of the same four in series (outline bars) has 0.5587%, the parts' spread times √(ρ + (1 − ρ)/4). Correlation that makes a ratio better makes a sum worse, by the same arithmetic. Networks, and how a solve is checked

What a ratio does not see

Parts from one reel, one wafer or one temperature share part of their error, and a circuit that uses them as a ratio cancels exactly the part they share. Four 1% resistors sharing ninety per cent of their error make a divider whose ratio spreads 0.091 per cent and a series chain whose total spreads 0.559 — the same four parts, one answer six times better than independent parts would give and one nearly twice as bad. The ratio's spread falls as √(1 − ρ), the total's rises as √(ρ + (1 − ρ)/4). From −40 to 85 °C the chain moves by up to 0.76 per cent and the ratio by 0.135, inside the ±25 ppm/K the coefficients differ by. And a resistor array whose parts share 99.5 per cent of their error puts a divider inside ±0.1 per cent every time that four discrete parts manage it one time in four.

 Routed along the glass weave, a length-matched pair with one conductor over a bundle and one over a gap builds 63.6 ps of skew in 20 cm — the skew nobody routed. computed by solving, not by drawing by integrating each conductor's delay along its route: a length-matched pair of 0.25 mm pitch over a glass weave of 0.5 mm pitch whose effective permittivity swings between 4.2 and 4.6 across it, positioned so that one conductor runs over a bundle and the other over the gap. Routed along the weave (dashed) the skew grows in proportion to the length, 0.318 ps a millimetre, to 63.62 ps in 200 mm. The slider rotates the route. Lines, where a wire has a length

The skew in the glass

Match a differential pair's two lengths exactly and it can still have tens of picoseconds of skew, because a circuit board is woven glass in resin and the two conductors need not sit over the same one. Over a weave of half a millimetre whose effective permittivity swings from 4.2 to 4.6, a pair of quarter-millimetre pitch with one conductor over a bundle and the other over a gap builds 0.318 ps of skew every millimetre — 63.6 ps in 20 cm, the equivalent of 13.6 mm of routed mismatch by 30 cm, and 68 per cent common mode at 2.5 GHz. Where the pair falls across the weave is a lottery from nothing to that. Rotate the route by ten degrees and the skew never exceeds 0.146 ps, however long the pair: each conductor crosses bundle and gap in turn, and the difference averages out.

 The trade, in the plane where both halves of it live. computed by solving, not by drawing. Leakage inductance across, interwinding capacitance up, both solved on the same cross-section with the same cells. The faint diagonals are lines of constant leakage-times-capacitance, so a design action that runs along one of them has bought nothing and only moved where the energy is kept. Interleaving runs at slope -0.78, which is nearly along them: six sections cut the leakage 21.3 fold and multiply the capacitance 11.0 fold, and the product moves by 1.94. What it does change is the winding's characteristic impedance, 95 ohms down to 6.2 — a factor of 15. Thickening the interlayer instead runs at -4.8, steeply across the diagonals, and moves the product 5.2 fold over the same sweep. It is the cheaper action by that measure and it is not free either: the millimetre it spends is a millimetre of window that is not copper. Two windings, and the band between them

Interleaving is a choice, not an improvement

Splitting a transformer's windings into six sections divides its leakage inductance by 21.4 and multiplies its winding-to-winding capacitance by 11.0. The product of the two — which is what sets the frequency the part stops being a transformer at — moves by 1.94, and the resonance it decides goes from 1.804 to 2.515 megahertz for all that work. What interleaving really changes is the winding's characteristic impedance, 95.2 ohms down to 6.2, and nobody quotes it.

 3 relaxations over two decades come back within 0.22% from readings good to 1e-5. computed by solving, not by drawing. A one-per-cent tail divided equally among 3 branches spread over two decades from 0.1 ms, read at 80 times with 1e-5 of seeded noise on each reading, and the amplitudes recovered by least squares with the time constants given to the fit. Dots: one recovery; bars: the rms spread over 200. The condition number of the fit is 8.29e+0, and the worst branch comes back with an rms error of 0.224 per cent. Before the steady state

How many relaxations a settling curve holds

A settling curve read at eighty times, each good to ten parts in a million of the step, returns the amplitudes of three relaxations spread over two decades to 0.22 per cent and of eight to 46 — even when the fit is handed every time constant exactly. The condition number of the inversion grows by a factor of 7.7 for each extra relaxation packed into a decade, so each tenfold improvement in the reading buys a little more than one more relaxation a decade and no more. A settling measurement can count the dielectric's slowest relaxations; it cannot resolve a crowded spectrum, and how crowded is readable before any amplitude is.

 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. Frequency, which is the same solve

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.

 The best foil thickness for 4 layers, for three currents with the same fundamental. computed by solving, not by drawing. The loss of a portion of 4 layers against foil thickness, with the loss weighted by the current in each harmonic rather than computed for one frequency. A sinusoid wants 0.6631 skin depths and lands at 1.3368 times the direct-current resistance — four thirds, the single-frequency constant, reproduced. A triangular ripple wants 0.6432, which is the same answer to within 3.0 per cent, so a winding carrying one needs none of this. A square current of the same fundamental wants 0.3838 — thinner by a factor of 1.728 — and lands at 1.8313, which is not four thirds and is not any constant the geometry knows. Building to the sinusoid's answer costs 16.8 per cent more loss. Two windings, and the band between them

The optimum a spectrum moves

The best foil thickness for a winding is derived for one sinusoid and quoted as a property of the geometry: a minimum at four thirds of the direct-current resistance, whatever the layer count. Weight the loss by the current in each harmonic instead and a square current of the same fundamental wants foil 1.728 times thinner and lands at 1.83, and a narrow pulse wants it 3.68 times thinner. Four thirds is a property of the current. The constant that replaces it for an ideal square edge is exactly two, and a real winding sits between them at a place its edge rate decides.

 Weighted towards the fast end as τ^(−1), the tail falls as a power of time and each decade of band costs 9.9× the time. computed by solving, not by drawing. A one-per-cent tail spread densely over five decades of time constant from 0.1 ms to 10 s, with each branch's amplitude weighted as τ to the power −m, for m = −0.5, 0, 0.5 and 1. The drawn weighting is m = 1: it settles to 10⁻³, 10⁻⁴, 10⁻⁵ and 10⁻⁶ in 972µs, 9.71ms, 96.3ms, 891ms. With m above zero the fast end carries more and the error between the ends is a power of time; at or below zero the slow end decides. Before the steady state

The tail with no slowest relaxation

Spread a one-per-cent tail densely over five decades of time constant and weight each relaxation as τ to the power −m. With m = 1 the tail falls as one over time and settles to 10⁻⁴ in 9.71 ms and to 10⁻⁶ in 891 ms — each tenfold tighter band costing ten times the time, not the fixed τ·ln 10 one relaxation charges. With even weighting it takes 14.4 s to reach 10⁻⁴; weighted towards the slow end, 26.5 s. The exponent decides whether any relaxation is the slowest one that matters: above zero none is, the settling is a power of the band with slope −1/m, and the spread's far end is never reached.

 With a 2 ps reference and a 0.1 ps-a-period oscillator the best loop has a time constant of 20.0 periods, where the clock's error is 0.453 ps — 90.9 dB on a 10 MHz tone. computed by solving, not by drawing, with seeded locked clocks as dots. A free-running oscillator whose period has a spread of 0.1 ps, locked by a first-order loop to a reference whose timing error, as the phase detector sees it each period, is 2 ps rms and white. The loop passes the reference's error below its corner and the oscillator's walk above it, so the stationary error is σₒ²/(1 − a²) + σᵣ²(1 − a)/(1 + a) with a = exp(−1/τ). The oscillator's part rises as √(τ/2)·σₒ and the reference's falls as σᵣ/√(2τ); they are equal, and their sum least, at τ = 20.00 periods, the ratio σᵣ/σₒ, where the clock's error is 0.4528 ps — √(σᵣ·σₒ). Seeded clocks of 65,536 periods: τ = 2, 1.000 ps; τ = 5, 0.656 ps; τ = 20, 0.452 ps; τ = 80, 0.649 ps; τ = 300, 1.187 ps; τ = 1000, 2.105 ps. Where a signal becomes a number

The loop a reference allows

A sampling clock is a free-running oscillator held by a loop to a reference, and the reference has an error of its own. With 0.1 ps of period spread in the oscillator and 2 ps of white error in the reference, a first-order loop of 20 periods leaves the clock 0.453 ps from ideal — 90.9 dB on a 10 MHz tone — and that is the best any first-order loop can do: the oscillator's share rises as the square root of the loop's time constant, the reference's falls as one over it, and they are equal at σᵣ/σₒ. The clock's error there is the geometric mean of the two, so a reference a hundred times worse costs ten. On a spectrum the rule reads as one instruction: put the loop's corner where the free oscillator's skirt meets the reference's floor. And it holds only for long captures. Below about seven loop time constants the free oscillator beats every loop.

 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. Frequency, which is the same solve

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.

