What a filter does not pass, it sends back

The two resistors a ladder was designed between found that a passive ladder filter is a two-port designed between two stated resistances, and that driving it from the wrong one changes its response by more than the tolerance of the resistor usually drawn there. The termination an even order cannot have found that an even-order Chebyshev cannot even be built between equal resistances. Both treated the terminations as something the filter’s transmission depends on. Neither looked at what the filter does with the power it does not transmit.

That question has a complete answer, and the answer turns the terminations from a sensitivity into a single quantity measured at the other port. An LC ladder dissipates nothing. Whatever power the source makes available either reaches the load or comes back to the source. So the transmission and the reflection are not two properties of the filter; they are one property, read at two ends.

Two readings of one solve #

The filter is the fifth-order Chebyshev of half a decibel of ripple the earlier essays on termination used, synthesised by continued fraction and scaled so that its half-power corner sits at a kilohertz, between a 1 kΩ source and a 1 kΩ load. At each frequency the circuit is solved once, and two numbers are taken from the solution without either being computed from the other.

The transmitted power is the transducer gain: the power in the load over the power the source could deliver into a matched load, $|t|^2 = (4R_s/R_L)\,|V_2/V_s|^2$. It is read from the load’s voltage.

The reflected power is $|\rho|^2$, with $\rho = (Z_{\text{in}} - R_s)/(Z_{\text{in}} + R_s)$ and $Z_{\text{in}}$ the impedance the filter presents at its input, read from the input node’s voltage and the source’s current.

Across 481 frequencies from direct current to 1.6 times the corner, the two sum to one to 2 × 10⁻¹⁵ — the last bits of the arithmetic. The transmission ripples between 1 and 0.8913 in the passband, and the reflection ripples between 0 and 0.1087 in exact complement: at every frequency where the load is short of a full share, the missing power is on its way back to the source. It reaches zero at direct current and at 0.555 and 0.898 of the corner, where the transmission is at its maxima, and it rises to meet one in the stopband, where the filter sends everything back.

This is Feldtkeller’s relation for a lossless two-port, and it is the reason a doubly-terminated ladder filter is insensitive to its components — the property a ladder is not a cascade measured as fifty-four times steadier than a buffered cascade. At a transmission maximum the reflection is zero, and a zero of $|\rho|^2$ is a minimum of a non-negative quantity: any component error can only raise it, so the transmission can only fall, and to first order it does not move. The insensitivity is not a separate virtue of the ladder filter. It is what complementarity looks like at a zero.

A circle drawn by the ripple #

Complementarity fixes the size of the reflection at every point of the passband by the ripple alone. An equiripple passband of ripple $r$ decibels has its troughs at a transmission of $10^{-r/10}$, which in the Chebyshev ripple factor $\varepsilon$ is $1/(1+\varepsilon^2)$. So at every trough

$$|\rho|^2 = \frac{\varepsilon^2}{1+\varepsilon^2}, \qquad |\rho| = \frac{\varepsilon}{\sqrt{1+\varepsilon^2}},$$

and nowhere in the passband can it be larger. Drawn in its own plane, the reflection coefficient is confined to a disk.

The locus starts at the centre, because at direct current the inductors are shorts, the capacitors are open and the filter presents exactly 1 kΩ. As the frequency rises it winds out to touch the circle of radius 0.3298 at the first ripple trough, 0.291 of the corner, returns through the centre at the first transmission maximum, touches the circle again at 0.764, returns through the centre again, and leaves towards the circle’s edge as the band ends at 0.944 of the corner.

A radio-frequency engineer would call that circle a return-loss specification: 9.64 dB, the reflection in decibels below the incident power. A filter designer calls the same thing half a decibel of passband ripple. They are two names for one number, and a designer who has specified one has specified the other whether they intended to or not.

The order does not enter #

The identity leaves one thing to check: that the worst return loss depends on the ripple alone, and not on how many sections the filter has.

It does. At every ripple from a hundredth of a decibel to three, orders three, five and seven return the same worst return loss to two decimal places, and all three sit on the closed form: 26.38 dB at 0.01 dB of ripple, 21.62 at 0.03, 16.43 at 0.1, 11.76 at 0.3, 9.64 at 0.5, 6.87 at 1, 4.33 at 2 and 3.02 at 3. The order decides how many times the reflection reaches its limit across the band and where; it has no say in how large the limit is.

The conversion is worth having in both directions, because the two communities that use these filters specify them in different currencies. A receiver’s front-end filter is often required to present at least 15 dB of return loss to the amplifier before it, so that the amplifier’s own stability and noise match are not disturbed. By the table that is a Chebyshev of at most about 0.14 dB of ripple, and a designer who picks the familiar half-decibel prototype for its sharper skirt has, without noticing, delivered a filter that reflects eleven per cent of the power at its troughs into an amplifier that was promised three.

