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7.7 — Transmission Lines and Impedance Matching
Coaxial cable is sold as "50 ohm" or "75 ohm". Measure a 10 m length with a multimeter and you find a fraction of an ohm between the conductors' ends and near-infinite resistance between inner and outer. Neither reading is 50 ohms.
The number is not a resistance you can measure with a meter. It is the ratio of voltage to current for a wave travelling along the cable, and understanding what that means is the whole of this chapter.
1. When a wire stops being a wire
Chapter 1.2 assumed the lumped element model: a wire has no length, and the voltage at one end equals the voltage at the other, instantly.
That is an approximation, and it fails when the wire is long enough that the signal's travel time matters.
\text{lumped model valid when} \quad \ell \lt \frac{\lambda}{10}
Worked examples of where the boundary falls:
| Signal | Wavelength | Limit |
|---|---|---|
| 50 Hz mains | 6000 km | 600 km |
| 1 MHz | 300 m | 30 m |
| 100 MHz | 3 m | 30 cm |
| 1 GHz | 30 cm | 3 cm |
| 5 GHz clock | 6 cm | 6 mm |
At 5 GHz, a 6 mm circuit board track is a transmission line. Which is why high-speed digital design is a microwave discipline in disguise, and why the layout of a fast board is as important as its schematic.
For a digital signal the relevant thing is the edge rate, not the clock rate. A 100 MHz clock with 1 ns edges contains energy up to roughly
f_{knee}=\frac{0.35}{t_{rise}}=350\ \text{MHz}
So the transmission-line limit is set by that, not by 100 MHz — and this is why slowing the edges deliberately (Chapter 3.6) fixes so many signal integrity problems.
2. The distributed model
A transmission line is modelled as an infinite chain of small sections, each with series inductance and resistance, and shunt capacitance and conductance — all quoted per unit length.
- L — inductance per metre (henries/m)
- C — capacitance per metre (farads/m)
- R — series resistance per metre
- G — shunt conductance per metre
For a lossless line (R = G = 0), which is a good approximation for a short good-quality cable:
\boxed{Z_0=\sqrt{\frac LC}}
Read that carefully. Z_0 is a ratio of an inductance to a capacitance, and its units work out to ohms even though nothing is dissipating.
The physical meaning. Launch a step of voltage into a very long line. The leading edge charges the line's capacitance as it advances, and the current that does the charging flows through the line's inductance. The ratio of the voltage to that current is Z_0, and it is constant while the wave is travelling.
So the source cannot tell the difference between an infinitely long 50 Ω line and a 50 Ω resistor — both draw the same current for the same voltage. The line is not dissipating the energy; it is storing it and carrying it away. But from the source's point of view they are identical.
The propagation velocity:
v=\frac{1}{\sqrt{LC}}=\frac{c}{\sqrt{\varepsilon_r}}
The velocity factor 1/\sqrt{\varepsilon_r}:
| Cable | \varepsilon_r | Velocity factor |
|---|---|---|
| Air-spaced | 1.0 | 1.00 |
| Foam polyethylene | 1.5 | 0.82 |
| Solid polyethylene | 2.3 | 0.66 |
| PTFE | 2.1 | 0.69 |
| FR4 circuit board | 4.2 | 0.49 |
Signals on a circuit board travel at about half the speed of light, which is 6.7 ps per millimetre. That number is the basis of every timing calculation on a fast board, and it is why matched-length routing exists — differences in track length become differences in arrival time.
3. Reflections
A wave travelling down a line meets the load. If the load impedance does not equal Z_0, some of the wave bounces back.
\Gamma=\frac{Z_L-Z_0}{Z_L+Z_0}
The three cases worth knowing by heart:
| Load | \Gamma | What happens |
|---|---|---|
| Z_L=Z_0 (matched) | 0 | nothing reflects, all power absorbed |
| Z_L=\infty (open) | +1 | full reflection, same polarity |
| Z_L=0 (short) | -1 | full reflection, inverted |
Why an open circuit reflects positively. At an open end the current must be zero. The incident wave carries current, so the reflected wave must carry an equal and opposite current to cancel it — and a wave with reversed current and unreversed voltage is a wave travelling backwards with the same voltage polarity. The voltage therefore doubles at the open end, which is a real and sometimes destructive effect.
