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

Quality factor — where it appears

The ratio of a reactance to the resistance beside it, read equivalently as the sharpness of a resonance: its centre frequency over its bandwidth. For a synthetic inductor it passes through infinity and comes back negative, which is a component becoming active rather than becoming excellent.

Named by 42 essays across 10 fields — each of them below, with the objects they name alongside it.

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

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

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

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

Resonance, and the bandwidth it sets exactly

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

frequency · Resonance The load at which a transformer becomes a resonant circuit, at k = 0.99. computed by solving, not by drawing. The peak output of the same 1:1 transformer against its load, as a multiple of what the turns ratio would give. Below about a kilohm the load damps the leakage resonance, the peak is the plateau, and the ratio is one: there is a band, and it is what the rest of this field measures. Above it the damping goes and the response peaks — 1.18× at 832 kHz into 1500 Ω, rising to 8.80× at the light end. A transformer with voltage gain is not a transformer behaving badly; it is a resonant circuit, and asking for "the band" of one returns the skirts of a resonance. So the measurement reports that the response is peaked, rather than returning two edge frequencies in the wrong order.

Where the band goes entirely

The three essays before this one measure a transformer's band, and all three assume there is one. Past a few hundred ohms of load there is not: the leakage that sets the upper edge is also what damps the resonance behind it, and a lightly loaded, well-coupled transformer peaks at 3.03 times its own turns ratio. A passive component with voltage gain is not a transformer behaving badly. It is a resonant circuit, and asking for its band returns the skirts of a resonance.

magnetics · Transformer Linearising an exponential at 125 °C, and what it costs. The linear model understates the gain by 1% at 9.69 mV and by 10% at 30.2 mV. The thermal voltage at this temperature is 34.3 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong.

The edges that move with the room

Every boundary in this collection is quoted at one temperature and most of them are functions of it. The small-signal edge is proportional to the thermal voltage, so it runs from 5.67 millivolts at −40 degrees to 9.69 at +125 — a factor of 1.71, the ratio of the absolute temperatures exactly. A realised Q is 1.54 per cent high at one end of that range and 2.58 at the other. The numbers are right; the condition attached to them was left off, and it is the same condition every time.

limits · Model edges A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees.

The pair that is worse than either

A bulk capacitor and a ceramic are fitted together because each is good where the other is not, and between them is a frequency at which the pair presents six times the impedance either one does alone. The peak is a parallel resonance between one part's inductance and the other's capacitance, its height is one over the series resistance every data sheet asks to be minimised, and at it the two capacitors exchange 5.87 amps for every amp the load draws.

power · Decoupling 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.

filters · Impedance scaling Where one pole goes when each component is 5% high. computed by solving, not by drawing. A series R–L–C, its poles recovered by rooting the determinant, and the derivative of the upper one with respect to each element taken exactly from the two null vectors at the pole. The dashed lines are the first-order prediction for a 5 per cent change; the filled circles are where the root actually goes when the element is changed and the determinant re-rooted. The three directions are the argument: the resistance moves the pole along a circle of constant radius, because the natural frequency does not contain it — its normalised sensitivity has a real part of 3.1e-16. The inductance and the capacitance each carry exactly −½ of the radius, and imaginary parts that are exact negatives. At 5 per cent the prediction is out by 0.122 per cent of the pole's own magnitude.

The derivative of a root

An earlier essay turned one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder filter stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder filter's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.

networks · Sensitivity Every node of a Chebyshev 5, and the one that clips first. computed by solving, not by drawing. The largest signal each node of the realised network ever carries, over the whole frequency sweep, relative to the input. The output reaches 1.000 times the input and F0b reaches 4.537, so on a ±15 V supply the input can be driven to 3.31 V rather than 15.00 before something clips — and the thing that clips is not the output. With the floor at 249.7 nV that is 142.44 dB of range against the 155.57 dB an instrument on the output would report, a difference of 13.14 dB that no measurement at the output can see.

