The end a lossy line looks like

The noise a pad adds is its temperature solved a matched attenuator resistor by resistor and found the law every radio-astronomy text quotes: a pad of loss $L$ at physical temperature $T$, fed from a source at $T_s$, presents $T_s/L + T(1 - 1/L)$ at its output. Read as transmission and emission, it says a pad passes $1/L$ of what arrives and adds what a body at its temperature emits, $1 - 1/L$ of a perfect emitter’s worth. At its own temperature the two balance and nothing can be told apart.

It closed on a line that is not at one temperature. A cable from a cold stage of a cryostat to the room outside is lossy along its whole length and passes through every temperature between its ends, so it is a chain of short pads each at its own temperature. Its output is each piece’s emission attenuated by every piece after it, and the question it left was which end that weighting favours, and by how much.

The answer is the end the line is read from. A lossy line looks like the temperature near its output, and how near depends on its loss; the same cable has two noise temperatures, one for each direction, and they can differ by a factor of two.

The line as a chain of pads #

The line is cut into $n$ equal pads, each with $1/n$ of the total loss in decibels and each at the temperature of its own centre on a profile running from 300 kelvin at the input to 4 kelvin at the output. Each pad applies the pad law: the noise temperature arriving is divided by the pad’s loss and the pad’s own emission is added. Applying it $n$ times in turn gives the output. Two of the forty pads’ laws — at 4 kelvin and at 150 — are held against the resistor-by-resistor sum the earlier essay built, every resistor’s Johnson noise carried to the pad’s open output, and agree to nine figures, so the chain is built from a law that has been checked rather than quoted.

The second route is the continuous line. With the loss spread uniformly at $A$ nepers of power along a unit length and a linear profile from $T_a$ at the input to $T_b$ at the output, the output is an integral of each element’s emission times the transmission from it to the end, and for a linear profile it has a closed form:

$$T_{out} = T_s e^{-A} + T_a(1 - e^{-A}) + (T_b - T_a)\left[(1 - e^{-A}) - \frac{1 - e^{-A}(1 + A)}{A}\right].$$

Forty pads give 67.7 kelvin at the cold end of a 20 decibel line; the closed form gives 67.63. The difference is the chain’s own resolution, and the next figure measures it.

A midpoint rule, and how fast it converges #

One pad at the line’s mean temperature, 152 kelvin, is out by 85.85 kelvin in either direction — the lumped model of a cable, and badly wrong. Sixteen pads are out by 0.439 kelvin and 256 by under two millikelvin; the error falls exactly as the square of the count. That is the signature of a midpoint rule: each pad evaluated at its centre’s temperature is the integral’s integrand sampled at the middle of each interval, and the closed form is the integral itself. The agreement of the two routes at that rate is what says the chain is the right model of the line rather than merely a model that converges somewhere.

It also says what a designer can get away with. A line through a cryostat is usually described by the temperatures of the stages it is anchored to, and treating each stretch between stages as one pad at its average is a model of a few pads. The figure says how wrong that is for a smooth gradient: tens of kelvin with one pad, a few with four.

Which pieces make the noise #

Every piece of the line is a resistance at a temperature, and the floor a resistor sets is the whole of what it contributes: a noise that depends on its temperature and on nothing about how it was made. A line’s dielectric loss and its conductors’ loss are different mechanisms and emit alike, so the chain needs only each piece’s loss and temperature, never its construction. That is what lets a cable be reduced to a profile and a number of decibels.

The hero figure divides the output into what each piece contributes. Each piece emits $(1 - 1/L_k)T_k$, the most at the warm end where $T_k$ is largest, and is then attenuated by every piece between it and the output — most at the warm end, which has the whole line ahead of it. At 20 decibels the attenuation wins: the warm half of the line, emitting nearly three times what the cold half does, delivers 18.0 kelvin to the output against the cold half’s 46.7. The pieces that contribute most are those a little way in from the cold end, warm enough to emit and close enough not to be attenuated away.

That is the resistor the noise comes from’s rule at the scale of a cable. A resistor’s noise is filtered by everything after it and by nothing before it; a piece of line’s noise is attenuated by everything after it and by nothing before it. The warm end of a line into a cryostat is the first resistor of a filter with a lot of filter after it.