 From a 10 Ω driver with a 1 ns edge, a CMOS input of 1.5 pF allows only 0.86 mm of track — less than the bare open end's 1.80 — and one of 100 pF allows 10.9 mm. computed by solving, not by drawing, bisected on the march by characteristics: the longest 50 Ω track a 1 ns edge from a 10 Ω driver can cross into an open end with a capacitance across it and overshoot by less than one per cent, against the capacitance. With none it is 1.80 mm. 0.3 pF: 1.28 mm, 0.5 pF: 1.04 mm, 1 pF: 0.88 mm, 1.5 pF: 0.86 mm, 2 pF: 0.88 mm, 3 pF: 0.96 mm, 5 pF: 1.15 mm, 10 pF: 1.68 mm, 20 pF: 2.71 mm, 50 pF: 5.70 mm, 100 pF: 10.94 mm. The limit falls to its least, 0.86 mm, near 1.5 pF, and from there rises nearly in proportion to the capacitance (slope 0.87 between 20 and 100 pF), because the driver's resistance damps the ring the track's inductance makes with a heavy load. The slider is the driver's resistance. Where the models stop

The heavier input that allows a longer track

Put a capacitance across a CMOS input and the longest track a 1 ns edge from a 10 Ω driver may cross inside one per cent of overshoot does not shrink steadily. It falls from the bare open end's 1.80 mm to 0.86 mm at 1.5 pF, and then it rises — 2.71 mm at 20 pF, 10.9 at 100 — nearly in proportion to the capacitance. A heavy load turns the track into the inductor of a series RLC whose damping, (Rₛ/2)·√(C/L), the driver's resistance supplies, and at every heavy load the one-per-cent length sits at the same damping, about 0.8. So the limit scales as Rₛ²·C: a slower edge buys a 20 pF input only the 0.4 power of its length, and a stronger driver — 2 Ω instead of 10 — cuts the track it may drive from 2.71 mm to 0.20. Into a heavy input, a weak driver is the clean one.

 With the weft counted as well as the warp, the skew a pair can reach is least at 45.0° to the warp — 128 fs — and turning further off the warp makes it worse again. computed by solving, not by drawing, to first order in the permittivity's swing, the summed bound held against an integration over 64 registrations of the pair to the cloth at six angles. A 0.25 mm pair over a cloth whose warp has a 0.5 mm pitch and whose weft has a 0.5 mm pitch, each swinging the permittivity by ±0.2 about 4.4. The warp's contribution falls as the route turns off it, as the earlier figures found; the weft's rises, because the pair's projection across the weft grows as d·sin θ while its yarns are crossed less often. The worst skew is least at 45.00°, 128.3 fs. At 10°, where the warp alone allows 291.4 fs over every registration — twice the earlier figure's 146 fs, which was for a route starting over bundle and gap — the weft adds 13.8 fs. Integrated: 5° 587.6 fs, 15° 215.1 fs, 30° 139.3 fs, 40° 126.3 fs, 60° 139.2 fs, 80° 304.9 fs. The slider is the weft's pitch. Lines, where a wire has a length

The angle two weaves allow

Turning a differential pair a few degrees off a board's glass weave stops its skew growing with length — but a cloth has two weaves, and turning off the warp turns onto the weft. Counting both, the worst skew over every registration of the artwork to the cloth is a sum of two bounds, one falling as the route leaves the warp and one rising as it meets the weft, and it is least at an angle the cloth decides: 45° and 128 fs for two identical half-millimetre weaves and a quarter-millimetre pair, against 305 at 10°; 52° for a finer weft; 67° for a flatter one. Doing the count properly also finds the earlier bound a factor of two low, and finds the obvious answer a trap: on a square weave's exact diagonal the two weaves' product stops averaging and the skew grows again, 2.2 fs a millimetre, while a degree either side it stays at 128.

 With a correlation of −0.3 between the two generators the bowl's bottom stays at 6.67 kΩ and its floor moves from 1.14 dB to 0.83. computed by solving, not by drawing, with the optimum found twice — in closed form and by a direct search of the noise factor, which agree to a part in a million. The 4 nV/√Hz, 0.6 pA/√Hz amplifier of the earlier essays, its two generators given a real correlation coefficient γ = −0.3. The noise figure against source resistance: the cross term 2γ·e·i·R adds the same fraction of the source's own noise at every resistance, so the bowl moves vertically and its bottom stays at e/i = 6.67 kΩ. The floor's excess over one scales as 1 + γ: 0.827 dB against 1.138 for independent generators. The slider is the correlation. The floor, which bounds from below

Two generators with one cause

The two-generator model of an amplifier assumes its voltage and current noise independent. Give them a correlation γ and the source sees a cross term, 2·Re(γ·e·i·Z*), linear in its resistance and its reactance. A real correlation leaves the quietest source resistance at e/i and moves the floor in proportion to 1 + γ: 0.83 dB instead of 1.14 at γ = −0.3. An imaginary one moves the optimum off the resistance axis — 5.77 kΩ with 3.33 kΩ of capacitive reactance at γ = 0.5j — and the noise figure's contours become circles about it; a resistive source cannot use it at all. If the two flicker components share one cause, the sign of the sharing decides the low-frequency floor: at 1 Hz, 3.9 dB or 18.0 instead of 15.1. And the correlation is measurable with three resistors, best with the middle one next to the optimum.

 Two 1% parts sharing 90% of their error: the divider's output spreads 0.129% at a half, 0.236% at ten to one and 0.258% at a hundred to one. computed by solving, not by drawing, over 20,000 seeded builds at each of seven ratios, every ratio computed exactly: a divider of two 1% parts, each with a uniform part's 0.577% spread and sharing a fraction ρ = 0.9 of it, with the top resistor N times the bottom. The output ratio's standard deviation (dots) against N, beside N/(N + 1)·√2·σ·√(1 − ρ) (line) and the same law for the other shared fractions (faint). N 1: 0.1286%; N 2: 0.1712%; N 5: 0.2145%; N 10: 0.2356%; N 20: 0.2477%; N 50: 0.2536%; N 100: 0.2581%. What is shared cancels at every ratio; what is not reaches the output with a leverage of N/(N + 1), which is a half for a halving divider and nearly one for a large ratio — so the same two parts make a ten-to-one divider 1.83 times worse. Networks, and how a solve is checked

The small leg of a large ratio

A divider of two 1% parts sharing ninety per cent of their error spreads 0.129 per cent at a half and 0.236 at ten to one — the same two parts, nearly twice as bad, because what they do not share reaches the output with a leverage of N/(N + 1) rather than one half. Correlation cancels at every ratio; the leverage does not. Built from equal units of one array, a divider of N to one spreads as σᵤ·√(N/M), so at any one area a ten-to-one ratio costs √10 against a half and needs ten times the units to match it, and the units it needs are the bottom leg's. And a ratio far from a half heats its own parts unequally: at 50 V across 11 kΩ the top of a two-part ten-to-one divider runs 46 K hotter than the bottom and the ratio moves 1,055 ppm with perfectly matched coefficients, where a chain of eleven equal units moves at most 29.

 A core walked from lossless to lossy, and the three straight lines it walks along. computed by solving, not by drawing. Loop area, measured coercivity and measured remanence against the threshold spread the model was handed, over five decades of it, driven sinusoidally to ±400 A/m. None of the three is an input: the area is ∮H dB round the marched loop, the coercivity is interpolated where the descending branch crosses zero, and the remanence is read at zero field. Over the lowest four decades all three are exactly proportional to the spread — 0.625961 joules per cubic metre per ampere-metre of spread, a coercivity 0.4499 of it and a remanence of 1.1304 millitesla per ampere-metre — and at 80 A/m the area is 2.69 per cent below the line and the remanence 16.42 per cent, because the pinned operators have reached the flat of the magnetisation curve. At zero the area is 9.8e-15 J/m³, which is the single-valued core of the anhysteretic model. Two windings, and the band between them

One dissipation, two exponents

The core this ladder built takes two numbers — a threshold spread and the fraction of the magnetisation that follows the field with no threshold — and seven rungs moved the first and left the second at 0.55 without ever saying why. The first decides how much loss there is: area, coercivity and remanence are all exactly proportional to it over four decades, at 0.625961 joules per cubic metre, 0.449775 and 1.130408 millitesla per ampere-metre of spread, the last two of which are closed forms. The second decides nothing about the loss at all — it is single-valued, so it contributes exactly zero to the loop area, to twelve digits — and it moves the Steinmetz exponent from 1.5042 to 2.9860.

 With a 300 ps skew and 5 mV of noise a core with 100 ns of headroom saturates after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. Seeded: 2,000 drives of an ideal core by the event map at β = 0.5, 5 mV of threshold noise and a comparator skew of 300 ps, each run until the summed high-minus-low time leaves ±100 ns. A histogram of the counts on logarithmic bins, against the closed-form first-passage distribution, between two barriers, of a walk with drift 0.200 ns a period and spread 1.89 ns per root period. Mean 501.9 against 499.9; quantiles 1% 184, 10% 275, 50% 463; the empirical and closed-form cumulative distributions differ by at most 0.014. Circuits that do a job, and the range they do it over

The count a core survives

A relaxation oscillator with a 300 ps comparator skew and 5 mV of threshold noise saturates an ideal core with 100 ns of volt-second headroom after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. The mean does not see the noise at all until the noise is large enough to saturate the core the other way first; the early tail sees nothing else. Every skew and noise level falls on one curve in one ratio, the skew's count over the count at which its ramp overtakes the walk. Read as a burst, a thousand periods with one failure in a thousand allows 150 ps of skew with no noise, 93.5 ps at a millivolt, and none at all above 2.41 mV.

