The cable that hides two things

Two essays in this collection own half of one decibel each and neither has ever met the other.

The mismatch that the cable hides measured what a length of lossy line does to a reading. The reflection seen at the near end is the load’s own attenuated by twice the one-way loss, so a four-to-one load at the end of twenty metres of ordinary coaxial cable reads 1.13:1 — 24.4 decibels of return loss, which would pass any acceptance test. The load is unchanged; the instrument is looking at it through ten decibels of attenuator.

The loss in front, counted twice measured what the same length does to a floor. A matched dissipative attenuator’s noise factor is its loss, exactly, so a decibel of cable ahead of an amplifier is a decibel on the noise figure of everything after it. It is the same decibel counted as an attenuation of the signal and as a noise source of its own, which is why it appears once in the answer rather than twice.

The first essay stated the join in a single sentence and priced nothing: “buy two decibels of match with one decibel of noise, and the exchange rate is fixed by physics rather than by the component.” The second named the condition that would break it and left it there: “if the attenuator is not terminated in its design impedance the available-power argument’s G is no longer the insertion loss, and the identity F = L stops being about the number on the label.”

Those are the same cable. The mismatch the first essay is hiding is exactly the thing the second essay says invalidates its own identity, and putting them on one axis is what this essay is.

The calibration, which is the reading alone #

The reading is the half of this that is already verified, and it is the instrument the rest is quoted against.

Two things about it are load-bearing here. The reflection is computed from the full complex hyperbolic tangent of $\gamma\ell$ rather than from the expression it is compared with, and the two agree to better than a part in a billion over two orders of length. And the boundary is a loss rather than a length, which is why it is also a frequency: the same cable hides a mismatch at three metres at ten gigahertz and needs thirty at a hundred megahertz.

That last point is what makes the join computable at all. If both halves are governed by $\alpha\ell$ and nothing else, then the ratio between them cannot contain a cable.

Two decibels per decibel, and it is exact #

Counted the way a link budget counts it — the cable is a matched attenuator, its noise factor is its loss, and Friis weights it by the unity gain in front of it — the exchange rate is exactly two, at every length on the axis and at every setting of the frequency.

Measured on the intervals between forty lengths spanning four hundred times: the departure from two is under $10^{-9}$. The reading rises at $2\alpha\ell$ and the noise figure at $\alpha\ell$, from two pieces of machinery that share the attenuation and nothing else — one walks a complex hyperbolic tangent and one sums a cascade formula. Two routes that disagree about nothing to nine decimals is the pairing one step, computed twice insists on, and it is worth having here because everything after this section is a departure from it.

That is the rule of thumb, it is exact, and it is the answer to a question nobody is asking. A link budget with a matched attenuator in it has no mismatch to hide.

The three columns, and the one decibel in all of them #

The reason two essays could each own half of this and never meet is not electrical. It is that the same decibel is written down in three different documents.

Insertion loss is on the passive component’s data sheet, beside its return loss and its power rating. Return loss is on the acceptance test, as a limit somebody has to pass. Noise figure is on the amplifier’s data sheet, beside its gain and its intercept point — and every row of a link budget without a noise-figure column is silently treated as having a noise figure of zero.

The correct entry for a passive row is not zero and it is not blank; it is the number already in the loss column, written again. That is the fix the loss in front, counted twice states as a habit rather than a theorem, and it is right as far as it goes. What this essay adds is the third column: the same number, doubled, is what the acceptance test is reading — so a decibel added to one column is subtracted from the second and added twice to the third, and the three are never on the same page.

Two of the three places it hides are worth naming because neither looks like an attenuator. A filter in front of the amplifier, put there to keep strong out-of-band signals out, has a pass-band insertion loss that goes straight onto the noise figure of everything after it, and what the filter in front costs prices the other side of that trade. And a connector, a switch or a relay contributes a few hundredths of a decibel that nobody budgets — negligible in fifty ohms and not negligible at all when the network is not matched, which is the same condition this essay turns on.

Where the two essays actually collide #

The identity $F = L$ is a statement about a two-port terminated in its design impedance. What is on the end of this cable is a four-to-one mismatch — that is the entire premise — so the available gain is not the insertion loss and the noise factor is not the loss on the label.

