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

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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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%.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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).

 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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