Remember Me?
Menu
Home
Search
Today's Posts
Home
Search
Today's Posts
RadioBanter
»
rec.radio.amateur
»
Homebrew
>
WR-90 X-Band aperature attenuator question??
LinkBack
Thread Tools
Search this Thread
Display Modes
Prev
Next
#
5
January 30th 05, 02:33 PM
Wes Stewart
Posts: n/a
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.
Reply With Quote
Thread Tools
Search this Thread
Show Printable Version
Search this Thread
:
Advanced Search
Display Modes
Switch to Linear Mode
Switch to Hybrid Mode
Threaded Mode
Posting Rules
Smilies
are
On
[IMG]
code is
On
HTML code is
Off
Trackbacks
are
On
Pingbacks
are
On
Refbacks
are
On
All times are GMT +1. The time now is
04:56 AM
.
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
Copyright ©2004-2025 RadioBanter.
The comments are property of their posters.
Contact Us
RadioBanter forum home
Privacy Statement
Copyright © 2017
LinkBack
LinkBack URL
About LinkBacks