Waveguides and 900 MHz

It was mentioned that a 6 inch diameter steel pipe might act as a
waveguide to a transmitter located 60 feet into the pipe. How would
this be calculated? It doesn't really matter since I'm gonna be doing
this. I'm just curious.
Eric

 

Hi,

A circular guide of only 6in. in diameter (15.2cm) isn't a good
choice for propagation at the wavelength corresponding to 900MHz
(33.3cm), which is above the guide cut-off wavelength. This, in the
dominant TE11 mode, is roughly equal to -

sqrt(3) x guide diameter = 26.4cm

and so you must expect some attenuation; the graphs also show that
this increases very rapidly above cut-off. You will have to factor in
material losses as well because you are expecting to use steel pipe.
The only formula I could find assumes a copper guide which wouldn't be
too accurate in this case.


Cheers - Joe

 

Thanks Joe, I understand it a little better now. I know a little about
light guides and how changing the refraction of the guide helps the
light travel the length of the guide. Fiber optics and GRIN lenses
work this way. There are layers in the guide so different diameters
refract light differently. But a metal tube is akin to a single type
of glass. How does the material the tube is made of affect the radio
waves. Light must pass through a material to be refracted differently
and so it is bent at different angles as it travels the length of the
guide. But if it reflects the only thing that material changes is how
well it reflects and how much light is absorbed. The angle is the same
for all materials as long as it is reflected. I thought that wave
guides also reflected the radio waves. Does the material absorbing the
radio waves make a big difference the way it does with light? And
since copper is a better conductor than steel wouldn't it absorb more
radio energy? Just shows how little I know about this.
Thanks again,
Eric

 

If I have all this right,

6" dia = 3" radius = 0.076 m

The longest wavelength mode is TE11, 3.41 * r = 0.259 m

Your wavelength is 0.33 m. So it's a waveguide beyond cutoff, and
losses will be extreme.

I think.

John

 

If I remember right, it's the other way round - because it's a better
conductor, it's better at carrying the induced currents inside
the inner surface of the guide that mirror the EM field back into the
space inside the guide, without dissipating energy through resistive loss.

 

Hi,

There is a continuous reaction between the E/M field and the
guide walls such that currents flow in them which then re-radiate
the wave. As a result any resistive losses there will absorb
energy from them increasing guide attenuation. Copper, being a
better conductor than steel, will thus afford lower attenuation.
A further consideration is the surface roughness of the guide as
any of that will make matters worse and I would expect commercial
tubing to be only fair in this regard.

Coupling into the TE11 mode is usually by a probe fitted into
the wall of the pipe with a similar one parallel to it at the
receiving end. For the impedances and tuning thereof, you will
have to look in the books.


Cheers - Joe

 

You can take conductivity to be an imaginary
component added to the permitivity. Then,
assuming monochromatic signal, you can put that
complex permitivity into Snell's law and compute
reflected and transmitted waves. You'll probably
need to look at an E&M text book to understand
the interpretation of the results, but the upshot
is that waves travel very slowly in a good conductor
so that in some sense, good conductors have very large
indexes of refraction.

 

Rectangular wave guides are easier to compute
and visualize than circular ones. Replace
your 6 inch pipe with a 6 inch square wave guide.
You won't get exactly the same numerical answers,
but you get close and the qualitative behavior
is similar.

 

The waves bounce off the walls of a waveguide, so stay inside the
pipe. Think of copper as a shiny mirror, and steel as a dirty one.

Radar absorbers are usually carbon or ferrite-loaded plastics, very
bad conductors.

John

 

Around here, they coat roofs with it.

John

 

what effect would the tube being grounded have?

Charles

 

An old book says that "Pitch" is good at bending microwaves. The stuff
seemed to figure in lots of electrical experiments late 1800's yet is not to
be seen nowadays.
regards
john

 

None. The waves are inside, and don't care about the outside. A mirror
doesn't need to be grounded.

John

 

Actually I was thinking the grounding could reduce the bounced energy.

Charles