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

Settling time — where it appears

How long a response takes to arrive inside a stated band of its final value and stay there, which is not a smooth function of the damping. It has a cliff in it just before the fastest damping, and a phase margin measured at crossover reports nothing about that cliff.

Named by 22 essays across 7 fields — each of them below, with the objects they name alongside it.

Two poles at ζ = 0.3, recovered from the matrix. The poles are at -477.5 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows.

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

transients · Poles How long a second-order step takes to arrive inside ±2%. computed by solving, not by drawing from the residue expansion at 260 damping ratios. The fastest is ζ = 0.780 at 3.60/ω₀; critical damping takes 5.83/ω₀, which is 62% longer. Between ζ = 0.775 and 0.780 the time falls by 33% in one step of the sweep, because which excursion is the last one outside the band changes there — the overshoot at the fastest damping is 1.99%, which is the band itself, and one step to the left it is larger. The faint curves are the other bands, each with its own step in a different place.

The cliff before the fastest settling

Settling time against damping is not a smooth curve with a minimum. It falls by a third in one step of a sweep of five thousandths, and the fastest damping sits on the edge of that step — so a design a hundredth of a damping ratio to the left of the optimum settles forty-eight per cent slower, with a waveform that looks no different.

transients · Damping A step on a series RLC at ζ = 0.079, and the two numbers read off H(s). computed by solving, not by drawing. A 50.3 kHz series RLC driven by a one-volt step, with the capacitor voltage and the inductor voltage drawn together. Two limits of the transfer function are two points of the waveform and neither needs the waveform: H(0) = 1.000000 is where the capacitor ends up, and H(∞) across the inductor is 1.0000, which is what it does at the first instant — the expansion gives 1.000000 for it at t = 0. Here the damping ratio is 0.0791, the response is inside ±2% after 7.6 cycles, and 100.0% of the last twenty-four cycles sit there. The poles are at a real part of -7.91e-2 of ω₀, which is the condition the final-value theorem actually has — not a property of H but of where sY(s) has its poles.

Two numbers without solving for the waveform

Where a step response starts and where it ends are two limits of the transfer function, and neither needs the waveform. Both are exact here — 1.000000000 volts at the end and the whole step at the first instant — and one of them is a lie waiting to happen: take the damping to zero and the final-value theorem still returns 1.000000 for a response that swings between 0 and 2 for ever. Its condition is not on the transfer function but on where the poles are, and the practical condition is narrower still: at five ohms the poles are safely in the left half-plane and sixty cycles is not enough time.

transients · Value theorems 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 The switches set the floor below 60.2 MHz and the amplifier sets it above. computed by solving, not by drawing. The two contributions to a switched-capacitor stage's noise floor, against the clock frequency. √(kT/C) is 63.3 µV on a 1 pF hold capacitor and does not move with the clock at all. The amplifier's own 4 nV/√Hz is white and is sampled, so all of it folds into the band: settling to 12 bits in half a clock period needs 8.32 time constants, the closed loop's noise bandwidth is then 8.32 clocks over two, and the number of folds is 8.32 — the same number, exactly. The marched route is a seeded white sequence put through the settling exponential and sampled once per clock, and it agrees with the closed form to 0.05% of variance. The two floors are equal at 60.2 MHz, above which a larger capacitor buys nothing.

The amplifier inside the sample

kT/C is exactly independent of the clock, of the capacitor ratio and of the switch resistance — two essays measured that and found it identically true rather than nearly so. The amplifier in the same loop behaves in the opposite way in every respect: its noise is white, it is sampled, and the number of times it folds into the band is exactly the number of time constants the settling needs. So the switches set the floor below sixty megahertz and the amplifier sets it above, and asking for two more bits of settling costs fifteen per cent more noise before anything else has changed.

noise · Kt over c The second path costs nothing at 12 pF and an order at 100 pF. computed by solving, not by drawing. Settling time to 0.01% of final value, marched on the closed loop, against the value of the second feedback path's capacitor — with the phase margin of the same circuit divided by ten drawn on the same axis so the two can be compared. The direct-current error the previous rung recorded as the isolation resistor's cost, 0.99% into 1 kΩ, falls to 1.20e-4% with the second path in. What the second path costs instead is a range: at 12 pF the circuit settles in 0.745 µs against the isolation resistor's 1.419 µs — faster than the thing it repairs — and at 100 pF it takes 9.18 µs, 12 times longer, at a phase margin of 47.9° that reports nothing whatever about it. What it is settling by there is one exponential of time constant 0.99 µs, which is the feedback network's own RC and contains no amplifier.

