Radio Propagation & Free-Space Attenuation

Radio propagation is the part of a link you cannot engineer — you can only measure it, model it, and buy margin against it. Everything else in the chain (power, gain, coding, modulation) is a budget line you control; propagation sets the number those lines have to beat. The starting point is the Friis transmission equation, and the single most useful thing to understand about it is where its frequency dependence actually comes from — because it is not absorption, free space does not absorb, and the widespread belief that "high frequencies don't travel as far" is a misreading of a term that describes the receiving antenna, not the medium.

Past that, a real link is never in free space. The ground puts a second ray into the receiver, terrain diffracts, oxygen and water vapour resonate, rain scatters, and below about 30 MHz the ionosphere reflects the signal back down. Each of these has a regime where it dominates and a boundary you can compute.

Friis, derived #

Two steps, no hand-waving.

Step 1 — spreading. A transmitter feeding power $P_t$ into an antenna of gain $G_t$ produces, at distance $d$ in the boresight direction, a power flux density

$$S = \frac{P_t\,G_t}{4\pi d^{2}} \qquad \left[\mathrm{W/m^2}\right]$$

This is conservation of energy on the surface of a sphere. Nothing is lost; the same total power crosses every sphere. $G_t$ is the factor by which the antenna concentrates that power relative to an isotropic radiator, so $P_t G_t$ is the e.i.r.p.

Step 2 — collection. A receiving antenna intercepts power in proportion to its effective aperture $A_e$, which is an area:

$$P_r = S \cdot A_{e,r}$$

The link between an antenna's gain and its effective aperture is the one genuinely non-obvious ingredient, and it is a theorem, not a definition:

$$A_e = \frac{G\,\lambda^{2}}{4\pi}$$

Substituting gives the Friis transmission equation:

$$\boxed{\;P_r = P_t\,G_t\,G_r\left(\frac{\lambda}{4\pi d}\right)^{2}\;}$$

Set both gains to unity and invert to get the free-space basic transmission loss — the loss between isotropic antennas, which is what ITU-R P.525 defines and names:

$$L_{bf} = \left(\frac{4\pi d}{\lambda}\right)^{2}, \qquad L_{bf}\,[\mathrm{dB}] = 20\log_{10}\frac{4\pi d}{\lambda}$$

The dB form, and its exact constants #

Rewrite in engineering units by expanding $\lambda = c/f$:

$$L_{bf}\,[\mathrm{dB}] = 20\log_{10} d + 20\log_{10} f + 20\log_{10}\frac{4\pi}{c}$$

With $c = 299\,792\,458$ m/s the last term is $-147.5522$ dB when $d$ is in metres and $f$ in hertz. Shifting units moves the constant by exactly 60 dB per factor of 1000:

ITU-R P.525-5 equation (6) prints the middle row rounded to one decimal:

$$L_{bf} = 32.4 + 20\log_{10} f_{\mathrm{[MHz]}} + 20\log_{10} d_{\mathrm{[km]}} \quad \mathrm{dB}$$

Free-space basic transmission loss in dB, computed from $20\log_{10}(4\pi d f/c)$. Two regularities are worth memorising: 20 dB per decade of distance and 20 dB per decade of frequency, in both cases because the term is a squared ratio. Doubling either adds 6 dB.

Why loss grows with frequency, when free space is lossless #

This is the point most treatments get wrong by omission. Trace the frequency dependence back through the derivation: it enters in exactly one place, the aperture relation $A_e = G\lambda^{2}/4\pi$, applied to the isotropic reference antenna at the receive end. Step 1, the spreading, has no $\lambda$ in it at all — a sphere's area does not care what frequency you illuminate it with.

An isotropic antenna, by definition of $G=1$, has an effective aperture of $\lambda^{2}/4\pi$. That area collapses with frequency:

The decisive test: hold physical aperture fixed at both ends instead of gain. Substituting $G = 4\pi A_e/\lambda^{2}$ at both ends into Friis:

$$P_r = P_t\,\frac{A_{e,t}\,A_{e,r}}{\lambda^{2} d^{2}}$$

The $\lambda^{2}$ has moved to the denominator. Received power now rises as $f^{2}$. With 1 m² apertures at both ends over a 10 km path:

Same dishes, same distance, same vacuum: 40 dB better at 100 GHz than at 1 GHz. This is why deep-space and satellite links climbed in frequency rather than down, and why mmWave cellular is viable at all — the array that fits in a handset has far more gain at 28 GHz than at 900 MHz. What actually costs you at mmWave is atmospheric absorption, rain, foliage and the total absence of diffraction, none of which appear in $L_{bf}$.

