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I have a question regarding Field Strength: I was told it is independent from the frequency, is that true? If so, why?

Moreover, I came accross this equation for converting the Field Strength to EIRP, but I do not get where the 104.8 comes from:

$$ E\ [dB \mu V/m] = EIRP\ [dBm] - 20 \log_{10}(d\ [m]) + 104.8 $$

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    \$\begingroup\$ The question "Why is A independent of B" doesn't make much sense if you can't explain why A should be dependent on B. \$\endgroup\$ – Marcus Müller Apr 26 '17 at 8:01
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    \$\begingroup\$ Instead of bluntly reproducing the equation and asking us about it, why don't you do some work yourself and try to relate certain the terms in the equation to the situation it described. The world is not "just formulas", we use formulas to describe certain situations. First describe the situation to which the formula applies. Without it, the formula is useless. \$\endgroup\$ – Bimpelrekkie Apr 26 '17 at 8:07
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Field strength has NOTHING to do with frequency - you can have a battery that generates a static electric field - no frequency involved, just DC. You can produce a constant magnetic field with a DC current and a coil. No frequency involved.

For the formula first consider this: -

enter image description here

Note that \$4\pi r^2\$ is the surface area of a sphere and \$120\pi\$ (377 ohms) is the impedance of free space.

So, if you plug in some numbers like 1 mW and 1 metre, E (volts per metre) is 0.173 volts per metre = 173,205 uV/m or 104.77 dBuV/m.

That is where the 104.8 comes from

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  • \$\begingroup\$ I must agree with you that E-field strength has nothing to do with frequency - it's a physical magnitude itself. However, the EIRP required from a transmitter to produce a given field strength at a given distance may depend on the radiating frequency. Thus, having the EIRP change with frequency will also change the PRODUCED E-field strength with frequency. It seems that's what the OP is confused about. \$\endgroup\$ – Enric Blanco Apr 26 '17 at 9:17
  • \$\begingroup\$ 5th line from Semtech should be E^2/120pi \$\endgroup\$ – philcolbourn Jan 8 at 4:44
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We start with the definition of radiated power density:

$$ P_{dens}\ [W/m^2]= \frac {EIRP} {4\pi d^2} $$

Which can also be expressed in terms of the radiated electric field and the impedance of free-space, \$ Z_{0,FS} \approx 120\pi \ \Omega \$:

$$ P_{dens}\ [W/m^2]= \frac {E^2} {Z_{0,FS}} $$

Thus:

$$ \frac {E^2} {Z_{0,FS}} =\frac {EIRP} {4\pi d^2} $$

We isolate E and take logarithms:

$$ \begin{align} 20 \log(E\ [V/m]) &= 10 \log(EIRP\ [W]) - 20 \log(d\ [m]) + 10 \log \left( \frac {Z_{0,FS}} {4\pi} \right) \\ E\ [dBV/m] &= EIRP\ [dBW] - 20 \log(d\ [m]) + 14.8 \end{align} $$

Now we can convert the units of E and EIRP to whatever we need. To suit your expression:

$$ \begin{align} dBV/m &= dB \mu V/m - 120 \\ dBW &= dBm - 30 \end{align} $$

Thus:

$$ \begin{align} E\ [dB \mu V/m] &= EIRP\ [dBm] - 20 \log(d\ [m]) + 14.8 + 120 - 30 = \\ &= EIRP\ [dBm] - 20 \log(d\ [m]) + 104.8 \end{align} $$

So that's where the 104.8 comes from.

It may seem that this equation is not frequency dependent, but that would be a false claim. Why? Because EIRP is the product of the transmitted power and the transmission antenna gain. In any practical transmitter BOTH magnitudes will be frequency dependent.

However, the transmitter (power amp and antenna) may be designed to be wideband enough so as to ignore the frequency dependency over its entire operating bandwidth.

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