Concept

Chebyshev — where it appears

The filter family whose passband ripples equally and which buys the steepest skirt of the three all-pole families for a given order. What it pays is delay variation and ringing, and its even orders cannot be realised between equal terminations at all.

Named by 13 essays across 4 fields — each of them below, with the objects they name alongside it.

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.

filters · Filter families

A single pole, and the brick wall that passes the same noise. computed by solving, not by drawing and integrated over 267 frequencies. The equivalent noise bandwidth is 1.5706 times the −3 dB point, and π/2 is 1.5708. A noise voltage computed with the corner frequency instead is 20.2% low. The area under the curve and the area of the rectangle are the same number.

The bandwidth noise sees

A single pole passes π/2 times as much noise power as a brick wall at its own corner frequency, so a noise voltage computed with the −3 dB point is twenty-one per cent low. Measured by integrating the solved response rather than taken from the table it usually comes from, the ratio is 1.5706 and π/2 is 1.5708. A five-pole Chebyshev's is 0.964 — less than one.

noise · Noise bandwidth

Several sections, and the band they buy

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

lines · Matching

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

The number that was wrong

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

lines · Matching

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.

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

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.

filters · Filter termination

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.

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

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.

filters · Filter termination

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.

filters · Filter families

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.

filters · Group delay

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

Three matches of one length, half a wavelength at f₀: two quarter-wave sections are bad again at 2f₀ and 4f₀; an exponential and a Klopfenstein taper are not bad again at all. computed by solving, not by drawing, as cascades of exact line input impedances. 50 Ω into 200 Ω through three matching sections each half a wavelength long at f₀: a two-section binomial transformer (dashed), an exponential taper (solid) and a Klopfenstein taper designed for 5% ripple (dotted), the tapers as four hundred thin slices. The stepped transformer is a band-pass: matched at f₀, and at 2f₀, where each section is half a wavelength and transparent, |Γ| is the bare mismatch, 0.600, again — and at 4f₀, and every even multiple. The tapers are high-pass: the exponential is 0.0194 at f₀ and under 0.131 everywhere above 1.6f₀; the Klopfenstein holds 0.050 from 1.1f₀ up.

The match with no upper edge

A stepped quarter-wave transformer is a band-pass in disguise: at twice its design frequency every section is half a wavelength and transparent, and a 50-to-200 Ω match reflects the bare 0.600 again. Let the steps multiply over a fixed length and the staircase of n steps returns the mismatch at n·f₀; in the limit the upper edge has gone, and what is left is a taper — a high-pass match. An exponential taper half a wavelength long is 0.019 at f₀ and never above 0.131 past 1.6f₀. The Klopfenstein profile, the Chebyshev design's continuous limit, holds 0.050 from 1.1f₀ up, needs half the exponential's length for |Γ| ≤ 0.1 and an eighth of it for 0.02, and is not continuous: it starts with a step from 50 to 52.6 Ω. And the small-reflection formula that defines all of them as a Fourier transform overstates a 10:1 taper's reflection by 41% — the line gives tanh of it.

lines · Matching

A fifth-order Chebyshev ladder built from parts of Q = 100: its passband sags by 0.94 dB near the edge and 0.38 dB at low frequency — 8.686·δ·τg, the group delay times the dissipation. computed by solving, not by drawing. A doubly terminated 0.5 dB Chebyshev ladder, fifth order, 1 kHz edge, 1 kΩ terminations, lossless (dashed) and with every inductor given a series resistance and every capacitor a shunt conductance for a uniform dissipation δ = ωc/Q = 62.8 s⁻¹ (solid), and the lossless response plus 8.686·δ·τg(f), with τg read off the lossless filter's phase (dots). The prediction holds to 3.6% of the extra loss across the passband. The extra is smallest where the delay is smallest, 0.383 dB at 38 Hz, and largest at 953 Hz near the edge, 0.943 dB, where the delay peaks at 1.79 ms: the ripple the design specified is rounded off where it was steepest.

The derivative that is a delay

A doubly terminated ladder filter is stationary in every element's value at its ripple peaks, and that insensitivity says nothing about the parts' losses. Give every reactance the same dissipation δ — a series resistance δL in each inductor, a shunt conductance δC across each capacitor — and every reactance has the impedance it had at s + δ, so the lossy ladder is exactly H(s + δ). By the Cauchy–Riemann equations the loss then rises by 8.686 times δ times the group delay, in decibels: the derivative of a ladder filter's response with respect to its parts' dissipation is its group delay. Solved with the resistors in, a fifth-order 0.5 dB Chebyshev of Q = 100 parts sags 0.383 dB at low frequency and 0.943 dB near its edge, where the delay peaks at 1.79 ms, the law holding to 3.6%. The sag falls as one over Q, grows with the order — 0.37 dB at third order, 2.74 at ninth — and is set by the mean of the inductors' and capacitors' dissipations.

networks · Sensitivity

Named alongside it

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

Reflection coefficient Bessel Butterworth Closed form Filter families Filter order Group delay Available power Doubly terminated ladder filter Impedance matching Insertion loss Quality factor

All concepts