Transmission Media: Copper, Fibre and Wireless

Physical and Data Link · 25 min

Core CS · Computer Networks

Two hundred decibels a kilometre, or two tenths of one

Copper, coaxial cable, glass and open air all carry the same bits. They charge for it in decibels per kilometre, and that one figure decides how far a run goes before somebody has to install equipment in the middle of it.

Pick a medium, set the run, count the regenerators
Cat 6 twisted pair loses roughly 200 dB per kilometre at 100 MHz, and more than that at the 250 MHz it is certified to. Single-mode fibre at 1550 nm loses roughly 0.2 dB per kilometre. That is a factor of one thousand, and almost every argument about cable in this course is downstream of it.

01 The idea

Every medium weakens the signal at a fixed rate per metre

Layer 1 has exactly one job: get a physical signal from one end of something to the other. That something is the transmission medium, and it is either a thing you install and the signal stays inside, or it is the open air and the signal goes wherever physics takes it. Copper pair, coaxial cable and optical fibre are the first kind. Radio, microwave, infrared and satellite are the second.

Whichever you pick, three things happen to the signal on the way and all three have precise names. Attenuation is the loss of signal power with distance, so what arrives is a weaker copy of what was sent. Noise is unwanted energy added to the signal from outside it, from a motor, a fluorescent ballast, the pair in the next slot of the same jacket, or the thermal agitation of the conductor itself. Distortion is the signal arriving changed in shape rather than merely smaller, because different frequency components travel at slightly different speeds and no longer line up at the far end.

Attenuation is the one that decides engineering. It is quoted in decibels per kilometre and it accumulates linearly with distance, so a run twice as long loses twice as many decibels. A receiver can only work with a signal above some floor, so there is a distance at which the signal has faded past what the receiver can read, and at that point somebody has to put a box in the middle of the cable that rebuilds the signal. The distance between those boxes is set by the attenuation figure of the medium and by nothing else.

That is why this topic is not a list of cables to memorise. It is one number per medium, and a set of prices you pay to get a better one. Fibre's number is a thousand times better than copper's, and fibre costs more to buy, more to install and much more to repair. Everything in the rest of this lesson is that trade, made concrete.

Every medium loses a fixed number of decibels for every unit of length. That one number sets how far you get before the signal has to be rebuilt, and it is the number on which fibre beats twisted pair by a factor of a thousand.
Guided mediaA physical path you install and the signal stays inside it: twisted pair, coaxial cable, optical fibre. You know where the signal goes, you can cost it per metre, and someone has to dig to change it.
Unguided mediaNo physical path. The signal is radiated into air or space as an electromagnetic wave: radio, microwave, infrared, satellite. Nothing confines it, which is what makes it cheap to deploy and impossible to keep private without encryption.
Decibel (dB)A power ratio on a logarithmic scale, 10 log₁₀ of output power over input power. Engineers use it because losses along a path add instead of multiplying. 3 dB of loss leaves half the power, 10 dB leaves one tenth, 20 dB leaves one hundredth.

02 Worked example

Two kilometres between two buildings, and nineteen repeaters

This is the run for the whole lesson, including the console in section 04 and every figure in the cheat sheet. A campus has to connect two buildings that are 2000 m apart. Somebody suggests pulling Cat 6 twisted pair, because Cat 6 is what is already in the walls and it is the cheapest thing on the shelf. Follow the signal along that cable and the suggestion answers itself.

One convention first, and the console uses the same one. This lesson works to a 20 dB budget: the receiver is assumed to still read a signal that has lost 20 dB and not one that has lost more. Real budgets are computed from a specific transmitter's output power and a specific receiver's sensitivity and land anywhere between about 9 and 30 dB. 20 is a round teaching figure, and it is not an arbitrary one, because it reproduces the real 100 m limit of Cat 6 exactly.

0 m · LaunchThe transmitter drives the pair at its full power. Nothing lost yet. 0 dB
AttenuationCat 6 loses about 0.2 dB per metre at 100 MHz. After 100 m that is 20 dB, and 20 dB down is one hundredth of the launched power.
Noise and distortionA press motor couples in energy the signal never had. Different frequencies arrive at slightly different times. Neither one shrinks as the signal does.
100 m · RegenerateA repeater decides what each bit was and transmits a brand new clean copy at full power. The loss counter goes back to 0 dB.
2000 m · ArrivalTwenty segments of 100 m each, so 19 repeaters sit between them, each needing power, a cabinet and somewhere dry.

