Between the antenna that sends and the antenna that receives, a radio signal spreads thin, gets chewed by walls and rain, bends around hills, bounces off the upper atmosphere and interferes with its own echoes. This chapter follows it all the way and adds up the losses in decibels, the way engineers do.
Common mix-up: in empty space, higher frequencies don't "fade faster" because the air eats them. Every frequency spreads out exactly the same way. The extra loss in the path-loss formula comes from the receiving antenna getting smaller as the wavelength shrinks. Real absorption by walls, water and oxygen is a separate effect, covered below.
Picture an antenna radiating equally in all directions. Its power flows outward through bigger and bigger spheres. A sphere's area is 4πr², so at twice the distance the same power covers four times the area, and each square meter gets a quarter as much. Nothing is absorbed; the energy is all still there, just thinner. That's the inverse-square law.
A 100 mW Wi-Fi radio gives a power density of about 80 microwatts per square meter at 10 m, and 0.8 µW/m² at 100 m. How much of that the receiver actually collects depends on its antenna's catching area, its effective aperture (see Antennas). For an ideal isotropic antenna that area is λ²/4π, about 12.4 cm² at 2.4 GHz.
Put those two facts together and you get free-space path loss (FSPL), the loss between two isotropic antennas with nothing in the way. For 2.4 GHz at 10 m it's 60 dB: a factor of a million. At 100 m it's 80 dB, at 1 km 100 dB. Every doubling of distance adds 6 dB; every tenfold adds 20 dB.
Those numbers are why engineers count in decibels. A decibel is ten times the base-10 logarithm of a power ratio: +3 dB is double, +10 dB is ten times, −20 dB is a hundredth. Losses that multiply become losses that add. dBm pins the scale to one milliwatt: 0 dBm = 1 mW, 20 dBm = 100 mW (a typical Wi-Fi router), 30 dBm = 1 W. A phone transmits at most around 23 dBm, about 200 mW. On the receive side, −50 dBm is a strong Wi-Fi signal, −70 dBm is getting weak, and a phone can still decode about −100 dBm, a ten-billionth of a milliwatt. GPS arrives at about −130 dBm, weaker than the background noise; the receiver digs it out by correlating against a known code.
Power density at distance d from an isotropic transmitter: S = Pt / 4πd². An isotropic receiver collects Ae = λ²/4π of it. So Pr/Pt = (λ/4πd)², and FSPL is the inverse, (4πd/λ)² = (4πdf/c)². Take 10 log₁₀ and the three terms fall out.
In the units link planners like: FSPL = 20 log₁₀ dkm + 20 log₁₀ fMHz + 32.44.
How weak is too weak? Every receiver hears thermal noise of kT per hertz of bandwidth, which at room temperature is −174 dBm/Hz. A 20 MHz Wi-Fi channel adds 10 log₁₀(2×10⁷) ≈ 73 dB, giving a floor near −101 dBm, and the receiver's own electronics add a few dB more (the noise figure). Signals must sit some decibels above that floor to be decoded, more for faster modulations.
The formula has a 20 log₁₀ f term, so at the same distance 5.5 GHz loses 7.2 dB more than 2.4 GHz. The popular conclusion is that "the air absorbs higher frequencies." In free space, that's simply not what's happening.
Look back at the derivation. The power density at distance d, P/4πd², doesn't mention frequency at all. A 5.5 GHz wave and a 2.4 GHz wave from equal transmitters are equally strong at 100 m. The frequency only enters through the receiving antenna: an isotropic (or dipole-like) antenna's catching area is proportional to λ², so at 5.5 GHz it catches (2.4/5.5)² ≈ 19% as much. The "loss" is a smaller bucket, not a leakier pipe.
The proof is what happens when you hold the antenna size fixed instead. A fixed-size dish has the same catching area at any frequency, so the received power stops depending on frequency. Put fixed-size dishes at both ends and higher frequency actually wins, because the transmitting dish also squeezes its beam tighter. Two 60 cm dishes 10 km apart deliver about 12 dB more at 24 GHz than at 6 GHz. That's why point-to-point microwave links and satellites climb to high frequencies.
So why does 5 or 6 GHz Wi-Fi really reach less far than 2.4 GHz in a house? Three real reasons: the small antennas in phones and laptops are roughly fixed in gain, not in size, so the λ² penalty applies; walls and people absorb more at higher frequencies; and shorter waves diffract less around obstacles. None of these is "the air."
Substitute G = 4πAe/λ² for both antennas and Friis becomes Pr/Pt = At Ar / (λ² d²). Now λ is in the denominator: shorter wavelength, more power, for the same physical apertures.
The worked example: a 60 cm dish at 60% efficiency has 29.3 dBi at 6 GHz and 41.4 dBi at 24 GHz. FSPL over 10 km is 128.0 dB and 140.1 dB. Net: 2 × 29.3 − 128.0 = −69.4 dB versus 2 × 41.4 − 140.1 = −57.3 dB. The higher band is about 12 dB better, before rain and water vapor take their share (24 GHz sits close to the 22 GHz water-vapor line).
