The collection

Every essay — page 2

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence. Page 2 of 6.

Feedback, and the margin

The site's strongest construction, and the subject supplies it free: a phase margin computed from the loop gain in the frequency domain, an overshoot measured from the step response in the time domain, one circuit, and a required agreement between two numbers that share no arithmetic.

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.

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.

7 figures
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.

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.

5 figures
What a feedback tee buys, and the single thing it charges. computed by solving, not by drawing. Two 50 kΩ resistors with a 6.250 kΩ tap give 500.0 kΩ of transimpedance, which is R₁(1 + R₂/R₃) + R₂ solved on the netlist. That expression has no term for the noise gain, and the tap sets it to 1 + R₁/R₃ = ×9.00 — against exactly one for a single resistor of any value, because at direct current the source is a capacitor. One quantity then does all three things at once. The signal and R₁'s own noise are multiplied together, so the tee buys no signal-to-noise ratio at all: 1846 against 5582 in a hertz at a nanoamp for one resistor of the same transimpedance, a factor of 3.02 which is the square root of the noise gain. The amplifier's own voltage noise and offset are multiplied where they were not before. And the phase margin RISES, from 1.8° to 15.9°, because starting the noise gain high shortens its climb to the peak — so the tee is a compensation that charges for itself in noise, and a capacitor across the feedback is the same compensation for nothing. With no tap the three effects are absent rather than small.

The tee that charges for its own compensation

A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.

5 figures
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.

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.

5 figures
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.

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.

6 figures
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.

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.

5 figures
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.

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.

6 figures
1 nF at the node lifts the inverting stage's virtual earth 1.28-fold at its resonance, the integrator's 1.09-fold. computed by solving, not by drawing. The impedance looking into the summing junction of an inverting stage (1 kΩ in, 10 kΩ feedback) and an integrator (1 kΩ in, 10 nF feedback) on the same 1 MHz amplifier, bare (dashed) and with 1 nF of sensor capacitance from the node to ground (solid). The inverting stage's node rises as an inductance, and a capacitance across an inductance is a resonance: its node reaches 909 Ω at 126 kHz, 1.28 times the bare node there. The integrator's node is a resistance, the shelf of 15.9 Ω, and a capacitance across a resistance is a corner: the loaded node is at most 1.086 times the bare one, and only where the amplifier's own output resistance is part of the loop.

The sensor an integrator does not see

A sensor's capacitance at an inverting stage's summing junction sits across a node that rises like an inductance, and makes a resonance: 10 nF lifts the node 2.50-fold at 39.8 kHz, the lift is the resonance's Q, and the loop's margin falls from 90° to 24.8°, and to 8.0° at 100 nF. The same capacitance at an integrator's node sits across a resistance — the shelf, 1/(2π·GBW·Cf) — and the shelf does not move, because the capacitance enters the open node and the loop gain together and cancels. With an ideal output the integrator's margin is 90.91° at every sensor from 1 pF to 100 nF. What it does see is its amplifier's output resistance: 50 Ω driving the feedback and sensor capacitors in series is a pole, worst when the two are equal, 58.3° at 10 nF, and it recovers on both sides.

5 figures
A 100 kΩ resistor across the capacitor: the node rises a decade per decade and stops at the shelf. computed by solving, not by drawing. The summing-junction impedance of the integrator (1 kΩ in, 10 nF feedback, 1 MHz amplifier) with 100 kΩ across its capacitor (solid), beside the ideal integrator (dashed) and the 10 kΩ inverting stage (dotted). At the lowest frequencies the resistor closes a direct-current loop and the node is 0.999 Ω; it rises a decade per decade, as a resistive stage's does, and stops on the integrator's shelf of 15.9 Ω at the feedback's own corner, 1/(2π·Rp·Cf) = 159 Hz. Its highest point below a tenth of the amplifier's bandwidth is 16.3 Ω.

The resistor that holds the bottom of the shelf

An integrator's summing junction sits on a flat shelf, 1/(2π·GBW·Cf), over its working band, and at direct current it is nearly its whole input resistor, because a capacitor closes no loop there. A resistor across the capacitor closes one. Below the feedback's own corner the node is then an inverting stage's, rising a decade per decade from Rp/A₀, and the rise meets the shelf exactly at that corner, 1/(2π·Rp·Cf), for any resistor and any amplifier — so the node never rises above its shelf at all. That holds for every resistor below 1/(2π·f₁·Cf), the one that puts the feedback's corner on the amplifier's own pole: 1.59 MΩ here. Above it the bottom climbs past the shelf. The same 1.59 MΩ is where a millivolt of offset becomes 1.58 V at the output, and the price below it is a flat gain of Rp/Rin and arctan(1/(2π·f·Rp·Cf)) of phase.

