The heat a standing wave leaves

The mismatch that the cable hides found that a lossy line improves any reflection it carries by twice its one-way loss: a 4:1 load at the end of twenty metres of ordinary coaxial cable reads a return loss of 24 dB at the near end and passes an acceptance test the load itself could never pass. The cable that hides two things priced the deception in noise figure — a decibel of floor for every two of apparent match. Both essays counted what reaches the instrument at the near end. Neither asked what happened to the reflected power that did not.

It went into the cable. A reflection that leaves the load and does not arrive back at the source has been dissipated on the way, and that dissipation is heat in the line’s conductor and dielectric, in addition to whatever the forward wave was already losing. So a mismatched load makes a lossy line lossier, and by an amount that has a closed form. The amount runs between two limits that say different things about what the line is doing: on a short line the standing wave multiplies the line’s own dissipation; on a long one it adds the load’s mismatch loss to it.

And the heat is not spread evenly. It stands where the current stands.

Forward, back, and what is left #

The bookkeeping is in waves. Let the line have a matched loss $a$, a power ratio, so that a wave travelling its length arrives with $1/a$ of the power it set out with. Let the load reflect a fraction $|\Gamma|^2$ of the power that reaches it. For one unit of power to be absorbed by the load, the forward wave must arrive with $1/(1 - |\Gamma|^2)$ and the reflected wave leaves the load with $|\Gamma|^2/(1 - |\Gamma|^2)$. Carry both to the input: the forward wave there is $a$ times larger and the reflected wave $a$ times smaller. On a line whose characteristic impedance is real, the net power flowing in at the input is the difference of the two:

$$\frac{P_{in}}{P_{load}} = \frac{a - |\Gamma|^2/a}{1 - |\Gamma|^2} = \frac{a^2 - |\Gamma|^2}{a\,(1 - |\Gamma|^2)},$$

against $a$ for a matched load. The ratio of the two is the extra loss the standing wave puts on the line. At a standing-wave ratio of 4, $|\Gamma| = 0.6$, on a cable of 3 dB of matched loss — $a = 2$ — it is $(4 - 0.36)/(2 \times 0.64) = 2.84$, which is 4.54 dB against the matched 3.01: 1.53 dB more.

The closed form assumes a real characteristic impedance, which a line with loss does not quite have. So the hero figure’s dots solve the line exactly: a 50 Ω line at a gigahertz whose loss is all series resistance, its propagation constant and complex characteristic impedance taken from the telegrapher’s equations without approximation, the input voltage and current found from the load by the line’s own two-port matrix, and the input power read directly. Across matched losses from 0.1 to 20 dB the exact line’s extra loss agrees with the closed form to a few hundredths of a decibel, the size of the low-loss approximation in the closed form.

Short lines multiply, long lines add #

The formula has two limits, and they describe different mechanisms.

On a short line with little loss, every bit of length carries both the forward wave and the reflected wave at nearly full strength. To deliver one unit, the forward wave carries $1/(1 - |\Gamma|^2)$ and the reflected wave $|\Gamma|^2/(1 - |\Gamma|^2)$, and the line’s resistance dissipates in proportion to the total, $(1 + |\Gamma|^2)/(1 - |\Gamma|^2)$. At $|\Gamma| = 0.6$ that is 2.12: a short cable feeding a 4:1 load runs more than twice as hot as the same cable matched, for the same power delivered. The figure’s left-hand values are that expression for every reflection on it.

On a long line the reflected wave does not survive the trip back. Nearly all of what the load reflects is dissipated before it reaches the source, so the line’s dissipation is its matched dissipation plus the reflected power itself, and the multiple falls to $1/(1 - |\Gamma|^2)$ — 1.56 at $|\Gamma| = 0.6$. In decibels the extra loss then equals the load’s mismatch loss, $10\log(1/(1 - |\Gamma|^2)) = 1.94$ dB, which is the hero figure’s ceiling: on a 20 dB cable the 4:1 load costs 1.94 dB, and the reflected power that a lossless line would have returned to the source has become heat instead.

That is the price of the deception the first essay found. The return loss read at the source improves by twice the cable’s loss because the reflected power is being dissipated; the same dissipation shows up as extra insertion loss for anyone who measures the power that reaches the load. A cable cannot make a mismatch disappear. It can only choose whether the reflected power arrives back at the source, where an instrument sees it, or stays in the cable, where it is heat.

The heat stands where the current stands #

The averages hide something a power rating cannot ignore. On a line whose loss is in its conductor, the dissipation per unit length is $R'|I(z)|^2/2$, and with a standing wave the current’s magnitude varies along the line between maxima and minima a quarter wavelength apart.

