What each element buys against the limit

The load for a source that will not hold still found that a conjugate match is exact at one frequency and poor across a band: held fixed while the frequency moves, it stays within a decibel of the available power over only $1.0176/x$ of the centre, where $x$ is the source’s reactance over its resistance. It found the best fixed load for a source whose reactance could be anywhere within $\pm\Delta x$ of a centre: cancel the centre exactly, and take a resistance of $\sqrt{1 + \Delta x^2}$ times the source’s. And it named what it had not done: with more elements the band can be widened towards the Bode–Fano limit for the source’s reactance. Solving the best two- and three-element ladder networks for a stated band would say how much each element buys, and how close to the limit a practical network gets.

The source, the load and the chain #

The source is a resistance in series with an inductance whose reactance at the centre frequency is twice the resistance — the $x = 2$ the earlier essay drew most. The load is a resistor whose value is free. Between them sits a chain of $n$ reactive elements, alternating series and shunt, starting in series next to the source; each element is an inductor or a capacitor according to the sign of the one number that describes it, its reactance or susceptance at the centre.

The best network for a band is the one whose worst power transfer anywhere in the band is largest — the minimax criterion the earlier essay used for an uncertain reactance, applied now to frequency. It is found by a Nelder–Mead search from forty seeded starts over the load and the $n$ element values. That is a search, and a search can miss: the numbers below are the best found, lower bounds on the best that exists, and every one is checked against the one number that bounds them from above.

Over ±25 per cent of the centre the worst transfer is 54.2 per cent with no network at all — the best plain resistor — 82.8 per cent with one element, 89.5 with two and 97.3 with three. The curves show why the minimax criterion shapes them as it does: each best network ripples across the band, touching its worst value at several frequencies, because a network that were better at one edge than the other could trade some of that surplus to lift the worse edge.

The first element is the earlier essay’s answer #

The single-element network found by the search is a series capacitor with a reactance of −1.875 at the centre, and a load of 1.414 times the source resistance. Both numbers have a reason.

With the capacitor in place the source’s net reactance is $2w - 1.875/w$ at a frequency $w$ times the centre: −1.0 at the band’s lower edge and +1.0 at its upper. The capacitor has not cancelled the centre; it has centred the swing, so that across the band the net reactance ranges over ±1. That is precisely an uncertain reactance of $\pm\Delta x$ with $\Delta x = 1$, and the earlier essay’s answer for it is a load of $\sqrt{1 + \Delta x^2} = \sqrt 2$ guaranteeing $2/(1 + \sqrt 2) = 82.8$ per cent. The search found the same load and the same guarantee without being told either.

The coincidence is not one. The earlier essay’s source had a reactance that was uncertain; this one has a reactance that is certain at every frequency but different at each, and a load that must serve every frequency in the band is in exactly the position of a load that must serve every reactance in a range. The minimax arithmetic cannot tell the two situations apart, because it only ever sees the range of reactances the load has to accept, and a band is one more way of producing a range.

So the conjugate match is not the one-element answer for a band. It cancels the centre, and a band does not care about its centre: it cares about the extremes. The best single capacitor sits slightly below full cancellation, where it balances the two edges, and the load rises from the source’s resistance to accommodate the reactance the band cannot escape.

The conjugate match, for comparison #

The conjugate match is the natural first thought and it is worth seeing why the band rejects it. It spends everything at one frequency: complete transfer at the centre, and a fall on either side as steep as the source’s reactance makes it. The load that takes the most is the rule it comes from, and at one frequency it is exactly right. Over ±25 per cent it delivers 74.6 per cent at the band’s lower edge and 83.2 at its upper, a worst case below the minimax single capacitor’s 82.8 — because the conjugate treats the centre as the only frequency that exists, and the minimax network treats the edges as the frequencies that decide.

That is the whole difference between matching a frequency and matching a band, and it is why the Bode–Fano bound is stated as an integral. A match is a budget spent over frequency; the conjugate spends all of it at one point, where a band needs it spread.

