Thierry, ON4SKY wrote:
"With -93 dBW for a S9+10 signal at receive, that`d mean that the
popwer`d be only P (W) = 10^ (dBW/10) = 0.7 watt?

I`m familiar with received carrier power expressed in dBm. -93 dBW is a
number, 30 dB less if expressed in dBm, or -123. This is a strong signal
for a narrow-band receiver.

Across 50 ohms, 1 watt is sq rt PR, or 7.07 volts.

93 dB is about 45,000 times, as a voltage ratio, so a received voltagw
of -93 dBw is about 0.00016 volts, or 0.16 millivolts, or 160
microvolts.

S9+10dB is a strong signal. So is 160 microvolts. I`ve read that each
"S-unit" is 6 dB. S-1 would be 18 dB less than S9+10 db, and that`s a
voltage ratio of 7.94. S-1 would be about 20 microvolts, and well above
the threshold of most receivers.

Best regards, Richard Harrison, KB5WZI