 With a 100 kHz flicker corner the 0.1 ps oscillator's ratio stops falling ten decibels a decade of capture and starts falling twenty, past 2,277 samples — 1.42 periods of the corner. computed by solving, not by drawing, from the exact expected residual of a summed process, with seeded clocks as dots (24 at each length to 16,384). A free-running oscillator with a 0.1 ps period spread and a flicker of its frequency whose timing spectrum meets the walk's at 100 kHz, sampling a 10.01 MHz tone at 160 MHz. Signal-to-timing-error after removing each capture's best delay and frequency: the walk alone falls ten decibels a decade, 85.7 dB at 1024, 67.6 dB at 65536; with the flicker, 84.1 and 52.9 dB. The flicker's share equals the walk's at 2,277 samples, 14.2 µs, which is 1.42 periods of the corner frequency; beyond it the ratio falls 19.5 dB a decade. Removing a quadratic as well buys 3.80 dB at 65,536 samples and changes no slope. Where a signal becomes a number

A frequency that wanders

Below its flicker corner an oscillator's phase noise falls thirty decibels a decade instead of twenty, because the oscillator's frequency itself wanders. A clock with 0.1 ps of period spread and a 100 kHz corner samples a 10 MHz tone at 84.1 dB over 1,024 samples and 52.9 over 65,536 — 14.7 dB worse than the walk alone — because past 2,277 samples its error grows as the capture rather than its square root. That crossing is 1.42 periods of the corner frequency for every corner. Fitting a parabola as well as a line buys 3.88 dB at every length and changes no slope; each further term buys a fixed fraction. And a first-order locking loop cannot stop it: the locked clock's variance grows with the logarithm of the record, 1.92 dB lost between 4,096 and 262,144 samples, until a loop with an integrator holds it at 87.5.

 A pair with 2.0% saturation-current mismatch, nulled at 300 K: trimmed by a constant it drifts 1.706 µV/K, trimmed by its load ratio 0.000 µV/K. computed by solving, not by drawing, at 31 temperatures — each a Newton solve of both junctions, a bisection for the tail current and a bisection for the input that nulls the output — a bipolar pair at 10 µA a side whose second transistor has 1.02 times the first's saturation current. Untrimmed (grey) it needs 511.9 µV at 300 K and 682.6 µV at 400 K. Trimmed at 300 K by subtracting a constant (dashed) it is zero there and 170.6 µV at 400 K, drifting 1.706 µV/K — the untrimmed offset's drift, untouched. Trimmed at 300 K by making the collector loads 1.02000 to one (solid) it is 0.0 µV at 400 K, drifting 0.000 µV/K. The load ratio acts inside the same logarithm the mismatch does, so it cancels the mismatch at every temperature, not at one. Devices, and the amplitude they stop being linear at

The trim that takes the drift with it

A bipolar pair's offset from a saturation-current mismatch is proportional to absolute temperature, so trimming it to zero at one temperature can trim its drift to zero too — but only if the trim acts inside the same logarithm the mismatch does. Trimmed by its collector-load ratio, a 2% mismatched pair is zero from 250 to 400 K; trimmed by subtracting a constant after it, it drifts the same 1.706 µV/K it drifted untrimmed. Add 20 kΩ of source resistance and a 10% current-gain mismatch, and a fifth of the offset follows the current gain instead: the load trim now leaves −0.574 µV/K, the constant trim 1.256, and above 34 kΩ the constant trim is the better one. A trim cancels the drift of the share whose temperature law it shares and converts the rest; two knobs, set for value and slope, leave 11 µV of curvature.

 At 20 steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9; the trapezoidal rule shows every Q, 1.6% high. computed by solving, not by drawing: a series resistor–inductor–capacitor resonance at 1 kHz, its quality factor set by the resistor, marched at 20 steps per undamped cycle with each rule; the quality factor the march shows — its decay per radian of its own ringing — against the circuit's, from 1 to 500. Lines are each rule's factor per step in closed form; dots are marched and measured from the record at Q = 3, 30, 300, agreeing to 0.3%. Backward Euler's shown Q flattens at 3.272 and BDF2's at 77.9, whatever the circuit's; the trapezoidal rule's sits 1.64% above the diagonal at every Q. Q 3: trapezoidal 3.05, BDF2 3.06, backward Euler 1.62; Q 30: trapezoidal 30.5, BDF2 22.5, backward Euler 2.97; Q 300: trapezoidal 305, BDF2 62.6, backward Euler 3.24. Before the steady state

The Q a rule can show

A resonance marched forward in time decays at the circuit's rate plus whatever the marching rule adds, and the rule's share has a ceiling built into it. At twenty steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9, whatever resistor the circuit has; the ceilings rise as N/2π and N³/4π³. The trapezoidal rule has no ceiling at all — it shows every Q 1.64 per cent high at twenty steps a cycle, a ratio fixed by the step alone. So the steps a cycle needed to read a resonance's Q to one per cent are about 26 for the trapezoidal rule at any Q, 99 for BDF2 at Q = 100 and 62,202 for backward Euler. Marched at Q = 50, the circuit has 0.533 V of ringing left after ten cycles, the trapezoidal rule 0.536 and BDF2 0.346.

 The longest track a 1 ns edge from a 10 Ω driver may cross inside 1% overshoot, with its load lumped at the end and spread over 32 taps. computed by solving, not by drawing: a network of line segments marched by its characteristics, the load spread over equal taps with the last at the far end, the length bisected at each total capacitance. Lumped at the end (faint, the earlier essay's curve) the limit falls from 1.83 mm to 0.88 mm and rises to 10.9 mm at 100 pF. Spread over 32 taps (solid) the valley bottoms at 1.14 mm, and the limit is 29.5 mm at 100 pF — 2.70 times the lumped length — and 74.2 mm at 300 pF. Where the models stop

The load spread along the track

A hundred picofarads at the end of a track ring with its inductance as a series circuit, and a 1 ns edge from a 10 Ω driver may cross only 10.9 mm inside 1% of overshoot. The same hundred picofarads spread over 32 inputs along the track allow 29.5 mm, and 300 pF spread allow 74.2. Spread capacitance is not a load at the end; it lowers the line's own impedance to Z₀ divided by the square root of one plus the ratio of the added to the line's own capacitance, and slows it, and once the loaded line's impedance falls below the driver's — 8.2 Ω for 150 pF over 30 mm — every reflection at the source returns with the same sign and the far end climbs a staircase that cannot overshoot. The valley the lumped load has at a picofarad or two survives, shallower: 1.14 mm at its lowest, against 0.88. Two taps are no better than one; from eight the track is a loaded line.

 Two junctions at 8 to 1 in area, at four currents from 10 µA to 10 mA. computed by solving, not by drawing: each drop is the diode law solved at its own temperature, from 200 to 400 K, with a saturation current that rises 15.0 orders of magnitude over the range. The two drops themselves sit 179 mV apart across the four currents; their difference does not move at all. It is Vₜ·ln 8 to 0.40 parts in a million at 10 mA and to 0.040% at the worst point, 10 µA at 400 K, where the larger junction's saturation current is no longer negligible against the current — 53.758 mV at 300 K, rising 179.19 µV/K — and the straight line through it reaches 0.001 µV at absolute zero (dashed). Devices, and the amplitude they stop being linear at

The voltage at which the slope is zero

Two junctions carrying one current, one of them eight times the other's area, differ by exactly the thermal voltage times ln 8 — 53.758 mV at 300 K, rising 179.19 µV/K along a line that reaches zero at absolute zero to a thousandth of a microvolt — while each drop alone moves 1.9 mV/K. Weighted by 10.72 and added to one drop, the difference cancels it, and the sum is then 1.171704 V. That voltage is not a coincidence of this design: every weighting of the sum leaves the band gap at absolute zero, so a reference's slope at 300 K is its distance from 1.1717 V divided by 300 K, and two hundred references built wrong three different ways sit on that one line to 0.16 nV/K. A reference is trimmed by its voltage, at one temperature, and the voltage it is trimmed to is a statement about how its current is made.

 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. Frequency, which is the same solve

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.

 A capacitor of 67% of the load's reactive power, moved along a feeder whose load is spread evenly over 10 km. computed by solving, not by drawing at each of 30 positions: one phase of an 11 kV feeder of 0.3 + j0.35 Ω/km, carrying 2 MW at 0.8 lagging spread over 30 equal loads. Uncorrected it loses 16.8 kW a phase; with every load corrected to unity at its own terminals, 11.1 kW — so 5.66 kW is the reactive current's share. This capacitor does best at 67% of the way out, where the loss is 11.8 kW and 87.6% of the reactive share is gone. At the substation end it removes nothing; at the far end it overshoots. Power, and the part that does no work

The capacitor two-thirds of the way out

A feeder's reactive current costs loss in every section it passes through, and on a feeder whose load is spread along its length that current is largest at the substation and falls to nothing at the far end. Correcting it at the substation removes no loss at all. One phase of an 11 kV feeder carrying 2 MW at 0.8 lagging over ten kilometres loses 16.8 kW, of which 5.66 kW is the reactive current's; the best single capacitor is about two-thirds of the load's reactive power placed about two-thirds of the way out, and it removes 87.6% of that share — against the eight-ninths the textbook rule gives for an idealised line. Two capacitors at 2/5 and 4/5 remove 95.6%. The same capacitor left in at light load costs more than it saves below half load, and the rule for a continuous load is a statement about many taps.

