Free-space path loss

The free-space path loss (FSPL) formula derives from the Friis transmission formula. The free-space path loss is the loss factor in this equation that is due to distance and wavelength, or in other words, the ratio of power transmitted to power received assuming the antennas are isotropic and have no directivity (D_t = D_r = 1): \begin{align} \mbox{FSPL} = \left ( \frac{4\pi d} \lambda \right )^2 \end{align} Since the frequency of a radio wave f is equal to the speed of light c divided by the wavelength, the path loss can also be written in terms of frequency: \begin{align} \mbox{FSPL} = \left({4\pi df \over c}\right)^2 \end{align} Beside the assumption that the antennas are lossless, this formula assumes that the polarization of the antennas is the same, that there are no multipath effects, and that the radio wave path is sufficiently far away from obstructions that it acts as if it is in free space. This last restriction requires an ellipsoidal area around the line of sight out to 0.6 of the Fresnel zone be clear of obstructions. The Fresnel zone increases in diameter with the wavelength of the radio waves. The concept of free space path loss is often applied to radio systems that don't completely meet these requirements, but these imperfections can be accounted for by small constant power loss factors that can be included in the link budget. ==Influence of distance and frequency==

, because the same amount of power spreads over an area proportional to the square of distance from the source. The free-space loss increases with the distance between the antennas and decreases with the wavelength of the radio waves due to these factors: • Intensity (I) – the power density of the radio waves decreases with the square of distance from the transmitting antenna due to spreading of the electromagnetic energy in space according to the inverse square law ==Derivation==

The radio waves from the transmitting antenna spread out in a spherical wavefront. The amount of power passing through any sphere centered on the transmitting antenna is equal. The surface area of a sphere of radius d is 4\pi d^2. Thus the intensity or power density of the radiation in any particular direction from the antenna is inversely proportional to the square of distance :I \propto {P_t \over 4\pi d^2} (The term 4\pi d^2 means the surface of a sphere, which has a radius d. Please remember, that d here has a meaning of 'distance' between the two antennas, and does not mean the diameter of the sphere (as notation usually used in mathematics).) For an isotropic antenna which radiates equal power in all directions, the power density is evenly distributed over the surface of a sphere centered on the antenna :I = {P_t \over 4\pi d^2} \qquad \qquad \qquad \text{(1)} The amount of power the receiving antenna receives from this radiation field is :P_r = A_\text{eff}I \qquad \qquad \qquad \text{(2)} The factor A_\text{eff}, called the effective area or aperture of the receiving antenna, which has the units of area, can be thought of as the amount of area perpendicular to the direction of the radio waves from which the receiving antenna captures energy. Since the linear dimensions of an antenna scale with the wavelength \lambda, the cross sectional area of an antenna and thus the aperture scales with the square of wavelength \lambda^2. The effective area of an isotropic antenna (for a derivation of this see antenna aperture article) is :A_\text{eff} = {\lambda^2 \over 4\pi} Combining the above (1) and (2), for isotropic antennas :P_r = \Big({P_t \over 4\pi d^2}\Big)\Big({\lambda^2 \over 4\pi}\Big) :\text{FSPL} = {P_t \over P_r} = \Big({4\pi d \over \lambda}\Big)^2 == Free-space path loss in decibels ==

A convenient way to express FSPL is in terms of decibels (dB): : \begin{align} \operatorname{FSPL}(\text{dB}) &= 10\log_{10}\left(\left(\frac{4\pi d f}{c}\right)^2\right) \\ &= 20\log_{10}\left(\frac{4\pi d f}{c}\right) \\ &= 20\log_{10}(d) + 20\log_{10}(f) + 20\log_{10}\left(\frac{4\pi}{c}\right) \\ &= 20\log_{10}(d) + 20\log_{10}(f) -147.55, \end{align} using SI units of meters for d, hertz (s−1) for f, and meters per second (m⋅s−1) for c, (where c=299 792 458 m/s in vacuum, ≈ 300 000 km/s) For typical radio applications, it is common to find d measured in kilometers and f in gigahertz, in which case the FSPL equation becomes :\operatorname{FSPL}(\text{dB}) = 20\log_{10}(d_{km}) + 20\log_{10}(f_{GHz}) + 92.45, an increase of 240 dB, because the units increase by factors of and respectively, so: :20\log_{10}(10^{3}) + 20\log_{10}(10^{9}) = 240. (The constants differ in the second decimal digit when the speed of light is approximated by 300 000 km/s. Whether one uses 92.4, 92.44 or 92.45 dB, the result will be OK as the average measurement instruments cannot provide more accurate results anyway. A logarithmic scale is introduced to see the important differences (i.e. order of magnitudes), so in engineering practice dB results are rounded) ==See also==