Check the arithmetic rather than trusting it. Over 2 km the pair loses 200 dB/km × 2 km = 400 dB. One segment may spend 20 dB, so a segment is 20 ÷ 200 = 0.1 km = 100 m. That gives 400 ÷ 20 = 20 segments, and a repeater sits at every join between segments, so there are 19 of them and not 20. Miscounting that by one is the classic error in this arithmetic.

Now put the same 2 km run on the other three media and read the last column.

Medium over the same 2 kmAttenuation, approximateLoss over 2 kmReach on a 20 dB budgetRegenerators
Cat 6 UTP, 100 MHz200 dB/km400 dB100 m19
RG-6 coaxial, 100 MHz66 dB/km132 dBabout 303 m6
OM3 multimode fibre, 850 nm3 dB/km6 dBabout 6.67 km0
OS2 single-mode fibre, 1550 nm0.2 dB/km0.4 dB100 km0

One caveat on the multimode row before you quote it at anybody: 6.67 km is what attenuation allows, and multimode almost never gets that far, because modal dispersion smears the pulses first. Run OM3 at 10 Gbps and you stop at about 300 m. That gap between the attenuation limit and the dispersion limit is exactly what section 03 unpacks, and it is the one place in this table where the decibel figure is not the binding constraint.

Nineteen boxes, six boxes, none, none. The single-mode run spends 0.4 dB of a 20 dB budget and has 19.6 dB left over for the connectors, the splices and the next twenty years of repairs. That is the entire case for fibre on a campus, and it is arithmetic rather than opinion.

One honest clause, because an interviewer will supply it if you do not. You could not actually build the copper version. Structured cabling standards cap a horizontal copper run at 100 m end to end, and classic Ethernet capped a repeated collision domain at four repeaters, so nineteen was never legal. The number is not a design; it is the price tag on a design nobody would sign off. Its job is to show what the 200 dB per kilometre actually buys you.

03 Mechanics

Nine media, and the six properties that decide between them

A medium is chosen on six things, and only six. Usable bandwidth, meaning the range of frequencies the medium carries without falling apart, which is what a data rate is then built on top of. Bandwidth is measured in hertz and throughput in bits per second, and they are not the same quantity. Distance before regeneration, which is the attenuation figure from section 02. Immunity to electromagnetic interference, how much of the outside world leaks in. Security against tapping, how hard it is to read the signal without touching the endpoints. Installation difficulty and cost. Read the guided table across a row and the pattern is the same every time: as you move down it, the first four get better and the last two get worse.

Guided mediumUsable bandwidthDistance before regenerationEMI immunityTap resistanceInstall and cost
UTP
unshielded twisted pair
16 MHz on Cat 3 up to 500 MHz on Cat 6A, and 2000 MHz on Cat 8 over a short rack link. 10 Gbps over 100 m at Cat 6A. 100 m per structured cabling standards Poor. The twist cancels a lot but there is no shield at all. Weakest. It radiates, and a clip-on inductive tap needs no cut. Cheapest of everything. RJ45 plugs, a crimp tool, no specialist.
STP / FTP
shielded twisted pair
Same categories as UTP, but holds its rating better in a noisy plant. 100 m, the same limit Good. A foil or braid intercepts interference and carries it to ground. Better than UTP. Much less radiates out of the shield. Stiffer, bulkier, and it only works if the shield is properly earthed.
Coaxial
RG-6, RG-58
Up to about 1 GHz on RG-6, which is what cable television is built on. Hundreds of metres. 10BASE5 allowed 500 m, 10BASE2 185 m. Good. The outer conductor completely surrounds the inner one. Harder. You have to break into the shield to reach the core. Heavy, poor bend radius, F or BNC connectors. Mid cost.
Multimode fibre
OM3, 50 µm core
Very high, but capped by modal dispersion: 10 Gbps to about 300 m on OM3. Hundreds of metres to a few kilometres, dispersion limited before attenuation is. Total. Glass carries light; there is no current for a field to induce. Very hard. Bending it to leak light shows up as a measurable loss. Needs polished or fusion-spliced terminations. The wide core is far more forgiving to align than single-mode.
Single-mode fibre
OS2, 9 µm core
Highest in practice. One fibre carries many wavelengths at once by wavelength multiplexing. 10 km to 80 km on ordinary Ethernet optics, with no repeater at all Total, for the same reason. Hardest of all. Any tap is an insertion loss somebody can measure. Dearest. A 9 µm core needs a fusion splicer and a laser transmitter, and every repair costs a technician visit.

