Hi all.

No doubt I could get this reading lots and lots of text books. It might
however be interesting to air a discussion on it..

Have been reading the free space pathloss formula from the ARRL antenna
handbook;

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

I have known about a "fixed K" loss in an antenna system for ages. It
even made mention I think in this NG recently when talking about a
passive repeater system. From my own exposure to path modelling
(EDX/Pathloss etc) I noted a very high dB loss per distance rate in the
first (say) 100 wavelengths when looking at graphs of same. (I wasnt
doing the actual job, just providing data to the engineer to compare
measured with predicted. Fascinating stuff!)

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant
possibly extract all of the energy from the wave as it goes past. I
would appreciate any input on this.

My initial forays (as a young ham) into LOS paths went through the
isotropic/point source radiators and looking at the surface area of the
covered "sphere" containing all of the radiated power idea. Then the RX
antenna "aperture" area was used to calculate the actual received power.
Needless to say it never met the actual measured values!

Cheers Bob VK2YQA (in W5)

"Bob Bob" wrote in message
...
Hi all.

No doubt I could get this reading lots and lots of text books. It
might
however be interesting to air a discussion on it..

Have been reading the free space pathloss formula from the ARRL
antenna
handbook;

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

I have known about a "fixed K" loss in an antenna system for ages.
It
even made mention I think in this NG recently when talking about a
passive repeater system. From my own exposure to path modelling
(EDX/Pathloss etc) I noted a very high dB loss per distance rate in
the
first (say) 100 wavelengths when looking at graphs of same. (I wasnt
doing the actual job, just providing data to the engineer to compare
measured with predicted. Fascinating stuff!)

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it
maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant
possibly extract all of the energy from the wave as it goes past. I
would appreciate any input on this.

My initial forays (as a young ham) into LOS paths went through the
isotropic/point source radiators and looking at the surface area of
the
covered "sphere" containing all of the radiated power idea. Then the
RX
antenna "aperture" area was used to calculate the actual received
power.
Needless to say it never met the actual measured values!

Cheers Bob VK2YQA (in W5)

For interest the K is 36 if the frequency is in MHz and the distance
in miles, or 32.5 if the distance is in Km.

Used it for years and it's never far out.


--
Woody

harrogate2 at ntlworld dot com

Hi,

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant
possibly extract all of the energy from the wave as it goes past. I
would appreciate any input on this.

Don't know that particular expression but plug in very low
values for 'f' and 'D' and you come up with something like
Loss(dB) = K(dB) -(~100dB) since the log of numbers less than 1 is
negative.

The formula then is simply of the form -

y = 20log(f) + 20log(D) + C

So when 'f' and 'D' are both equal to 1, the path loss = 96dB
and, when much less than that, it approaches zero.


Cheers - Joe

Also, for frequencies above 5 to 6 GHz the effects of atmospheric
absorption become much higher and non linear with frequency. It has to
do with molecular resonance and atomic cross sections of the molecules.

Bob Bob wrote:

Hi all.

No doubt I could get this reading lots and lots of text books. It might
however be interesting to air a discussion on it..

Have been reading the free space pathloss formula from the ARRL antenna
handbook;

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

I have known about a "fixed K" loss in an antenna system for ages. It
even made mention I think in this NG recently when talking about a
passive repeater system. From my own exposure to path modelling
(EDX/Pathloss etc) I noted a very high dB loss per distance rate in the
first (say) 100 wavelengths when looking at graphs of same. (I wasnt
doing the actual job, just providing data to the engineer to compare
measured with predicted. Fascinating stuff!)

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant
possibly extract all of the energy from the wave as it goes past. I
would appreciate any input on this.

My initial forays (as a young ham) into LOS paths went through the
isotropic/point source radiators and looking at the surface area of the
covered "sphere" containing all of the radiated power idea. Then the RX
antenna "aperture" area was used to calculate the actual received power.
Needless to say it never met the actual measured values!

Cheers Bob VK2YQA (in W5)

"Bob Bob" wrote in message
...
Hi all.

No doubt I could get this reading lots and lots of text books. It might
however be interesting to air a discussion on it..

