Two solves that add, and the one that does not

Superposition is the first theorem anybody meets in this subject and it is almost always stated about a circuit: turn off all the sources but one, solve, repeat, add. Stated that way it invites a question it does not answer, which is add what. The theorem is not about circuits. It is about a class of quantities, and the class has a sharp edge that runs straight through the middle of the things a designer actually wants to know.

Inside the class: every node voltage, every branch current, every charge, every flux. Outside it: every power, every root-mean-square value, every efficiency — everything quadratic in the solution. And the second list is not “approximately inside”. The error has a closed form, it can be larger than either contribution, and it can carry either sign.

The network, and the three solves #

Two independent sources, both ten volts root-mean-square at a kilohertz. The first reaches a node through a hundred ohms; so does the second; and a hundred-ohm load runs from that node to ground. Six components with the load, five without, and nothing about it is contrived — it is the shape of two supplies feeding one rail, or two antennas feeding one receiver, or two amplifiers driving one speaker.

Three solves are done at every point on the sweep, and the middle one is worth stating carefully because it is where the theorem is usually mis-stated. “Set the other source to zero” does not mean delete it. A voltage source with no voltage in it is a short circuit and it stays in the netlist, contributing its own equations, because the topology must not change between the three solves. The site’s solver takes a source value of zero and does exactly that; nothing is removed and no node disappears.

With both sources at ten volts and in phase, the node sits at 6.6667 V. Each source alone puts 3.3333 V there, which is ten volts divided by three: with the other source shorted, its hundred ohms sits in parallel with the load’s hundred, so the live source drives 100 Ω into 50 Ω. The two contributions add.

The worst departure between the combined solve and the sum of the two single-source solves, across eighty-one settings of the second source and all seven angles the slider offers, is $3.8\times10^{-16}$ of the voltage. That is one unit in the last place of a double. There is nothing left to measure: the two computations produced the same number and differ only in which intermediate rounded which way.

The same holds for the current through every element, checked the same way and coming out at $1.1\times10^{-16}$. So superposition is exact here in the strongest sense the arithmetic allows, and no part of what follows is a criticism of it.

It is worth pausing on how weak an assumption that took. Nothing was said about the frequency, the element values, the topology or the number of sources. Superposition needs one property and one only — that every element’s law be a linear relation between its own voltage and its own current — and every element in this collection except the diode and the transistor has one. That is why the theorem is the load-bearing wall under Thévenin equivalents, transfer functions, phasors, Fourier analysis and the whole of the frequency field: each is superposition applied to a different decomposition of the same input.

Where the powers go #

Now ask the same three solves for the power in the load.

With both sources on, the node is 6.6667 V and the load takes 444.44 mW. With the first source alone it takes 111.11 mW; with the second alone, the same. Their sum is 222.22 mW. The truth is twice the sum.

Nothing has gone wrong. Power is $|I|^2R$ and the current is a sum, so

$$P = |I_1 + I_2|^2 R = |I_1|^2R + |I_2|^2R + 2\,\mathrm{Re}(I_1 I_2^{*})\,R$$

and the third term is the whole of the discrepancy. The figure asserts that identity rather than bounding the disagreement: the measured gap and $2\,\mathrm{Re}(I_1 I_2^{*})R$ agree to $6\times10^{-16}$ at every point on every sweep. This is the site’s usual habit — a residual with a closed form is a measurement, and a residual without one is a disappointment.

The last row is the one to sit with. Two sources, each of which would deliver a ninth of a watt on its own, deliver nothing whatever together. The load is not being under-driven; there is no current in it at all, to thirty digits. Adding the powers gives 222 milliwatts of heat that does not exist.

The angle at which the shortcut is exact #

The cross term is $2\,\mathrm{Re}(I_1 I_2^{*})R$, and the real part of a product of two phasors is their magnitudes times the cosine of the angle between them. So it vanishes when they are ninety degrees apart — not approximately, and not on average, but identically, at every amplitude ratio on the axis.

That is the one case in which “the powers add” is a theorem rather than a mistake, and the figure tests it directly: at ninety degrees, the two curves agree across the whole sweep to $1\times10^{-14}$. Two sources in quadrature contribute independently to the heat.

