I had an interesting experience along these lines back in 2006 for the
CQ WW 160 CW contest. Following the lead of K6SE and others, I got
permission from the Bureau of Land Management to operate from the salt
evaporation ponds on Koehn dry-lake near Mojave, California. I spent all
day Thursday setting up the transmit antenna and when I started
listening Thursday night my heart sank. The static crashes were 20 over
S-9 and it was difficult to hear much of anything. I was worried that
the weekend might be a bust. I had the parts with me for my home-brew RX
4-square, but it wasn't installation friendly and there was way too much
to do on Friday morning to have any hope of getting it up and installed
in time for the start of the contest. Fortunately, as it turned out, 160
was a completely different band the next two nights. All of the static
crashes were gone and the band was amazing just listening on the TX
antenna. I worked over 100 Asian stations that weekend (mostly JAs). So
yeah, using the transmit-antenna for receive at a location with low
ambient ground-wave noise can work, but as George aptly points out it's
a crap-shoot.
73, Mike W4EF.............
On 12/25/2022 7:36 AM, GEORGE WALLNER wrote:
In my experience, a DXpedition does not need an RX antenna on 160 for
the first two to three days. During that time they are working the big
guns with strong signals. The RX antenna is needed to give the weaker
stations a chance once the big guns are out of the way.
In case of Bouvet, most of the signals will be coming from the NE or
NW direction. Their "back" will be towards Antarctica, which is not a
major source of noise. The F/B of an RX antenna will contribute little
(but not zero).
The bigger problem will be that Bouvet Island is in the southern
hemisphere, where it is summer this time of the year. Late-afternoon
or evening thunderstorms taking place south of the Equator will
generate a lot of noise (think of Africa, Amazon). Most of that noise
will be coming from the same direction as the the NA and EU signals,
in which case an RX antenna may not help (much). Their biggest
challenge will be working the Far East (JA especially). That is where
a good RX antenna, pointed in the right direction, could help because
most of the noise would be coming from its side. With marginal TB
QSO-s even a fraction of a dB in S/N can make the difference . Also,
they wont have much darkness to work with. It will be tough. I would
make the loop bigger to improve the S/N ratio and improve the JA-s'
chances.
73,
George,
AA7JV/C6AGU
On Fri, 23 Dec 2022 23:03:36 -0600 wrote:
I was one of the 160m ops at "nearby" FT5XO. I can tell you that when
the
WX was good the TX antennas worked very well for receiving. When the
WX turned bad, such as during the snow storm we worked through, the 160m
TX antennas
were very noisy....just as noisy as those anywhere on the planet.
Don't get me started about those awful 160m fishing buoy transmitters!!
73,
Charlie, N0TT
On Fri, 23 Dec 2022 19:45:28 -0700 Wes <wes_n7ws@triconet.org> writes:
All interesting. But let me ask (and standby for flames) what is
wrong with them simply listening on the TX antenna?
I know, I know, conventional wisdom says that you can't possibly
work 160 DX without a separate RX antenna. I'll confess that I am a
little pistol and will never be on the TB Honor Roll, but I got on
the band just to add another DXCC band to my collection (now nine).
I'm now at 144 confirmed, running just 500W and a 55' inverted-L on
both TX and RX. Generally speaking I hear better that I get out.
Looking at my chances of working 3Y the optimum time is their
sunrise (~3:30Z) when I am in complete darkness and straight across
the terminator. They will have the sunlit ocean to their rear and
the S. American landmass toward me. Maybe someone can enlighten me,
but I fail to see how a directional antenna will improve the SNR of
my signal at their end.
Wes N7WS
On 12/23/2022 6:46 PM, JC wrote:> Hi topband lovers>> >> Some
friends contact me with deep concerns about the next Bouvet DX
expedition receiver antenna called SALAD>> >>
<
http://www.lz1aq.signacor.com/docs/active-wideband-directional-antenna.p
hp>
Salad antenna>> >> I understand the concerns, Bouvet on 160m is a
lifetime opportunity for most top-banders!>> >> When Doug NX4D, me
N4IS and Dr Dallas started to try to understand the limitation of
the new Waller Flag, the first big question was;>> >> How small a
loop antenna can be to receive weak signal on 160, or MW?>> >> Dr.
