Hello.

I took some measurements of an AM broadcast from multiple points up to
about half a mile from the transmitter. The transmitter was supposed
to be operating at 20W. I don't know what kind of pattern the AM
antenna had, but can I just use the field strength readings that I got
to predict how far the signal can go out? I only was able to measure 3
points, but I want to know if that is sufficient to make an estimate
under the same conditions.

If so, how do I go about starting this? Are there any formulas? Also,
on average AM receivers, what would be considered a minimum field
strength value that would yield an audible sound?

I also read that frequencxies less than 2MHz rely on surface waves. Is
there a site that I can visit that will give me an idea on how to take
this into account given that there are different ground
characteristics? Thanks!

"MRW"
I took some measurements of an AM broadcast from multiple points up to
about half a mile from the transmitter. The transmitter was supposed
to be operating at 20W. I don't know what kind of pattern the AM
antenna had, but can I just use the field strength readings that I got
to predict how far the signal can go out? I only was able to measure 3
points, but I want to know if that is sufficient to make an estimate
under the same conditions.

An estimate is possible, but assumptions will need to be made. If you can
post the readings you got, the path lengths, and the general location for
the transmit site I can give you some idea about that.

Of course if you can post the call letters of the station then the FCC data
for that station will show they should be.

If so, how do I go about starting this? Are there any formulas? Also,
on average AM receivers, what would be considered a minimum field
strength value that would yield an audible sound?

I also read that frequencxies less than 2MHz rely on surface waves. Is
there a site that I can visit that will give me an idea on how to take
this into account given that there are different ground
characteristics? Thanks!

The FCC publishes charts that may be used for this purpose. As for minimum
usable daytime field for a typical, cheap indoor receiver in an urban,
residential area, a value of 1 or 2 mV/m typically is needed.

RF http://rfry.org

On Feb 22, 10:58 am, "Richard Fry" wrote:
An estimate is possible, but assumptions will need to be made. If you can
post the readings you got, the path lengths, and the general location for
the transmit site I can give you some idea about that.

Of course if you can post the call letters of the station then the FCC data
for that station will show they should be.

The readings that I got were 3.1 mV/m at 15 meters away. 1 mV/m at
0.32 km away. Finally, 0.71 mV/m at 0.80 km away. We traveled a
straight path that was parallel to the direction of the antenna. It's
not an FCC station, but my professor got permission to test it briefly
at 1.79MHz. I don't know the exact area that we were located, but it
is somewhere east of Magdalena, NM.

The FCC publishes charts that may be used for this purpose. As for minimum
usable daytime field for a typical, cheap indoor receiver in an urban,
residential area, a value of 1 or 2 mV/m typically is needed.

RF http://rfry.org

Thanks! I was curious about AM stations at one time and looked up some
FCC data. I noticed that some stations have lower power output during
night time compared to daytime. For example, one station had 1 kW
during daytime and decreased it to 300W (I think) during night time.
Does that indicate that night time propagation is better?

"MRW" wrote in
oups.com:

On Feb 22, 10:58 am, "Richard Fry" wrote:
An estimate is possible, but assumptions will need to be made. If
you can post the readings you got, the path lengths, and the general
location for the transmit site I can give you some idea about that.

Of course if you can post the call letters of the station then the
FCC data for that station will show they should be.

The readings that I got were 3.1 mV/m at 15 meters away. 1 mV/m at
0.32 km away. Finally, 0.71 mV/m at 0.80 km away. We traveled a
straight path that was parallel to the direction of the antenna. It's
not an FCC station, but my professor got permission to test it briefly
at 1.79MHz. I don't know the exact area that we were located, but it
is somewhere east of Magdalena, NM.

The FCC publishes charts that may be used for this purpose. As for
minimum usable daytime field for a typical, cheap indoor receiver in
an urban, residential area, a value of 1 or 2 mV/m typically is
needed.

RF http://rfry.org

Thanks! I was curious about AM stations at one time and looked up some
FCC data. I noticed that some stations have lower power output during
night time compared to daytime. For example, one station had 1 kW
during daytime and decreased it to 300W (I think) during night time.
Does that indicate that night time propagation is better?

Night propagation is complicated by the existence of a strong sky-wave
reflection from the ionosphere's E and F layers. In the daytime, the D
layer absorbs emissions at these angles. Hence the need to reduce power
at night to avoid interference with more distant stations' ground wave
coverage.

