a calculation of the associated
free space and ionosheric
losses in a major DXpedition
The QSO
An original message from the UKSMG Announcement Page:
"QRP DX...I met Peter PY5CC at ~1700z yesterday (3 April) with a 59+20 signal. He was using only 80W rf. After some chat about VP8 on the band I asked him to try a 2W QSO. I lowered my power to minimum and he reported 59 with my signal. A 59 signal on my meter is 30+db above the noise and I asked him to try to copy me at 1mW (1 milliwatt). Inserting 3+10+20 db of fixed HP coaxial attenuators the output power was at approx. 1mW level and Peter was still able to copy my signal at 51 !! All the test was recorded by him and during his playback my 1mW signal was R5. With a distance of 10000km between us, it seems that during the opening some kind of ionospheric focusing was taking place."
Costas SV1DH - Tuesday, April 04, 2000 at 06:58:59 (GMT/BST)
This QSO must have been the QRP world record! The message triggered me to make some calculations with the formula for basic transmission loss in free space. Was the six-metre ionospheric signal stronger than the one calculated for free-space loss?
Free-space loss
This is the formula:
L /dB = 20 log(4d/) = 32,5 + 20 log f/MHz + 20 log d/km
For 50 MHz follows:
L /dB = 66,5 + 20 log d/km
For d = 10000 km follows
L = 146,5 dB
This is the value for isotropic antennas (0 dBi gain). Antenna gains have to be subtracted from this value to get to the right attenuation figure from one antenna connector to the other. PY5CC and SV1DH gave me their antenna gain figures; one dB cable loss assumed on both systems gave a sum of both antenna gains of 29 dBi.
A free space loss of L = 117.5 dB results.
What would the theoretical signal level in free space operating with 1 milliwatt (= 0 dBm) be?
The result is very simple:
L(rx) = -117.5 dBm
Recalculated in voltage level this gives V(rx) = 0.3 V (Most radio amateurs can think in microvolt levels rather than in dBm). So that should be a readable signal. But let’s consider this a little bit further.
What is a typical 50MHz receiver noise floor?
Thermal noise = -174 dBm/Hz = - 140 dBm /2.5 kHz (SSB bandwidth). This is the noise power coming out of a dummy load at room temperature. Connecting an antenna instead will cause the noise level to increase. Let us assume the total input noise power is 10 dB higher than the thermal noise. The additional noise is a combination of of galactic noise, atmospheric noise received by ground wave and sky wave, local man made noise and man made noise received via the ionosphere. It is all added to the receiver input noise floor.
So the actual receiver noise level would be NL = -130 dBm/SSB bandwidth. The -117.5 dBm signal should be well received with 12.5 dB signal to noise level, free space loss assumed.
Signal loss for ionosheric propagation
1. Results of the 1mW QSO
According to PY5CC the actual signal was 3 to 5 dB above noise. This corresponds to a signal level of -127 dBm to -125 dBm. So the 50 MHz signal was only about 8.5 dB weaker than what would be expected in free-space propagation.
This is remarkable. 8.5 dB would be about the best value expected for a single-hop ionosheric refraction loss on HF. Why should we get a lower figure on Six than on HF?
Peter, PY5CC thinks that two hops were involved in the propagation to Greece. This implies about 4dB additional loss per ionospheric refraction.
2. Results of the 2W QSO
Two watts resulted in an S9 report at PY5CC. While S-meter readings are all but accurate the S9 indication is mostly set by the factory, so an S9 signal level can plausibly be used to calculate the propagation loss. On VHF, S9 is normally equal to 5 V or -93 dBm (in fact the s-meter of my IC575 is S9 with -92 dBm). Two watts is equal to +33 dBm.
So the signal loss was 126 dB. This is incredibly close to the value calculated from the 1mW QSO.
How accurate are these results? We have to consider several measurement uncertainties.
Measurement Uncertainties
There are several uncertainties:
• the receiver S-meter: that can be quite accurate.
• antenna gain figures: the free space gain can be precisely calculated.
• additional antenna gain by ground reflection: up to 6 dB of additional gain can be observed at certain elevation angles, although the value can also be below the free space gain!
• measuring or guessing the signal-to-noise ratio of a weak signal: a skilled operator or an experienced RF engineer can ‘guess’ this quite accurately.
The real antenna gain could have been higher during the tests because of ground-gain. This would result in a higher measured ionospheric refraction loss than the 4.5 dB assumed (which would make sense). I would guess the uncertainty to be 3 dB but this statement would certainly be rejected by a specialist.
Can S-Meter readings tell the radiated power?
Yes, we can make a rough calculation of the radiated power of a station heard. This works best with signal strengths above S9 and a calibrated attenuator, assuming that the S-meter has been set to S9 at -93 dBm.
The precondition of course is a propagation peak. For that, let us assume that the ionospheric diffraction loss is 6dB per hop.
Let us talk about sporadic-E contacts. For a distance of 1500 km the formula gives a free space loss of 130 dB. With one ionospheric hop the attenuation for isotropic antennas would be 136 dB. With a 10dBi antenna on the receiving side the adjusted signal loss is 126 dB.
A practical measurement. The strongest TV carrier signal ever received at DL7AV was Spanish TV on 48.250 at -43 dBm. A vertical HB9CV antenna was used which might have a 6 dBi gain, resulting in a net signal loss of 130 dB. The addition of both values results in a radiated power of +87 dBm, equal to 500 kW. This number is 3 dB above the radiated power of that station which I found later in a "TV-List 44-108 MHz TV-Stations Worldwide, Edition 2.0, August 1993" by AGDX Germany. I don’t think we can expect much better accuracy from such measurements.
What can the maximum signal strength of an amateur signal be? An average six-metre DXer might have 100 watts and a 13 dBi antenna. This results in a radiated power of +63 dBm. The receiver signal strength in a distance of 1500 km under the conditions mentioned above (126 dB signal loss) might peak +63 –126 = -63 dBm. This is S9 + 30dB by the way with a real S-meter. Stronger indications on your S-meter imply that the meter is not correctly calibrated or that the other station is using high power.
Conclusions
With optimum propagation and low angles of radiation, surprisingly low path losses can be observed on 50MHz. But in the case discussed, the measured path loss was still above free space loss, so there is no proof that ionospheric focusing was taking place. The formula given allows a rough calculation of the radiated power of the station received, subject to the accuracy of the S-meter used.