The category numbers on twisted pair are not a list to memorise either. Each step is a higher frequency the cable is certified to carry over a full 100 m channel, and the Ethernet rate is a consequence of that frequency rather than a separate fact. Read this table as one number going up.

CategoryCertified toWhat it carries over 100 m
Cat 316 MHz10BASE-T, and the telephone cabling of an entire generation of offices.
Cat 5e100 MHz1000BASE-T. The first category that carried a gigabit the full distance.
Cat 6250 MHz1000BASE-T over 100 m, and 10GBASE-T only to about 55 m.
Cat 6A500 MHz10GBASE-T over the full 100 m. The A is for augmented, and it is the current office default.
Cat 82000 MHz25 and 40 Gbps, but only to about 30 m. A data-centre rack cable, not a building cable.

Unguided media are chosen on a different set of questions, because there is no cable to cost and no route to dig. What matters is whether the two ends can see each other, what else is transmitting nearby, and what the weather does.

Unguided mediumTypical bandRangeLine of sight neededWhat degrades itWhat it costs you
Radio
broadcast, Wi-Fi
3 kHz to about 1 GHz for classic radio; Wi-Fi sits at 2.4, 5 and 6 GHz. Tens of metres indoors, kilometres for a broadcast tower. No. Long wavelengths bend around and pass through building-scale obstacles. Everything else on the same band. It is shared, and it is omnidirectional. Zero privacy. Anyone in range receives the signal, so encryption is mandatory rather than optional.
Terrestrial microwave
dish to dish
Roughly 1 GHz to 300 GHz. Short wavelengths focus into a narrow beam. Tower to tower, typically 40 to 60 km, set by mast height and the curvature of the earth. Yes. The two dishes must physically see each other. Rain absorbs it badly above about 10 GHz, which is called rain fade. Two masts and planning permission, but no trench between them.
Infrared
remote controls, IrDA
Roughly 300 GHz to 400 THz, just below visible light. One room. Metres, not kilometres. Yes, or a bounce off the ceiling. Walls, doors and partitions are opaque at these wavelengths. Sunlight, which is a very strong infrared source and swamps it outdoors. Almost nothing, and the containment is a genuine security property rather than a limitation.
Satellite
geostationary relay
Microwave, commonly the C, Ku and Ka bands from about 4 to 40 GHz. A third of the planet from one satellite. Yes, to the satellite. A GEO dish needs a clear view of the equatorial sky. Rain fade, and an unavoidable propagation delay that no bandwidth reduces. The altitude. 35 786 km up and the same back down is about 239 ms one way before a byte is processed.

Why the pair is twisted, and why the twist is the entire design. A receiver on a twisted pair does not measure either wire against ground. It measures the difference between the two wires, which is what differential signalling means. Now run the two wires straight and parallel past a motor: the nearer wire picks up more induced voltage than the further one, the difference between them changes, and that change is indistinguishable from signal. Twist them, and each wire spends half the run on the near side of the motor and half on the far side, so both accumulate the same induced voltage. The receiver subtracts one from the other, the common part cancels, and only the real signal survives. This is also why cabling standards limit how much you may untwist at a jack, to roughly half an inch: every untwisted millimetre is a millimetre where the cancellation stops. And it is why the four pairs inside one jacket have different twist rates, so that adjacent pairs do not stay in step with each other and couple, which is the interference called crosstalk.

UTP or STP, and the way the shield backfires. UTP relies on the twist alone. STP adds a foil or braid, either around each pair or around the whole bundle, that intercepts interference and carries it to ground before it reaches the conductors. The catch is the phrase to ground. The shield must be earthed, and it must be earthed properly, which normally means at one end only. Earth it at both ends of a long run between two buildings whose earths sit at slightly different potentials and a current flows continuously along the shield: a ground loop, which is itself a noise source. A badly installed STP link is worse than a plain UTP one, so the shield is a decision and not an upgrade.

