M12-L1 assembled one end-to-end link out of everything the course had built: a source, a modulator, an upconverter, an amplifier, an antenna, a channel that attenuates and fades, and a receiver that undoes all of it while fighting its own thermal noise. This lesson looks at what is being proposed next — sub-terahertz carriers, low-orbit satellite Internet, reflecting surfaces, radios that sense as well as talk, learned air interfaces — and it is the lesson in this course most at risk of becoming a press release.
So here is the rule for reading it, and it is the only rule you need: every one of these technologies is a physical trade, and the price is quoted in the same decibels as the benefit. Not one of them repeals Friis, Shannon, kTB or Doppler. Each takes a resource the course has already taught you to count — bandwidth, aperture, transmit power, time, geometry, or someone else’s coordination — and spends more of it than the previous generation did. Your job in this lesson is not to memorise the promises. It is to price them with the six tools you already own: FSPL (M8-L1), the link budget (M9-L4), Shannon (M6-L1), the −174 dBm/Hz noise floor (M9-L1), Doppler (M8-L3), and array gain and aperture (M7-L2, M7-L4). Every number below is computed with one of those and shown, so you can check it.
A generation of mobile radio is a process before it is a product, and the process runs in a fixed order: a vision document that names capabilities, a spectrum decision at a World Radiocommunication Conference, study items in 3GPP, then specifications, then silicon, then networks. As an approximate account of published milestones: the ITU-R agreed a framework recommendation for IMT-2030 — the formal name of what everyone calls 6G — which sets out usage scenarios and capability targets rather than any air interface; candidate spectrum is on the agenda of the next World Radiocommunication Conference; 3GPP began 6G study work around Release 20 with the first specifications expected in the release after; and the whole plan aims at deployment around 2030, which is where the “2030” in the name comes from.
Read that list for what is not in it. There is no 6G standard, no 6G specification, no 6G chipset and no 6G channel. Every specific 6G performance figure you have seen is a target in a requirements document or a vendor projection, not a measurement. That does not make the research uninteresting — it makes it research, which is a different and more honest thing to call it.
One target is worth pricing immediately, because it prices half the lesson. Take the headline figure of 1 Tbit/s. Shannon says C = B log₂(1 + SNR), so a rate is a bandwidth multiplied by a spectral efficiency. Suppose you are wildly generous and assume 20 bit/s/Hz aggregated across all spatial layers — better than anything 5G achieves in the field. Then B = 10¹²/20 = 50 GHz of bandwidth. At a more sober 4 bit/s/Hz it is 250 GHz. There is no unallocated 50 GHz anywhere below 100 GHz. So a terabit claim is not a modulation claim or a coding claim; it is a claim about operating above 100 GHz — and the next section computes what that costs in range. Hold both numbers in mind at once, because no marketing slide will.
M11-L3 already made this climb once, from 3.5 GHz to 28 GHz, and found that the 20 log₁₀(8) = 18.1 dB of extra path loss was cancelled almost exactly by the 10 log₁₀(64) = 18.1 dB of gain from a 64-element panel. Do the same arithmetic one decade higher.
Now buy it back. At 300 GHz, λ = c/f = 2.998×10⁸/3×10¹¹ = 0.999 mm, so call it 1 mm and half-wavelength element spacing is 0.5 mm. Array gain is 10 log₁₀N (M7-L4), so recovering 20.6 dB needs N = 102.06 = 114.8 elements — take an 11 × 11 grid of 121, worth 10 log₁₀(121) = 20.83 dB. Those 121 elements span 10 × 0.5 = 5.0 mm between the outer centres, so the whole panel is about 5.5 mm square. The identical 11 × 11 array at 28 GHz, where λ = 10.7 mm, would span 10 × 5.36 = 53.6 mm — a panel 115 times larger in area for the same gain. This is M8-L1’s honest statement, quoted exactly: at a fixed antenna gain, higher frequencies lose more; at a fixed antenna size, higher frequencies win.
