Andy asks:
I have a quick question about the number of microwatts that an
antenna will extract from the air and put into a known load.

Assume a 70 ohm perfect antenna used for receiving a signal....

Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq
mtrs)

Is this the power that the 70 ohm antenna will put into a 70
ohm resistor connected to the antenna (as the load), OR is half
this power dissipated in the antenna and the other half sent to
the load ( considering the antenna as a Thevenin generator)....?????

Thanks for any discussion on this. I know it is simple, but I
would like to hear what others have to say about it....

Andy W4OAH

AndyS wrote:
Andy asks:
I have a quick question about the number of microwatts that an
antenna will extract from the air and put into a known load.

Assume a 70 ohm perfect antenna used for receiving a signal....

Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq
mtrs)

Is this the power that the 70 ohm antenna will put into a 70
ohm resistor connected to the antenna (as the load), OR is half
this power dissipated in the antenna and the other half sent to
the load ( considering the antenna as a Thevenin generator)....?????

Thanks for any discussion on this. I know it is simple, but I
would like to hear what others have to say about it....

Andy W4OAH

The short answer is that the product of impinging wave power density and
capture area equals the power dissipated by the load, assuming a
lossless antenna and matched load.

A lossless antenna won't dissipate any power applied to it. So it
certainly won't dissipate half, or any, of the power from the wave.

You won't find "capture area" in many texts, but when you do, it'll be
another term for the much more common "aperture". Among variations of
this are "aperture", "effective aperture", "maximum effective aperture",
"scattering aperture", "loss aperture", "collecting aperture", and
"physical aperture". The aperture is the ratio of power delivered to the
load, to the power density of the impinging wave. If the antenna is
lossless, terminated in the complex conjugate of its transmitting
feedpoint impedance, and oriented for maximum response to the wave's
polarization, the effective aperture, maximum effective aperture, and
scattering aperture are all equal, and represent the cross section of
the power in the impinging wave delivered to the load.

Kraus's _Antennas_ is a good source of information about this topic.

Roy Lewallen, W7EL

"AndyS" wrote in news:1175314795.342922.213610
@o5g2000hsb.googlegroups.com:

Andy asks:
I have a quick question about the number of microwatts that an
antenna will extract from the air and put into a known load.

Assume a 70 ohm perfect antenna used for receiving a signal....

Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq
mtrs)


Further to Roy's notes, keep in mind that for most antennas, you cannot
determine the Effective Aperture Area (or Capture Area) using a ruler.
This effect is demonstrated if you consider that a short low loss dipole
with low loss matching has an Effective Aperture Area nearly as large as
the much larger half wave dipole.

For many purposes, you would determine the Effective Aperture Area by
determining Gain from Directivity and Loss, thence Effective Aperture
Area from Gain and Wavelength.

Owen

"Owen Duffy" wrote in message
...
"AndyS" wrote in news:1175314795.342922.213610
@o5g2000hsb.googlegroups.com:

Andy asks:
I have a quick question about the number of microwatts that an
antenna will extract from the air and put into a known load.

Assume a 70 ohm perfect antenna used for receiving a signal....

Preceived = Pwr density ( watts/sq mtr) X Capture Area (sq
mtrs)


Further to Roy's notes, keep in mind that for most antennas, you cannot
determine the Effective Aperture Area (or Capture Area) using a ruler.
This effect is demonstrated if you consider that a short low loss dipole
with low loss matching has an Effective Aperture Area nearly as large as
the much larger half wave dipole.

For many purposes, you would determine the Effective Aperture Area by
determining Gain from Directivity and Loss, thence Effective Aperture
Area from Gain and Wavelength.

Owen

and hence if you start with the capture area by definition all the power is
sent to the matched load. basically you measure the capture area by
measuring the power in the load and dividing by the power density.

Owen Duffy wrote:
Further to Roy's notes, keep in mind that for most antennas, you cannot
determine the Effective Aperture Area (or Capture Area) using a ruler.
This effect is demonstrated if you consider that a short low loss dipole
with low loss matching has an Effective Aperture Area nearly as large as
the much larger half wave dipole.

For many purposes, you would determine the Effective Aperture Area by
determining Gain from Directivity and Loss, thence Effective Aperture
Area from Gain and Wavelength.

Owen

Andy writes:
Thanks to all for the discussion....... I have used the results of
these
calculations for many years but wanted to see what explanations I
would
get from others who also have experience in these matters.....

For capture area, I use " Gee lambda squared over four pi ",
which
is the standard definition for a well behaved antenna with a main
lobe, and it has always worked well for me. I don't remember doing
the
derivation for this, but I'm sure I must have done it in the past
(olden times)...

Yet, I was not comfortable with what seemed like a discrepancy
between
"all the extracted power goes to the load" and the proposition that
the
cosmos and antenna acted like a generator and therefore had an
internal
impedance which must be Thevenin matched for max extracted power....
I resolved it, in my mind, by ignoring the latter explanation
(grin)..... and
apparently it was the correct thing to do...

So, thanks again, guys........

Andy W4OAH

On 31 Mar 2007 04:57:23 -0700, "AndyS" wrote:

Yet, I was not comfortable with what seemed like a discrepancy
between
"all the extracted power goes to the load" and the proposition that
the
cosmos and antenna acted like a generator and therefore had an
internal
impedance which must be Thevenin matched for max extracted power....
I resolved it, in my mind, by ignoring the latter explanation
(grin)..... and
apparently it was the correct thing to do...

