The stage that overshoots a little

The delay a chain adds read a chain of one-pole stages in time and found that it leaves its settling time faster than it leaves its flat band. The chain’s step is the sum of $N$ exponential lags, so its delay adds as $N$ and its spread as $\sqrt N$, and four stages take 10.0 time constants to settle within one per cent against one stage’s 4.61. Its closing question was about the kind of stage. A second-order stage with a little overshoot settles a single step sooner than its rise suggests, and in a chain its overshoot is filtered by the next stage. Whether a chain of slightly underdamped stages settles faster than the same chain of single poles of equal bandwidth is the measurement that would decide how a chain’s stages should be compensated.

There is a strong theoretical reason to expect yes. A one-pole stage has a spread as large as its delay, and a second-order stage can have far less spread than that, down to none at all. The arithmetic says so and a long chain is what the spread governs, so a long chain of these stages should be a large improvement. The improvement is real, and measuring it is worth doing because it runs out. At about ten stages the supposed improvement has become a cost.

What “the same bandwidth” holds fixed #

A comparison between stages needs something to hold fixed, and the choice is not innocent. The stage here costs a certain half-power bandwidth, $\omega = 1$, and the time axis is in units of the time constant of a one-pole stage with that bandwidth. Every stage has its natural frequency set so that this is true. A stage of damping $\zeta$ has

$$H(s) = \frac{1}{1 + 2\zeta s/\omega_n + s^2/\omega_n^2},$$

and its $\omega_n$ is chosen for a half-power point at one. That is the sensible fixed quantity because it is what an amplifier and its gain supply: the edge of a chain used the same bookkeeping for how to spend a gain–bandwidth product. A stage with a fixed bandwidth can have whatever damping its compensation gives it. So a comparison at equal stage bandwidth asks, with the same resources, how the stage should be built.

The chain is marched by fourth-order Runge–Kutta, with two states a stage, and at $\zeta = 1$ the march has a closed form to answer to: two equal poles a stage make an Erlang distribution of order $2N$ in $\omega_n$. They agree within 0.03 of a time constant at every length used. The one-pole chain beside it is the closed Erlang form of the essay this one is continuing, so nothing about it is approximated either.

The fastest stage, at every length #

For one stage the comparison has been made before. The cliff before the fastest settling showed a second-order step’s settling against damping: it drops suddenly where an overshoot peak falls inside the band. With a 1 per cent band and equal bandwidths, one stage settles in 3.53τ at a damping of 0.833, against the one-pole stage’s 4.61, a saving of 23 per cent.

The chain has that cliff too, and it stays in nearly the same place. The best damping is 0.842 for two stages, 0.833 for four and eight, and 0.817 for sixteen, and in every case the chain’s overshoot is just under the band: 0.841, 0.988, 0.793 and 0.957 per cent. So the best stage has the same character at every length, an overshoot the chain can only just stay inside. Below the cliff the overshoot leaves the band, and the settling time jumps upward by a whole swing of ringing. In a chain that jump is larger than in one stage.

It is the height of the curve that is the surprise. At two stages the best settling is 0.83 of the one-pole chain’s, at four it is 0.89, at eight 0.98, and at sixteen 1.05. The second-order stage is better up to about ten stages, and after that it is worse, even at its best damping. The chain that seemed to have the most to gain has the least.

Two numbers a stage adds #

The reason lies in the two quantities that made the one-pole chain’s analysis so simple. Independent stages add their first two moments: the delay of a chain is the sum of its stages’ delays, and its variance is the sum of its stages’ variances. For a one-pole stage both are 1 in the appropriate units. For this stage they are

$$d = \frac{2\zeta}{\omega_n}, \qquad v = \frac{4\zeta^2 - 2}{\omega_n^2}.$$

The squared spread has a striking property: it is zero at $\zeta = 1/\sqrt 2$ and negative below it. A Butterworth stage adds delay and no second moment at all. Its first two moments match a pure delay. That is what makes it look like the ideal stage for a chain, since a chain of pure delays would have a perfectly sharp step. A negative “variance” is not a variance, of course, because the stage’s impulse response is negative for part of its length and is not a probability distribution; and that negative stretch is exactly the overshoot on the step. So the central-limit reasoning that served the one-pole chain so well, with its Gaussian shape, has no meaning at or below that damping. A long Butterworth chain does not become a Gaussian. It settles into a shape set by its third moment, with an overshoot that does not go away. At eight stages it is 10.7 per cent.

