Ed,
In my simple minded way of looking at things in terms of units, Amp/Watt
= Amp/((Amp)(Volt)) = 1/Volt. Hence the quantity is neither
conductance, nor admittance.
Jim
------------------------------------------------------------------------
---------------------------
Dr. Jim Knighten
NCR
17095 Via del Campo
San Diego, CA 92127
Telephone: 619-485-2537
Fax: 619-485-3788
e-mail: jim.knigh...@sandiegoca.ncr.com
----------
From: ed.pr...@cubic.com [SMTP:ed.pr...@cubic.com]
Sent: Monday, June 29, 1998 5:36 PM
To: edpr...@pacbell.net; emc-p...@majordomo.ieee.org
Cc: atpar...@solar-emc.com; chad.a...@express.corp.cubic.com;
colo...@gdesystems.com; ddlu...@ssd.bna.boeing.com; ozziee...@aol.com;
te...@eglin.af.mil
Subject: Method CS114
Hi Group!
I need some help unravelling the calculations associated with the
MIL-STD-462D Method CS114 Loop Circuit Impedance Test.
Specifically, once you have a table of Forward Applied RF Power (in dBm)
and Observed Injected Current (in dBuA), how do you arrive at the
Normalized Amps per Watt?
And what is an "Amp per Watt"? Is that a "mho", a unit of conductance?
Or is it a unit of admittance, a Siemen?
Anybody out there ever really done this science project?
Ed
--------------------------
Ed Price
ed.pr...@cubic.com
Electromagnetic Compatibility Lab
Cubic Defense Systems
San Diego, CA. USA
619-505-2780
List-Post: emc-pstc@listserv.ieee.org
Date: 06/29/98
Time: 16:36:12
--------------------------
