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

[Michael Tope]

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 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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