Module 7 ended one step short on purpose. You know that a transmitting antenna concentrates power into a direction and that a receiving antenna behaves as a collecting area of Ae = λ²G/4π square metres. What nobody has yet written down is the sentence that joins them: given a transmitter of so many watts, two antennas of such-and-such gain, and a gap of d metres, how many watts arrive? That sentence is the Friis transmission equation, and it is the single most-used formula in radio engineering. We are going to build it in two steps rather than quote it, because the two steps are what make its most notorious consequence — the apparent frequency dependence of path loss — make sense.
Put a transmitter in the middle of empty space and let it radiate Pt watts equally in all directions. Nothing is absorbed, so after the wave has travelled a distance d, all Pt watts are still there — but they are now smeared over the surface of a sphere of radius d, and that surface is 4πd² square metres. The power density, in watts per square metre, is what you get by dividing. A directional antenna changes nothing about the accounting; it just multiplies the density in its favoured direction by its gain Gt and divides it everywhere else, exactly as M7-L1 insisted.
This is the inverse square law you met informally in M1-L3, now written as an equation. Double d and the same power covers four times the area, so the density falls to a quarter. It is worth being clear about why, because the intuition survives into every later model: nothing has been lost. Free space does not consume energy. The power is thinner, not smaller.
Now put the receiver in that thin wash of power. From M7-L1 you already have the piece you need: a receiving antenna of gain Gr intercepts the arriving wave over an effective aperture Ae = λ²Gr/4π. Received power is therefore just density multiplied by area — watts per square metre times square metres — and the units cancel to watts without any fudge factor. Substitute one expression into the other and the whole of Friis falls out in a single line of algebra.
Read the result and notice what it does not contain. There is no absorption coefficient, no material property, no atmosphere. Every term is either something you built (power, two gains) or pure geometry (distance, wavelength). That is why this is called the free-space equation: it is the answer for a universe containing nothing but two antennas.
Where the λ actually came from. Trace it back and the wavelength entered in exactly one place: the receive antenna’s aperture. It is not in the spreading term, because a sphere does not care what frequency crosses it. Hold that thought — it is the whole answer to “why do high frequencies lose more?”, and almost every wrong explanation of that question comes from forgetting that λ arrived through an antenna rather than through the space between them.
Engineers rarely use Friis in the form above. It is more useful to split the link into things you chose and things nature imposed, so the geometric factor is pulled out, inverted so that it is a number greater than one, and given its own name: free-space path loss. Inverting is a bookkeeping convention, nothing more — it lets you write “minus 100 dB of loss” instead of “plus a gain of 10⁻¹⁰”.
FSPL is a ratio of two powers, so it goes into decibels with the 10·log₁₀ rule from M2-L4 — and because the ratio is a squared quantity, the square comes out front as a factor of two, giving the 20·log₁₀ you see below. This is not the amplitude-versus-power confusion M2-L4 warned about; it is a power ratio that happens to be squared. Taking logs of the constant 4π/c gives 20 log₁₀(4π/2.998×10⁸) = 20 log₁₀(4.192×10⁻⁸) = −147.55 dB.
Nobody types frequencies in hertz, so two rescaled constants do the everyday work — and they are the same equation, not three different ones. Change d from metres to kilometres and you add 20 log₁₀(1000) = 60 dB; change f from hertz to megahertz and you add 20 log₁₀(10⁶) = 120 dB. So −147.55 + 60 + 120 = +32.45. Rescale the frequency once more, from megahertz to gigahertz, and you add another 60 dB: 32.45 + 60 = +92.45. Any of the three gives the same answer to the same problem, and checking one against another is the fastest way to catch a units error.
