I would like to extrapolate my current router wifi radiation from my phone. If I know that my router is transmitting at 300mW and my phone displays the strength in -dbm (from 0 to -100 scale) if I have -50dbm strength does it mean that the current radiation is 150mW? If not how can I extrapolate the radiation from -dbm scale?

  • $\begingroup$ Could you clarify what you're asking: If the phone registers -50dBm, that means that the received strength at the phone is -50dBm. Is your question about how to convert this to milliwatts? $\endgroup$ – twistor59 Jan 1 '13 at 18:20
  • $\begingroup$ Exatcly, I would like to convert it in mW/cm2 but I am no sure if the transmission power counts... PS. I know that -60dbm is 10^(-60/10) mW Thank you $\endgroup$ – Yannis Assael Jan 1 '13 at 18:22
  • $\begingroup$ But I think you've answered your own question if you substitute 50 for 60. You don't need to know the transmission power - just use the mobile reading directly. $\endgroup$ – twistor59 Jan 1 '13 at 18:28
  • $\begingroup$ What do you mean extrapolate? Extrapolate for with respect to what? You already know how to convert from dmb to mW. $\endgroup$ – unsym Jan 1 '13 at 18:36
  • $\begingroup$ Is this that simple? Doesn't the transmission power count? I want to calculate the mW/cm2 as I have only DBm Thank you all very much for your effort $\endgroup$ – Yannis Assael Jan 1 '13 at 18:37

You wish to estimate the flux density (power per unit area) at a location based on a reading of the received power at a mobile device situated there. The signal transmitted from the router will result in an energy density of radiated power, decreasing with distance. The mobile device has a receive antenna which collects some of this RF power in its vicinity. This receive antenna is characterized by a parameter called its "effective area", or "effective aperture". This has the dimensions of an area, and if its value is $A \ cm^2$, then at a location where the RF power had a flux density of $F \ mW/cm^2$, the power captured by the antenna would be $FA \ mW$

Now the effective aperture is related to a parameter called the antenna gain by $$ G(\theta, \phi) = \frac{4\pi A(\theta, \phi)}{\lambda^2} $$ where $\lambda$ is the wavelength. Both the antenna gain and the effective aperture are functions of direction (specified by the two parameters $\theta$, $\phi$). Antennas have varying degrees of directionality (for example a parabolic reflector is highly directional). The formula shows that the direction in which the gain is maximum also maximises the effective aperture.

Now for a few numbers - say the frequency $f = 2.4GHz$ (depends which WiFi band you're using), we get a wavelength $\lambda = 12.5cm$ Antennas on mobile devices are actually quite lossy, so lets assume a gain of -3dB in the given direction (i.e. 0.5 in linear terms). This gives us $A = 6.2cm^2$

Your reading of -50dBm corresponds to $10^{-5} mW$, so the flux density is $10^{-5}/6.2 = 1.6^{-6}mW/cm^2$

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