The unphysical free-space path loss

Propagation of wave energy to distance d, 2d, 3d
 showing how the size of the surface that the energy
spreads over increases in proportion to 
d², (2d)², and (3d)²

Many places it is stated that the lower the frequency, the smaller the free space loss. 

This is counterintuitive, because if one solves the wave equation, the solution is independent of frequency. The physics of the problem is that intensity just falls with distance squared according to how the surface of a sphere increases with the radius as shown in the figure.

The Friis formula used in link budgets says something else. It gives received power, Pr, as a function of transmitter power, Pt, receiver and transmitter antenna gains, Gr and Gt; wavelength, lambda, and distance, d:

The inverse of the squared term in brackets is usually called free-space path loss: 

The mystery is that it also varies with wavelength. The smaller the wavelength (i.e. higher frequency), the larger the loss. By the way, strictly speaking, this is not "loss", but "attenuation". In the limit as one approaches light and wavelengths get into the nanometer range, attenuation should increase much.

Voyager 1

As an example, consider Voyager 1 which at the time of writing is about four times the distance to Pluto, or about 25 billion km (25e12 m) away. Light takes more than 23 hours to travel from Voyager 1 to Earth. Free space loss according to the formula above is:

• FSPL = 308 dB (5.8e30) at 2.3 GHz, wavelength 13 cm
• FSPL = 416 dB (3.9e41) for light at 500 nm 

Why more than 108 dB difference? With such a disadvantage for light, one would think that observation of distant stars many light-years away would be impossible. But that's not the case!

In reality, waves spread according to the size of a sphere falling with distance squared, independent of wavelength, as shown above. Intensity for an isotropic (omnidirectional) transmitter will spread according to 

This factor amounts to 279 dB in the Voyager I example. The remainder in the free space loss formula is this factor:

It turns out that this is the effective area of an omnidirectional antenna at the wavelength in question. At 2.3 GHz an omnidirectional antenna is about half a wavelength, or 6-7 cm squared. At light frequencies, if we had treated light in the same way as radio waves, it would be 2-300 nm squared. The lower the frequency, the larger the antenna, and the larger the part of the expanding sphere of energy it will pick up. That is what the second factor tells us.

Interestingly, Danish-American Harald T Friis' original formulation from 1946 was different as shown below, although it also had the wavelength squared term.

It is very unfortunate for the understanding of the physics of wave propagation that one lumps the distance and the wavelength terms together and calls it free space loss.

Images of formulas from Wikipedia.

This blog post first appeared on the LA3ZA blog.