The start a step takes from infinity

Two numbers without solving for the waveform read two points of a step response off its transfer function. Where the step ends is $H(0)$, and where it starts is $H(\infty)$ — the initial-value theorem — and both are exact without the waveform in between. That essay spent most of its length on the final value, because the final-value theorem is the one that lies: it returns a number for a lossless circuit that never settles, and the condition that makes the number true is a condition on where the poles are, not on the transfer function.

The initial-value theorem does not lie, but it says very little on its own. For almost every network anyone steps, $H(\infty)$ is zero: a capacitor across the output, an inductor in series, any low-pass anything. The theorem then says the step starts at zero, which is true and uninformative.

It can be made to say much more by being applied again. The initial value of a waveform’s derivative is the initial-value theorem applied to the derivative’s transform, and each derivative multiplies the transform by $s$. So the whole of how a step begins — at what power of time, with what coefficient — is written at the other end of the frequency axis from where it ends.

The theorem, where it was left #

The earlier figure reads both limits at opposite ends of one network: the inductor’s voltage starts at the whole step, which is $H(\infty) = 1$ for that output, and the capacitor’s starts at nothing, $H(\infty) = 0$. The capacitor’s voltage is where this essay starts. It is zero at the first instant, and so — it turns out — is its slope, because the capacitor is charged through an inductor whose current also starts at zero. Its curvature is not zero. The step across the capacitor begins as a parabola, and the theorem says so if asked about $s^2H$.

The start of a step has been in view on this subject several times without being the subject. Where the behaviour is written down located everything a second-order step does in its poles, which is true of the whole response and silent about its first instants. A ladder is not a line found a lumped chain of inductors and capacitors whose far end moves before the wave could arrive, and the power of time that early movement follows is this essay’s relative degree; the sections a wavelength needs put the same chain in the frequency domain. And the energy that arrives first moved a zero across the plane and watched when a network delivers its energy, which is the integrated form of the question asked here at a single instant.

Asking the theorem about higher powers of s #

A step response $y(t)$ has the transform $H(s)/s$. Its $k$-th derivative has the transform $s^k \cdot H(s)/s$, less terms from the earlier derivatives’ initial values, which are all zero if the earlier derivatives start at zero. The initial-value theorem then gives

$$y^{(k)}(0^+) = \lim_{s\to\infty} s^k H(s).$$

If $H$ has $r$ more poles than zeros — a relative degree of $r$ — then $s^k H(s)$ goes to zero for every $k < r$ and to a finite number for $k = r$: the ratio of the numerator’s leading coefficient to the denominator’s. So the first $r - 1$ derivatives of the step are zero at the start, the $r$-th is that ratio, and the step begins as

$$y(t) \approx \frac{b_m}{a_n}\cdot\frac{t^r}{r!}.$$

Two facts about a network set the start of its step, and neither is a pole’s damping or a corner frequency. The power is the relative degree: how many more poles than zeros. The coefficient is the behaviour at infinite frequency, where $H(s) \approx (b_m/a_n)\,s^{-r}$ — the network’s high-frequency asymptote.

The whole of the start, not just its first term, comes from the same place. Expand $H$ about infinity as a series in $1/s$, $H(s) = \sum_k h_k s^{-k}$, and the step is $y(t) = \sum_k h_k t^k/k!$ term by term. The coefficients $h_k$ — sometimes called the network’s Markov parameters — are what a long division of the numerator by the denominator produces. The figures compute them from the transfer function recovered from the network, and use them where the residue expansion is poorly conditioned: at small times, where the residues of several poles cancel to a tiny difference.

Sections in a row #

The networks here are chains of buffered RC sections, the $k$-th with a time constant of $\tau/k$, so that the poles are distinct and the residue expansion applies. A chain of $r$ sections has relative degree $r$, and the figure checks that the transfer function recovered from each network has exactly that. The high-frequency asymptote is the product of the poles’ magnitudes, $r!/\tau^r$; the figure reads $\lim s^r H(s)$ off the network solved at ten thousand times the slowest corner and checks it against that product, and checks the leading term of the expansion about infinity against it to a part in a million.

On logarithmic axes each step starts as a straight line whose slope is the number of sections, and the figure fits the slope a thousandth of a time constant in and checks it: one, two, three, four, five. The expansion about infinity and the residue expansion agree to a part in a million at the times where both are well conditioned.

These particular time constants have a closed form, which makes a third route with no poles in it at all. A step through sections of $\tau, \tau/2, \ldots, \tau/r$ is exactly $(1 - e^{-t/\tau})^r$ — it is the probability that $r$ exponential waits with rates $1/\tau, 2/\tau, \ldots$ have all ended, which is also the distribution of the longest of $r$ equal waits. The figure checks both expansions against it at five times from a hundredth of $\tau$ to three. And it makes the start transparent: for small $t$, $1 - e^{-t/\tau} \approx t/\tau$, so the step begins as $(t/\tau)^r$, with the coefficient $r!/\tau^r$ and the $1/r!$ cancelling.