The ceiling is not at the output

A fifth-order Chebyshev's noisiest node is its output and its largest signal is not. F0b carries 4.537 times the input, so on ±15 V the range is 142.44 decibels and not the 155.57 an instrument on the output reports. Across fifteen realised filters the output reading spans 3.97 dB and the range they actually have spans 22.34 — and one family of the three has no internal peaking at all, at any order.

noise · Dynamic range 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.

filters · Impedance scaling 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.

filters · Gyrator 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.

filters · Gyrator The bank reaches 1.500 mΩ and the load sees 141.5 mΩ at that same frequency. computed by solving, not by drawing. One bulk part and 20 ceramics, with a nanohenry of mounting loop each and two nanohenries of plane between the bank and the load, solved once per frequency and read at both nodes. The dashed curve is the bank's own node — what a probe on the parts measures. The solid one is the load. The bank's least impedance is 1.500 mΩ at 11.3 MHz, and at that frequency the load sees 141.5 mΩ, which is 94.3 times more, against 141.5 mΩ of plane reactance at that frequency. Whatever the parts do, the load's reading cannot fall below the reactance of the copper in front of them, and the parts reach their best by moving up the frequency axis into it.

The floor and the ceiling move apart

A decoupling bank is judged by two numbers — the lowest impedance it reaches and the highest frequency at which it still meets its target — and with no copper between the parts and the load both improve together as capacitors are added, 5.000 milliohms down to 1.500 and 19.8 megahertz up to 162. Three nanohenries of ordinary board separate them. The bank's own floor still falls 3.33 times while the load's falls 1.49, and the ceiling read at the parts climbs to 82.4 megahertz while the load's peaks at 8.06 and falls to 5.05. At twenty parts the two nodes disagree by a factor of 94 about the same solve.

power · Decoupling Correcting a 20 Ω, 50 mH load to unity, and what it stores. computed by solving, not by drawing. A capacitor across a 230 V, 50 Hz supply is swept from nothing to 300 µF against a load drawing 1636 W and 1285 var. The reactive power falls through zero at 77.31 µF and keeps going; the energy stored in the installation rises from 2044.9 mJ to 4089.7 mJ at that point — exactly twice, because unity power factor means the two stores are equal — and goes on rising afterwards. Only the cable current has a least value, 7.113 A against 9.044 A.

The energy a unity power factor doubles

Reactive power can be computed three ways, and their agreement is easily called a verification. Two of the three are one theorem written twice and cannot disagree about anything; only the third is independent, and what it computes is a difference. Correcting a 20 Ω, 50 mH load to a power factor of 1.000000 takes its reactive power from 1,285 var to nothing and takes the energy stored in the installation from 2,044.9 mJ to 4,089.7 mJ — exactly twice, at every load and every frequency.

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

The arrow that goes past where it settles

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

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

How long a sweep waits at each step

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

frequency · Phasors A core loss of 100 kΩ gives the resonance a floor at −66 dB, and by 10.0 GHz the choke is worth 0.02 dB. computed by solving, not by drawing. The common mode at the load, in decibels against the same circuit with no choke in it, with 5 pF across each winding and a core loss of 100 kΩ across each, drawn over the lossless curve. At the winding's resonance, 1.59 MHz, the inductance and the capacitance cancel and what is left is the loss, so the deepest point is 66.02 dB — 20·log(1 + (Rp/2)/(Zs + Zl)), with no inductance and no capacitance in it. Above the resonance the part is a capacitor across the path it was fitted to block, and by 10.0 GHz it attenuates 0.02 dB.

The depth a resonance does not have

The inductor one mode cannot see printed its choke's best attenuation as 102 decibels at 1.59 megahertz. Swept with 1,999, 2,000 and 2,001 samples the same lossless part reports 110.0, 96.3 and 103.6 decibels, because a resonance with nothing to dissipate has no bottom and a sweep reports how near its nearest sample fell. Give each winding a core loss and the depth is 20·log(1 + (Rp/2)/(Zs + Zl)) — 66.02 decibels at a hundred kilohms, twenty more per decade of loss, with no inductance and no capacitance in it, and half of it belonging to the circuit the choke sits in.

lines · Common-mode choke 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.

filters · Q enhancement 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.

filters · Q enhancement 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.

filters · Gyrator The resistance that just stabilises it is 17 times smaller than the one that damps it. computed by solving, not by drawing. Two consequences of one base resistor, against how much of it there is, for a follower fed through 100 nH of wire with 47 pF on its emitter. The falling curve is the Q of the worst pole pair, rooted from the determinant so that no frequency grid is involved; the rising one is the output impedance the stage presents at low frequency. Bisecting on the SIGN of the pole's real part returns 69.7 Ω — at which the pair is stable with a Q of 1.7e+15, which is to say no damping and a peak whose height belongs to the arithmetic. A Q of one needs 1.15 kΩ, 16.6 times more, and that resistor takes the output impedance from 5.66 to 13.10 Ω — exactly R/(β+1) added, which is the bare follower's own expression with the base resistance in the place of the source resistance.

