 called line-of-sight propagation. The power of the radio wave is reduced with distance just as a cable attenuates propagating electrical signals. The attenuation of a radio wave, free-space loss, on the line-of-sight path is due to the spherical dispersion of the radio wave. Here we assume that both antennas are isotropic antennas, which radiate to and receive from all directions equally. The transmitted power from isotropic antennas is distributed over a spherical surface and the radiated power per unit area decreases in proportion to the square of the radius because the area of the spherical surface increases in proportion to the square of the radius. The area of the spherical surface follows the equation A = 412, where / is the radius. The power density flow Fthrough the surface of a sphere at distance l from isotropic antenna becomes F = PT /(4rl2) [W/m2] as shown in Figure 4.11 [3]. The receiving antenna is able to receive the power that passes through its effective aperture area or capture area [4]. The effective aperture area of the receiving isotropic antenna is proportional to the square of the wavelength according to Aei = 22/(4), and received power becomes PR = Aei F [[W]. From these two facts we can easily derive that the free-space loss, that is, the ratio of transmitted power and the received power in the case of isotropic antennas, where antenna gains gT = gR = 1, is Introduction to Telecommunications Network Engineering Effective aperture area of the Transmitting receiving antenna isotropic antenna PR=Ae F Figure 4.11 Radio wave loss with isotropic transmitting antenna. (4.3) where A is the wavelength, fis the frequency of the signal, C is the speed of light, and / is the transmission distance (Figure 4.11). We usually prefer to describe attenuation or loss in decibels instead of by the absolute value as given in the previous equations. We obtain the for- mula that gives decibel values by taking Ldb = 10 logio L. Now if we express the frequency f in gigahertz ( f = fGHz 10° ) and l in kilometers 