The generalisation is the one that makes this useful rather than a curiosity. Two sources at different frequencies are orthogonal over any interval containing a whole number of both periods, so their cross term averages to zero and their mean powers add exactly. Driving the same hundred-ohm resistor with 3.3333 V at 1 kHz and 3.3333 V at 1.6 kHz, averaged over five milliseconds — five cycles of the first and eight of the second — gives 222.222 222 2 mW against a sum of 222.222 222 2 mW, agreeing to $3.5\times10^{-14}$.

This is why noise powers add, why the harmonics of a distorted current can be summed in quadrature to get its heating, and why an equivalent noise bandwidth is a meaningful thing to compute at all. Every one of those is the quadrature case of this identity, and each is used daily by people who would say, if asked, that powers do not superpose.

How far apart the sources have to be #

The practical question is not whether the shortcut is exact but when it is close enough, and there is a clean answer. Writing $r$ for the ratio of the two sources, in phase, the fractional error of the sum against the truth is

$$\frac{|P - (P_1 + P_2)|}{P} = \frac{2r}{(1 + r)^2}$$

which is a half at $r = 1$, falls away on both sides, and reaches one per cent at $r = 0.005\,051$. Bisected on the solved network rather than on that expression, the figure returns $0.005\,051$ — one source has to be a hundred and ninety-eight times the other before adding their powers is right to a per cent.

That number is the reason this is worth an essay. A designer’s intuition is that the shortcut fails when two contributions are comparable and works when one dominates, which is true; the intuition about how much domination is needed is wrong by two orders of magnitude. Twenty decibels of separation between two in-phase sources still leaves the sum of their powers twenty per cent short.

What happens when the network is not linear #

Everything above rests on the network being linear, and it is worth seeing what the failure looks like rather than being told that it happens.

Take the same two-source arrangement — five volts through a kilohm, three volts through another kilohm, meeting at a node — and put a diode from that node to ground through a hundred ohms. Solve the whole thing with Newton’s method on the netlist, then solve it twice more with one source zeroed each time.

The added node voltage is 43.2% high. The added diode drop, 1.35 V, is a voltage no silicon diode has ever had. And the added current is 19.6% low — the true current is 1.244 times the sum of the two — because the exponential is convex and the sum of the responses to two inputs is less than the response to their sum.

Two things are worth noticing. The failure is not small, and it is not in a consistent direction: the voltages come out high and the current comes out low, because the two quantities sit on opposite sides of a curve. And nothing in the arithmetic complains. Each of the three solves converges, each satisfies the current law rebuilt from the exponentials to a part in $10^{15}$, and adding three correct numbers gives a wrong one.

Three places this shows up on a bench #

Adding up interference. Several sources of hum and pickup arrive at one input, each measured alone. If they are at the same frequency — and mains-related pickup is — the cross terms are real and the total can be anywhere between the difference and the sum of the amplitudes. Only when the sources are genuinely independent in time does the root-sum-square that everybody uses become correct.

Two supplies sharing a load. Two regulators paralleled onto one rail deliver currents that add; their dissipations do not, and the one that happens to be a few millivolts higher takes very much more than half the heat. The current is a linear quantity and the heat is not, which is why current sharing needs a mechanism and voltage sharing does not.

Reading a distorted waveform. The heating in a wire is the sum of the heating from each harmonic, exactly, because the harmonics are orthogonal. The heating from a fundamental plus an identical-frequency reflection is not the sum of the two, for exactly the reason the table above gives. Same instrument, same arithmetic, two different answers, and which one applies depends on a property of the signal rather than of the circuit.

What this essay does not claim #

That the cross term is a nuisance. It is the signal, in half the circuits in this collection. A mixer, a phase detector, a lock-in amplifier and a power meter all exist to measure $2\,\mathrm{Re}(I_1I_2^{*})$ and would have nothing to do if powers superposed.

That superposition is slow. It is not, and the three-solve route is often the fastest way to understand a network even when one solve would do — separating a supply’s contribution from a signal’s is exactly this construction. What is being warned about is one step further on, where the separated results are recombined into something quadratic.

That the diode result generalises to a bound. It does not. The 43% above is a property of that netlist at those two source values; a different bias gives a different number, and the only general statement is the one the semiconductors field makes, that the error is second order in the drive about any operating point.