The path that buys the error back

An isolation resistor restores a capacitively loaded amplifier's phase margin and costs it the thing feedback was for: the loop stops regulating the node the load is on, and a kilohm of load pulls the output down by a per cent. The standard repair is a second feedback path, and its cost is not an error or a margin — it is a range. At twelve picofarads it settles to a hundredth of a per cent in 0.745 microseconds, faster than the circuit it repairs; at a hundred it takes 9.18, and the phase margin there is better.

feedback · Capacitive load Subtracting removes kT/C entirely and doubles the amplifier — worth 31× at a megahertz and a loss above 60 MHz. computed by solving, not by drawing. The noise on one sample of a switched-capacitor stage, and on the difference of two samples taken a settled interval apart, against clock frequency. The reset level is the same number in both samples and cancels exactly; the amplifier's own noise is two independent samples and its variance doubles, measured at 2.000 against the 2 the correlation predicts. At a megahertz that is 63.8 µV down to 11.53 — 31 times in power. The two curves cross at 60.2 MHz, which is where the amplifier's own noise equals kT/C, and above it the subtraction costs more than it removes.

The sample that is subtracted

Three rungs of this argument have measured floors that no gain moves and no filter reaches, because both arrive as numbers already sampled. One of them can be subtracted: the reset level a capacitor holds is the same number in two consecutive samples and cancels exactly. What it costs is that the amplifier's own noise is not — two samples of it are independent, so its variance doubles. That is thirty times better at a megahertz, a loss above sixty, and the crossing is the one the rung below computed for a different question.

noise · Kt over c A switched capacitor is a resistor below a ratio, not below a frequency. computed by solving, not by drawing. The exact response of a capacitor shuttled between the input and a holding capacitor at 1.00 MHz — a difference equation with one pole, evaluated on the unit circle — against the continuous R–C its equivalent resistance is supposed to make. The corner is 3.15 kHz against the model's 3.18 kHz, 0.99% out, and the discrepancy is set by the capacitor ratio alone: one per cent needs a ratio under 0.0201, which is a clock 315 times the corner. The second difference has no counterpart at all — the sampled response repeats at the clock, so the image rising on the right of this plot is signal at 997 kHz arriving as though it were at the corner. The third is settling: 1 pF charged through 1 kΩ gets 500.0 time constants a half period at this clock, and being short of full charge raises the equivalent resistance by 0.000%, which puts a ceiling on the clock at 108 MHz.

A resistor made of a clock

A capacitor shuttled between two nodes at a megahertz behaves as a megohm, and a tenth of a picofarad shuttled at ten kilohertz behaves as a gigohm — which is how a filter with a one-hertz corner fits on a chip. What the equivalence costs is three conditions, and they bind on three different quantities: a capacitor ratio under 0.0201, a signal below half the clock, and a clock below the frequency at which the charge stops arriving.

filters · Switched capacitor The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range.

What the second path costs at the floor

The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

feedback · Capacitive load The load sees 10.0 Ω, 1.2e-3 Ω or 1.2e-3 Ω at direct current, and the peak is lowest for the arrangement with both paths. computed by solving, not by drawing. The impedance at the load node of all three arrangements, measured by grounding the input and driving a unit current into the load. Feedback from the amplifier leaves the load looking at the isolation resistor — 10.0 Ω, with no loop gain in it at all. Feedback from the load gives 1.2e-3 Ω, and the two-path arrangement has the same, which is what its direct-current path is for. All three resonate with the load capacitance near 3.2 MHz, and the two-path arrangement's peak is the lowest — 27.6 Ω against 37.4 and 59.2. What it gives up is between: above the 159 kHz handover it has let go of the load node.

What the load sees looking back

Four rungs of this argument have measured what the amplifier does to the signal — the margin, the settling, the error, the noise. None has asked the question from the other end. A load that draws its own current sees an impedance looking back, and with the feedback taken from the amplifier that impedance is the isolation resistor, with no loop gain in it whatever: ten ohms, and a load step leaves an error that never goes away. The two-path arrangement recovers to a thousandth of it and charges for that in a quantity none of the four rungs below measured.

feedback · Capacitive load The charge that comes back: a 0.2% dielectric, 10 s shorted, read at 900 s. computed by solving, not by drawing. The capacitor is charged to 10 V until every relaxation is complete, shorted for 10 seconds, then opened and watched. It climbs back to 20.00 millivolts — 0.2000 per cent of where it was — and the shape is the finding: it is a straight line on a logarithmic time axis, gaining 0.097 per cent of the charging voltage per decade. There is no time constant after which it is over, because there is no single time constant: one branch of the model comes to equilibrium per decade, for as many decades as the dielectric has. A decade before the reading it was at 0.1033 per cent.

