X-GM-THRID: 1196684404245296047 X-Gmail-Labels: rsgb lf X-Gmail-Received: 3b915dc9955b87a45d9f44396fb7859f89c1f48b Delivered-To: daveyxm@gmail.com Received: by 10.54.71.12 with SMTP id t12cs55187wra; Sat, 4 Mar 2006 15:36:10 -0800 (PST) Received: by 10.66.164.4 with SMTP id m4mr2118575uge; Sat, 04 Mar 2006 15:36:09 -0800 (PST) Return-Path: Received: from post.thorcom.com (post.thorcom.com [193.82.116.20]) by mx.gmail.com with ESMTP id m1si1798730ugc.2006.03.04.15.36.08; Sat, 04 Mar 2006 15:36:09 -0800 (PST) Received-SPF: neutral (gmail.com: 193.82.116.20 is neither permitted nor denied by best guess record for domain of owner-rsgb_lf_group@blacksheep.org) Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1FFgGl-0002q0-I4 for rs_out_1@blacksheep.org; Sat, 04 Mar 2006 23:34:39 +0000 Received: from [193.82.59.130] (helo=relay2.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1FFgGk-0002pr-IP for rsgb_lf_group@blacksheep.org; Sat, 04 Mar 2006 23:34:38 +0000 Received: from imo-m20.mx.aol.com ([64.12.137.1]) by relay2.thorcom.net with esmtp (Exim 4.51) id 1FFhDt-0003nf-4S for rsgb_lf_group@blacksheep.org; Sun, 05 Mar 2006 00:35:46 +0000 Received: from MarkusVester@aol.com by imo-m20.mx.aol.com (mail_out_v38_r7.3.) id l.1da.4dde0529 (25508); Sat, 4 Mar 2006 18:34:25 -0500 (EST) From: MarkusVester@aol.com Message-ID: <1da.4dde0529.313b7e00@aol.com> Date: Sat, 4 Mar 2006 18:34:24 EST To: rsgb_lf_group@blacksheep.org CC: dl4yhf@freenet.de MIME-Version: 1.0 X-Mailer: 8.0 for Windows sub 6104 X-Spam-Flag: NO X-Spam-Score: -2.2 (--) X-Spam-Report: autolearn=disabled,AWL=0.084,BAYES_00=-2.599,HTML_20_30=0.226,HTML_MESSAGE=0.001,MIME_QP_LONG_LINE=0.039,NO_REAL_NAME=0.007 Subject: Re: LF: LORAN spurious emission levels Content-Type: multipart/mixed; boundary="part1_1da.4dde0529.313b7e00_boundary" X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: * X-Spam-Status: No, hits=1.2 required=5.0 tests=HTML_MESSAGE, HTML_TAG_BALANCE_HTML,NO_REAL_NAME,WEIRD_QUOTING 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 Status: O X-Status: X-Keywords: X-UID: 6228 --part1_1da.4dde0529.313b7e00_boundary Content-Type: multipart/related; boundary="part1_1da.4dde0529.313b7e00_rel_boundary" --part1_1da.4dde0529.313b7e00_rel_boundary Content-Type: multipart/alternative; boundary="1da.4dde0529_alt_bound" --1da.4dde0529_alt_bound Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: quoted-printable Hi Peter, went through your calculation and basically reproduced your result, with onl= y=20 minor differences: Following your reasoning, the power in the last bin would be 10^(-59.54/10)=20= *=20 1.37^2 =3D 2.087e-6 units. For the power sum of all bins I got 0.464 units,=20 resulting in 4.50 (instead of 4.62) ppm in the 137 kHz bin. With 2.47 kW ave= rage,=20 the average power per line would be 41.3 =B5W - perfect. I do not think this is too low. This EMRP should produce about 0.5 =B5V/m at= =20 100 km. My best daytime noise background was on the order of -27 dB=B5V/m in= 1 Hz,=20 so when viewed with in QRSS3 (1.5*0.34 Hz noise bandwidth), the line would=20 obtain 24 dB SNR. Here at 630 km from Sylt, I can only rarely see hints of L= oran=20 in QRSS3. Using MathCad's symbolic solver, the Fourier integration of the analytical=20 formula yields (if i did it right) =20 using omega =3D 2 pi (f/kHz - 100) * 0.065, and with the result normalized t= o 1=20 at center. It falls off only at -40 dB/decade, as expected for the click of=20= a=20 t^2 pulse start. The following plot shows how the formula fits the spectrum=20 