The same line, read from the other end #

A line’s emission depends on where it is read, and reversing the direction reverses the weighting.

At three decibels the two ends read 67.4 and 84.3 kelvin, not far apart and both near the line’s mean temperature times its loss. At ten they read 89.7 and 183.9: the same cable, the same temperatures, a factor of two between its two noise temperatures. At thirty they read 46.5 and 257.2, and both are heading for the temperature of the end they are read from — 4 and 300 — because a very lossy line is opaque, and a reader at one end of an opaque line sees only the stretch nearest it.

The cold-end reading is not monotonic in the loss, and that is worth seeing as well as knowing. It rises from 67.4 kelvin at three decibels to 89.7 at ten and then falls, to 64.6 at twenty and 46.5 at thirty. A lossier line into the cold can be quieter, because past a certain loss the extra loss mostly attenuates the warm end’s emission while adding cold emission. It approaches 4 kelvin, the temperature of the end it is read from, only slowly: the closed form’s last term falls as one over the loss in nepers, so even a hundred decibels reads 16.9 kelvin, the cold end seen through a stretch of line just warm enough to show.

Where twenty decibels should go #

A cryostat’s input line usually has attenuators added on purpose, to stop room-temperature noise reaching a cold amplifier or a cold device, and the line’s lesson says where they belong.

All twenty decibels at room temperature deliver exactly 300 kelvin: an attenuator at the source’s temperature removes the source’s noise and replaces it with its own, which is the same, so it achieves nothing. All twenty at the 4 kelvin stage deliver 6.96: three kelvin of the source left after a hundredfold attenuation and 3.96 of the attenuator’s own. Every split between those is worse than the second, and a split that puts any of the twenty at room temperature wastes that part entirely — ten at 300 kelvin and ten at 4 deliver 33.6, five times the all-cold arrangement.

The rule is the pad law applied once: a decibel of loss is worth most where the pad is coldest, because there it removes a large share of what arrives and adds almost nothing of its own. The only reason not to put every decibel at the coldest stage is that an attenuator dissipates the signal’s power and the coldest stage has the least cooling power to absorb it — which is a thermal budget, outside this figure, and is why practical lines spread their attenuation across stages at all.

The profile between the same two ends #

The two ends do not fix the answer. A line whose temperature stays near room temperature until its last stretch — a cable thermally anchored poorly, or one whose conductivity carries heat down into the cryostat — delivers 101.9 kelvin into the cold at twenty decibels. One that cools early, anchored near its warm end, delivers 27.4. The ratio is 3.7, between two lines with the same ends and the same loss, and it is set by the profile near the cold end, which is the part of the line the cold end can see.

At a decibel each line delivers close to its own mean temperature times its loss, because a nearly transparent line weights all its pieces nearly equally. The profile begins to matter only as the loss grows and the weighting tilts toward the output, and by twenty decibels the line’s mean temperature has stopped being a useful description of it at all.

The noise is a flow of heat #

A line between two temperatures is not in equilibrium, and the noise it delivers is a net flow of power. At its own temperature the cold stage emits 4 kelvin’s worth back up the line, $k \times 4$ kelvin a hertz of bandwidth, and receives the line’s 67.6 kelvin’s worth; the difference, $k \times 63.6$ kelvin a hertz, is power arriving at the cold stage that it must absorb. Which way the noise goes found exactly this between two resistors at different temperatures: a net flow of $k\Delta T$ a hertz at a match, from the warm one to the cold. A line along a gradient is that exchange with the line’s own loss in between, and the figure the cold end reads is the warm side of the exchange.

Over a gigahertz of bandwidth $k \times 63.6$ kelvin is under a picowatt, which no cryostat notices as heat. It is not negligible as noise: a device at 4 kelvin whose input is meant to be at 4 kelvin is being driven by a source at 67.6, sixteen times hotter, and that is the number a qubit, a bolometer or a cold amplifier’s input sees. The attenuator at the cold stage is the device that reduces the flow; it does so by being a resistor at 4 kelvin that absorbs most of what arrives and replaces it with its own, nearly cold, emission.