Coaxial cable, and what the geometry buys. Coax is a centre conductor, a dielectric spacer, an outer conductor of foil or braid, and a jacket. The outer conductor is doing two jobs at once: it is the return path for the current, and it is a shield that completely surrounds the inner conductor along the whole length. External fields terminate on the outside of the braid and never reach the core, and the signal's own field is confined inside the dielectric instead of radiating away. That is why coax carries far higher frequencies far further than an open pair, and why cable television has run gigahertz down it for decades. Two impedances are standardised and you should know both: 50 Ω for radio, instrumentation and the old 10BASE5 and 10BASE2 Ethernets, and 75 Ω for video and cable broadband.

Total internal reflection, stated so it survives a follow-up. A fibre is a core of glass surrounded by a cladding of glass with a lower refractive index. Light travelling in the core meets the boundary with the cladding at some angle. When the angle of incidence measured from the normal exceeds the critical angle, none of the light crosses into the cladding: all of it is reflected back into the core. That is total internal reflection, and it happens only because the cladding index is lower. The cladding is not a coating and not a mirror; it is the thing that makes the core guide at all. Put a number on it. With a core index of 1.48 and a cladding index of 1.46, the critical angle is arcsin(1.46 / 1.48) = 80.6° from the normal, which is only 9.4° away from straight down the fibre. Rays must travel nearly along the axis, which is exactly why the core is narrow and why the light you couple in has to arrive within a small cone. The same index also fixes the speed: 3 × 10^8 / 1.48 = 2.03 × 10^8 m/s, which is where the roughly 2 × 10^8 m/s used everywhere in this course comes from.

Single-mode against multimode, honestly. Multimode has a wide core, 50 µm on OM3 and OM4 or 62.5 µm on older OM1. Wide enough that light can travel along several distinct paths, called modes: one straight down the axis, others zig-zagging at steeper angles. The zig-zag path is physically longer, so its light arrives later, and one pulse launched cleanly arrives smeared across time. That is modal dispersion, and smear it enough and consecutive pulses overlap until the receiver cannot separate them. It is dispersion, not attenuation, that caps multimode distance in practice. Single-mode has a core of about 9 µm, narrow enough that only one path can propagate at the wavelength used, so there is no modal dispersion to accumulate; it needs a laser, because you cannot usefully couple a broad LED into a 9 µm target, and it runs tens of kilometres. Both types have a 125 µm cladding, so the two are physically identical from the outside and the connectors are interchangeable. That is precisely why they get mixed up on site.

A repeater is not an amplifier, and the difference is noise. An amplifier multiplies everything it receives by a gain, signal and the noise already riding on it alike, so the ratio between them is unchanged and every amplifier adds a little noise of its own. Chain ten of them and the noise has compounded ten times. A repeater is digital: it decides what each bit was, discards the waveform entirely, and transmits a brand new clean bit at full power. Noise does not accumulate across a chain of repeaters, because nothing is carried forward except the decision. That is why the console counts regenerators rather than boosters, and it is why a long analogue chain degrades and a long digital one does not.

Two delays on every link, and the bits-versus-bytes trap that ruins them. Sending a frame costs two separate times that are computed from completely different quantities. Transmission delay is frame size divided by link rate: it is how long the sender takes to clock the bits out. Propagation delay is distance divided by the speed of the signal in the medium: it is how long the first bit takes to fly. Work the 2 km fibre run at 1 Gbps with a 1518-byte frame. Convert first: 1518 bytes × 8 = 12 144 bits. Link rates are powers of ten, so 1 Gbps = 10^9 bits per second, never 2^30 and never bytes. Transmission delay is 12 144 / 10^9 = 12.14 µs. Propagation is 2000 m / (2 × 10^8 m/s) = 10 µs. The last bit lands 22.14 µs after the first one left. Two habits save you here: multiply bytes by 8 before anything else, and never use 3 × 10^8 m/s inside copper or glass, because that is the speed in a vacuum and a real medium runs at about two thirds of it.

What a budget is actually spent on. The 20 dB in this lesson is spent on two different kinds of loss and only one of them grows with length. The per-kilometre part is the attenuation coefficient, and it is what the console models. The fixed part is charged once per component regardless of route length: roughly 0.3 to 0.75 dB for every mated connector pair and about 0.1 dB for every fusion splice. On a short link the fixed part can be most of the budget, which is why a link that passes on paper can fail as built. Signal levels themselves are quoted in dBm, the same logarithmic scale referenced to one milliwatt, so 0 dBm is 1 mW and −20 dBm is 0.01 mW. Because it is logarithmic, signal-to-noise ratio in dB is a subtraction: a signal at −10 dBm over a noise floor at −60 dBm is an SNR of 50 dB.