Why a short reflects negatively. At a short the voltage must be zero, so the reflected wave must cancel the incident voltage — the polarity inverts.
Ringing on a digital line
A fast digital signal driven into an unterminated track:
- The driver launches a step. Its own output impedance and the track's Z_0 form a divider, so a partial step travels down.
- At the open receiver end, \Gamma=+1, so it reflects and doubles.
- The doubled wave travels back to the driver, where the low source impedance gives \Gamma\approx-1, so it inverts.
- Round and round, decaying as losses take their toll.
The result is ringing on every edge, and if the overshoot exceeds the receiver's absolute maximum input it damages the chip, while undershoot below the threshold can cause a false extra clock edge.
Termination, three ways:
Series termination. A resistor at the driver, chosen so that R_{source}+R_{series}=Z_0. The reflection from the far end is absorbed on its return. Costs nothing in static power, and it is the standard choice for point-to-point digital lines.
Parallel termination. A resistor of Z_0 at the receiver to ground. Perfect absorption, and it burns DC current continuously — for a 3.3 V signal into 50 Ω that is 66 mA, which is unacceptable in most designs.
Thevenin termination. Two resistors, to supply and ground, whose parallel combination is Z_0. Halves the current and biases the line to mid-rail, which suits some logic families.
AC termination — a resistor in series with a capacitor — passes no DC at all but adds a time constant.
A reflected wave meeting the wave still arriving does not simply add up to a mess. It produces a stationary pattern, and that pattern is what an instrument measures.
The mathematics is Volume IV Chapter 2.4's: adding a rightward and a leftward wave of equal frequency gives 2A\sin(kx)\cos(\omega t), in which position and time sit in separate factors, so the shape stops travelling and only breathes.
On a cable this is not a curiosity, it is a fault report. A perfectly matched load absorbs everything, nothing comes back, and there is no standing wave at all — the voltage envelope along the cable is flat. A mismatch sends part of the wave back, the two interfere, and the envelope develops peaks and troughs. The ratio of the biggest envelope voltage to the smallest is the standing wave ratio, and it is the number on the meter every radio installer watches:
\text{SWR} = \frac{V_{max}}{V_{min}} = \frac{1+|\Gamma|}{1-|\Gamma|}
An SWR of 1 means a perfect match and no reflection. An SWR of 3 means a substantial fraction of your transmitter power is coming back down the cable instead of leaving the antenna.
That returning energy has to go somewhere, and in a high-power transmitter the place it goes is the output stage — the last set of transistors before the cable, which were designed to deliver power rather than absorb it. This is why transmitters have SWR protection that reduces output when the reflection rises. Without it, a disconnected antenna destroys the transmitter within seconds.
4. Standing waves
When a reflected wave travels back along the line, it adds to the forward wave. At some points they reinforce and at others they cancel, producing a stationary pattern of maxima and minima — a standing wave.
\text{SWR}=\frac{V_{max}}{V_{min}}=\frac{1+|\Gamma|}{1-|\Gamma|}
The pattern repeats every half wavelength, because that is the distance over which the round trip adds a full cycle of phase.
Why standing waves matter, in three concrete ways:
Voltage stress. At a maximum the voltage can be much higher than the source produced. A cable rated for 2 kV can arc internally with a 500 V transmitter and a badly mismatched antenna, and the failure is not at the antenna but somewhere along the cable at a voltage maximum.
Extra loss. The reflected wave travels back along the line and suffers the cable's loss again, so a mismatched line loses more than a matched one carrying the same power.
Impedance transformation. The impedance seen looking into a line depends on where you are along it. Section 5 makes this useful.
5. Line length as a component
Z_{in}=Z_0\frac{Z_L+jZ_0\tan\beta\ell}{Z_0+jZ_L\tan\beta\ell}, \qquad \beta=\frac{2\pi}{\lambda}
A messy formula whose special cases are extremely useful.
A half-wave line (\ell=\lambda/2): \tan(\pi)=0, so
Z_{in}=Z_L
A half wavelength of line repeats whatever is on the end, regardless of Z_0. Useful for placing a measurement point where the load is inaccessible.
