What are pyramidal RAM geometry? (on Anechoic chamber)

[STINGERX]

i found that in Anechoic chamber there are pyramidal structures over the walls,
what are the geometries of the height and base of those pyramids?
W and H on the picture:

 

[Jeff Rosenbury]

I'll take a stab. H/W = 35/12. This is what NASA used in one of it's chambers. (70" height, 2' square base). This was a commercially supplied material.

Thinking it through:

First we need to know the reflection coefficient (gamma, but don't confuse with gamma function or you'll get string theory. ) and index of refraction. These are all frequency dependent. The index of refraction will be a complex number with the imaginary part representing the lossiness of the material at that frequency (i.e make it as big as possible for the relevant frequencies).

The incident wave will impinge at various angles primarily dependent on where the radio source is in the room. (Such echos that exist will come in at different angles). Only a few tiles will have a wave perpendicular to the wall.

This is a guess: For waves >> than W the surface might be considered a corrugated surface. I don't know the equations for corrugated surfaces, but we will want to match impedance with air (377Ω). (Corrugated surfaces have differing impedances than smooth ones.) I don't know how the impedance of the surface affects the equations compared to the assumed perfect electric conductor, but I'm sure it does. Note this applies to surface waves. I suspect some of those would develop unless suppressed.

For waves << than W, we are going to want to bounce the wave off the sides of the pyramid as many times as we can. Each time some of the wave will pass into the RAM (depending on the reflection coefficient) and be attenuated while traveling through to the far wall where it will reflect again. If it's the back wall of the chamber the reflection coefficient will be different. This would argue for a shallow angle C.

For waves on the order of W, try antenna modeling software?

This isn't a great response. I hope someone can do better.

 

[STINGERX]

can you be a little bit more specific about W?

 

[Jeff Rosenbury]

W is the dimension of the base in your drawing. I assume it stands for width.

How EM waves react to conductors (even poor, lossy conductors) near their wavelengths is more complex than those which are much larger or smaller than the wavelength. Electromagnetics deals with this as a subject. Answering this particular question by solving the field equations (for several octaves) would likely take me some months of study and some nice modeling software. Sorry, I would only be willing to do that for way too much money.

Perhaps someone has already done the work and would like to chip in?

 

[STINGERX]

is there someone who know something about this? or maybe used to do something on anechoic chamber with pyramidal RAM?

 

[STINGERX]

really need this information, please help
thank you for any information.

 

[tech99]

I'll take a stab. H/W = 35/12. This is what NASA used in one of it's chambers. (70" height, 2' square base). This was a commercially supplied material.

Thinking it through:

First we need to know the reflection coefficient (gamma, but don't confuse with gamma function or you'll get string theory. ) and index of refraction. These are all frequency dependent. The index of refraction will be a complex number with the imaginary part representing the lossiness of the material at that frequency (i.e make it as big as possible for the relevant frequencies).

The incident wave will impinge at various angles primarily dependent on where the radio source is in the room. (Such echos that exist will come in at different angles). Only a few tiles will have a wave perpendicular to the wall.

This is a guess: For waves >> than W the surface might be considered a corrugated surface. I don't know the equations for corrugated surfaces, but we will want to match impedance with air (377Ω). (Corrugated surfaces have differing impedances than smooth ones.) I don't know how the impedance of the surface affects the equations compared to the assumed perfect electric conductor, but I'm sure it does. Note this applies to surface waves. I suspect some of those would develop unless suppressed.

For waves << than W, we are going to want to bounce the wave off the sides of the pyramid as many times as we can. Each time some of the wave will pass into the RAM (depending on the reflection coefficient) and be attenuated while traveling through to the far wall where it will reflect again. If it's the back wall of the chamber the reflection coefficient will be different. This would argue for a shallow angle C.

For waves on the order of W, try antenna modeling software?

This isn't a great response. I hope someone can do better.

At the lowest frequency of operation, I would expect maximum current (maximum absorption) to be obtained in the cone wall at a height of a quarter wave above the ground plane and where the cone has a length of side of half a wavelength. Below this frequency the cone will not absorb much energy. The taper needs to be gentle to obtain smooth absorption across the spectrum, maybe 30 degree apex angle, rather like a log periodic antenna. Maybe you can find manufacturer's data.

 

[berkeman]

really need this information, please help
thank you for any information.

Have you read the wikipedia article on Anechoic Chambers? It's reasonably good...

http://en.wikipedia.org/wiki/Anechoic_chamber

 

[the_emi_guy]

There are two mechanisms to consider here:

First, when RF enters the absorber material from inside the chamber it is attenuated by the lossy RAM as it propagates toward the reflective wall to which it is attached. Whatever energy reaches the reflective wall will experience more loss as it reflects back and completes the round trip back into the chamber. The idea here is to have enough absorber to ensure that the energy returned back into the chamber via the two trips through the RAM is below your attenuation goal. Now, it turns out that for a given loss tangent, the attenuation constant, which is the attenuation per unit distance, is proportional to frequency. The lower the frequency, the more distance through RAM we need to obtain a given attenuation. So the overall size of the absorber (both W and H), needs to be larger to be effective at lower frequencies. 10GHz absorbers can be a couple inches high, 30MHz absorbers can be 7 feet high.

Now the second mechanism (which I think is more toward your question):
Although RAM is designed with fairly low dielectric constant foam, it still presents a mismatch with air. Thus, not all of the incident RF enters the RAM, some will reflect at the RAM/air interface. The tapered surface provides a means to minimize reflections. There is really nothing magic about the specific geometry, ratio of W to H, or the angle C. It is a tradeoff between effectiveness and size. A longer taper the better the impedance match, but the material becomes physically larger and more fragile. This tapering impedance matching technique can be seen in the "AT&T Archives: Similarities of Wave Behavior" around minute 24 (but by all means watch the entire video).



His tapered section reduced the reflection but did not eliminate it entirely. He could improve the match by increasing the length of the taper, but at some point it would no longer fit in the room.