What each further element buys #

In mismatch loss the ±25 per cent band reads 2.66 dB with no network, 0.82 with one element, 0.48 with two, 0.12 with three and 0.12 with four. The first element buys 1.8 dB. The second buys a third of a decibel and the third buys more than the second did, 0.36 dB. The two-element network the search finds is a series capacitor and a shunt capacitor into a load of 2.1 times the source, and the third element, a series inductor, lets the load come down to half the source instead: a different transformation, not a refinement of the same one. The fourth element, in a chain of single elements, buys two thousandths of a decibel.

The narrower band tells the same story more dramatically. Over ±10 per cent the losses are 2.33, 0.16, 0.10, 0.004 dB for zero to three elements: one element comes within a sixth of a decibel, three within four thousandths. The wider band tells it less kindly. Over ±50 per cent they are 3.18, 2.09, 1.15 and 0.59 dB, and three elements are still six-tenths of a decibel short.

The limit, and why it cannot be reached #

The Bode–Fano limit is the reason the curves flatten, and it is worth stating because it bounds every network, not only the chains searched here. A source that is a resistance $R$ in series with an inductance $L$ has, for any lossless network between it and any load, a reflection coefficient $\Gamma$ satisfying

$$\int_0^\infty \ln\frac{1}{|\Gamma(\omega)|}\,d\omega \;\le\; \frac{\pi R}{L}$$

The integral is a budget. Spending it on a band of width $B\omega_0$ at a constant reflection gives $|\Gamma| \ge e^{-\pi/(xB)}$ with $x = \omega_0L/R$, and so a transfer no better than $1 - e^{-2\pi/(xB)}$: 99.81 per cent over ±25 per cent here, 95.7 over ±50, and so close to one over ±10 that the figure reads 100.0.

Two things make it a limit rather than a target. Reaching a constant reflection across the band and none outside it needs a network whose response is a perfect rectangle, which needs infinitely many elements. And any reflection spent outside the band — which every finite network spends, since its response rolls off rather than stopping — is budget taken from inside. So the best finite network is always short of the bound, by an amount that falls as elements are added: 0.47 dB short with two elements over ±25 per cent, 0.11 with three.

What the bound does say sharply is which sources can be matched over which bands. The exponent $2\pi/(xB)$ is the whole of it: a source with more reactance, or a wider band, spends the budget faster. Doubling $x$ is exactly as bad as doubling the band. A source whose reactance is eight times its resistance, matched over ±25 per cent, is bounded at 79 per cent however clever the network.

The narrow band is the regime most matching is actually done in, and it is the regime in which the earlier essay’s single-element rule is nearly the whole answer.

The source’s own Q is the exchange rate #

The exponent in the bound, $2\pi/(xB)$, contains the source’s reactance over its resistance, and that ratio has a name: it is the source’s own quality factor at the centre frequency, the Q a resonator built from it would have. So the bound says something general about matching a reactive source over a band: the source’s Q and the band’s fractional width multiply, and their product is what the matching network is fighting. A source of Q two over ±25 per cent is the same problem as a source of Q one over ±50 per cent, or a source of Q ten over ±5 per cent.

That makes the bound a relative of resonance and its bandwidth: a resonator of Q $x$ has a bandwidth of $1/x$ of its frequency, and a band much wider than that cannot be matched well, whatever the network. The figures bear it out. The ±10 per cent band is two-fifths of the source’s own bandwidth of one half, and one element matches it within a sixth of a decibel; the ±50 per cent band is twice it, and three elements are still six-tenths of a decibel short.

It is also why small reactive sources are hard to match. A source whose reactance is large against its resistance — a short antenna is the textbook case, and a piezoelectric transducer another — has a high Q, and its Bode–Fano budget over any useful band is small. No network changes that; a network can only spend the budget well.

What a matching network in a real instrument does #

The three families of result on this page — the single element that centres the swing, the few elements that approach the bound, and the bound itself — correspond to three stages of a practical design.

