# What would be the typical received power of an “S9” signal?

Obviously, I would expect the amount of power at a receiver would vary greatly based on distance from, and the power/efficiency of, the transmitter. But I don't really have a good sense for what "ballpark" the power would even be in for a "good quality" signal. Is it — milliwatts? picowatts?

The answer for a literal "S9" value is actually pretty simple! According to the table in this helpful article on S-meters, S9 corresponds to 50 picowatts of input power.

This is actually by definition: a standard S9 value was defined as 50 μV in the HF range (but 5 μV for frequencies above 30 MHz) — and at 50 ohms impedance, such voltage is developed by that approximately 50 pW of power. To avoid a dependency on impedance, the standard has apparently been updated directly in terms of power. The Wikipedia S Meter article cites:

IARU Region 1 Technical Recommendation R.1 defines S9 for the HF bands to be a receiver input power of -73 dBm. This is a level of 50 microvolts at the receiver's antenna input assuming the input impedance of the receiver is 50 ohms.

All other standard S-values are relative to this S9 definition [which I ended up finding in the IARU Region 1 HF Manager Handbook, chapter 8.2.1 rather than Wikipedia's citation]. Granted, the signal meter on any given receiver may or may not actually give readings according to this full standard, but contrary to rumors, at least there is a standard.

Since I had started to work out my own estimate via Free-space path loss equations before I thought to simply search for the S-value definitions, I thought it'd be interesting to see what sort of distance S9 would then correspond to given a 100W transmission between two dipoles.

Re-arranging the Friis transmission equation, and neglecting all other factors, this works out to:

$$R = \lambda / 4 \pi \cdot 10^{\large (G_r + G_t + P_t - P_r)/20}$$

For the following values:

Pr = -73 dBm (S9 power at receiver relative to millwatts)
Pt = 50 dBm (100W output power from transmitter)
Gt = Gr = 2.15 dBi (half-wave dipole gain over isotropic)
λ = 40 meters (is a fun band)


Works out to:

$$40/(4\cdot\pi) \cdot 10^{\large (2.15+2.15+50+73) / 20} = 7,376,496.273$$

So given an otherwise lossless system, but tempered by fairly low-gain antennas, you would, on the 40m band, receive an S9 signal from a 100W transmitter when it was 7375 kilometers away. A longer distance than I imagined! But I guess that's reasonable given complete transmitter system efficiency and no atmospheric attenuation — if calling mere picowatts of energy a "strong signal" is "reasonable" in the first place!

• If it were me, I'd avoid putting the voltage figures up front, since voltage can always be transformed so it doesn't represent something inherent to the signal like power does. – Kevin Reid AG6YO Dec 15 '16 at 22:53
• On HF, the noise floor is already high, and if you are on SSB the power is spread over a relatively large bandwidth compared to CW. I'd suppose in practice it requires quite a lot more power than a strictly defined S9 to achieve a subjective S9 in these conditions. – Phil Frost - W8II Dec 16 '16 at 3:52
• Friis equation really doesn't work at all outside of free space and shouldn't be used for range estimations on Earth at all. – AndrejaKo Dec 16 '16 at 9:29
• In case of mountain tops, tall buildings and similar situations, you have the issues with Fresnel zones. Read up a bit about them. As a rule of thumb, on 2 m, first Fresnel zone is around 230 m and on 430, it's around 130 m at the widest point. Therefore, you'd need to be very high above local terrain for them to be insignificant. Things do change at microwave frequencies though, so if you have links on 1.2 GHz, 2.3 GHz or up, influence is not so high, but they still need to be taken into account in say WiFi planning or similar. – AndrejaKo Dec 17 '16 at 21:13
• Playing with VOACAP you can get more realistic estimates of path loss. You'll also see that as a general rule, the received power is higher at lower frequencies, but so is the noise floor. – Phil Frost - W8II Mar 21 '17 at 22:38