Hello,

Let's say I have a transmitter with output power T dBm using an
antenna with A dBi. Then we have a receiver, with gain B on the
antenna. How does this affect the received signal strength? I guess
the field strength at a point (in the transmitter antenna lobe) would
be something like T+A-PL where PL is the path loss at the point.

But what does the receiver get when it uses an antenna? Is there a
notion of reciever gain of an antenna? In my mind I can't see that
anything else than antenna area would be relevant. I mean, a reciever
antenna shouldn't be able to suck in radio waves from the sides...
Can anyone sort this out?

Regards,
Daniel

In article
,
Daniel wrote:

But what does the receiver get when it uses an antenna? Is there a
notion of reciever gain of an antenna?

Daniel-

I think the easiest way to see this is by looking at the transmitted
power having a value at the receive antenna, of so many Watts per square
meter power density. The receive antenna presents an "effective" area
of so many square meters. Its value depends on things like frequency,
antenna size, gain and direction.

Multiply transmitter power density at the receive antenna, by the
receive antenna's effective area to get received power.

But how do you determine your antenna's effective area? That isn't so
easy! I understand a half wave dipole has an effective area of about
one half wavelength times one quarter wavelength in the direction of the
antenna's major lobes. On 40 Meters, that would be 200 square meters.

Once all the factors are known, you can reduce the calculations to
Decibels.

Fred
K4DII

Daniel wrote in news:0ba6e6da-f856-4e7b-a272-
:

Hello,

Let's say I have a transmitter with output power T dBm using an
antenna with A dBi. Then we have a receiver, with gain B on the
antenna. How does this affect the received signal strength? I guess
the field strength at a point (in the transmitter antenna lobe) would
be something like T+A-PL where PL is the path loss at the point.


Daniel,

Look up the Friis Transmission equation on Google.

Here is a little application for solving the Friis Transmission equation:
http://www.vk1od.net/software/fsc/ .

Another concept that is important is reciprocity, but understand that
limit of the scope of the concept (eg it doesn't capture differences in
impedance mismatch from tx to rx).

Owen

But what does the receiver get when it uses an antenna? Is there a
notion of reciever gain of an antenna? In my mind I can't see that
anything else than antenna area would be relevant. I mean, a reciever
antenna shouldn't be able to suck in radio waves from the sides...
Can anyone sort this out?

Regards,
Daniel

Daniel wrote:
Hello,

Let's say I have a transmitter with output power T dBm using an
antenna with A dBi. Then we have a receiver, with gain B on the
antenna. How does this affect the received signal strength? I guess
the field strength at a point (in the transmitter antenna lobe) would
be something like T+A-PL where PL is the path loss at the point.

Yes, that's correct. But remember that what counts is the
signal-to-noise ratio at the receiver. At VHF and above, the majority of
the noise comes from the receiver, but at HF and below, from the
atmosphere. So at HF and below the receive antenna gain magnifies both
the signal and noise in the same proportion and doesn't help the S/N
ratio. The transmitter antenna gain, however, does, because it magnifies
only the signal and not the noise.

But what does the receiver get when it uses an antenna? Is there a
notion of reciever gain of an antenna?

Yes. It's exactly the same as the gain the same antenna has when used
for transmitting. This principle is called "reciprocity".

In my mind I can't see that
anything else than antenna area would be relevant. I mean, a reciever
antenna shouldn't be able to suck in radio waves from the sides...
Can anyone sort this out?

Sure, it's been sorted out for more than a hundred years. Antennas do
indeed suck radio waves in from the sides, unless you're talking about
antennas with dimensions of many wavelengths on each side. Waves aren't
little tiny things like BBs, but big things that spread out over a large
amount of space and interact with antennas in complex ways. Passing
waves induce currents in an antenna which then creates waves of its own
that interact with the original field. Ordinary intuition doesn't work
well for thinking about this, unless you took physics in high school and
got to play with a ripple tank. Lacking that, spend some time at a
harbor and see how water waves interact with pilings and docks. Of
course, there's always the option of reading some books on antenna theory.

