Thanks, Roy for your and everyone's participation. I think I will
bow out here also. Hope all this hasn't been a waste of space.
"Thinking" usually has some value.

Ron W4TQT

Roy Lewallen wrote:
Ron wrote:

. . .
By "focal point" I meant the center of the sphere where the rays
converge and where the antenna would be located.


I have to admit, I was looking at this a more of a problem of equal
signals arriving from all directions, rather than at the middle of some
sort of convergence. Of course, any rays reaching the center would
continue on through, Cecil's unique theories notwithstanding. I don't
have the spare time to contemplate what the end field distribution would
be like at the center of the antenna or its periphery.

When an antenna intercepts one watt from a field having a power density
of one watt per square meter, it's said to have an "effective aperture"
or "capture area" of one square meter. The higher the gain of an antenna
in some particular direction, the larger its effective aperture in that
direction. Consequently, a high gain antenna would "capture" more power
from a wave arriving in its favored direction than an isotropic antenna
would. It would, of course, capture less from other directions, but
assuming equal efficiency, both antennas would capture equal amounts
overall.



In the unusual field defined in my example, the algebraic sum of all
the rays collected by the antenna would be higher in the isotropic
antenna than a high gain antenna.


It's not obvious to me why that would be.

Think of the front to back ratio of the high gain antenna which would
result in very little output from the rays behind and on the sides of
the antenna.


That's true. But the output would be higher in reponse to the rays
arriving from the front. We call that "gain". Another way to express it
is that it intercepts a field from a larger area of the wave front.

Therefore, the isotropic would have a higher output which is
indicative of higher gain.


You're right that higher output means higher gain. I maintain that both
antennas have the same total gain, i.e., the same total interception of
power from all directions. This follows directly from the reciprocity
principle.

I do not understand what you mean by "capture equal amounts overall".
Energy which may strike the antenna but does not result in any output
power isn't "captured".


The field you're creating comes from something and goes somewhere. If
you subtract the total amount going from the total amount generated,
you'll get the amount dissipated in the load connected to the antenna.
That is the amount of energy "captured" or "intercepted" by the antenna.
And that's what I thought you were talking about all along.

The "capture area" isn't some physical region with boundaries -- it's
simply a way of expressing how much power is extracted from a field
having a given power density. In other words, it's just another way
of expressing antenna gain.



How about a dish antenna? Isn't the capture area proportional to the
physical area of the dish?


Indeed it is, in the front direction. But how about a dipole? The
capture area (or gain) broadside to an infinitesimal dipole is just
slightly less than that of a half wavelength dipole. And wire diameter
makes almost no difference.

Sorry, the theoretical construct is just a little too much like
Calvinball to hold my interest. I'll bow out now. Best luck in sorting
it out.

Roy Lewallen, W7EL

For your conceptual purposes, your question would be similar to this?:

In a deep focal point of parabolic dish two antenns are mounted...

Which of them it does pick up more energy?

An antenna with 180 degree beamwidht or an highly directional antenna
with 0,1 degree beamwidth (both pointed to dish, of course)?

(In focal point of dish there are convergent frontwaves also). (We
could think in a sperical dish, also).

73's

Miguel Ghezzi (LU 6ETJ)

Miguel Chezzi, LU6ETJ wrote:
"In a deep focal point of parabolic dish two antennas are mounted...

Which of them does pick up more energy?
An antenna with 180 degree beamwidth or a highly directional sntenna
with 0.1 degree beamwidth (both pointed to dish, of course)?"

I`ll risk being the fool. We sometimes test for illumination of a
reflector. We would not be concerned were it not advantageous to do so.

With 180-degree radiation, we fill the dish, using all its surface.
With 0.1-degree illumination, we might as well remove all but the
illuminated area. It would save dead load and wind loading.

My answer: The 180-degree radiation angle will receive a larger area of
the plane-wavefront and extract more watts from the wave with a given
number of watts per square area.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Miguel Chezzi, LU6ETJ wrote:
"In a deep focal point of parabolic dish two antennas are mounted...

Which of them does pick up more energy?
An antenna with 180 degree beamwidth or a highly directional sntenna
with 0.1 degree beamwidth (both pointed to dish, of course)?"

I`ll risk being the fool. We sometimes test for illumination of a
reflector. We would not be concerned were it not advantageous to do so.

With 180-degree radiation, we fill the dish, using all its surface.
With 0.1-degree illumination, we might as well remove all but the
illuminated area. It would save dead load and wind loading.

My answer: The 180-degree radiation angle will receive a larger area of
the plane-wavefront and extract more watts from the wave with a given
number of watts per square area.

Best regards, Richard Harrison, KB5WZI

The question you have to ask yourself is, does it intercept all of the
energy reflected toward it, or only some fraction of it.

We should always be cognizant of the limits imposed by the absence of a
free lunch.

ac6xg

Jim Kelley, AC6XG wrote:
"The question you have to ask yourself is, does it intercept all of the
energy reflected toward it, or only some fraction of it."

Nothing is perfect as Jim observes. "Imperfections" are sometimes
exploited to improve an antenna pattern. To a first approximation
though, we assume that all the parallel rays intercepted by a dish are
focused on the radiator and aid, adding in-phase. Received carrier power
excites the antenna and this causes a minimum of 50% of this power to be
re-radiated if the antenna is perfectly matched to to the receiver load.
The antenna`s radiation resistance in this case becomes the Thevenin`s
source resistance for the receiver load on the antenna. This requires a
conjugate match between the antenna and receiver input impedances.

50% of the received power to the receiver is the best that can be done
under optimum conditions, that is , with a perfect match. With a 100%
mismatch, a short-circuit, 100% of the intercepted power is re-radiated
by the antenna. If the antenna is open-circuited, it accepts none of the
power focused upon it.

Best regards, Richard Harrison, KB5WZI

See comment below.
--
J. Mc Laughlin; Michigan U.S.A.
Home:

"Richard Harrison" wrote in message
...
Jim Kelley, AC6XG wrote:
snip

To a first approximation
though, we assume that all the parallel rays intercepted by a dish are
focused on the radiator and aid, adding in-phase. Received carrier power
excites the antenna and this causes a minimum of 50% of this power to be
re-radiated if the antenna is perfectly matched to to the receiver load.
The antenna`s radiation resistance in this case becomes the Thevenin`s
source resistance for the receiver load on the antenna. This requires a
conjugate match between the antenna and receiver input impedances.

----- ... not if one wishes to maximize SNR. Best SNR requires a (slight)
mismatch. Of course, this issue is only significant if SNR is due to the
noise figure (NF) of the receiver and the SNR is small.
73 Mac N8TT