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

Superposition — where it appears

The property of a linear network that every voltage and every current is the sum of those produced by each source acting alone. It holds for the solution and not for its square, so two sources' powers do not add even though their voltages do.

Named by 11 essays across 7 fields — each of them below, with the objects they name alongside it.

Five steps, each divided by its own size, from an amplifier limited to 0.50 V/µs. A linear circuit would put these five curves exactly on top of each other. The 20.0 mV step is linear; everything above 79.6 mV is not, and the largest step takes 16.0 µs to travel a distance the linear model says takes 0.159 µs.

The step that is too big

A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

transients · Slew rate 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.

networks · Superposition Which resistor the noise of a Chebyshev 5 actually comes from. computed by solving, not by drawing, one solve per resistor. Each bar is that resistor's share of the noise power at the output, found by splitting its node, putting a source of √(4kTR) in series with it and re-solving the whole network — so what is drawn is not how much noise each resistor makes but how much of it arrives. The largest contributor is F2R at 64.1 per cent, the smallest F0R at 9.0, and the shares add to 1.000000000000 because noise powers add. A resistor's share of the noise is not its share of the resistance: the largest departure between the two is 3.9 percentage points.

The resistor the noise comes from

Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.

noise · Johnson noise Where 8 channels leak to: 91.0 MHz from buffered sources, 901 kHz from 50 Ω. computed by solving, not by drawing. The frequency at which the open channels of a multiplexer built from the 0.5 Ω, 100 MΩ, 5 pF switch leak 1% of the signal onto the shared output, against the impedance of the source driving the selected channel, into 1 MΩ. 2 channels: 637 MHz buffered, 6.30 MHz from 50 Ω; 8 channels: 91.0 MHz buffered, 901 kHz from 50 Ω; 16 channels: 42.4 MHz buffered, 420 kHz from 50 Ω. Each falls as the reciprocal of the source impedance plus the on-resistance, and the single switch's own band closes at 6.43 MHz.

Where an open switch leaks to

A lone switch has a band whose width no load can change, because its open state leaks into the load. In a multiplexer the seven open channels leak into a node the selected channel holds, so the load leaves the answer and the source takes its place: one per cent of leak at 91.0 megahertz from buffered sources and 901 kilohertz from fifty ohms, moving as the first power of the tolerance rather than the second. Adding the channels' capacitance into one forty-picofarad switch puts it at 804 kilohertz, near the fifty-ohm figure by coincidence and a hundred and thirteen times low for a buffered one.

limits · Ideal switch The ripple 1000 µF leaves, through the regulator: 60.62 mV rather than 21.11 mV. One settled cycle of the reservoir's output — 1.331 V peak to peak across 1000 µF — split into 1000 Fourier lines, each passed through the solved regulator's rail-to-output response, and summed. The output is 60.62 mV peak to peak. The ripple times the rejection at 100 Hz, -36.00 dB, gives 21.11 mV, which is 2.872 times too little: the rejection worsens at twenty decibels a decade, so each harmonic of the sawtooth arrives at nearly the size of the first. The 100 Hz line carries 23.4% of the output's mean square and the lines above 1 kHz 10.4%; the first three arrive at 7.731 mV at 100 Hz, 7.200 mV at 200 Hz, 6.388 mV at 300 Hz. The dashed curve is the 100 Hz line alone.

The ripple that arrives as a comb

The rejection essay multiplied two numbers: 1.331 volts of reservoir ripple and the regulator's 36.0 decibels of rejection at a hundred hertz, for 21 millivolts at the output. The ripple is a sawtooth, a comb of lines at every multiple of a hundred hertz, and the rejection worsens at twenty decibels a decade — so the second line arrives at nearly the size of the first, and the third too. Summed with their phases, the output is 60.62 millivolts, 2.87 times the estimate, and the hundred-hertz line carries only 23.4 per cent of it. A reservoir twenty-one times larger cuts the rail's ripple 14.8 times and the regulated ripple 5.95.

