I can receive a calibrated decibel milliwatt signal level from the output of any antenna at a given frequency.

With a given antenna gain (dbi) in the direction I want to receive the signal (including a 1/4 wave vertical or a loopstick antenna etc...), how can I calculate the field strength in µV / meter?


2 Answers 2



  • the receiving antenna's gain in dBi, $G_\mathrm{dBi}$
  • the power received by the antenna in dBm, $P_\mathrm{dBm}$
  • the frequency in Hz, $f$
  • the impedance of free space in ohms, $Z_0$
  • the speed of light in meters per second, $c$

then the field strength in V/m is:

$$ \begin{align*} V &= \:\sqrt{ 0.004\, \pi\, Z_0 } \:f/c \:10^{(P_\mathrm{dBm}-G_\mathrm{dBi}) / 20} \\ &\approx 7.258\times 10^{-9}\: f \:10^{(P_\mathrm{dBm}-G_\mathrm{dBi}) / 20} \end{align*}$$

If the frequency is in MHz and the result is in µV/m then the constant becomes more manageable:

$$ V_\mathrm{\mu V/m} \approx 7258 \: f_\mathrm{MHz} \: 10^{(P_\mathrm{dBm}-G_\mathrm{dBi}) / 20} $$

Note that the receiving antenna must match the polarization of the voltage field being measured. In situations where the polarization isn't known, an isotropic receiver can be used, which is usually an orthogonal arrangement of short dipoles in 3 axes. Such an antenna can measure a voltage gradient in any direction.

I find it's easier to understand why this is true by thinking about power density rather than field strength.

If you have a 1 watt isotropic antenna at the center of a sphere, the entire 1 watt of that transmitted power must pass through the sphere. The power density is then the transmitted power, divided by the surface area of the sphere. The result is a number in units of watts per square meter.

The receiving antenna, assuming it is identically polarized, is then a "net" for this energy flux. The size of the net is the antenna aperture, which can be calculated from its gain.

$$ A = {\lambda^2 \over 4 \pi} 10^{G_\text{dBi}/10} $$

For more detail, see What does the Friis transmission equation represent and how is it derived?

In the far field, power density and the field strength are related by the free space impedance, which is approximately 377Ω. Converting between power density and field strength is very much like Ohm's law. This can be derived from Poynting's theorem:

$$ \text{power density} = {\text{field strength}^2 \over \text{free space impedance}} $$

Or solved for field strength:

$$ \text{field strength} = \sqrt{\text{power density} \times \text{free space impedance}} $$

Thus, if you can measure the power density at some point, say by measuring the feedpoint power from an antenna of known gain, then you can calculate the field strength.

For example, say the antenna has a gain of 3 dBi at a wavelength of 2 meters, and you measure a power at the feedpoint of 0.1μW.

The aperture is:

$$ {(2\:\mathrm{m})^2 \over 4 \pi} 10^{3\:\mathrm{dBi}/10} = 0.635\:\mathrm{m}^2 $$

The power density is then:

$$ {0.1\:\mathrm{\mu W} \over 0.635\:\mathrm{m}^2} = 0.157\:\mathrm{\mu W}/\mathrm{m}^2 $$

And the corresponding field strength:

$$ \sqrt{0.157\:\mathrm{\mu W/m^2} \times 377\:\Omega} = 7705\:\mathrm{\mu V/m}$$

Put these three equations together and simplify to get the equation at the top.

For more detail, Semtech has a good paper on field strength and many related topics.

  • $\begingroup$ You need the 1 mW reference to convert from $dB_{1 mW} $ to Watts $\endgroup$
    – tomnexus
    Commented Apr 20, 2017 at 0:36
  • $\begingroup$ @tomnexus To do the math, yes. But you don't need a calibrated, 1mW transmitter if the receiving antenna gain is already known, which I think is what the OP was talking about. $\endgroup$ Commented Apr 20, 2017 at 13:01
  • $\begingroup$ Though now that I read it again, maybe not... $\endgroup$ Commented Apr 20, 2017 at 13:02

The antenna factor gives, for each frequency, the ratio between electric field strength and voltage at the antenna terminals.

In the linked article you'll find the relation between AF and antenna aperture, and antenna gain.


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