And it is worth seeing why the two cancel, because it is not a coincidence and it names the real cost. Gain is inverse to beam solid angle. For a uniformly illuminated aperture of width D, M7-L2 gives the beamwidth as roughly 51 λ/D degrees. Fix D at 1 cm. At 28 GHz that is 51 × 10.7/10 = 54.6° — barely directional. At 300 GHz it is 51 × 1.0/10 = 5.1°. The solid angles are in the ratio (54.6/5.1)² = 114.6, which is (300/28)² = 114.8 to within rounding. The gain that pays the frequency penalty is the narrowed beam. You have not found free power; you have agreed to know where the other end is to within a few degrees, and to re-find it every time anything moves.
Fold it into M9-L4’s machinery and get a range. Assumptions, all stated so you can disagree with them: B = 10 GHz of channel; receiver noise figure 10 dB, which is pessimistic at 3.5 GHz and optimistic at 300 GHz; required SNR 10 dB; transmit power +10 dBm, because solid-state power amplifiers get dramatically weaker as frequency rises and a watt at 300 GHz is not a thing you buy; and the 20.83 dB array at each end.
Invert FSPL for the distance. At 300 GHz, 92.45 + 20 log₁₀(300) = 92.45 + 49.54 = 141.99 dB at one kilometre, so 20 log₁₀dkm = 105.7 − 141.99 = −36.3 dB, giving d = 10−1.815 km = 15.3 m. And Shannon at that operating point: C = 10×10⁹ × log₂(1 + 10) = 10¹⁰ × 3.4594 = 34.6 Gbit/s. That pair of numbers is the whole sub-THz proposition: tens of gigabits per second, across a room. Both halves are real. Quoting either one alone is the dishonesty.
Two follow-up checks are more instructive than the result. First, notice what the 10 GHz of bandwidth cost: relative to M11-L3’s 400 MHz millimetre-wave carrier, the noise floor rose by 10 log₁₀(10¹⁰/4×10⁸) = 10 log₁₀(25) = 14.0 dB, exactly as M9-L1 promised, and that 14 dB came straight out of the range. Bandwidth is never free; it is the cheapest thing on the menu, because Shannon is linear in B and logarithmic in SNR, but you pay for it in decibels. Second, ask what it would take to reach 100 m instead of 15.3 m: 20 log₁₀(100/15.3) = 16.3 dB, or about 8.2 dB more at each end, which means multiplying each array by 6.6 — roughly a 28 × 28 grid of 784 elements. At 300 GHz that panel is still only 13.5 mm square. The cost of sub-THz is not panel area. It is 784 phase shifters, 784 amplifiers and the power they burn, all of which sit in the parts of the design this course has treated as free.
The folklore says terahertz signals travel a few metres because the air swallows them. That is not what the physics says, and the reason matters. Molecular absorption by oxygen and water vapour adds a loss that is linear in distance — so many decibels per kilometre — while spreading loss is logarithmic in distance, 20 dB per decade. Two different shapes, and which one dominates depends entirely on the range.
Take approximate published clear-air figures (order of magnitude, from the ITU-R gaseous-attenuation model at sea level with about 7.5 g/m³ of water vapour): a fraction of a dB/km below 10 GHz; roughly 15 dB/km in the 60 GHz oxygen band; under a dB/km in the window near 100 GHz; tens of dB/km on the water-vapour lines near 183 GHz; and around 10 dB/km in the window near 300 GHz. Now apply them to the link just computed. Over 15.3 m, 10 dB/km costs 10 × 0.0153 = 0.15 dB — invisible. Over 100 m, 1 dB. Over a kilometre, 10 dB, which finally matters. Even the ferocious 60 GHz oxygen band costs only 15 × 0.1 = 1.5 dB across 100 m, which is why 60 GHz WiGig links work perfectly well inside a room.
So absorption is not what limits a short sub-THz link. Spreading loss, weak transmitters and a wide noise bandwidth are, and all three are in the budget above. Absorption is what stops sub-THz from ever becoming a coverage layer: it puts a hard, distance-linear wall a few hundred metres to a few kilometres out that no amount of array gain climbs, because every 10 dB of gain you add buys only 10 ÷ 10 = 1 km more of absorbing air. The two losses fail differently, and confusing them is how “THz only goes 3 metres” and “THz will replace cellular” both get published.
