# Free-space path loss

In telecommunication, free-space path loss (FSPL) is the loss in signal strength of an electromagnetic wave that would result from a line-of-sight path through free space (usually air), with no obstacles nearby to cause reflection or diffraction. It does not include factors such as the gain of the antennas used at the transmitter and receiver, nor any loss associated with hardware imperfections. A discussion of these losses may be found in the article on link budget.

## Free-space path loss formula

Free-space path loss is proportional to the square of the distance between the transmitter and receiver, and also proportional to the square of the frequency of the radio signal.

The equation for FSPL is

\begin{align} \mbox{FSPL} &= \left ( \frac{4\pi d}{\lambda} \right )^2 \\ &= \left ( \frac{4\pi d f}{c} \right )^2 \end{align}

where:

• $\ \lambda$ is the signal wavelength (in metres),
• $\ f$ is the signal frequency (in hertz),
• $\ d$ is the distance from the transmitter (in metres),
• $\ c$ is the speed of light in a vacuum, 2.99792458 × 108 metres per second.

This equation is only accurate in the far field where spherical spreading can be assumed; it does not hold close to the transmitter.

## Free-space path loss in decibels

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

\begin{align} \mbox{FSPL(dB)} &= 10\log_{10}\left(\left(\frac{4\pi}{c}df\right)^2\right) \\ &= 20\log_{10}\left(\frac{4\pi}{c}df\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}

where the units are as before.

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

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

For $\ d,f$ in meters and kilohertz, respectively, the constant becomes $\ -87.55$ . For $\ d,f$ in meters and megahertz, respectively, the constant becomes $\ -27.55$ . For $\ d,f$ in kilometers and megahertz, respectively, the constant becomes $\ 32.45$ .

## Physical explanation

The FSPL expression above often leads to the erroneous belief that free space attenuates an electromagnetic wave according to its frequency. This is not the case, as there is no physical mechanism that could cause this.

The expression for FSPL actually encapsulates two effects. Firstly, the spreading out of electromagnetic energy in free space is determined by the inverse square law, i.e.

$\ S = P_t \frac{1}{4 \pi d^2}$

where:

• $\ S$ is the power per unit area or power spatial density (in watts per metre-squared) at distance $\ d$,
• $\ P_t$ is the total power transmitted (in watts).

Note that this is not a frequency-dependent effect.

The second effect is that of the receiving antenna's aperture, which describes how well an antenna can pick up power from an incoming electromagnetic wave. For an isotropic antenna, this is given by

$\ P_r = S \frac{\lambda^2}{4 \pi}$

where $\ P_r$ is the received power. Note that this is entirely dependent on wavelength, which is how the frequency-dependent behaviour arises.

The total loss is given by the ratio

$\ \mathrm{FSPL} = \frac{P_t}{P_r}$

which can be found by combining the previous two expressions.