free space

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, 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{4pi d}{lambda} right )^2 &= left (frac{4pi d f}{c} right )^2 end{align}


This equation is only accurate in the far field; it does not hold close to the transmitter.

Free-space path loss in Wireless Communication

This formula applies to Wireless Communications as shown below. In highways and in rural areas where there are at most 2 multipaths (the LOS-Line of Sight and/or the ground reflected ), this is part of the 2-ray path loss model

As depicted in the figure above, it assumes that the distance between transmitting and receiving antennas 'd' is much greater than the wavelength of the EM (Electromagnetic ) waves and the wavefronts impinging the receiving antenna are planar and not spherical and this distance is large enough to neglect the curvature of earth. The relation between the transmitted and the received power (Pu and Pr ) in Watts is given as :

begin{align} P_{r} &= P_{u}left (frac{G_{l}lambda}{4pi d} right )^2 end{align}


  • Gl is the product of transmit and receive antenna radiation patterns

Derivation of the formula can be found in

Free-space path loss in decibels

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

begin{align} mbox{FSPL(dB)} &= 10log_{10}left(left(frac{4pi}{c}dfright)^2right) &= 20log_{10}left(frac{4pi}{c}dfright) &= 20log_{10}(d) + 20log_{10}(f) + 20log_{10}left(frac{4pi}{c}right) &= 20log_{10}(d) + 20log_{10}(f) - 147.56 end{align}

where the units are as before.

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

mbox{FSPL(dB)} = 20log_{10}(d) + 20log_{10}(f) + 32.44

For d in statute miles, the constant becomes 36.6 .

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}


  • S is the power per unit area (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.

See also

Further Reading

External references

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