Hi folks,

I needed to produce some referance power levels that were 'down' from
my calibrated 0DbM generator. (It's output power is auto leveled, and
non-adjustable) 9445mhz, center of one of the maritine radar bands.

I built two little aperature plates that look like a waveguide cover,
but have a hole drilled in the exact center. A .185" hole provides about
45db of reduction, and a .310" hole provides about 25db of reduction.

There are some differances of exact gain reduction depending on where
in the waveguide system I place these plates. (Close to source, or close
to receiver) Perhaps up to about 3 db differance, not a big worry for me.

Do the numbers sound correct? Should I have attenuated in a differant
way? Like perhaps a slot, instead of a round hole??

If I want to make some more of these plates, is there some math to help
me predict the size of the hole for 10db?

Thanks for your help and comments. I am exploring some new areas here.

You might google for "waveguide vane attenuator". That is the device
used to measure fade margin on microwave links. A company called ARRA
makes them, so did HP at one time. There is a lossy vane which rotates
into the waveguide, turned by a calibrated knob and scale. Might be
worth your while to check e-bay to see if you can find exactly what you
want. Also some old microwave radios had these attached to the output
waveguide. Of course you need to find X-band.
The-RFI-EMI-GUY

gudmundur wrote:

Hi folks,

I needed to produce some referance power levels that were 'down' from
my calibrated 0DbM generator. (It's output power is auto leveled, and
non-adjustable) 9445mhz, center of one of the maritine radar bands.

I built two little aperature plates that look like a waveguide cover,
but have a hole drilled in the exact center. A .185" hole provides about
45db of reduction, and a .310" hole provides about 25db of reduction.

There are some differances of exact gain reduction depending on where
in the waveguide system I place these plates. (Close to source, or close
to receiver) Perhaps up to about 3 db differance, not a big worry for me.

Do the numbers sound correct? Should I have attenuated in a differant
way? Like perhaps a slot, instead of a round hole??

If I want to make some more of these plates, is there some math to help
me predict the size of the hole for 10db?

Thanks for your help and comments. I am exploring some new areas here.





--
Joe Leikhim K4SAT
"The RFI-EMI-GUY"

The Lost Deep Thoughts By: Jack Handey
Before a mad scientist goes mad, there's probably a time
when he's only partially mad. And this is the time when he's
going to throw his best parties.

I needed to produce some referance power levels that were 'down' from
my calibrated 0DbM generator. (It's output power is auto leveled, and
non-adjustable) 9445mhz, center of one of the maritine radar bands.

I've never worked in this band, but recall that some lower-frequency signal
generators use a "waveguide beyond cutoff" attenuator. If you can find some
waveguide sections designed for a sufficiently higher band, the attenuation
will be proportional to their length.

This doesn't address impedance mismatch, but neither does your method.

73, Fred, K4DII

On 28 Jan 2005 19:01:02 GMT, unicate (Fred McKenzie)
wrote:

I needed to produce some referance power levels that were 'down' from
my calibrated 0DbM generator. (It's output power is auto leveled, and
non-adjustable) 9445mhz, center of one of the maritine radar bands.

I've never worked in this band, but recall that some lower-frequency signal
generators use a "waveguide beyond cutoff" attenuator. If you can find some
waveguide sections designed for a sufficiently higher band, the attenuation
will be proportional to their length.

This doesn't address impedance mismatch, but neither does your method.

73, Fred, K4DII

but another case is that while normal attenuation may be 0.15dB/m for
a waveguide within the specified band, it may become 0.1dB/m below the
cutoff frequency, it seems that the lower limit is not for
attenuation, but for phase relationship, so it may be difficult to
predict phase along the cable and it is a question of narrowband
contra wideband applications

73, LA8AK
---
J. M. Noeding, LA8AK, N-4623 Kristiansand
http://home.online.no/~la8ak/c.htm

On Fri, 28 Jan 2005 05:26:09 -0000, (gudmundur)
wrote:

Hi folks,

I needed to produce some referance power levels that were 'down' from
my calibrated 0DbM generator. (It's output power is auto leveled, and
non-adjustable) 9445mhz, center of one of the maritine radar bands.

I built two little aperature plates that look like a waveguide cover,
but have a hole drilled in the exact center. A .185" hole provides about
45db of reduction, and a .310" hole provides about 25db of reduction.

There are some differances of exact gain reduction depending on where
in the waveguide system I place these plates. (Close to source, or close
to receiver) Perhaps up to about 3 db differance, not a big worry for me.

Do the numbers sound correct? Should I have attenuated in a differant
way? Like perhaps a slot, instead of a round hole??

According to the one reference I can lay my hands on at the moment
(MIT Rad Lab Report No.43, Feb. 1944), a round aperature in a W/G is
equivalent to a shunt inductance.

The reference gives the data in graphical form. I put a few data
points into Excel and let it solve for the normalized susceptance v.
iris diameter.

For WR-90 guide (0.4" x 0.9") at 9.445 GHz, the formula is:

-Y' ~ 0.2838 * (d/0.9) ^ -3.1635

Where Y' is the normalized susceptance and with the iris centered in
the W/G "d" is the iris diameter in inches.

The reference doesn't specify a thickness of the iris plate, but I
would assume it's for plates lambda.

After writing the foregoing, I remembered Steve Adam's book,
"Microwave Theory and Applications" (I should have remembered it
sooner, he signed it for me).

He references H. A****er's, "Introduction to Microwave Theory" for the
following equation:

Z/Zo = j (2 * pi * d^3) / (3 * a * b * lambda_g)

Where for WR-90:

Zo = W/G impedance
a = 0.9
b = 0.4
d = iris diameter in inches
lambda_g = guide wavelength.

Whether these two references give the same answer will be, as they
say, left for an exercise for the reader.

What you're getting with these reactive elements is mismatch loss. If
you want true loss in your W/G, you must introduce resistance.