05 Cheat sheet

The numbers they ask you to produce

Every row is something you can be asked to state or compute in under ten seconds. The right-hand column is the specific wrong answer that gets given, not a general warning.

What they askThe answerThe trap
Horizontal copper run limit100 m end to end: 90 m of solid cable in the wall plus up to 10 m of patch cords100 m of cable and then patch cords on top of it
Fibre core and cladding sizessingle-mode 9 µm core, multimode 50 or 62.5 µm, cladding 125 µm on bothAssuming single-mode fibre is physically thinner. Only the core differs, which is why the connectors are interchangeable.
Condition for total internal reflectioncore index higher than cladding index, and the angle at the boundary above the critical anglesaying the cladding reflects the light like a mirror
What limits multimode distancemodal dispersion, not attenuationAnswering attenuation, which is true of copper and false of multimode fibre.
Why fibre is immune to EMIit carries light through glass, a dielectric, so there is no current for a magnetic field to induceSaying "because it is shielded". It has nothing to shield.
Signal speed in copper or fibreabout 2 × 10^8 m/s, roughly two thirds of the vacuum speedusing 3 × 10^8 m/s inside a cable
Geostationary altitude and one-way delay35 786 km; up and back down at 3 × 10^8 m/s is about 239 msUsing 2 × 10^8 for the satellite hop. That path is near-vacuum, so the vacuum speed is the right one here.
Repeater against amplifiera repeater re-decides the bits so noise does not accumulate; an amplifier raises signal and noise togetherTreating them as the same box with two names.
1 Gbps expressed in bytes10^9 bits per second = 125 × 10^6 bytes per seconddividing by 1024, or forgetting the factor of 8 entirely
dB on the power scale3 dB is half, 10 dB is one tenth, 20 dB is one hundredthAdding decibels as if they were percentages. They are logarithmic, which is exactly why losses add.
Classic coax Ethernet segment lengths10BASE5 thicknet 500 m, 10BASE2 thinnet 185 mReading the 2 in 10BASE2 as 200 m. It is a rounding in the name, not a specification.
The two standard coax impedances50 Ω for radio and legacy Ethernet, 75 Ω for video and cable broadbandQuoting one and assuming it covers both. A 75 Ω cable on a 50 Ω system reflects power back at the transmitter.
Attenuation sets the spacingDecibels per kilometre is the only figure that decides how far a run goes before regeneration. Bandwidth, cost and installation difficulty decide whether you can afford the medium; attenuation decides whether the link is even possible.
Fibre wins four, loses twoBandwidth, distance, EMI immunity and tap resistance all go to fibre by wide margins. Cost and installation difficulty go to copper. There is no property on which fibre is worse at carrying the signal, only properties on which it is worse to own.
Bandwidth never buys back delayPropagation delay is distance divided by speed and contains no bandwidth term anywhere. A gigabit geostationary link and a megabit one have identical delay, which is why a satellite connection feels sluggish however fast the brochure says it is.

06 Where & why

Where these figures are somebody's product specification

None of these numbers were invented for exams. Each one is a clause in a published standard or a line on a datasheet that somebody is contractually held to.

TIA-568 · Cat 6A
The 100 m rule is an attenuation budget with a number on it

Every structured cabling job in every office is built to a 100 m horizontal channel: up to 90 m of solid-core cable in the wall plus up to 10 m of stranded patch cord at the two ends. Cat 6A is certified to 500 MHz and carries 10GBASE-T over that full channel, where plain Cat 6 at 250 MHz manages 10 Gbps only to about 55 m. The 100 m is not a convention somebody chose; it is roughly where insertion loss reaches the receiver's limit at the frequencies the standard tests.

ITU-T G.652 · single-mode fibre
Amplifier spacing is the 20 dB span you just calculated

G.652 is the standard single-mode fibre: a 9 µm core in a 125 µm cladding, and the fibre under oceans, along railways and into homes as FTTH. The standard sets a ceiling on attenuation and production fibre comfortably beats it, landing near 0.2 dB/km in the 1550 nm window. Submarine systems put an optical amplifier roughly every 50 to 100 km, which at 0.2 dB/km is a span of 10 to 20 dB. Note the honest wrinkle: those are amplifiers rather than repeaters, so noise does accumulate and is managed by other means.

