The derivative that is a delay

Every derivative, and the one that is zero tested the claim the ladder-filter essays had leaned on from the start: that a doubly terminated ladder filter’s response is stationary in every element at its passband maxima. The adjoint method returned two parts in ten billion. The next four essays took that result apart — the poles are not stationary, the second derivatives are all negative, three of five directions in tolerance space do nothing, and the one that survives the whole band is the ripple depth. All five measured how the response moves when an element’s value moves.

A real element also has a loss. An inductor’s winding has resistance, a capacitor’s dielectric leaks, and both are usually quoted as a quality factor. Losses are not value errors: an inductor of exactly the specified inductance with a winding resistance is still the specified inductance. So the filter’s celebrated stationarity — a zero first derivative with respect to every value — does not obviously extend to them, and the question is what their first derivative is.

It has a name. Give every part the same dissipation and the derivative of the filter’s loss with respect to it is the filter’s group delay — the quantity the delay that is a centre of mass identified with where an impulse response balances. A ladder sags exactly where its delay is largest, which for a Chebyshev design is the band edge.

A loss that is a shift in s #

The construction is exact and short. Let every reactance in the filter have the same dissipation $\delta$: each inductor $L$ a series resistance $\delta L$, each capacitor $C$ a shunt conductance $\delta C$. Then the inductor’s impedance is $sL + \delta L = (s + \delta)L$ and the capacitor’s admittance is $(s + \delta)C$. Every reactance has exactly the impedance it would have had at the complex frequency $s + \delta$, and the terminating resistors, having no frequency in them, are unchanged. So the lossy ladder’s transfer function is the lossless one evaluated at a shifted frequency:

$$H_{lossy}(s) = H(s + \delta).$$

On the imaginary axis that is the lossless function evaluated a distance $\delta$ to the right. Expanding $\ln H$, which is analytic, to first order:

$$\ln|H(j\omega + \delta)| = \ln|H(j\omega)| + \delta\,\frac{\partial \ln|H|}{\partial\sigma}.$$

The Cauchy–Riemann equations say that the real part’s derivative along $\sigma$ equals the imaginary part’s derivative along $\omega$ — $\partial\ln|H|/\partial\sigma = \partial\varphi/\partial\omega$ — and that is minus the group delay. So the magnitude falls by $\delta\,\tau_g(\omega)$ nepers, which is $8.686\,\delta\,\tau_g(\omega)$ decibels. The dissipation’s derivative is the delay.

A fixed $\delta$ is a quality factor that rises with frequency, $Q = \omega/\delta$, and that happens to be what a real inductor’s winding does: a series resistance that does not change with frequency is exactly $\delta L$ with $\delta = R/L$, a constant. So for inductors whose loss is their copper’s direct-current resistance, the uniform-dissipation model is not an idealisation but the circuit itself; it is skin effect and core loss, which make the resistance rise with frequency, that break it. With $\delta$ chosen so that every part has a quality factor $Q$ at the band edge, $\delta = \omega_c/Q$, which is 62.8 s⁻¹ for $Q = 100$ at 1 kHz. The hero figure solves the filter with those resistors and conductances in the netlist — no shift, no expansion — and adds the prediction to the lossless response as dots. Across the passband the prediction holds to 3.6 per cent of the extra loss.

Where it sags, and why there #

The sag is not uniform, because the delay is not. A Chebyshev ladder’s group delay is smallest at low frequency and rises to a sharp peak just inside the band edge, where the poles closest to the axis put their phase change. So the extra loss is 0.383 dB at 38 Hz, where the delay is under a millisecond, and 0.943 dB at 953 Hz, where it peaks at 1.79 ms. The design’s half-decibel ripple, flat-topped by construction, is rounded off exactly where it was steepest, and the band edge — which a specification usually writes as the frequency where the loss reaches the ripple — moves inward.

Divided by $8.686\,\delta$, the extra loss at forty frequencies across the band lies on the lossless filter’s delay for every Q tried. The scatter about the diagonal shrinks as the parts improve — 10.5 per cent at Q = 30, 3.6 at 100, 1.24 at 300 — because the law is the first term of an expansion in $\delta$, and the next term is the delay’s own derivative, largest where the delay peak is sharpest. A specification on passband loss, then, is a specification on one product: the parts’ dissipation times the filter’s delay. Nothing else about the circuit enters at first order — not which element is where, not the element values, not the impedance level.

