The load for a source that will not hold still

The load that may be complex found the two answers to matching a source that has a reactance. The conjugate load cancels the reactance and takes all the available power; a resistor alone, the best one being the magnitude of the source’s impedance, takes $2/(1 + \sqrt{1 + x^2})$ of it, with $x$ the reactance over the resistance. The reactance that is not free put a loss in the cancelling component and found that the best design then cancels only part of the reactance, and none of it below a golden-ratio threshold of quality factor.

Every number in both essays is at one frequency. The earlier of the two named what that leaves out: the claim that the conjugate match holds over roughly $1/x$ of its centre frequency is a quality-factor argument rather than a measurement, the resistor-only answer is sometimes described as broadband without anyone saying over what, and a source whose reactance is not known exactly cannot be conjugate-matched at all. This page measures all three.

Two loads, and a frequency that moves #

The source is the one the earlier essays used: 50 Ω of resistance and an inductance whose reactance is 100 Ω at 1 kHz, so $x = 2$ there. Its conjugate load for 1 kHz is 50 Ω and a capacitor of −j100 Ω; its resistor-only load for 1 kHz is 111.8 Ω. Both are then held fixed, which is what a built circuit does, and the frequency is swept across two decades.

The two curves have entirely different shapes. The conjugate match is a resonance: the source’s inductance and the load’s capacitor cancel exactly at 1 kHz and nowhere else, and away from it the uncancelled reactance grows in both directions. The resistor-only load has no resonance at all. Its power falls monotonically with frequency, because the only reactance in the loop is the source’s inductance, which grows with frequency, and at 100 Hz it takes 0.851 of the available power where at its design frequency it takes 0.618.

So the conjugate match wins over a band around its centre, from 730 Hz to 1.84 kHz, and loses on both sides of it: below, because the resistor-only load improves as the inductance shrinks; above, because the conjugate’s capacitor stops cancelling anything and the mismatch it leaves is worse than the resistor-only load’s.

The crossings have a practical reading. A signal whose useful content lies between 100 Hz and 1 kHz — a band that sits mostly below the match — is better served by the resistor-only load over most of it, even though the conjugate match was designed for the top of that band and takes all the available power there. Which load delivers more energy to a broadband signal depends on where the signal’s power is, and a match designed for one frequency can lose that comparison at most of the frequencies the signal occupies.

The one-decibel band, exactly #

The conjugate match’s band has a closed form that is the quality-factor argument made exact. At frequency $f$ the uncancelled reactance is $X_S(f/f_0 - f_0/f)$, and the power falls a decibel below the available when that reactance reaches $2R_S\sqrt{10^{0.1} - 1}$, about 1.02 times the source’s resistance. The two frequencies at which it does are reciprocal about the centre, and their difference, as a fraction of the centre, is

$$\frac{\Delta f}{f_0} = \frac{2\sqrt{10^{0.1} - 1}}{x} = \frac{1.0176}{x}.$$

That is the “roughly 1/x” of the argument, with its constant: the loaded quality factor of the matched loop is $x/2$, its half-power band is $2/x$ of the centre, and a decibel is about half of three.

Across a sixteenfold range of source reactance the solved one-decibel band lies on $1.0176/x$ to six figures. A reactance of eight times the source’s resistance leaves a band of 12.7 per cent; a reactance of half the resistance leaves one wider than the centre frequency itself.

The band in which the conjugate beats the resistor-only load narrows more slowly. At $x = 2$ it spans a ratio of 2.52 in frequency; at $x = 8$, 1.62. The reason is that the resistor-only load also gets worse as the reactance grows — at $x = 8$ it takes only 0.221 of the available power at its own design frequency — so the conjugate has further to fall before it is beaten, even though it falls faster.

A small reactance and a large one #

With a small reactance there is little to choose between the two. The resistor-only load takes 94 per cent of the available power at its design frequency, and the conjugate match’s gain over it, though it spans a wide band, is at most six per cent. Here the resistor-only load is as close to broadband as it ever gets: within a few per cent of its best from direct current to past its design frequency.