The equilibrium check is the one the earlier essay leaned on. With the whole line and the source at one temperature, the closed form returns that temperature exactly — $T_s e^{-A} + T(1 - e^{-A})$ with $T_s = T$ — and the chain does the same pad by pad. Every departure in the figures is the gradient’s and nothing else.

A receiver behind the line #

The usual use of a cold line is to bring a signal into a cold amplifier, and there the line’s own noise adds to the amplifier’s. A cold amplifier with a noise temperature of 5 kelvin behind a bare 20 decibel line along the linear gradient sees the line’s 64.6 kelvin of emission plus the source’s attenuated 3, for a system temperature near 73 kelvin: the line, not the amplifier, is the receiver’s noise. With the same 20 decibels as one attenuator at the 4 kelvin stage and an otherwise lossless line, the system temperature is 6.96 plus 5, twelve kelvin, six times better. The signal is attenuated by the same twenty decibels either way; only the noise moves.

That is the Friis reading the loss in front, counted twice gave, with the cable’s temperature now a variable. Loss in front of an amplifier costs its own noise at its own temperature; a cold loss costs little, a warm loss a great deal, and a loss along a gradient costs what the cold end can see of it.

What this says about a cable in front of an amplifier #

The loss in front, counted twice found a decibel of cable before an amplifier costing a decibel of signal and a separate decibel of noise figure, at room temperature. The cable that hides two things found it improving a return-loss reading by twice its loss while raising the noise by once. Both assumed the cable at one temperature, where the pad law reduces to the familiar $(L - 1)T$ of added noise referred to the input. A cable along a gradient has no single temperature to put in that expression; the closed form here is what replaces it, and it can be used as a pad of the right effective temperature: a 20 decibel line into the cold from room temperature acts as a pad at about 65 kelvin, where its mean temperature would say 152.

That effective temperature is also how a cryogenic receiver’s noise budget should treat its input line. The line’s contribution at the amplifier’s input is its own emission at the cold end, not $(1 - 1/L)$ times the average of its two ends, and the two differ by the factor of two or more the direction figure shows.

What the chain leaves out #

The loss is uniform along the line and independent of temperature. A real coaxial cable’s loss falls as it cools, because its conductors’ resistance falls, so a cable into a cryostat is lossier at its warm end than at its cold end. That moves the weighting further toward the warm end, which emits most, and raises the cold-end reading; the chain handles it directly, by giving each pad its own loss, and the closed form does not.

The line is matched everywhere and the pads are ideal attenuators. A mismatch at a stage’s thermal anchor reflects part of the noise travelling toward the cold end back up the line, where it is attenuated twice, and part of the cold end’s own emission back down; only the real part is warm and which way the noise goes are the essays whose arithmetic that would need.

The number to keep #

A line from 300 K into a 4 K stage is a chain of pads, each at its own temperature, and its output is each piece’s emission attenuated by everything after it. At 20 dB, fed from 300 K, its cold end is at 67.6 K, more than two thirds of that from the cold half of the line. Read from its two ends the same line’s own noise is 89.7 and 183.9 K at 10 dB: a lossy line looks like the end it is read from. Twenty decibels of attenuation at 4 K deliver 6.96 K; at 300 K, nothing at all.

Still open: the loss that changes with temperature, thermometry, and the reflection at a thermal anchor #

A loss that falls as the line cools. A coaxial line’s attenuation per metre at 4 kelvin is a fraction of its value at room temperature. Giving each pad the loss its own temperature implies would tilt the weighting toward the warm end, and how much that raises the cold-end reading for a stated cable is a measurement the chain is already built to make.

The line as a thermometer of itself. A line’s output noise depends on its whole profile, so a measured output noise, read at both ends, constrains the profile. Two readings cannot fix a profile, but they can fix a two-parameter one; how well, given the noise of the readings themselves, would say whether a cable’s noise can tell where it is warm.

A reflection at a thermal anchor. Every stage a line is anchored to is a connector, and every connector reflects a little. A small reflection at the 4 kelvin anchor sends part of the cold end’s own emission back up the line and returns part of what was travelling down, and whether that matters beside the gradient’s own factor of two is the measurement that would finish the account of a real input line.