From: Erik Daniel (Daniel.Erik@mayo.edu)
Date: Mon Jan 03 2000 - 06:33:55 PST
I assume you're referring to attenuation of incident EM radiation as it
traverses the hole in the chassis. Several folks have responded to this
question with formulas or rules of thumb indicating cutoff frequencies
proportional to the wavelength. I don't think this is the whole story.
Although diffraction certainly forces the incident radiation to spread out
when the wavelength is significantly larger than the aperture, "significant"
(defined in relative terms, of course) amounts of radiation can still escape.
In the case of an ideal conductor with a circular hole of radius a, when the
wavelength is much greater than the aperture size, I believe the transmitted
power is proportional to (a/l)^2 where l is the wavelength (this seems to be
the result given in Jackson's EM book). I suppose it is up to you to determine
whether this defines "significant" transmission.
If I remember correctly, however, in the case of a non-ideal conductor, the
problem is more interesting, and a sharper cutoff of the transmitted energy
arises, related to the skin depth. In short, I seem to remember that when the
aperture size becomes comparable to the skin depth, the transmitted energy is
sharply attenuated. Can anyone out there confirm this?
On Dec 30, 9:45am, Doug Brooks wrote:
> Subject: [SI-LIST] : Chassis hole opening and frequencies
> As I recall, there is a relationship between a hole in a chassis and the
> frequencies that can pass through that opening. I recall that the longest
> dimension of the hole defines the wavelength, or quarter wavelength, or
> something, of the lowest frequency than can conveniently enter or escape
> through the opening.
> Can anyone give me the correct relationship?
> And Happy New Year to All...........
> Doug Brooks
> and all of us here at UltraCAD
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