The ground is always there: two-ray and the d⁴ regime #

Put both antennas above a flat reflecting surface and the receiver sees two rays: the direct one and a ground-reflected one. Their path-length difference, for $d \gg h_t, h_r$, is

$$\Delta d = \sqrt{d^{2}+(h_t+h_r)^{2}} - \sqrt{d^{2}+(h_t-h_r)^{2}} \;\approx\; \frac{2 h_t h_r}{d}$$

giving a phase difference $\Delta\phi = 4\pi h_t h_r / (\lambda d)$. At grazing incidence the ground reflection coefficient tends to $\Gamma \to -1$ for both polarisations, so

$$P_r = P_t G_t G_r \left(\frac{\lambda}{4\pi d}\right)^{2}\left|1 - e^{-j\Delta\phi}\right|^{2} = P_t G_t G_r \left(\frac{\lambda}{4\pi d}\right)^{2} \cdot 4\sin^{2}\!\frac{\Delta\phi}{2}$$

Two regimes fall out:

• Close in, $\Delta\phi$ sweeps through multiples of $2\pi$ and the received power oscillates between four times free space (rays in phase) and a deep null (rays anti-phase). This is the fringing pattern every drive test shows.
• Far out, $\Delta\phi \ll 1$, so $\sin x \approx x$ and everything collapses:

$$P_r \;\approx\; P_t G_t G_r \left(\frac{h_t h_r}{d^{2}}\right)^{2} \qquad\Longrightarrow\qquad L\,[\mathrm{dB}] = 40\log_{10} d - 20\log_{10}(h_t h_r)$$

The wavelength has cancelled out entirely. Plane-earth loss is frequency-independent and falls at 40 dB/decade, not 20 — and it depends on antenna heights, which is why raising a base station 6 dB-worth (doubling its height) is a real coverage lever in a way that has no free-space analogue.

The crossover is the breakpoint distance, where the last in-phase maximum occurs ($\Delta\phi = \pi$). ITU-R P.1411-12 equation (3) states it directly, and models below-rooftop LoS propagation as $20\log_{10} d$ before it and $40\log_{10} d$ after:

$$R_{bp} = \frac{4 h_1 h_2}{\lambda}$$

Note the direction of the trend: the breakpoint moves further out as frequency rises, because $R_{bp} \propto 1/\lambda$. A 28 GHz link stays in the 20 dB/decade regime for kilometres, and if it survives absorption it degrades with distance much more gently than a 900 MHz link past 540 m.

For the 900 MHz macrocell above, the two models diverge fast:

Inside the breakpoint the plane-earth formula is not merely optimistic, it is inapplicable — its asymptotic derivation assumed $\Delta\phi \ll 1$.

flowchart LR
  TX["Transmitter"] --> SP["Spherical spreading, 1/d^2"]
  SP --> GR["Ground reflection: d^4 past breakpoint"]
  SP --> DF["Terrain and building diffraction"]
  SP --> GA["Gaseous absorption: O2 60 GHz, H2O 22 GHz"]
  SP --> RN["Rain scattering above ~5 GHz"]
  SP --> IO["Ionospheric reflection below ~30 MHz"]
  SP --> DU["Tropospheric ducting and scatter"]
  GR --> RX["Receive aperture, G lambda^2 / 4 pi"]
  DF --> RX
  GA --> RX
  RN --> RX
  IO --> RX
  DU --> RX

Obstacles: Fresnel zones and diffraction #

Line of sight in the optical sense is not sufficient for a radio link, because energy travels in a volume around the ray, not along a line. The Fresnel ellipsoids are the loci of points where the two-hop path A→M→B exceeds the direct path AB by $n\lambda/2$. ITU-R P.526-15 gives the radius of the $n$-th ellipsoid at a point splitting the path into $d_1$ and $d_2$:

$$R_n = \sqrt{\frac{n\,\lambda\,d_1 d_2}{d_1+d_2}}$$

and in practical units (equation 3), with $f$ in MHz, $d_1, d_2$ in km, result in metres:

$$R_n = 550\sqrt{\frac{n\,d_1 d_2}{(d_1+d_2)\,f}}$$

ITU-R P.530-18 equation (3) states the same thing for $n=1$ in GHz units — $F_1 = 17.3\sqrt{d_1 d_2/(f\,d)}$ metres — and the two agree, since $550/\sqrt{1000} = 17.39$.

The dimensionless diffraction parameter for a single knife edge (P.526-15 equation 26) is

$$\nu = h\sqrt{\frac{2}{\lambda}\left(\frac{1}{d_1}+\frac{1}{d_2}\right)}$$

with $h$ the obstacle-top height above the direct ray (negative below it). Since $F_1 = \sqrt{\lambda d_1 d_2/d}$, the two are related by $\nu = h\sqrt{2}/F_1$ — clearance expressed in Fresnel radii maps onto $\nu$ by a factor of $\sqrt 2$. The loss follows from the complex Fresnel integral (equation 30), with the closed-form approximation $J(\nu) \approx 6.9 + 20\log_{10}\!\left(\sqrt{(\nu-0.1)^2+1}+\nu-0.1\right)$ valid for $\nu > -0.78$:

Positive clearance = obstacle below the ray. Values from the exact Fresnel-integral form of P.526-15 eq. (30); the ripple above 0.6 F₁ is the same interference that produces the two-ray fringes.

