dBW in S-unit and power strenght
 

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) ?

Thanks in advance

Thierry, ON4SKY
http://www.astrosurf.com/lombry

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

Reverse calculation:

dbW / 10 10^x = P

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

Regards,

Ralf

--
Vy 73 es 55 de Ralf, DL2MRB
E-Mail:
Board: www.hamradioboard.de

"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

--
Vy 73 es 55 de Ralf, DL2MRB
E-Mail:
Board: www.hamradioboard.de

On Wed, 15 Sep 2004 12:56:16 +0200, "Thierry" to answer direct see
http://www.astrosurf.com/lombry/post.htm wrote:

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) ?

Hi Thierry,

S-9 is the reference point for a 50µV signal into the receiver's 50
Ohm input. This is -73dBM (level below one milliwatt). Translating
to levels below one Watt is performed by subtracting another 30 (the
milli) dB - hence -103dBW.

Power is a curious thing. Transmitter efficiency as measured by the
target population that receives the signal is pitiful. If every
person on the planet had a radio that picked up your 100W signal with
S-9 level, you would be pumping 99.9% of your 100W into the void
without anyone missing it anywhere. Efficiency 0.1%

Given that you are NOT heard by everyone (they don't care to listen),
then transmitter efficiency plunges at least 7 or 8 more orders of
magnitude. Efficiency 0.00000001%

73's
Richard Clark, KB7QHC

"Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm
wrote:

Does not exist ! But well dBW = 10 Log P, this is not the same excepting
reading from right to left, Hi!

OK, hi...

Your explanation is not one. refer my link for a correct explanations.

Taka a look at:

http://hyperphysics.phy-astr.gsu.edu.../sound/db.html

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...

This calculations aren't depend on frequency or wavelength.

Regards,

Ralf

--
Vy 73 es 55 de Ralf, DL2MRB
E-Mail:
Board: www.hamradioboard.de

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

"Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm
wrote in message ...
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) ?
Thanks in advance
Thierry, ON4SKY
http://www.astrosurf.com/lombry



As others pointed out, your calculation is wrong and that .7 W is way too
large for received power. We receive powers in the -123 dBm (-153 dBW)
range easily and that's ok

However, it is not clear from your post just what you want to do/ask/learn.
What is your number too low for? What are you wanting?
--
Steve N, K,9;d, c. i My email has no u's.

I 've got te answer

I wonder first if dBW depended or not on the QRG below 30 MHz.
It doesn't (excepting that over 30 MHz there is less noise temperature and
often transverter in front of TX showing a 20 dB gain, and this figure is
thus till less).
In addition dBm and co are defined by IARU recommendations where I found all
equivalence dBm-S-point-dBW.

Thanks for your answers but I never imagine that at receive the true power
was so low. I add well a signal report and equivalent in dB or dBW but I
never really make this relation.

Thierry, ON4SKY


"Steve Nosko" wrote in message
...

"Thierry" to answer direct see http://www.astrosurf.com/lombry/post.htm
wrote in message ...
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) ?
Thanks in advance
Thierry, ON4SKY
http://www.astrosurf.com/lombry



As others pointed out, your calculation is wrong and that .7 W is way too
large for received power. We receive powers in the -123 dBm (-153 dBW)
range easily and that's ok

However, it is not clear from your post just what you want to
do/ask/learn.
What is your number too low for? What are you wanting?
--
Steve N, K,9;d, c. i My email has no u's.

"Richard Clark" wrote in message
...
On Wed, 15 Sep 2004 12:56:16 +0200, "Thierry" to answer direct see
http://www.astrosurf.com/lombry/post.htm wrote:

...

Hi Thierry,

S-9 is the reference point for a 50µV signal into the receiver's 50
Ohm input. This is -73dBM (level below one milliwatt). Translating
to levels below one Watt is performed by subtracting another 30 (the
milli) dB - hence -103dBW.

Power is a curious thing. Transmitter efficiency as measured by the
target population that receives the signal is pitiful. If every
person on the planet had a radio that picked up your 100W signal with
S-9 level, you would be pumping 99.9% of your 100W into the void
without anyone missing it anywhere. Efficiency 0.1%

Given that you are NOT heard by everyone (they don't care to listen),
then transmitter efficiency plunges at least 7 or 8 more orders of
magnitude. Efficiency 0.00000001%

You really put the finger on something that I ignored during 25 years of ham
radio, hi !...
I know the correspondence between power expressed in dB or dBm, dBW but I
never see that under that angle...

Thanks
Thierry, ON4SKY

73's
Richard Clark, KB7QHC

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