M8-L1 already computed a geostationary link: 35 786 km, 12 GHz, and 205.2 dB of free-space path loss. Geostationary satellite Internet has existed for decades and nobody liked it, for one reason that has nothing to do with bandwidth. Put the two altitudes side by side and divide by the speed of light.
Overhead is the best case. At a realistic minimum elevation of 25°, the slant range to a 550 km shell is d = √(r² − R²cos²ε) − R sinε with R = 6371 km and r = 6921 km, which is √(47.90 − 33.34)×10⁶ − 2692.5 = 3815.8 − 2692.5 = 1123 km, or 3.75 ms — twice the overhead figure. Low orbit does not mean short path; it means short path sometimes.
There is a second prize besides latency, and it is the larger one in engineering terms. FSPL at 11 GHz to 550 km is 92.45 + 20 log₁₀(550) + 20 log₁₀(11) = 92.45 + 54.81 + 20.83 = 168.1 dB. To GEO it is 92.45 + 91.08 + 20.83 = 204.4 dB. The difference is 20 log₁₀(35 786/550) = 20 log₁₀(65.07) = 36.3 dB. (Cross-check against M8-L1: its 204.4 dB figure at 11 GHz becomes 204.4 + 20 log₁₀(12/11) = 204.4 + 0.8 = 205.2 dB at 12 GHz ✓.) Thirty-six decibels is the difference between a 2.4 m dish on a pole and a flat panel on a roof, and it is why LEO changed who can buy satellite Internet, not merely how it feels.
A satellite in a 550 km circular orbit moves at v = √(GM/r) = √(3.986×10¹⁴/6.921×10⁶) = √(5.759×10⁷) = 7.59 km/s, and completes an orbit in 2π√(r³/GM) = 95.5 minutes. Everything expensive about LEO follows from that one speed.
Compare that with the terrestrial case the course already worked. M11-L3 computed a car at 100 km/h at 3.5 GHz as fd = 324 Hz. The LEO figure is about 790 times larger — partly speed, partly the higher carrier. And measure it the way M11-L3 taught, as a fraction of the subcarrier spacing: 256 kHz against the widest 5G numerology of 120 kHz is 213% — more than two entire subcarriers of offset. An OFDM receiver cannot absorb that; inter-carrier interference would destroy the symbol. This is precisely why 3GPP’s non-terrestrial network work does not treat Doppler as an impairment to be tracked but as a quantity to be pre-compensated: the satellite’s position and velocity are predictable from published ephemeris, so both ends compute the shift and the timing advance in advance rather than discovering them.
The second cost is handover, and it comes from the same 7.59 km/s. With a 25° elevation mask, the geocentric angle over which one satellite is visible is arccos((R/r) cos 25°) − 25° = arccos(0.8343) − 25° = 33.44° − 25° = 8.44°, so the whole pass spans 16.9° of a 360° orbit: 16.9/360 × 95.5 min = 4.5 minutes, and that is a pass straight over your head. Drop the mask to the horizon and it stretches to about 12 minutes; take a pass off to one side and it shrinks. A terrestrial cell can hold a stationary user indefinitely. A LEO terminal hands over every few minutes, forever, whether or not it has moved, and every handover is a beam re-acquisition at a new Doppler and a new timing advance.
The third cost is the constellation itself, and it can be derived rather than looked up. One satellite covers a spherical cap of half-angle 8.44°, whose fraction of the Earth’s surface is (1 − cos 8.44°)/2 = (1 − 0.98918)/2 = 0.00541, or 0.54%. Perfect tiling with no overlap would therefore need 1/0.00541 = 185 satellites for one-fold global visibility, and real orbital-plane geometry with usable overlap pushes that to several hundred. Published figures for the operational Starlink fleet are in the thousands — the exact number changes constantly, so treat any specific count as a snapshot. The gap between 185 and thousands is the answer to a question people ask backwards: constellation size is not set by coverage, which a few hundred satellites achieve. It is set by capacity. Each satellite has a finite bandwidth to divide among every user in its 0.54% of the planet, and adding satellites is the only way to shrink the cell.
