1
\$\begingroup\$

How to calculate the field strength of a shielded theoretical isotropic transmitter, if there is a hole in the shield for example.

I am looking for a complex formula that also takes into account the hole in the shield and possibly the skin depth of the shield if possible.

enter image description here

If we have a simple transmitter/receiver system, then the field strenght can be both calculated at the source (EIRP) or at the Pr which is the receiver.

Pr= Pt / (4* PI * r^2)

Pt= (E^2*r^2)/30

What I need is a formula for this scenario:

enter image description here

Where we have an isotropic transmitter inside a shield. The shield has a hole, or perhaps multiple holes (if there is formula for that) with a diameter z, and also if possible to take into account the skin depth where the thickness of the shield is x. Let's say the material of the shield is a variable so that we can calculate the skin depth for this formula for different materials.

What I need is the field strength at Pr in V/m or dBμV/m units that is emmitted from the shielded environment. Or vice-versa the EIRP of the transmitted signal outside the shield (obviously different than the original transmitter power inside the shield), measured in W.

Obviously z and x and all other variables are in SI units.

\$\endgroup\$
4
  • \$\begingroup\$ One of your formulas appears to be in error. Pr = Pr etc... If Pt was an omni directional LED could you work out the power received that flowed through the hole. I'm not saying that gives an easy way to get to your answer, I'm just trying to understand your level on stuff like this. \$\endgroup\$
    – Andy aka
    Commented Apr 3, 2017 at 18:03
  • \$\begingroup\$ What do you mean by Pr=Pr? You mean Pt=Pr? Yes Pt is equal Pr since the energy moves on to infinity (the em radiation), but I was referring to the fact that there are 2 points of view, from the transmitter and from the receiver. So I guess either formula is ok with me, but we need to take into account the attenuation of the shield, so I believe it's easier to put a formula together for Pr. Sorry if it was ambiguous. \$\endgroup\$
    – user138887
    Commented Apr 4, 2017 at 6:53
  • \$\begingroup\$ You wrote this: Pr= Pr / (4* PI * r^2) and this has to be wrong. \$\endgroup\$
    – Andy aka
    Commented Apr 4, 2017 at 8:22
  • \$\begingroup\$ @Andyaka, indeed, sorry for typo, it's Pr= Pt / (4* PI * r^2) \$\endgroup\$
    – user138887
    Commented Apr 4, 2017 at 8:53

1 Answer 1

Reset to default
0
\$\begingroup\$

You must research into Shielding Effectiveness, which is quite a broad topic itself.

As a starting point, take a look at this 2-part article about shielding of boxes and enclosures by Michel Mardiguian: part 1, part 2.

It takes into account both skin effects and reradiation due to apertures, which is probably what you need:

Combined shielding effectiveness

\$\endgroup\$

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.