5 figures
The gain's sensitivity to the device and to the divider add to one at every frequency, and at crossover each is exactly one half (34.9° of margin). computed by solving, not by drawing, three times at each of 241 frequencies: nominal, with the amplifier's gain raised by one part in 10⁵, and with the feedback fraction raised by one part in 10⁵. The fraction of the device's change reaching the closed-loop magnitude is 9.97 × 10⁻⁴ at 10.0 Hz and 1.60 at its peak near 7.94 kHz; the fraction of the divider's change is 0.99899 at 10.0 Hz and falls to −0.60 at 7.94 kHz. They add to one within 1.16 × 10⁻⁵ everywhere, and at the loop's crossover, 5.73 kHz, they are 0.5000 and 0.5000: the real parts of 1/(1 + T) and T/(1 + T), which are one half at |T| = 1 for any phase. The cut loop, a second route, agrees to 9.55 × 10⁻⁶.

Half to the amplifier, half to the resistors

A feedback amplifier is supposed to hand its accuracy to the resistors, and at low frequency it does — 99.9 per cent of a change in the divider reaches the gain and 0.1 per cent of a change in the device. The two fractions are measured here by nudging each and solving again, and they add to one at every frequency, to five parts in a hundred thousand. At the loop's crossover each is exactly one half, whatever the margin: the gain's magnitude answers the amplifier and the resistors equally. As complex numbers both are 1/(2 sin(φ/2)) there, above one for any margin under sixty degrees, and around the crossover lies a band where the gain is more sensitive to both than it would be with no feedback at all.

5 figures
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.

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.

6 figures
A tee of factor 10 turns 1 µV/K of offset drift into 65.1 pA of error at −40 °C, 10 times the single resistor's. computed by solving, not by drawing. The direct-current error of a 10 MΩ photodiode amplifier calibrated at 25 °C, as the input current that would give the same output, against temperature: the part from the amplifier's offset drift of 1 µV/K (solid) and the part from its input bias current, 1 pA at 25 °C and doubling every 10 K (dashed), for a tee of factor 10 (heavy) and a single resistor (light). The offset reaches the output multiplied by the direct-current noise gain — one for the single resistor, since the source is a capacitor, and 10 for the tee — while the bias current sees the same transimpedance through both. The bias term overtakes the offset term at 85 °C with the tee and 40 °C with one resistor; below that the tee's error is the offset's, 10 times as large.

The drift a tee multiplies

A feedback tee reaches a large transimpedance with small resistors, and on a photodiode its price has been counted in noise: at direct current its noise gain is its own factor, where a single resistor's is one. The same noise gain multiplies the amplifier's offset. Calibrated at 25 °C and read at −40, a 10 MΩ amplifier with 1 µV/K of offset drift is 6.56 pA wrong through one resistor and 65.1 pA through a tee of factor ten — ten times, exactly, while the input bias current's error is identical through both. So the tee's drift penalty is hidden wherever the bias current dominates, which for a field-effect input doubling every 10 K is the hot end, and fully exposed at the cold end. The worst error over −40 to 85 °C comes within a factor of two of the single resistor's only above 10 MΩ for a tee of ten, and above 100 MΩ for a tee of a hundred.

5 figures
A 10 µs delay in the loop: the peak rises to 13.53 dB and the sensitivity above crossover becomes a comb, and the two areas are still equal — 7.14 kHz·Np each side. computed by solving, not by drawing, from the loop's closed form, held against the netlist's cut loop to a part in a billion. The three-pole amplifier of the area essay (second pole 4 kHz) with a pure delay of 10 µs in its loop. Its margin falls from 34.9° to 14.3° at an unchanged crossover of 5.73 kHz, and the peak of |1/(1 + T)| rises from +5.70 to +13.53 dB. Above crossover the delay keeps turning the loop's phase, so the sensitivity rises and falls about zero with a period of 100 kHz. The area below zero is 7.144 kHz·Np, the area above 7.144, net 9.00 × 10⁻⁵ Hz·Np — integrated to 50 MHz on a grid that takes sixty steps a tooth.

The phase a delay never stops adding

Put a pure delay in a feedback loop and Bode's sensitivity integral does not notice: with 10 µs in the three-pole amplifier's loop the logarithm of the sensitivity still has equal areas above and below zero, 7.14 kHz·Np each side. What changes is where they are. The delay keeps turning the loop's phase above crossover, so the sensitivity becomes a comb with a tooth every 100 kHz; the desensitisation shrinks from 6.74 to 6.48 kHz·Np, the main peak rises from 5.70 dB to 13.53 and overpays it by 7%, and the comb hands the excess back below zero. The margin reads the peak from below by about 1.4 dB at every delay. Past 16.9 µs the loop is unstable and the integral stops balancing — by exactly the growth rate of the pole pair that has crossed.

5 figures
Sampled at 100 kHz, the loop repays its whole debt below 50.0 kHz: 6.60 kHz·Np each side, with a peak of 8.72 dB — the continuous loop had repaid 89% by then. computed by solving, not by drawing, from the loop's exact zero-order-hold equivalent. The three-pole amplifier of the area essay (second pole 4 kHz), closed through a sampler and hold at 100 kHz. The device sensitivity, in decibels, against linear frequency from zero to half the sample rate, where the discrete response ends. Margin 24.7° at a crossover of 5.71 kHz, against 34.9° for the continuous loop; peak +8.72 dB at 6.45 kHz. The area below zero is 6.598 kHz·Np and the area above, all of it inside the band, 6.598; net 1.84 × 10⁻² Hz·Np. The continuous loop, by 50.0 kHz, had repaid 88.6% of its own.

