"Ralf Ballis - DL2MRB" wrote in message
...
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.

One S unit referencing 6 dB.
Some receivers has meter in dB over 1 micro V.

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 ?

"W" mean level in Watt.

IMHO this power is much to low... What is wrong in this relations (or
in
my interpretation) ?

P log * 10 = dbW

Does not exist ! But well dBW = 10 Log P, this is not the same excepting
reading from right to left, Hi!
Your explanation is not one. refer my link for a correct explanations.
Anyway, how do you calculate the dBW on 20m with such a formula. Someone
calculated dBW on 10m, so we must be able to do the same at 20m...

But how ?

Thierry

Reverse calculation:

dbW / 10 10^x = P

-93 / 10 10^x = 5,011e-10 = 0,0000000005011 Watt

Regards,

Ralf

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