So the step through one section reaches one per cent of its final value at 10.1 microseconds, and through five sections at 508 — fifty times later, for a network whose slowest pole is the same millisecond. Every additional section adds a factor of $t/\tau$ to the start, and a factor of $t/\tau$ is a large penalty when $t$ is a hundredth of $\tau$.

The start moves later than the middle #

Adding sections delays the whole step, but not uniformly. The middle of the step — its fifty per cent point — follows the sum of the time constants, which for these sections is τ times the harmonic series and grows only as a logarithm: 693 microseconds for one section, 2.49 milliseconds for eight. The start grows much faster. The one-per-cent point is at 10.1 microseconds for one section and 826 for eight, eighty-two times later.

As a fraction of the fifty-per-cent time the start moves from 1.4 per cent to 33.2 per cent. The step changes shape as sections are added — a slow start and a steep middle instead of an abrupt start and a long exponential tail — though its finish stays longer than its start: for these time constants the step is the distribution of the longest of $r$ equal exponential waits, and that distribution approaches the skewed shape of the largest of many rather than a symmetric one. Nothing about the start needs the poles; the relative degree predicts it, and the sum of the time constants predicts the middle.

The closed form puts exact numbers on it. The time to a fraction ε is $-\tau\ln(1 - \varepsilon^{1/r})$. At $r = 2$ the one-per-cent time is the one-section ten-per-cent time, $-\tau\ln 0.9$, 105 microseconds, because $0.01^{1/2} = 0.1$; at $r = 4$ it is the two-section ten-per-cent time, 380 microseconds. A chain of $2r$ sections reaches one per cent when a chain of $r$ reaches ten.

A zero takes the start #

The relative degree is a count of poles against zeros, so a single zero anywhere lowers it by one — and changes the power of time the step starts with, however far away the zero is.

The two-section network alone has relative degree two and starts as $(t/\tau)^2$. Add a zero with a time constant of $\tau_z$ — built here as an ideal differentiator summed with the network’s output, so that the network becomes $(1 + s\tau_z)/((1 + s\tau)(1 + s\tau/2))$ — and the relative degree becomes one. The figure checks it. The step now starts linearly, with a slope of $\lim sH(s) = \tau_z \cdot 2/\tau^2$, which the figure checks against the expansion about infinity: $2 \times 10^{-3}$ per millisecond for a zero at a thousandth of a millisecond.

The linear term is the larger until the quadratic catches it, which is at $t \approx 2\tau_z$: two microseconds for the smallest zero, two hundred for the largest. After that the step looks like the network without the zero. A zero at a thousandth of the slowest pole’s time constant is invisible on any ordinary plot of the step and decides entirely how it spends its first two microseconds.

That is a small effect with consequences in exactly the places where a start is measured. A delay measured at a low threshold, a comparator that trips on the first millivolt, a digital input that switches at ten per cent of a rail: each of these reads the start of a step, and the start is set by the network’s behaviour at the frequencies furthest from anything the network was designed to do — by the zero a layout parasitic contributes, or by the feedthrough capacitance nobody drew.

The damping that the start does not contain #

The earlier essay’s series RLC makes the point in its most surprising form. Its capacitor’s voltage, for a step, has the transfer function $1/(LCs^2 + RCs + 1)$: relative degree two, leading coefficient $1/LC$. So the step across the capacitor starts as $\omega_0^2 t^2/2$, where $\omega_0$ is the natural frequency — and the resistance is not in it.

That resistance decides almost everything else about the response. The earlier essay swept it from zero to five hundred ohms and took the damping ratio from nothing to 0.79, the overshoot from a hundred per cent to almost none, the settling from never to six-tenths of a cycle. None of that reaches the first two derivatives at the start. The resistance first appears in the third: the expansion about infinity gives $y'''(0^+) = -\omega_0^2 R/L$, so the damping enters as a fractional correction of $(R/L)\,t/3$ to the parabola.

For the network at fifty ohms, with a millihenry of inductance, that correction is eight per cent by a quarter of a cycle of its 50.3-kilohertz ringing, five microseconds in — the moment the parabola’s rise is about to turn over. So for the rise to the first peak the capacitor’s voltage is very nearly the same curve whatever the resistor is, and a step observed only over that rise says the natural frequency and very little about the damping. The two quantities a second-order design is specified by arrive at different times: the natural frequency at once, in the second derivative, and the damping a derivative later, growing in proportion to the time the step has had to feel the resistor.