“NTN” is the name for putting the cellular standards themselves — not a proprietary satellite protocol — onto a platform that is not on the ground: geostationary or low-orbit satellites, and high-altitude platforms. Two payload architectures, and the distinction decides everything else:
And one honest comparison, since the course has the tools. Over long distances a satellite relay can beat terrestrial fibre on propagation delay, because glass is slow: light in fibre travels at c/n with n ≈ 1.47, that is 2.04×10⁸ m/s, so 1000 km of fibre costs 4.90 ms against 3.34 ms in vacuum. The LEO route pays a fixed 3.67 ms to climb up and come back down, and then saves 1.56 ms for every 1000 km, so it breaks even at 3.67/1.56 × 1000 = about 2350 km and wins beyond it. That is genuine physics and it is why long-haul low-latency routing is a real application. It also ignores routing, queueing and scheduling delay, all of which hurt the satellite; and it ignores the fact that real fibre routes are considerably longer than the great-circle distance, which helps it. Treat 2350 km as the order of the crossover, not a number to quote.
A reconfigurable intelligent surface is a flat panel of many small passive elements — typically half-wavelength patches — each of which can be electronically switched between a few reflection phases. Set the phases correctly and the surface reflects an incoming wave not specularly, the way M8-L2’s mirror does, but towards a chosen direction, with all N elements arriving in phase at the target. It is a mirror you can aim. The proposition is that you hang them on buildings and fill in coverage holes without power, backhaul or a base station licence, and it is the most oversold idea in this lesson.
So price it. The single fact that governs everything is that a passive surface is not an amplifier: it re-radiates only the power its own aperture happened to intercept. That intercepted fraction is set by the first hop, and the re-radiated wave then spreads over the second hop. Two spreading losses, one after the other — and they multiply.
Now put numbers in it, at 3.5 GHz where λ = 85.7 mm and a half-wavelength element occupies (λ/2)² = 18.4 cm². Take a 16 × 16 surface, N = 256: a panel 16 × 42.9 mm = 0.69 m square, about the size of a road sign, with A = 0.470 m² and Gs = 10 log₁₀(4π × 0.470/0.0857²) = 10 log₁₀(805) = 29.05 dBi. That is a large, respectable gain — nearly thirty decibels, and it appears twice. Then compare three routes from a transmitter to a receiver 100 m apart:
| Route | Arithmetic | Total loss | vs. direct |
|---|---|---|---|
| Direct, 100 m | 32.45 + 20 log₁₀(3500) + 20 log₁₀(0.1) | 83.3 dB | — |
| RIS midway, 50 m + 50 m | 77.31 + 77.31 − 2(29.05) | 96.5 dB | 13.2 dB worse |
| RIS near the receiver, 85 m + 15 m | 81.92 + 66.85 − 2(29.05) | 90.7 dB | 7.3 dB worse |
| Amplify-and-forward relay midway | worse hop only: 77.31 | 77.3 dB | 6.0 dB better |
Read the table slowly, because it contains the entire argument. A 0.69 m surface with 58 decibels of round-trip aperture gain still loses to doing nothing at all, by 13 dB in the worst placement and by 7 dB in a good one. And the last row is why: a relay receives, so its end-to-end quality is set by the worse of two 50 m hops — a maximum of decibels. The surface never receives anything; its two spreading losses add in decibels, which is to say the distances multiply. Halving 100 m into 50 + 50 helps a relay and hurts a surface, and that asymmetry is not a detail of the implementation. It is the difference between an active and a passive device, and no amount of engineering removes it.
The product d₁²d₂² also tells you where the surface must go. For a fixed total d₁ + d₂ = 100 m, a product is largest when the factors are equal, so the midpoint is the single worst place to mount a RIS — the opposite of where you would put a relay. Push it to 85 + 15 and the product falls from 2500 to 1275, worth 20 log₁₀(2500/1275) = 5.8 dB; push it to 98 + 2 and the product is 196, worth 22.1 dB against the midpoint. Anyone quoting a RIS gain without quoting d₁ and d₂ has told you nothing.