| Units of d | Units of f | Constant | Formula |
|---|---|---|---|
| metres | hertz | −147.55 | 20log(d_m) + 20log(f_Hz) − 147.55 |
| kilometres | megahertz | +32.45 | 32.45 + 20log(d_km) + 20log(f_MHz) |
| kilometres | gigahertz | +92.45 | 92.45 + 20log(d_km) + 20log(f_GHz) |
Take a Wi-Fi carrier at 2.4 GHz and an unobstructed 100 m — a clear car park, say. Using the gigahertz form, d = 0.1 km and f = 2.4 GHz:
Now check it the long way, from the definition. At 2.4 GHz, λ = 3×10⁸/2.4×10⁹ = 0.125 m, so 4πd/λ = 12.566 × 100 / 0.125 = 10,053, and 20 log₁₀(10,053) = 80.05 dB. The two routes agree, which is exactly the check worth building the habit of. And the result is worth pausing on: 80 dB is a factor of 100 million. Of the 100 mW a typical access point radiates, one part in a hundred million reaches a receive antenna 100 m away — about 1 nW — and that is on the best day physics allows.
Once path loss is in decibels, the inverse square law becomes something you can do in your head. Doubling the distance multiplies the loss by four, and a factor of four in power is 20 log₁₀(2) = 6.02 dB — the same 6 dB you would get from 10 log₁₀(4), because both describe the same power ratio. Ten times the distance is 20 log₁₀(10) = 20 dB. So free-space loss climbs at a steady 20 dB per decade of distance, and on a log-distance axis it is a straight line.
Six decibels per doubling sounds gentle and is brutal. Going from 100 m to 1 km costs 20 dB — a hundredfold drop in received power — and from 100 m to 36,000 km costs 20 log₁₀(360,000) = 111 dB on top of what you already had. It is also the reason cell radii are what they are: coverage does not shrink gracefully as you push a link, it falls off a cliff that is exactly 6 dB steep per doubling.
A limit the formula does not respect. FSPL passes through 0 dB when 4πd = λ, that is at d = λ/4π, which at 2.4 GHz is 9.95 mm. Closer than that the formula returns a negative loss — a gain — which is obviously nonsense. Nothing is broken: you have simply walked inside the near field, where M7-L1’s far-field assumptions and therefore this entire derivation stop applying. Treat FSPL as valid only in the far field of both antennas.
The best way to feel 20 dB per decade is to go as far as radio routinely goes. A geostationary satellite sits 35,786 km above the equator — call it 36,000 km, which is the figure everyone quotes — and a Ku-band television downlink uses roughly 12 GHz. Path loss first, in the gigahertz form: 20 log₁₀(36,000) = 91.13 dB, 20 log₁₀(12) = 21.58 dB, and 92.45 + 91.13 + 21.58 = 205.16 dB, which we will carry as 205.2 dB.
Two hundred and five decibels is a ratio of 3×10²⁰. Written out, roughly one part in three hundred quintillion of the radiated power arrives. No amount of transmitter is going to brute-force that, which is why both ends of a satellite link are almost entirely antenna. Here is the whole link as one line of addition, in the M2-L4 house style — dBW this time rather than dBm, because satellite people work in watts:
| Term | Value | Where it comes from |
|---|---|---|
| Satellite transmit power | +20 dBW | 100 W travelling-wave-tube amplifier |
| Satellite antenna gain | +35 dBi | Shaped spot beam; EIRP is therefore 55 dBW |
| Free-space path loss | −205.2 dB | 36,000 km at 12 GHz, computed above |
| Atmosphere and pointing | −1.5 dB | Clear-sky gases plus imperfect dish aim |
| Dish gain, 1 m at 12 GHz | +40 dBi | 0.65 × (πD/λ)² = 10,280, i.e. 40.1 dBi |
| Feed and connector loss | −0.5 dB | Short run to the low-noise block downconverter |
Does it close? That depends on the noise it has to beat, which is Module 9’s subject, but the shape of the answer is already available. Thermal noise power is kTB. For a system noise temperature of 150 K and a 27 MHz transponder, 10 log₁₀(k) = −228.6 dBW/K/Hz, 10 log₁₀(150) = 21.8 dB and 10 log₁₀(27×10⁶) = 74.3 dB, so the noise floor is −228.6 + 21.8 + 74.3 = −132.5 dBW, or −102.5 dBm.
