The delay that is a centre of mass

Flat magnitude, unflat delay measured group delay across the passbands of the three classical filter families and found it varying by half across the band of the two that are usually chosen. Flat delay, bought with more delay then flattened it with an all-pass section and paid in a longer delay everywhere. Both treated group delay as a function of frequency, $-d\phi/d\omega$, to be read off a curve.

There is one point on that curve with a meaning that does not depend on frequency at all, and it is the one designers most often use and least often check. The group delay at zero frequency is usually quoted as “the filter’s delay” and used as though it were the time at which a step arrives. It is not that. It is something more exact and in one respect more useful: the centre of mass of the filter’s impulse response. And the step’s arrival, read as the time it crosses half its final value, is a different number that can be earlier or later than the centre of mass depending on the filter’s family.

This essay measures both, by routes that share nothing but the netlist, and finds which one a designer is actually computing when they quote a filter’s delay.

Two routes that share only the netlist #

The filters are the ladder networks of the earlier measurements: Bessel, Butterworth and half-decibel Chebyshev low-pass filters, each with its half-power frequency at one kilohertz, built as doubly-terminated ladder networks and solved as netlists.

The first route is the frequency domain. The network is solved at a frequency a thousandth of the corner, and at two frequencies a part in a thousand either side; the phase difference over the frequency difference is the group delay there, which at a thousandth of the corner is its zero-frequency value to many figures.

The second route never evaluates a phase. The network’s determinant is rooted for its poles, the step response is built from the residues, and the impulse response’s first moment is computed from it — $\int t\,h(t)\,dt / \int h(t)\,dt$, the time at which the impulse response would balance on a pivot. Integrating by parts turns that into the area between the step and its final value,

$$\bar t = \int_0^\infty \left(1 - \frac{y(t)}{y(\infty)}\right) dt,$$

which is what is integrated, over forty periods of the corner, with the final value taken from the network at zero frequency rather than from the last sample.

The two routes agree. On the fifth-order Chebyshev they give 0.7090 milliseconds both.

The agreement is a theorem #

That agreement is not a property of the Chebyshev, and the measurement shows it on everything available.

Twenty-one filters, from a single pole to a seventh-order Chebyshev, sit on the diagonal with the worst disagreement 5.7 parts in a million, which is the finite-difference phase route’s own error. The reason is a line of algebra. The transfer function is the Laplace transform of the impulse response, $H(s) = \int h(t)e^{-st}dt$, so its derivative at $s = 0$ is $-\int t\,h(t)\,dt$. Divide by $H(0) = \int h\,dt$ and the result is minus the centre of mass. On the other side, at zero frequency the phase of $H$ is zero and its logarithmic derivative is exactly $-\tau_g(0)$. The two are the same derivative of the same function, taken once in time and once in frequency.

That makes the zero-frequency group delay unusually robust. It needs no assumption about the filter’s shape, no symmetry and no minimum phase: any stable linear network with a non-zero gain at DC has a low-frequency delay equal to its impulse response’s balance point. It is also additive: the centre of mass of two filters in cascade is the sum of their centres, because the delays add, which is why a chain’s total delay can be budgeted stage by stage.

Half-way is a different question #

What it does not equal is the time at which the step crosses half its final value, and that is the number an oscilloscope shows and a timing diagram uses.

For a single pole the step is an exponential, its centre of mass is the time constant, and half-way is reached at $\ln 2$ of it, 0.693. Its impulse response is a decaying exponential, maximally lopsided — all of its weight at the start, with a long tail that pulls the balance point late. Half-way comes early.

For a Bessel filter the impulse response grows more symmetric with every order, and the half-way time climbs toward the centre of mass: 0.900 at order two, 0.989 at five, 0.996 at seven. A symmetric impulse response has its half-way point and its balance point at the same place, and a Bessel filter’s delay is flat across its band precisely because its impulse response is close to a symmetric bell.

Butterworth and Chebyshev filters go the other way. From order two their half-way times are later than their centres of mass — by 1.3 to 8 per cent for Butterworth and up to 16.9 per cent for a second-order Chebyshev. Their impulse responses overshoot and swing negative before they settle, and a negative lobe late in the response pulls the balance point earlier than the bulk of the weight, since negative mass at a late time counts as weight removed from there. The step rises late and then overshoots, so its half-way crossing falls after the balance point.