The capacitor that remembers

Charge a capacitor, short it for ten seconds, open it, and it climbs back to a fifth of a per cent of where it was. Nothing leaked and nothing was gained: some of the dielectric had not finished discharging. The same defect measured as an admittance says the part is 0.593 per cent more capacitance at a tenth of a millihertz than at a kilohertz, and measured in a sample-and-hold it says a millisecond of hold costs a hundred parts per million — thirteen bits, on a part specified at nothing.

transients · Dielectric absorption The growth per cycle, and the form that is a fifth low at the top of the range. computed by solving, not by drawing. The factor the envelope is multiplied by each cycle, against the gain. The solid curve is exp(π(k−3)/√(1 − ((k−3)/2)²)), which is what the characteristic equation gives and what the netlist's own poles return to twelve digits; the dashed one is exp(π(k−3)), which drops the denominator. The circles are the marched envelope, fitted over the cycles that are still small — 80 of them at k = 3.01 and 4 at k = 3.2, and none at all above that. The two expressions differ by 3.9e-7 at k = 3.01 and by 20.462% at k = 3.8, so the approximation fails exactly where nothing is left to check it against.

Two exponentials, and where they meet

An oscillator's envelope grows by exp(π(k−3)/√(1 − ((k−3)/2)²)) a cycle, and the form usually quoted drops the denominator — exact to four parts in ten million at a hundredth above three, and 20.462 per cent low at 3.8, which is precisely where too few small cycles are left to measure it. Where the growth stops is the diode's own exponential: 108.5 millivolts of amplitude for every decade of saturation current, proportional to the ideality to four parts in a thousand. Above 60.121 nanoamperes the limiter is already conducting at zero signal and there is no oscillation at all.

applied · Oscillator Three cliffs, not one, and the fastest damping is on the last of them. computed by solving, not by drawing. Settling time against damping for a third-order response — a complex pair at unit natural frequency and a real pole at 3 — with the second-order case behind it. Both are staircases: the settling time is set by the last excursion outside the band, so there is one step for each excursion that stops happening, and there are 3 of them between 0.3 and 0.98. They are at 0.378, 0.522, 0.773, with jumps of 1.24, 1.30, 1.42. The rung below found the last and largest of them and did not look below it. The fastest damping is 0.775, sitting on the edge of the last step, and a design a hundredth to the left of it settles 42 per cent slower.

Three cliffs, and where they are

The rung below sweeps a second-order step's damping, finds the settling time falling by a third in one step of a five-thousandth sweep, and calls it the cliff. There are three of them between 0.3 and 0.98, one for each excursion that stops leaving the band, and adding a third pole moves all three left and makes all three shallower — so the classic 0.78 for fastest two per cent settling is a second-order number, and at a third pole one and a half times the natural frequency the answer is 0.745 and 0.78 is on the wrong side of the step.

transients · Damping Which limit binds is a property of the load, and they change places near 22 nF. computed by solving, not by drawing. Each limit measured on its own, as the departure of its march from the linear one, at a load step of half the output stage's rating. The input pair's departure falls with load capacitance — a bigger reservoir holds the node while the loop responds, which is the sixth rung's own result — and the output stage's does not fall nearly as fast, because what it has to supply is the charge the capacitor wants. Below about 22 nanofarads the thermal voltage decides the answer and above it the output stage does, and nothing about the amplifier changed.

The current above which there is no impedance

The sixth rung found the impedance leaving at 10.6 mA, where the input pair's own tanh takes over and the slew rate is twice the thermal voltage times the gain-bandwidth in radians, with no design choice in it. A real output stage has a second limit that is nothing but design choice, and the two do not bind at the same load: at 0.47 nF the input pair's departure is 19.4 per cent against the output stage's 4.2, at 22 nF it is 0.9 against 2.3, and above the output stage's rating the excursion does not come back at all — 2,254 Ω for a quantity that was 37.

feedback · Capacitive load 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 true-RMS reading of a sine: ripple a second filter removes, and a bias it cannot. An explicit converter — square, average through a one-pole of τ = 100 ms, take the root — in steady state on a sine of unit root-mean-square value, integrated exactly over a period at 91 frequencies and by a fourth-order march of its own equation at 6, which agree to 1.9e-8. The upper curve is half the ripple on the reading and the lower one the amount by which its mean is low. The reading is low at every frequency, because the square root is concave; it is 1% low below 1.86 Hz, while the ripple is inside ±1% only above 39.8 Hz. The dashed curve is the small-ripple form, an eighth of the averaged square's ripple power, which the bias approaches as the ripple shrinks.