Have been reading the free space pathloss formula from the ARRL antenna
handbook;

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

I have known about a "fixed K" loss in an antenna system for ages. It even
made mention I think in this NG recently when talking about a passive
repeater system. From my own exposure to path modelling (EDX/Pathloss etc)
I noted a very high dB loss per distance rate in the first (say) 100
wavelengths when looking at graphs of same. (I wasnt doing the actual job,
just providing data to the engineer to compare measured with predicted.
Fascinating stuff!)

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant possibly
extract all of the energy from the wave as it goes past. I would
appreciate any input on this.

My initial forays (as a young ham) into LOS paths went through the
isotropic/point source radiators and looking at the surface area of the
covered "sphere" containing all of the radiated power idea. Then the RX
antenna "aperture" area was used to calculate the actual received power.
Needless to say it never met the actual measured values!

Cheers Bob VK2YQA (in W5)

Take the ratio of the destination antenna aperture (0.13*lambda^2 for a half
wave dipole), to the surface area of a sphere, of a radius equal to the
distance between the source and destination antennas (i.e. Pr/Pt). The
assumption is that the source is isotropic. Massaging the above expression
will produce a constant and the sum of two logs.

Regards,

Frank

Bob Bob,
The formula for pathloss of a signal is

= 10*log10( [lambda/(4*pi*D)]^2 ) {See any myriad of texts}

where log10 is log base 10, lambda is the wavelength, D is the LOS
distance from one station to another, and of course pi is ~ 3.14. For
this formula to work, lambda and D need to be in the same units. Using
logarithm theorems we can rewrite it as

= 20*log10( lambda/(4*pi*D) )

Using more log theorems, this equals

= 20*log10(lambda) - 20*log10(4*pi*D)
= 20*log10(lambda) - [ 20*log10(4) + 20*log10(pi) + 20*log10(D) ]
= 20*log10(lambda) - 20*log10(4) - 20*log10(pi) - 20*log10(D)


lambda = c/f where c is the speed of light and f is the frequency (in
the same unit system)

so

= 20*log10(c/f) - 20*log10(4) - 20*log10(pi) - 20*log10(D)
= 20*log10(c)- 20*log(f) - 20*log10(4) - 20*log10(pi) - 20*log10(D)

Now... we need to make sure we are using the same unit convention. If
D is to be in MILES, c needs to be in miles per second.

c ~= 3*10^8 m/s
and in miles per second
c ~= 186411.4 mi/s

= [20*log10(c) - 20*log10(4) - 20*log10(pi)] - 20*log10(D) - 20*log(f)
plug into your handy-dandy calculator
= (83.4 dB) - 20*log10(D) - 20*log(f)

However, f can be refered to in GHz rather than Hertz so

= (83.4 dB) - 20*log10(D) - 20*log(F_in_GHz*10^9)
= (83.4 dB) - 20*log10(D) - [20*log(F_in_GHz) + 20*log10(10^9)]
= (83.4 dB) - 20*log10(D) - 20*log(F_in_GHz) - 20*log10(10^9)
= (83.4 dB) - 20*log10(D) - 20*log(F_in_GHz) - 180

and...

= (-96.6 dB) - 20*log10(D) - 20*log(F_in_GHz)

Note that to be mathematically consistent, when you "add" loss to a
system, you subtract a number therefore your value is actually
negative. It's mostly semantics. When you say "loss" you mean
subract.

Hope this helps!

Dario


Bob Bob wrote:
Hi all.

No doubt I could get this reading lots and lots of text books. It might
however be interesting to air a discussion on it..

Have been reading the free space pathloss formula from the ARRL antenna
handbook;

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

I have known about a "fixed K" loss in an antenna system for ages. It
even made mention I think in this NG recently when talking about a
passive repeater system. From my own exposure to path modelling
(EDX/Pathloss etc) I noted a very high dB loss per distance rate in the
first (say) 100 wavelengths when looking at graphs of same. (I wasnt
doing the actual job, just providing data to the engineer to compare
measured with predicted. Fascinating stuff!)

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant
possibly extract all of the energy from the wave as it goes past. I
would appreciate any input on this.

My initial forays (as a young ham) into LOS paths went through the
isotropic/point source radiators and looking at the surface area of the
covered "sphere" containing all of the radiated power idea. Then the RX
antenna "aperture" area was used to calculate the actual received power.
Needless to say it never met the actual measured values!

Cheers Bob VK2YQA (in W5)

Thank you all for your feedback/comment. So much for an easy answer!

grin)

Bob Bob wrote: