Topband: Small Loop does not receive weak signal on 160m BOUVET RX SPOILER

[GEORGE WALLNER]

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 at 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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