That two frequencies are always orthogonal. They are orthogonal over an interval containing a whole number of both periods. Over an arbitrary window they are not, which is the whole content of spectral leakage, and the digital field measures what it costs.

Where superposition is refused #

Voltages add and powers do not, and this collection meets the same distinction three more times. Exact outside and wrong within is a reduction exact in the voltages and silent about the dissipation. What a network answers, and how the answer is checked is where energy counted two ways is one of the two standing checks. A bias point is a solution, not a choice is where superposition fails outright rather than only for powers, and The current that does no work is the power field’s own version of the same accounting.

The gate #

Superposition is asserted on the solution, not on the answer. Every node voltage and every branch current from the combined solve is compared with the sum of the two single-source solves at eighty-one amplitude ratios and seven angles, and required to agree to $10^{-14}$ of the larger contribution.

That normalisation is deliberate and it is the second time this site has needed it. At 180° and equal amplitudes the result is zero by cancellation, and a relative error taken against the result reports $10^{-14}$ of nothing as a catastrophe — which is what the first version of this check did, and what the feedback field’s sensitivity figure did a phase earlier where its own curve passes through zero. The scale of a difference between two numbers is the size of the numbers.

The power gap is asserted to equal its closed form, $2\,\mathrm{Re}(I_1I_2^{*})R$, to $10^{-12}$ — not merely to exist and not merely to be bounded.

The quadrature case is asserted separately, at every amplitude ratio, because it is the one setting of the slider at which the essay’s claim reverses, and a figure whose slider makes its own headline false at one stop needs that stop tested rather than avoided.

And the hundred-and-ninety-eight-to-one figure is bisected on the netlist, not evaluated from $2r/(1+r)^2$. The expression is in the prose because a reader can check it; the number in the caption came from solving.

Every quadratic quantity in the collection has the same hole #

Superposition failing on power is one instance of a general fact — a quantity quadratic in the sources is not a sum of per-source answers — and it is worth listing the other places that fact decides a result, because in each of them the quadratic quantity is the one somebody actually wanted.

Exact outside and wrong within is the same failure wearing a different hat: six elements reduce to one source and one resistor that no load can distinguish from them, to the last bit of a double, and the reduction is wrong about the heat by a factor of forty-three — reporting nothing dissipated with nothing connected while the network burns 48 milliwatts. A reduction and a superposition are both linear manipulations, and both are exact about voltages and silent about power.

The current that does no work is where the collection makes the distinction structural rather than incidental: a solved network reports its own real power on every page as the second of two checks, and what that check discards is the imaginary half — power that flows out to a reactance and back, does nothing, and is still carried by the cable and still on the bill.

And the load that takes the most is the quadratic quantity being optimised rather than merely computed, with the answer — a load equal to the source resistance, at exactly fifty per cent efficiency — being a statement about a product of two linear quantities that no superposition of either would produce.

The rule they share is short. Solve for voltages and currents, add those, and compute every power at the end. Every failure on this page comes from doing the last two steps in the other order.

Two sources that are not two sources #

The arrangement measured here — two independent sources in one network — is rarer in practice than the arrangement it stands in for, which is one source and one disturbance. Three of those appear elsewhere in this collection and each is a superposition being relied on without being named.

What gets through from the rail is a superposition exactly: the reference is set to zero and the rail is driven, which is only a measurement of anything because the two contributions add. The essay’s whole result — sixty decibels between two topologies with identical loops — is a statement about one term of a sum.

The millivolts in the wire is the same construction with the disturbance being somebody else’s load current: half a millivolt added to a ten-millivolt reading, which is an addition and not an interaction, and which is why subtracting the two ends of the conductor removes it at all.

And the rail the load moves is where the superposition is what is being measured: one amplifier’s load step reaching a second amplifier that shares nothing with it but a wire, at 3.84 microvolts per ampere or 17.3 millivolts depending on where one capacitor returns.

In all three the linearity is doing real work and nobody writes it down, which is the reason this essay exists. What none of the three could do is add the powers — the disturbing channel’s dissipation and the disturbed channel’s are not separable that way, and any thermal argument built on summing them is making the mistake measured here at 222 milliwatts. Which is a mistake with no residual to warn about it: both per-source solves are exact, their sum is exact, and only the quantity assembled from them afterwards is wrong.