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The following table represents the FCC radiated emission limits for class A devices at a distance of 10 m : enter image description here

How do we convert this table for a distance of 5 m?

Edit : There is an approximation in which we can calculate the values at that distance. The new value of the field intensity at 5m (absolute value) = absolute value at 10 m *10/5. I don't know if it is a good approximation or not and also I don't understand how it was derived

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    \$\begingroup\$ note that for a 30 MHz signal, no statement about the relationship between the field strength at 10m and 5m can be made, since you'd still be in the reactive near field of the emitter – 30 MHz's wavelength is 10m. \$\endgroup\$ Commented Mar 1, 2019 at 21:15

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You can't "convert it", there is a reason why that's how the standard is specified.

If you assume far field applies (at 10m it is barely at the edge of where it might start to become a valid approximation but at 5m it definitely isn't), then you can simply assume a power decrease with the square of the distance. That is you will have 4 times the power closer to the source (20dB).

In more general case you would use the Path Loss equation:

\$ L = 10 * n * \log_{10}(d) + C \$

Where n = 2 for free space far field (what I stated above) but it can be as high as 6 for near field.

There are other considerations such as the characteristics of the measuring antenna and also other requirements such as the conductive surfaces around the equipment.

The most you can do is to get a qualitative idea of what is going on, not a quantitative one. You can use this to compare to actual testing to standards and correct as as appropriate.

The far-field approximation becomes more and more valid as you increase the frequency. So you can get a better approximation on that section of the table.

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  • \$\begingroup\$ There is an approximation in which we can calculate the values at that distance. The new value of the field intensity at 5m (absolute value) = absolute value at 10 m *10/5. I don't know if it is a good approximation or not and also I don't understand how it was derived. \$\endgroup\$
    – John adams
    Commented Mar 1, 2019 at 22:50
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How do we convert this table for a distance of 5 m?

I don't know if FCC standards allow using different measurement distance but there is an equation in the EN standards as following to convert the limit for any distance.

L2 = L1 + 20 log(d1/d2)

where L1 is the specified limit in dBmV/m at the distance d1, L2 is the new limit for distance d2.

So in the case of using 5 m measurement distance, the limit given for 10 m should be increased approximately 6 dB.

Please consider far-field issues. According to EN 55032 standard, the measurement distance for radiated emission measurement for frequencies below 1 GHz shall be 3 m and for frequencies above 1 GHz shall be 1 m at least.

source: EN 55032

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