Real paths aren't empty. Every wall takes a bite, and the bites add in decibels. Rough figures at 2.4 GHz: drywall about 3 dB, a clear glass window about 2 dB, a brick wall about 10 dB, a concrete wall 10–20 dB, a concrete floor more. Metal is close to a wall of mirror. Energy-saving low-E glass has a thin metal coating and can be far worse than plain glass. These are approximate; construction varies a lot, and losses usually rise with frequency. The table further down collects them.
Water is the big absorber. Its molecules are electric dipoles that try to twist with the field and dissipate energy as heat. People are mostly water, so a person standing in the path can cost several decibels; a crowd in a stadium is a Wi-Fi engineer's nightmare. Trees in leaf do the same. Microwave ovens use this effect at 2.45 GHz, though not because of any special water "resonance" there (see The 2.4 GHz band).
The air itself is nearly transparent below about 10 GHz, losing around 0.01 dB per km. Higher up there are absorption peaks: water vapor at 22.2 GHz (about 0.2 dB/km at sea level) and oxygen near 60 GHz, where the loss reaches about 15 dB per kilometer. That sounds bad and is sometimes useful: 60 GHz links are short-range by nature, so neighbors can't interfere with each other.
Rain matters once raindrops are not tiny compared with the wavelength. In a heavy downpour of 25 mm per hour, the loss is roughly 1 dB per km at 12 GHz and roughly 5 dB per km at 30 GHz. A thunderstorm cell several kilometers deep between you and a satellite is why satellite TV (around 12 GHz) blinks out in a storm, and why higher-band satellite links need bigger margins.
Gas molecules absorb at frequencies that match their quantum transitions. Water vapor has a rotational line at 22.235 GHz and a much stronger one at 183.3 GHz. Oxygen, unusually, has a magnetic dipole moment, and a cluster of its transitions near 60 GHz merges into one broad band at sea-level pressure, with another line at 118.75 GHz. Between the peaks are "windows," like the band near 94 GHz used by imaging radars.
Rain attenuation is usually modeled as γ = k Rα dB/km, with R the rain rate in mm/h and k and α tabulated by frequency and polarization (ITU-R Recommendation P.838). Both grow quickly through the tens of gigahertz.
Radio doesn't only travel in straight lines. Reflection off metal is nearly total, off ground and water partial, and almost total again at grazing angles. Refraction bends a wave when its speed changes: the air's refractive index falls slightly with height, so radio paths curve gently downward and reach a little beyond the visual horizon. Occasionally a temperature inversion forms a "duct" that carries VHF and UHF signals hundreds of kilometers.
Diffraction lets waves bend around edges, and how much depends on the obstacle's size compared with the wavelength. An AM wave at 1 MHz is 300 m long; a 100 m hill is small next to it, and the signal flows around. FM at 3 m bends less. A 5 GHz wave, 6 cm long, casts sharp shadows behind a building.
At low and medium frequencies, a vertically polarized wave also clings to the Earth as a ground wave, following its curvature. Loss depends on the ground's conductivity, so seawater is best and dry sand worst. That's how a big AM station covers a radius of 100 km or more by day, and how very low frequency transmitters reach submarines across oceans.
Above 60 km, sunlight's ultraviolet and X-rays ionize the thin air into the ionosphere: the D layer (about 60–90 km), the E layer (about 90–150 km) and the F layer (about 150–500 km, splitting into F1 and F2 by day). Free electrons there turn back radio waves below a certain frequency. Shortwave signals bounce off the F layer and come down 2,000–4,000 km away, and can hop several times around the world. Between the end of the ground wave and the first landing there is often a skip zone where nothing is heard.
Day and night change everything. By day the low D layer absorbs medium-wave signals, so AM stations reach only as far as their ground wave. After sunset the D layer fades within hours, the AM skywave reflects off the layers above, and stations 1,000 km away come booming in. That's why many US AM stations must cut power or change their antenna pattern at night.
An ionized gas has a natural oscillation frequency set by its electron density N (electrons per m³): fp ≈ 8.98 √N Hz. A wave below fp can't propagate in it and is turned back. With N = 10¹² per m³, typical of the daytime F2 layer, fp ≈ 9 MHz.
A wave hitting the layer obliquely only needs to be turned through a smaller angle, which is why the secant law raises the usable frequency at low elevation angles, to around 30 MHz on long daytime paths near solar maximum. The 11-year sunspot cycle moves all of these numbers.
The D layer absorbs rather than reflects because it's low and dense with neutral molecules: electrons set oscillating by the wave collide and lose the energy as heat. Absorption falls roughly as 1/f², so it hits AM hard and higher shortwave lightly.
Above about 30 MHz the ionosphere mostly lets signals through and ground waves die quickly, so VHF, UHF and microwave links need a line of sight. The Earth's curvature sets the limit. Thanks to the gentle downward bending in the atmosphere, engineers pretend the Earth is 4/3 its real size and get a radio horizon of about 4.12√h km for an antenna h meters up, about 15% farther than the visual horizon (3.57√h).