Dallas Lankford III (SK), measured the internal noise of a small
loop. 15x15 FT on his quiet QTH, and wrote a paper with the
derivation necessary to calculate the thermal noise of a small loop.
The study most important point was:>> >> The sensitivity of small
loop antennas can be limited by internally generated thermal noise
which is a characteristic of the loop itself. Even amplifying the
loop output with the lowest noise figure preamp available may not
improve the loop sensitivity if manmade noise drops low enough>>
>> The noise on Bouvet
island will be very low, < -120 dBm at 500Hz, and for sure the
internal thermal noise of the prosed RX antenna will limit the
reception of weak signals on 160m, it may work on 80 and above, but
for 160 m, it will be a set up for failure.>> >> Why not a single,
trustable beverage antenna over the ice or snow?? Or a proved K9AY or
a DHDL??>> >> Below is the almost good transcript of the original
pdf Flag Theory, for the long answer.>> >> >> 73?s>> JC>> N4IS>>
>> Flag Theory> Dallas Lankford, 1/31/09, rev. 9/9/09>>> The
derivation which follows is a variation of Belrose's classical
derivation for ferrite rod loop antennas,> ?Ferromagnetic Loop
Aerials,? Wireless Engineer, February 1955, 41? 46.>>> Some people who
have not actually compared the signal output of a flag antenna to
other small antennas have expressed their opinions to me that the
signal output of a flag antenna has
great attenuation compared to those other small antennas, such as
loops and passive verticals. Their opinions are wrong. One should
never express opinions which are based, say, on computer simulations
alone, without actual measurements. The development below is based on
physics (including Maxwell's equations), mathematics, and
measurements.>>> Measurements have confirmed that the flag signal to
noise formula derived below is approximately correct despite EZNEC
simulations to the contrary. For example, EZNEC simulation of a 15'
square loop at 1 MHz predicts its gain is about +4 dbi, while on the
other hand EZNEC simulation of a 15' square flag at 1 MHz predicts its
gain is about ?46 dBi. But if you construct such a loop and such a
flag and observe the signal strengths produced by them for daytime
groundwave MW signals, you will find that the maximum loop and flag
signal outputs are about
equal. Although somewhat more difficult to judge, the nighttime sky
wave MW signals are also about equal.>>> Also, the signal to noise
ratio formula below for flag arrays has been verified by manmade noise
measurements in the 160 meter band using a smaller flag array than the
MW flag array discussed below. Several years ago a similar signal to
noise ratio formula for small un-tuned (broadband) loop antennas was
verified at the low end of the NDB band.>>> The signal voltage es in
volts for a one turn loop of area A in meters and a signal of
wavelength ? for a given radio wave is>> >> es = [2pA Es /?]
COS(?)>> >> where Es is the signal strength in volts per meter and ?