That is the result that sits beside the stationarity the earlier essays measured. With respect to every element’s value at a ripple peak, the first derivative of the response is zero; with respect to every element’s dissipation, it is the group delay, which at the edge is the largest number in the passband. The ladder filter is insensitive to tolerance and fully sensitive to loss, and the two properties have nothing to do with each other. The tolerance that can only take away found that the second-order effect of tolerance at a peak is always a loss; dissipation is a first-order loss everywhere.

The same fact in the time domain, and at the poles #

The shift has two other readings, and both make the result less surprising.

In time, $H(s + \delta)$ is the transform of $h(t)\,e^{-\delta t}$. Uniform dissipation multiplies the lossless filter’s impulse response by a decaying exponential, exactly, and so it takes most from whatever part of the response arrives late. The frequencies whose energy arrives late are, by definition, the ones with a large group delay: the delay that is a centre of mass found the low-frequency delay to be exactly the centroid of the impulse response, and a part of the response centred at time $\tau$ is multiplied by roughly $e^{-\delta\tau}$, which is $1 - \delta\tau$ for small $\delta\tau$ — the delay law again, read off a clock instead of a phase. A Chebyshev filter’s edge rings longest because its edge poles are its highest-Q poles, and the exponential takes most from the ringing.

At the poles, the shift moves every pole of the filter a distance $\delta$ to the left, the same for all of them. That is a large change for a pole near the axis and a small one for a pole far from it: a pole pair at a distance $\sigma$ from the axis has its quality factor reduced by the factor $\sigma/(\sigma + \delta)$. The derivative of a root found the filter’s poles no more than twice as robust to tolerance as a cascade’s; to dissipation every realisation’s poles are equally robust, because the shift is the same for every pole whatever circuit puts it there. The poles nearest the axis set both the delay peak and the edge’s ringing, and they are the ones a fixed δ moves furthest in proportion.

One over Q, with the delay as its constant #

Swept over the parts’ quality, the sag falls as one over Q with a slope of −0.986: 0.943 dB near the edge at Q = 100, 0.480 at 200, 0.097 at 1000. The constant in front of $1/Q$ is $8.686\,\omega_c\,\tau_g$, so the edge sags 2.7 times as much as the frequency at 300 Hz, close to the ratio of their delays, 2.8. For an LC filter built from ordinary inductors, whose Q at a kilohertz is a few tens to a hundred, that is the better part of a decibel at the band edge — more than the ripple the design was chosen for.

At low Q the law overstates the loss: at Q = 10 it predicts 9.77 dB at the edge where the solved filter loses 7.45. The reason is visible in the hero figure at a lower Q: the losses that move the response also flatten the delay peak, and a first-order law computed from the lossless delay is using a peak the lossy filter no longer has. The Q the components allow found that a resonator’s quality factor is limited by its worst component’s; for a ladder filter the equivalent statement is that its passband flatness is limited by its parts’ Q times its own delay.

Higher order, longer delay, larger sag #

The delay at a Chebyshev ladder’s edge grows with its order, since more poles crowd towards the axis there.

At Q = 100 the worst passband sag is 0.370 dB at third order, 0.942 at fifth, 1.752 at seventh and 2.742 at ninth, and in each case it sits near the edge at the delay peak — 0.69, 1.79, 3.44 and 5.58 ms. The ninth-order ladder filter, chosen for its steep skirt, has lost five times its own specified ripple at the edge. The first-order law tracks the trend and overstates the ninth by 11 per cent, where $\delta\tau_g$ approaches a third and the loss has begun to reshape the delay it multiplies.

The trade is the familiar one between selectivity and delay, with loss added as a third party. A designer who adds sections for a steeper skirt adds delay near the edge, and with lossy parts that delay is paid for in passband sag that no choice of element values can recover — the filter’s insensitivity to values does not reach it. A ladder is not a cascade chose the ladder filter over the active cascade on its tolerance behaviour; the dissipation result is the same for both, since a cascade realising the same transfer function with the same lossy reactances is the same $H(s + \delta)$. Loss does not distinguish between realisations; tolerance does.

When the dissipation is not uniform #

The construction needs every reactance to have the same $\delta$. Real filters do not: an inductor’s Q is usually tens, a capacitor’s thousands.

Putting the same total dissipation all into the inductors or all into the capacitors splits the loss unevenly, and the figure shows how. At low frequency an inductor’s series resistance sits directly in the signal’s path between the terminations, so the inductors-only case loses 0.530 dB at 20 Hz, while a capacitor’s shunt leakage barely loads a node whose reactance is enormous there, so the capacitors-only case loses 0.224. Across the band, though, the two average to the uniform case to 2.1 per cent. To first order the law holds for the mean dissipation of the inductors and the capacitors, and the difference between them adds a correction that depends on how the filter’s stored energy is divided between its electric and magnetic fields at each frequency — information the delay alone does not carry.