What the cure at the base costs

Bisecting for the smallest base resistor that stops an emitter follower oscillating gives 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.

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

The circle a peak sits on

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

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

A tolerance is an arrow

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

frequency · Frequency response 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.

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

A sweep that never stops

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

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

Two parts that drift together

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

frequency · Frequency response At 20 steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9; the trapezoidal rule shows every Q, 1.6% high. computed by solving, not by drawing: a series resistor–inductor–capacitor resonance at 1 kHz, its quality factor set by the resistor, marched at 20 steps per undamped cycle with each rule; the quality factor the march shows — its decay per radian of its own ringing — against the circuit's, from 1 to 500. Lines are each rule's factor per step in closed form; dots are marched and measured from the record at Q = 3, 30, 300, agreeing to 0.3%. Backward Euler's shown Q flattens at 3.272 and BDF2's at 77.9, whatever the circuit's; the trapezoidal rule's sits 1.64% above the diagonal at every Q. Q 3: trapezoidal 3.05, BDF2 3.06, backward Euler 1.62; Q 30: trapezoidal 30.5, BDF2 22.5, backward Euler 2.97; Q 300: trapezoidal 305, BDF2 62.6, backward Euler 3.24.

The Q a rule can show

A resonance marched forward in time decays at the circuit's rate plus whatever the marching rule adds, and the rule's share has a ceiling built into it. At twenty steps a cycle backward Euler cannot show a quality factor above 3.27 and BDF2 none above 77.9, whatever resistor the circuit has; the ceilings rise as N/2π and N³/4π³. The trapezoidal rule has no ceiling at all — it shows every Q 1.64 per cent high at twenty steps a cycle, a ratio fixed by the step alone. So the steps a cycle needed to read a resonance's Q to one per cent are about 26 for the trapezoidal rule at any Q, 99 for BDF2 at Q = 100 and 62,202 for backward Euler. Marched at Q = 50, the circuit has 0.533 V of ringing left after ten cycles, the trapezoidal rule 0.536 and BDF2 0.346.

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

The part that decides the match

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

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

The Q a pi network makes a designer choose

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

frequency · Series-parallel A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV².

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

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

Five readings of one Q

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

frequency · Resonance 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 The Q a 10 mH coil with 1.2 Ω of winding and 8 pF across it shows, against frequency. computed by solving, not by drawing, the coil's impedance at each frequency and its reactance over its resistance. The straight ωL/R line (pale) rises without limit; the coil's Q follows it at first, peaks at 11340 at 325 kHz — 0.5773 of the 563 kHz self-resonance, which is 1/√3 — and falls to zero at the resonance. The peak is 0.3849 of what the line reaches at resonance, 2/(3√3).

The Q that peaks before resonance

A coil's quality factor is quoted as ωL/R and drawn rising with frequency, and its own winding capacitance turns it over: Q peaks at the self-resonance over √3 and falls to zero at the resonance. Eighteen coils from 1 mH with 300 pF to 100 mH with 1 pF all peak at 0.5774 of their resonance, at 2/(3√3) of the line's value there. Skin effect moves the peak down toward the resonance over √5. And the falling Q is a property of the coil read alone: in a tank tuned at 0.8 of its self-resonance, the same coil gives the full ωL/R, 23,570 against the 8,485 a meter reads off the bare part.

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

The ring a drive leaves behind

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

frequency · Phasors The Q of a 10 mH coil with 1.2 Ω of winding and 8 pF across it, with cores whose own Q is 1000, 300, 100 and 50. computed by solving, not by drawing: the core's loss a resistance ωL·Qc across the inductance, constant loss tangent, the coil solved at each frequency. Without a core loss the Q peaks at 11340 at 325 kHz, the self-resonance over √3. With the cores: 825.6 at 136 kHz for Qc = 1000, 274.9 at 94.0 kHz for Qc = 300, 95.9 at 66.2 kHz for Qc = 100, 48.7 at 52.8 kHz for Qc = 50. The peak falls to just under the core's own Q for a poor core and further under it for a good one, and moves down from the self-resonance over √3.