is the angle between the plane of the loop and the radio wave. It is
well known that if an omnidirectional antenna, say a short whip, is
attached to one of the output terminals of the loop and the phase
difference
between the loop and vertical and the amplitude of the whip are
adjusted to produce a cardioid patten, then this occurs for a phase
difference of 90 degrees and a whip amplitude equal to the amplitude
of the loop, and the signal voltage in this case is>> >> es = [2pA
Es /?] [1 + COS(?)]>> .> Notice that the maximum signal voltage of the
cardioid antenna is twice the maximum signal voltage of the loop (or
vertical) alone.>> A flag antenna is a one turn loop antenna with a
resistance of several hundred ohms inserted at some point into the one
turn. With a rectangular turn, with the resistor appropriately placed
and adjusted for the appropriate value, the flag antenna will generate
a cardioid pattern. The exact mechanism by which this occurs is not
given here. Nevertheless, based on measurements, the flag antenna
signal voltage is approximately the same as the cardioid pattern given
above. The difference between an actual flag and the cardioid pattern
above is that an actual flag pattern is not a perfect cardioid for
some cardioid geometries and resistors.>> >> In general a flag pattern
will be>> es = [2pA Es /?] [1 + kCOS(?)]>> >> where k is a constant
less than or equal to 1, say 0.90 for a ?poor? flag, to 0.99 or more
for a ?good? flag. This has virtually no effect of the maximum signal
pickup, but can have a significant effect on the null depth.>>> 1- The
thermal output noise voltage en for a loop is>> >> en = v(4kTRB)>>
>> where k (1.37 x 10^?23) is Boltzman's constant, T is the absolute
temperature (taken as 290), (Belrose said:) R is the resistive
component of the input impedance, (but also according to Belrose:) R =
2pfL where L is the loop inductance in Henrys, and B is the receiver
bandwidth in Hertz.>> >> When the loop is rotated so that the
signal is maximum, the signal to noise ratio is>> >> SNR = es/en =
[2pA Es /?]/v(4kTRB) = [66Af/v(LB)]Es .>> >> The point of this
formula is that the sensitivity of small loop antennas can be limited
by internally generated thermal noise which is a characteristic of the
loop itself. Even amplifying the loop output with the lowest noise
figure preamp available may not improve the loop sensitivity if
manmade noise drops low enough.>> >> Notice that on the one hand
Belrose said R is the resistive component of the input impedance, but
on the other hand R = 2pL. Well never mind. Based on personal on hands
experience building small loops, I believe R = 2pL is approximately
correct. What I believe Belrose meant is that R is the magnitude of
the output impedance. For a flag antenna rotated so the signal is
maximum, the signal to noise ratio is>> >> SNR = es/en = 2[2pA
Es/?]/v(4kTv((2pfL)^2 + (Rflag)^2)B) = [322Af/v(v((2pfL)^2 +
(Rflag)^2)B)]Es .>>> Now let us calculate a SNR. Consider a flag 15'
by 15' with inductance 24 ?H at 1.0 MHz with 910 ohm flag> resistor,
and a bandwidth of B = 6000 Hz. Then A = 20.9 square meters and SNR =
2.86x10^6 Es . If Es is in> microvolts, the SNR formula becomes SNR =
2.9 Es .>>> Any phased array has loss (or in some cases gain) due to
the phase difference of the signals from the two> antennas which are
combined to produce the nulls. This loss (or gain) depends on (1) the
separation of the two> antennas, (2) the arrival angle of the signal,
and (3) the method used to phase the two flags. Let f be the phase>
difference for a signal arriving at the two antennas. It can be shown
by integrating the difference of the squares> of the respective cosine
functions that the amplitude A of the RMS voltage output of the combiner
given RMS> inputs with amplitudes e is equal to ev(1 ? COS( f)) where
e is the amplitude of the RMS signal, in other> words,> A=? 1> 2p?> 0>
2p> 2 e2?cos?t?-cos?t?f??2dt=e?2?1-cos?f?>>> The gain or loss for a
signal passing through the combiner due to their phase difference is
thus v(1 ? COS( f)).> Let us consider the best case, when the signal
arrives from the maximum direction. For a spacing s between the>
centers of the flags, if the arrival angle is a, then the distance d
which determines the phase difference between> the two signals is d =
s COS(a). If s is given in feet, then the conversion of d to meters is
d = s COS(a)/3.28.>>> The reciprocal of the velocity of light
1/2.99x10^8 = 3.34 nS/meter is the time delay per meter of light (or
radio> waves) in air. So the phase difference of the two signals above
