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

Harmonic content — where it appears

The amplitudes of the harmonics a periodic waveform carries, which for anything with a step in it fall as one over the harmonic number. They decide what the waveform heats, because harmonics are orthogonal and their powers add exactly; they do not decide what it peaks at, because a maximum depends on their phases as well.

Named by 3 essays across 3 fields — each of them below, with the objects they name alongside it.

A switching load's comb in a 10 mV reading: every tooth the same height. computed by solving, not by drawing. The harmonics of a 50-per-cent trapezoid of 100 mA at 100 kHz with a 10 ns edge, each multiplied by the shared conductor's impedance at its own frequency. Below 79.6 kHz the conductor is 5.00 mΩ of resistance and the teeth fall as 1/n. Above it the conductor rises as n while the harmonics still fall as 1/n, so the two cancel and every tooth from 239 kHz to 10.6 MHz is 0.828 mV — flat to 5.4 per cent across 52 of them. The comb stops at 31.8 MHz, which is the edge's own corner, so the total of 12.6 mV is decided by how many teeth there are: fitted over five edge rates it grows as the edge time to the power -0.497, against the −½ that a quadrature sum of equal teeth gives. A faster edge raises no harmonic and adds teeth. What a slow measurement keeps is a different term entirely: the zeroth harmonic, which is the 100 mA average through 5.00 mΩ of resistance — 500.0 µV, with no inductance in it and no edge, against the comb's 12.6 mV.

Every tooth the same height

A shared return conductor's impedance rises a decade per decade above 79.6 kilohertz and a switching load's harmonics fall a decade per decade, so the two cancel exactly: every harmonic of a hundred-milliamp square wave puts 0.828 millivolts into a ten-millivolt reading, from the conductor's own corner up to the edge's. The comb is flat rather than falling, so the total is decided by how many teeth there are — and since they add in quadrature it grows as the square root of the edge rate, 39.7 millivolts at a nanosecond against 4.03 at a hundred.

instruments · Common-impedance 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.

networks · Superposition The best foil thickness for 4 layers, for three currents with the same fundamental. computed by solving, not by drawing. The loss of a portion of 4 layers against foil thickness, with the loss weighted by the current in each harmonic rather than computed for one frequency. A sinusoid wants 0.6631 skin depths and lands at 1.3368 times the direct-current resistance — four thirds, the single-frequency constant, reproduced. A triangular ripple wants 0.6432, which is the same answer to within 3.0 per cent, so a winding carrying one needs none of this. A square current of the same fundamental wants 0.3838 — thinner by a factor of 1.728 — and lands at 1.8313, which is not four thirds and is not any constant the geometry knows. Building to the sinusoid's answer costs 16.8 per cent more loss.

The optimum a spectrum moves

The best foil thickness for a winding is derived for one sinusoid and quoted as a property of the geometry: a minimum at four thirds of the direct-current resistance, whatever the layer count. Weight the loss by the current in each harmonic instead and a square current of the same fundamental wants foil 1.728 times thinner and lands at 1.83, and a narrow pulse wants it 3.68 times thinner. Four thirds is a property of the current. The constant that replaces it for an ideal square edge is exactly two, and a real winding sits between them at a place its edge rate decides.

magnetics · Winding

Named alongside it

The objects these essays reach for when they reach for this one.

Model rangeCrest factorCommon-impedanceConvergence orderDesign tradeoffMeasurement conditionMeasurement errorNumerical errorParasiticsProximity effectQuadratureReal power

All concepts