Circuit Response
every model has an edge, and the figure carries it
Series

Power transfer — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The load that takes the most power, and the load that wastes the least. computed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, which is the magnitude of the source impedance to six figures, and the efficiency there is 0.500000 — the source dissipates as much as the load receives. Ninety per cent efficiency needs a ratio of 9.7 and delivers 34% of the available power.

    The load that takes the most

    A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.

    part 1 · power
  2. The load that may be complex, and what a resistor alone gives up. computed by solving, not by drawing, with the best load searched over BOTH of its parts on the solved network rather than substituted. Against a source of 50 Ω + j100 Ω the search returns 50.00 − j100.0 Ω — the conjugate — which delivers the available 500.00 mW at exactly 50.00% efficiency, and does so at every source reactance. A load that may only be a resistance takes 2/(1 + √(1 + x²)) of that, matching the solved answer to 4.4e-16 at ten reactances — and at x = 2 that fraction is exactly the golden ratio less one, 0.618034. The resistor-only load is the MORE efficient of the two at every non-zero reactance, rising towards one while the conjugate match sits at a half for ever, so the familiar "maximum power at fifty per cent" belongs to the conjugate and not to the load.

    The load that may be complex

    Freed of the constraint that it be a resistance, the best load is the source's conjugate — found here by a two-dimensional search on the solved network rather than assumed — and it takes the available power at exactly fifty per cent efficiency whatever the source's reactance. A load that may only be a resistance takes 2/(1 + √(1+x²)) of that, and at a source reactance of twice its resistance that is exactly the golden ratio less one, 0.618034. The resistor-only load is also the MORE efficient of the two, rising towards one while the conjugate sits at a half for ever.

    part 2 · power
  3. A lossy cancelling capacitor beats a resistor alone only above Q = 3.236 if it cancels everything, and only above 1.618 at all. computed by solving, not by drawing. A source of 50 Ω + j100 Ω, its reactance cancelled by a capacitor whose quality factor is Q, so that the capacitor carries a series resistance of its reactance over Q. Cancelling fully, with the load resistor matched to the source and capacitor resistances together, the load takes 1/(1 + x/Q) of the available power, at exactly half efficiency; a resistor alone takes 0.6180, 2/(1 + √(1 + x²)). They are equal at Q = 3.2361, twice the golden ratio. Cancelling only part of the reactance does better than either: it helps at all above Q = (√(1 + x²) + 1)/x = 1.6180, and at Q = 3 takes 0.6650 against 0.5971 cancelled fully.

    The reactance that is not free

    A conjugate match cancels the source's reactance with an opposite one and takes all the available power, where a resistor alone takes 2/(1 + √(1 + x²)), 0.618 of it when the reactance is twice the resistance. The cancelling capacitor is not free. With quality factor Q it carries a resistance of its reactance over Q, and cancelling fully then delivers 1/(1 + x/Q). That beats the resistor alone only above Q = 2(√(1 + x²) + 1)/x, 3.236 here, twice the golden ratio. But full cancellation is not the best use of a lossy capacitor. Cancelling part of the reactance beats both, and helps at all above Q = (√(1 + x²) + 1)/x, the golden ratio itself. Below that, the right amount to cancel is none.

    part 3 · power
  4. Matched at 1 kHz, the conjugate load wins only from 730 Hz to 1.84 kHz; the resistor-only load is a low-pass, not a broadband answer. computed by solving, not by drawing. A source of 50 Ω and an inductance of 100 Ω at 1 kHz, with the conjugate load for 1 kHz — 50 Ω and a capacitor of −j100 Ω — and the resistor-only load for 1 kHz, 111.8 Ω, both held fixed while the frequency moves. The conjugate match takes all the available power at 1 kHz and stays within a decibel of it from 777 Hz to 1.29 kHz, a band of 50.9% of the centre, 1.0176/x. The resistor-only load takes 0.6180 at 1 kHz, 0.8509 at 100 Hz, and less at every frequency above: a low-pass whose peak is at direct current. The conjugate load is the better of the two only between 730 Hz and 1.84 kHz.

    The load for a source that will not hold still

    A conjugate match is exact at one frequency. Held fixed while the frequency moves, it stays within a decibel of the available power over 1.0176/x of the centre: 51 per cent of it when the source's reactance is twice its resistance, 12.7 per cent at eight times. It beats the resistor-only load only between 0.73 and 1.84 of the centre frequency. The resistor-only load is not a broadband answer, as it is usually called. It is a low-pass whose best frequency is direct current. And for a source whose reactance could be anywhere within ±Δx of a centre, the fixed load that guarantees the most power cancels the centre exactly and takes a resistance of √(1 + Δx²) times the source's. That guarantees 2/(1 + √(1 + Δx²)): the resistor-only rule, applied to the uncertainty instead of to the reactance.

    part 4 · power
  5. Over ±25% of the centre, three elements deliver 97.3% at worst; Bode and Fano allow 99.8%. computed by solving, not by drawing. The fraction of the available power a source of 2 times its resistance in inductive reactance delivers to a resistive load through the best matching network of 0, 1, 2 and 3 reactive elements for a band of ±25 per cent about the centre, each network the one whose worst transfer in the band is largest; the vertical lines mark the band. The worst in-band transfer is 54.2%, 82.8%, 89.5%, 97.3%; the Bode–Fano limit for any lossless network, reached only with infinitely many elements, is 99.81%.

    What each element buys against the limit

    A source whose resistance sits behind twice as much inductive reactance can deliver, over ±25 per cent of its centre frequency, 54.2 per cent of its available power to the best plain resistor and at worst 82.8 per cent through the best single capacitor. That single-element answer is exactly the fixed load for an uncertain reactance: the capacitor makes the reactance swing ±1 across the band and the load is √2 times the source. Two elements lift the worst case to 89.5, three to 97.3, and a fourth, in a chain of single elements, to 97.4. Bode and Fano bound every lossless network at 99.81 per cent over that band — a limit three elements come within a tenth of a decibel of and no finite network reaches.

    part 5 · power

All series