# Free-space path loss

(Redirected from Free space loss)

In telecommunication, free-space path loss (FSPL) is the attenuation of radio energy between the feedpoints of two antennas that results from the combination of the receiving antenna's capture area plus the obstacle free, line-of-sight path through free space (usually air).[1] The "Standard Definitions of Terms for Antennas", IEEE Std 145-1993, defines "free-space loss" as "The loss between two isotropic radiators in free space, expressed as a power ratio."[2] Despite this name and definition, the FSPL includes a receiving antenna aperture component in the total attenuation.[1] It does not include any loss associated with hardware imperfections, or the effects of any antenna gains. A discussion of these losses may be found in the article on link budget. The FSPL is rarely used standalone, but rather as a part of the Friis transmission formula, which includes the gain of antennas.[3]

## Free-space path loss formula

The free-space path loss (FSPL) formula derives from the Friis transmission formula[3] that states power gain of an antenna system thus...

${\displaystyle {\frac {P_{r}}{P_{t}}}=D_{t}D_{r}\left({\frac {\lambda }{4\pi d}}\right)^{2}}$

The FSPL formula expresses a loss value that is the reciprocal of gain and assumes the directivity for the transmit and receive antennas are isotropic and therefore unity. Both modifications simplify the equation to...

{\displaystyle {\begin{aligned}{\mbox{FSPL}}=\left({\frac {4\pi d}{\lambda }}\right)^{2}\end{aligned}}}

where:

• ${\displaystyle \ \lambda }$ is the signal wavelength,
• ${\displaystyle \ d}$ is the distance between the antennas,
• ${\displaystyle \ \lambda }$ and ${\displaystyle \ d}$ are in the same unit of length,
• ${\displaystyle \ d\gg \lambda }$ such that both antennas are in the far field of each other.[4]

Substitute ${\displaystyle c/f}$ for ${\displaystyle \lambda }$ to calculate from frequency where:

• ${\displaystyle \ c}$ is the speed of light (in metres/s),
• ${\displaystyle \ f}$ is frequency (in Hz).

## Antenna effective aperture and free space loss

Despite the misleading name, the free-space path loss formula inherits two important effects from the Friis Transmission Formula:

• Intensity (${\displaystyle I}$) - the power density due to path loss by spreading of the electromagnetic energy yielding loss proportional to the square of distance;[1]
• Antenna Capture Area (${\displaystyle A_{eff}}$) - the capability of the receiving antenna to capture electromagnetic energy yielding loss inversely proportional to the square of wavelength or proportional to the square of frequency.[1]

Hence the equation for FSPL is:

{\displaystyle {\begin{aligned}{\mbox{FSPL}}\propto {\frac {1}{I\cdot A_{eff}}}\end{aligned}}}

Friis implied the antennas are isotropic.[3] Hence the power density ${\displaystyle I}$ from the isotropic transmit antenna vs. distance from the antenna ${\displaystyle d}$ is...

{\displaystyle {\begin{aligned}|I|\propto {\frac {1}{4\pi d^{2}}}\end{aligned}}}

The effective area of the isotropic antenna is...

{\displaystyle {\begin{aligned}A_{eff}={\frac {\lambda ^{2}}{4\pi }}\end{aligned}}}

Combining the two yields...

{\displaystyle {\begin{aligned}{\mbox{FSPL}}&=\left({\frac {4\pi d^{2}}{1}}\right)\left({\frac {4\pi }{\lambda ^{2}}}\right)\\&=\left({\frac {16\pi ^{2}d^{2}}{\lambda ^{2}}}\right)\\&=\left({\frac {4\pi d}{\lambda }}\right)^{2}\end{aligned}}}

## Free-space path loss in decibels

A convenient way to express FSPL is in terms of dB:

{\displaystyle {\begin{aligned}{\mbox{FSPL(dB)}}&=10\log _{10}\left(\left({\frac {4\pi df}{c}}\right)^{2}\right)\\&=20\log _{10}\left({\frac {4\pi df}{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{aligned}}}

where the units are as before.

For typical radio applications, it is common to find ${\displaystyle \ f}$ measured in units of GHz and ${\displaystyle \ d}$ in km, in which case the FSPL equation becomes

${\displaystyle \ {\mbox{FSPL(dB)}}=20\log _{10}(d)+20\log _{10}(f)+92.45}$

For ${\displaystyle \ d,f}$ in meters and kilohertz, respectively, the constant becomes ${\displaystyle \ -87.55}$ .

For ${\displaystyle \ d,f}$ in meters and megahertz, respectively, the constant becomes ${\displaystyle \ -27.55}$ .

For ${\displaystyle \ d,f}$ in kilometers and megahertz, respectively, the constant becomes ${\displaystyle \ 32.45}$ . [5]