Free Space Path Loss Calculator
Work out the free space path loss for a radio link, then check the received power and the margin against your receiver.
What it does
Every radio link has a budget. The transmitter puts out a certain power, the antennas add a few decibels of gain, the signal spreads out as it travels and the receiver needs a certain power to work. The gap between what arrives and what is needed is the margin, and the first thing to work out is how much the signal loses on the way. That is what the free space path loss gives you: the loss in an empty world with nothing in the way.
The International Telecommunication Union sets out the method in Recommendation P.525, the recommendation for calculating free space attenuation. For a point to point link between two isotropic antennas the basic transmission loss works out to 20 log of 4 pi d divided by lambda, which written with the frequency and distance in convenient units becomes 32.4 plus 20 log of the frequency in megahertz plus 20 log of the distance in kilometres, in decibels. The same recommendation is what the radio engineering texts build on.
Two features of that equation explain most of what people find surprising about radio. The loss grows with the square of the distance, so each doubling of the range costs 6 dB and each tenfold costs 20 dB. It also grows with the square of the frequency, so moving from 900 MHz to 2.4 GHz costs about 8.5 dB at the same distance, which is why the higher bands do not reach as far.
The page takes the frequency and the distance, applies the standard formula, and then finishes the link budget. Enter the transmit power, the gain of each antenna and the sensitivity of the receiver, and it returns the received power in dBm and the margin left over. The wavelength is shown too, because it tells you whether the link is short compared with the size of the antenna that is doing the work.
One point about the constant. The recommendation uses 32.4, which comes from taking the speed of light as 300 million metres per second. Work the same equation with the exact value of 299792458 and the constant becomes 32.45, so the two differ by about 0.05 dB, well below the accuracy of any real radio measurement. This page follows the recommendation exactly.
How to use
- Choose metric or imperial for the distance.
- Enter the operating frequency in megahertz.
- Enter the distance between the two antennas.
- Enter the transmit power in watts and the gain of the transmit and receive antennas in dBi.
- Enter the receiver sensitivity in dBm, taken from the data sheet.
- Read the received power and the margin. A positive margin means the link should work in free space, before anything else is taken into account.
Formula
- f
- Frequency in megahertz.
- d
- Distance between the antennas in kilometres.
- Pt
- Transmit power in watts.
- Gt and Gr
- Antenna gains in dBi, where 0 dBi means an isotropic reference antenna.
- Lbf
- Free space basic transmission loss in decibels.
The formula gives the loss between two isotropic antennas in free space and nothing else. A real link also carries cable losses, polarisation mismatch, the loss through walls and foliage, and the extra loss from diffraction around obstacles, so the margin you calculate here is the best case. The 32.4 constant matches Recommendation ITU-R P.525; using the exact speed of light gives 32.45 and a difference of roughly 0.05 dB.
Worked example
Example: a 2.4 GHz link over half a kilometre
Defaults: metric units, 2400 MHz, 0.5 km, 0.1 W of transmit power, 2 dBi on each antenna and a receiver that needs -110 dBm.
- Wavelength: 299792458 / 2.4 x 10 to the power of 9 = 0.1249 m.
- Frequency term: 20 log10(2400) = 67.604 dB.
- Distance term: 20 log10(0.5) = -6.021 dB.
- Path loss: 32.4 + 67.604 - 6.021 = 93.98 dB.
- Transmit power in dBm: 10 log10(0.1) = -10 dBm.
- Received power: -10 + 2 + 2 - 93.98 = -99.98 dBm.
- Link margin: -99.98 - (-110) = +10.02 dB, and the tool shows 93.98 dB, -99.98 dBm, +10.02 dB and 0.1249 m.
Reference table
| Frequency | 0.1 km | 0.5 km | 1 km | 5 km | 10 km |
|---|---|---|---|---|---|
| 433 MHz | 65.13 dB | 79.11 dB | 85.13 dB | 99.11 dB | 105.13 dB |
| 868 MHz | 71.17 dB | 85.15 dB | 91.17 dB | 105.15 dB | 111.17 dB |
| 2400 MHz | 80.00 dB | 93.98 dB | 100.00 dB | 113.98 dB | 120.00 dB |
| 5800 MHz | 87.67 dB | 101.65 dB | 107.67 dB | 121.65 dB | 127.67 dB |
Limits
- Free space only. Walls, floors, foliage, rain and the shape of the ground all add loss that this calculation does not see.
- The link budget is a single hop between two antennas, so it does not cover reflections, multipath or the fading that a real link sees.
- Antenna gains are taken at face value and ignore misalignment, polarisation mismatch and the efficiency of the antenna itself.
- Both antennas are assumed to sit in the open with a clear line of sight.
Common mistakes
- Using kilometres when the formula expects kilometres and entering miles. Convert before you take the logarithm, or the answer is wrong by a constant.
- Mixing dBm and watts. A transmit power of 0.1 W is -10 dBm, and dBm is what the whole link budget is written in.
- Forgetting that dBi and dB are different things. dBi is an absolute gain against an isotropic antenna, while dB is a plain ratio.
- Expecting the same answer at twice the frequency. Path loss rises with the square of the frequency, so the higher band loses more.
- Treating the free space figure as the whole story. Any real path through buildings or foliage adds several decibels that the margin has to cover.
Try it yourself
Quick checks. Answers sit tucked under each question.
What is the free space path loss at 2400 MHz over a distance of 1 km?
100.00 dB, from 32.4 + 20 log10(2400) + 20 log10(1) = 32.4 + 67.604 + 0.
How much more loss do you get if you double the distance?
6.02 dB more. Path loss grows with the square of the distance, so each doubling of the range costs 20 log10(2) = 6.02 dB.
What power reaches the receiver in the worked example?
-99.98 dBm. The transmitter puts out -10 dBm, the two antennas add 4 dB, and the path loss takes away 93.98 dB.
Questions
What is free space path loss?
The reduction in signal power caused by the signal spreading out as it travels through empty space. It is the baseline loss that every real radio link starts from.
Why does the loss grow so fast with distance?
Because the signal spreads over the surface of a sphere, so the power per square metre falls with the square of the distance. Each doubling of the range costs 6 dB and each tenfold costs 20 dB.
What does dBi mean?
Gain in decibels against an isotropic reference antenna, the theoretical point source that radiates equally in every direction. 0 dBi is the reference, 2 dBi is a modest gain, and 9 dBi is a small patch antenna.
What is link margin?
The difference between the power arriving at the receiver and the power the receiver needs. A positive margin means the link should work, and a healthy design keeps several decibels in hand.
Does this cover a link through buildings?
No. Free space path loss is the best case. Walls, floors and people add attenuation on top, and those losses have to be measured or looked up for the materials involved.
Where does the 32.4 constant come from?
From the frequency in megahertz and the distance in kilometres, using a speed of light of 300 million metres per second. Using the exact value gives 32.45, a difference of about 0.05 dB.
Sources
- ITU-R Recommendation P.525: calculation of free space attenuation
- ITU-R P.525-5 full text with the point to point loss equation
- Georgia Tech wireless systems lecture notes: free space loss and the 32.4 constant