The cross terms that outnumber the sources

Two solves that add measured the power’s failure to superpose on the smallest network that can show it: two sources, one load, one cross term. The cross term is $2\,\mathrm{Re}(I_1 I_2^{*})R$, the figure requires that identity rather than bounding the disagreement, and the two extreme cases are the memorable ones — in phase the load takes twice what adding gives, and in antiphase it takes nothing at all while adding gives 222 milliwatts of heat that does not exist.

Two is the smallest number of sources that has a cross term and it is not the usual number. A power rail has several supplies on it, an antenna array has many elements, a noise floor is the sum of contributions from every resistor in a circuit. With $N$ sources there are $N(N-1)/2$ cross terms, and what they do in the two limits turns out to be two results that are normally taught in different subjects.

The coherent limit, which is the count #

Put $N$ equal sources in phase. Each reaches the load through its own resistor, so with the others zeroed — left in the netlist as short circuits, which is the care the essay before it insists on — one source alone puts $v_1$ on the load node. With all $N$ live the node is exactly $N v_1$, because voltages superpose and the divider is the same in both solves.

The power is therefore $N^2$ times one source’s, and the sum of the separate powers is $N$ times one source’s. So:

$$\frac{P_\mathrm{coherent}}{\sum P_k} = N$$

exactly, at every count, to a part in $10^{12}$:

Two columns in that table are worth reading together, because their opposite trends are the arithmetic of the whole essay. “One alone” falls with the count — a hundred and eleven milliwatts at two sources and 0.918 at thirty-two — because each source’s own hundred ohms is loaded by every other source’s hundred ohms to ground, so one source acting alone into thirty-two paths delivers very little. And “all in phase” rises towards a limit, because with every source live the parallel combination is a source rather than a load.

So the ratio is $N$ and neither of its two terms is. A designer reading “adding the powers is wrong by the number of sources” would expect the true power to grow as $N$; it grows towards 940 milliwatts, which is what a stiff source delivers into a hundred-ohm load from a hundred-ohm internal resistance, and the load that takes the most is where that maximum is located. The sum of the separate powers falls as $N$ grows, and the ratio between them is the count.

Why one source alone delivers so little #

The table’s first column falls by a factor of 121 across the sweep, and it is worth stopping on because it is the part of the arithmetic that a reader would get wrong from the ratio alone.

With $N$ sources on the node, one source acting alone drives its own hundred ohms into the parallel combination of the load’s hundred and the other $N-1$ sources’ hundred ohms each — because zeroing a voltage source leaves it in the netlist as a short circuit, so every other source’s resistor is a path to ground. At thirty-two sources that is a hundred ohms into 3.23 ohms, so almost nothing reaches the load.

This is that essay’s care about zeroing doing real work rather than being a technicality. Deleting the other sources instead of zeroing them would leave one source driving a hundred ohms into a hundred, so “one alone” would be a hundred and eleven milliwatts at every count and the table’s first column would be flat. The sum of the separate powers would then be $N$ times that, growing with the count, and the ratio would come out as one rather than as $N$ — so the whole result inverts on a step that looks like bookkeeping.

That is worth knowing as a warning about the shape of the arithmetic. The two quantities being compared are both computed on the same $N$-source network, and a superposition that changes the network between the solves does not merely lose accuracy — it reverses the conclusion. The source that must not be zeroed is the same hazard for a controlled source, where zeroing removes the element the answer depends on; here it is for an independent one, where deleting rather than zeroing removes a loading path.

The practical reading of the first column is also useful in its own right. A rail with thirty-two supplies on it is a rail where any one supply, tested alone, appears to be barely able to drive the load — and that is not a fault. It is the other thirty-one supplies’ output impedances acting as a load, which is the same shared conductance the node that does not care how many measures on a multi-input summing junction, and it disappears the moment they are all turned on.

The incoherent limit, which is one #

Now give the same $N$ sources random phases. Every cross term is $2|I_j||I_k|\cos(\varphi_j - \varphi_k) R$ with a cosine of a random angle, so each averages to nothing — and the load takes the sum of the separate powers exactly.

Measured on seeded phases, sixty-four runs at each count:

Every one inside the standard error, and it is worth being precise about what that error is, because it is not the $\sqrt{2/n}$ this site usually quotes for a variance. The squared magnitude of a sum of many unit phasors at random phases is exponentially distributed, whose relative standard deviation is one — so the mean of $n$ runs has a relative standard error of $1/\sqrt n$, which for sixty-four runs is 12.5 per cent. Eight runs would be fifty per cent, which is not a check, and the first version of this used eight.