A start as a measurement #

Read in the other direction, the start of a measured step is an instrument. A step observed on a logarithmic time axis, rising as a straight line on logarithmic axes, has a slope that is the relative degree of whatever lies between the source and the probe — including the parts nobody drew. A schematic with two poles and a measured start with a slope of three has a third pole somewhere above the measurement’s own bandwidth, and the intercept of the line gives the product of all three poles’ magnitudes.

The same reading has a floor, and it is set by the instrument. An oscilloscope’s own response is a chain of poles too, and its step starts with its own power of time; what the display shows is the start of the whole chain, instrument included. So the slope counts poles until the times being read are shorter than the instrument’s rise time, and then it counts the instrument’s. The start of a step is a precise measurement of relative degree down to the instrument’s own time constant, and a precise measurement of the instrument below it.

Where a march starts wrong #

The power of time also says what a marched simulation gets wrong first. One step, computed twice found the trapezoidal rule’s error falling as the square of the step, but that is a statement about the error accumulated over a response. At the very first step, a network of relative degree $r$ has an exact output of order $h^r$, and the march’s first value is built from a rule that is exact only for the first few terms of that series — so at the start the relative error of a march is of order one whenever the true output is smaller than the rule’s own local error.

For a network of relative degree five and a step of a hundredth of a time constant, the true output after one step is ten to the minus ten. No rule whose local error is of order $h^3$ gets that value right to more than its sign. The start of a step is the one place where a march and the exact response disagree completely, and it is also the one place where the expansion about infinity is exact, which is why the figures here use it there.

The two marching essays on this subject meet the start from the numerical side. The ringing that belongs to the rule found the trapezoidal rule flipping a pole faster than its step, which is what a discontinuity looks like to a rule — an output whose start is a jump rather than a power of time; the phase the rule loses is the same rule on the poles it resolves. And the cancellation that leaves a tail put a zero beside a pole and found the tail it leaves; a zero far above every pole, as measured above, leaves its mark at the other end of the step instead.

The theorem’s own condition, and where it fails #

The final-value theorem had a condition that was easy to miss. The initial-value theorem’s condition is kinder but real: the limit must exist, which for a rational $H$ means the relative degree is not negative. A network with more zeros than poles — an ideal differentiator, a capacitor driven by an ideal source and asked for its current — has an $H$ that grows at infinite frequency, and its step response has an impulse at $t = 0$ that no power of time describes. The figures here never build one; the zero was added to a network that had two poles to spare.

A relative degree of zero is the other boundary. Then $H(\infty)$ is not zero and the step jumps at the first instant to that value, as the inductor voltage in the earlier essay did. Every derivative after the jump is again a limit of $s^k$ times what remains after the jump is subtracted, and the same expansion about infinity carries on from there.

The undamped case makes the difference between the two theorems’ conditions concrete. With no resistance the final-value theorem returns a number the waveform never reaches — the condition on the poles has failed — while the initial-value theorem, asked about $s^2H$, still returns the curvature of the start exactly, because its only condition is that the limit at infinite frequency exists, and it does whether the poles are damped or not.

And every statement here is about an ideal network. A real chain of sections has parasitics that add poles far above everything drawn, which raise the relative degree and change the power of time the step starts with — at times so short that no oscilloscope sees them. The measurement is honest about a model; the model’s start is only as physical as its highest-frequency behaviour, which is the part of any model least likely to be right.

Still open: the delay a threshold measures, the parasitic pole, and the start of a sampled step #

Delay at a threshold. A digital receiver defines propagation delay at a threshold, often fifty per cent and sometimes ten, and the middle is the usual place to quote it. The start’s power law says a delay measured at ten per cent moves with the number of poles much more than one at fifty. Measuring the two thresholds’ delays through a real interconnect model — a line’s LC ladder of sections, say — would say how much of a quoted delay is a property of the threshold rather than the path.

The parasitic pole. A single pole added far above a network’s bandwidth raises its relative degree by one and changes its start from $t^r$ to $t^{r+1}$, for times shorter than that pole’s time constant. A zero, as measured above, does the reverse. Which of the two a real layout adds more of — and so whether a real step starts faster or slower than its schematic’s — is a question about the parasitics of a specific board, and the measurement it needs is the expansion about infinity with them included.

A sampled step. A step applied by a digital-to-analogue converter is a staircase, not a step, and its first sample holds for a whole period. The start of the analogue response is then the network’s response to a pulse followed by further pulses, and its early power law is set by the relative degree of the network and the hold. Whether the hold’s own $(1 - e^{-sT})/s$ changes the power a real reconstructed step starts with is unmeasured here.