 The Chebyshev 5's sections, each node's peak before and after equal-maxima scaling. computed by solving, not by drawing. Unscaled (lighter), section by section the largest signal anywhere inside reaches 4.537×, 3.109×, 1.000× the input, so the ceiling is 4.537× at F0b. With a gain of 0.220 at the input, 1.000 and 1.602 between sections and 2.833 at the output, every section peaks at 1× (darker) and the response is unchanged. The floor rises from 249.7 to 883.8 nV, so the range gains 2.16 dB of the 13.14 dB the internal peak was costing. The floor, which bounds from below

The ceiling a gain gives back

A fifth-order Chebyshev loses 13.14 dB of its range because its first section's capacitor node carries 4.537 times the input. The textbook repair is to scale: divide the signal in front of the loud section, restore it later, so that every node peaks at the same level. Done exactly, with silent gains of 0.220, 1.000, 1.602 and 2.833 and the response unchanged to the last digit, it lowers the ceiling to the input and raises the floor from 249.7 to 883.8 nV — and gives back 2.16 dB of the 13.14. Every section between the division and the restoration has its noise multiplied by what the signal was divided by. On the even-order Chebyshevs scaling loses half a decibel; on the seventh-order it recovers 3.77 of 22.45. Reordering the sections and scaling them together gives 4.75 dB. The rest of the ceiling belongs to the high-Q section and no gain moves it.

 Fifth-order filters between Bessel (m = 0) and Butterworth (m = 1), all at one corner: the step at m = 0.5. computed by solving, not by drawing: each pole set interpolated between the Bessel and Butterworth poles of the same rank, renormalised to a 1 kHz half-power point, built as a network and its step taken from the residues. m 0: 0.77% overshoot, 0.68 periods to 1%, 14.1 dB at 2 kHz; m 0.25: 2.19% overshoot, 0.94 periods to 1%, 16.6 dB at 2 kHz; m 0.5: 4.53% overshoot, 1.27 periods to 1%, 20.7 dB at 2 kHz; m 0.75: 7.99% overshoot, 1.58 periods to 1%, 25.7 dB at 2 kHz; m 1: 12.78% overshoot, 2.21 periods to 1%, 30.1 dB at 2 kHz. Every quantity moves monotonically from one end to the other: the families are the two ends of one line, not two choices. Filters, measured not tabulated

The line from Bessel to Butterworth

Bessel, Butterworth and Chebyshev are taught as three choices, and measured at one order and one corner they are points on one line. Interpolate each Bessel pole towards its Butterworth partner and every quantity moves smoothly: a fifth-order filter's step overshoots 0.77% and settles to 1% in 0.68 periods of the corner as a Bessel, 4.53% and 1.27 periods halfway, 12.78% and 2.21 periods as a Butterworth, while its attenuation at twice the corner climbs from 14.1 to 30.1 dB. The Chebyshevs carry the same line on to 47.0 dB at 4.97 periods, and every step along it buys attenuation with settling time — a frontier, with no member beaten on both. Overshoot is not the price: past 0.05 dB of ripple a Chebyshev overshoots less while ringing longer. And a specification of 25 dB at twice the corner within 1.6 periods is met first at order five, by a member nobody names.

 The gain a Wien oscillator needs, when one of its four parts is off its value. computed by solving, not by drawing: for each part moved alone from −5% to +5%, the gain at which the poles reach the axis, bisected on the netlist, against 1 + Rs/Rp + Cp/Cs (to ten digits at every point). A series resistor or shunt capacitor 1% high raises the gain needed to 3.0100; a shunt resistor or series capacitor 1% high lowers it to 2.9901. Every part moves the frequency the same way, by half its error: 1% high on any one of them, 1.58 kHz against 1.59 kHz. The gain needed moves by the error itself, a third of it relative to three. Circuits that do a job, and the range they do it over

The gain a tolerance demands

A Wien oscillator needs a gain of exactly three only when its four network parts are exactly equal. In general it needs 1 + Rs/Rp + Cp/Cs — bisected on the netlist to ten digits — so a series resistor 1% high asks for 3.0100 and a shunt resistor 1% high for 2.9901. Four thousand seeded oscillators from 1% parts set to the textbook's comfortable 3.02 leave a quarter of them silent; 99.9% start only at 3.072, and at that setting the median unit distorts 2.6% — the limiter has to absorb a margin the tolerance forced. The frequency spreads by one part's own spread, 0.58% for 1% parts, as large as the limiter's own 0.771% shift. Parts from one array, tracking to 0.05%, need only 3.0035, because the gain depends on ratios and the frequency on products.

 A 1 ns edge from a 10 Ω driver, a trunk, and a fork into two branches each ending in 2 pF. computed by solving, not by drawing: the network of trunk and two branches marched by its characteristics, the branch length bisected for 1% overshoot at either end. Forked at the driver, each branch may be 2.33 mm, against 1.05 mm for one track carrying both inputs at its end (dashed). Every millimetre of trunk costs more than a millimetre of branch: 0.0 mm of trunk, 2.33 mm branches; 0.2 mm of trunk, 1.89 mm branches; 0.4 mm of trunk, 1.43 mm branches; 0.6 mm of trunk, 1.01 mm branches; 0.8 mm of trunk, 0.57 mm branches; 1.0 mm of trunk, 0.11 mm branches. Where the models stop

The track that forks

Two 2 pF inputs on one track from a 10 Ω driver with 1 ns edges may sit only 1.05 mm away inside 1% overshoot if both hang from its end. Fork the track at the driver and each branch may run 2.33 mm — more than twice as far — because two fifty-ohm branches in parallel are a 25 Ω line, half as mismatched to the driver. Every millimetre of trunk before the fork costs more than a millimetre of branch: with a millimetre of shared trunk the branches may run only 0.11 mm. Equal branches are the only good ones: a 2.2 mm branch overshoots 0.91% beside an equal partner and 1.7% beside a short one. And a star reaches further the more inputs it serves — 2.33 mm for two, 11.85 for four, where four branches in parallel make 12.5 Ω — while a daisy chain of the same inputs reaches 2.54.

 A sinusoid sampled 8 times a cycle, at the phase that puts its peak between two samples. computed by solving, not by drawing: the samples of a unit sinusoid at 8 a cycle, synchronous, at the sampling phase that hides its peak worst. 8 samples a cycle is 4 times the Nyquist rate and holds the waveform completely; the largest sample is still 0.9239, 7.61% below the peak, which is 1 − cos(π/8). Read straight off the samples the peak needs 22.2 samples a cycle for 1% and 70.2 for 0.1%. Where a signal becomes a number

The rate each reading needs

Twice the frequency is the rate at which a set of samples holds a sinusoid completely, and it is not the rate at which any particular property can be read off them. Eight samples a cycle — four times the Nyquist rate — hold the waveform, and their largest is still 7.61% below the peak; the largest sample needs 22.2 a cycle for 1% and 70.2 for 0.1%, because its error falls only as the square of the rate. A parabola through the three largest gets 1% at 8. A zero crossing found by a straight line between two samples is a hundredth of a cycle right at 5 a cycle, its error falling as the cube, because a sine is straight at its zero. The rms of whole cycles is exact from three a cycle — and at exactly two it reads anything from zero to √2. A slow rate with a long record of a steady signal does what a fast rate does, once the samples have slipped a whole interval past it: 42 cycles at 4.0236 a cycle.

 One channel of a summing amplifier whose junction is shared by 8 inputs. computed by solving, not by drawing. Every input has its own 10 kΩ resistor into one junction and a gain of −1 at a kilohertz; the 10 MHz part closes its loop against a noise gain of 9. Input 1's −3 dB point is 1.17 MHz with 8 inputs, against 6.43 MHz with one: 1: 6.43 MHz, 2: 3.95 MHz, 4: 2.21 MHz, 8: 1.17 MHz, 16: 603 kHz. No other input has to be driven for this to happen; it is enough that its resistor is there. Feedback, and the margin

The input that costs every other input

A summing amplifier gives each of its inputs a gain of −1 and closes its loop against one more than the number of inputs, so every input added divides every channel's bandwidth and multiplies the amplifier's own noise — 6.43 MHz with one input, 1.17 MHz with eight, 603 kHz with sixteen, on a 10 MHz part. Above 5.4 inputs the amplifier's 5 nV/√Hz, multiplied by n + 1, outgrows the resistors' Johnson noise, which rises only as its square root. A muted channel whose source is grounded still costs: a sixteen-input mixer is 11.5 dB louder with two channels in use than one whose unused resistors are disconnected. A channel of gain 0.1 beside one of gain 10 is 13.7 dB noisier than alone. A tree of smaller summers doubles the bandwidth and does not make the sum any quieter.