analyzer data: 73 and have a nice weekend Markus In einer eMail vom 03.03.2006 10:56:07 Westeurop=E4ische Normalzeit schreibt= =20 peter.martinez@btinternet.com:=20 > From G3PLX: >=20 > I calculated the radiated power of a LORAN line now, but I would like othe= r=20 > people to check if I did it right, because the result is surprisingly smal= l. >=20 > The spectrum measurement made at Rugby gives a chart of dB levels for 2kHz= =20 > steps over 70-130kHz, going from -8dB in the centre to -53dB at 130kHz, an= d=20 > I extrapolated by eye to estimate the levels at three more steps up to=20 > 136kHz. Here are the numbers themselves... >=20 > -52.00,-49.95,-47.49,-45.03,-43.10,-41.42,-40.24,-38.42,-35.72,-31.13,-28.= 75, > -22.49,-17.82,-13.84,-8.01,-8.46,-13.29,-16.76,-18.44,-21.41,-27.43,-29.16= ,-3 > 3.89,-40.21,-40.09,-43.32,-45.47,-44.45,-48.63,-51.86,-52.94,-55.24,-57.44= ,-5 > 9.54 >=20 > Notice that there are two 'largest' figures, so although the chart shows=20 > these two as being at frequencies of 98 and 100kHz, I am guessing that the= y=20 > are really figures for the centre of bands 98-100 and 100-102kHz. The last= =20 > three are my guesses for the values for the bands 132-134, 134-136, and=20 > 136-138 kHz. From the other information in the spreadsheet I was sent, I=20= am=20 >=20 > quite certain that this spectrum was measured at an antenna current sampli= ng=20 >=20 > point. >=20 > For each of these figures I calculated the relative "current-squared" valu= es=20 >=20 > from 10^(dB/10) then multiply each one by (F/100)^2, where F is the=20 > frequency in kHz. This gives a figure for the relative radiated power in=20 > each 2kHz slot. The square-frequency factor arises because the radiation=20 > resistance will be proportional to F-squared, and I am seeking to calculat= e=20 > the I-squared-R for each 2kHz slot. The figure I get for the 136-138kHz sl= ot=20 >=20 > is 0.0021. >=20 > Next I add all these figures to give me a total of 454.6338 units of power= .=20 > I don't know what the units are because I don't have enough information=20 > about the transmitter or the antenna, but I can divide 0.0021 by it to giv= e=20 > me the fraction of the total mean power that is radiated in the 136-138kHz= =20 > slot. This is 4.62e-6. >=20 > For the Lessay chain I know the group repetition interval is 67.31mS which= =20 > means there are 14.85 pulse groups per sec. Because half the pulses are=20 > constant polarity and half are alternating, I know that half the energy go= es=20 >=20 > into spectral lines at this spacing and half goes into lines at odd=20 > multiples of half this figure. Effectively there are twice as many spuriou= s=20 > lines as the GRI says, spaced at 7.43Hz. There are about 270 of these=20 > (2000/7.43) in a band of 2kHz wide. >=20 > Therefore the proportion of the total mean power which is radiated in a=20 > single spectral line at 136kHz is 4.62e-6 / 270 =3D 1.71e-8. >=20 > If we estimate (from the info given by Marco) that each LORAN transmitter=20 > radiates 7.5kW mean power, then the radiated power in each LORAN line is=20 > 1.71e-8 * 7500 =3D 0.128 mW. >=20 > I don't know if I should believe this! Can anybody find a mistake or at=20 > least check my calculations? All the figures needed for the calculations=20 > are in this email. Does anyone have any evidence that any of these figures= =20 > are wrong? >=20 > 73 > Peter G3PLX >=20 >=20 >=20 --1da.4dde0529_alt_bound Content-Type: text/html; charset="ISO-8859-1" Content-Transfer-Encoding: quoted-printable Hi Peter,