So the carrier-to-noise ratio is −112.2 − (−132.5) = 20.3 dB. A QPSK digital-television carrier needs something like 7 dB to decode, so the link closes with about 13 dB in hand — and it needs that margin, because heavy rain at 12 GHz can absorb 6 to 10 dB and take it away. Choosing how much margin to hold is fade margin, which is M9-L4’s job; the point here is that FSPL was the single largest term in the budget by an enormous distance, and everything else was a correction to it.
Look at the dB formula and the frequency term is right there: +20 log₁₀(f). Go from 2.4 GHz to 12 GHz, a factor of five, and you pay 20 log₁₀(5) = 14.0 dB more loss over the same distance. This is repeated everywhere, usually with an explanation attached — that high frequencies are absorbed more, or blocked more easily, or scatter more. In free space every one of those explanations is wrong, and the derivation above shows why: there is nothing in the spreading term to absorb anything, and the geometry of a sphere has no idea what frequency is crossing it.
What actually happened is the callout above. The only λ in the equation came from Ae = λ²Gr/4π. Holding gain fixed while raising frequency shrinks the physical size of the antenna that delivers that gain, so it shrinks the collecting area, so less power is captured. The loss is not in the channel at all — it is in the implicit assumption that your antennas stayed the same electrically while shrinking physically. Say the assumption out loud and the puzzle dissolves.
Which suggests the obvious experiment: hold the antennas’ physical size fixed instead, and let their gain do whatever physics says. A dish of diameter D has gain G = η(πD/λ)², which rises as f². Two of those, one at each end, contribute f⁴, against the f² that path loss takes away — so received power goes up as f². Written in apertures rather than gains, Friis makes this plain, and it is the form satellite and microwave engineers actually think in:
Put numbers on both readings, over the same 10 km hop, and the symmetry is striking. With 6 dBi at each end — small fixed-gain antennas — moving from 2.4 GHz to 12 GHz costs 14.0 dB. With a 1 m dish at each end and 65% aperture efficiency, the identical move gains 14.0 dB. Same channel, same distance, same physics, opposite conclusions, because the two experiments held different things constant.
| 10 km link | 2.4 GHz | 12 GHz | Change |
|---|---|---|---|
| FSPL | 120.05 dB | 134.03 dB | +13.98 dB worse |
| Fixed gain, 6 dBi each end | 6 + 6 − 120.05 = −108.05 dB | 6 + 6 − 134.03 = −122.03 dB | 13.98 dB worse |
| Fixed 1 m dish each end | 26.14 + 26.14 − 120.05 = −67.77 dB | 40.12 + 40.12 − 134.03 = −53.79 dB | 13.98 dB better |
The last column is 20 log₁₀(5) = 13.98 dB in all three rows, with the sign set by what you fixed. So the honest statement is not “high frequencies propagate worse” but this: at a fixed antenna gain, higher frequencies lose more; at a fixed antenna size, higher frequencies win. Millimetre-wave 5G is a live example — it survives its enormous FSPL precisely because a 28 GHz array packs dozens of elements into a fingernail and claws the aperture back, as M7-L1 foreshadowed. Where high frequencies genuinely do suffer from the medium is oxygen and water-vapour absorption, rain, and poor diffraction around obstacles, and none of those is in this equation.
Everything above assumed a universe with two antennas in it and nothing else: one unobstructed path, no ground, no walls, no atmosphere, no rain, no reflections. That universe does not exist. FSPL is therefore a lower bound on loss — an optimistic floor. Real links are worse, sometimes by tens of decibels, and just occasionally momentarily better, when a reflection happens to add in phase with the direct path. Any measurement that comes out below FSPL over a long average is a measurement error, not a discovery.
Where FSPL is close to true is where it earns its keep: satellite links above the atmosphere, radio astronomy, microwave hops on towers built tall enough to clear everything, deep-space telemetry, and a first estimate for any line-of-sight link. Everywhere else it is the starting point that the rest of Module 8 corrects. M8-L2 adds reflection, diffraction and scattering, so the wave arrives by more than one route. M8-L3 lets those routes interfere and gives you multipath and fading. M8-L4 abandons closed-form physics for measured models — log-distance, Okumura-Hata, COST-231 — that fit real cities. And M9-L4 assembles the complete budget with noise and fade margin included. All four begin by computing FSPL and then adding to it.