Re: Antennas 101 for Androcles


jmorriss@xxxxxxxxxxx wrote:
Randy Poe wrote:
Androcles is currently cluttering multiple threads with his
assertion that beams from parabolic antennas experience no
1/r^2 free-space loss. Rather than continue to clutter those
threads, let me put the relevant information into one place.

Engineers characterize every antenna by its gain, which
relates the power to omnidirectional power. A perfectly
omnidirectional antenna has a gain of 1, or 0 dB. Gain is
related to directivity. An antenna with a gain of 30 dB
(a factor of 1000) is highly directional. If receiving, it
gets its energy from about 1/1000-th of the sphere. If
transmitting, it is putting its energy out in 1/1000 of the
sphere and therefore the intensity is 1000 times higher
in that direction than if it tried to spread the energy in
all directions equally.

Good parabolic antennas have gains in the neighborhood
of 50 dB. Aside from shape, the most important parameter
affecting gain is D/lambda, the ratio of diameter to
wavelength. If the antenna is not circular, it is some other
characteristic length/lambda. If you take a given antenna
shape and double its size, but then double the wavelength,
you'll get the same gain.

The area of a sphere grows as r^2 where r is the radius
of the sphere. An antenna like Cassini's with a gain of
50 dB is transmitting its energy into 1/100000 of the
sphere. 1/100000 of a thing growing as r^2 is still growing
as r^2.

Now if you do simple ray tracing, it looks like every ray
leaving the focus and bouncing off a parabolic mirror should
exit parallel. So why is there a beamwidth?

First of all, it is because light is a wave, and the reflection
and focusing of waves is not so simple as ray-tracing. The
pattern from a parabolic antenna is very similar to the diffraction
pattern from a small circular hole, a pattern which also has
angular width ~ D/lambda.

Secondly, the focus is a point, and real antenna feeds
have a volume, so the energy is coming from places
that aren't exactly at the focus.

Third, the shape will never be exactly parabolic. In real
life, there are always variations.


OK, but surely diffraction is a minor factor in spreading,isn't it?

It all comes down to D/lambda. You can easily get D/lambda in the
hundreds of thousands when dealing with light. With RF,
you get much more modest numbers, in the hundreds.

Consider these two scenarios:

We have a 10 metre radius opaque sphere with a 10 cm diameter hole in
it and a very bright, small lamp at the centre of the sphere. The
light intensity observed outside the sphere will drop off by an inverse
square law as we move away, staying in line with the lamp and the hole.
When we get 10 metres from the surface of the sphere, the intensity
will be 0.25 of the intensity at the surface of the sphere. When we
get to 90 metres from the surface, the intensity will be 0.01 of the
surface intensity, u. s. w....

I hope this is OK so far...

Now, we replace the very bright lamp with a merely bright lamp at the
focus of a fast parabolic mirror

"fast"?

located at the centre of the sphere,
with parabola's axis aligned on the hole. We also cleverly adjust the
lamp brighness so that the intensity at the surface of the sphere is
the same as in the previous example.

Is it your position that the light intensity measured along the
radiated light beam will be the same in these two cases?

No, it is brighter because the divergence angle is smaller.

Inverse square drop off?

There still is an inverse square law, but the distance is not the
distance
to the center of the sphere.

The beam from the mirror is very gently diverging. If you
trace the edges of that diverging beam back, they converge
at a point far behind the mirror. So the beam follows an
inverse square law in terms of distance from that point, not
from the center of the sphere.

Let us say the edges of the beam are 0.1 degree away
from parallel, and the mirror is 1 m in diameter, 0.5 m
in radius. The convergence point is then 0.5/tan(1 deg)
= 28.6 m behind the mirror. The surface of the sphere is
38.6 m from this point.

10 m from the surface, the relative intensity is (38.6/48.6)^2
compared to the surface. With a point source that ratio
is (10/20)^2. 0.6 compared to 0.25.

90 m from the surface, it is (38.6/128.6)^2 compared to
(10/100)^2, 0.09 compared to 0.01.

x m from the surface, it is [38.6/(x+38.6)]^2, which asymptotically
approaches 38.6^2/x^2 at large x.

For the point source, it is [10/(x+10)]^2, asymptotically approaching
100/x^2 at large x. In the far field, the parabola is about 14.9
times as bright as the point source. The reason for this
constant difference is that same energy is spread over a
larger angle in the point source case.

Or have I merely moved to a wavelength/diamter regime where diffraction
is much less important?

It's always present, but of course it's less significant for light
of wavelength 500 nm than for RF energy of wavelength
4 cm.

- Randy

.

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