Hi Andy,

It is a bit of a stumble reading through that, and it gives the
impression that you have learned the incorrect thing to do (although I
find it hard to imagine how you could be "doing" anything when the
mechanics of radiation and reception take care of themselves).

73's
Richard Clark, KB7QHC

On 31 mar, 13:57, "AndyS" wrote:
{text deleted}
Andy writes:

Thanks to all for the discussion....... I have used the results of
these
calculations for many years but wanted to see what explanations I
would
get from others who also have experience in these matters.....

For capture area, I use " Gee lambda squared over four pi ",
which
is the standard definition for a well behaved antenna with a main
lobe, and it has always worked well for me. I don't remember doing
the
derivation for this, but I'm sure I must have done it in the past
(olden times)...

Yet, I was not comfortable with what seemed like a discrepancy
between
"all the extracted power goes to the load" and the proposition that
the
cosmos and antenna acted like a generator and therefore had an
internal
impedance which must be Thevenin matched for max extracted power....
I resolved it, in my mind, by ignoring the latter explanation
(grin)..... and
apparently it was the correct thing to do...

So, thanks again, guys........

Andy W4OAH

Hi Andy,

As you mentioned, Thevenin does apply, and your effective area formula
is correct (gain w.r.t. isotropic radiator). So a thin halve wave
dipole antenna has an internal resistance of about 70 Ohms, and half
the power is lost, but not as heat in the antenna. The driving
(incident) EM-field induces voltage with causes a current to flow in
your 75 Ohms load (and in the antenna).

The current that flows into the antenna structure (two bars) radiate.
So part of the incident field is reradiated by the dipole. It doesn't
matter whether the antenna current is caused by a transmitter or
incident field. As your dipole delivers 1mW to the load, also 1mW is
reradiated into space. Behind the antenna, the reradiated field
cancels (partially) the incident field (causing a shadow effect).

When you remove the load, the output voltage at the antenna doubles
and the current in the dipole reduces to a low value, hence less power
is reradiated. When you short circuit the dipole, the current will
double (w.r.t. to the matched situation). This will also happen to
the current in the bars forming the dipole. As the current has been
doubled, the reradiated power is 4 times higher (w.r.t. the matched
situation).

The reradiated power can be detected. Radar and passive UHF RFID use
the reradiated power. For example in UHF RFID, the chip on the RF tag
changes the load to the dipole (that receives power from the
interrogator), and controls the reradiated power. In this way the ID
of the tag is transmitted from the tag to the interrogator (the tag
has no battery present).

The more gain you have in your antenna, the higher the reradiated EIRP
in the direction of maximum gain.

I hope this clarifies the antenna internal resistance.

Best Regards,

Wim
PA3DJS

As you mentioned, Thevenin does apply, and your effective area formula
is correct (gain w.r.t. isotropic radiator). So a thin halve wave
dipole antenna has an internal resistance of about 70 Ohms, and half
the power is lost, but not as heat in the antenna.

If I may add just a simple observation to the exhaustive Wim's explanation, I
would say that we generally tend to consider resistors as things that
necessarily dissipate power, turning it into heat. That is not always the case.



As a matter of fact, having current in phase with voltage (what we usually call
resistance) signify the transformation of electrical power into any form of
power, not just heat. For instance:



- an ideal electrical engine (no ohmic loss, no friction) fully
transforming the applied electrical power into mechanical power. It looks like a
resistor, but no heat is generated anywhere

- an ideal antenna (no ohmic loss) fully transforming the applied
electrical power into electromagnetic power. Again it looks like a resistor, but
no heat is generated anywhere

..

73



Tony I0JX

Wimpie wrote: I hope this clarifies the antenna internal resistance.

Best Regards,

Wim
PA3DJS

Andy replies:
Thanks Wim and Tony, for a very well presented explanation of
the Thevinen .... I KNEW it had to coincide with the CAxPdens
numbers, but your post pretty well cleared it up....for me, at
least...

Regarding RFID using re-radiation --- I was the designer of
the TIRIS toll tag system for T I ( at 915 mhz) and the method I used
was
to open and short a dipole in the TAG to send data replies to the
Reader.... The specification required an RCS
(Radar Cross Section) change of 180
sq cm at 915, and that was easy to obtain with a dipole and a
pin diode . ( Actually, the RCS changed too much for the spec,
and I had to pad it down a bit )......

I hadn't considered that an antenna delivering to a matched load
lost it's power by re-radiation .... DUHHHH!!!.... Been building stuff
for 50 years , and that obvious fact just never dawned on me....
So, thanks again....

73, Andy W4OAH

Antonio Vernucci wrote:
- an ideal electrical engine (no ohmic loss, no friction) fully
transforming the applied electrical power into mechanical power. It
looks like a resistor, but no heat is generated anywhere

No, it looks like a resistance, not a resistor. There is one
IEEE definition for resistor. There are two IEEE definitions
for resistance. A resistor with current dissipates power. A
resistance may or may not dissipate power. One definition of
resistance in the IEEE describes a dissipationless resistance.
There is no such thing as a dissipationless resistor.
--
73, Cecil http://www.w5dxp.com