Rune Allnor wrote:
> Jerry Avins wrote:
> 
> 
>>There will always be a difference between what can be
>>measured and what can be calculated from theory. That difference is a
>>common cause of discussions at cross purposes here on comp.dsp.
> 
> 
> Well, yes, and that's how it has to be. No theories fit the real world
> perfectly, and no real-world measurement comply 100% with any given
> theory. Knowing to what extent a theory fits, and the nature of the
> discrepancy, is the key to get anything done what "practical" DSP
> is concerned.

One can calculate that the sagitta of a kilometer-long chord across a 
wavefront is a few �ngstroms, but such a result defies measurement.

>>I, for
>>one, have grown gun shy and try to dot the i's and cross the t's.
> 
> 
> Don't know what you mean by that, but it is always a good idea to
> make clear what the basis of one's arguments is.

It means that (when I remember) I try to make constraints and 
assumptions explicit. It can seem boring and pedantic -- some would have 
it anal retentive -- but the extra words usually save words in the long 
run (this occasion being an exception).

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins wrote:

> There will always be a difference between what can be
> measured and what can be calculated from theory. That difference is a
> common cause of discussions at cross purposes here on comp.dsp.

Well, yes, and that's how it has to be. No theories fit the real world
perfectly, and no real-world measurement comply 100% with any given
theory. Knowing to what extent a theory fits, and the nature of the
discrepancy, is the key to get anything done what "practical" DSP
is concerned.

> I, for
> one, have grown gun shy and try to dot the i's and cross the t's.

Don't know what you mean by that, but it is always a good idea to
make clear what the basis of one's arguments is.

Rune

John Monro wrote:
> Jerry Avins wrote:
> 
>> John Monro wrote:
>>
>>> Jerry Avins wrote:
>>>
>>>> I meant that the transmitter seems like a point source to the 
>>>> receiver, and that it is sufficiently remote that the bearing to it 
>>>> is the same from every point on the receiving array. If these 
>>>> conditions are met, they would also be met if the role of receiver 
>>>> and transmitter were interchanged. "Plane wave" encompasses these 
>>>> conditions, but doesn't serve well as a definition. "Point source" 
>>>> and "plane wave" are in fact mutually contradictory, but serve well 
>>>> as local approximations. A true plane wave doesn't have 
>>>> inverse-square intensity.
>>>>
>>>
>>>
>>> Jerry,
>>> Another way of looking at it is this:
>>>
>>> Those two terms: "point source" and "plane wave" are not mutually 
>>> contradictory if you take into account the 'optical' location of the 
>>> source.
>>>
>>> If you trace back the 'nearly' plane-wave radiated by a large dish or 
>>> array, the wave appears to be coming from a point that is located 
>>> well behind the antenna.  The less curvature you have on that 
>>> wave-front the further behind the antenna the source appears to be.
>>>
>>> In the case of a theoretical-perfect plane-wave, the source appears 
>>> to be located at an infinite distance behind the antenna aperture.  
>>> As the apparent distance between the signal source and the receiving 
>>> antenna is already infinite it is not possible to change this 
>>> distance to any significant degree by changing the physical 
>>> separation between the antennas.  The inverse-square law is working 
>>> correctly when it shows us that there is no change in received signal 
>>> under these circumstances.
>>
>>
>>
>> Sure. I'll say it two more ways.
>>
>> When the curvature of the wavefront is infinitesimal, the distance to 
>> the point radiator is infinite.
>>
>> The radius of curvature is the reciprocal of the curvature.
>>
>> Jerry
> 
> 
> Because of the fact that: "when the curvature of the wavefront is 
> infinitesimal, the distance to the point radiator is infinite", it is 
> incorrect to say that "a true plane wave doesn't have inverse-square 
> intensity."

Here on earth, the light from Betelgeuse might as well be plane, but we 
(think we) know the distance to the star and can calculate the departure 
from planarity. There will always be a difference between what can be 
measured and what can be calculated from theory. That difference is a 
common cause of discussions at cross purposes here on comp.dsp. I, for 
one, have grown gun shy and try to dot the i's and cross the t's.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins wrote:
> John Monro wrote:
> 
>> Jerry Avins wrote:
>>
>>> I meant that the transmitter seems like a point source to the 
>>> receiver, and that it is sufficiently remote that the bearing to it 
>>> is the same from every point on the receiving array. If these 
>>> conditions are met, they would also be met if the role of receiver 
>>> and transmitter were interchanged. "Plane wave" encompasses these 
>>> conditions, but doesn't serve well as a definition. "Point source" 
>>> and "plane wave" are in fact mutually contradictory, but serve well 
>>> as local approximations. A true plane wave doesn't have 
>>> inverse-square intensity.
>>>
>>
>>
>> Jerry,
>> Another way of looking at it is this:
>>
>> Those two terms: "point source" and "plane wave" are not mutually 
>> contradictory if you take into account the 'optical' location of the 
>> source.
>>
>> If you trace back the 'nearly' plane-wave radiated by a large dish or 
>> array, the wave appears to be coming from a point that is located well 
>> behind the antenna.  The less curvature you have on that wave-front 
>> the further behind the antenna the source appears to be.
>>
>> In the case of a theoretical-perfect plane-wave, the source appears to 
>> be located at an infinite distance behind the antenna aperture.  As 
>> the apparent distance between the signal source and the receiving 
>> antenna is already infinite it is not possible to change this distance 
>> to any significant degree by changing the physical separation between 
>> the antennas.  The inverse-square law is working correctly when it 
>> shows us that there is no change in received signal under these 
>> circumstances.
> 
> 
> Sure. I'll say it two more ways.
> 
> When the curvature of the wavefront is infinitesimal, the distance to 
> the point radiator is infinite.
> 
> The radius of curvature is the reciprocal of the curvature.
> 
> Jerry

Because of the fact that: "when the curvature of the wavefront is 
infinitesimal, the distance to the point radiator is infinite", it is 
incorrect to say that "a true plane wave doesn't have inverse-square 
intensity."