So the number quoted as a filter’s delay depends on which number was meant. For a Bessel filter the two agree to within a per cent above order four, and the distinction is academic. For a fifth-order Butterworth they differ by eight per cent, and a timing budget that uses $\tau_g(0)$ as the step’s arrival is eight per cent early.

The RC tree, where the centre of mass has a name #

The centre of mass has a much older name in a different setting. In 1948 Elmore proposed estimating the delay of a wideband amplifier, and later of a chain of resistors and capacitors, as the first moment of its impulse response — and gave a formula for an RC tree that needs no solve at all: the sum over every capacitor of its capacitance times the resistance it shares with the path to the output.

The Elmore sum, computed by hand from the component values, matches the solved centre of mass to five figures at every section count — it is the theorem above, evaluated without a solver. As the ladder network is divided more finely it approaches the distributed RC line’s $\tfrac12RC$, reaching 0.5156 at 32 sections.

The half-way time falls faster, from 0.693 of the delay for one section to 0.757 at 32: a long RC line’s step crosses half at about three-quarters of its Elmore delay. That ratio is the correction every interconnect timing tool that uses Elmore’s estimate has to carry, and it is a fixed number only for a uniform line; a tree with a different shape has a different impulse response, a different asymmetry and a different ratio. Elmore’s delay is exact about the centre of mass and approximate about everything a waveform-reading engineer actually looks at. The ladder network built for the rest of itself measures a ladder network from the other direction, as a divider loaded by its own continuation, and the two together are the static and the transient account of one network.

The second moment is the rise time #

If the first moment is the delay, the second is the spread of the impulse response about it — its standard deviation — and that should be related to how long the step takes to rise.

The second moment comes from the same step, weighted by time: $\int t(1 - y/y_\infty)\,dt$ is half the impulse response’s second moment, and with the first moment it gives the spread. For a Gaussian impulse response the step is an error function, and its 10-to-90 per cent rise is $2 \times 1.2816$ standard deviations, 2.563. A Bessel filter’s ratio rises once, to 2.777 at order three, and then closes on the Gaussian value order by order, reaching 2.671 at seven.

That is the second half of why Bessel filters are prized for edges. Their delay is a centre of mass that their step genuinely arrives at, and their rise time is a spread that their step genuinely takes; as the order grows the step becomes an error function delayed by its centroid, which is the cleanest edge a linear network can deliver. The line from Bessel to Butterworth measured what is traded along the way to sharper selectivity, and the moments say what that trade does to the two numbers a timing diagram reads.

Why delays add and rise times add in quadrature #

The moments explain a rule every engineer with an oscilloscope uses, and say exactly when it is right. When two filters are cascaded the impulse response of the chain is the convolution of the two, and convolution does two simple things to moments: it adds the centres of mass and it adds the variances. That is not an approximation; it is true of any two impulse responses whose moments exist, for the same reason that the mean and the variance of a sum of independent random quantities add.

So the delay of a chain is the sum of the stages’ zero-frequency group delays, exactly, and the chain’s spread is the square root of the sum of the stages’ squared spreads, exactly. The familiar rule that rise times add in quadrature — an edge of rise time $t_1$ through an instrument of rise time $t_2$ displays $\sqrt{t_1^2 + t_2^2}$ — is the second of these statements with rise time standing in for spread. It is exact only where rise time is proportional to spread with the same constant for every stage, which the figure above says is true only of Gaussian edges. The instrument’s own rise time measured how far the rule departs for one-, two-, three- and four-pole front ends, and the departures are this essay’s ratio moving from stage to stage: a one-pole instrument’s rise time is a different multiple of its spread from a Bessel edge’s.

The first-moment half of the rule has no such caveat. Delays add exactly, whatever the shapes, provided “delay” means the centre of mass. A designer aligning a signal path with a reference path through different filters can match their zero-frequency group delays and be certain that slow signals arrive together, regardless of how differently the two paths ring. What the moments cannot promise is that a fast edge arrives together, because an edge’s half-way time is not a moment.

