How many sources the cross terms count

The cross terms that outnumber the sources put $N$ equal sources on one load and counted what adding their separate powers gets wrong. With every source in phase the load takes $N$ times the sum, exactly, because all $N(N-1)/2$ cross terms are positive and equal. With random phases it takes the sum itself, because the cross terms average away — which is why noise powers add and a root-sum-square is the right arithmetic for uncorrelated interference. The factor between the two limits is the count.

That essay ended with the case it had not drawn. The quantity $(\sum|a_k|)^2/\sum a_k^2$ is the coherent penalty for sources of any size, and it is the one a real power budget needs, because a rail with one large supply and several small ones is the ordinary case and its penalty is much less than the count. Sweeping amplitude distributions rather than counts turns $N$ into an effective number of contributors — the quantity other fields call a participation ratio. This essay measures it on the network, for the distributions that occur, and checks that the random-phase answer does not care.

The formula, solved #

The network is the earlier essay’s: each source reaches one 100 Ω load through its own 100 Ω, at a kilohertz. The coherent power is solved with every source in phase, and each source’s power alone is solved with every other source zeroed — left in the netlist as a short, which is what zeroing a voltage source means, and the distinction the source that must not be zeroed turned on. The penalty is the one over the sum of the others.

With amplitudes $a_k$, the load current is proportional to $\sum a_k$ when they are in phase and to each $a_k$ alone, so the ratio is

$$N_{eff} = \frac{\left(\sum_k |a_k|\right)^2}{\sum_k a_k^2},$$

the sum of the amplitudes squared over the sum of their squares. For equal amplitudes it is $N$. For one nonzero source it is one. In between it counts how many sources are participating, weighted by how much.

The solves agree with the formula to ten figures at every one of the twenty-five networks, which is the check that the network is doing nothing the algebra does not know about, in the way two solves that add, and the one that does not checked superposition itself to the last bit — the sources’ own series resistances divide every current equally and cancel from the ratio.

A falling series has a ceiling #

The first distribution is a geometric one: each source a fixed fraction $q$ of the one before. It is the shape of many real budgets, where contributions are ranked and each is smaller than the last by roughly a constant factor.

Equal sources, $q = 1$, give the count — 2, 4, 8, 16, 32. Any $q$ below one gives less, and more importantly gives a ceiling. With each source 0.8 of the one before, the penalty rises through 1.98, 3.77, 6.41 and 8.51 and reaches 8.99 at thirty-two sources, against a limit of

$$N_{eff} \to \frac{1 + q}{1 - q} = 9.$$

With each source half the last the limit is 3, reached to three figures by thirty-two sources. The sum of the amplitudes converges, to $1/(1 - q)$ of the first, and the sum of their squares converges faster, to $1/(1 - q^2)$; their ratio is the ceiling. So a budget whose contributions fall geometrically has a coherent penalty that stops growing, however long the tail — the tail adds little amplitude and less power.

The slider on the figure at the head of the page steps $q$ from equal to a fifth, where the ceiling is 1.5: the largest source dominates so completely that lining the others up with it adds only half again.

One large source and many small ones #

The second case is the one the earlier essay named: a rail with one large contributor and $N - 1$ small ones, each a fraction $\varepsilon$ of the large one’s amplitude.

The penalty is $(1 + (N - 1)\varepsilon)^2/(1 + (N - 1)\varepsilon^2)$, and it is much less than the count. Thirty-one small sources at a tenth of the large one’s amplitude give 12.8, not 32. At a hundredth they give 1.71.

The shape of the curve at small $\varepsilon$ is the part to carry away. The penalty’s excess over one is $2(N - 1)\varepsilon$ to first order — proportional to the small sources’ amplitudes, not their powers — and the solves confirm the first-order form to two per cent. That is why small contributions are so easy to underestimate. Each of thirty-one sources at a tenth of the amplitude carries a hundredth of the large one’s power, so added they are 1.31 times its power; lined up in phase they deliver 16.8 times it. A budget that ranks its contributions by power and drops the ones below a per cent has dropped the ones that, in phase, add most.

The practical reading is a test: when the small contributors might be correlated with the large one — harmonics of the same switching frequency, pickup from the same clock — budget them by amplitude. When they cannot be, budget them by power. The difference at thirty-one sources a tenth the size is a factor of 12.8.

Random amplitudes, a fixed fraction #

The third distribution is random: amplitudes drawn from a spread, as the contributions of many nominally similar parts are.