This is why noise powers add, and it is worth saying that plainly because the two facts are normally learned separately and neither is usually presented as the other’s limit. Every resistor in a circuit contributes a noise voltage whose phase is unrelated to every other’s, so their cross terms average away and the total is the root sum of squares — which is exactly the arithmetic the floor a resistor sets uses to combine sources, and exactly the arithmetic the earlier measurement says is not generally available for powers. The exception it relies on is orthogonality, and random phase is a way of being orthogonal on average rather than exactly.

Two results that are one identity #

The factor between the two limits is the source count, and it grows without bound. That is the essay’s own finding and it has two names in two different subjects.

A phased array’s gain. $N$ elements driven coherently deliver $N^2$ times one element’s power into a matched load, or equivalently $N$ times what the same elements deliver independently. The $N^2$ is the figure an antenna engineer quotes and the $N$ is the figure this essay’s ratio column shows, and they are the same statement divided by different things.

A noise sum. $N$ uncorrelated contributions deliver $N$ times one contribution’s power, which is $\sqrt N$ times its voltage. That is the figure a noise budget uses and it is the same identity with the cosines averaged instead of set to one.

The two are not analogies. They are the two ends of $\sum_j\sum_k \mathrm{Re}(I_j I_k^{*})R$ — a double sum whose diagonal is the sum of the separate powers and whose off-diagonal terms are all $+1$ in one limit and average to $0$ in the other. Nothing else changed: the network is the same, the elements are the same, and the sources’ amplitudes are the same. Only their phases differ.

Which sharpens that essay’s warning into something usable. It said that adding powers is within one per cent of the truth only when one source is a hundred and ninety-eight times the other — a specific number for two sources, and an alarming one. Generalised, the statement is that the error in adding powers is not a bounded mistake; it is a factor equal to how many things are being added, when they are coherent, and nothing at all when they are not. So the question to ask of any power budget is not how large the contributions are but whether they are correlated, and that is a question about the sources rather than about the network.

Where the answer is neither limit #

Real cases sit between, and the between is where the arithmetic stops being simple in an informative way.

Sources at the same frequency from one clock. Several converters synchronised to one oscillator have fixed relative phases — not zero, and not random. Their cross terms are constant and non-zero, so the total is a fixed number somewhere between $\sum P_k$ and $N\sum P_k$ decided by a set of phases nobody chose. The essay below’s own slider is this case for two sources, and its table runs from 444 milliwatts at 0° to nothing at all at 180°.

Sources at slightly different frequencies. Two converters at nominally the same frequency and actually a few parts per million apart have a relative phase that drifts through the whole range, so the total drifts between the two extremes at the beat frequency. A power measurement averaged over a long time reads the incoherent answer; one taken over a short time reads whatever the phase happened to be; and the peak reads the coherent answer, which is $N$ times the average. That is the failure mode behind a supply that runs warm intermittently with nothing about its load changing.

And harmonics of one current. The essay before it establishes that two components at different frequencies have cross terms averaging to zero over any whole number of both periods, so a distorted current’s harmonics add in quadrature exactly rather than on average. That is the strongest of the three cases and it is the one an instrument relies on: a conductor’s heating from a distorted current is the sum of its harmonics’ heatings, with no phase term at all, which is why a spectrum is a complete description for that purpose. Every tooth the same height uses exactly that to sum a comb of interference in quadrature, and the sum is the one operation in that essay that needs no justification.

The measurement a bench would take, and why it reads the wrong limit #

A power budget is checked by measuring, and the measurement has a bias worth knowing about.

Put a power meter on the load with all $N$ sources running at unrelated phases and it reads the incoherent answer, because a power meter averages. Turning the sources on one at a time and adding the readings gives the same number, because that is the sum of the separate powers and the two limits agree. So a bench measurement of an incoherent set of sources confirms the arithmetic that would be wrong if they were coherent, and gives no warning at all about the case it does not cover.

Three ways out, in increasing order of what they cost.

Measure the peak rather than the average. An incoherent set has a peak power $N$ times its average — the coherent answer arriving momentarily whenever the phases happen to line up — so a peak-reading instrument distinguishes the two cases where an averaging one cannot. The distribution is exponential, which the scatter on the figure shows, so the peak over a long observation is several times the mean rather than exactly $N$ times it.

Change one source’s frequency deliberately. Two sources at exactly the same frequency have a fixed cross term and two at slightly different frequencies sweep through the whole range, so shifting one source by a few parts per million converts a fixed unknown into a measurable swing. The swing’s size is the cross term, and its absence is the evidence that the sources are genuinely uncorrelated.

Or measure the temperature. A dissipation that is $N$ times the average for a fraction of the time is a thermal problem rather than an electrical one, and whether it matters depends on a thermal time constant against a beat period. That is the honest form of the question and it is the one an averaging power meter cannot be made to answer, because the quantity it needs is a distribution rather than a mean.