The average a square root pulls low

A true-RMS converter squares, averages and takes the root, and the root of a quantity that ripples averages below the root of its mean. With a hundred-millisecond averager a sine is read one per cent low below 1.86 hertz, where the ripple is still ±20 per cent — and a second filter that steadies the display takes the ripple away and leaves the reading exactly as low as it was. A square wave is read exactly at any averaging time; a rectifier current conducting for twenty degrees needs 2.41 times the averaging a sine does. The implicit converter is the explicit one at half the time constant, and a reading falls 1.38 times slower than it rises.

power · RMS and average The fastest damping is a surface, and the band is worth 5 times the third pole. computed by solving, not by drawing. Each point is the last settling cliff, bisected — the damping at which the first overshoot's peak lands exactly on the band's edge, which is where the fastest settling is. Across the five bands the optimum moves by 0.231 of damping ratio; across a third pole from 1.5 times the natural frequency out to a second-order response it moves by 0.047. The two axes are worth 5.0 to one, and the expensive one is the specification rather than the parasitic. The classic 0.78 for fastest two per cent settling is the second-order curve's value at ±2%, 0.7797; at ±1% the same response wants 0.8261.

The best damping is not the one to build

The fastest settling damping is the right-hand limit at a discontinuity, so two thousandths below it costs 41 per cent and two thousandths above it costs 0.34 — a ratio of 120 in the penalty for the same error. With ±2 per cent on the damping ratio the nominal that minimises the worst case is 0.7927 rather than the optimum's 0.7734, and it guarantees 4.243/ωₙ against 5.943. The band moves the optimum by 0.231 of damping ratio and the third pole by 0.047, and 0.78 is exact at ±2% and 55 per cent slow at ±1%.

transients · Damping Three averagers passing the same noise, and three different half-power points. Integrated by eight-point quadrature on every lobe, with the tail past the last lobe in closed form. A mean over 20 ms has the response sin(πfT)/(πfT), and the area under its square is 25.000000 Hz against the 25 Hz of 1/(2T) — the brick wall drawn shaded. A mean over the window passes half its power at 22.147 Hz, so the noise bandwidth is 1.12880 times that frequency; A one-pole averager passes half its power at 15.915 Hz, so the noise bandwidth is 1.57080 times that frequency; Two means in cascade pass half their power at 23.919 Hz, so the noise bandwidth is 1.04521 times that frequency. Every curve drawn encloses the same area; they differ in where they spend it.

The filter an average is

A mean taken over a window is a filter, and the area under its squared response is exactly one over twice the window — 25 hertz of noise bandwidth for twenty milliseconds, passing half its power at 22.15. Built to the same noise, a one-pole averager passes half its power at 15.92 hertz and takes 2.33 times as long to settle to one per cent, and two means in cascade pass half at 23.92 and take 1.25 times as long. Between its nulls a mean rejects the mains no better than the one-pole does, and one per cent off a null it rejects it by forty decibels however many cycles the window holds.

noise · Noise bandwidth Two large-signal limits, each alone and then both, at 20 mA and half of it. computed by solving, not by drawing. Four marches of one netlist at each load: neither limit, the input pair's tanh alone, the output stage's 20 mA alone, and both, driven by a 10 mA step. The three curves are each limit's departure from the linear march and the departure with both present; the faint line is the two singles added. Both lies on the sum and a little above it — 1.112 times it at 0.47 nF and 1.022 at 47 nF — so the limits are present together rather than taking turns. Where the two singles cross, near 10 nanofarads, the pair costs 1.88 times what the worse of them costs alone.