With a large reactance the picture inverts. The conjugate match is sharp — 12.7 per cent of the centre within a decibel — and inside that band it takes four and a half times what the resistor-only load does, which manages only 22 per cent at its own design frequency and 39 at 100 Hz. A source with a large reactance forces a choice between a narrow band of nearly all the power and a wide band of a fifth of it. There is no load that gives both, which is the Bode–Fano limit on matching a reactive source in its simplest instance.

The limit named there is a theorem about exactly this trade. For a source that is a resistance in series with an inductance, no lossless matching network of any complexity can hold the power transfer above a given fraction over a band wider than a bound set by the ratio of the inductance to the resistance; the better the match demanded, the narrower the band allowed, and the two-element conjugate load is the crudest network that the bound constrains. The figures here measure the two ends of the trade for that crudest network. The theorem says how far a better network could push the same trade, and that it can never escape it.

What “broadband” meant #

The resistor-only answer is sometimes described as broadband, and the figures say what that amounts to. It has no resonance, so it has no band edge on the high side the way the conjugate match does; but its delivered power falls with frequency from the start, because the source’s inductance is not being cancelled at any frequency and grows with every hertz. Its best frequency is direct current, where the inductance vanishes and the load’s only fault is that it is larger than the source’s resistance. Its design frequency is simply the one at which it is the best resistor; at every other frequency a different resistor would do better, and below the design frequency a smaller one would.

So the honest description is a low-pass whose corner is set by the source’s own inductance and the load’s resistance, $f = (R_S + R_L)/(2\pi L)$ — 1.62 kHz for the $x = 2$ source here, which is the golden ratio times the matching frequency, since the load resistance is the magnitude $|Z_S|$ and $(R_S + |Z_S|)/X_S$ is φ at $x = 2$. The matching frequency sits on the low-pass’s shoulder, already 1.4 decibels down from its value at direct current. It is broad in the sense of being gently sloped, not in the sense of being flat.

The resistor-only load’s design frequency is also a choice that could be made differently. Chosen for 1 kHz it is the magnitude of the source’s impedance there, 111.8 Ω; chosen for 300 Hz it would be smaller, and would take more at low frequencies and less at high ones, the two curves crossing somewhere above 300 Hz. A resistor-only load has one parameter, and every choice of it is a different low-pass with a different shoulder. None is flat, because the source’s inductance is in the loop at every frequency and no resistance can remove it.

A source whose reactance is uncertain #

The frequency sweep changes the source’s reactance and the load’s together, since the conjugate load’s capacitor is a reactance that moves with frequency too. The third question the earlier essay left is different: a source whose reactance varies — an antenna whose feed impedance drifts across its band, a transducer whose reactance moves with temperature — matched at a single frequency by a load that cannot move. The designer knows only that the reactance lies somewhere between two values, and must choose one load for all of them.

The question then is which fixed load guarantees the most power against the worst reactance in the range. It is a two-dimensional search over the load’s resistance and reactance, with the objective the minimum over the range, and it has a closed-form answer that the search finds exactly.

The best fixed load has two parts, each with an interpretation. Its reactance is exactly minus the centre’s: whatever the range, the load cancels the middle of it. Its resistance is $\sqrt{1 + \Delta x^2}$ times the source’s, which is the magnitude of the source’s resistance combined with the half-range of reactance — exactly the resistor-only rule of the earlier essay, applied not to the source’s reactance but to the part of it that the load cannot know. What it guarantees is the same formula: $2/(1 + \sqrt{1 + \Delta x^2})$ of the available power, 0.8284 for a range of ±1 and 0.6180 — the golden ratio less one again — for a range of ±2.

That is a clean division of labour. The part of the reactance that is known is cancelled; the part that is uncertain is treated as if it were a reactance the load may only answer with a resistance, because against an unknown sign that is all a fixed load can do. The two earlier results — conjugate for a known reactance, resistor-only for one that cannot be cancelled — are the two limits of one rule.

Why it cancels the centre can be seen without the search. Against a range of reactances, the worst cases are at the range’s two ends, since the delivered power falls monotonically as the leftover reactance grows in either direction. A load whose reactance is off the centre leaves more leftover at one end than the other, and that end becomes the worst case and is worse than it needed to be; moving the load’s reactance back to the centre equalises the two ends. With the centre cancelled, both ends face a leftover of exactly Δx, and the best resistance for a known leftover reactance is the resistor-only rule’s magnitude. The search confirms both halves: it finds the centre’s conjugate reactance to a part in a thousand and the magnitude resistance to the same.