How a step starts, and how it finishes #

The first two moments describe the bulk of a step. They say nothing about its very beginning or its very end, and both have exact statements of their own that complete the picture. The start a step takes from infinity reads the beginning off the transfer function at infinite frequency: a network with r more poles than zeros starts its step with its first $r - 1$ derivatives zero, so a seventh-order filter’s output leaves the axis as the seventh power of time. The end is set by the slowest pole, and for the Chebyshev filters here that pole is lightly damped, which is why their steps take several centres of mass to settle even though their centres of mass come earlier than their half-way times.

Between those limits, the moments are what a single number can say. The zero-frequency group delay is the step’s balance point, the half-way time is where it is half done, and their ratio is a direct measure of the impulse response’s asymmetry: below one for a lopsided, tail-heavy response like an RC network’s, near one for a symmetric bell like a high-order Bessel’s, and above one for a response with a negative lobe. The boundary that is a rise time shows what an edge’s shape decides on a circuit board, where a track’s reflections sit inside or outside the edge depending on exactly these times.

Where the distinction is spent #

The practical consequence is a choice of which number to put in a timing budget. A designer who needs the time at which a digital receiver after a filter will see a transition should use the half-way time, and for any filter other than a high-order Bessel that is not $\tau_g(0)$: it is earlier for one pole and later for the filters that ring. A designer who needs to align two signal paths — a channel and its timing reference, a sensor and its compensation — and whose signals are slow compared with the filters’ bandwidth should use the centre of mass, because for slowly varying signals every filter delays by exactly $\tau_g(0)$, and the delays add down a chain.

The theorem also gives a measurement method that does not need a phase meter. The centre of mass of a filter’s response can be measured from its step response alone, as the area between the step and its final value; a digitised step and a sum give the low-frequency delay to the precision of the record, with no differentiation of a noisy phase. The energy that arrives first uses the same impulse response as a schedule for its energy rather than for its mass, and a step recorded once gives both.

What the measurement leaves out #

The filters are ideal ladder networks with lossless reactances; losses change the poles slightly and the delays with them, but not the theorem, which holds for any stable linear network. Every filter is a low-pass with a gain at DC. A band-pass has no DC gain, its impulse response integrates to zero, and its centre of mass is undefined in this form; the corresponding statement for a band-pass is about the envelope of its response, and its delay is the group delay at the centre frequency rather than at zero.

The half-way time assumes a step that is perfectly sharp. A real edge has a rise time of its own, and the measured half-way time of the output is then the input’s half-way time plus a delay that lies between the filter’s half-way time and its centre of mass, approaching the centre of mass as the input edge becomes slow compared with the filter.

The number to keep #

A filter’s group delay at zero frequency is exactly the centre of mass of its impulse response — measured equal on twenty-one filters to 5.7 parts in a million by routes that share only the netlist — and Elmore’s RC-tree delay is the same quantity computed by hand. It is not the step’s half-way time: that is $\ln 2$ of the delay for one pole, approaches it from below for Bessel filters, and passes it for filters that ring, by eight per cent for a fifth-order Butterworth and up to 16.9 per cent for a Chebyshev. On a uniform RC line half-way is 0.757 of the Elmore delay. The second moment is the rise time, and a Bessel step’s rise closes on a Gaussian’s 2.563 standard deviations.

Still open: the band-pass’s envelope, the negative delay, and the cascade of moments #

The band-pass, where the delay is carried by an envelope. A band-pass has no DC delay; its delay belongs to the envelope of a tone at its centre, and the group delay at the centre should be the envelope’s delay exactly when the envelope is slow enough. Marching tone bursts through one would test that, and say how slow is slow.

A negative group delay. A notch has a group delay that is negative at its centre. If the envelope theorem holds there too, the envelope of a smooth burst through a notch arrives before the input’s — a measurement that looks impossible and should not be.

Moments down a chain. Means and variances add for cascaded filters, so a chain’s delay and rise time can be predicted from each stage’s first two moments. How quickly the chain’s step becomes Gaussian, by a central-limit argument applied to impulse responses, is measurable stage by stage.