A quarter-wave line (\ell=\lambda/4): \tan(\pi/2)=\infty, and taking the limit:
\boxed{Z_{in}=\frac{Z_0^2}{Z_L}}
A quarter wave inverts the impedance. A short becomes an open; an open becomes a short.
And it is a transformer. To match Z_L to Z_{in}:
Z_0=\sqrt{Z_{in}Z_L}
Worked example. Match a 100 Ω antenna to a 50 Ω system:
Z_0=\sqrt{50\times100}=70.7\ \Omega
A quarter wavelength of 75 Ω cable is close enough, and 75 Ω cable is a standard product. A matching network made of nothing but a length of cable.
The quarter-wave short is worth its own note. A quarter-wave line shorted at the far end presents an open circuit at the near end — at the design frequency only. At DC it is a dead short. So a quarter-wave shorted stub across a feed line passes DC and blocks the signal frequency, which is exactly what a lightning-protection stub on a radio mast does: it shorts static and lightning to earth while being invisible to the signal.
The Smith chart is a graphical device for solving these impedance transformations, invented by Phillip Smith in 1939. It maps the entire infinite impedance plane onto a finite disc, so any impedance can be plotted, and moving along a transmission line becomes a rotation about the centre. Before computers it was indispensable; it survives because it makes the behaviour visible in a way that a numerical answer does not.
6. Loss
Real lines lose power to two mechanisms.
Conductor loss comes from resistance, and it grows as \sqrt f because of the skin effect: at high frequency, current crowds into a thin layer at the conductor's surface.
\delta=\sqrt{\frac{\rho}{\pi f\mu}}
Worked example. Copper at 1 GHz: $\delta = 2.1\ \mu$m. The current flows in a layer thinner than a human hair, so the effective cross-section is tiny and the resistance is far higher than the DC value.
This is why high-frequency conductors are silver-plated — only the surface carries current, so plating the surface with a better conductor works. It is also why a thick solid wire is no better than a thin tube at radio frequencies.
Dielectric loss comes from the insulator absorbing energy, and it grows linearly with frequency.
Total loss is therefore dominated by conductor loss at lower frequencies and by dielectric loss above a crossover, and this is why cable specifications list loss per 100 m at several frequencies rather than a single number.
Typical figures for RG-58, a common thin coaxial cable, per 100 m:
| Frequency | Loss |
|---|---|
| 10 MHz | 4.5 dB |
| 100 MHz | 16 dB |
| 1 GHz | 60 dB |
Sixty decibels is a factor of a million. A 100 m run of RG-58 at 1 GHz delivers one millionth of the power fed into it, which is why microwave links use waveguide, or put the electronics at the antenna, or use much better cable.
7. Waveguide
Above a few gigahertz, coaxial cable's loss becomes intolerable and a hollow metal pipe works better.
A waveguide has a cutoff frequency below which nothing propagates at all:
f_c=\frac{c}{2a}
for the dominant mode in a rectangular guide of width a.
Worked example. WR-90 guide, 22.86 mm wide:
f_c=\frac{3\times10^8}{2\times0.02286}=6.56\ \text{GHz}
Its usable band is 8.2 to 12.4 GHz — comfortably above cutoff and below the frequency where the next mode would start propagating and cause trouble.
Why the cutoff exists. The wave propagates by bouncing between the walls at an angle, and the geometry only works if the guide is wide enough for the wave to fit sideways. Below cutoff the wave cannot satisfy the boundary conditions and decays exponentially instead of propagating.
Waveguide's advantages: very low loss, since there is no centre conductor to heat and often no dielectric at all, and it handles enormous power. Its disadvantages: rigid, bulky, expensive, and it works only over its band.
Where you meet it: radar, satellite earth stations, and — most familiarly — the short rectangular tube carrying energy from the magnetron to the cooking cavity in a microwave oven.
8. Why 50 and 75 ohms
Two standards, and both come from a genuine physical optimum for coaxial cable.
For air-filled coax, working through the geometry:
- Minimum loss occurs at a ratio of outer to inner diameter giving 77 Ω.
- Maximum power handling — the highest voltage before the air breaks down — occurs at 30 Ω.