The load that neither limit owns

Nine rungs of this argument asked which of an amplifier's two large-signal limits binds, and drew the load capacitance where the answer changes hands. Both are present at every load: the excursion with both in the netlist is the two departures added and between 2 and 12 per cent more, never the larger of them. So the crossing is not a handover but a maximum — at 12 nanofarads the pair costs 2.084 times what the worse of them costs alone, against 1.35 at 2.2 nanofarads and 1.07 at 47 — and the same peak sits on the resistance axis at 20 ohms and the gain-bandwidth axis at 50 megahertz.

feedback · Capacitive load 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 A true-RMS converter reads noise low by 1/(16Bτ): 0.600% at Bτ = 10 and 672 ppm at 100, where a sine at fτ = 100 is 0.0396 ppm. Seeded, and measured. An explicit true-RMS converter — square, one-pole average of time constant τ, root — reading Gaussian noise of unit power, against the product of the noise's bandwidth and τ, from 1 to 100. Each point is 2²¹ samples at eight times the bandwidth, with the reading compared against the record's own root-mean-square and its standard error from batch means. Bτ = 1: 4.550% ± 150 ppm (band from zero), 4.081% (band of the same width about 3B). Bτ = 2: 2.598% ± 114 ppm (band from zero), 2.411% (band of the same width about 3B). Bτ = 5: 1.149% ± 75.4 ppm (band from zero), 1.098% (band of the same width about 3B). Bτ = 10: 0.600% ± 53.1 ppm (band from zero), 0.576% (band of the same width about 3B). Bτ = 20: 0.309% ± 39.1 ppm (band from zero), 0.295% (band of the same width about 3B). Bτ = 50: 0.127% ± 26.3 ppm (band from zero), 0.119% (band of the same width about 3B). Bτ = 100: 672 ppm ± 19 ppm (band from zero), 623 ppm (band of the same width about 3B). The dashed line is 1/(16Bτ), an eighth of the averaged square's variance, which the readings approach above Bτ = 10 and fall short of below it. A sine read by the same converter at the same product of frequency and τ is low by 3.96 ppm at 10 and 0.0396 ppm at 100 — the square of the noise's rate rather than its first power.

The noise a true-RMS meter reads low

A true-RMS converter reads a sine low by an amount that falls as the square of its frequency, and for anything but a slow sine that amount vanishes: 3.96 parts per million at ten times the averager's corner. Noise is not a sine. Its square fluctuates at every frequency down to zero, and the averager passes a share of that set by its own bandwidth against the noise's, so the reading is low by 1/(16Bτ) — the first power, not the second. Measured on seeded noise at Bτ = 10 it is 0.600 per cent low against a predicted 0.625; at 100, 672 parts per million. One reading scatters by eighteen times that, so no single reading shows the bias and the mean of a few hundred is nothing but bias.

power · RMS and average A step through r sections starts as (t/τ)^r: it reaches 1% at 10.1 µs, 105 µs, 243 µs, 380 µs, 508 µs for r = 1 to 5. Solved, and expanded two ways. The step response of buffered RC sections of time constants τ, τ/2, … τ/r, with τ = 1 ms, on logarithmic axes. The relative degree of the recovered transfer function is r, so the first r − 1 derivatives of the step are zero at the start and the r-th is lim s^r·H(s) = r!/τ^r, read off the network solved far above its poles and off the expansion of H about infinity; the step therefore starts as (t/τ)^r, a straight line of slope r. The expansion about infinity and the residue expansion agree to a part in a million where both are well conditioned. The output reaches 1% at 10.1 µs (r = 1), 105 µs (r = 2), 243 µs (r = 3), 380 µs (r = 4), 508 µs (r = 5), and half its final value at 693 µs, 1.23 ms, 1.58 ms, 1.84 ms, 2.04 ms. For these time constants the whole step is (1 − e^(−t/τ))^r, checked against both expansions, so the time to a fraction ε is −τ·ln(1 − ε^(1/r)).

The start a step takes from infinity

The initial-value theorem reads where a step starts off H at infinite frequency. Apply it again to s·H, s²·H and on, and it reads how the step starts: the first r − 1 derivatives are zero for a network r degrees more poles than zeros, and the r-th is the ratio of the leading coefficients. So a step through r sections begins as a power of time — for sections of τ, τ/2, … τ/r, exactly (t/τ) to the r — and reaches one per cent at 10.1 µs through one section, 105 µs through two and 508 µs through five. Put a zero anywhere, even a thousand times above every pole, and the step starts linearly instead, with a slope of twice the zero's time constant over τ² that is the larger term for the first two of them.

transients · Value theorems

Named alongside it

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

Design tradeoffModel rangePolesCapacitive loadResiduesDamping ratioOvershootAveragingEquivalent noise bandwidthMarchingMeasurement conditionOutput impedance

All concepts