For many sources the penalty divided by the count settles at a property of the distribution, $(\overline{|a|})^2/\overline{a^2}$: the squared mean over the mean square. For amplitudes uniform between nothing and a maximum that is 0.75; for amplitudes log-uniform over a decade, 0.711; for Rayleigh amplitudes, $\pi/4 = 0.785$. The seeded networks give 0.752, 0.715 and 0.798 at thirty-two sources, within the scatter of thirty-two draws. At two sources the ratio is higher, because $(\sum|a|)^2/\sum a^2$ includes each source’s square as well as the cross terms, and the cross terms’ share grows with the count.

So random amplitudes do not change the earlier essay’s conclusion; they scale it. The coherent penalty still grows in proportion to the number of sources, reduced by a fraction between about 0.7 and 0.8 for any reasonable spread. It is only a spread with structure — a few large and many small, or a falling ranking — that stops it growing.

With random phases the amplitudes do not matter #

The incoherent limit was the sum of the separate powers for equal sources. It should be the sum for any amplitudes, since each cross term $2a_j a_k\cos(\varphi_j - \varphi_k)$ averages to nothing over random phases whatever its size.

It is. Over sixty-four seeded phase draws, sixteen sources give 1.093, 0.963 and 0.956 times the sum of their powers for the three amplitude shapes, each within its standard error of one. What the amplitudes do change is the scatter. The relative spread of the power over random phases is $\sqrt{1 - \sum a^4/(\sum a^2)^2}$ — 0.968 for equal sources, where a sum of many equal random phasors is nearly exponentially distributed in power, 0.883 for a 0.8 series and 0.632 for a halving one, because a sum dominated by one source is nearly that source and cannot fluctuate much.

That is a second participation ratio in disguise. $(\sum a^2)^2/\sum a^4$ is the effective count for the power’s scatter, as $(\sum|a|)^2/\sum a^2$ was for its coherent penalty. The two agree for equal sources and differ for unequal ones, and a budget can use either depending on whether it is asking about a worst case or a spread.

Equal sources are the worst case #

The earlier essay’s count is not only the equal-amplitude case; it is the largest the penalty can be. For a fixed total of separate powers $\sum a^2$, the Cauchy–Schwarz inequality bounds the square of the amplitudes’ sum by $N\sum a^2$, with equality only when every amplitude is the same. So among all the ways of dividing a given power among $N$ sources, the equal division maximises the coherent penalty, and every unequal division scores less. The count $N$ is an upper bound on $N_{eff}$, attained only by the most democratic distribution.

That turns the participation ratio into a practical correction rather than a curiosity. A designer who knows only the count and assumes coherence has budgeted for the worst possible distribution; one who knows the amplitudes can budget for the actual one, and the saving is the ratio of the two — nine against thirty-two for the 0.8 series, 12.8 against 32 for the dominated rail. The bound is also the reason the coherent penalty behaves so differently from the incoherent one: the incoherent sum does not depend on the distribution at all, while the coherent one depends on it as strongly as the inequality allows.

The ratio of a coherent peak to a root-sum-square is a crest factor of sorts, the same kind of quantity what a meter multiplies by is built on, where a meter calibrated for one waveform’s ratio of peak to rms misreads every other. Here the waveform is a sum of phasors, and three voltages that close on one is the picture of what coherence means for them: arrows laid end to end in one direction make the longest possible sum.

A ripple budget, worked #

The case that makes this concrete is a supply rail’s ripple, which is a comb of lines at the harmonics of a switching or rectifying frequency. The ripple that arrives as a comb found a regulator’s output ripple arriving as exactly such a comb, with the second and third lines nearly as large as the first, and found the summed output 2.87 times the estimate from the fundamental alone. The lines of a comb are harmonics of one waveform, so their phases are fixed relative to one another: whether they add coherently at the peak is a property of the waveform’s shape, not of chance.

Take a comb whose lines fall by half each — 20, 10, 5 and 2.5 mV. Summed as powers, which is the arithmetic for independent contributions, they give an rms of 23.0 mV. Their amplitudes sum to 37.5 mV, which is the peak when they line up, as the harmonics of a train of narrow pulses do at every pulse. Whether a given comb lines up is a property of the waveform that made it, and a measured one can be checked by comparing its peak with its rms. The ratio of the two is $\sqrt{N_{eff}}$, the square root of the participation ratio — 1.63 here, for $N_{eff} = 2.65$ against the four lines. A budget that treats the lines as independent and quotes their root-sum-square is 38 per cent low on the peak, and the participation ratio says so from the amplitudes alone, without knowing the phases.

The same budget with lines falling by 0.8 each would have a ceiling of nine rather than three as more harmonics were added, and the root-sum-square estimate would be low by a factor approaching three. The slower the comb falls, the more the coherent peak outruns the power sum.

