Friis transmission equation

Path loss = 20 log (4 * pi * r / lambda)

[Note: use log base 10]

where,

Path loss = signal attenuation. Unit: dB

pi = 22 / 7

r = distance between transmitter and receiver. Unit: m

lambda = wavelength of signal. Unit: m

Wavelength = C/f

where,

C = speed of electromagnetic waves in free space

= 299792458. Unit: m / s

f = frequency of signal. Unit: Hz

Let r = R * (2 ^ x)

Path loss = 20 log (4 * pi * R * L / lambda)

= 20 log (4 * pi * R / lambda) + 20 log (2 ^ x)

= 20 log (4 * pi * R / lambda) + 6x

The above derivation implies range(R) doubles every 6dB of path loss.

The path loss is +8.519dB more over a given range for the 2.4 GHz compared to 900MHz for the same range. In other words operating at 900 MHz exhibits a significantly longer range than is possible at 2.4 GHz.

links for 2007-03-14