The band a sampled loop repays in

Close the three-pole amplifier's loop through a sampler and hold at 100 kHz and Bode's sensitivity integral still balances — but only over the band below 50 kHz, because that is all the band a sampled loop has. The whole debt, 6.60 kHz·Np, is repaid there; the continuous loop had repaid 88.6% of its own by the same frequency and spends decades above on the rest. Below a tenth of the sample rate the sampled loop is the continuous loop with half a sample of delay, 24.7° of margin against 24.6°. And the finite band turns the equality into a floor: hold the sensitivity 20 dB down over a third of the band and no loop can peak below 10 dB. This loop's floor is low and its margin sets its peak; the sample rate at which the peak is 3 dB above the continuous loop's is 18 times the crossover.

6 figures
A gain of −1 on a part stable only at gains of ten, with 1.25 kΩ across its inputs. computed by solving, not by drawing. A 100 MHz amplifier whose second pole is at 20 MHz, wired as an inverter with two 10 kΩ resistors. Its signal gain is one at a kilohertz on every curve. Its noise gain is two with nothing across the inputs (dashed): 35.0° of margin and 4.4 dB of peaking. A resistor from the summing junction to ground raises the noise gain without touching the signal: at a noise gain of 10 the margin is 65.6°, the peak is 0.00 dB and the −3 dB point is 14.1 MHz.

The gain added to buy a margin

An amplifier sold as stable only at gains of ten — 100 MHz of gain–bandwidth, its second pole at 20 MHz — has 35.0° of margin and 4.4 dB of peaking as a gain of −1, because that gain closes its loop against two. A 1.25 kΩ resistor from the summing junction to ground raises the noise gain to ten without touching the signal gain, which stays one to a part in ten thousand, and the margin becomes 65.6°: still 14.1 MHz of bandwidth, 2.19 times what a unity-stable part from the same die gives at its natural gain. The price is everything the amplifier contributes, multiplied by ten instead of two — 64.0 nV/√Hz at the output against 20.5 — and a capacitor in series with the resistor buys back the low-frequency half of it and none of the wideband total.

6 figures
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.

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.

6 figures
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.

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.

5 figures
A gain of ten round an amplifier with a second pole: its half-power bandwidth and its noise bandwidth, each over what the gain–bandwidth product predicts. computed by solving, not by drawing: the stage as a netlist round a two-pole amplifier of 1 MHz, the second pole placed at r times GBW/G. The half-power bandwidth agrees with the second-order closed form √r·g(√r/2) to two parts in a thousand at every placement: 1.000 at r = 0.5, 1.272 at r = 1, 1.414 at r = 2, 1.231 at r = 5, 1.110 at r = 10, 1.010 at r = 100 times GBW/G. It is 1.414 times the product's prediction at r = 2 and within ten per cent only beyond r = 10. The noise bandwidth is (π/2)·GBW/G to half a per cent at every r: the second pole moves the one and leaves the other where it was.

The bandwidth the product still gets right

Gain times bandwidth is a constant for an amplifier with one pole, and real amplifiers have a second. Put it at r times the closed-loop bandwidth the product predicts and the half-power bandwidth becomes √r·g(√r/2) times GBW/G — 41.4 per cent wide at r = 2, within ten per cent only beyond r = 10 — so a follower round a 1 MHz part with its second pole at 2.5 MHz runs 40 per cent wider than the rule says. The noise bandwidth does not move at all. A second-order loop's noise bandwidth is set by one coefficient the second pole cannot touch, and it stays (π/2)·GBW/G at every r and every gain, to half a per cent on the netlist. A third pole is what breaks it.

5 figures
One loop, cut at its output wire with 160 nF on it: the voltage injection crosses unity twice, the current injection reads 57 dB too much at low frequency, and together they are the loop gain. computed by solving, not by drawing. A three-pole amplifier closed for a gain of a hundred, with 100 Ω of output resistance and 160 nF of load. The loop gain read at the controlled source's own input, which draws no current (solid); read by a voltage injected in series with the output wire (dashed); and by a current injected into the same wire (dotted). The true loop crosses unity at 5.31 kHz with 8.9° of margin. The voltage reading first crosses at 5.04 kHz with 15.1°, and rises through it again at 10.6 kHz, because above 9.95 kHz the load is a lower impedance than the source. The current reading crosses at 5.82 kHz with 19.8°. Combined through 1/(1 + T) = 1/(1 + Tv) + 1/(1 + Ti) (dots), the two give the controlled source's reading to 3e-9 at every frequency.

Two wrong readings and one loop gain

A loop gain measured on a bench is measured at a wire, and a wire has an impedance on each side. Cut an amplifier's output with 100 Ω behind it and 160 nF ahead, inject a voltage in series, and the reading is the loop gain plus (1 + T)·Zs/Zl — 15.1° of margin where the loop has 8.9°, and a second unity crossing at 10.6 kHz that the loop does not have. Inject a current into the same wire and the error is (1 + T)·Zl/Zs: 19.8°, and 57 dB too much loop gain at low frequency. The two errors are reciprocal, so the readings combine exactly: 1/(1 + T) = 1/(1 + Tv) + 1/(1 + Ti) returns the loop gain at the controlled source's own input to a few parts in a billion, at every frequency and every load.