The honest quantity is $F = 1/G_{\text{av}}$, and for a matched lossy line driven by a source of reflection $\Gamma$ that is

$$G_{\text{av}} = \frac{u\,(1-\Gamma^2)}{1-\Gamma^2 u^2}, \qquad u = 10^{-\alpha\ell/10}$$

which reduces to the loss when $\Gamma$ is zero and exceeds it otherwise. The excess is $(1-\Gamma^2u^2)/(1-\Gamma^2)$ exactly: 0.028 decibels at five centimetres of cable, and 1.938 decibels — the load’s own mismatch loss, the power it never accepts — once the cable is long enough that nothing comes back. It is monotone in the length, so the rule of thumb flatters the cable everywhere.

Which inverts the exchange rate near the short end, and the closed form for it is worth having because of what is missing from it.

The exchange rate is a function of the reading and nothing else. Not of the cable, not of its length, not of the frequency. Three cables differing by ten times in attenuation, at lengths differing by ten times, that happen to read the same standing-wave ratio buy it at the same rate — checked at a hundred and eighty-three points across three decades of frequency, agreeing with the closed form to a part in ten thousand.

And it runs the wrong way from every intuition about attenuators. The first decibel is the expensive one. At the load’s own 4.0:1 a decibel of noise figure buys 0.94 decibels of apparent match rather than two; by the time the reading is 1.05:1 the rate has reached 1.998 and the rule of thumb is right. A three-decibel pad on the front of a conditionally stable amplifier — the standard defence, and the example the first of these two essays reaches for — is exactly the case where the rule of thumb is worst, because a pad is short and the thing it is defending against is badly mismatched by construction.

What an acceptance limit costs #

Both halves are governed by $\alpha\ell$, so the cable divides out and what is left is a statement about two reflection coefficients.

Reaching a reading $\Gamma_{\text{read}}$ from a load $\Gamma_L$ needs $2\alpha\ell = 20\log_{10}(\Gamma_L/\Gamma_{\text{read}})$, so the noise figure it costs is $10\log_{10}(\Gamma_L/\Gamma_{\text{read}})$ counted naively, and that plus the difference of the two mismatch losses counted honestly. There is no cable, no length and no frequency in either.

That is the essay’s number and it is worth stating as a sentence about a specification rather than about a cable. An acceptance limit written at the instrument, on a load that cannot meet it, is a noise-figure budget in disguise, and the conversion is a logarithm of two reflection coefficients:

• read 3.00:1 on a 4.0:1 load — 1.481 decibels;
• read 2.00:1 — 3.979;
• read 1.50:1 — 6.532;
• read 1.20:1 — 10.098.

Nothing in that list mentions what the cable is made of. A test written as “1.5:1 at the connector” and met by moving the connector further away has spent six and a half decibels of system noise figure, and it has spent them whether or not anybody was measuring.

What the reading was supposed to be a proxy for #

The whole reason anybody measures a return loss is that it is a cheap stand-in for delivered power, and the substitution is what fails here.

At twenty metres and a gigahertz the input accepts 99.6 per cent of the power available from a matched source — which is what 24.4 decibels of return loss means — and delivers ten decibels less of it to the load, because ten decibels went into the cable. The arrangement is superbly matched and delivers a tenth. That much the mismatch that the cable hides already measured.

The receiving direction is where the second half lands, and it is exact by definition. A noise figure is the decibels of signal-to-noise ratio a stage costs, so the chain’s 13.92 decibels against the amplifier’s own 2.00 says the arrangement has given away 11.92 decibels of signal-to-noise ratio at twenty metres — not as an estimate but as the meaning of the number. The return-loss reading improved by 20.0 decibels over the same twenty metres. One quantity went up by twenty and the other went down by twelve, and only the first of them appears on a test report.

That is what makes the substitution dangerous rather than merely approximate. A proxy that fails by a constant factor can be corrected; this one fails in opposite directions, and the direction it fails in is the one that reads as success. It is the shape every model has an edge collects — a model conservative in a quantity nobody pays for and optimistic in the quantity they do — arriving here as an instrument rather than as a model.

The reading at which the cure passes the disease #

The mismatch is not free either, and that gives the comparison a crossing.

A load reflecting $\Gamma$ never accepts $\Gamma^2$ of the power available to it, which for 0.60 is 1.938 decibels. So there are two costs on the table — the mismatch’s own, which is fixed, and the cable’s, which grows with how good a reading is demanded — and they are equal at one reading.

Bisected on the chain rather than solved from the expression, that reading is 2.762:1, and the closed form it agrees with is a root of $\Gamma x^2 + x - \Gamma$, which again has no cable in it. Below 2.762:1 — that is, demanding a reading better than that — the cure costs more noise figure than the mismatch was costing in received power. Above it, less.

