Return-Path: Received: from mtain-mj02.r1000.mx.aol.com (mtain-mj02.r1000.mail.aol.com [172.21.164.86]) by air-df06.mail.aol.com (v128.1) with ESMTP id MAILINDF062-5ef74bb4a563163; Thu, 01 Apr 2010 09:53:39 -0400 Received: from post.thorcom.com (post.thorcom.com [193.82.116.20]) by mtain-mj02.r1000.mx.aol.com (Internet Inbound) with ESMTP id 585713800008D; Thu, 1 Apr 2010 09:53:37 -0400 (EDT) Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1NxKot-0007BU-UA for rs_out_1@blacksheep.org; Thu, 01 Apr 2010 14:52:27 +0100 Received: from [193.82.116.32] (helo=relay1.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1NxKot-0007BL-Cf for rsgb_lf_group@blacksheep.org; Thu, 01 Apr 2010 14:52:27 +0100 Received: from smtp819.mail.ukl.yahoo.com ([217.12.12.248]) by relay1.thorcom.net with smtp (Exim 4.63) (envelope-from ) id 1NxKoq-0006P3-SL for rsgb_lf_group@blacksheep.org; Thu, 01 Apr 2010 14:52:27 +0100 Received: (qmail 15574 invoked from network); 1 Apr 2010 13:52:18 -0000 Received: from MJUPC (mike@86.133.180.191 with login) by smtp819.mail.ukl.yahoo.com with SMTP; 01 Apr 2010 13:52:17 +0000 GMT X-Yahoo-SMTP: 7XGQHgGswBDgkLLWIBu6pqNs.6tUpXu5kYQqaG06DJ4lPfgH_WfGxPLi X-YMail-OSG: oqoah28VM1mHNbrgg5zEY0jehX_fFbrDk9Bda7.W88W19a0GrF2tqD2LlfdBy6hygObOr_FQmpFGBV_.Tp06zeZZawEPYOanFIna8JEW_IYs4hT3dMkd5oa2RIwBN_fO5d29PNJeLrjWDf4Fto7RKqUnNL766jMoKykgruxy2aj_YQsYj7ln7XDidt2CziQ.q1J8tx6p8bvI2JRfwECeNYdZyj7lgnvnNpzyIYL6s3bTEq9gItOAFtFqvShqVmeIK5V.4.mli4Q- X-Yahoo-Newman-Property: ymail-3 Message-ID: From: "Mike Underhill" To: References: <001d01cad190$aa92e590$0901a8c0@lark> In-Reply-To: Date: Thu, 1 Apr 2010 14:51:49 +0100 MIME-Version: 1.0 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Windows Mail 6.0.6002.18005 X-MimeOLE: Produced By Microsoft MimeOLE V6.0.6002.18005 X-Spam-Score: 0.0 (/) X-Spam-Report: autolearn=disabled,HTML_MESSAGE=0.001,UNPARSEABLE_RELAY=0.001 Subject: Re: LF: Re: E-Field Probe calibration question Content-Type: multipart/alternative; boundary="----=_NextPart_000_019C_01CAD1AA.DBB13630" X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.6 required=5.0 tests=HTML_20_30,HTML_MESSAGE, MISSING_OUTLOOK_NAME autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false x-aol-global-disposition: G x-aol-sid: 3039400c89a44bb4a5611a72 X-AOL-IP: 193.82.116.20 ------=_NextPart_000_019C_01CAD1AA.DBB13630 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Andy If or when all else has failed why not try the (my?) "two identical an= tenna method"? Set up a pair of two identical short whips about 30 to= 50m apart on reasonably uniform ground. they should be identically= loaded /matched preferably to 50 ohms. Do not worry about how much lo= ss you have for each of these. It factors out. Measure x (dB)=3D the= path loss in dB between them, with Tx on one and Rx on the other. = You should use a source of sufficient power to give a good SNR at the= receive end. Then place your vehicle with whip at the half wave poin= t. Put a known power Pin (dBW) into the Tx whip. The power density= or erp at the halfway point is then Pin(dBW) - x/2(dB). Measure the= signal strength on the car whip. You may then convert this into volts= per metre if you wish. This method should work for any of the LF ban= ds. In theory it should also work at 9kHz!? Hope this turns out to be useful. Mike - G3LHZ ----- Original Message -----=20 From: Andy Talbot=20 To: rsgb_lf_group@blacksheep.org=20 Sent: Thursday, April 01, 2010 2:05 PM Subject: Re: LF: Re: E-Field Probe calibration question Hi Alan + All... Actually its a bit more complicated on its own... The physical len= gth of the element is 0.75m, which is appreciably longer than the elec= trical effective length. And I'm betting a chunk of that difference= is made up by having a series resistor in the output circuitry to giv= e a 50 ohm source impedance and help stability but reducing voltage ga= in to below unity. If 50R in series, there is an immediate -6dB halv= ing the length. If my original dipole concept were to apply, then= the other half of the leg caused by the car would also be affected by= the attenuation. But it appears from your comments so far the argument doesn't apply= in which case the 0.15m length still holds. How do the EMC measurem= ent community do it, as their measurements have to be valid in law and= could be held up to scrutiny? Now, what was the voltage from an untuned loop, again ;-? Andy www.g4jnt.com Mostly retired inguneer On 1 April 2010 12:44, Alan Melia wrote: Hi Andy maybe I dont read it right, but if that is an E-field prob= e it is not a "dipole" The input capacitance senses the voltage at two poi= nts on the wavefront. The ground plane is, I believe, irrelevant. You just ne= ed another capacitance separated from the "probe" to reference to voltage to.= See Renato's balanced probe (www.vlf.it) In your case this is the car= .....but it could be the size of a dinky toy.The area of the wavefront inte= rcepted by the probe or the ground reference does not affect the terminal voltage.....as I see it.................of course I might just be= wrong cos i am really a scientist not an enjuneir!! :-)) Alan ----------------------------------------------------------------------= -------- No virus found in this incoming message. Checked by AVG - www.avg.com=20 Version: 8.5.437 / Virus Database: 271.1.1/2782 - Release Date: 03/3= 1/10 18:32:00 ------=_NextPart_000_019C_01CAD1AA.DBB13630 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
Andy
If or when all else has failed why not try the (my?) "two identic= al antenna=20 method"?  Set up a pair of two identical short whips about 30 to= 50m apart=20 on reasonably uniform ground.  they should be identically loaded= /matched=20 preferably to 50 ohms. Do not worry about how much loss you have= for each=20 of these.  It factors out.  Measure x (dB)=3D the path= loss in dB=20 between them, with Tx on one and Rx on the other.   You should us= e a source=20 of sufficient power to give a good SNR at the receive end.  Then= place your=20 vehicle with whip at the half wave point.  Put a known power Pin= (dBW) into=20 the Tx whip.  The power density or erp at the halfway point is th= en=20 Pin(dBW) - x/2(dB).  Measure the signal strength on the car whip.= You may=20 then convert this into volts per metre if you wish.  This method= should=20 work for any of the LF bands.  In theory it should also work at= =20 9kHz!?
Hope this turns out to be useful.
Mike - G3LHZ
 