Don't get hung up on an antenna's physical area, unless you're dealing
with antennas that are many wavelengths across, like parabolic
reflectors and horn antennas. An infinitesimally short, lossless dipole
has nearly the same aperture ("effective area" or "capture area") as a
half wavelength dipole. A fair size loop is about the same as a dipole.
Making a dipole's wire diameter several times larger makes no
significant difference to its aperture. Aperture or "capture area" is
simply an alternate way of stating gain -- if you know one you know the
other.

Roy Lewallen, W7EL

On 6 Mar, 19:44, Fred McKenzie wrote:
I think the easiest way to see this is by looking at the transmitted
power having a value at the receive antenna, of so many Watts per square
meter power density. *The receive antenna presents an "effective" area
of so many square meters. Its value depends on things like frequency,
antenna size, gain and direction.

By using the word "effective", do you suggest that the word area
should not be interpreted in it's literal sense of the size-property
of a surface in physical space? So it is not simply a matter of taking
the product of a height and a width (of a e.g. a patch antenna)? I
understand that the direction is a factor to deal with; that's
obvious. I also understand there might be losses if the frequency is
different than what the antenna designer had in mind. But I don't
understand how the gain makes any difference when receiving. Here I
use gain in the sense of transmission gain (radiation intensity in the
lobe compared to an ideal isotropic antenna).

Multiply transmitter power density at the receive antenna, by the
receive antenna's effective area to get received power.

I would totally agree unless I was confused about the notion of
"effective area". I guess my mental picture of a real physical surface
exposed to incident radiation is naïve and stops me from
understanding, but I simply don't know what image to replace it with.

But how do you determine your antenna's effective area? *That isn't so
easy! *I understand a half wave dipole has an effective area of about
one half wavelength times one quarter wavelength in the direction of the
antenna's major lobes. *

I assume by definition the height of a half wave dipole antenna is one
half wavelength. But what about the width? Why one quarter wavelength?

On 40 Meters, that would be 200 square meters.

This I don't understand. Again, it seems to me to be something that
would only be meaningful when transmitting, not when receiving.

Once all the factors are known, you can reduce the calculations to
Decibels.

Yeah, I look forward to that day!! Thanks for your reply,

Regards,
Daniel

On 6 Mar, 20:31, Roy Lewallen wrote:
But what does the receiver get when it uses an antenna? Is there a
notion of reciever gain of an antenna?

Yes. It's exactly the same as the gain the same antenna has when used
for transmitting. This principle is called "reciprocity".

Ok, at least that makes the definition of receiver gain clear!

In my mind I can't see that
anything else than antenna area would be relevant. I mean, a reciever
antenna shouldn't be able to suck in radio waves from the sides...
Can anyone sort this out?

Sure, it's been sorted out for more than a hundred years. Antennas do
indeed suck radio waves in from the sides, unless you're talking about
antennas with dimensions of many wavelengths on each side. Waves aren't
little tiny things like BBs, but big things that spread out over a large
amount of space and interact with antennas in complex ways. Passing
waves induce currents in an antenna which then creates waves of its own
that interact with the original field. Ordinary intuition doesn't work
well for thinking about this, unless you took physics in high school and
got to play with a ripple tank. Lacking that, spend some time at a
harbor and see how water waves interact with pilings and docks. Of
course, there's always the option of reading some books on antenna theory.

I *have* done one experiment as a kid: I placed a dipole permanent
magnet under a piece of paper and poured iron filings on the paper to
see the pattern created when the filings arranged itself in the
magnetic field. So I kinda have some understanding how field phenomena
can behave. I just didn't connect it to antennas. Is it more correct
to say that a field is established between the two antennas rather
than saying that something is transmitted in a direction from one end
to the other? (Of course on another level, *information* is
transmitted in a direction, but I am referring to the level of the
electromagnetic field).


Don't get hung up on an antenna's physical area, unless you're dealing
with antennas that are many wavelengths across, like parabolic
reflectors and horn antennas. An infinitesimally short, lossless dipole
has nearly the same aperture ("effective area" or "capture area") as a
half wavelength dipole.

Ok, good to know!