in terms of time is T = 3.34 s COS(a)/3.28 nS> when s is in meters. The
phase difference in degrees is thus f = 0.36Tf = 0.36 f x 3.34 s
COS(a)/3.28 where f is> the frequency of the signals in MHz. If
additional delay T' is added (phase shift to generate nulls or to
adjust the> reception pattern), then the phase difference is f =
0.36(T + T')f = 0.36f(T' + 3.34 s COS(a)/3.28) . If the> additional
delay is implemented with a length of coax L feet long with velocity
factor VF, then the phase delay is>>> f = 0.37f(L/VF + s COS(a))>>
>> where f is the frequency of the signal in MHz, s is in feet, L is
in feet, and a is the arrival angle.>>> 2->>> In the case of the flag
array above in the maximum direction there are two sources of delay,
namely 60.6 feet of> coax with velocity factor 0.70, and 100 feet of
spacing between the two flag antennas. The phase delay at 1.0> MHz for
a 30 degree arrival angle is thus>>> f = 0.37 x 1.0 x (60.6/0.70 + 100
COS(30)) =
64.1 degrees.>>> Thus the signal loss in the maximum direction at 30
degree arrival angle due to spacing and the phaser is>>> v(1 ? COS(
64.1)) = 0.75 or 20 log(0.75/2) = ?8.5 dB.>>> Now comes the
interesting part. What happens when we phase the WF array with
dimensions and spacing given> above? The flag thermal noise output
doubles (two flags), and the flag signal output decreases (due to
spacing> and phaser loss), so the SNR is degraded by 14.5 dB to SNR =
0.55 Es .>> >> So a signal of 1.8 microvolts per meter is equivalent
to the thermal noise floor of the flag array.>>> On some occasions,
when manmade noise drops to very low levels at my location, it
appeared to fall below the> thermal noise floor of the WF array. By
that I mean that the characteristic ?sharp? manmade noise changed>
character to a ?smooth? hiss. To determine whether this was the case,
I measured the manmade noise
at my> location for one of these low noise events at 1.83 MHz.>>> To
measure manmade noise at my location I converted one of the flags of
my MW flag array to a loop. The loop was 15' by 15', or 20.9 square
meters. I used my R-390A whose carrier (S) meter indicates signals as
low as ?127 dBm. The meter indication was 4 dB. Then I used an
HP-8540B signal generator to determine the dBm value for 4 dB on the
R-390A meter. It was ?122 dBm. Now the fun begins. The RDF of a loop
for an arrival angle of 20 degrees (the estimated wave tilt of manmade
noise at 1.83 MHz) was 4 dB. So now manmade noise after factoring out
the loop directionality was estimated as ?118 dBm.>> >> Field
strength is open circuit voltage equivalent, which gives us ?112 dBm.
I measured MM noise on the R-390A with a 6 kHz BW. The conversion to
500 Hz is>> >> ?10 log(6000/500) = ?10.8,>> >> which gives us
?122.8 or
?123 dBm.>> >> The conversion to 500 Hz was necessary in order to be
consistent with the SNR above which was calculated for a 500 Hz BW.>>
>> The loop equation is es = 2pAEs/lambda = 0.41 Es, and 20
log(0.41) = ?7.7, rounded off to - 8, so we have -115 dBm, or 0.40
microvolts per meter for my lowest levels of manmade noise at 1.83 MHz
in a 500 Hz bandwidth.>> >> This seemed impossibly low to me until I
came across the ITU graph at right. Manmade noise at quiet rural
locations may be even lower than 0.40 microvolts per meter at 1.83
MHz. But what about the MW band? From the CCIR Report 322 we find that
the manmade noise field strength on the average is about 10 dB higher
at 1.0 MHz than 1.83 MHz, which would make it 1.26 microvolts per
meter at 1.0 MHz. Another 4 dB is added because of impedance mismatch
between the R-390A and the loop, which brings manmade noise up to 2.0
microvolts per meter at 1.0 MHz. The RDF of one of these flags is
about 7 dB, which lowers the manmade noise to 0.89 microvolts per
meter. Observations in the 160 meter band do not seem to agree exactly
with this analysis because flag thermal noise has never been heard on
the MW flag array. But it would not surprise me at all if the flag
array thermal noise floor were only a few dB below received minimum
daytime manmade noise and that measurement error (for example,
calibration of my HP 8640B) accounts for the difference between
measurement and theory. Also, observations with a flag array having
flag areas half the size of the MW flag elements in the 160 meter band
do confirm the signal to noise ratio formula; in this case, flag
thermal noise does dominate minimum daytime manmade noise at my
location (0.40 microvolts per meter field strength measured as
described above.>> >> >>
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