24GHz Wireless

Very fast wireless that claims real throughput of more than 1Gbps. For example, the airFibre from Ubiquiti. Ofcom has this to say about the part of the 24GHz radio spectrum used by this equipment in the UK Bandplan:

  1. Non-government low power devices in the radiolocation services are limited to:

    a. Portable and fixed applications between 24.15-24.25 GHz; and

    b. Mobile applications between 24.25-24.35 GHz on a NIB to the radionavigation service

    Power flux-density at 10 metres from the system antenna in the direction of maximum radiation is not to exceed \(1.5 mW/m^2\) without approval

  2. 24.05-24.25 GHz is used by the Amateur service. The part of the allocation between 24.05 and 24.150 GHz may only be used with written consent of the Secretary of State.

    Home Office/Office of The Scottish Executive for the Emergency Services between 24.05-24.15 GHz

    ISM apparatus may use the band 24.0-24.25 GHz.

Also, the UK Spectrum Strategy 2000 (appendix B) allows this frequency range to be used for “Technology Development” for low-power fixed and portable devices. In this context, “low-power” means less than \(1.5 mW/m^2\).

The regulations are confusingly laid out, it would appear from the arrangement of the text that the sentence “The part of the allocation between 24.05 and 24.15 GHz may only be used…” pertains only to the amateur service, but it seems to be a global restriction, particularly since another Ofcom document on Vehicle Mounted Radar Detectors mentions that the police use this part of the band for radar speed meters.

So generally it is allowed to use this equipment in the UK, it is just necessary to check the output power and keep away from the bottom half of the band. The airFibre gives values in \(dBm\) and the government, gives the permitted radiation energy per unit area at 10 meters from the system. So we need to do a bit of arithmetic.

The datasheet says that the beam-width is less than 3.5°. Let’s make a simplifying assumption that it is exactly that width and further that the cross-section of the main lobe of radiation is circular. Generally the radius of the cross-section will be,

Radiation Cone

[ r = \ell sin(\frac{\theta}{2}) = 10m \times sin(1.25°) = 0.22m ]

The area of this circle is, \(\pi r^2\ = 0.15m^2\).

So now, how much power goes through one square meter when transmitting at 33 dBm?

[ p = \frac{10^{\frac{\beta}{10}}}{a} = \frac{10 ^ {33dBm/10}}{0.15m^2} = \frac{1995mW}{0.15m^2} = 13.3 W / m^2 ]

This would seem to be well in excess of what is allowed by the rules…

Is this number, \(13 W / m^2\), reasonable?