5 figures
A stage with its second pole at 3 times the bandwidth the product predicts settles within 0.1% in 5.06τ, against 6.91τ for the one-pole stage the product describes. computed by solving, not by drawing: the step of a stage whose amplifier has a second pole at r = 3 times GBW/G (solid), against the one-pole stage of the same product (dashed). Time is in units of τ = G/(2π·GBW). Settled within 0.1% at 5.055τ, against ln(1000) = 6.908τ, with the netlist of a gain-of-ten stage agreeing at 5.057τ. The stage with the second pole is usually faster, whatever it does on the way.

The gain that stops making a stage faster

The gain–bandwidth product says a stage of gain G settles in time proportional to G, and a stage with a second pole does something else. Below r = 4 the closed loop's poles are a complex pair whose real part is half the sum of the open loop's, π times the second pole's frequency, with neither the gain nor the product in it: a 1 MHz amplifier with a 0.5 MHz second pole takes 4.10 µs to settle within 0.1 per cent at unity gain, 4.04 at two and 4.22 at five, where the product says 1.10 to 5.50. Above the knee at G = 4·GBW/f₂ the product returns. In between lies the fastest stage, r = 3.32 for a 0.1 per cent band, which settles in 3.44 time constants against the one-pole stage's 6.91: half the time, and not at the r that gives the widest band.

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Two loops with exactly 45° of margin and the same crossover: with 100% of the extra lag a delay, the step overshoots 28.7%, against 23.3% for two poles. computed by solving, not by drawing: the closed-loop step of a loop K·exp(−sD)/(s(1 + s/p)) with its crossover at one, its phase margin exactly 45°, and 100% of the 45° beyond the integrator supplied by the delay (solid), against the same margin from a lag alone (dashed). Delay 0.785, lag's pole absent. Overshoot 28.7% against 23.3%; settled within 2% at 7.03 against 7.06; gain margin 6.0 dB against no limit. Marched with a history buffer, and at δ = 1 checked against the exact series of the method of steps.

The margin a delay spends

A phase margin can be made of a lag or of a delay, and the number is the same either way. Forty-five degrees from a lag overshoots 23.3 per cent, and forty-five degrees from a delay overshoots 28.7; at sixty degrees the sign reverses, 8.77 against 5.57, and the two cross near 54 degrees. The overshoot is the smaller difference. A two-pole loop's gain margin has no limit at all, and five per cent of its extra lag as delay brings that to 25.5 dB; with all of it delay the gain margin is 90°/(90° − margin), which is 6.0 dB at 45°, and nothing else is in the expression. Across twenty-one such loops the margin alone misses the overshoot by 13.1 points. The closed-loop peak, which already reads the whole band, misses by 1.2.

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Filters, measured not tabulated

Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number. Each family's poles are computed from its definition, built as a network, and then the ripple, the skirt, the delay variation and the ringing are measured on the network that results.

Three filter families at order 5, all with the same half-power point. At three times the corner the Chebyshev is -64.0 dB down, the Butterworth -47.7 dB and the Bessel -28.3 dB. The inset is the passband at forty times the vertical magnification, which is the only place the Chebyshev's half-decibel of ripple is visible at all.

Three families, one corner

Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number, so the table they appear in cannot answer the question anybody has. Here each family's poles are computed from its definition, built as an actual network, and then measured — starting with the step every comparison skips.

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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%.

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.

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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.

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.

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

The zeros that buy an order

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

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One per cent on one component, in two realisations of the same order-5 filter. computed by solving, not by drawing. The two realisations agree to 3e-14 dB before anything is moved. Moving each element in turn by 1%, the worst deviation at the 2 ripple peaks is 0.1621 dB for the cascade and 0.0030 dB for the LC ladder — 54 times smaller. Across the whole passband the two are within 3% of each other, because near the band edge both are dominated by the response's own steepness rather than by the realisation.

A ladder is not a cascade

The same fifth-order Chebyshev, built two ways. A one per cent component moves the buffered cascade's passband by 0.162 dB at the ripple peaks and the doubly terminated LC ladder's by 0.003 dB — fifty-four times less. The number is not the claim: the claim is the exponent. Fitted over two decades of tolerance the cascade's error grows as the 0.99 power and the LC ladder's as the 2.00 power, because at maximum power transfer the response is stationary in every element it contains.

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A fifth-order chebyshev delay, and 1 all-pass section in front of it. computed by solving, not by drawing at 60 frequencies across the passband. The delay varies by 891 µs from end to end; 1 all-pass section, with pole pair found by search rather than taken from a table, brings that to 498 µs — a factor of 1.79. The price is that everything is later: 722 µs more at direct current, which is more than the 392 µs of variation removed.

Flat delay, bought with more delay

An all-pass section has a magnitude of one at every frequency — measured here on a solved network as 1 to within 9 × 10⁻¹⁶ over six decades — which makes it the only thing that can change a filter's delay without touching its magnitude response. It flattens by adding. A fifth-order Chebyshev's 891 microseconds of delay variation comes down to 498, and everything leaves 722 microseconds later than it did.

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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.

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.

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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.

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.