The N² in the numerator is the reason the field is interesting rather than dead, so give it its due. Loss falls as N², so quadrupling the element count buys 20 log₁₀(4) = 12 dB — twice the return per element that an ordinary array gets, again because the surface is counted at both ends. Ask for break-even against the direct 100 m path with the surface midway: you need 2Gs = 2(77.31) − 83.33, so Gs = 35.6 dBi, hence N = 103.5645/π = 1168 elements. A 34 × 34 grid of 1156 gives 35.60 dBi and a cascaded loss of 83.4 dB — equal to the direct path to within a tenth of a decibel. That panel is 34 × 42.9 mm = 1.46 m square, and it has bought you exactly nothing over an empty street.
Which is the point: a RIS is never worth deploying on a path that already works. It is worth deploying when the direct path does not exist. Block that 100 m link with a brick wall — M8-L4 prices brick or block at 10–15 dB — and the direct route becomes 98.3 dB. Now the same 256-element sign at 85 + 15 m, with clear line of sight to both ends, delivers 90.7 dB and is 7.7 dB better than going through the wall. The honest claim is not “a RIS adds gain”. It is “a RIS creates a second path where geometry denied you one”, and it is competing against diffraction and penetration loss (M8-L2, M8-L4), not against free space.
Three costs remain, and the third is the one that decides whether any of this ships. First, near field. The product law above is a far-field result, and M7-L2’s Rayleigh distance 2D²/λ for the 1.46 m panel is 2(1.456)²/0.0857 = 49.5 m. A receiver 15 m from that surface is inside its near field, where the formula does not apply and the true loss is better than it predicts — which is precisely why large-surface near-field operation is an active research direction and not a rounding error. Quote the far-field product law as the pessimistic bound it is. Second, the surface is not free. N elements need N controllable phase states, a bias network, a control link and a power supply for the controller; “passive” describes the RF path, not the bill of materials. Third and worst, channel knowledge. To set 256 phases you must know both hops’ complex channels — 2N coefficients — for a device that cannot transmit a pilot or receive one, because it has no receiver. M10-L3 already identified pilot overhead as the binding constraint on massive MIMO with one channel per element; a RIS has the same problem with the measurement instrument removed. And M8-L3’s coherence time gives you 1.3 ms at 100 km/h to solve it in. That, not the decibels, is why surfaces have not left the laboratory.
Integrated sensing and communication observes that a radar and a communication radio are the same hardware — an upconverter, a power amplifier, a steerable array, a coherent receiver — pointed at different problems, and proposes to run both from one waveform in one band. A base station that already sweeps beams across a street could, at no extra silicon, report that a pedestrian is crossing it. The idea is sound and the hardware overlap is real. The trade is entirely in the link budget, and it is a brutal one.
Work an example on the mid-band cell the course keeps returning to. Take 3.5 GHz (λ = 85.7 mm), a 25 dBi array used for both transmit and receive, σ = 10 m² for a car, and a target at 100 m. Then Pr/Pt = G²λ²σ/((4π)³d⁴), which in decibels is a loss of 74.3 dB. Compare it with the one-way communication loss over the same 100 m with the same two antennas: 83.3 − 25 − 25 = 33.3 dB. The echo is 41 dB weaker than a comms path to a phone standing where the car is. Sensing is not a free by-product of a communication link; it is a 41 dB harder problem at the same range.
Turn that into a sensing range with M9-L4’s machinery, so the answer is a distance rather than an adjective. At +23 dBm of transmit power and 100 MHz of bandwidth, the noise floor is −174 + 10 log₁₀(10⁸) + 7 dB of noise figure = −87 dBm. Ask for 10 dB of detection SNR and add 3 dB of implementation margin, and the tolerable two-way loss is 23 + 87 − 13 = 97 dB. Since the loss goes as 40 log₁₀d, the range is 100 × 10(97−74.3)/40 = 369 m. That same 97 dB budget on a one-way comms link, where loss goes as 20 log₁₀d, reaches 482 m before antenna gain is even counted — and M12-L1 got a 340 m cell radius with realistic margins. So a cell-sized sensing range is achievable, and that is the genuinely encouraging result in this section — but only for a 10 m² car. Drop to σ = 1 m² for a pedestrian and the echo loses 10 dB, which at 40 dB per decade shrinks the range by 1010/40 = 1.78×, to 207 m.