In numbers for a practical case: a source whose reactance may lie anywhere from 50 to 150 Ω — twice its resistance, give or take one resistance — is best served by 70.7 Ω and −j100 Ω, which guarantees 83 per cent of the available power at every reactance in the range. The conjugate of the centre, 50 Ω and −j100 Ω, guarantees 80 per cent and does better only near the centre; the resistor-only load for the centre, 111.8 Ω, guarantees 46.

Both simpler choices do worse. The conjugate of the centre guarantees 0.8000 for ±1 and 0.5000 for ±2, because at the ends of the range it is left with the full half-range of reactance and a load resistor sized for none. The resistor-only load for the centre guarantees only 0.4593 and 0.3379, because it cancels nothing and so faces the whole of the reactance at the far end of the range.

What the three results say together #

A matched load is a bet on the source. The conjugate match bets everything on one frequency and one reactance, and wins everything there; its band is 1.0176/x of the centre, which for a strongly reactive source is narrow. The resistor-only load bets on nothing, gives up a fraction of the power everywhere, and is best at direct current rather than at any particular frequency. The minimax load hedges exactly: it cancels what is certain and applies the resistor-only rule to what is not.

Resonance, and the bandwidth it sets exactly measured a resonator’s bandwidth as its centre over its quality factor; the conjugate match is that resonance with the load as its loss, and its band is that result in the units of delivered power. The Q the components allow measured what the components’ own losses do to such a band. The load that takes the most began the whole question with a resistive source, where every one of these distinctions vanishes: with no reactance there is nothing to cancel, nothing to bet on, and the one answer is exact at every frequency.

The same division runs through all four essays on this question. When the source is a pure resistance, there is nothing to cancel and the only decision is the resistance. When its reactance is known and the cancelling part is lossless, cancel all of it. When the cancelling part is lossy, cancel most of it and let the load’s resistance absorb the rest. When the reactance is uncertain, cancel what is certain and let the resistance absorb what is not. In every lossless case the resistance that is left to choose is the magnitude of whatever impedance the load could not cancel, and the power it delivers is $2/(1 + \sqrt{1 + y^2})$ with $y$ the uncancelled part over the source’s resistance — which is why the golden ratio keeps appearing whenever that part is twice the resistance.

How the numbers were obtained #

Every power is a solve of a netlist — a 10 V source, 50 Ω, an inductance, and the load’s resistor and capacitor or resistor alone — at the stated frequency, with the power read from the load resistor’s current. The one-decibel band edges and the crossings between the two loads are bisected on the solved power, and the band is compared with its closed form. The minimax load is found by a coarse grid over the load’s resistance and reactance followed by a pattern search, with the objective the least power over twenty-one reactances spread evenly across the range, and its resistance, reactance and guaranteed power are compared with the closed forms to a part in a thousand or better.

What it leaves out #

Frequency and uncertainty together. The minimax load is for a source whose reactance is uncertain at one frequency; a source whose reactance changes with frequency across a band is matched by a load whose reactance changes too, and the best fixed network for a band is a harder problem whose answer — the Bode–Fano limit — says how much power any lossless network can guarantee across it.

Loss in the load’s components, which the previous essay measured at one frequency and which would both lower the conjugate’s peak and widen its band.

And a load with more than two elements. A two-element load can cancel one reactance at one frequency; a ladder network of several can approximate a cancellation across a band, and how much of the Bode–Fano limit each added element recovers is the question broadband matching is about.

Still open: the best network for a band, the lossy components across it, and the uncertain resistance #

The best fixed network for a band. With a two-element load the band is 1.0176/x; with more elements it 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.

Lossy components across the band. A lossy cancelling capacitor lowers the conjugate’s peak and broadens it. Whether the partial cancellation that is best at the centre frequency is also best across a band, or whether a band favours cancelling more or less, is the previous essay’s question with frequency added.

An uncertain resistance as well. The minimax result here holds the source’s resistance fixed and lets its reactance move. A source whose resistance is uncertain too — a sensor whose losses drift — adds a second dimension to the range, and whether the best fixed load still cancels the centre and applies a magnitude rule to the rest is the natural extension.