 A 10 mH winding read at 1 kHz while a 40 mH winding coupled to it at k = 0.99 is loaded from a short to an open. computed by solving, not by drawing at 61 loads for each coupling. Open, the first winding reads 10.0000 mH, its own inductance. Shorted, it reads 199.00 µH — L₁(1 − k²), the leakage seen from this side, because the second winding's current cancels all of the flux the two share. The change happens around a load equal to the second winding's reactance, 251 Ω. At k = 0.5, 0.9, 0.99 and 0.999 the shorted readings are 7500 µH, 1900 µH, 199.00 µH, 19.99 µH. Two windings, and the band between them

The leakage a series reading cannot see

The series-aiding and series-opposing readings recover a coupling coefficient to fifteen figures on a netlist, and on a bench with a 0.1% meter they recover it to about 0.001 at every coupling — which is useless exactly where transformers live. At k = 0.999 the leakage, L₁(1 − k²), is two thousandths of the winding, and the series route leaves it 133% uncertain: 88 of 400 seeded benches report a negative leakage. Shorting the other winding reads the leakage directly, to the meter's own 0.1%, and recovers k 943 times more precisely at 0.999. The short is a short only where the shorted winding's reactance dwarfs its resistance by 10k/√(1 − k²), and the winding to short is the one with the longer time constant.

 A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV². Before the steady state

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

 A 230 V transformer's primary switched on at 0° of the mains, secondary open. computed by solving, not by drawing, marched with the core's saturating flux law and checked at the voltage zero against the netlist march. Settled, the primary draws a magnetising current of 111 mA peak. Switched on at 0° it draws 87.6 A in the first three cycles — 791 times the settled peak. At the voltage zero the first peak is 87.6 A, 791 times; at the voltage peak there is no inrush at all. Power, and the part that does no work

The instant a transformer is worst at

A capacitor-input rectifier draws its worst inrush when it is switched on at the peak of the mains. A transformer's own magnetising branch is the other way round: switched on at the voltage zero, a 230 V primary worked at 1.4 T draws 87.6 A in its first half-cycle, 791 times its settled magnetising current, and at the voltage peak it draws nothing extra at all. The core's flux has to start from zero, the voltage does not, and how close the working flux sits to half of saturation decides the size.

 Five readings of one passive ladder network, each taken forwards and backwards. computed by solving, not by drawing; every reading built as its own netlist, never converted from another. Transimpedance (current in, open-circuit voltage out), transadmittance (voltage in, short-circuit current out) and the reverse voltage gain against minus the forward current gain all give a ratio of one, to 1.0e-13. The voltage gain forwards against backwards does not: 1.000 at 100 Hz, 2.129 at 1 MHz. Nor does the current gain. Networks, and how a solve is checked

The readings that come back the same

Reciprocity is usually stated as 'swap the source and the meter and the reading is unchanged', and that is true of some readings and false of others. On one passive ladder network the transimpedance, the transadmittance, and the reverse voltage gain against the forward current gain each come back the same to 10⁻¹³, while the voltage gain forwards and backwards differ by 13.1 times at 82.5 kHz — exactly z₂₂/z₁₁. A 600-to-50-ohm pad has voltage gains of 0.0426 and 0.511 and one power gain both ways. A controlled source breaks all three reciprocal pairs by the same factor.

 Near-end crosstalk from a 200 ps edge, 100 mm of coupled track against 10 mm. computed by solving, not by drawing, the pair solved mode by mode and returned to time from its harmonics. Over 100 mm the near end rises with the edge, holds a plateau of 12.51 mV per volt for twice the line's delay, 1.33 ns, and falls. Over 10 mm the round trip is shorter than the edge and the pulse never reaches the plateau: 8.34 mV. The plateau is Kb, a quarter of the sum of the two couplings, on the 0.5 V the driven line carries: 12.50 mV. Lines, where a wire has a length

The near end that stops growing

Near-end crosstalk is a length only until the coupled run's round trip equals the edge's rise time; past that it is a constant. Solved mode by mode and returned to time, a 200 ps edge on 100 mm of coupled track gives a plateau of 12.51 mV per volt — Kb, a quarter of the sum of the two couplings, on the driven line's half-volt — held for the 1.33 ns round trip. The knee is at v·tr/2: 7 mm for a 100 ps edge, 75 mm for 1 ns. Past it a longer run adds width and not height, and the pulse's area keeps growing as Kb times the round trip.

 A chebyshev low-pass of order 5: where its impulse response balances, and where its step crosses half. computed by solving, not by drawing, the step built from the network's poles and the impulse response its derivative. The impulse response (dark, scaled to its peak) balances at 0.7090 ms; the group delay at a thousandth of the corner is 0.7090 ms — the same number by a different route. The step (pale) crosses half its final value at 0.7238 ms, 2.1% after it. Filters, measured not tabulated

The delay that is a centre of mass

A filter's group delay at zero frequency is not an approximation to when its step arrives; it is exactly where its impulse response balances. Across three families and orders one to seven the two agree to 5.7 parts in a million by routes that share only the netlist. The step's half-way point is a different thing: ln 2 of the delay for one pole, approaching it from below for a Bessel filter, and passing it — by up to 16.9 per cent — for Butterworth and Chebyshev filters whose impulse responses swing negative. Elmore's RC-tree delay is the same centre of mass, and a uniform ladder network crosses half at 0.757 of it.

 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. Frequency, which is the same solve

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.

 The quantisation error's power as a function of where the input sits, for four dithers. computed by solving, not by drawing; the error's mean square given the input, integrated over each dither's density through the converter's own quantiser, and checked against seeded draws to 6 × 10⁻³. Undithered it is the square of a sawtooth, x² up to half a step and (1 − x)² beyond. One uniform step makes it x(1 − x): zero at a code, a quarter of a step² between codes — the error's power is carried by the signal. Two steps (triangular) hold it at ¼ everywhere, and one step subtracted after the converter at 1/12 everywhere. Averaged over the input the prices are 3.01, 4.77 and 0.00 dB against the undithered 1/12. Where a signal becomes a number

One step of dither per moment

A whole step of uniform dither makes a quantiser's error average to zero wherever the input sits, and leaves its power at x(1 − x) steps²: zero at a code and a quarter between codes, so the noise rises and falls with the signal. Two steps — a triangular dither — hold the power at a quarter everywhere and leave the third moment swinging by 0.096 steps³; three steps fix that too. Each added step fixes one more moment of the error and costs a twelfth of a step² of power: 3.01 dB for one, 4.77 for two. Subtracted after the converter, one step fixes the error's whole distribution for 0.00 dB.

 Where a chain of 4 stages stops obeying its model, against where one stage does. computed by solving, not by drawing. amplifier, 1%: one stage 14.2 kHz, 4 stages 7.06 kHz, a ratio of 0.498; Kirchhoff, 1°: one stage 3.97 MHz, 4 stages 993 kHz, a ratio of 0.250; small-signal, 1%: one stage 7.30 mV, 4 stages 7.26 µV, a ratio of 0.000995. The amplifier's edge moves as N to the −½ because its error grows as the square of frequency; the board's as N to the −1 because a phase grows in proportion; and the small-signal edge collapses, because the last stage sees the gain of all the others. Where the models stop

The edge of a chain

A system built from stages that are each inside their own edge is not inside its own: the stages share one error budget. Four ×10 stages from a 1 MHz amplifier are 1% from ideal at 7.06 kHz, not at the single stage's 14.2 kHz — the edge moves as N to the −0.502. A signal path through four ten-centimetre boards is a degree long at 993 kHz, a quarter of one board's edge, as N to the −1. The chain's exponent is minus the edge's own exponent in its tolerance, measured on the line y = −x. And the small-signal edge of a chain with gain belongs to its last stage: four junction stages of ten are 1% off at 7.26 µV, a thousandth of one stage's 7.30 mV.

 The supply's mean output against its load, with no leakage inductance, 0.3 mH and 1 mH in the winding. computed by solving, not by drawing, marched with the diodes in the netlist: 17 V peak into 1000 µF, 50 mΩ of winding throughout. Without inductance the output falls 1.69 V from 200 Ω to 25 Ω, an average slope of 3.43 Ω. With 0.3 mH it falls 0.62 V, 1.14 Ω, from a higher start; with 1 mH, 0.42 V, 0.77 Ω. An element that carries the load's current and drops no mean voltage makes the regulation better, not worse. Circuits that do a job, and the range they do it over

The inductance that flattens the supply

An unregulated supply's output resistance counts each resistance by the mean current it carries, and an inductance in the winding carries the whole mean current and drops no mean voltage — so by that rule it should count for nothing. Marched, it counts for less than nothing: 0.3 mH of leakage takes the output resistance at 100 Ω from 3.88 Ω to 0.67, and the mean output falls 0.62 V from 200 Ω to 25 Ω instead of 1.69. The effect has an interior optimum — 0.03 mH makes it slightly worse, 0.3 mH best, 3 mH back to 2.86 Ω — because the inductance first rings the reservoir up and then turns the rectifier toward choke input.