The exact solution shows it directly. Two wavelengths of line with a fifth of a decibel of loss, all in its conductor, feeding 200 Ω and delivering the same power as the matched line beside it: the matched line’s dissipation is flat along its length, falling by the line’s own small attenuation; the mismatched line’s has peaks every half wavelength at 4.04 times the matched value and troughs at 0.250. The peaks are the standing-wave ratio, 4.00, to the small correction the line’s loss makes over two wavelengths, and the troughs its reciprocal. The average along the line is 2.15, against $(1 + |\Gamma|^2)/(1 - |\Gamma|^2) = 2.125$, the small difference again the attenuation along two wavelengths. Integrated, the dissipation profile equals the power that went in less the power that came out to four parts in ten billion — the two routes through the exact solution, one a difference at the two ends and one a sum along the way, agreeing.

The reason the peak is exactly the standing-wave ratio is short. At a current maximum the forward and reflected currents add, so $|I|^2$ is $(1 + |\Gamma|)^2$ times the forward wave’s; the forward wave itself must carry $1/(1 - |\Gamma|^2)$ times the delivered power; and the product $(1 + |\Gamma|)^2/(1 - |\Gamma|^2)$ is $(1 + |\Gamma|)/(1 - |\Gamma|)$, the standing-wave ratio.

What a power rating has to be derated by #

A cable’s power rating is a temperature limit: the inner conductor may not exceed some temperature, and the rating is the matched power at which it reaches it. With a mismatch, two numbers now describe the heat.

The hottest point runs at the standing-wave ratio times the matched line’s dissipation, and the exact line confirms it at ratios of 1.5, 2 and 4 to within two per cent. The average runs at $(1 + |\Gamma|^2)/(1 - |\Gamma|^2)$: 1.25 at a ratio of 2, 2.12 at 4, 5.05 at 10. Which of the two sets the rating depends on how heat moves along the conductor. A copper inner conductor conducts heat well along its length, and over the few centimetres between a current maximum and a minimum at a gigahertz the peaks and troughs partly average out; at tens of megahertz the peaks are metres apart and each one is its own hot spot. A rating that holds for a matched line therefore needs derating by somewhere between the average and the standing-wave ratio, and the conservative figure — the one that cannot be wrong — is the ratio itself: a 4:1 load halves the margin the average suggests.

Dielectric loss has the same structure on the voltage’s standing wave, whose maxima sit a quarter wavelength from the current’s. On a line whose loss is shared between conductor and dielectric, the peaks of the two kinds of heat fall at different places and partly fill each other’s troughs, which is one reason the derating curves that cable makers publish lie between the two lines on this figure.

A standing-wave ratio that changes along the line #

There is a trap in reading any of this off an instrument at the near end, and it follows from the first essay’s result. On a lossy line the reflection shrinks by twice the loss between the load and the source, so the standing-wave ratio is not one number: it is largest at the load and smallest at the source. A 4:1 load at the end of a 3 dB cable has a ratio of 4.00 at its terminals and, with $|\Gamma|$ halved by the round trip to 0.3, a ratio of 1.86 at the instrument.

The hot spots follow the local ratio, so the cable runs hottest near the load and coolest near the source — up to four times a matched cable’s dissipation at the far end, under twice it at the near end. A user who measures 1.86 at the transmitter and derates the cable by 1.86 has derated it correctly for the end that was measured and by half too little for the end that was not. The resistor at the wrong end found a transmission line looking respectable at its two ends and wrong in its middle; here the measurement is made at one end and the damage is done at the other.

The same structure appears in time. The staircase in time drew a mismatched line’s response to a step as a series of reflections bouncing between its ends; on a lossy line each bounce is smaller than the last, and the energy the bounces lose is the heat this essay’s figures add up. A pulse into a mismatched lossy line heats it more near the end that reflects, for the same reason a sine wave does.

Matching, priced in heat #

The usual case for matching a load to a line is about power delivered — the load that takes the most derived the conjugate match that maximises it — and about the reflection a source sees. The heat in the line is a third argument, and on a high-power feed it is often the deciding one. A transmitter feeding an antenna through a long cable at a standing-wave ratio of 2 loses, on a 1 dB cable, 0.20 dB more than matched; that is a few per cent of the power, of no consequence to the link budget. But the cable’s dissipation near the antenna runs at twice its matched value, and if the cable was sized to its matched rating with little margin it is now running above it there. The case for a matching network at the antenna end, rather than at the transmitter, is that it removes the standing wave from the cable altogether, where a match at the transmitter only makes the transmitter happy. It is the radio-frequency version of the choice terminated at both ends weighed for a digital line — which end gets the resistor — with heat in place of edge quality as the thing being protected.