--
Dave Oldridge+
ICQ 1800667

"MRW"
The readings that I got were 3.1 mV/m at 15 meters away. 1 mV/m at
0.32 km away. Finally, 0.71 mV/m at 0.80 km away. We traveled a
straight path that was parallel to the direction of the antenna. It's
not an FCC station, but my professor got permission to test it briefly
at 1.79MHz. I don't know the exact area that we were located, but it
is somewhere east of Magdalena, NM.
_____________

Your description resembles a "Part 15" type installation more than the
broadcast system of your first post. Here are some general statements about
how a compliant Part 15 AM system might perform.

Antenna engineering textbooks, NEC calculations, and thousands of field
strength measurements made in the broadcast industry over the last 75+ years
show that the maximum field strength in the horizontal plane that is
produced by a vertical, 1/4-wave monopole with 1 kW of applied power over an
almost perfectly conducting, flat ground plane is about 300 millivolts/meter
(mV/m) at a radius of 1 km (0.62 miles).

A legal Part 15 AM tx that was 100% efficient, and used with the above
antenna system would generate a field at 1 km that would be reduced by the
square root of the power difference, or to a field of 3 mV/m in this case.

But a 3-meter, ground-mounted Part 15 antenna system is only about 1% as
efficient as a 1/4-wave broadcast radiator system. So instead of radiating
100 mW, the Part 15 antenna system radiates around 1 mW. That leads to a
further reduction in the field at 1 km by the square root of 100, bringing
it to about 300 microvolts/meter (µV/m).

Note that all of these fields assume an almost perfectly-conducting ground
over the propagation path. Typical ground conditions are far from perfect,
so the fields at 1 km would not be even this high.

By broadcast standards, a 300 µV/m field is very marginal in providing a
usable signal to a typical, cheap AM receiver located inside a home. And
every doubling of the distance decreases the received field by more than 50%
(including ground losses).

From this information it can be seen that claims of "legal" Part 15 AM
coverage extending for a radius of 2, 3 and 4 miles cannot be realistic,
unless the system is not meeting Part 15 limits.

RF

On Feb 22, 1:07 pm, "Richard Fry" wrote:
Your description resembles a "Part 15" type installation more than the
broadcast system of your first post. Here are some general statements about
how a compliant Part 15 AM system might perform.

Don't know much about "Part 15" is that something that I can find in
the CFR 47? http://www.access.gpo.gov/nara/cfr/c...le-search.html


Antenna engineering textbooks, NEC calculations, and thousands of field
strength measurements made in the broadcast industry over the last 75+ years
show that the maximum field strength in the horizontal plane that is
produced by a vertical, 1/4-wave monopole with 1 kW of applied power over an
almost perfectly conducting, flat ground plane is about 300 millivolts/meter
(mV/m) at a radius of 1 km (0.62 miles).

A legal Part 15 AM tx that was 100% efficient, and used with the above
antenna system would generate a field at 1 km that would be reduced by the
square root of the power difference, or to a field of 3 mV/m in this case.

How did you get 3 mV/m? I tried 300 mV/m / (sqrt(1kW - 20W)).


But a 3-meter, ground-mounted Part 15 antenna system is only about 1% as
efficient as a 1/4-wave broadcast radiator system. So instead of radiating
100 mW, the Part 15 antenna system radiates around 1 mW. That leads to a
further reduction in the field at 1 km by the square root of 100, bringing
it to about 300 microvolts/meter (µV/m).

Note that all of these fields assume an almost perfectly-conducting ground
over the propagation path. Typical ground conditions are far from perfect,
so the fields at 1 km would not be even this high.

Are there any existing resources that talk about how much attenuation
can each ground type contribute?

By broadcast standards, a 300 µV/m field is very marginal in providing a
usable signal to a typical, cheap AM receiver located inside a home. And
every doubling of the distance decreases the received field by more than 50%
(including ground losses).

From this information it can be seen that claims of "legal" Part 15 AM
coverage extending for a radius of 2, 3 and 4 miles cannot be realistic,
unless the system is not meeting Part 15 limits.

RF

Thanks, Richard!

I don't have the experience background, yet. But in your opinion, is a
field strength of 1mV/m the absolute minimum for decent AM reception?
Also, is the fact that AM more prone to interference the reason why
its field strength requirements are higher than FM?

"MRW"
How did you get 3 mV/m? I tried 300 mV/m / (sqrt(1kW - 20W)).

Other things equal, the far field varies by the square root of the power
change.
If 1 kW into a given antenna system produces 300 mV/m at 1 km, then 0.1 W
into that same system produces SQRT(0.1W / 1000W) x 300 mV/m at 1 km,
which is 3 mV/m.

Are there any existing resources that talk about how much attenuation
can each ground type contribute?

The FCC charts probably are the best resource. Surface wave propagation
losses are dependent on frequency as well as ground conditions.