A fair size loop is about the same as a dipole.
Making a dipole's wire diameter several times larger makes no
significant difference to its aperture. Aperture or "capture area" is
simply an alternate way of stating gain -- if you know one you know the
other.

As far as I can understand, the notion of "effective area" is quite
artificial and only meaningful if one wants to fit a complex wave
phenomena into a form where one can think of it in the naive way I
have expressed in my earlier posts... Anyways, thanks for a great
reply!

Regards,
Daniel

Daniel wrote:
. . .
I *have* done one experiment as a kid: I placed a dipole permanent
magnet under a piece of paper and poured iron filings on the paper to
see the pattern created when the filings arranged itself in the
magnetic field. So I kinda have some understanding how field phenomena
can behave. I just didn't connect it to antennas.

Your magnet experiment showed you one thing about a *static* field.
Time-varying fields behave much differently, so it's a mistake to think
that you now understand radio waves.

Is it more correct
to say that a field is established between the two antennas rather
than saying that something is transmitted in a direction from one end
to the other? (Of course on another level, *information* is
transmitted in a direction, but I am referring to the level of the
electromagnetic field).

No, it's not. One antenna creates a field whether or not the other
antenna is there. The field propagates at the speed of light away from
the transmit antenna.

. . .

As far as I can understand, the notion of "effective area" is quite
artificial and only meaningful if one wants to fit a complex wave
phenomena into a form where one can think of it in the naive way I
have expressed in my earlier posts... Anyways, thanks for a great
reply!

Yes and no. "Effective area" or "effective aperture" describes the cross
section of the impinging field which contains the amount of energy the
antenna captures and delivers to its load. It's most useful in the
analysis of antennas that are very large in terms of wavelength, such as
parabolic reflectors and horn antennas. For those, the effective
aperture is on the order of, and closely related to, the physical area.
When dealing with simpler and smaller antennas, though, there isn't any
direct relationship between effective aperture and physical size, which
leads to a lot of misunderstanding.

Roy Lewallen, W7EL

Owen Duffy wrote:
Look up the Friis Transmission equation on Google.

Here is a little application for solving the Friis Transmission equation:
http://www.vk1od.net/software/fsc/ .


Andy asks:

The Friis equation at the above website states that the distance
exponent may be a number different than "2".

Can anyone here explain why a number other than "2" could be
used, and under what conditions ??

Thanks,

Andy W4OAH

AndyS wrote:

Andy asks:

The Friis equation at the above website states that the distance
exponent may be a number different than "2".

Can anyone here explain why a number other than "2" could be
used, and under what conditions ??

Thanks,

Andy W4OAH

The exponent of 2 assumes no dissipative path loss. That is, the
reduction in field strength is due solely to the power being spread out
over an area which increases with distance from the source, and not to
any reduction in the total power at any distance. A larger exponent
would be appropriate when the path is lossy, for example when ground
wave attenuation is present. A path through air could also get lossy at
microwave frequencies due to water in the air in liquid or gas form, and
at extremely high frequencies due to absorption by various atmospheric
gases.

Roy Lewallen, W7EL

AndyS wrote in
:



Owen Duffy wrote:
Look up the Friis Transmission equation on Google.

Here is a little application for solving the Friis Transmission
equation: http://www.vk1od.net/software/fsc/ .


Andy asks:

The Friis equation at the above website states that the distance
exponent may be a number different than "2".

I don't think I say that, Friis uses an exponent of 2, but the calculator
can also solve the equation with an exponent other than 2.

Can anyone here explain why a number other than "2" could be
used, and under what conditions ??

An example is that the FCC stipulates exponents of other than 2 for some
field strength interpolation / extrapolation, up to 4 IIRC, and this is
for application in real situations at closer distances than radiation far
field conditions.

The use of the exponent is to account for some other effects. In some
cases the value of the exponent has been challenged. If I understand the
case correctly, the ARRL has challenged some instances used for BPL and
the court has required the FCC to scientifically substantiate its
exponent, or use another which it can scientifically substantiate.

The ability to specify the exponent is principally provided for peforming
those interpolation / extrapolations.

Owen


Thanks,

Andy W4OAH