 A field-effect input against a bipolar one biased for each source, at 25 °C. computed by solving, not by drawing. The field-effect input (dark) has 1 mA of drain current and a gate leakage that is 1 pA at 25 °C, from the junction's own law; the bipolar input (pale) is at its best current for each source. It wins between 2.83 kΩ and 5.14 GΩ. At 25 °C the ceiling was 5.14 GΩ; it falls in proportion to the leakage, which the slider's temperature multiplies. The floor, which bounds from below

The window that closes from the top

A field-effect input beats a bipolar one biased for each source between 2.83 kΩ and 5.14 GΩ at 25 °C, and the top of that window is twice the thermal voltage over √β times the transistor's gate leakage: it falls as fast as the input junction leaks. Carried through the junction's own law, the ceiling is 39.5 GΩ at 0 °C, 913 MΩ at 50 and 5.79 MΩ at 150 — while the floor does not move from 2.83 kΩ until 200 °C. A 1 GΩ sensor leaves the window at 48.6 °C, a 10 GΩ one at 16.4, about 33 kelvin a decade. The leakage's doubling interval is not ten kelvin but 7.5 at 0 °C and 15.0 at 150, so the rule of thumb is right at one temperature only.

 An integrator of 100 kΩ and 1 µF with its input grounded: what its own input currents do to it. computed by solving, not by drawing, marched for ten seconds from rest. With 50 nA flowing out of each input and the + input grounded, the output ramps at 50.0 mV/s — the bias current over the capacitance, to six figures — and reaches 0.500 V. With 100 kΩ in the + input to balance the resistances, only the 5 nA of offset current is left and the ramp is 5.00 mV/s. Neither stops: an integrator has no gain to quote, because its gain is time. Measurement, which is a circuit on a circuit

The current a capacitor adds up

An amplifier's input current in a resistor is an offset; in a capacitor it is a rate, and time is the gain. An integrator of 100 kΩ and 1 µF with its input grounded ramps at 50.0 mV/s from 50 nA, and balancing the inputs leaves the 5.00 mV/s of the offset current, still unbounded. A leak resistor bounds it at the current times the leak and costs the lowest frequency the circuit integrates. A hold capacitor's droop times its acquisition time is fixed, so 50 nA through a 100 Ω switch needs 655 nF and 0.77 ms for half a 16-bit step. A charge amplifier's offset times its low corner is Ib/2πCf, whatever its bleed.

 A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it. Before the steady state

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

 The second harmonic in each cycle of a transformer's inrush, as a share of the fundamental. computed by solving, not by drawing, the harmonics taken per cycle of the marched current. The first cycle carries 79% second harmonic and the second 93%. The share falls below the 15% a differential relay is commonly set to restrain at only in cycle 30, when the peak is 205 mA against a first peak of 88 A — the harmonic outlasts the inrush it identifies. Power, and the part that does no work

The harmonic that names an inrush

A transformer's differential relay sees an inrush as current going in and nothing coming out, which is what an internal fault looks like, and it tells them apart by the second harmonic. Marched cycle by cycle, an inrush starts at 79% second harmonic and stays above the usual 15% restraint for thirty cycles — long after its current has fallen from 88 A to 205 mA, because the current unwinds in a few cycles and the flux offset under it takes seconds. A fault's offset is a smooth exponential and peaks at 14.6% only when X/R is as low as three.

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

The envelope that leaves early

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

 What balancing the source resistances is worth, against how correlated the two input currents are, for a bias-cancelled input. computed by solving, not by drawing; 400 seeded parts at each correlation, each solved unbalanced and balanced at a 1 MΩ source, input currents of zero mean and 1 nA spread. The ratio of rms errors (dots) follows 1/√(2(1 − ρ)) (line): 0.72 for independent currents — balancing makes it 38% worse — one at ρ = ½, and 6.5 at 0.99. The cure removes what the two currents share, and is a cure only if they share more than half. Measurement, which is a circuit on a circuit

The cure that needs one current

Balancing the resistances an amplifier's inputs look into removes whatever its two input currents have in common and leaves their difference. On a plain bipolar input the two are one current at two pins, correlated at 0.991, and the cure is worth ten. On an input whose bias current has been cancelled they are two residues of either sign, correlated at 0.10, and over 400 seeded parts the cure makes the error 38 per cent worse. It is worth 1/√(2(1 − ρ)) for a zero-mean current, nothing at ρ = ½, and in general exactly the input current over the offset current.

 Three loops with exactly 45° of phase margin, and their step responses. computed by solving, not by drawing; three lags spaced by a ratio r, the amplifier's gain bisected in each for a 45.000° margin, the step from the closed loop's residues. Lags a hundredfold apart need a loop gain of 141 and overshoot 23.6%. Spaced by three they need 4.7 and overshoot 34.5%; coincident, 2.83 and 41.0%. The same margin, and overshoots 1.73 times apart. Feedback, and the margin

One margin, and the gain under it

A phase margin of 45° is quoted as an overshoot of 23 per cent, and on a loop with plenty of gain below its crossover it is: three lags a hundredfold apart, closed at 45°, overshoot 23.6%. Bring the same three lags together and the loop reaches 45° at a loop gain of 2.83 instead of 141, and overshoots 41.0% — the same margin, 1.73 times the ringing. The closed-loop peak follows the overshoot where the margin does not: predicted from the peak the worst error over 27 loops is 3.1 points, from the margin 22.2. The gain margin moves with it, from 37.2 dB to 9.0.

 A second-order loop at a ratio of 64, with a quantiser of one to five bits: the ratio in the band against the input amplitude. computed by solving, not by drawing: the loop whose noise transfer is exactly (1 − z⁻¹)², marched with a sine inside the band, the error measured through a squared Hann window. At half scale, 1 bit 69.4 dB, 2 bits 82.5 dB, 3 bits 90.3 dB, 4 bits 96.8 dB, 5 bits 102.0 dB. The first extra bit adds 13.1 dB because a one-bit quantiser's error is larger than a step's uniform error; after that each adds six or seven. The amplitude each does best at moves up with the bits — 0.6 for 1, 0.6 for 2, 0.8 for 3, 0.85 for 4, 0.95 for 5 — because a fine quantiser is never driven past its last level by the shaped error. Where a signal becomes a number

The levels the loop cannot shape

Give a second-order noise-shaping loop a quantiser of several bits instead of one and every bit beyond the second buys six or seven decibels, the best input amplitude climbs from 0.6 of full scale to 0.95, and loops of third and fourth order that will not run on one bit run on three or four. What it charges is outside the loop's reach. The levels the quantiser feeds back enter where the input does, so their errors are passed rather than shaped: a three-bit loop loses its lead over a one-bit one when each level is out by 0.027 per cent, its error stops falling with oversampling, and the error arrives as harmonics. One bit is immune because two levels always lie on a line.

 One, two, three and four means of twenty milliseconds in cascade: the response, and the band each holds forty decibels down. computed by solving, not by drawing: |sin(πfT)/(πfT)|ⁿ with T = 20 ms, held against running means on samples at three frequencies. Each mean nulls 50 Hz, so n of them null it to order n, and their first sidelobe sits at −13.26 dB a mean. Forty decibels down, one mean holds 49.50–50.51 Hz, two 45.40–55.69, three 40.71–69.49 — its sidelobe at −39.78 dB just above the line — and four everything above 36.90 Hz: the null has become a stopband. The floor, which bounds from below

The mean that makes a stopband

A cascade of n means of one mains period holds, at a depth D, exactly the band a single mean holds at D/n — to a part in a million, because the cascade's response is the single mean's raised to the nth power. That one identity answers whether the third mean pays. Forty decibels down, two means hold 45.4 to 55.7 Hz, three 40.7 to 69.5, and four hold everything above 36.9 Hz: once D/n is shallower than one mean's first sidelobe, 13.26 dB, the null stops being a null and becomes a stopband. The price is a mains period of settling per mean and a noise bandwidth that exceeds a single mean of the same length by √(3n/π) — a factor that never stops growing.

 The droop of the ceramic regulator against the rise time of a 100 mA load, with the capacitor at full value, half and a fifth. computed by solving, not by drawing, marched from the operating point: 10 µF, 5 mΩ of series resistance, 630 pF across the upper divider resistor, the 100 mA arriving as a straight ramp. A step droops 103 mV at 100%, 137 mV at 50%, 203 mV at 20% of the capacitance. A ramp of a millisecond droops 2.98 mV, 2.56 mV, 2.56 mV, and the slowest ramps approach the loop's inductance, 25.2 µH, times the ramp's rate (dashed) — a quantity with no capacitance in it — times 1.18, 1.03, 1.03, the share the loop's ringing adds. The full-value and fifth-value curves cross at 23 µs; slower than that, the larger capacitor droops more. Circuits that do a job, and the range they do it over

The load that rises slower than the loop

Every droop a regulator is sized against is a load step with an infinitely fast edge, and a ceramic regulator droops 103 millivolts on its full 10 µF and 203 on a fifth of it. Give the same 100 mA a rise time and the ordering does not survive: past 23 microseconds the full capacitor droops more than the fifth, and a slow ramp's droop is the loop's own 25.2 µH times the load's rate of change — a number with no capacitance in it, raised 18 per cent on the full capacitor by the impedance peak its slower loop leaves. For a 10 mV budget the load must take 298 µs to arrive on the full capacitor and 253 on the fifth.

