Circuit Response
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Superposition — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Superposition holds for the solution and not for its square. computed by solving, not by drawing, at 81 settings of the second source. Two ten-volt sources reach one hundred-ohm load through a hundred ohms each. Every node voltage and every branch current is the sum of the two single-source solves to 3.8e-16 of itself — superposition, exactly. The power is not: the difference is 2·Re(I₁·conj(I₂))·R to 1.0e-15, and at 0° between the sources and equal size it is 444.4 mW against the 222.2 mW that adding gives. Adding the two powers is within one per cent of the truth only when one source is below 0.00505 of the other. The slider is the angle between them: at 90° the two curves coincide to the last bit, and at 180° the true power falls to zero while the sum does not.

    Two solves that add, and the one that does not

    Every node voltage and every branch current in a linear network is the sum of the per-source solves, here to the last bit of a double at eighty-one settings. The power is not, and the gap is not a correction: two equal sources in antiphase put nothing at all into a load while adding their powers gives 222 milliwatts, and the sum is within one per cent of the truth only when one source is two hundred times the other.

    part 1 · networks
  2. One rms value, and a peak that moves 41 per cent with a phase. A 10 V fundamental and a 3.33 V 3th harmonic, summed, with the harmonic's relative phase swept. The two are orthogonal over a period, so the root-mean-square value is the quadrature sum 7.45356 V at every phase — to 6.2e-15, which is the one case superposition allows and is a theorem rather than an approximation. The peak is not a sum of anything: it runs from 9.428 V to 13.33 V, a factor of 1.414, against a bound of 13.33 V that it reaches exactly at 180° — where this harmonic's own crest lands on the fundamental's, which for n ≡ 3 (mod 4) is that end of the sweep and not the other. At the far end it flattens the crest to 9.428 V, and there the peak has no closed form at all. So one signal has one heating and a headroom requirement that varies by 41 per cent, decided by a phase that no magnitude spectrum carries.

    The peak that only has a bound

    Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.

    part 2 · networks
  3. Zeroing the dependent source too: an answer that does not depend on the gain. computed by solving, not by drawing. Two independent sources reach a node through resistors; a voltage-controlled source of gain −10 is driven from that node and fed back to it through 10 kΩ. Solved whole, the output is -25.0000 V. Solved by superposition with the dependent source LEFT IN each of the two solves it is -25.0000 V — the same number to 1.1e-16, so superposition is exact here as it is everywhere else in a linear network. Zeroing the dependent source along with each independent one gives nought volts at the output and 4.06250 V at the node, against the node's true 2.50000 V — 62.5 per cent out — because a controlled voltage source set to zero is a short circuit and the three solves are then of three different circuits. What makes it unmistakable is that the wrong answer is 4.06250 V at EVERY gain from one to three hundred: the route has deleted the only element that knows what the gain is, while the true answer moves by more than a factor across the same range.

    The source that must not be zeroed

    Superposition says to zero all the sources but one, and the care usually taken over that instruction is that zeroing is not deleting — a voltage source with no voltage is a short circuit and it stays in the netlist. A dependent source needs the opposite care: it must not be zeroed at all, because it is an element whose law involves another node rather than a source the theorem is about. Zeroed, it is a short circuit, and the answer that comes back is 4.0625 volts at every gain from one to three hundred — while the true answer moves from 3.82 volts to 0.206.

    part 3 · networks
  4. Adding powers is wrong by the number of sources, when they are coherent. computed by solving, not by drawing, at five source counts. 8 ten-volt sources reach one hundred-ohm load through a hundred ohms each. With every source in phase the load takes 790.1 mW against the 98.77 mW that adding their separate powers gives — exactly 8 times, at every count, to a part in 10¹², because all 28 cross terms are positive and equal. With seeded random phases it takes 0.9865 times the sum over 64 runs, against a standard error of 0.125 — the cross terms average to nothing, which is the reason noise powers add and a root-sum-square is the right arithmetic for uncorrelated interference. The factor between the two limits is the source count itself, so it grows without bound: adding powers is not wrong by a bounded amount, it is wrong by how many things are being added.

    The cross terms that outnumber the sources

    With two sources the power in a load has one cross term, and adding the separate powers is wrong by a factor of two. With thirty-two there are four hundred and ninety-six of them, and the factor is thirty-two: 940 milliwatts delivered against the 29.4 that adding gives. With the same thirty-two sources at seeded random phases it is 26.9 milliwatts — the sum itself, inside the standard error of sixty-four runs, because every cross term averages to nothing. One identity, two limits, and the two are a phased array's gain and a noise sum's.

    part 4 · networks

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