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One Sallen-Key design at 10 kΩ, and the band of impedance levels it survives. computed by solving, not by drawing. A 10.0 kHz unity-gain Sallen-Key section 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 1.0e-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 31.6 Ω to 31.6 kΩ, with the least departure of 0.0133 dB at 1000 Ω; at this setting it is 0.036 dB at 20.0 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.

The same filter a thousand times larger

Multiply every resistance by a thousand and divide every capacitance by a thousand and the response does not change — not approximately, but to a part in ten to the fifteenth, which is the last bits of a double. So a designer has a free parameter that the design says nothing about, and what decides it is the two quantities that refuse to scale: fifty ohms of amplifier output resistance at one end and two picofarads of stray at the other. Between them the realisation survives over three decades of impedance level and nowhere else.

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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.

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.

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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.

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.

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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×.

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.

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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.

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.

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A one-henry inductor with nothing magnetic in it, good for 3.6 decades. computed by solving, not by drawing. The impedance at the input of an Antoniou impedance converter, read as an inductance: a current source drives the node and the voltage is solved for, and the imaginary part divided by ω is what is plotted. Five components — four resistors of 10 kΩ and a 10 nF capacitor — behave as 1000 mH, which as a wound coil would be several henries of wire. It is that inductance to within one per cent from 1.00 Hz to 3.65 kHz, 3.56 decades, and the upper edge belongs to the amplifiers rather than to the arrangement: with ideal ones in the same netlist the inductance is exact everywhere drawn. Nothing in it stores energy in a magnetic field — the current lags because an amplifier is holding a capacitor's voltage somewhere else in the loop.

The inductor that is an amplifier

Four resistors and a capacitor, arranged around two amplifiers, present one henry at a node — an inductance with nothing magnetic in it, which as a wound coil would be several henries of wire. It is that inductance to within one per cent over three and a half decades, and its series resistance goes negative at 63 hertz, which is well inside the band where it is still an excellent inductor. A resonator built around it there does not have a high quality factor; it has a negative loss, and starts on its own noise.

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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.

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.

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A switched-capacitor low-pass driven past half its own clock. computed by solving, not by drawing. The clock is 1.00 MHz and the corner the earlier essay fitted is 1.59 kHz. The falling dashed curve is that continuous model, which knows nothing about a clock and goes on falling. The circles are the marched circuit, read at the frequency the output actually appears at. They part company past half the clock and by 992 kHz the model is 42.1 decibels wrong — an input just below the clock arrives just above direct current, in the middle of the passband, with the passband's own gain. The third curve is the exact discrete transfer function evaluated at the folded frequency, and it agrees with the march to 0.26 decibels, which is what says the march is measuring the folding rather than an artefact of itself.

The filter that samples

The rung below built a resistor out of a clock and measured two ways it is not one: a settling time, and a corner that is a capacitor ratio rather than an R–C product. Both are errors in a value and both get smaller as the design gets better. This is an error of a different kind — the arrangement is not a continuous system at all, and nothing below half the clock shows it. An input at 992 kilohertz arrives at 7.8 kilohertz with the passband's own gain, where the continuous model the rung below fitted says it is 56 decibels down.

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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.

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.

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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.

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.

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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.

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.

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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.

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.

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

Where the Q comes from

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

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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.

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.

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The sign change follows the 0.51 power of the amplifier, not the inductor. computed by solving, not by drawing. Both of the arrangement's frequency boundaries against the gain–bandwidth of the two amplifiers in it, over three decades. The lower curve is the frequency at which the series resistance changes sign, bisected on the sign of the real part; the upper one is where the inductance leaves one per cent. The crossing grows as the 0.513 power of the gain–bandwidth, and the dashed prediction over it is ½√(fc·fₚ) — half the geometric mean of the arrangement's own corner r/2πL = 1.59 kHz and the amplifier's open-loop pole f_t/A₀ — which is inside one per cent while that pole is at least fifteen times below the corner and 8.7 per cent out at the top of the sweep, where it is not. The two boundaries stay between 1.61 and 2.30 per cent of one another throughout, so a faster amplifier moves the active region rather than removing it.

The boundary that improves when the part gets worse

A synthetic inductor's series resistance changes sign at 63.0 Hz with one-megahertz amplifiers, and that frequency is not a property of the inductor. It is half the geometric mean of the arrangement's own corner and the amplifier's open-loop pole — half the square root of their product, which the bisection confirms to a part in a thousand — and it therefore falls as the amplifier's direct-current gain rises, from 686 Hz at a gain of a thousand to 19.9 Hz at a million. The quantity that decides whether a resonator starts does not move at all: it is the transition frequency over four times the Q, 2.50 kHz for a tank of a hundred.

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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.

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.

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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.

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.

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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.

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.

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On a 300 kHz amplifier the stage lags 27.9° at the null: the Q stays at 25.4 and the peak moves 1.02% below it. computed by solving, not by drawing. The 10 kΩ, 10 nF twin-T, exact, as the feedback network of a stage of gain 100 whose own bandwidth is the amplifier's gain–bandwidth over 100, so that at the null, 1591.5 Hz, it lags 27.95° (solid); dashed is an ideal stage. The quality factor is 25.440 against 25.247 — hardly moved — and so is the peak's height, 99.7 against 100.0. What moves is where the peak is: 1.024% below the null, where 2·f₀/GBW is 1.061%. At the null itself the gain is 88.3, 0.886 of the peak. With the shunt elements 1% out, the Q is 33.72 against 33.50.

