Free space attenuation

Free-space path loss (FSPL ) in dB as a function of distance for different frequencies

The free space attenuation summarizes two terms of the power transmission balance of a radio link : the reduction of the power density according to the square law of distance and the effective area of a receiving antenna which shrinks with the frequency without antenna gain . It is the simplest model for path losses and does not take into account any attenuation caused by the propagation medium .

In radio technology, the free space attenuation factor is usually expressed logarithmically as the free space attenuation factor in dB .

calculation

Power density on a spherical surface

Since the antenna gain of the transmitting and receiving antenna appears separately in the power balance and relates to the (theoretical) isotropic radiator , its directional characteristic is used here. For the transmitter, this means that its high-frequency power is  evenly distributed in all directions. As a result, areas with the same power density form  spheres around the radiator. As the spherical radius  increases, the power is distributed over a larger area ( sphere ) around the radiator, the power density decreases quadratically: ${\displaystyle P}$ ${\displaystyle S}$ ${\displaystyle r}$ ${\displaystyle A=4\pi r^{2}}$

(1) ${\displaystyle S={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}}$

The approximately at the receiving plane wave removes the receiving antenna performance

(2) ${\displaystyle P_{\mathrm {r} }=SA_{\mathrm {W} }}$

The isotropic antenna is to be used for the effective area . It only depends on the wavelength : ${\displaystyle A_{\mathrm {W} }}$ ${\displaystyle \lambda }$

(3) ${\displaystyle A_{\mathrm {W} }={\frac {\lambda ^{2}}{4\pi }}}$

Inserting (1) and (3) into (2) it follows:

${\displaystyle P_{\mathrm {r} }={\frac {P}{4\pi r^{2}}}{\frac {\lambda ^{2}}{4\pi }}=\left({\frac {\lambda }{4\pi r}}\right)^{2}P}$

The relationship

${\displaystyle F={\frac {P}{P_{\mathrm {r} }}}=\left({\frac {4\pi r}{\lambda }}\right)^{2}}$

of the two powers is referred to as free space attenuation and is also given as a function of frequency : ${\displaystyle f={\frac {c}{\lambda }}}$

${\displaystyle F=\left({\frac {4\pi rf}{c}}\right)^{2}}$

at the speed of light  . ${\displaystyle c}$

The frequency dependence of the free space attenuation results from the fact that a power is emitted, but a power density is considered at the receiving location . Therefore, a unit of area must be included in the equation, the dimension of which can be given as a multiple of the wavelength (a sequence from equation 3). The wavelength can in turn be expressed in terms of the frequency, which creates the frequency dependence. The free space attenuation itself is dimensionless, since the unit of area is set in relation to the spherical surface. At a higher frequency, the considered unit area becomes  smaller and the ratio to the spherical surface  deteriorates. ${\displaystyle \lambda ^{2}}$ ${\displaystyle 4\pi r^{2}}$

The addition of the antenna gain and the transmission power to this equation to form a power transmission balance is called the Friis transmission equation .

Free space attenuation

The amount of free space attenuation can be derived directly from the above equation. When taking the logarithm , exponents become factors and factors become summands:

${\displaystyle {\begin{alignedat}{2}F{\text{ in dB}}&=10\log _{10}(F)\\&=10\log _{10}\left({\frac {4\pi rf}{c}}\right)^{2}&&=10\log _{10}\left({\frac {4\pi r}{\lambda }}\right)^{2}\\&=20\log _{10}\left(r\cdot f\cdot {\frac {4\pi }{c}}\right)&&=20\log _{10}\left(r\cdot \lambda ^{-1}\cdot 4\pi \right)\\&=20\log _{10}(r/\mathrm {m} )+20\log _{10}(f/\mathrm {Hz} )+20\log _{10}\left({\frac {4\pi }{c}}\mathrm {\frac {m}{s}} \right)&&=20\log _{10}(r/\mathrm {m} )-20\log _{10}(\lambda /\mathrm {m} )+20\log _{10}\left(4\pi \right)\\&=20\log _{10}(r/\mathrm {m} )+20\log _{10}(f/\mathrm {Hz} )-147{,}55&&=20\log _{10}(r/\mathrm {m} )-20\log _{10}(\lambda /\mathrm {m} )+21{,}984\end{alignedat}}}$

Examples

With the transceiver of a car key with (corresponding to the wavelength  ) and a power of about 4 mW (corresponding to 6 dBm), a distance of 5 m should be achieved. The free space attenuation is approx. 54 dB. Antenna gains should not be included because of the omnidirectional characteristics sought on both sides . The reception level is thus −48 dBm, corresponding to 13 nW. ${\displaystyle f=2{,}4\,\mathrm {GHz} }$ ${\displaystyle \lambda =125\,\mathrm {mm} }$

Frequency f	Distance r	Free space attenuation
		Factor F	Measure in dB
27 MHz	300 m ( RC model )	10 5	51 dB
100 MHz	100 km ( VHF radio )	1.6 · 10 12	112 dB
13 GHz	30 km ( directional radio )	3 · 10 14	144 dB
1575 MHz	25,000 km ( GPS  L1)	3 · 10 18	184 dB
15 GHz	38,000 km ( broadcast satellite )	6 · 10 20	208 dB
2.1 GHz	384,000 km (moon-earth, Apollo program )	10 21	211 dB
5 GHz	405,000,000 km (Earth - Rosetta space probe )	10 28	280 dB

literature

• Jürgen Detlefsen, Uwe Siart: Basics of high frequency technology. 2nd Edition. Oldenbourg Verlag, Munich / Vienna 2006, ISBN 3-486-57866-9
• Hans Lobensommer: Handbook of modern radio technology. 1st edition. Franzis Verlag, Poing 1995, ISBN 3-7723-4262-0

Individual evidence

1. ↑ Manfred Thumm, Werner Wiesbeck, Stefan Kern: High frequency measurement technology. Methods and measuring systems Springer DE, 1998, ISBN 3-519-16360-8 , p. 245
2. ↑ Bernhard Walke: Cellular networks and their protocols 1 Springer DE, 2001, ISBN 3-519-26430-7 restricted preview in the Google book search
3. ↑ Jerry C. Whitaker: The Electronics Handbook, Second Edition CRC Press, 2012, ISBN 0-8493-1889-0 , p. 1517f limited preview in the Google book search
4. ^ Ulrich Freyer: Media technology. Basic knowledge of communications technology, terms, functions, applications 2013, ISBN 978-3-446-42915-4, limited preview in the Google book search