# Signal strength in telecommunications

(Redirected from ΜV/m²)

In telecommunications, particularly in radio frequency, signal strength (also referred to as field strength) refers to the transmitter power output as received by a reference antenna at a distance from the transmitting antenna. High-powered transmissions, such as those used in broadcasting, are expressed in dB-millivolts per metre (dBmV/m). For very low-power systems, such as mobile phones, signal strength is usually expressed in dB-microvolts per metre (dBµV/m) or in decibels above a reference level of one milliwatt (dBm). In broadcasting terminology, 1 mV/m is 1000 µV/m or 60 dBµ (often written dBu).

Examples
• 100 dBµ or 100 mV/m: blanketing interference may occur on some receivers
• 60 dBµ or 1.0 mV/m: frequently considered the edge of a radio station's protected area in North America
• 40 dBµ or 0.1 mV/m: the minimum strength at which a station can be received with acceptable quality on most receivers

## Relationship to average radiated power

The electric field strength at a specific point can be determined from the power delivered to the transmitting antenna, its geometry and radiation resistance. Consider the case of a center-fed half-wave dipole antenna in free space, where the total length L is equal to one half wavelength (λ/2). If constructed from thin conductors, the current distribution is essentially sinusoidal and the radiating electric field is given by

Current distribution on antenna of length ${\displaystyle \scriptstyle {L}}$ equal to one half wavelength (${\displaystyle \scriptstyle {\lambda /2}}$).
${\displaystyle E_{\theta }(r)={-jI_{\circ } \over 2\pi \varepsilon _{\circ }c\,r}{\cos \left(\scriptstyle {\pi \over 2}\cos \theta \right) \over \sin \theta }e^{j\left(\omega t-kr\right)}}$

where ${\displaystyle \scriptstyle {\theta }}$ is the angle between the antenna axis and the vector to the observation point, ${\displaystyle \scriptstyle {I_{\circ }}}$ is the peak current at the feed-point, ${\displaystyle \scriptstyle {\varepsilon _{\circ }\,=\,8.85\times 10^{-12}\,F/m}}$ is the permittivity of free-space, ${\displaystyle \scriptstyle {c\,=\,3\times 10^{8}\,m/S}}$ is the speed of light in a vacuum, and ${\displaystyle \scriptstyle {r}}$ is the distance to the antenna in meters. When the antenna is viewed broadside (${\displaystyle \scriptstyle {\theta \,=\,\pi /2}}$) the electric field is maximum and given by

${\displaystyle \vert E_{\pi /2}(r)\vert ={I_{\circ } \over 2\pi \varepsilon _{\circ }c\,r}\,.}$

Solving this formula for the peak current yields

${\displaystyle I_{\circ }=2\pi \varepsilon _{\circ }c\,r\vert E_{\pi /2}(r)\vert \,.}$

The average power to the antenna is

${\displaystyle {P_{avg}={1 \over 2}R_{a}\,I_{\circ }^{2}}}$

where ${\displaystyle \scriptstyle {R_{a}=73.13\,\Omega }}$ is the center-fed half-wave antenna’s radiation resistance. Substituting the formula for ${\displaystyle \scriptstyle {I_{\circ }}}$ into the one for ${\displaystyle \scriptstyle {P_{avg}}}$ and solving for the maximum electric field yields

${\displaystyle \vert E_{\pi /2}(r)\vert \,=\,{1 \over \pi \varepsilon _{\circ }c\,r}{\sqrt {P_{avg} \over 2R_{a}}}\,=\,{9.91 \over r}{\sqrt {P_{avg}}}\quad (L=\lambda /2)\,.}$

Therefore, if the average power to a half-wave dipole antenna is 1 mW, then the maximum electric field at 313 m (1027 ft) is 1 mV/m (60 dBµ).

For a short dipole (${\displaystyle \scriptstyle {L\ll \lambda /2}}$) the current distribution is nearly triangular. In this case, the electric field and radiation resistance are

${\displaystyle E_{\theta }(r)={-jI_{\circ }\sin(\theta ) \over 4\varepsilon _{\circ }c\,r}\left({L \over \lambda }\right)e^{j\left(\omega t-kr\right)}\,,\quad R_{a}=20\pi ^{2}\left({L \over \lambda }\right)^{2}.}$

Using a procedure similar to that above, the maximum electric field for a center-fed short dipole is

${\displaystyle \vert E_{\pi /2}(r)\vert \,=\,{1 \over \pi \varepsilon _{\circ }c\,r}{\sqrt {P_{avg} \over 160}}\,=\,{9.48 \over r}{\sqrt {P_{avg}}}\quad (L\ll \lambda /2)\,.}$

## Cellphone signals

Although there are cell phone base station tower networks across many nations globally, there are still many areas within those nations that do not have good reception. Some rural areas are unlikely ever to be effectively covered since the cost of erecting a cell tower is too high for only a few customers. Even in high reception areas it is often found that basements and the interiors of large buildings have poor reception.

Weak signal strength can also be caused by destructive interference of the signals from local towers in urban areas, or by the construction materials used in some buildings causing rapid attenuation of signal strength. Large buildings such as warehouses, hospitals and factories often have no usable signal further than a few metres from the outside walls.

This is particularly true for the networks which operate at higher frequency since these are attenuated more rapidly by intervening obstacles, although they are able to use reflection and diffraction to circumvent obstacles.

### Estimated received signal strength

The estimated received signal strength in a mobile device can be estimated as follows:

${\displaystyle dBm_{e}=-113.0-40.0\ \log _{10}\left({\frac {r}{R}}\right)}$

More general you can take the path loss exponent into account:[1]

${\displaystyle dBm_{e}=-113.0-10.0\ \gamma \ \log _{10}\left({\frac {r}{R}}\right)}$
Parameter Description
dBme Estimated received power in mobile device
−113 Minimum received power
40 Average path loss per decade for mobile networks
r Distance mobile device - cell tower
R Mean radius of the cell tower
γ Path loss exponent (average value of 4 for mobile networks)

If the mobile device is at cell radius distance from the cell tower the received power is estimated as −113 dBm. The effective path loss is depending on the frequency, the topography, and the environmental conditions.

Actually one could use any known signal power dBm0 at any distance r0 as a reference:

${\displaystyle dBm_{e}=dBm_{0}-10.0\ \gamma \ \log _{10}\left({\frac {r}{r_{0}}}\right)}$

### Number of decades

${\displaystyle \log _{10}(R/r)}$ would give an estimate of the number of decades, which coincides with an average path loss of 40 dB/decade.

### Estimate the cell radius

When we measure cell distance r and received power dBmm pairs, then we can estimate the mean cell radius as follows:

${\displaystyle R_{e}=\operatorname {avg} [\ r\ 10^{(dBm_{m}+113.0)/40.0}\ ]}$

Specialized calculation models exist to plan the location of a new cell tower, taking into account local conditions and radio equipment parameters. Take also into consideration that mobile radio signals have line-of-sight propagation, unless reflexion would occur.