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We have a storage area, that is surrounded by corrugated sheet metal. (Appears to be galvanised steel, and probably 2-3mm thick).

Inside the sheet metal, we have a LTE/4G router, on the Telstra network.

These are the bands that Telstra uses:

  • 2G
    • N/A
  • 3G
    • 850MHz (B5)
    • 2100MHz* (B1)
  • 4G
    • 700MHz (B28)
    • 900MHz (B8)
    • 1800MHz (B3)
    • 2100MHz (B1)
    • 2600MHz (B7)

The throughput inside the storage cage is not very good - 22Mbps up, 5 Mbps down.

I assume the sheet metal surrounded the cage isn't helping.

My question is - if we created an aperture in the sheet metal - what is the minimum size of a gap that would be needed to penetrate through the sheet metal?

Using λ = C/f, I believe 2100 Mhz has a wavelength of 0.14 metres (14 cm), whilst 700 Mhz has a 0.42 metres wavelength. Does that have any bearing on the aperture (is that the right term?) size I need?

Or how do you figure it out?

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    \$\begingroup\$ Move the LTE box outside and run Ethernet to a router inside. \$\endgroup\$
    – Chris Stratton
    Commented Oct 8, 2020 at 11:26
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    \$\begingroup\$ Even if you made an aperture just large enough for the lowest frequency to pass through, reception and range would very likely be completely terrible due to too little power from the base station reaching your router, too little power from your router reaching the base station and signals reflecting inside this shielded room. \$\endgroup\$
    – Bimpelrekkie
    Commented Oct 8, 2020 at 11:32
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    \$\begingroup\$ do you have the option to move the antenna outside? \$\endgroup\$
    – Deepak
    Commented Oct 8, 2020 at 12:12

1 Answer 1

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This answer, regarding metal shields, included math from Richard Feynmann's lectures.

He spoke of wire mesh with specific spacings between wires.

For the field inside the mesh (aperature in a metal box), the field drops off exponentially:

Why are many IR receivers in metal cages?

The useful point about this (years old) answer is this

  • you need to place the Receiver (or Transmitter) close to the hole in the shield, or the severe attenuation ruins the communication link

Summary: if the hole is X size, and the router is X distance INSIDE the metal box, the attenuation is

e ^ (X/X) * 2 * PI = e^6.28 = 1/533 (21 dB if power, 42 dB if voltage)

Thus you either need a BIG hole, or locate the router to be very close to the hole.

Because of "reciprocity" of transmitter fields and receiver fields, if you have 15 dB attenuation of TX fields, you will also have 15 dB attenuation of RX fields.

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