Two consequences follow directly from $F_1 \propto \sqrt{\lambda}$. First, clearance requirements shrink with frequency, which is why an 80 GHz hop over a rooftop needs 58 cm and a 900 MHz link over the same geometry needs metres. Second, diffraction fill-in disappears with frequency — a knife edge that costs a VHF signal a few dB will cost a mmWave signal tens of dB, because $\nu$ scales as $1/\sqrt{\lambda}$ for a fixed obstruction height, which is why mmWave cells are line-of-sight or nothing and why a human body is an opaque obstacle at 60 GHz.

The atmosphere: oxygen at 60 GHz, water vapour at 22 GHz #

Above roughly 10 GHz the medium stops being transparent. ITU-R P.676 models this line by line: 44 oxygen resonances and 35 water-vapour resonances, with specific attenuation $\gamma = 0.1820\,f\,N''(f)$ dB/km.

Specific attenuation in dB/km at sea level (1013.25 hPa, 15 °C, water-vapour density 7.5 g/m³), evaluated from the ITU-R P.676-13 Annex 1 line-by-line method with the Recommendation's own spectroscopic tables.

Two features dominate the picture:

• The 60 GHz oxygen band. Near 60 GHz dozens of individual O₂ magnetic-dipole lines are pressure-broadened at sea level into one continuous band peaking at ≈ 15 dB/km. That is 15 dB you pay per kilometre on top of spreading. It is why the 57–71 GHz unlicensed band exists: the absorption guarantees spatial reuse, so dense short-range links (WiGig, 802.11ad/ay) do not interfere with each other a few blocks away.
• The 22.235 GHz water-vapour line. Much weaker — 0.18 dB/km at 7.5 g/m³ — but it scales linearly with humidity, so a tropical 30 g/m³ day is roughly four times worse, and it sits inconveniently close to the Ka satellite uplink band.

The atmospheric windows between the lines are where systems cluster: 38 GHz (0.12 dB/km), 77 GHz automotive radar (0.34 dB/km), 94 GHz imaging (0.40 dB/km). Note the ordering — 77 and 94 GHz are quieter than the 60 GHz band despite being higher in frequency, because absorption here is resonance physics, not a monotonic function of frequency.

Rain fade, and why Ku/Ka links carry weather margin #

Raindrops are comparable in size to a centimetre-scale wavelength, so they scatter and absorb. ITU-R P.838-3 models specific attenuation as a power law in rain rate:

$$\gamma_R = k\,R^{\alpha} \quad \mathrm{[dB/km]}, \qquad R \ \mathrm{in\ mm/h}$$

with $k$ and $\alpha$ tabulated per frequency and polarisation. Evaluating P.838-3 Table 5 directly (horizontal polarisation, the worse case — flatter, more oblate drops present a larger horizontal cross-section):

Specific attenuation in dB/km, horizontal polarisation, from ITU-R P.838-3 Table 5 coefficients. Vertical polarisation runs roughly 10–20% lower in this range; circular is between the two.

ITU-R P.530-18 puts the practical boundary bluntly: rain attenuation "can be ignored at frequencies below about 5 GHz", and above that "its importance increases rapidly". The numbers show why C band survived as the tropical-broadcast workhorse and why Ka-band VSATs need adaptive coding and modulation: at 30 GHz a 50 mm/h cell costs almost 10 dB per kilometre of rain traversed.

Two design consequences that fall out of the same table. Uplink is worse than downlink on both Ku and Ka because the uplink sits higher in frequency (14 vs 12, 30 vs 20 GHz) — hence uplink power control. And satellite fade depends on elevation angle, because a low-elevation path spends more kilometres inside the rain layer; the same site can be 3 dB worse at 20° than at 55° for identical weather.

Beyond line of sight #

Three mechanisms carry signals past the geometric horizon: ionospheric reflection at HF, tropospheric refraction and ducting from VHF upward, and forward scatter from tropospheric turbulence.

Ionospheric reflection, below ~30 MHz. Free electrons make the ionosphere a plasma with a refractive index below unity. The plasma frequency is

$$f_p = \frac{1}{2\pi}\sqrt{\frac{N e^{2}}{\varepsilon_0 m_e}} \approx 8.98\sqrt{N} \ \ \mathrm{Hz}, \qquad N \ \mathrm{in\ el/m^3}$$

At vertical incidence a wave is reflected if $f < f_p$ at the layer peak, and passes through otherwise — the critical frequency $f_0F2$. Typical F2 peak densities of $10^{11}$–$10^{12}$ el/m³ put $f_0F2$ between about 2.8 and 9 MHz. Oblique incidence extends this: the secant law gives a maximum usable frequency of $f_0F2\sec\varphi$, so a low-angle long hop can use roughly three times the vertical critical frequency, which is where the classic HF ceiling around 30 MHz comes from. Below the reflecting layers, the D region absorbs rather than reflects, with loss scaling as $\sec i / f^{2}$ — which is why AM broadcast dies out to a groundwave-only service in daylight and comes back at night when the D region recombines.