- Maximum voltage breakdown for a given size is around 60 Ω.
75 Ω was chosen for receiving and broadcast applications, where minimising loss matters and power is small. It also happens to be close to a dipole's 73 Ω impedance, which is convenient.
50 Ω is roughly the geometric mean of the loss optimum and the power optimum:
\sqrt{30\times77}=48\ \Omega
A compromise between low loss and high power handling, and it became the universal standard for transmitting equipment, test instruments and radio systems.
93 Ω existed for low-capacitance applications and is now essentially extinct.
The practical warning. Mixing 50 and 75 Ω systems gives \Gamma = 0.2, an SWR of 1.5 and 4% reflected — about 0.18 dB. Usually harmless, and worth knowing so you do not spend an afternoon worrying about it. What is not harmless is the connector: a 75 Ω BNC pushed into a 50 Ω socket damages both, because the centre pin diameters differ.
9. Time-domain reflectometry
A genuinely useful measurement that falls straight out of this chapter.
Send a fast pulse down a cable and watch what comes back. Any impedance discontinuity reflects a portion, and the timing tells you where it is:
d=\frac{v\,t}{2}
with the factor of 2 for the round trip.
Worked example. A reflection arrives 340 ns after the pulse is sent, on a cable with a velocity factor of 0.66:
d=\frac{0.66\times3\times10^8\times340\times10^{-9}}{2}=\frac{67.3}{2}=33.7\ \text{m}
The fault is 33.7 m along the cable.
And the reflection's polarity identifies what kind of fault it is. Positive means an increase in impedance — a break, a corroded joint, a stretched cable. Negative means a decrease — a short, water ingress, a crushed section.
This is how buried cables are located and diagnosed without digging them up, how faults are found in undersea cables, and — with a much faster pulse — how signal integrity problems are traced on a circuit board. The same technique with radar pulses instead of cable pulses is ground-penetrating radar.
Part 8 takes everything from Part 7 and follows it into the systems you use every day — how a call reaches a tower, how GPS works without any internet at all, and what actually happens when you tap a card on a reader.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Transmission lines
Lumped model valid when \ell \lt \lambda/10.
Digital knee frequency:
f_{knee}=\frac{0.35}{t_{rise}}
The edge rate, not the clock rate, decides whether a track is a transmission line.
Z_0=\sqrt{\frac LC} \qquad\text{(lossless)}, \qquad Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}} \qquad\text{(general)}
v=\frac{1}{\sqrt{LC}}=\frac{c}{\sqrt{\varepsilon_r}}
Coaxial cable:
Z_0=\frac{60}{\sqrt{\varepsilon_r}}\ln\frac ba
Input impedance of a terminated line:
Z_{in}=Z_0\frac{Z_L+jZ_0\tan\beta\ell}{Z_0+jZ_L\tan\beta\ell}, \qquad \beta=\frac{2\pi}{\lambda}
Half wave (\beta\ell=\pi): Z_{in}=Z_L — repeats the load.
Quarter wave (\beta\ell=\pi/2): Z_{in}=\dfrac{Z_0^2}{Z_L} — inverts, and acts as a transformer with
Z_0=\sqrt{Z_{in}Z_L}
Skin depth:
\delta=\sqrt{\frac{\rho}{\pi f\mu}}
Copper at 1 GHz: 2.1 µm. Conductor loss grows as \sqrt f; dielectric loss grows as f.
Waveguide cutoff:
f_c=\frac{c}{2a}
Time-domain reflectometry:
d=\frac{v\,t}{2}
Positive reflection = impedance increase (break, corrosion). Negative = decrease (short, water).
Why 50 and 75 ohms: for air coax, minimum loss at 77 Ω, maximum power at 30 Ω, and \sqrt{30\times77}=48\ \Omega — so 50 Ω is the compromise, and 75 Ω is the loss optimum used where power is small.
Fourteen worked problems next.
What the next chapter fixes
Part 7 has built the components of a radio link in isolation. Part 8 assembles them into the systems you actually use, and asks the questions the theory alone cannot answer: how a phone call follows you between towers, how a satellite tells you where you are without ever hearing from you, and why your headphones and your router share a band and mostly cope.