The peak the amplifier moves

A twin-T around a stage of gain A makes a peak of quality factor (A + 1)/4, and a real stage has a bandwidth of its own, so its gain at the null is complex. The expectation is that the lag turns the loop's gain and the Q rises by 1/cos φ, doubling at 60°. It does not: a single pole that turns the gain by φ also shrinks it by cos φ, and the two cancel in the width of the peak — the Q holds to about one part in the gain up to 60°, and the loop still oscillates at the same shunt error, 4.1% at a gain of 100, to three per cent. What the stage's bandwidth moves is the peak itself: 2·f₀/GBW below the null, with the gain nowhere in it. At 45° of lag the frequency the circuit was built to select sits on its own −3 dB edge.

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A bridged-T with R₁/R₂ = 100 bottoms out at 34.2 dB, and in a loop of gain 1000 it makes a Q of 4.76. computed by solving, not by drawing. A bridged-T of two 10 nF capacitors, R₁ = 100 kΩ bridging them and R₂ = 1.00 kΩ from their junction to ground, which puts its minimum at the twin-T's 1591.5 Hz: its own transfer (dashed), bottoming at 0.0196, 34.2 dB, rather than a null; and the loop it makes as the feedback network of a gain of 1000 (solid), a peak of 48.5 with a Q of 4.760. Dotted, for scale, is the exact twin-T in a loop of gain 100: a Q of 25.2.

The residue that has only one sign

A bridged-T puts its minimum where a twin-T puts its null, with one fewer part, and its residue there is exactly 2R₂/(2R₂ + R₁): real, like the twin-T's, but positive for any values the parts can take. In a loop of gain A the two networks follow one law — the Q falls by 1 + A·ρ — and they sit on opposite sides of it. The twin-T's residue is zero with exact parts and either sign with real ones, so its loop's Q can be anything, up to oscillation. The bridged-T's loop is always damped: it rises with the gain and stops at a ceiling near √(R₁/R₂)/2, 5.00 for a ratio of 100, with no gain up to a million able to make it oscillate. What the ceiling buys is steadiness: at a Q of five, ±1% parts move the twin-T's loop by 9.1% and the bridged-T's by 2.0%.

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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.

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.

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The dual bridged-T is the same function: at a ratio of 100 both floor at 0.0196 and both loops make a Q of 4.76. computed by solving, not by drawing. The transfer of an RC bridged-T — two 10 nF capacitors bridged by 100 kΩ with 1 kΩ from their junction to ground — and of its dual — two 10 kΩ resistors bridged by 1 nF with 100 nF from their junction to ground — over two decades about 1591.5 Hz. The two curves are one to 5.7e-8 of each other, which is the solver's own conditioning: the dual's capacitor ratio plays the resistor ratio's part exactly, the residue is 2/(2 + n) for both, and in a loop of gain 1000 both make a Q of 4.760. What differs is which components carry the ratio.

The dual that is the same function

Swap a bridged-T's resistors for capacitors and its capacitors for resistors and the transfer does not change: two 10 kΩ resistors bridged by 1 nF with 100 nF to ground are the RC network's 100 kΩ and 1 kΩ around two 10 nF capacitors, to 5.7×10⁻⁸, which is the solver's own conditioning. The residue is 2/(2 + n) for both and the loop's ceiling √n/2 for both. So the choice between them is not about the response; it is about which components carry the ratio. Near the ceiling the Q moves by half the ratio's error, and a ratio of two capacitors at ±5% spreads the Q by 10.0% where two resistors at ±1% spread it by 2.1% — and drifts 2.0% over a hundred kelvin where the resistors drift 0.25.

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

The residue that falls with its slope

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

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With inductors of Q 100 the filter absorbs up to 10.5% of the available power, and the loss rate times its delay predicts the curve's shape — 11.2% at the same frequency. computed by solving, not by drawing. The fifth-order 0.5 dB Chebyshev ladder filter of the reflection essay, each inductor given a series resistance ωL/Q at the corner, so that each loses energy at the rate σ = 2πfc/Q. The power absorbed in the inductors, as a fraction of the available power, against frequency; beside it the lossless ladder filter's delay, D = |t|²τ₂₁ + |ρ|²τ₁₁, times σ — the stored energy over the available power, times the loss rate. At Q 100 the absorption peaks at 10.53% at 0.957 of the corner, where σ·D is 11.24%; between 0.3 and 0.95 of the corner the two agree to 7%, the rest being the split between magnetic and electric energy and the terms of second order in the loss. The slider is the inductors' Q.

The heat a delay predicts

A lossy LC ladder filter absorbs power where it stores energy, and a lossless ladder filter stores energy in exact proportion to its delay — the transmitted delay weighted by what it transmits plus the reflected delay weighted by what it reflects. So the droop of a lossy filter is its loss rate times that delay: with equal losses in inductors and capacitors, to 0.09% at Q 10,000. With inductors of Q 100 in a fifth-order Chebyshev, 10.5% of the available power is absorbed at the band edge against 11.2% predicted. Inductor loss alone misses only by the split between magnetic and electric energy — near one in the passband, 4.6 to one at 1.6 times the corner — and the dual ladder filter mirrors the split. The law is first order, and it overstates the band-edge peak by 17.5/Q.