Above the critical frequency the ionosphere is transparent but not inert. ITU-R P.531-14 tabulates the residual effects for an Earth-space path, at 1 GHz, ~30° elevation, one-way, assuming a vertical TEC of 10¹⁸ el/m²:

The group delay is the same quantity GNSS receivers spend enormous effort removing: $\Delta t = 1.345\,N_T f^{-2}\times 10^{-7}$ seconds ($f$ in Hz, $N_T$ in el/m²), which at GPS L1 works out to metres-to-tens-of-metres of range error — hence dual-frequency receivers, which cancel it by exploiting exactly this $1/f^2$ law. The Faraday rotation figure is why VHF/UHF satellite links use circular polarisation: 108° of rotation would otherwise cost a linearly polarised link most of its signal (cross-polar discrimination is $-20\log_{10}\tan\theta_F$).

Tropospheric refraction and ducting, at VHF and above. The lower atmosphere's refractive index decreases with height, bending rays gently downward. The standard trick is the effective-Earth-radius factor $k$: replace the real Earth with one of radius $ka$ and let the rays travel straight. P.530-18 recommends $k = 4/3$ in the absence of data; ITU-R P.453-14 gives the reference exponential profile $N(h) = 315\exp(-h/7.35\,\mathrm{km})$ behind it.

P.453-14 also defines modified refractivity $M(h) = N(h) + 157h$ (h in km), which is the whole ducting theory in one line: 157 N-units/km is exactly the gradient at which a ray's curvature matches the Earth's. Where $\mathrm{d}N/\mathrm{d}h < -157$ N-units/km — a temperature inversion, a marine evaporation layer, subsidence over a warm sea — $\mathrm{d}M/\mathrm{d}h$ goes negative and the atmosphere traps energy in a waveguide. Ducts produce two symmetrical problems: your link fades because the duct steers energy off the receiver, and someone else's link interferes, because signals reach hundreds of kilometres past the horizon. Separately, forward scatter from refractive-index turbulence in the troposphere supports beyond-horizon links of a few hundred kilometres at a large penalty relative to free space; ITU-R P.617 is the prediction method for those.

Empirical models and the ranges you may not leave #

None of the above predicts the loss through a city, because a city is not a knife edge, a sphere or a gas. Empirical models fill that gap by curve-fitting measurement campaigns — which means they are only valid inside the parameter box that was measured.

Hata (1980), fitted to Okumura's Tokyo measurements, for urban areas:

$$L_{50} = 69.55 + 26.16\log_{10} f - 13.82\log_{10} h_b - a(h_m) + \bigl(44.9 - 6.55\log_{10} h_b\bigr)\log_{10} d$$

with $f$ in MHz, heights in metres, $d$ in km, and the mobile-height correction for small/medium cities

$$a(h_m) = \bigl(1.1\log_{10} f - 0.7\bigr)h_m - \bigl(1.56\log_{10} f - 0.8\bigr)$$

Suburban and open-area variants subtract $2\left[\log_{10}(f/28)\right]^{2}+5.4$ and $4.78(\log_{10}f)^{2}-18.33\log_{10}f+40.94$ respectively.

COST 231-Hata re-fits the same structure for the DCS/PCS bands:

$$L_{50} = 46.3 + 33.9\log_{10} f - 13.82\log_{10} h_b - a(h_m) + \bigl(44.9-6.55\log_{10}h_b\bigr)\log_{10} d + C_m$$

with $C_m = 0$ dB for medium cities and suburbs, 3 dB for metropolitan centres.

The distance term is the interesting part: $(44.9 - 6.55\log_{10} h_b)/10$ is the path-loss exponent, so $n = 3.52$ for a 30 m mast and $n = 2.98$ for a 200 m one. Urban clutter turns a free-space $n=2$ into $n\approx3.5$, and raising the antenna buys back part of the exponent rather than just an offset.

Fifty dB of excess loss at 10 km is what "urban" costs, and it is a median — shadow fading adds a log-normal spread on top (standard deviations of roughly 4–10 dB depending on environment; 3GPP TR 38.901 is commonly quoted at 4 dB for urban-macro LoS and 6 dB for NLoS, though those two figures come from a secondary summary here, not from the TR itself), which is what actually drives the fade margin in a link budget.

Picking the right model #