DOCSIS 3.1 · cable broadband
Hybrid fibre-coax is this entire lesson turned into a business

Cable operators run fibre from the headend out to a neighbourhood node, then coax for only the last few hundred metres into each house. DOCSIS 3.1 uses that coax plant to about 1.2 GHz downstream and reaches multi-gigabit rates on cable that was laid for television. The architecture is the section 02 table made into a budget: fibre wherever the distance is long, coax wherever the distance is short and the cable is already in the ground.

Starlink · low earth orbit
They could not fix the delay, so they moved the satellite

A geostationary satellite sits at 35 786 km because that is the altitude whose orbital period is one day, which is what lets a rooftop dish be aimed once and never moved again. The price is the round trip through that altitude. Starlink's main shell sits at roughly 550 km instead, which cuts the propagation part of the hop by a factor of about 65, at the cost of needing thousands of satellites and a phased-array antenna that tracks them. Measured latency is higher than the raw propagation figure because of ground stations and routing, but the change that mattered was the orbit and not the radio.

Two sentences to be able to defend on the spot. Attenuation is what sets repeater spacing, so the decibels-per-kilometre figure of a medium is the first number to ask for and the last one to compromise on. And fibre's advantage is physical, not a feature: it carries light through a dielectric, so interference has nothing to couple into and a tap has nothing to touch without being seen.

07 Interview questions

What they actually ask

Transmission media come up early in a networking round because they are easy to ask about and merciless about whether you understood or memorised. Expect to be asked why after every answer, and expect at least one question where the honest answer is that copper is still the right choice.

