What's the easiest/fastest way to calculate received power out of a very
short non-resonant antenna? Frequencies are in the HF and VHF ranges, TX
powers in the kilowatt range, and distance from transmitting antenna a few
miles. Thanks for any help.

George K6GW

On 23 jun, 05:22, "George" wrote:
What's the easiest/fastest way to calculate received power out of a very
short non-resonant antenna? Frequencies are in the HF and VHF ranges, TX
powers in the kilowatt range, and distance from transmitting antenna a few
miles. Thanks for any help.

George K6GW

Hello George,

I think this question relates to a previous posting regarding power
extraction?

First step is to find path attenuation.

Second step is to convert the Fieldstrength to EMF out of your
antenna.

Third step, converting EMF into power.


You may use ITU propagation curves for frequency planning or use two-
ray propagation formula as a ballpark.

EMF out of your antenna (optimal orientation) will be about
0.5*Efield*length.

The capacitance of your whip will be about some pF. This limits the
available power (non resonant).

Example:

A broadcast transmitter (100 MHz) with ERP = 10kW produces about 30mV/
m at 3 miles and h=5 ft.

This would produce about 4.5mV out of your whip.

When we assume that the capacitance with respect to the ground plane
is about 5 pF (-j320 Ohm), you may extract about 32nW of AC power
(without resonance).

When the antenna has no descent "counterpoise" and / or is not
oriented optimal, the output will be significantly less. Terrain
features will also affect output power.

Hope this will help you a bit,

Wim
PA3DJS
www.tetech.nl

George, K6GW wrote:

"What is the esiest/fastest way to calculate received power out of a
very short non-resonant antenna?"

Field strengths are given in volts per meter. This is related to power
by e squared / r.
The dissipationless resistance of free-space (a ratio between voltage
and current) is 377 ohms. Attenuation due only to reduction of power per
square meter as the radio wave envelope expands in space is 6 dB every
time distance from the transmitter doubles. That means getting only 1/4
the previous power every time distance from the transmitter doubles.

Volts per meter decline linearly with distance from the transmitter. At
10x the distance, the signal strength or volts per meter is 1/10 the
previous value. 1/10 the voltage is 1/100 of the power.

We have a participant in this newsgroup, Art Unwin, who disdains book
contents, but the subject is not nearly exhausted and I don`t intend to
exhaust it. Instead, I`ll recommend one of the best books:
"Radio-Electronic Transmission Fundamentals" by B. Whitfield Griffith,
Jr. Its 2nd edition has just been released by Scitech Publishing, Inc.
Whatever it costs it is well worth the price.

Best regards, Richard Harrison, KB5WZI

I wrote;
'That means getting only 1/4 the previous power every time the distance
from the transmitter doubles."

That`s true, but it`s only part of the story. A receiving antenna can
only deliver 50% of the power it receives and that can happen only when
it is perfectly matched to its receiver. Otherwise, it reradiates more
than 50% of its received power.

There is another fly in the ointment. A former colleague of mine, Peter
N. Saveskie compiled an interesting book, "Radio Propagation Handbook".
On page 826 Pete writes:
"Most of the attenuation of "free-space" loss occurs close-in to the
adiating antenna. The loss between two isotropic antennae separated by
one wavelength is 22.0 dB, thereafter increasing by 6.0 dB for each
doubling of the inter-antenna distance.'

This sort of loss in the first wavelength is one problem in trying to
use antennas back to back as passive repeaters. Enormous receiver
sensitivity makes the passive repeater practical sometimes in spite of
the path loss.

Pete`s path loss formula has worked well for me.

Best regards, Richard Harrison, KB5WZI

On 23 Jun, 13:34, (Richard Harrison) wrote:
George, K6GW wrote:

"What is the esiest/fastest way to calculate received power out of a
very short non-resonant antenna?"

Field strengths are given in volts per meter. This is related to power
by e squared / r.
The dissipationless resistance of free-space (a ratio between voltage
and current) is 377 ohms. Attenuation due only to reduction of power per
square meter as the radio wave envelope expands in space is 6 dB every
time distance from the transmitter doubles. That means getting only 1/4
the previous power every time distance from the transmitter doubles.

Volts per meter decline linearly with distance from the transmitter. At
10x the distance, the signal strength or volts per meter is 1/10 the
previous value. 1/10 the voltage is 1/100 of the power.

We have a participant in this newsgroup, Art Unwin, who disdains book
contents,
snip
Not so.
It is just that parrots are unreliable when quoting from books
Art

Best regards, Richard Harrison, KB5WZI

On 23 jun, 22:34, (Richard Harrison) wrote:
George, K6GW wrote:

"What is the esiest/fastest way to calculate received power out of a
very short non-resonant antenna?"

Field strengths are given in volts per meter. This is related to power
by e squared / r.
The dissipationless resistance of free-space (a ratio between voltage
and current) is 377 ohms. Attenuation due only to reduction of power per
square meter as the radio wave envelope expands in space is 6 dB every
time distance from the transmitter doubles. That means getting only 1/4
the previous power every time distance from the transmitter doubles.