In the other direction, the ripple a filter can tolerate is bounded by what its source and load can tolerate being reflected at them. Terminated at both ends found a line’s series and parallel terminations costing half the swing to buy the absence of reflection; a filter whose passband is equiripple buys its sharp skirt by spending reflection, in a controlled amount, at chosen frequencies.

When the filter is not lossless #

The identity is exact for reactances, and real inductors are not reactances. A coil has a resistance, usually stated through its Q, and a filter built from coils absorbs some of the power the identity said had only two destinations. The shortfall is a third route to the same budget.

With a series resistance of $\omega_c L/Q$ added to each of the three inductors, the transmitted and reflected powers no longer sum to one. The shortfall was computed separately, as the power dissipated in the added resistances from their own currents, and the three quantities — transmitted, reflected, absorbed — sum to the available power to 10⁻¹⁰ at every frequency and every Q. The budget still closes; it has a third line in it.

The absorbed power is not spread evenly. At Q 400 its largest value is 2.79 % of the available power; at Q 100 it is 10.5 %, at Q 50 19.5 % and at Q 25 33.9 %, and in every case it peaks near the band edge — 0.952 of the corner at Q 50 — which is where a Chebyshev LC ladder stores the most energy per unit of power it passes, and where its group delay peaks. Loss that is proportional to stored energy lands where the energy is.

Two consequences follow. The first is that a lossy filter’s passband droops towards its edge rather than rippling evenly, which is the familiar rounding of a real filter’s corner, and it is the loss rather than any error in the values that does it. The second is less familiar: the reflection is also changed, because a resistance inside the filter moves its input impedance. A filter measured only for its transmission can look merely lossy while its return loss has moved too, and the complementarity check, $|\rho|^2 + |t|^2$, is the measurement that shows how much of the shortfall is absorption. On a bench that is a network analyser’s two readings added together: anything short of one is heat.

Where the stopband goes #

The same budget answers a question that is easy to get wrong: what happens to the power a filter rejects. The intuitive answer is that a filter removes it — that the stopband is somewhere the unwanted signal is lost. For a passive LC ladder that is almost entirely false.

At 1.4 times the corner, the Q 50 filter transmits almost nothing, absorbs 3.40 % of the available power and reflects 96.4 %. At Q 400 it absorbs 0.43 % and reflects 99.3 %. The rejected power is not removed; it is sent back to the source, and the stopband is a mirror. A filter that provides 40 dB of rejection does so by reflecting 99.99 % of the incident power at that frequency, and the source has to do something with it.

For a signal generator with a resistive output that is harmless: the source’s own resistance absorbs what comes back. For a power amplifier followed by a harmonic filter it is the whole design problem. The harmonics the filter rejects arrive back at the amplifier’s output, where they are reflected again by its impedance, which is usually neither the design value nor resistive, and the result is a standing wave between the amplifier and the filter at the harmonic frequencies. The filter’s rejection in a data sheet is measured between matched resistances, which is the one condition under which the reflected harmonics are absorbed rather than recirculated.

That is also why three families, one corner could compare Butterworth, Chebyshev and Bessel responses by their transmission alone and be complete about the load: in each family the stopband’s reflection is fixed by its transmission. The family decides how steeply the reflection rises from the passband’s controlled level to near one; it does not decide whether it does.

Loss flatters a return loss #

The loss figure contains a warning for anyone measuring a filter’s match. Loss inside a two-port reduces what comes back, for the reason the mismatch that the cable hides found in a lossy line: a reflection that has to travel through loss twice arrives smaller than it left. The same happens inside a filter. The worst reflection inside the ripple band is 10.87 % of the available power for the lossless filter, 10.76 % at Q 400, 10.45 % at Q 100, 10.12 % at Q 50 and 9.70 % at Q 25: the lossiest filter reads a return loss of 10.1 dB against the lossless design’s 9.64. A lossy filter looks better matched at its input than a lossless one of the same design, and the improvement is not a property of the match — it is heat.

The complementarity check is what separates the two. A lossless filter’s reflection and transmission sum to one; a lossy one’s fall short, and the shortfall is exactly the dissipated power, to 10⁻¹⁰ in the figure. A return-loss measurement quoted without the transmission beside it cannot say which of the two it is looking at, and at the band edge, where a Chebyshev’s absorption peaks, the difference can be a fifth of the power.