Where a matched source’s power goes #

The two limits are easiest to see as a budget. A matched source launches one unit of power into the line; the line takes some going out, the load takes some, the line takes some of the reflection coming back, and the source absorbs what returns.

With no loss, 64 per cent reaches the load and 36 per cent — the load’s $|\Gamma|^2$ — returns to the source and is absorbed there. Add loss and both the returning power and the delivered power fall, but the returning power falls twice as fast in decibels, since it crosses the line twice: at 3 dB of cable, 9.0 per cent returns and 9.0 per cent is burnt on the way back; at 10 dB, 0.36 per cent returns and 3.24 per cent is burnt. The source sees the returning sliver and reports an excellent match. The cable has quietly absorbed nine-tenths of the reflection. And on the exact line, at 3 dB, the load receives 35.2 per cent of the power that entered, against the bookkeeping’s 35.3 — the same accounting, with the line’s slightly complex impedance in it.

The termination that runs out counted a termination’s standing power, the 75.6 mW a parallel resistor burns holding a line at its level; a mismatched lossy line has its own standing power of a different kind, burnt not in a resistor at one end but along the whole length of the conductor.

The cable’s loss is a function of its load #

The practical lesson is that a cable’s loss figure is a measurement condition, not a property. A data sheet gives attenuation in decibels per metre with both ends matched — the only condition under which the number means anything — and a user who multiplies it by a length and subtracts it from a transmitter’s power has computed what reaches a matched load. Into a 4:1 antenna on a 3 dB cable the same arithmetic is 1.53 dB optimistic, and into the same antenna on a 1 dB cable, 0.82 dB. The error is largest, relative to the cable’s own loss, on short low-loss runs, where it is the factor $(1 + |\Gamma|^2)/(1 - |\Gamma|^2)$ on the cable’s dissipation rather than a few tenths of a decibel on top of a long line’s large loss.

The mismatch that the cable hides said that a long lossy cable makes a bad load look good. This essay says the cable pays for the illusion: the reflected power is heat in the cable, the cable runs hotter than its matched rating assumes, and the delivered power is lower than the matched loss figure predicts. A measurement at the near end sees the improvement; a thermometer on the cable, or a power meter at the far end, sees the cost.

The model leaves out two things worth naming. The loss is taken as a fixed fraction per unit length, which for a conductor at a gigahertz follows the square root of frequency through the skin effect, as the conductor made causal too required it to; a standing wave changes nothing about that law, only about where the current is. And the load is a fixed impedance. A load that heats, or a cable whose resistance rises as it does, has a loss that depends on the power through it, and a standing wave concentrates that dependence at its peaks.

The numbers #

For a load reflecting |Γ| through a line of matched loss a, the line must be fed $(a^2 - |\Gamma|^2)/(a(1 - |\Gamma|^2))$ per unit delivered. A 4:1 load costs 0.109 dB extra on 0.1 dB of cable, 0.82 on 1 dB, 1.53 on 3 dB and 1.94 — the mismatch loss — on 20 dB. A short line’s dissipation is multiplied by (1 + |Γ|²)/(1 − |Γ|²), 2.12 at |Γ| = 0.6; a long line’s by 1/(1 − |Γ|²), 1.56. Solved exactly, the heat on a conductor-loss line peaks at the standing-wave ratio times a matched line’s, 4.04 against 4.00, and averages 2.15, and the integral of the heat equals the power in less the power out to four parts in ten billion. Through 10 dB of cable, 3.24 per cent of a matched source’s launched power is burnt by the reflected wave and 0.36 per cent returns.

Still open: a mixed loss, a hot spot that moves, and the noise a mismatched line adds #

A loss shared between conductor and dielectric. The current’s peaks and the voltage’s are a quarter wavelength apart, so a line with both kinds of loss has a dissipation profile whose peak depends on the split. The peak against the conductor’s share of the loss, at a fixed standing-wave ratio, is the curve a cable maker’s derating chart is an approximation to, and it has a closed form nobody has drawn here.

A hot spot with a thermal length. Whether the conductor’s own thermal conduction averages the peaks depends on the half wavelength against the length over which heat spreads along the conductor. The frequency below which the hot spots are separate, for a given cable, is a number with a thermal conductivity in it.

The noise of a mismatched lossy line. The cable that hides two things found a matched cable adding noise in proportion to its loss. A mismatched cable dissipates more, and a dissipative element at a temperature radiates in proportion to what it absorbs, so a mismatched cable should be noisier than its matched loss figure says, by a factor this essay’s bookkeeping already contains.