The general shape is one these essays keep meeting: an instrument that integrates reports the quantity that superposes, and the quantity that does not superpose is the one that breaks things. The peak that only has a bound is the same distinction inside one waveform rather than across several sources, and its conclusion is the same — a root-mean-square value is exactly additive and a maximum has only an inequality.

What is not in the figure #

The sources are equal. Unequal amplitudes change both limits: the coherent ratio becomes $(\sum|a_k|)^2/\sum a_k^2$, which is $N$ for equal sources and less otherwise, and reaches one when a single source dominates. That is that essay’s hundred-and-ninety-eight-to-one result generalised, and it is the reason a power budget with one large contributor is safer than one with many comparable ones.

The phases are drawn once per run. A real set of drifting phases is correlated in time, so a measurement’s average depends on how long it is and on the drift rate, which is a bandwidth question rather than a counting one. The distribution measured here is the long-time one.

And the load is resistive. With a reactive load the cross term picks up the load’s own angle, so the coherent limit is not $N$ but $N$ times the cosine of something — which is the reactive-power arithmetic the current that does no work is about, and which this figure deliberately avoids by making the load a hundred ohms of resistance.

The count that is not a count #

One reading of this essay is that the penalty for adding powers is the number of sources, and that is right for equal sources and misleading for any real set. It is worth writing down what replaces it, because the replacement is a quantity with a use.

The coherent power is $(\sum|a_k|)^2$ times a constant and the sum of the separate powers is $\sum a_k^2$ times the same constant, so the ratio is

$$\frac{P_\mathrm{coherent}}{\sum P_k} = \frac{\left(\sum_k |a_k|\right)^2}{\sum_k a_k^2}$$

which is $N$ when every amplitude is equal and less otherwise. At the other extreme — one source much larger than the rest — it is one, and adding the powers is right.

That expression is an effective number of contributors, and it behaves the way one would want. Four equal sources give four. One source with three others a tenth its size gives 1.51. Sixteen sources whose amplitudes halve down a geometric series give 3.86. So a power budget’s exposure to coherence is not how many terms it has but how nearly equal its largest ones are, which is a much less alarming statement than “the number of sources” and a more useful one.

It also says which budgets to worry about. Several identical supplies paralleled onto one rail is the worst case available, because they are equal by construction and often synchronised by design. A noise budget with one dominant resistor is the best, because its effective count is near one whatever the phases do. And the case in between — a handful of comparable contributors from separate equipment — is where the coherent factor is several and nobody knows the phases, which is exactly where a design ought to allow for the worst and usually allows for the average.

The essay before it reached this from the two-source end and found the same thing in different words: adding the powers of two in-phase sources is within one per cent of the truth only when one is a hundred and ninety-eight times the other, which is an effective count of 1.01. Twenty decibels of separation between two contributors still leaves the sum of their powers twenty per cent short, and the effective count there is 1.2.

Still open: the unequal amplitudes, and the drift that makes the peak #

The coherent ratio for unequal sources. $(\sum|a_k|)^2/\sum a_k^2$ is a closed form and it is not drawn, and it is the quantity a real power budget needs — because a rail with one large supply and several small ones is the ordinary case and its coherent penalty is much less than the count. Sweeping an amplitude distribution rather than a count would turn this essay’s $N$ into the effective number of contributors, which is the quantity every other field calls a participation ratio and this one has not needed a name for yet.

The drift, and the peak it produces. The middle case above says a slowly drifting set of phases makes the total wander between the two limits, so the peak power is the coherent answer however uncorrelated the sources are on average. That is a thermal statement rather than an electrical one — a dissipation $N$ times the average, for a duration set by the beat period — and whether it matters depends on a thermal time constant against a beat frequency. Both are computable and the pair is not drawn.

And the same count with the sources in series. Everything here is $N$ sources in parallel into one load, which is a rail. $N$ sources in series is an array’s other arrangement and its arithmetic is not the same, because the currents rather than the voltages are constrained to be equal — so which of the two limits is favourable ought to invert, and that is worth checking rather than requiring.

What is checked #

The coherent ratio is required to be the count, at every count, to a part in $10^{12}$. That is the essay’s headline and it is an identity rather than a measurement, so it is stated as one — a tolerance would have been the wrong shape of check.

The incoherent ratio is held against its own standard error, three of them, rather than against a fixed tolerance. The error is $1/\sqrt n$ and not $\sqrt{2/n}$ because the squared magnitude of a sum of random phasors is exponentially distributed rather than a sum of squared normals — which is a factor of $\sqrt2$ and, more importantly, is the reason sixty-four runs are used where eight would have been a fifty per cent check.

And the ratio between the two limits is required to grow with the count by more than a factor of eight across the range, because the essay’s claim is that the error is unbounded rather than large. A figure that required only the two limits at one count would be a figure about two numbers.