0
\$\begingroup\$

According to the EN 55022:2006 standard max electric field strength should be:

  • 40 dBμV/m for 30–230 MHz
  • 47 dBμV/m for 230–1000 MHz

I am a little bit confused, I don't know how to calculate the field strength for this frequency range as a function of the frequency. I can't measure it, I just want a theoretical estimation.

The basic field strength formula is:

$$ E = \frac{\sqrt{30*P}}{r} (V/m)$$

or

$$E = \log_{10}\bigg(\frac{\sqrt{30*P}}{ r}*10^6\bigg)*20 \frac{dBμV}{m}$$

So we have the field strength as a function of the distance measured (in M) and the power of the antenna (in W).

But how can we transform this formula to include the frequency. I believe the unit is dBμV/(m · MHz) , but I don't know the formula. Please help.

\$\endgroup\$
4
  • \$\begingroup\$ I think defining what "P" is explicitly will actually help yourself answer the question quite a bit! \$\endgroup\$ Mar 31, 2017 at 8:21
  • \$\begingroup\$ and I think you're slightly overthinking this: your power is happening at some (hopefully rather known) frequencies. So just check that the amount of power in these two bands doesn't exceed the limits given by your formula \$\endgroup\$ Mar 31, 2017 at 8:22
  • \$\begingroup\$ @MarcusMüller the transmitter power, also noted as Pt sometimes. What do you mean by explicitly define? \$\endgroup\$
    – user138887
    Apr 1, 2017 at 6:04
  • \$\begingroup\$ @MarcusMüller that's the point, if we know the electric field strenght at 1 frequency, how to calculate from that the field strenght at another frequency, without measuring? \$\endgroup\$
    – user138887
    Apr 1, 2017 at 6:09

1 Answer 1

1
\$\begingroup\$

Here's what it says in the standard: -

enter image description here

I don't know how to calculate the field strength for this frequency range as a function of the frequency

These are limits imposed by the standard - they are not related to frequency other than at the break point of 230 MHz.

So, for instance, at 300 MHz the actual volts per metre will be 223.87 uV/m.

This was calculated by raising 10 to the power 47/20 (\$10^{\frac{47}{20}})\$ i.e. converting from decibels back to real numbers.

Because the impedance of free space is (approximately 377 ohms) the power will be the field strength squared divided by 377 = 132.9 pico watts per metre squared.

This basically tells you that at 10 metres distant from some piece of electronics it must not produce a power density greater than 132.9 pW/m^2.

Given that a sphere of radius 10 m has a surface area of \$4\pi r^2\$ metres (1257 sq metres), the power produced at the point of generation is approximately 0.167 uW.

But how can we transform this formula to include the frequency

You don't - it's not a formula - it's just a set of legal limits that happen to have a step change at 230 MHz.

\$\endgroup\$
2
  • \$\begingroup\$ Is there a formula where I can convert the frequency you specified (I suppose arbitrary) to other frequency. So if a device emits 223.87 uV/m field strength at 300 Mhz, how much does it emit at say 900 Mhz? Is there any way to caculate this, if we assume that the device is an isotropic transmitter. \$\endgroup\$
    – user138887
    Apr 1, 2017 at 6:02
  • \$\begingroup\$ It emits the same. There is no relationship. The "per metre" part is not related to wavelength and the impedance of free space is also fixed regards frequency. \$\endgroup\$
    – Andy aka
    Apr 1, 2017 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.