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

Stopband attenuation — where it appears

How far a filter's response has fallen at and beyond a stated frequency, which is what an order is usually bought for. It is measured on the built network here, where a realisation's own stray capacitance puts a floor under it that the design does not contain.

Named by 13 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

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

What the filter in front costs

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

digital · Anti-alias 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.

filters · Filter families 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.

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

The ratio that does not walk to one

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

noise · Noise bandwidth 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.

filters · Filter tradeoff An all-pole filter's stopband floor: -120 dB, from 10 fF nobody drew. computed by solving, not by drawing. A butterworth 8 realised at a 20 kHz passband, drawn twice: from its poles, which fall at 48 decibels an octave for ever, and from the same netlist with 10 femtofarads of stray capacitance from its input to its last shunt node — a fraction of a picofarad between two tracks. The second stops falling at -120.0 dB at 136 kHz, because the stray bypasses every pole in front of it and what is left is its ratio to the shunt capacitance it divides against. The poles alone predict -293 dB a decade further out, which is 173 decibels lower. So an all-pole design has a stopband floor exactly as an elliptic one does — -140 dB at 1 fF, -130 dB at 3 fF, -120 dB at 10 fF, -110 dB at 30 fF, -100 dB at 100 fF, which is 20.0 decibels per decade of stray — and above that frequency a faster sample rate buys nothing at all, which is the refusal recorded for the elliptic row and not for this one.

The floor every filter has

The elliptic design was refused because its equiripple stopband is a floor and no sample rate reaches past a floor, while an all-pole design falls at six decibels an octave per pole for ever — so a faster clock is always an answer for one and never for the other. On paper. Ten femtofarads of stray capacitance from a filter's input to its output, which is a fraction of a picofarad between two tracks, puts the eighth-order Butterworth's stopband at −120 decibels from 136 kilohertz onwards where its poles predict −293. Every all-pole design has a floor, and it is a layout rather than a design.

digital · Anti-alias 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.

filters · Null depth 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.

filters · Null depth 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%.

filters · Null depth 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.

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

The mean that makes a stopband

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

noise · Noise bandwidth 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.

filters · Filter families

Named alongside it

The objects these essays reach for when they reach for this one.

Transmission zeroDesign tradeoffComponent toleranceFilter familiesFilter orderThe quality factorElliptic filterModel rangeParasiticsPassband rippleSelectivityAnti-alias filter

All concepts