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

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.

 The harmonics of one phase of three balanced rectifier loads conducting 60°, each sorted into the sequence its three phasors form. computed by solving, not by drawing: each harmonic of the three phases decomposed into zero, positive and negative sequence. Every one is a single sequence to the arithmetic's floor: h1 positive, h3 zero, h5 negative, h7 positive, h9 zero, h11 negative, h13 positive, h15 zero. A balanced set shifts harmonic h by h × 120° between phases, so the third and its multiples are not shifted at all — zero sequence — the seventh is shifted as the fundamental is, and the fifth the other way round. Relative to the fundamental, the zero-sequence harmonics together are 72.5%, the negative 24.1% and the positive 16.7%. Power, and the part that does no work

The sequence a harmonic belongs to

Symmetrical components are defined on phasors, so they describe one frequency, and a rectifier load's current has many. Taken harmonic by harmonic, three balanced loads give a single sequence at every harmonic, to the arithmetic's floor: the third and its multiples zero sequence, the seventh, thirteenth and nineteenth positive, the fifth, eleventh and seventeenth negative — because harmonic h is shifted h × 120° between phases. The sequence decides where each harmonic can go. Connected line to line, the loads put no triplen in a line at all: 3.32 A of them circulates round the delta and never leaves it. Fed through a delta–wye transformer beside a second group, the fifth and seventh cancel in the supply and the eleventh and thirteenth remain, because the transformer turns the two sequences opposite ways.

 A milliohm read four-terminal with a drifting thermoelectric EMF: the error against the test current, read three ways, and the shunt's own heating. computed by solving, not by drawing: an EMF of 0.5 µV settled plus 1 µV relaxing with a minute's time constant, read thirty seconds after a disturbance with readings a second apart; the shunt 20 ppm/K and 20 K/W. At 10 mA one polarity is out by 11.1%, a reversal by 501 ppm, the three-reading delta by 4.14 ppm. Every EMF error falls as 1/I; the heating, 0.400 ppm at an ampere, rises as I². The best current is 11 A for one polarity (149 ppm in all), 1.8 A reversed (4.08 ppm) and 0.37 A with the delta (0.167 ppm). Measurement, which is a circuit on a circuit

The voltage four wires carry in

Four wires remove a resistance measurement's leads by keeping the sense loop free of current, and that loop is exactly where a thermoelectric voltage lives. Ten milliamps through a milliohm is ten microvolts, so every microvolt where a copper lead meets a brass terminal moves a single-polarity reading by a tenth. Reversing the current removes any EMF that holds still and leaves the change over one interval; the + − + delta leaves the second difference; four readings leave the third — 501 ppm, 4.14 and less at 10 mA while a disturbed junction settles. Against the shunt's own heating each scheme has a best current, and the least error rises only as the cube root of the shunt's temperature coefficient: the reading method moves the error by orders of magnitude, the alloy by a cube root.

 Averaging an input that drifts 1 µV a cycle through a 1 pF switched-capacitor averager: too short a memory keeps the noise, too long a lag, and the best is 16.5 cycles at 27.2 µV. computed by solving, not by drawing, and seeded. A switched-capacitor exponential averager — 1 pF sampling capacitor, an equal damping capacitor for unity gain, memory set by the feedback capacitor — fed a ramp of 1 µV a cycle. The rms error against the input (solid) is the lag (dashed), (a/(1 − a))·r, and the switch noise (dotted), 2kT(Cs + Cd)/(Cf²(1 − a²)), together; seeded marches of sixty thousand cycles (dots) agree to 5%. The best memory is 16.55 cycles against (kT/(Cs·r²))^(1/3) = 15.88, with 27.2 µV of error: 15.5 µV of lag and 22.3 µV of noise. The floor, which bounds from below

The leak worth building in

An integrator that never forgets cannot follow an input that moves. Give a switched-capacitor integrator a deliberate leak — a damping capacitor that samples the output each cycle and takes its charge away — and it becomes an exponential average with a memory of Cf/Cd cycles. Fed an input drifting 1 µV a cycle, through a 1 pF sampling capacitor, too short a memory passes the switch noise and too long a memory lags the drift; the best, on the exact closed form and on seeded marches of sixty thousand cycles, is 16.5 cycles, for 27.2 µV of error against 89.5 µV in a single sample. The best memory follows a cube-root law, the cube root of kT/(Cs·r²), ten times the drift for a memory 4.64 times shorter, and at it the noise variance is twice the squared lag. And the leak is not free: the damping capacitor is switched too, and at unity gain it samples half of the averager's noise.

 A zero at 1ωₙ moves the fastest-settling damping from 0.782 to 0.873 and the fastest settling from 3.62 to 2.38/ωₙ. computed by solving, not by drawing. The 2% settling time of a second-order step against its damping ratio, with no zero (grey), a zero at 2ωₙ (dashed) and at 1ωₙ (solid), each located by residues on the recovered polynomial and refined by bisection; the step by residues equals the plain step plus (1/b) times its derivative to 2e-16. zero none: fastest at ζ = 0.7825, 3.623/ωₙ; zero 2ωₙ: fastest at ζ = 0.7975, 3.028/ωₙ; zero 1ωₙ: fastest at ζ = 0.8725, 2.376/ωₙ. The zero adds the step's own derivative, which adds overshoot, so more damping is needed to keep the last excursion inside the band — and it speeds the rise, so the best settling is faster than with no zero at all. The cliffs in the curve are still there, moved. Before the steady state

The damping a zero asks for

The fastest-settling damping of 0.78 belongs to a step with no zero in it. Put a zero at the natural frequency and the step becomes the plain one plus its own derivative: the best damping moves to 0.873 and the best settling falls from 3.62/ωₙ to 2.38/ωₙ. A zero five times further out moves nothing, and a zero at three times moves the optimum by only 0.005, yet keeping 0.78 there costs 52 per cent, because the optimum sits on a cliff. A zero in the right half-plane moves the best damping by 0.015 and costs 20 per cent in settling that no damping can buy back.

 Forty thousand periods at β = 0.5: each period's error against the same period's duty error, correlated by 0.008. Seeded: forty thousand periods of the event map, β = 0.5, 5 mV of threshold noise; one in twenty is drawn, each scaled by its own standard deviation (3.4 ns for the period, 2.49 ns for the high-minus-low time). A period is three draws with coefficients A, A + B, B and its duty error the same three with A, B − A, −B, so their covariance is σ²[A² + (A + B)(B − A) − B²], which is zero at every β. Measured, the correlation is 0.0080, inside the ±0.0100 that forty thousand samples resolve. With normal draws, uncorrelated is independent: the cloud is round. Circuits that do a job, and the range they do it over

Every draw shared, and nothing in common

A relaxation oscillator's period error and its volt-second error are made of the same threshold draws, and they are exactly uncorrelated. A period weighs its three draws A, A + B and B; the high half less the low weighs the same three A, B − A and −B; the products sum to A² + B² − A² − B², which is zero at every hysteresis. Over forty thousand periods the measured correlation is 0.008 against a resolution of 0.010. What the two do share is one draw with the period on either side, ±0.157 at β = 0.5, and over any whole burst those two lags cancel term for term. Four thousand bursts judged against twenty-five pairs of limits fail both ways exactly as often as the product of the two rates says: the margins specified separately double-count nothing.

 A power ferrite driven one way to 0.3 T: ungapped, every pulse starts at 90.8 mT; with a 0.1 mm gap, at 22.0 mT. computed by solving, not by drawing, marched on the hysteretic core. A ferrite with 90.6 mT of remanence and a coercive field of 38 A/m, 100 turns on 100 mm² and 60 mm of path, its flux raised to 0.3 T by a pulse and then released by letting the winding current fall to zero, cycle after cycle until the starting flux stops moving. Ungapped (solid) the release ends at H = 0 on the descending branch, 90.8 mT: each pulse can add only 209.2 mT. With a 0.1 mm gap (dashed) the current is zero where le·H + lg·B/μ₀ = 0 — the shear line (dotted) — at H = −29.0 A/m and 22.0 mT, under the bound μ₀·Hc·le/lg = 30.2 mT; each pulse can add 278.0 mT. Two windings, and the band between them

The flux the reset leaves behind

A transformer driven one way only — a forward converter, a single-ended pulse transformer — is reset each cycle by letting its magnetising current fall to zero, and zero current is not zero flux. Marched on the hysteretic core, a power ferrite with 90.6 mT of remanence driven to 0.3 T starts every pulse but the first at 90.8 mT, and keeps 34.9% of the swing a symmetric drive would get, where half is the most a one-way drive could; a square-loop core starts at 212 mT and keeps 14.7%. A gap shears the loop: the current is zero where H = −lg·B/(μ₀·le), so the start is bounded by μ₀·Hc·le/lg and falls onto the bound as the gap grows — 22.0 mT at 0.1 mm, recovering 69 mT of swing for 239 mA more magnetising current. The first pulse from a demagnetised core is the only one that starts at zero, and it is the one a bench test sees.