The first number is the bad news, and it is easy to overlook because attention is on the second. The delay of a stage at $\zeta = 0.83$ is about 1.4 times a one-pole stage’s. A second-order stage’s bandwidth comes from a flat response and a steep roll-off, and that costs a good deal of delay against the first-order stage, whose response starts falling at once but keeps a small phase. That is an old result in another form: the phase of a minimum-phase stage follows from its magnitude, as the phase that comes from the magnitude and its neighbours measure, and holding bandwidth fixed while flattening the band has to raise the delay.

Now the arithmetic of a long chain is plain. Settling is a delay plus a multiple of a spread:

$$t_s \approx N d + z\sqrt{Nv}.$$

For the one-pole chain, $d = v = 1$. For the second-order stage at its best damping, $d \approx 1.4$ and $v$ is about half. At small $N$ the second term matters and the second-order stage wins. At large $N$ the first term dominates and the 40 per cent of extra delay can never be recovered. The crossover is at about ten stages at this band, and it has nothing to do with overshoot at all. A designer’s argument for second-order stages is good and it stops at a definite length.

How far two moments go #

That sum is a sketch rather than a law, and it is worth checking how far it can be trusted before leaning on it. For the one-pole chain it needs a multiple $z$ of the spread, and the chain’s own numbers supply it: sixteen one-pole stages have a delay of 16τ and a spread of 4τ, and they settle in 26.74τ, so $z$ is 2.69. For sixteen second-order stages at a damping of 0.817, the delay is about 22.2τ and the spread about 2.8τ, and they settle in 28.01τ, so $z$ is about 2.1. For eight at 0.833 the delay is about 11.1τ and the spread about 2τ, and they settle in 15.61τ, a $z$ of about 2.25.

So the multiple is not one number. It is smaller for the second-order stage, because the stage’s response is less skewed: a one-pole chain’s step has a long exponential tail that a Gaussian’s multiple underestimates, while a second-order chain has been arranged to end sharply, with its overshoot sitting right at the edge of the band. Part of what the best damping is doing is that: it saves on the multiple as well as on the spread, by truncating the tail. That is a third-moment effect and two moments cannot show it. It is also why the sum cannot put an exact number on the crossover, and why the figure below has to be measured rather than predicted.

What the sum gets right is the shape of the argument. Over a chain from eight to sixteen stages the one-pole time rises by 10.74τ, and the second-order time by 12.40τ. Those increments are dominated by the delay’s 8τ and the 11τ it costs the second-order stage, and the spread’s contribution is a small correction on top of them. By that length a chain’s settling is a question about how much delay each stage has. A one-pole stage has the least delay of any stage with a given bandwidth, and that is the whole of why it eventually wins.

What the best stage is worth #

Putting the three lines together gives what a designer would ask for.

The best stage’s ratio to the one-pole chain runs 0.77, 0.83, 0.89, 0.98, 1.05 for 1, 2, 4, 8 and 16 stages. Critically damped stages, the conservative choice, are 0.93, 0.97, 1.02, 1.08 and 1.12, so a designer who removes the overshoot altogether loses the advantage by about three stages. Butterworth stages, the flat-band choice, are the worst throughout, 1.27 to 1.46 of the one-pole chain, because their overshoot is well outside a 1 per cent band and they spend their settling time ringing.

So a chain’s stages have a practical rule at this tolerance: below about eight stages, compensate each stage to about 0.83 of critical damping, which gives a modest saving that shrinks with length. Above about ten, the stage’s compensation has stopped mattering very much, and one-pole stages of the same bandwidth are at least as fast. They also have the virtue of being robust, since the cliff is close above the best damping and a component tolerance can put a stage off it, as the best damping is not the one to build argued.

The frequency edge asks for something else #

The same stages have another number to report, and it does not agree. The edge of a chain measured how soon a chain leaves its flat band as a function of its stages, and a Butterworth stage holds a flat band for far longer than a one-pole stage does.

For eight stages, the flattest is at a damping of 0.7, 1 per cent flat out to 0.269 of a stage’s bandwidth, more than five times the one-pole chain’s 0.050. It also settles in 23.3τ, 46 per cent longer. The fastest settling in this sweep is at 0.85, flat only to 0.066, which is barely better than the one-pole chain, in 15.9τ. And the two are separated by a real trade: at a damping of 0.75 the chain is flat to 0.105 and settles in 19.5τ. There is no stage that does both, and they are asked for in different places.

