Voltage or power, a dB is a dB.  Constructive interference between the
direct and reflected ray (given a perfectly conductive ground plane) can
increase the field intensity by a factor of two if the phase difference
between the two is a half-wavelength.  That is 6 dB, whether you speak of
the field intensity, or the signal power at the antenna port measured at the
EMI receiver.
 
Ken Javor

Phone: (256) 650-5261



From: "Sundstrom, Michael" <michael_sundst...@overheaddoor.com>
Date: Thu, 22 Dec 2011 07:54:21 -0600
To: Bill Owsley <wdows...@yahoo.com>, "EMC-PSTC@LISTSERV.IEEE.ORG"
<EMC-PSTC@LISTSERV.IEEE.ORG>
Conversation: [PSES] Semi-Anechoic Chamber Question - Correction Factor
Subject: RE: [PSES] Semi-Anechoic Chamber Question - Correction Factor

Bill,
I¹d guess 6dB as in voltage, with dBuV being used.
 

 
Michael Sundstrom
OHD / TREQ Dallas
Electronic Lab Analyst, EMC Lead
2170 French Settlement Rd, Suite B
Dallas, Texas  75212
(214) 579 6312
(940) 390 3644c
KB5UKT
 
Albert Einstein once said, "The definition of insanity is doing the same
thing over and over again and expecting different results".
 

From: Bill Owsley [mailto:wdows...@yahoo.com]
Sent: Wednesday, December 21, 2011 11:02 PM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Semi-Anechoic Chamber Question - Correction Factor
 

Go back, way back, to the paper by pate, german, smith, on the NSA.
Go through all the details and calculations and surmise that there is the
direct path, and the reflected path.
The reflected path adds, or not, to the direct path, depending on wavelength
(phase at receive antenna) and distance between antennas.
A scan up to 4 meters will generally cover the apparent phase shift at the
lower frequencies such that a "maximum" is recorded.
This scan will certainly cover the apparent phase shift of higher freq's.
The reflected wave verses the direct wave, will have a longer distance and
thus a little more loss. In the worst case, it can be neglected.
It will also suffer some loss at the reflecting surface, generally assumed
to be a "perfect" boundary since it is unknown but defined as metal.
Thus, the received voltage received via a direct path added with a reflected
path, assuming no distance or reflection loss, would be 6 dB higher than the
direct path alone.  This direct path alone is the assumed field measured in
a FAC, neglecting any chamber anomalies.
Or is it 3 dB higher?? as in power?

 


From: "Grasso, Charles" <charles.gra...@echostar.com>
To: Jim Hulbert <jim.hulb...@pb.com>; "EMC-PSTC@LISTSERV.IEEE.ORG"
<EMC-PSTC@LISTSERV.IEEE.ORG>
Sent: Wednesday, December 21, 2011 4:29 PM
Subject: RE: Semi-Anechoic Chamber Question - Correction Factor


Professor Leferink published a nice paper that proposed a different
correction
factor for a SAC to a FAC. I tried to track it down but had no luck.

 

Best Regards
Charles Grasso
Compliance Engineer
Echostar Communications
(w) 303-706-5467
(c) 303-204-2974
(t) 3032042...@vtext.com <mailto:3032042...@vtext.com>
(e) charles.gra...@echostar.com <mailto:charles.gra...@echostar.com>

(e2) chasgra...@gmail.com <mailto:chasgra...@gmail.com>

 

From: emc-p...@ieee.org [mailto:emc-p...@ieee.org] On Behalf Of Jim Hulbert
Sent: Monday, December 19, 2011 2:27 PM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: RE: Semi-Anechoic Chamber Question

 

Thank you everyone for the helpful feedback.  The consensus is clearly: Save
my back and leave the absorber and ferrite tiles in place.  Compare radiated
emissions measurements in the chamber to those on the OATS (several
suggested a comb generator for this purpose).  Assign an adjustment factor
to the chamber measurements to correlate as closely as possible to the OATS
measurements (6dB has been suggested as a reasonable factor, although I need
to confirm this through my own measurements).

 

Jim

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