 At 1.00 A forward the silicon-carbide Schottky hands the switch more charge than the silicon PN diode above 245 V. computed by solving, not by drawing. The charge each diode returns through the switch when it is forced to block, against the reverse voltage, at 1.00 A forward and 100 A/µs. The silicon-carbide Schottky's is its depletion capacitance charged to the rail, closed form checked against Simpson's rule: 11.8 nC at 100 V, 25.0 nC at 400 V, 30.9 nC at 600 V, the same at any current. The ultrafast silicon diode's is its stored minority charge, 18.1 nC at this current from the recovery's closed form, plus 1.59 nC of its own capacitance at 400 V. The two are equal at 245.3 V. Before the steady state

The recovery a Schottky does have

A Schottky diode stores no minority charge and is quoted as having no reverse recovery. It has a depletion capacitance instead, and to block it must be charged to the rail through the switch: a 650 V silicon-carbide part holds 25.0 nC at 400 V whatever current it was carrying. An ultrafast silicon diode with a 20 ns lifetime returns 19.7 nC at 1 A, so below 1.42 A forward the diode said to have no recovery hands the switch more charge, and at 125 °C the boundary falls to 632 mA. In energy the capacitive charge costs 64.9 per cent of V·Q rather than all of it, and the boundary is 799 mA.

 A 4:1 load on a cable of 3 dB matched loss costs the cable 1.53 dB more: the standing wave makes the line lossier, by up to the mismatch loss. computed by solving, not by drawing. The extra loss — power into the line over power reaching the load, beyond the matched line's — against the matched loss, for loads giving a standing-wave ratio of 1, 1.5, 2 and 4: 10·log((a² − |Γ|²)/(a(1 − |Γ|²))) − 10·log a (lines), with a 200 Ω load on a 50 Ω line solved exactly as a line whose loss is all series resistance (dots). At a ratio of 4 the extra is 0.109 dB on a 0.1 dB line, 0.82 on a 1 dB line, 1.53 on 3 dB and 1.94 on 20 dB — rising towards the mismatch loss 10·log(1/(1 − |Γ|²)) = 1.94 dB, which it reaches when the reflected wave is burnt before it can return. Lines, where a wire has a length

The heat a standing wave leaves

A cable improves a mismatch's apparent return loss by twice its own loss because the reflection is burnt on the way back, and the burning is not free: it is extra loss in the cable. For one unit delivered to a load reflecting |Γ| through a line of matched loss a, the line must be fed (a² − |Γ|²)/(a(1 − |Γ|²)). A 4:1 load on a 3 dB cable costs 1.53 dB more than a matched one; on a 20 dB cable, 1.94 dB, which is exactly the mismatch loss, because by then every reflected watt is heat. Short lines multiply their dissipation by (1 + |Γ|²)/(1 − |Γ|²), long ones by 1/(1 − |Γ|²). Solved exactly on a line whose loss is in its conductor, the heat stands where the current does: the hottest point runs at the standing-wave ratio times a matched line's, 4.04 against 4.00, so a power rating limited by temperature falls by the ratio, not by the average.

 Asked for 105% of the nose power, a tap changer that blocks on a wrong-signed step stops at 9.759 V and holds it; one that does not keeps raising, and the voltage keeps falling. Marched. The 10 V source behind 1 Ω of the regulator essays, its load drawing 105% of the line's 25 W nose power at the 10 V setpoint, regulated by a tap changer that raises the ratio by 1.25% every 5 s while the load is more than 20 mV low. With no blocking rule (dashed) it climbs past the nose ratio √(Rₗ/R) = 1.9518 and keeps going, to n = 3.250 and 8.61 V by 900 s, still falling; the integrating regulator of the earlier essay (dotted) collapses the same way, faster. With the rule (solid) the tap at n = 1.9625 is the first to lower the voltage, the regulator stops at 385 s, and the load holds 9.759 V — within a tap of the most this line can give it, 9.759 V. The setpoint is out of reach either way; the rule converts a collapse into a stall. Power, and the part that does no work

The rule that reads the sign of the gain

A regulator that raises a ratio whenever its load is low collapses the voltage once the load asks for more than the line's nose power, because past the nose raising the ratio lowers the voltage. The standard protection is a rule: after each tap, compare the voltage with the one before, and if it fell, stop. On a 10 V line asked for 105% of its 25 W, a 1.25% tap changer with the rule stops at n = 1.9625, the first tap past the nose, and holds 9.759 V — the most the line can give — where without it the ratio climbs to 3.25 and the voltage is 8.61 V and still falling. The rule is exact without noise and fragile with it, because it reads a difference that vanishes at the nose: with 3 mV of measurement noise it wrongly stops a raise that could have reached the setpoint 0.3% of the time at 98% of the nose and 15% at 99.5%. Trusting it to 1% costs margin — 97.9% of the nose at 3 mV, 79.9% at 10 mV — or coarser taps.

 Three waveforms at 16 samples a period, sampled at the worst instant: the largest sample of a sine is 1.92% short, of a triangle 12.5%, of a clipped sine 0. computed by solving, not by drawing: three waveforms with the same period and the same peak, sampled 16 times a period at the phase that puts the peak midway between two samples (dots). The sine leaves its peak as the square of the time, the triangle in proportion to it, and the sine clipped at nine-tenths of its height does not leave it at all for as long as it stays flat. The largest sample therefore falls short by 1.92% for the sine, 12.5% for the triangle, 0 for the clipped sine, at one rate, for three waveforms with the same period. Where a signal becomes a number

The peak that is a corner

The largest of a set of samples falls short of a sine wave's peak as the square of the sample interval, and that law belongs to the sine. Measured over every sampling phase, a triangle's corner falls short as the first power, 12.5 per cent at sixteen samples a period against the sine's 1.92, a peak whose curvature vanishes as the fourth power, and a sine clipped at nine-tenths of its height reads exactly above seven samples a period. The order in which a waveform leaves its peak is the whole of the law, and it also fixes how much better the average reading is than the worst. A real corner is round at some scale, so the law bends where the samples resolve its top harmonic, and a parabola through three samples helps a round peak, barely touches a corner and makes a flat one worse.

 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. Frequency, which is the same solve

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.

 Paralleling a 4 nV/√Hz amplifier on a 50 Ω source, summed through 100 Ω from each: the best is 1.19 dB at 140 parts, against 1.14 dB at 134 parts with a perfect summing node. computed by solving, not by drawing: the noise figure from a 50 Ω source against the number of amplifiers in parallel, each 4 nV/√Hz and 0.6 pA/√Hz, with a perfect summing node (dashed) and with their outputs averaged through 100 Ω each into one node (solid), which adds a thermal 4kTR/N at the node. The summing resistors' noise falls as 1/N exactly as the amplifiers' voltage noise does, so their ratio is the same at every N, and a summing resistor comparable with the resistance an amplifier's voltage noise is worth is a constant factor on that term. The floor, which bounds from below

The resistors that sum a hundred amplifiers

A hundred amplifiers in parallel across a 50 Ω source are an ideal transformer of ratio ten, and the summing network between them and the output is where the comparison was left. Its resistors turn out to be the least of it. N resistors averaging N outputs make the noise of R/N, falling exactly as the amplifiers' voltage noise does, so the resistor equal to a 4 nV/√Hz amplifier is about a kilohm at every N, and at 133 parts 200 Ω costs only a tenth of a decibel. The stage that was forgotten is whatever reads the sum. Averaging leaves it alone, so a following amplifier as good as one of the parts costs 12.14 dB, and each part needs a gain of 26 to bring that under a tenth. Parallel amplifiers then match a transformer with one ohm of winding.

 A 0.2% part sampling 0 V after long at 10 V, holding 1 ms: the first reading is 109 ppm of the step, and the hundredth still 4.5 ppm, every one of them made from the same branches. computed by solving, not by drawing: the reading at each sample, as a fraction of the step, when a sample-and-hold with a 1 µF part of 0.2% absorption samples 0 V after a long time at 10 V, acquiring for 100 µs and holding for 1 ms. The first reading agrees with a march of the netlist. It is 109 ppm, the tenth 46.7 ppm, the hundredth 4.5 ppm and the four-hundredth 0.904 ppm. The earlier samples are still being read, and for far longer than an acquisition suggests, because each acquisition can reach a branch only in proportion to the fraction of its time constant it lasts. Before the steady state

The sample before the sample

A sample-and-hold whose capacitor has 0.2 per cent of dielectric absorption reads a 10 V step 109 parts per million wrong on the first sample after it, 46.7 on the tenth, and 4.5 on the hundredth, because every earlier sample is still in the dielectric. That error has been described as distortion, since it grows with the signal. It is not. The state is linear and one period is a single matrix, so the reading is a fixed weighted sum of earlier samples, a filter with a very long tail: a sine reads 28.1 parts per million low and its largest harmonic is 1.5 × 10⁻¹² of the fundamental. A filter can be undone. Subtracting fifty earlier samples takes a stepped signal from 51.6 parts per million to 1.61, and with the absorption known to 20 per cent it stops at 5.47.

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