What actually separates guided media from unguided media, and which one is Wi-Fi?
Guided media confine the signal inside something you installed: twisted pair, coaxial cable, optical fibre. Unguided media radiate it into air or space with nothing confining it: radio, microwave, infrared, satellite. Wi-Fi is unguided, and the consequence is not academic. Because nothing confines the signal, every device in range receives it, so confidentiality has to come from encryption rather than from the medium, and capacity has to be shared with everyone else transmitting on that band.
Why is twisted pair twisted?
To cancel induced noise. The receiver measures the difference between the two wires, not either wire against ground. Two parallel wires sit at different distances from an interfering source, so they pick up different amounts and the difference changes, which is indistinguishable from signal. Twisting makes each wire spend half the run nearer the source and half further away, so both pick up the same voltage and the subtraction cancels it. It also explains two rules that look like fussiness: you may only untwist about half an inch of pair at a jack, and the four pairs in one jacket deliberately use different twist rates so they do not couple into each other.
UTP or STP? When would you actually pay for the shield?
Pay for it when the environment is electrically hostile and you cannot move the cable away from what is causing it: a plant floor with motors and variable-speed drives, a run alongside mains conduit, a hospital imaging suite. Otherwise UTP, because the shield is not a free upgrade. It has to be earthed, normally at one end only, and earthing a long shield at both ends between buildings whose earths sit at different potentials creates a ground loop current that becomes its own noise source. A poorly installed shielded link is worse than a plain unshielded one.
What does coaxial cable get from its geometry that a twisted pair does not?
The outer conductor completely surrounds the inner one along the entire length, and it is doing two jobs at once: it is the return path for the current and it is a continuous shield. External fields terminate on the outside of the braid and never reach the core, and the signal’s own field stays inside the dielectric instead of radiating away. That is why coax carries far higher frequencies far further than an open pair, which is why cable television has run gigahertz down it for decades. Know both standard impedances: 50 Ω for radio and legacy Ethernet, 75 Ω for video and cable broadband.
State the condition for total internal reflection in an optical fibre.
Two conditions, and both are needed. The core must have a higher refractive index than the cladding, and the ray must strike the boundary at an angle from the normal greater than the critical angle, which is arcsin(n_cladding / n_core). With a core at 1.48 and cladding at 1.46 that critical angle is about 80.6 degrees from the normal, so light can only be about 9 degrees off the fibre axis and still stay trapped. The cladding is not a mirror and not a protective coating: it is the lower-index layer without which the core would not guide light at all.
Single-mode or multimode. What is the real difference?
The core width, and everything follows from it. Multimode has a 50 or 62.5 µm core, wide enough that light travels several distinct paths at once, and because the steeper paths are longer their light arrives later and one clean pulse arrives smeared. That is modal dispersion, and it is what caps multimode at hundreds of metres rather than kilometres. Single-mode has about a 9 µm core, narrow enough that only one path propagates, so there is no modal dispersion and it runs tens of kilometres; it needs a laser rather than an LED because you cannot usefully couple a broad source into a 9 µm target. Both have a 125 µm cladding, so they look identical from the outside and are constantly mixed up on site.
Define attenuation, noise and distortion, and say which one you design around.
Attenuation is the loss of signal power with distance, so what arrives is a weaker copy. Noise is unwanted energy added from outside the signal, from motors, adjacent pairs or the thermal agitation of the conductor. Distortion is the signal arriving changed in shape rather than merely smaller, because different frequency components travel at slightly different speeds and stop lining up. You design around attenuation, because it is the one quoted as a fixed rate per unit length, so it is the one that turns directly into a distance and then into equipment somebody has to buy and power.
Is there a difference between a repeater and an amplifier?
Yes, and it is the whole reason digital links scale. An amplifier multiplies whatever it receives by a gain, so it raises the signal and the noise already riding on it by the same factor and adds a little noise of its own; chain several and the degradation compounds. A repeater is digital: it decides what each bit was, throws the received waveform away entirely, and transmits a brand new clean bit at full power, so nothing is carried forward except the decision and noise does not accumulate. If somebody says "amplifier" about a digital link, that is usually the follow-up question waiting to happen.
Which unguided media need line of sight, and what decides it?
The wavelength relative to the size of the obstacle. Classic radio below about 1 GHz has a wavelength of metres, comparable to or larger than doors and walls, so it diffracts around them and passes through them and needs no line of sight. Microwave at 10 GHz has a wavelength of about 3 cm, so it behaves much more like light: it is blocked by buildings and terrain and the two dishes must physically see each other, which is also what lets you focus it into a narrow beam. Infrared is shorter still and is stopped by any ordinary wall or partition, which confines a link to one room.
Infrared cannot pass through a wall. Is that a limitation or a feature?
Both, and interviewers ask it to see whether you can hold two ideas at once. It is a limitation because it forces line of sight or a ceiling bounce, caps you at one room, and leaves you helpless outdoors where sunlight is a very strong infrared source that swamps the receiver. It is a feature because the same opacity means the signal physically cannot leave the room, so an infrared link cannot be intercepted from the corridor or the car park the way a radio link can. Containment you get from physics is stronger than containment you get from configuration.
Why does a satellite dish bolted to a roof never need to move?
Because the satellite is geostationary. At an altitude of 35 786 km above the equator the orbital period is one day, so the satellite keeps pace with the rotating earth and appears fixed at one point in the sky. Aim the dish once at installation and it never moves again, and the same satellite covers about a third of the planet. What you pay for that convenience is the altitude itself: every hop climbs 35 786 km and comes back down, and that distance is fixed by the physics of the orbit rather than by anything an engineer can tune.
Fibre beats copper on bandwidth, distance, interference and tapping. So why is there still copper in every office?
Because the last hundred metres is the one place copper wins. Almost every desk is well inside the 100 m limit, so fibre’s distance advantage buys nothing there, and Cat 6A already carries 10 Gbps over that full run. Copper is cheaper per metre, terminates with a crimp tool instead of a fusion splicer, survives being trodden on and re-terminated by anybody, and it delivers power over the same cable with Power over Ethernet, which is how the access points, cameras and phones on that floor are fed. Fibre owns the backbone, the risers and everything between buildings. Copper owns the last hop, and that is an economic answer rather than a technical one.

08 Practice problems

Six runs to work out

For every one: write down the units before you write down a number. Bits or bytes, metres or kilometres, powers of ten or powers of two, and which speed applies in which medium. Almost every wrong answer in this topic is correct arithmetic performed on two quantities that were never measuring the same thing.