I don't have the experience background, yet. But in your opinion, is a
field strength of 1mV/m the absolute minimum for decent AM reception?
Also, is the fact that AM more prone to interference the reason why its
field strength requirements are higher than FM?

A good AM receiver in the absence of natural/man-made interference can
produce useful (but not highly "competitive") performance in fields of less
than 100 µV/m.

At a given value of received field strength, an FM broadcast is received
with less noise than AM because of the lower noise level at VHF frequencies,
and due to the ability of an FM receiver to the reject the a-m noise that is
present.

RF

On Feb 22, 12:10 pm, "MRW" wrote:
On Feb 22, 1:07 pm, "Richard Fry" wrote:

Your description resembles a "Part 15" type installation more than the
broadcast system of your first post. Here are some general statements about
how a compliant Part 15 AM system might perform.

Don't know much about "Part 15" is that something that I can find in
the CFR 47?http://www.access.gpo.gov/nara/cfr/c...le-search.html

Antenna engineering textbooks, NEC calculations, and thousands of field
strength measurements made in the broadcast industry over the last 75+ years
show that the maximum field strength in the horizontal plane that is
produced by a vertical, 1/4-wave monopole with 1 kW of applied power over an
almost perfectly conducting, flat ground plane is about 300 millivolts/meter
(mV/m) at a radius of 1 km (0.62 miles).

A legal Part 15 AM tx that was 100% efficient, and used with the above
antenna system would generate a field at 1 km that would be reduced by the
square root of the power difference, or to a field of 3 mV/m in this case.

How did you get 3 mV/m? I tried 300 mV/m / (sqrt(1kW - 20W)).

But a 3-meter, ground-mounted Part 15 antenna system is only about 1% as
efficient as a 1/4-wave broadcast radiator system. So instead of radiating
100 mW, the Part 15 antenna system radiates around 1 mW. That leads to a
further reduction in the field at 1 km by the square root of 100, bringing
it to about 300 microvolts/meter (µV/m).

Note that all of these fields assume an almost perfectly-conducting ground
over the propagation path. Typical ground conditions are far from perfect,
so the fields at 1 km would not be even this high.

Are there any existing resources that talk about how much attenuation
can each ground type contribute?

By broadcast standards, a 300 µV/m field is very marginal in providing a
usable signal to a typical, cheap AM receiver located inside a home. And
every doubling of the distance decreases the received field by more than 50%
(including ground losses).

From this information it can be seen that claims of "legal" Part 15 AM
coverage extending for a radius of 2, 3 and 4 miles cannot be realistic,
unless the system is not meeting Part 15 limits.

RF

Thanks, Richard!

I don't have the experience background, yet. But in your opinion, is a
field strength of 1mV/m the absolute minimum for decent AM reception?
Also, is the fact that AM more prone to interference the reason why
its field strength requirements are higher than FM?


With regard just to your last question, atmospheric (mainly from
lightening around the world), galactic, and typical man-made noise are
all much higher--40dB or more, usually--at 1MHz than at 100MHz. You
need signals above the noise for practical communications.
(Bandwidths also enter into the picture, but I believe the noise is
the key issue.)

Cheers,
Tom

On Feb 22, 3:43 pm, "Richard Fry" wrote:
Other things equal, the far field varies by the square root of the power
change.
If 1 kW into a given antenna system produces 300 mV/m at 1 km, then 0.1 W
into that same system produces SQRT(0.1W / 1000W) x 300 mV/m at 1 km,
which is 3 mV/m.

So in terms of the 20W test, this value at 1km would be 42.4 mV/m? If
so, how would I translate this value to another distance (e.g. 0.4 km
away)? I'm assuming that it would be greater than 42.4mV/m.

A good AM receiver in the absence of natural/man-made interference can
produce useful (but not highly "competitive") performance in fields of less
than 100 µV/m.

Would a typically AM receiver be able to pick up signals 50 uV/m in
field strength? I'm assuming that it wouldn't since the components in
the AM receiver have noise levels probably close to this, huh?

On Feb 22, 3:52 pm, "K7ITM" wrote:
With regard just to your last question, atmospheric (mainly from
lightening around the world), galactic, and typical man-made noise are
all much higher--40dB or more, usually--at 1MHz than at 100MHz. You
need signals above the noise for practical communications.
(Bandwidths also enter into the picture, but I believe the noise is
the key issue.)

Cheers,
Tom

Thanks, Tom! I did not know about this noise factor in this lower
frequency range. In a way, if I were operating at 100MHz using AM,
then I should be able to work with lower field strength values?