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In the stopband the series-first ladder filter is an inductance — 5.24 R₀ at three times the corner — and its dual a capacitance, 0.191 R₀: the same filter, opposite loads. computed by solving, not by drawing. The input impedance of the order-5 0.5 dB Chebyshev ladder filter realised with a series inductor first and with a shunt capacitor first, both terminated in the design resistance R₀, against frequency from a tenth of the corner to ten times it. In the passband both sit near R₀ and are nearly resistive. Above the corner they part: at 1.2 times the corner 1.33 R₀ at 89.4° against 0.749 R₀ at −89.4°; 2 times the corner 3.19 R₀ at 90.0° against 0.314 R₀ at −90.0°; 3 times the corner 5.24 R₀ at 90.0° against 0.191 R₀ at −90.0°; 5 times the corner 8.90 R₀ at 90.0° against 0.112 R₀ at −90.0°. The product of the two is R₀² at every frequency to 2e-15, which is what duality means for an input impedance. The slider is the order.

The load a filter is in its stopband

A fifth-order Chebyshev ladder filter can begin with a series inductor or, as its dual, with a shunt capacitor, and between its design resistances the two transmit identically. In the stopband they are opposite loads: at three times the corner the first is 5.24 R₀ of inductance and the second 0.191 R₀ of capacitance, their product R₀² at every frequency. That stops mattering only while the source is the design resistance. From a tenth of R₀ the shunt-first ladder filter rejects 55.9 dB at three times the corner instead of 64.0, draws 14 times the passband current at five times the corner against the series-first ladder filter's 0.24, and with a lead inductance of a tenth of R₀ over the corner's angular frequency grows a hole in its stopband — −33.8 dB where a matched source gives −54.0. A soft source does the mirror image to the series-first ladder filter. The first element has to be the one the source cannot short or open.

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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.

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.

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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.

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.

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

The envelope that leaves early

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

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The group delay of fifth-order filters along the line from Bessel to Butterworth, and of a 0.1 dB Chebyshev, all with one half-power point. computed by solving, not by drawing: the delay as a sum over each pole set's poles, held to the slope of each realised network's own phase at four frequencies. At zero frequency the delay is 0.386, 0.413, 0.449, 0.488, 0.515, 0.633 periods of the corner; across the passband it varies by 0.40%, 5.7%, 16.4%, 34.1%, 59.6%, 122.7% of that. The Bessel is flat to the corner and beyond; every step toward Butterworth bends the delay up into a peak near the corner, and the Chebyshev's peak is the largest.

The attenuation a flat delay cannot buy

Along the fifth-order line from Bessel to Butterworth the group delay's spread across the passband grows at every step, from 0.40 per cent to 59.6, and on through the Chebyshevs to 191 — monotone, like overshoot and settling, and lost fastest in the first eighth of the way. The third axis turns the frontier into something sharper than a trade. More order flattens a Bessel's delay four to seven times an order and buys it no attenuation at all: 14.17 dB at twice the corner at order six is the most any Bessel reaches, and from there the attenuation falls toward the 12.04 dB of the Gaussian every delay-flat filter tends to. A flat delay and a steep skirt cannot both come from poles.

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A 63.7 Ω resistor in series with a gyrator's capacitor moves its series resistance's sign change from 63 Hz to 754 Hz, and 70.0 Ω removes it below the amplifiers' edge. computed by solving, not by drawing. The one-henry synthetic inductor of four 10 kΩ resistors, a 10 nF capacitor and two 1 MHz amplifiers with a gain of 10⁵, its series resistance against frequency with a resistor r₄ in series with the capacitor of 0 (grey), 0.9, 1 and 1.1 times 4/(ωₜ·C) = 63.66 Ω. Without it the resistance is −0.153 Ω at 100 Hz and −25.2 Ω at 1 kHz, negative from 63.0 Hz. r₄ adds ω²·L·C·r₄, the same frequency law, so at exactly 4/(ωₜ·C) the resistance is 0.099 Ω at 100 Hz and −0.075 Ω at 1 kHz and changes sign only at 754 Hz; at 1.1 times it stays positive throughout: 2.44 Ω at 1 kHz.

The resistor that pays the amplifiers back

A synthetic inductor's negative resistance grows as the square of frequency, and so does the real part a resistor in series with the gyrator's capacitor adds: Zᵢₙ = sL/(1 + sCr₄) ≈ sL + ω²LCr₄. Fitted on solves across six designs, the resistor that cancels the amplifiers' term is 4/(ωₜ·C) to 1.8% — 63.7 Ω for 10 nF and 1 MHz amplifiers, whatever the resistance level or the open-loop gain. Put in, it moves the sign change from 63 Hz to 754 Hz; a tenth more removes it below the amplifiers' edge and gives a quality factor of 2,579 at 1 kHz where the bare inductor had −250. The catch is that the cancellation is as good as the gain–bandwidth is known: amplifiers 10% slower than the resistor was chosen for leave −2.9 Ω at 1 kHz. Chosen a tenth over the slowest amplifier a ±30% spread allows, the inductor's Q at 1 kHz runs from 312 to 1,847 and is never negative.