----- Original Message -----
From:=20 Andy Talbot
Sent: Thursday, April 01, 201= 0 2:05=20 PM
Subject: Re: LF: Re: E-Field= Probe=20 calibration question

Hi Alan + All...
 
Actually its a bit more complicated on its own...  = The=20 physical length of the element is 0.75m, which is appreciably longer= than the=20 electrical effective length.   And I'm betting a chunk of= that=20 difference is made up by having a series resistor in the output circ= uitry to=20 give a 50 ohm source impedance and help stability but reducing volta= ge gain to=20 below unity.  If 50R in series, there is an immediate -6dB = ; halving=20 the length.    If my original dipole concept were= to=20 apply, then the other half of the leg caused by the car would also= be affected=20 by the attenuation.
 
But it appears from your comments so far the argument doesn't= apply in=20 which case the 0.15m length still holds.   How do the EMC= =20 measurement community do it, as their measurements have to be valid= in law and=20 could be held up to scrutiny?
 
Now, what was the voltage from an untuned loop, again ;-?
 
Andy
Mostly retired inguneer
 
 
On 1 April 2010 12:44, Alan Melia <alan.melia= @btinternet.com>=20 wrote:
Hi Andy maybe= I dont read it right, but if that is an=20 E-field probe it is
not a "dipole" The input capacitance senses= the=20 voltage at two points on the
wavefront. The ground plane is, I= believe,=20 irrelevant. You just need another
capacitance separated from th= e "probe"=20 to reference to voltage to. See
Renato's balanced probe (www.vlf.it)  = ;In your case=20 this is the car.....but
it could be the size of a dinky toy.The= area of=20 the wavefront intercepted by
the probe or the ground reference= does not=20 affect the terminal
voltage.....as I see it.................of= course I=20 might just be wrong cos
i am really a scientist not an=20 enjuneir!!
 :-))
Alan


No virus found in this incoming message.
Checked by AV= G -=20 www.avg.com
Version: 8.5.437 / Virus Database: 271.1.1/2782 - Re= lease=20 Date: 03/31/10 18:32:00
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