The second half of the equation block is the sensing side of a trade the course has already met from the other direction. Range resolution is Δr = c/2B — the factor of 2 is the round trip — and it says the only way to resolve two objects in range is bandwidth. At 100 MHz, Δr = 2.998×10⁸/(2×10⁸) = 1.50 m: enough to separate two cars, not enough to separate two people standing together. At 400 MHz it is 37.5 cm, and at the 10 GHz of sub-THz channel from earlier in this lesson it is 1.5 cm. This is the same statement as M6-L2’s: bandwidth buys resolution in time, and a delay is a range. It is also why automotive radar sits at 76–81 GHz, where 4 GHz of allocation gives Δr = 3.7 cm.
Now name the bills, because both functions pay one:
The honest summary is that ISAC is cheap where it is easy and expensive where it is interesting. Coarse presence and motion detection over tens of metres, using beams that were being swept anyway, is nearly free — WiFi sensing already does a crude version of it by watching how the channel estimates of M10-L2 wobble when someone walks through the room, and 802.11bf standardises exactly that. Centimetre-resolution imaging of a person at 300 m needs gigahertz of bandwidth, 120-odd decibels of self-interference cancellation and a scheduler willing to donate 10 ms beams. Both get called ISAC.
This is the topic where the gap between the phrase and the content is widest, so begin by clearing away what is not new. The radio has been adaptive for thirty years. M6-L4 described adaptive modulation and coding as a run-time decision; M5-L4 listed the ladder it climbs; M6-L4’s LTE section has the handset reporting a 4-bit channel-quality indicator once per 1 ms subframe and the scheduler choosing a modulation and coding scheme from it. M11-L2 traced the idea back to GPRS in 2000, picking one of four coding schemes per connection. A scheduler that measures, predicts and decides every millisecond, per user, is not a 6G proposal. It has been shipping since 3G, it is a control loop, and calling a control loop “AI” retroactively is how a research programme becomes a press release.
What is actually being proposed is narrower and more interesting: replacing individual signal-processing blocks whose optimal form is unknown or intractable with functions learned from data. Three have real evidence behind them, and each has a cost the course can name.
And then the costs, which are the part the vision documents skip. Interoperability is the hard one: a standard exists so that any handset works with any base station, and a learned encoder in the phone paired to a learned decoder in the network is two halves of one model that must be trained together, versioned together and updated together, by two companies that compete. This is why the standardisation work is largely about how to describe and manage a model rather than about the model itself. Generalisation is next: a function trained in one city, band and antenna configuration has no guarantee off its training distribution, and a radio meets weather, crowds and construction it never saw. Energy and latency are third — a per-symbol inference must finish inside the 1 ms scheduling interval and within a handset’s power budget, which rules out most of what the word “model” evokes. And verification is last and worst: M9-L2 measured bit error rate as a probability, and demonstrating one error in 10⁵ (M11-L3’s reliability target) requires an argument about the tails, which is exactly where a learned function’s behaviour is least characterised and hardest to bound.
The test to apply to any AI-native claim is the one this whole lesson uses. Ask what physical resource is being spent differently. If the answer is “fewer feedback bits for the same channel knowledge” or “fewer measurements to find the same beam”, that is a real and quantifiable saving in overhead. If the answer is “higher throughput” with no bandwidth, no power and no antennas added, then it is claiming to beat Shannon, and M6-L1 settled that in 1948.
Five technologies, five trades, and not one of them repealed anything the course taught. Sub-THz buys 34.6 Gbit/s and pays 15.3 m and 784 phase shifters. LEO buys 36.3 dB and 470 ms of latency back, and pays ±256 kHz of Doppler, a handover every 4.5 minutes and hundreds of satellites. A RIS buys a path where geometry denied one, and pays a product of distances that makes it lose to an empty street. ISAC buys sensing from hardware you already own, and pays 41 dB and the amplifier back-off. Learned blocks buy overhead, and cannot buy capacity. The pattern is the lesson: when the next announcement arrives, find the resource being spent — bandwidth, aperture, power, time, geometry or coordination — and price it in the same decibels as the benefit. If the announcement does not name a cost, it has not finished being engineering.