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A stray of a tenth of the sampling capacitor moves a one-plate integrator's coefficient by 9.85%; switched at both plates, by 0.14%, nearly all of it the amplifier's own gain error. computed by solving, not by drawing. A switched-capacitor integrator with a 1 pF sampling capacitor, 4 pF of feedback and 0.5 pF at the amplifier's input, and a stray capacitance to ground at the sampling capacitor's switched plate; the error of the per-cycle coefficient against Cs/Cf, against the stray as a fraction of Cs. With one plate switched (solid) the stray is charged to the input and dumped with the capacitor, so it is part of the capacitor: the error is the stray's fraction less the amplifier's small gain error, 9.85% at a tenth. With both plates switched (dashed, A = 1000; dotted, A = 10⁵) the plate that meets the summing node moves between ground and −Vo/A, so the stray's charge carries a factor 1/A: its sensitivity falls from 1/Cf to Cs/(A·Cf²), 4000 times smaller, and what is left, 0.14%, is the amplifier's gain error (1 + (Cx + Cs)/Cf)/A, which the one-plate arrangement has as well.

The capacitance at the switched plate

A switched-capacitor filter's corner is a ratio of two capacitors, and every plate on a chip has a capacitance to the substrate that is on no schematic. Switch one plate of the sampling capacitor and that stray is charged to the input and dumped with it: a stray a tenth of the capacitor is a tenth of error, and a bottom plate's fifteen per cent is fifteen per cent at any size. Switch both plates and integrate into a virtual ground and the stray at the plate that meets the summing node moves only between ground and −Vₒᵤₜ/A — its sensitivity falls from 1/Cf to Cs/(A·Cf²), four thousand times less at a gain of a thousand, and what is left is the amplifier's own gain error, 0.14%. The stray has not vanished: it has moved out of the coefficient that sets the corner and into the integrator's leak, where the direct-current gain becomes A·Cs/(Cs + Cp).

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A follower section of Q = 5 on an amplifier 100 times the corner: +4.79% as designed from the formula; designed for that amplifier, exact at it, and +1.64% and −0.966% on parts 30% slow and fast. computed by solving, not by drawing: the realised Q of a unity-gain Sallen–Key section against the actual amplifier's gain–bandwidth, relative to the typical value it was chosen for, from the poles of the netlist. Designed from the textbook formula (solid): +6.51% at 0.7, +4.79% at 1, +3.78% at 1.3. Predistorted, with its capacitance ratio and resistors found by iteration on the netlist so that the typical part is exact (dashed): +1.64% at 0.7, exact at 1, −0.966% at 1.3. Predistortion moves the line down without changing its slope much: it removes the error the amplifier has in common and leaves what differs between amplifiers.

The Q designed for the amplifier

A follower section's Q comes out 4.79 per cent high at Q = 5 with an amplifier a hundred times the corner, and nothing stops a designer asking for 4.771 so that the real circuit lands on 5.000. That removes the error the amplifier has in common and leaves what differs between amplifiers: parts 30 per cent slow and fast read +1.64 and −0.97 per cent. So the exchange rate is the spread itself, and it is large. One per cent of Q across ±30 per cent asks for 725 times the corner from the formula and 196 designed for the part. The design runs out at a definite ratio, 7.2 times at Q = 5, below which no value of its parts reaches the Q wanted. On a whole Chebyshev 5 the passband error falls from 1.43 dB to nothing on a typical part and 0.55 on a slow one.

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Frequency, which is the same solve

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

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

One solve, read four ways

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

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

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

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

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

Resonance, and the bandwidth it sets exactly

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

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

The capacitor that is an inductor

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

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

The Q the components allow

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

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

The same part written two ways

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

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

The resistor that is only a resistor

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

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

The phase the magnitude already knows

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

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

The straight lines, and where they are not the curve

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

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

The capacitance that is not one number

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

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

The coefficient that is about one reading

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

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

The reading a data sheet does not take

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

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

Two parasitics, and the resonance neither of them has

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

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

The phase a decibel buys

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

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

The margin the straight lines report

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

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

The resistor that is right in size and wrong in angle

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

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

The arrow that goes past where it settles

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

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

How long a sweep waits at each step

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

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

The energy that arrives first

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

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

The match with no knob

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

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

The efficiency a fixed Q costs

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

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

The gain margin the straight lines get exactly wrong

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

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

The corner error a filter hides in its sections

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

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

The zero that lifts the lines

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

8 figures
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.

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.

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

The level the phase cannot know

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

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

The two mirrors that change places

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

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

The resistor that is a helix

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

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

The circle a peak sits on

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

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

A tolerance is an arrow

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

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

A sweep that never stops

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

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

Two parts that drift together

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

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

The part that decides the match

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

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

The Q a pi network makes a designer choose

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

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

Five readings of one Q

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

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

The zero the lines put on the wrong side

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

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

The ring a drive leaves behind

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

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

The zeros that have no side

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

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