A phone at 1.5 m sees a radio horizon of 5.0 km; a 30 m cell tower, 22.6 km; and the two together can just reach each other at about 27.6 km. A 300 m TV mast reaches about 71 km. An airliner at 10,000 m sees about 412 km, which is why air traffic VHF radio works so far.
"Line of sight" is not a thin line, though. Energy travels through a football-shaped region around the direct path called the first Fresnel zone: the set of points where a detour adds no more than half a wavelength. Obstacles that poke into it cause loss even when you can see the other end. For a 2.4 GHz link 1 km long, the zone is 5.6 m in radius at the middle; planners try to keep at least 60% of it, about 3.4 m, clear. At 5 km it's 12.5 m, which is why long links go on towers even over flat ground.
The distance to the horizon from height h on a sphere of radius R is √(2Rh + h²) ≈ √(2Rh). With R = 6,371 km that's 3.57√h km. Multiply R by k = 4/3 to account for refraction and it becomes 4.12√h km. Two antennas can see each other out to the sum of their horizons.
For a point at distance r off the line, the detour is about r²(d₁ + d₂)/(2d₁d₂). Set it equal to λ/2 and solve for r: that's r₁. The next zone (detour up to one full wavelength) reflects energy that arrives out of step, which is why a reflection from flat ground can cancel the direct signal.
Indoors, and in cities, a signal reaches the receiver by many routes at once: straight from the transmitter and bounced off floors, walls, cars and buildings. Each copy arrives with its own delay and phase. Where they arrive in step they add; where they're out of step they cancel. Move the receiver half a wavelength (6 cm at 2.4 GHz) and the sum can swing by 20–30 dB. That's fading, and it's why nudging a laptop a few centimeters sometimes fixes a bad connection. When no single path dominates, the statistics are called Rayleigh fading.
A car at 100 km/h drives through this pattern fast: at 2 GHz the fades are 7.5 cm apart, so it crosses about 370 of them per second. Echoes with long delays also smear one symbol into the next, which modern radios defeat with OFDM and guard intervals (see Putting information on a wave).
Radios fight back with diversity: two antennas half a wavelength apart rarely fade at the same moment, so the receiver uses whichever is stronger. MIMO goes further and turns multipath into capacity. Because each transmit antenna reaches each receive antenna by a different mix of paths, the receiver can untangle several data streams sent on the same frequency at the same time. Without the echoes, there would be nothing to tell them apart.
Finally, the bookkeeping that ties the chapter together: the link budget. Start with transmit power in dBm, add both antenna gains, subtract path loss and every obstacle, and compare what's left with the receiver's sensitivity. The difference is the margin, and you want 10–20 dB of it to ride out fades. Example: a router at 20 dBm with a 2 dBi antenna, a laptop with 0 dBi, 15 m away through two drywalls at 2.4 GHz: 20 + 2 + 0 − 63.6 − 6 ≈ −47.6 dBm. Against a −82 dBm sensitivity, that's 34 dB of margin, plenty for the fastest rates.
What decides the speed is signal-to-noise ratio. With a 6 dB noise figure the 20 MHz noise floor is about −95 dBm, so the example above has an SNR of about 47 dB. Shannon's limit, C = B log₂(1 + SNR), gives about 315 Mb/s for one 20 MHz stream at that SNR. Real radios fall well short of Shannon: Wi-Fi 6's best 20 MHz single-stream rate is about 143 Mb/s, and it needs a strong signal to use it.
Each modulation needs a minimum SNR. The 802.11n standard writes this as minimum receiver sensitivities for a 20 MHz channel: −82 dBm for its slowest rate (6.5 Mb/s, BPSK) up to −64 dBm for 65 Mb/s (64-QAM). 802.11ac adds 256-QAM at about −59 dBm. Good receivers beat these minimums by several dB. Instrument 1 uses this ladder.
Set the transmitter, antennas, frequency and obstacles. Drag the marker on the plot (or use the distance slider) to move the receiver; the curve is received power at every distance.
A router 1.2 m above a reflecting floor, in a room 4 m wide. Drag the receiver anywhere. The background shows the signal everywhere, relative to the direct path alone: teal where the echo helps, red where it cancels.
Pick day or night, a frequency and a launch angle. Heights are stretched several times over so the layers are visible; the hop distances in the readout use true geometry.
| Obstacle | ≈ 2.4 GHz | ≈ 5 GHz | Notes |
|---|---|---|---|
| Drywall | 3 dB | 4 dB | Studs, pipes and wiring inside add more |
| Clear glass | 2 dB | 3 dB | Metal-coated low-E glass can be 10–25 dB or worse |
| Wooden door | 3 dB | 4 dB | A metal fire door is close to a wall |
| Brick wall | 10 dB | 13 dB | Varies with thickness and moisture |
| Concrete wall | 10–20 dB | 15–25 dB | Rebar and thickness dominate |
| Concrete floor | 15–25 dB | 20–30 dB | Why one router rarely covers two storeys well |
| Human body | 3–6 dB | 5–8 dB | Mostly water |
| Heavy rain, per km | < 0.1 dB | ≈ 0.1 dB | ≈ 1 dB/km at 12 GHz, ≈ 5 dB/km at 30 GHz |