went through your calculation and basically reproduced your result, with onl= y minor differences:

Following your reasoning, the power in the last bin would be 10^(-59.54/10)=20= * 1.37^2 =3D 2.087e-6 units. For the power sum of all bins I got 0.464 units= , resulting in 4.50 (instead of 4.62) ppm in the 137 kHz bin. With 2.47 kW a= verage, the average power per line would be 41.3 =B5W - perfect.

I do not think this is too low. This EMRP should produce about 0.5 =B5V/m at= 100 km. My best daytime noise background was on the order of -27 dB=B5V/m i= n 1 Hz, so when viewed with in QRSS3 (1.5*0.34 Hz noise bandwidth), the line= would obtain 24 dB SNR. Here at 630 km from Sylt, I can only rarely see hin= ts of Loran in QRSS3.

Using MathCad's symbolic solver, the Fourier integration of the analytical f= ormula yields (if i did it right)



using omega =3D 2 pi (f/kHz - 100) * 0.065, and with the result normalized t= o 1 at center. It falls off only at -40 dB/decade, as expected for the click= of a t^2 pulse start. The following plot shows how the formula fits the spe= ctrum analyzer data:


73 and have a nice weekend

Markus


In einer eMail vom 03.03.2006 10:56:07 Westeurop=E4ische Normalzeit schreibt= peter.martinez@btinternet.com:

From G3PLX:

I calculated the radiated power of a LORAN line now, but I would like other=20=
people to check if I did it right, because the result is surprisingly small.=

The spectrum measurement made at Rugby gives a chart of dB levels for 2kHz <= BR> steps over 70-130kHz, going from -8dB in the centre to -53dB at 130kHz, and=20=
I extrapolated by eye to estimate the levels at three more steps up to
136kHz. Here are the numbers themselves...

-52.00,-49.95,-47.49,-45.03,-43.10,-41.42,-40.24,-38.42,-35.72,-31.13,-28.75= ,-22.49,-17.82,-13.84,-8.01,-8.46,-13.29,-16.76,-18.44,-21.41,-27.43,-29.16,= -33.89,-40.21,-40.09,-43.32,-45.47,-44.45,-48.63,-51.86,-52.94,-55.24,-57.44= ,-59.54

Notice that there are two 'largest' figures, so although the chart shows these two as being at frequencies of 98 and 100kHz, I am guessing that they=20=
are really figures for the centre of bands 98-100 and 100-102kHz. The last <= BR> three are my guesses for the values for the bands 132-134, 134-136, and
136-138 kHz.  From the other information in the spreadsheet I was sent,= I am
quite certain that this spectrum was measured at an antenna current sampling=
point.

For each of these figures I calculated the relative "current-squared" values=
from 10^(dB/10) then multiply each one by (F/100)^2, where F is the
frequency in kHz. This gives a figure for the relative radiated power in each 2kHz slot. The square-frequency factor arises because the radiation resistance will be proportional to F-squared, and I am seeking to calculate=20=
the I-squared-R for each 2kHz slot. The figure I get for the 136-138kHz slot=
is 0.0021.

Next I add all these figures to give me a total of 454.6338 units of power.=20=
I don't  know what the units are because I don't have enough informatio= n
about the transmitter or the antenna, but I can divide 0.0021 by it to give=20=
me the fraction of the total mean power that is radiated in the 136-138kHz <= BR> slot. This is 4.62e-6.

For the Lessay chain I know the group repetition interval is 67.31mS which <= BR> means there are 14.85 pulse groups per sec. Because half the pulses are
constant polarity and half are alternating, I know that half the energy goes=
into spectral lines at this spacing and half goes into lines at odd
multiples of half this figure. Effectively there are twice as many spurious=20=
lines as the GRI says, spaced at 7.43Hz. There are about 270 of these
(2000/7.43) in a band of 2kHz wide.

Therefore the proportion of the total mean power which is radiated in a
single spectral line at 136kHz is 4.62e-6 / 270 =3D 1.71e-8.

If we estimate (from the info given by Marco) that each LORAN transmitter radiates 7.5kW mean power, then the radiated power in each LORAN line is 1.71e-8 * 7500 =3D 0.128 mW.

I don't know if I should believe this!  Can anybody find a mistake or a= t
least check my calculations?  All the figures needed for the calculatio= ns
are in this email. Does anyone have any evidence that any of these figures <= BR> are wrong?

73
Peter G3PLX




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