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

Transmission zero — where it appears

A frequency at which a network passes nothing, produced by a resonance in a series arm and infinitely deep only in the absence of loss. Its depth is what a stray capacitance across the section fills, so a null in a realisation is a measurement of the layout rather than of the design.

Named by 14 essays across 4 fields — each of them below, with the objects they name alongside it.

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 A 1.0% doublet: 0.078 dB in the magnitude, 36× the settling time. computed by solving, not by drawing. Above, the magnitude of a fast circuit followed by a pole and a zero that were meant to cancel and miss by 1.00%, against the same circuit with the cancellation exact: the worst disagreement anywhere up to the fast corner is 0.0777 dB. Below, the error left in the step response, in units of the tail's own amplitude of 0.909%. Settling to 0.10% takes 245.2 fast time constants against 6.9 with the cancellation exact, and the closed form τ·ln(A/B) gives 245.2 — a time that contains nothing of the fast circuit at all.

The cancellation that leaves a tail

A pole and a zero placed on top of each other disappear from the response. Miss by one per cent and the magnitude changes by 0.078 decibels, which no measurement would report as a fault, while the time to settle to a thousandth goes from 6.9 time constants to 245 — thirty-six times longer. The settling time has a closed form containing neither the fast circuit nor the doublet's separation as such, and its consequence is blunt: settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is.

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

The three tolerances that do nothing

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

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

The direction a response is most sensitive to

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

networks · Sensitivity 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 What widens a null: the order of the zero, not the placement of two. Integrated by quadrature on the union of both windows' nulls, at a fixed total length of 40 ms so that every arrangement settles in the same time. One mean of 40 ms holds -40 dB over 2.002 per cent of 50 Hz and passes 12.50 Hz of noise. Two means of 20 ms in cascade — a second-order null in the same place — hold it over 20.59 per cent — a factor of 10.3, which is 10^(−dB/40) because a first-order null rises linearly through the target and a second-order one quadratically — for 1.333 times the noise bandwidth. Splitting the two lengths puts two first-order nulls either side instead, and the widest trough is at a split of 10.12 per cent — 17.98 and 22.02 ms — holding 29.93 per cent for 16.53 Hz — which is exactly √2 times the equal pair's band, at every depth, for slightly LESS noise. It is also a cliff: a split of 10.118 per cent — 0.01 per cent further — leaves no band at 50 Hz at all, because the peak between the two nulls has climbed through the target.

The order of the null, not the number of them

A mean rejects the mains exactly and rejects it one per cent off frequency by forty decibels, whatever the window length. Two means in cascade of the same total length settle in the same time, cost a third more noise bandwidth and hold that forty decibels over ten times the band — the factor being ten raised to minus the depth in decibels over forty, because a first-order null rises linearly through a target and a second-order one quadratically. Splitting the two lengths to straddle the frequency buys a further factor of exactly root two, for no noise at all, and puts the design two parts in ten thousand from a cliff.

noise · Noise bandwidth 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 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.

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 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.

filters · Null depth 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.

filters · Group delay 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

Named alongside it

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

Component toleranceStopband attenuationDesign tradeoffThe quality factorLoop gainModel rangeSettling timeFilter orderIntegrationSelectivityAdjoint networkAveraging

All concepts