applied · Regulator Two of the four errors are divided by the gain; the best the instrument gets is 110.1 dB, at ×776. computed by solving, not by drawing. The four mechanisms that limit a three-amplifier instrumentation amplifier's common-mode rejection, each measured alone against the gain of its input stage and then all four together, at 50.0 Hz with 1 kΩ of imbalance between the source resistances and 10 pF at each input. The difference stage's four resistors and the difference amplifier's own rejection are injected after the gain, so their common-mode gain is a constant — 1.998 mV/V and 0.0100 mV/V — and the rejection they allow rises decibel for decibel with the gain. The input pair's mismatch and the source's time-constant gap are injected before it, so they are amplified by exactly the gain the signal is and the rejection they allow is flat. The four add as complex numbers: the two largest are real and of opposite sign, they cancel at a gain of 776, and what is left there is the source's 3.142 µV/V, which is purely imaginary because it is ωΔτ. The instrument's best is 110.07 dB against the source's own 110.06, and above that gain more of it buys nothing.

The errors that arrive before the gain

Four earlier measurements each found a different owner of one instrument's common-mode rejection and each measured it alone. Solved together, the four are complex numbers that add — to 1.25 parts in ten thousand — and two of them carry a factor of the first stage's gain while two do not. The two that do cancel the two that do not at a gain of 776, and what is left there is 110.066 decibels, which is exactly the number the cable sets. Better-matched amplifiers move that gain from 93 to 3392 and do not move the ceiling by a hundredth of a decibel.

instruments · Common-mode rejection The two sequences a neutral current says nothing about. computed by solving, not by drawing at 61 imbalances. Three 20 Ω loads on a 230 V, 50 Hz star supply, one of them raised by a fraction of itself, with the neutral in place. The zero-sequence current is the one the neutral carries three times and is the only one a neutral ammeter reads; the negative sequence is a balanced set of three phasors rotating the other way. At 30.0 per cent imbalance it is 0.8846 A against 10.6154 A of positive sequence, 8.333 per cent, against 8.333 per cent from x/(3 + 2x). Two per cent arrives at 6.250 per cent imbalance, bisected on the network.

The half the neutral does not carry

A star load unbalanced in one phase produces two things, not one. The neutral carries three times the zero-sequence current, which is the half a neutral ammeter reads; the other half is a negative-sequence set of exactly the same size, rotating backwards, that the neutral never sees. With 0.5 Ω of line in front of 20 Ω loads, losing a phase entirely puts 50.00 per cent negative sequence in the current and 0.8265 per cent in the voltage a switchboard meter reads.

power · Three-phase 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 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.

networks · Superposition 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.

networks · Superposition With each source 0.8 of the one before, 32 coherent sources cost 8.99 times the sum of their powers, and never more than 9.00. computed by solving, not by drawing, at five counts and five amplitude ratios. Sources of 10 V, 10q V, 10q² V … each reach one 100 Ω load through 100 Ω; with all in phase the load's power over the sum of each source's power alone is (Σ|a|)²/Σa² to ten figures. Equal sources (q = 1) give the count itself, 2, 4, 8, 16, 32. Falling amplitudes give less, and a limit: for q = 0.8, 1.98, 3.77, 6.41, 8.51, 8.99, tending to (1 + q)/(1 − q) = 9; for q = 0.5, 3.000 against 3. The coherent penalty counts sources by their participation, not their number.

How many sources the cross terms count

N equal coherent sources on one load deliver N times the sum of their separate powers. Sources of unequal size deliver (Σ|a|)²/Σa² times it — solved on the network to ten figures — which is the number of equal sources that would carry the same penalty: a participation ratio. Amplitudes falling by 0.8 a source cap it at (1 + q)/(1 − q) = 9 however many there are; one large source and thirty-one at a tenth of its amplitude give 12.8, not 32, and for small sources the excess is 2(N − 1)ε — their amplitudes, not their powers, so thirty-one sources carrying a hundredth of the large one's power each turn 1.31 times its power into 16.8 times it when they line up. Random amplitudes take a fixed fraction off the count, (mean)²/(mean square): 0.75 for uniform ones. And with random phases none of it matters: the load takes the sum.

networks · Superposition

Named alongside it

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

Model rangeVerificationDesign tradeoffQuadratureReal powerImbalanceLoop gainMeasurement conditionModified nodal analysisNonlinearitySource impedanceCoherence

All concepts