For the usual filter, with lossy inductors and nearly lossless capacitors, that means the delay law with $\delta$ set to half the inductors’ dissipation gives the passband loss to within the split’s correction: a ladder of Q = 50 inductors and ideal capacitors sags about as much as one built from Q = 100 parts across most of the band, and more at low frequency, where the inductors’ resistance is in series with the signal.

Delay, flatness and loss, three faces of one design choice #

Flat magnitude, unflat delay found that the flattest magnitude responses have the most peaked delay at their edges, and that the two cannot be traded independently. The dissipation law adds a third face to that trade. A filter chosen for a flat passband and a sharp edge — a high-order Chebyshev — has a delay peak at its edge, and its lossy realisation has a loss peak there in proportion. A filter chosen for a flat delay — a Bessel — has no such peak, and its lossy realisation sags uniformly across the band, by $8.686\,\delta\,\tau_0$ with $\tau_0$ its nearly constant delay: a loss, but a flat one, which a gain stage can make up exactly.

So the choice of filter family, made for reasons of magnitude or delay, also decides the shape of the passband error that real parts will add. A Chebyshev’s ripple specification and its loss with real parts both live at the edge, and they add; a Bessel’s error is a level shift. For a filter that must be built from ordinary inductors and still hold a tight passband tolerance, that is an argument for a lower-ripple design than the skirt alone would ask for, and it is a number: the tolerance has to absorb $8.686\,\omega_c\,\tau_g(\omega_{edge})/Q$ at the edge.

What it means for a passband specification #

The practical reading is that a filter’s passband loss specification, written in decibels across the band, should be checked against the delay curve before it is checked against anything else. The delay is computed from the lossless design, which every filter tool produces; the parts’ Q is on their data sheets; the product, times 8.686, is the sag at each frequency to a few per cent. A designer who finds the sag unacceptable has three moves, and the law ranks them. Better parts reduce it as $1/Q$. A lower order reduces it roughly in proportion to the delay. And predistortion — designing the lossless filter with its ripple pre-shaped to cancel the expected sag, which is what the classical insertion-loss design method does — moves the sag rather than removing it, since the delay, and therefore the loss, is still largest at the edge; it buys flatness at the price of absolute loss.

The same law reads backwards as an instrument. A lossy ladder filter’s measured passband loss, minus the lossless design’s, divided by the delay, is the parts’ mean dissipation, measured in the circuit at the frequencies it works at. Because the delay varies across the band and the dissipation is assumed constant, the scatter of that quotient across frequency says how well the uniform-dissipation model fits the real parts.

The law, collected #

A doubly terminated ladder filter whose every reactance has dissipation δ is exactly H(s + δ), and its loss rises by $8.686\,\delta\,\tau_g(\omega)$ decibels to first order — the derivative with respect to dissipation is the group delay, where the derivative with respect to value is zero at a ripple peak. A fifth-order 0.5 dB Chebyshev of Q = 100 parts sags 0.383 dB at low frequency and 0.943 dB at its edge, the law holding to 3.6 per cent, 1.24 per cent at Q = 300. The sag falls as 1/Q with a slope of −0.986, grows with order from 0.37 dB at third to 2.74 at ninth, and for unequal inductor and capacitor losses follows their mean.

Still open: the second-order term, a ladder with zeros, and the delay used as a sensor #

The term after the delay. The law is first order, and it overstates the loss where $\delta\tau_g$ is not small, near the edge of a high-order filter. The second term in the expansion is the derivative of the delay with respect to frequency, times $\delta^2/2$; whether adding it closes the gap at Q = 10 and at ninth order is a check on whether the whole loss is the analytic continuation it appears to be.

A ladder with transmission zeros. An elliptic ladder has tuned shunt branches whose resonances put zeros on the axis in the stopband. Uniform dissipation moves those zeros off the axis, which fills the stopband nulls — a loss of a different kind, in the band where the delay is not defined. How far a given Q fills a null, against the null’s depth, would complete the picture for the filters the zeros that buy an order built.

The split, computed. The correction for unequal inductor and capacitor dissipation depends on the division of stored energy between the two at each frequency, which the adjoint machinery of the first essay in this sequence can compute directly. Putting it beside the delay law would turn the half-the-inductors’-dissipation rule into a number with its own error bar.