Volts per meter decline linearly with distance from the transmitter. At
10x the distance, the signal strength or volts per meter is 1/10 the
previous value. 1/10 the voltage is 1/100 of the power.

We have a participant in this newsgroup, Art Unwin, who disdains book
contents, but the subject is not nearly exhausted and I don`t intend to
exhaust it. Instead, I`ll recommend one of the best books:
"Radio-Electronic Transmission Fundamentals" by B. Whitfield Griffith,
Jr. Its 2nd edition has just been released by Scitech Publishing, Inc.
Whatever it costs it is well worth the price.

Best regards, Richard Harrison, KB5WZI

Hi Richard,

That E is proportional to 1/r is valid for free space propagation.
George's question relates to the frequency range where there is no
free space (obstacle free) propagation in most cases. For his case,
the most important obstacle is mother earth.

As in practical cases (distance about several miles, height of
receiving antenna about 5 ft), the earth reflected ray cancels most
part of the direct ray. Fast methods to get a result are the two-ray
formula and/or propagation curves for frequency planning. Add about
10dB loss for heavily populated areas (buildings).

Using one of the above methods, will result in E is proportional to 1/
r^3...to 1/r^4. This matches measurements I did myself. Therefore 1/r
gives a far from realistic result (too optimistic).

Best regards,

Wim
PA3DJS
www.tetech.nl

"Richard Harrison" wrote
George, K6GW wrote:

"What is the easiest/fastest way to calculate received power out
of a very short non-resonant antenna?"

.. The dissipationless resistance of free-space (a ratio between voltage
and current) is 377 ohms. Attenuation due only to reduction of power per
square meter as the radio wave envelope expands in space is 6 dB every
time distance from the transmitter doubles. That means getting only 1/4
the previous power every time distance from the transmitter doubles.
(etc)
__________

The answer to the way I read the original question does not involve knowing
what the field strength is, or how it varies with distance or other factors,
but knowing the power it will produce via the receive antenna system.

Here is a paste from a spreadsheet I set up to calculate the field needed to
produce a defined power level at the input terminals of a receiver,using the
equation

µV/m = 10^[(dBmW - G + 20*(log F) + L + 75)/20]

where

-66.89 dBmW = signal level at receiver input in dB with respect to 1 mW
-2.14 G = receiving antenna gain in dBd
0.1 L = line loss in decibels
98 F = frequency in MHz


Solution for the values shown at the left margin above:

322.65 µV/m field required.

The equation could be rewritten to solve for power, given a field strength.

RF http://rfry.org

"Wimpie" wrote
A broadcast transmitter (100 MHz) with ERP = 10kW produces
about 30mV/m at 3 miles and h=5 ft.
____________

Although you didn't define the height of the transmit antenna, just picking
a height of 984 feet to be on the high side, and referring to the FCC's
F(50,50) propagation curves shows that for 10 kW ERP a field of about 42
mV/m will be generated at a radius of 3 miles and an elevation of 9 meters
(29.5 feet) above the earth, assuming flat terrain.

Field strength at lower elevations for these conditions varies about in an
inverse relationship with height, so at 5 feet it would be 5/29.5, or 17% of
42 mV/m, which is about 7 mV/m.

This is quite a bit different than your number, so I was curious as to your
method.

RF

Correcting myself...

Field strength at lower elevations for these conditions IS REDUCED about
in a DIRECT relationship with height, ....

On 24 jun, 16:36, "Richard Fry" wrote:
"Wimpie" wrote A broadcast transmitter (100 MHz) with ERP = 10kW produces
about 30mV/m at 3 miles and h=5 ft.

____________

Although you didn't define the height of the transmit antenna, just picking
a height of 984 feet to be on the high side, and referring to the FCC's
F(50,50) propagation curves shows that for 10 kW ERP a field of about 42
mV/m will be generated at a radius of 3 miles and an elevation of 9 meters
(29.5 feet) above the earth, assuming flat terrain.

Field strength at lower elevations for these conditions varies about in an
inverse relationship with height, so at 5 feet it would be 5/29.5, or 17% of
42 mV/m, which is about 7 mV/m.

This is quite a bit different than your number, so I was curious as to your
method.

RF

Hi Richard,

I was a little bit lazy to look into the propagation curves and do the
correction for terrain roughness. I live in the middle part of the
Netherlands where the country is really flat. Therefore I just took
the two-ray formula (and checked that the result is below free space
propagation). Normally spoken when I do not have reliable propagation
curves at hand, I use the two ray formula and add about 10..20 dB to
the loss.

The calculation example was just to show how (the procedure) someone
can calculate the received power from a non-resonant whip (that was
the original question). When using the FCC 50% curves, the output
would be even less (as your example showed).


In my previous posting I said:
"Using one of the above methods, will result in E is proportional to 1/
r^3...to 1/r^4. This matches measurements I did myself. Therefore 1/r
gives a far from realistic result (too optimistic)."

That had to be be:
"Using one of the above methods, will result in Preceived is
proportional to 1/r^3...to 1/r^4. This matches measurements I did
myself. Therefore 1/r^2 gives a far from realistic result (too
optimistic)."

Best regards,

Wim
PA3DJS.
www.tetech.nl