The practical reading of that number is unflattering to a common procedure. Nobody writes an acceptance limit at 2.76:1. Limits are written at 2:1 and at 1.5:1, which are 2.05 and 3.37 times the disease, so a cable long enough to pass a normal acceptance test on a bad load has always cost more than the load was costing. The improvement is entirely in the measurement and the loss is entirely real — which is the first essay’s own conclusion, now with the other half’s number attached to it.

What is checked #

Five assertions, and every one of them is an expression rather than a value.

That the reading is the load’s own return loss plus twice the one-way loss, against a line solved with its loss in, to better than $10^{-6}$ decibels over the lengths at which the reflection is measurable. Lengths past that are excluded and the reason is arithmetic rather than physics: the hyperbolic tangent forms its answer as a difference of two exponentials of $\gamma\ell$, and comparing there would be measuring the cancellation rather than the claim.

That the naive exchange rate is exactly two, to $10^{-9}$, over forty intervals — the reading from a transmission-line solve and the noise figure from a cascade formula, sharing the attenuation and nothing else.

That the honest count exceeds it by $(1-\Gamma^2u^2)/(1-\Gamma^2)$ at every length, monotonically, reaching the load’s own mismatch loss at the long end.

That the honest exchange rate is $2(1-\Gamma^2)/(1+\Gamma^2)$ in the reading, to a part in ten thousand, across a hundred and eighty-three points spanning three decades of frequency and four hundred times in length — which is the claim that the rate contains no cable, made as a sweep over three cables rather than as an algebraic remark.

And that the noise figure a stated reading costs is the same for three different cables: the length that produces a given reading is bisected on the solved line at a hundred megahertz, a gigahertz and ten, the three lengths come out in the ratio $\sqrt{10}$ each, and the chain noise figure at all three agrees to six decimals.

What this does not say #

That the amplifier’s own noise figure is unchanged. It is not. A real amplifier’s noise figure depends on the source impedance presented to it, and the impedance at the near end of this cable rotates and shrinks with length. That is a two-parameter description this collection does not carry, and the number quoted here is the cable’s contribution with the amplifier held at a fixed 2 decibels. The direction of the missing term is not known without the amplifier’s noise parameters, which is the honest limit rather than a small one.

That $F = 1/G_{\text{av}}$ needs no condition of its own. It needs the two-port to be passive and at 290 kelvin, and the second half is not automatic — the same essay that owns the identity measures what happens when it is not, and a cable on a roof or in a rack is neither. Nor is the bandwidth the figure is integrated over the corner frequency it is quoted at, which the bandwidth noise sees measures at π/2 times it for a single pole. The temperature clause is the one that occasionally decides a design, and it is not exercised here.

That any of this is about a bad cable. The attenuation used is ordinary small coaxial cable at 0.5 decibels a metre at a gigahertz, rising as the root of frequency in the regime that decides it. Nothing here is a defect; the loss is doing exactly what loss does, twice, in two different budgets.

Where the same shape appears #

A quantity that improves a measurement while degrading the thing measured is not rare in this collection, and the family is worth naming because the tell is always the same.

A quarter wave, and the path the current takes back is the case where a match is bought rather than faked, and the price is a band rather than a floor — which is the honest version of the same purchase.

Every model has an edge collects the general failure: a model that is conservative in a quantity nobody pays for and optimistic in the quantity they do. Here the conservative quantity is the return loss, which improves, and the optimistic one is the noise figure, which nobody reads off a passive component’s data sheet because that column does not exist.

And the staircase in time is where the reflection this whole argument is about comes from — a wave that does not know what is at the far end for one propagation delay, whose return is what a return-loss reading is a frequency-domain summary of.

The number worth carrying #

$10\log_{10}(\Gamma_L/\Gamma_{\text{read}})$, plus the difference of the two mismatch losses.

That is what a return-loss specification costs in noise figure when it is met by putting distance between the instrument and the problem, and it contains no cable, no length and no frequency because both halves of it are governed by the same $\alpha\ell$. On a 4.0:1 load it is 3.98 decibels to read 2:1 and 6.53 to read 1.5:1, against a mismatch that was costing 1.938 — so the ordinary acceptance limits are two to three times the disease.

The habit is about which of two numbers a measurement is being asked to support. A reflection coefficient measured at the wrong end of a lossy line is a statement about power accepted and not about power delivered, and the difference between accepted and delivered is the loss — which is also, to the decibel, the noise figure. One quantity, read as an improvement at one end of the cable and as a cost at the other, and the only way to tell which reading is being quoted is to ask where the instrument was standing.