The Q a core allows

A coil's Q, limited by its winding and its own capacitance, peaks at the self-resonance over √3. Put a core with a loss of its own across the inductance and the peak falls and moves: with a core whose own Q is 100, the 10 mH coil that peaked at 11,340 near 325 kHz peaks at 95.9 near 66 kHz. The peak sits where x³ = a/2 — x the frequency over the self-resonance, a the core's Q over the winding's — and reaches the core's Q less three times the two-thirds power of a/2 of it. At that peak the core carries 97 per cent of the loss, so a better winding hardly moves it. A core whose loss is a fixed resistance instead puts the peak at the geometric mean of two corners, as high as half the square root of the core's resistance over the winding's; a loss tangent that rises with frequency drags the peak down to the material's corner.

magnetics · Real inductor The bank's impedance peaks at 0.966 Ω at 7.12 MHz; a one-ampere load step moves the supply by 132.4 mV. computed by solving, not by drawing. The bulk-and-ceramic pair of the decoupling essays — 10 µF with 5 mΩ and 5 nH, 100 nF with 30 mΩ and 1 nH — standing on a regulator of 2 mΩ and 20 nH, and the voltage at the bank when the load's current steps by one ampere, from the impedance's own step response by residues (solid). The impedance this bank presents peaks at 0.966 Ω at 7.12 MHz (upper line), the parallel resonance of the bulk part's inductance with the ceramic. The step rings at that frequency and its largest swing is 132.4 mV, 31.3 ns after the step — about a quarter period — which is 14% of the peak and 59% of the resonance's characteristic impedance √(L/C) = 0.224 Ω (lower line).

The peak a load step does not see

A decoupling bank is specified by its impedance against frequency, and the number that fails a target is the anti-resonance peak — 0.966 Ω at 7.12 MHz for the bulk-and-ceramic pair standing on a regulator. A load that steps by one ampere moves that supply by 132 mV, 14% of what the peak suggests: a step rings the resonance once, and its first swing is set by the resonance's characteristic impedance √(L/C) = 0.224 Ω, not by its Q. The peak is the voltage per ampere of a load that sits at the resonance; switched at 7.12 MHz the same load builds over six cycles to ±611 mV, the peak times the current's fundamental. So damping, the cure for the peak, is a cure for that load only: sixteen times the ESR cuts the peak 8.9-fold and leaves a step's swing between 120 and 155 mV. What decides a step is its edge — 132 mV for anything faster than a few nanoseconds, 28 mV for a microsecond.

power · Decoupling 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.

filters · Gyrator 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 A tank with its loss in its coil, Q = 3: the voltage is largest at 0.9972 of 1/√(LC) and in phase with the current at 0.9428, and neither is where the poles sit. computed by solving, not by drawing, by closed form and by a netlist of L, Rs and C. The impedance of a coil of inductance L and series resistance Rs across a capacitor, against frequency, for Q = √(L/C)/Rs = 3. The poles are the series circuit's, at 1/√(LC) with a damping of Rs/2L. The coil's own zero at −Rs/L moves the peak to ωₘ = 0.9972·ω₀ and the zero-phase point to ωᵣ = 0.9428·ω₀, and the ring is at 0.9860·ω₀. Four frequencies, which a series circuit's voltage across its resistor would have as one.

The reading a coil leaves nearly exact

On a series resonator three readings of Q agreed exactly and two did not. A parallel tank whose loss is in its coil has no exact readings at all, and the order is not the one its schematic suggests. Its poles are the series circuit's, but the coil's own zero puts the peak, the zero-phase point and the ring at three different frequencies, and all of them below 1/√(LC). At Q = 3 the phase slope reads 16.2 per cent low, the stored energy 5.72 low and the peak height 5.42 high, while the bandwidth reads only 0.307 low: the loss that a parallel resistance stands in for tilts across the band, and the tilt is nearly antisymmetric about the centre, so the reading that spans it cancels it. The instrument's own source then decides what is being read.

frequency · Resonance 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.

filters · Q enhancement

Named alongside it

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

ResonanceDesign tradeoffModel rangeComponent sensitivityComponent toleranceGain–bandwidth productHalf-power bandwidthMeasurement conditionPole pairDamping ratioEquivalent series resistanceInsertion loss

All concepts