Every source from one impedance #

The last figure returns to where the first essay on this filter began: the wrong source resistance. There, the response at each mismatched source was found by solving the circuit again with that source. Complementarity says that is unnecessary.

The filter’s input impedance does not depend on the source at all — it is the impedance looking into the filter with its load attached — so it can be computed once, from the design. Change the source to $R_s'$ and the power the load receives, as a fraction of what that new source makes available, is still $1 - |\rho'|^2$, with $\rho'$ now measured against $R_s'$:

$$|t'|^2 = 1 - \left|\frac{Z_{\text{in}} - R_s'}{Z_{\text{in}} + R_s'}\right|^2 .$$

The two routes agree to 10⁻¹² at every frequency and every source. So the whole family of mismatched responses was already contained in one curve, the input impedance, and the first essay’s window — the range of source resistances inside which the passband holds its shape — is a statement about how far that impedance wanders from the design value across the band.

The same identity gives the worst case in closed form. The source’s own mismatch is a reflection coefficient too, $\Gamma = (1-m)/(1+m)$ for a source $m$ times the design value, and two reflections compose by the rule $(a + b)/(1 + ab)$ at their worst relative phase — the same bilinear rule by which velocities add in relativity, and for the same reason, since both keep a quantity inside the unit disk. With the filter’s own reflection at most 0.3298:

• a source of 0.8 or 1.25 times the design has $|\Gamma| = 1/9$, and the worst the band can deliver is 0.8191 of the available power; the solves give 0.8221 and 0.8205;
• a source of 0.5 or 2 times has $|\Gamma| = 1/3$, and the bound is 0.6431; the solves give 0.6487 and 0.6459.

The bound depends on the mismatch only through $|\Gamma|$, so a source too low by a factor and a source too high by the same factor have the same worst case, and the measured responses land within half a per cent of it on both sides. They do not land on it exactly, because the bound assumes the two reflections line up in phase somewhere in the band and the filter’s reflection sweeps through only some of the phases. How closely they line up is the one thing the input impedance’s shape decides.

How the numbers were obtained #

The filter is synthesised from the Chebyshev prototype by continued-fraction expansion, scaled so that its measured half-power frequency is 1 kHz, and built as a netlist with a 1 V source, exactly as in the essays before this one. Each frequency is one complex nodal solve. The transducer gain is taken from the load node’s voltage and the reflection from the input node’s voltage and the source resistor’s current; the two are never formed from each other. The ripple band’s edge is the highest frequency below the corner at which the transmission is still at its ripple floor, found by a scan and a bisection. With loss, each inductor is placed in series with a resistance $\omega_c L/Q$ at the corner, and the absorbed power is the sum over those resistances of $|\Delta V|^2/R$, normalised by the source’s available power.

What it leaves out #

Loss in the capacitors. A capacitor’s dielectric loss is a conductance across it rather than a resistance in series, and at audio frequencies it is usually far smaller than the inductors’ — but it lands at the same band-edge frequencies for the same reason, and it would add its own term to the budget.

A source that is not a resistance. An amplifier’s output impedance is often inductive at the frequencies a filter cares about. The same identity holds with a complex source impedance, but “available power” and the reflection then have to be defined against the conjugate, and the load that takes the most from such a source is not the one that matches its resistance.

The group delay, which complementarity does not mention. A ladder filter like this one has every transmission zero on the frequency axis and is minimum-phase, so its phase, and with it the delay that peaks at the band edge, is fixed by its magnitude — the phase the magnitude already knows is that argument. But the reflection’s phase is not a transmission phase, and how the delay moves when the source is wrong has not been drawn here.

Still open: the band edge, the stopband and the filter that is matched on purpose #

Where the loss concentrates. The absorption peaked at the band edge because the stored energy does, and the stored energy is proportional to the group delay. Whether the absorbed fraction is, to first order, the group delay times the loss rate — a single formula for the droop of any lossy filter — is a comparison of two measured curves this filter already provides.

The stopband’s reflection. In the stopband an LC ladder reflects almost everything, and what it reflects goes back into the source. A source that is an amplifier sees a nearly reactive load at those frequencies, and whether that load is capacitive or inductive decides whether the amplifier is stable into it. The input impedance’s phase above the corner, and how it depends on whether the filter begins with a shunt capacitor or a series inductor, is the question the match with no knob asked of a single L-section.

Equal ripple in reflection rather than transmission. A diplexer or a multiplexer splits a band between several filters sharing one input, and there the specification is a return loss across the whole band at the common port. Since reflection and transmission are complementary, a filter designed for equiripple return loss is a filter designed for equiripple transmission, and the question is what a shared input does to the identity when two filters each reflect what the other passes.