That claim — the peak over the root-sum-square is $\sqrt{N_{eff}}$ — is the network result read from a waveform instead of a load, and the figure makes the check the long way round: it builds each comb as a sum of cosines, samples a period at four thousand points, takes the largest magnitude, and divides by the root-sum-square of the amplitudes. With every line in phase the answer agrees with the square root of the participation ratio to six figures at every count and every rate of fall — 1.627 for the four-line example, 3.00 for thirty-two lines falling by 0.8, and 5.66, the square root of thirty-two, for thirty-two equal ones. With the phases drawn at random the same combs peak at 1.33, 1.59 and 2.02 on average. The amplitudes set how far a comb’s peak can outrun its power sum; the phases decide whether it does, and a comb whose lines are harmonics of one pulse train is the case where they all do.

The same number, in other fields #

The ratio $(\sum w)^2/\sum w^2$ is not particular to circuits, and it is worth recognising where it appears because the intuition transfers. A survey statistician computes it from a sample’s weights as the effective sample size — how many equally weighted observations the unequally weighted sample is worth. A physicist computes the same form from a wavefunction’s amplitudes to say over how many sites a state is spread. In each case it answers the question asked here: of these many contributors, how many are really taking part?

The scatter’s effective count, $(\sum a^2)^2/\sum a^4$, is also familiar under another name. Written in terms of each source’s share of the total power, $p_k = a_k^2/\sum a^2$, it is $1/\sum p_k^2$, the effective number of parties a political scientist computes from vote shares. Coherent penalty and power scatter are the same idea applied to amplitudes and to powers: the first asks how many sources are participating in the sum of voltages, the second how many in the sum of powers. For equal sources they are both the count; for any other distribution the first is the larger, since amplitudes are less concentrated than their squares.

What a designer should take #

When sources may line up, count them by participation: $(\sum|a|)^2/\sum a^2$, computed from their amplitudes. It is the count for equal sources, stops growing for a falling series at $(1 + q)/(1 - q)$, is $1 + 2(N - 1)\varepsilon$ for many small sources beside one large one, and is about three-quarters of the count for random amplitudes. When they cannot line up, their powers add and none of this matters.

And rank contributions by amplitude, not power, whenever coherence is possible. The contributions a power ranking drops as negligible are the ones whose amplitudes, in phase, add most — a hundredth of the power is a tenth of the amplitude, and thirty-one of those beside one source are sixteen times its power when they align.

The peak that only has a bound found that a waveform’s peak has no closed form, only a bound, and the bound is the coherent sum. The participation ratio is how large that bound is in power, relative to the sum a budget would otherwise use, and it is what a peak-power rating has to be checked against.

How the numbers were obtained #

Each network is the earlier essay’s: $N$ phasor voltage sources, each through 100 Ω to a common node, and a 100 Ω load from that node to ground, solved at a kilohertz. The coherent power sets every phase to zero; each source’s power alone is a separate solve with every other source’s voltage set to zero. The random amplitude draws and random phases come from a seeded generator, thirty-two draws for each distribution and count and sixty-four phase draws for each amplitude shape, and the incoherent means are checked against the standard error the amplitudes themselves imply. The closed forms are checked against the solves and not used to draw.

What it leaves out #

Sources with their own frequencies. Every source here is at one frequency, so coherence is a matter of phase. Sources at slightly different frequencies drift through coherence at their beat frequency, and the power swings between the coherent and incoherent limits at that rate; the average is the incoherent sum and the peak is the coherent one, and which matters is a question about how long the load can sustain the peak.

Unequal source impedances. Every source has the same series resistance, which cancels from the ratio. Sources behind different impedances deliver currents in proportion to their voltage over their impedance, and the participation ratio is then of those currents rather than of the voltages.

Correlation that is partial. Sources are either perfectly coherent or independent here. Partly correlated sources — pickup that shares some of its path — give a penalty between the two, set by the correlation coefficients as well as the amplitudes.

Still open: the beat, the partial correlation, and the sources in series #

The drift and the peak it produces. Sources at nearly the same frequency pass through the coherent sum once per beat period, and the load dissipates $N_{eff}$ times the average for a fraction of it. Whether that matters is a thermal question — the peak’s duration against the load’s thermal time constant — and the pulse that ends before the heat has the arithmetic for a pulse that is shorter than the part it heats.

Partial correlation. With a correlation coefficient $\rho$ between each pair, the expected power is $\sum a^2 + \rho\sum_{j\ne k}a_j a_k$, which moves smoothly from the sum to the coherent limit. Measuring it on a network of sources that share a common component would say how much correlation it takes before a budget must stop adding powers.

The same count with the sources in series. Everything here is sources in parallel on one load, which is a rail. Sources in series — an array’s other arrangement — share a current rather than a voltage, and whether the participation ratio carries over, or whether the favourable and unfavourable limits change places, is a solve on the dual network.