A narrowband instrument, operating over a few per cent, needs one element chosen by the minimax rule and is done. A broadband one needs a few elements chosen together, and the search here is the kind of computation a design tool makes; its result is a network whose response ripples across the band, as the hero figure’s does. And any design, before it begins, can read the bound to decide whether the band is achievable at all — a single line of arithmetic, $1 - e^{-2\pi/(xB)}$, that says whether a quarter of a decibel is available over the band before any element is chosen.

A quarter wave and the path back meets the same three stages for resistive terminations joined by transmission lines, where the band is limited by the lines’ own frequency dependence rather than by a reactive source. The two problems share the structure of the answer and differ in what limits it: here the source’s Q, there the lines’ electrical length.

Two routes to each network’s transfer #

Each network’s transfer is computed from its chain matrix — the product of each element’s two-by-two matrix — which is what makes a search over thousands of candidate networks affordable. The chain-matrix route is checked against the netlist: the same source, elements and load assembled as a circuit and solved by nodal analysis at a frequency inside the band, for every network in every figure. The two agree to a part in a million, which is the check that the sign convention that turns one number into an inductor or a capacitor means the same thing in both. A convention that differed would not fail loudly: it would quietly swap an inductor for a capacitor, and the network drawn would be a different network from the one whose transfer was printed.

The search is checked by the bound. Every network found is required to fall below the Bode–Fano limit for its band, and each added element is required never to do worse than the network without it. The first is a property of physics the search cannot violate if it is computing transfers correctly; the second is a property of the search, since a network with one more element can always set that element to nothing.

What the numbers say about practical matching #

One element is worth having whenever the band is modest. Over ±10 per cent it recovers all but a sixth of a decibel of a 2.3 dB loss. It should be chosen for the band, not the centre: slightly less capacitance than cancellation, and a load above the source’s resistance, by the arithmetic of the uncertain reactance.

Three elements are worth having when the band is wide, and two are a halfway house that the third completes. That is the pattern several sections and the band they buy finds for a transmission-line transformer: each section widens the band, with the steps unequal and the limit approached rather than reached.

Past three, in a chain of single elements, look elsewhere. The fourth element buys almost nothing here, which says the chain topology has run out, not that the bound has. Networks that approach the Bode–Fano limit closely use resonators as their elements — series and shunt LC pairs — and they are the next thing to try, not a fourth single element.

Lossless elements, a local search and a resistive load #

The elements are lossless. A real inductor has resistance, and the reactance that is not free is what a lossy cancelling element does to the single-element answer; a three-element network carries three such losses, and at high Q they are what limits it before the bound does.

The search is local. Forty starts of Nelder–Mead are not a global optimiser, and the numbers above are the best networks found. That they fall below the bound and rise with the element count is what can be checked; that they are the best of their size is not proved. The caution has a precedent: the number that was wrong is a multi-section transformer whose printed band came from a search that had converged to the wrong answer, found only when a second method disagreed. Here the second method is the bound, which catches a search that claims too much but not one that finds too little; a network that beats these figures would not contradict anything on this page.

And the load is a pure resistance. A load with its own reactance has its own Bode–Fano budget, and the network must then match two reactive terminations at once, which is the harder problem and the usual one in practice.

Still open: resonators as elements, the lossy three-element network, and the uncertain resistance #

Ladders of resonators. Replacing each single element with a series or shunt LC pair doubles the parameters and gives the network the band-pass shape a Chebyshev matching network has. How close two and three such resonators come to the Bode–Fano bound — and whether the fourth single element’s failure here is a property of the topology — is the natural next search.

Loss, element by element. Each element added to approach the bound adds its own loss, so there is a number of elements past which the added loss outweighs the approach. For a stated inductor Q that number is a measurement, and it is the practical end of the chain.

An uncertain resistance as well. The minimax load here and in the earlier essay holds the source’s resistance fixed. A source whose resistance also drifts — a sensor whose losses change with temperature — adds a second dimension to the range, and whether the best fixed network still centres the swing and applies the uncertain-reactance rule is the extension the load that may be complex would need.