A run that does not divide evenly

Easy
A 750 m link is to be built from RG-6 coaxial cable, which loses about 6.6 dB per 100 m, on the 20 dB budget this lesson uses. Give the total loss over the run in dB, how far one segment reaches, how many segments the run needs, and how many regenerators that means.
Follow-up
The division does not come out whole, and which way you round it decides the answer. The count you are asked for at the end is also not the count you produced in the step before it.
Show the hint
Round the segment count in the direction that leaves no stretch of cable without a receiver that can read it, then ask where in the run a regenerator physically sits.

Frame time against flight time

Easy
A 9000-byte jumbo frame is sent at 1 Gbps over 300 m of OM4 fibre, where the signal travels at about 2 x 10^8 m/s. Give the transmission delay and the propagation delay in microseconds, say which one dominates and by what factor, and then give the cable length at which the two would be equal.
Follow-up
The two delays are computed from completely different quantities, and only one of them changes if you make the frame bigger while only the other changes if you make the cable longer. The last part is not a lookup; it is the same two expressions set equal to each other.
Show the hint
Convert bytes to bits before you divide by anything, and remember that 1 Gbps means 10^9 bits per second rather than a power of two or a byte count. For the last part, solve for the length.

The two dB that were not in the calculation

Medium
A 6.5 km link is planned on OM3 multimode fibre at 3.0 dB/km against a 20 dB budget, and on that basis it is signed off. The link as built has two patch panels and three fusion splices contributing 2 dB in total, and none of that was counted. Give the reach the original calculation implied, give the reach once the 2 dB is accounted for, say whether the 6.5 km link works, and say by how many dB it is over budget.
Follow-up
The loss that breaks this link is the smallest number in the problem and the only one that does not depend on the length at all, which is exactly why it was left out of a calculation that looked complete.
Show the hint
Split the 20 dB into the part that grows with every kilometre and the part that is spent once, and subtract the fixed part before you divide by anything.

Two dishes on two hills

Medium
A terrestrial microwave link uses two masts, each 80 m tall. The distance from a mast to its own horizon is approximately d = 3.57 x the square root of h, with h in metres and d in kilometres, and each mast contributes its own horizon to the link. Give the maximum separation of the two masts, give the new maximum if one of them is rebuilt at 40 m, and then explain why the same link would survive heavy rain at 2 GHz but not at 30 GHz.
Follow-up
Height sits under a square root, so halving one mast does not halve anything. The rain part has nothing to do with line of sight and is answered by comparing two lengths that are not distances between masts.
Show the hint
Add the two horizon distances rather than doubling one. For the last part, work out the wavelength at each frequency and compare it with the physical size of a raindrop.

Where the noise starts to hurt

Medium
A signal is launched at 0 dBm on a medium that attenuates 0.2 dB per metre, into an environment with a constant noise floor of -60 dBm. Give the signal-to-noise ratio in dB at 10 m, 50 m and 100 m. Then, if the receiver needs 45 dB of SNR to work, give the longest run that still meets it, and give the longest run if the noise floor is -50 dBm instead.
Follow-up
The noise floor in the second half moves by only 10 dB and the answer moves by far more than 10 percent, because the quantity that has to clear the floor is shrinking at a fixed rate for every metre while the floor is not shrinking at all.
Show the hint
On a logarithmic scale a ratio is a subtraction, so work out the signal level at each distance first and only then take the difference. For the last two parts, find the lowest signal level that still clears the requirement and convert the loss you are allowed into metres.

Eight thousand kilometres, two ways

Hard
A firm must move data 8000 km. Option A is a new single-mode fibre route of 8000 km at 0.2 dB/km on a 20 dB budget per span, with the signal travelling at 2 x 10^8 m/s. Option B is a satellite relay whose spacecraft sits 8000 km above the ground, with radio travelling through near-vacuum at 3 x 10^8 m/s. Give the one-way propagation delay of each, give the number of in-line amplification points the fibre route needs, say which option wins and by how many milliseconds, and finally give the satellite altitude at which the two options would tie.
Follow-up
The satellite’s signal travels 50 percent faster than the fibre’s, and speed is not the only thing that decides a delay. Settle which quantity actually differs between the two routes before you compute anything, and notice that neither answer contains a bandwidth term anywhere, so raising either link from 1 Gbps to 100 Gbps changes nothing at all.
Show the hint
Write both delays as distance divided by speed, and be careful that the satellite path is not the same as the satellite altitude. For the final part, set the satellite expression equal to the fibre answer and solve for the height.