Thierry wrote:

Hi,

A question about dBW and power at receive.
Among scales used in power measurement, there is the signal strength or
noise level estimation, also known as the "dB below W" (dBW or SDBW). Its
relation is dBW = 10 Log P, where P is the power expressed in watt :
That means that 100 W is 20 dBW into 50 ohms.

But usually, in propagation program (VOACAP, etc), the power strength at
receive expressed in dBW is far below such values. I read somewhere the next
equivalences :
- (minus) 93 ~ S9+10, -103 = S9, -127 = S5, -151 = S1.

With -93 dBW for a S9+10 signal at receive, that 'd mean that the power 'd
be only P(W) = 10^ (dBW/10) = 0.7 watt ?
IMHO this power is much to low... What is wrong in this relations (or in my
interpretation) ?


Your calculation of 0.7 W is wrong by many
orders of magnitude.

dBW = 10 Log(P)

-93 = 10 Log(P)
-9.3 = Log(P)
Alog(-9.3) = P
ALog (-9.3) = 5.012 * 10^-10 W
(5.0118723362727228500155418688495e-10 by win2k calculator)
So the problem is much, much worse than you thought.

Try this scenario:-

You are in Belgium, I'm in the west of the U.K. about
400 km away. You transmit using 100W and I get an S9+10
signal. All seems reasonable? Say this is on 80 metres
so both our antennas are omnidirectional. My antenna
will only receive power proportional to the solid
(stereo) angle it subtends at your QTH. So the power
of 5e-10 W seems about right to me. Just think how
many other 80 metre dipoles would fit in around that
400 km circle. Then all the ones that would fit onto
a 400km radius hemisphere. They would all get a tiny
share of your 100 W!!!

Receivers are very, very sensitive or radio wouldn't
work at all.

A signal of 1 microvolt into 50 ohms is the same as
a power of 2 x 10^-14 W !!! (P = V^2/R)


Hope this helps.

vy 73

Andy, M1EBV