That disagreement is a useful thing to find, because the two edges have usually been treated as different views of one quantity. The opening essay in this group put edges on one axis and took each edge as a property of the circuit. Here one circuit, at one damping, has its frequency edge improve when its time edge worsens. A designer who chooses compensation by reading a chain’s flatness has made a choice about time without knowing it. And a chain that is required to pass a band has to be built differently from one that is required to settle, and may simply have to be two chains.

The frequency edge has one more thing to say. A 1 per cent flat edge at damping 0.7 is a narrow target: 0.65 reaches 0.084, and 0.75 reaches 0.105, so the flatness is a sharp peak at a particular damping, much sharper than the settling’s cliff, and with a tolerance on the damping most of the headline number disappears.

Why the overshoot does not filter away #

The original intuition was that a following stage filters the overshoot of the one before it. Eight stages in a row at 0.8 should each smooth what the previous stage’s overshoot did, and the chain’s overshoot should be less than a stage’s. In fact a single stage at 0.8 overshoots 1.5 per cent and the chain 2.02. Overshoot is not an isolated high-frequency feature of the stage, and each following stage has the same response that produced it. The chain’s magnitude response has a roll-off that sharpens with length, and an increasingly sharp roll-off produces more ringing, not less, at a fixed damping. At ζ = 0.7 a single stage overshoots 4.6 per cent and eight of them 11.8. So the chain concentrates the overshoot.

The cascaded section and the literature of active filters use exactly this fact deliberately: cascades of different sections are arranged so that no section’s peaking adds to another’s. A chain of identical stages has no such arrangement, and with a different Q in each stage there would be better answers than any of these. That would be a filter, however, rather than a chain, and it would have to be designed as one.

What the march assumes #

The stages are ideal second-order sections that do not load each other, and their dampings and frequencies are identical. They are linear, so a large step would meet the slew limit of each stage in a way that has nothing to do with this, as the step that is too big measures. Everything is at a 1 per cent band, and the best damping would move with the band as it does for one stage, although the structure would not. A finer band is a stronger test of the delay, and it would move the crossover to fewer stages, since the second term in the arithmetic above carries $z$ and the spread is what this stage is short of. That effect has not been computed here.

The comparison is also at equal stage bandwidth. If the stage’s resource were something else, a stated gain-bandwidth product with a closed-loop gain chosen for the second pole, the comparison would change, since a higher gain in each stage with the same product would lower its bandwidth directly. The gain that stops making a stage faster gives the relation for that case.

The numbers to keep #

At equal bandwidth and within 1 per cent, a second-order stage at its best damping beats a one-pole stage by 23 per cent alone, 17 in a chain of two, 11 in four and 2 in eight, and it loses by 5 per cent in a chain of sixteen. The best damping stays near 0.83 at every length. A stage’s squared spread, $(4\zeta^2 - 2)/\omega_n^2$, vanishes at $1/\sqrt 2$, but its delay is about 1.4 times a one-pole stage’s and a chain’s delay adds as $N$. So the spread that the stage removes is worth less at every further stage, and its delay is worth the same. The flattest eight-stage chain, $\zeta = 0.7$, is 1 per cent flat to five times the one-pole chain’s frequency and takes 46 per cent longer to settle.

Still open: unequal stages, a finer band, and errors that cancel #

Unequal stages. A chain whose stages have different bandwidths or different dampings can be designed so that their overshoots do not accumulate, and that is where a filter’s section ordering comes from. Which stage to make fastest, for a fixed total of bandwidth, is a budget question with a closed answer in the normal limit for a one-pole chain, and with no such limit for second-order ones.

A finer band. At a tighter settling band the spread’s multiple rises as the logarithm of the tolerance and the delay’s contribution does not. So the crossover should move, and it should move in the direction of fewer stages being better served by second-order sections at coarse bands. A measurement across bands would decide whether the rule of about ten stages is a rule or an accident of 1 per cent.

Errors that cancel. Alternating stages whose amplitude errors have opposite signs can be flatter than any one of them. In time the delays do not cancel. Whether a flatter chain also settles faster, in an arrangement where the stages differ, is the comparison that would say which of the two edges a designer can trade against the other. The figure above suggests they are nearly independent for identical stages, and alternating stages could change that.