Pardon my ignorance here if this has already been answered. I'm not fully familiar with the English terminology here so my searches came up empty. If there is any keyword I’m missing or I’m not using correctly, please inform me.

I see several examples of guided wave attenuation for holes in metal boxes to increase the attenuation to/from inside to/from outside of high frequencies. I need to create a fairly large hole here which will most likely cause EMC issues, but not larger than that a small tube could mitigate the problem. Other solutions could exist here as well, but I would still like to understand more about guided wave attenuation.

What would happen if half the length (d/2) of the tube would protrude on the outside and half (d/2) on the inside, instead of all online examples I can find where it's a flush panel on one side and the entire length (d) sticking out on the other?enter image description here

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Since the tube now functions like a waveguide, I would argue that the shield need not be considered as all you need to be worried about is the radiation reflected and transmitted by the tube. So the tube in either configuration can be approximated by waveguide laws, and the effects from the position of the conductor would not affect the radiation passing through the waveguide.

$$ \lambda (meters) = \frac{300m/\mu s}{Frequency(MHz)}$$

As the frequency increases, its wavelength gets smaller. Once the signal frequency increases to the point where its half-wavelength (l/2) is smaller than the spacing between the transmission-line conductors, a new propagation mode can exist, and the conductors can be configured as a hollow tube or waveguide. At these frequencies and higher, the RF energy can propagate through a hollow tube, essentially a single-conductor transmission line.

Although there are several popular waveguide configurations, the rectangular cross-section is the easiest to visualize. In this case, the waveguide’s cutoff frequency ( Fco ) is determined by its width, which must be slightly wider than l/2.

The height is independent of frequency, but a 2:1 width-to-height ratio is very popular. At this ratio, the RF energy entering the waveguide reflects at an angle from side to side and travels through the tube with very little loss.

A waveguide is very efficient at passing all frequencies above its Fco, acting like a high-pass filter. For example, at 10-GHz, transmission-line losses in a 2-in. rectangular copper waveguide range from approximately 0.02 to 0.12 dB per meter length, depending on height.

There are a number of possible RF transmission modes, but from a shielding perspective, it is the waveguide’s behavior below cutoff for its lowest frequency propagation mode, such as the TE10 mode, that is of interest. Below its Fco, a waveguide is not a good transmission line, and it has very high attenuation.

For a circular waveguide:

$$ F_{co}(MHz) = \frac{6920in/\mu s}{D}$$

Below the Fco, the attenuation per unit length (inches) at any frequency (F) is given by:

$$ A_{UL}(db/in) = 0.00463 F \sqrt{(F_{co}/F)^2-1}$$

Source: https://evaluationengineering.com/features/0101emc.htm

  • \$\begingroup\$ Why would you use inches to obfuscate what would otherwise be a straight-forward equation in either meters or, better yet, wavelengths? \$\endgroup\$ – Edgar Brown Jan 9 at 20:50
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    \$\begingroup\$ The question is why would the website do this... I'm to lazy to convert it. Feel free to edit. \$\endgroup\$ – laptop2d Jan 9 at 21:05

This equation is just for waveguide attenuation and, being a passive system, its attenuation and emission characteristics must be the same. This would mean that you just have to combine the absorption losses of the two different sections together, which is equivalent to just having all of the length on one side of the wall.

However, the specifics of the application are very important here. Why do you need a large aperture on, what I assume to be, a Faraday cage?

If you are intending to thread any type of conductors through it, this equation does not apply anymore. As the conductors would completely remove the low-frequency cutoff of the aperture by allowing propagation modes that don't exist in a waveguide.

If it is for ventilation purposes, multiple smaller apertures would be more effective at reducing RF than a larger one.

  • \$\begingroup\$ Optical fiber into said faraday cage. I have limited space on both inside and outside for any protrusion, hence the question if I can split the difference. \$\endgroup\$ – winny Jan 9 at 21:25
  • \$\begingroup\$ It is just a waveguide. It does not have to have a circular section, it doesn’t have to be straight, it doesn’t have to stand away from the wall, it doesn’t have to have a constant cross-section. Just use the largest cross-section dimension for the equation (or divide in different sections) and you would be calculating for the worst case scenario. \$\endgroup\$ – Edgar Brown Jan 9 at 21:32
  • \$\begingroup\$ So no edge cases or what you call them? I’m thinking about the analogy to a bass reflex port on a speaker. The equation does not hold true if you are too close to a wall. \$\endgroup\$ – winny Jan 9 at 21:51
  • \$\begingroup\$ The similarities are mostly superficial. That equation completely ignores the effects due to the conductive wall (which will affect coupling into and out of the waveguide) and only considers the attenuation of the waveguide itself. \$\endgroup\$ – Edgar Brown Jan 9 at 22:09

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