Regards,
John

John Monro wrote:
> Jerry Avins wrote:
> 
>> I meant that the transmitter seems like a point source to the 
>> receiver, and that it is sufficiently remote that the bearing to it is 
>> the same from every point on the receiving array. If these conditions 
>> are met, they would also be met if the role of receiver and 
>> transmitter were interchanged. "Plane wave" encompasses these 
>> conditions, but doesn't serve well as a definition. "Point source" and 
>> "plane wave" are in fact mutually contradictory, but serve well as 
>> local approximations. A true plane wave doesn't have inverse-square 
>> intensity.
>>
> 
> 
> Jerry,
> Another way of looking at it is this:
> 
> Those two terms: "point source" and "plane wave" are not mutually 
> contradictory if you take into account the 'optical' location of the 
> source.
> 
> If you trace back the 'nearly' plane-wave radiated by a large dish or 
> array, the wave appears to be coming from a point that is located well 
> behind the antenna.  The less curvature you have on that wave-front the 
> further behind the antenna the source appears to be.
> 
> In the case of a theoretical-perfect plane-wave, the source appears to 
> be located at an infinite distance behind the antenna aperture.  As the 
> apparent distance between the signal source and the receiving antenna is 
> already infinite it is not possible to change this distance to any 
> significant degree by changing the physical separation between the 
> antennas.  The inverse-square law is working correctly when it shows us 
> that there is no change in received signal under these circumstances.

Sure. I'll say it two more ways.

When the curvature of the wavefront is infinitesimal, the distance to 
the point radiator is infinite.

The radius of curvature is the reciprocal of the curvature.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Brenneman wrote:
>>I meant that the transmitter seems like a point source to the receiver,
>>and that it is sufficiently remote that the bearing to it is the same
>>from every point on the receiving array. If these conditions are met,
>>they would also be met if the role of receiver and transmitter were
>>interchanged.
> 
> 
> Thank you for the explanation: I understand exactly what you mean now
> and the distinction you make is an important one. Since the case to
> which you are referring is the one of interest to me let me go back to
> your first post now. When you say that the "DOA makes sense" under
> these conditions, what do YOU mean by DOA: the two spherical coordinate
> angles (phi,theta)?

Matt,

I'm sorry I didn't get back to you sooner. Two linear antenna arrays are 
needed to get two angles. (Of course, a plane structure serves also.) 
The best illustration of the One-dimensional situation I can think of is 
the usual freshman physics explanation of a diffraction grating. That 
explanation also makes clear why an extended line of closely spaces 
antennas reduces ambiguity (sidelobes).

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins wrote:

> I meant that the transmitter seems like a point source to the receiver, 
> and that it is sufficiently remote that the bearing to it is the same 
> from every point on the receiving array. If these conditions are met, 
> they would also be met if the role of receiver and transmitter were 
> interchanged. "Plane wave" encompasses these conditions, but doesn't 
> serve well as a definition. "Point source" and "plane wave" are in fact 
> mutually contradictory, but serve well as local approximations. A true 
> plane wave doesn't have inverse-square intensity.
> 


Jerry,
Another way of looking at it is this:

Those two terms: "point source" and "plane wave" are not mutually 
contradictory if you take into account the 'optical' location of the 
source.

If you trace back the 'nearly' plane-wave radiated by a large dish or 
array, the wave appears to be coming from a point that is located well 
behind the antenna.  The less curvature you have on that wave-front the 
further behind the antenna the source appears to be.

In the case of a theoretical-perfect plane-wave, the source appears to 
be located at an infinite distance behind the antenna aperture.  As the 
apparent distance between the signal source and the receiving antenna is 
already infinite it is not possible to change this distance to any 
significant degree by changing the physical separation between the 
antennas.  The inverse-square law is working correctly when it shows us 
that there is no change in received signal under these circumstances.

Regards,
John

Brenneman skrev:
> >
> > I meant that the transmitter seems like a point source to the receiver,
> > and that it is sufficiently remote that the bearing to it is the same
> > from every point on the receiving array. If these conditions are met,
> > they would also be met if the role of receiver and transmitter were
> > interchanged.
>
> Thank you for the explanation: I understand exactly what you mean now
> and the distinction you make is an important one. Since the case to
> which you are referring is the one of interest to me let me go back to
> your first post now. When you say that the "DOA makes sense" under
> these conditions, what do YOU mean by DOA: the two spherical coordinate
> angles (phi,theta)?

Don't know about Jerry, but *I* think of the pair of angles (phi,
theta) as 
the DoA in the case of the planar array. 

Rune

>
> I meant that the transmitter seems like a point source to the receiver,
> and that it is sufficiently remote that the bearing to it is the same
> from every point on the receiving array. If these conditions are met,
> they would also be met if the role of receiver and transmitter were
> interchanged.

Thank you for the explanation: I understand exactly what you mean now
and the distinction you make is an important one. Since the case to
which you are referring is the one of interest to me let me go back to
your first post now. When you say that the "DOA makes sense" under
these conditions, what do YOU mean by DOA: the two spherical coordinate
angles (phi,theta)?

Thanks,

Matt

Brenneman skrev:
> >
> > DoA makes sense when the receiving antenna subtends an infinitesimal
> > angle at the source and the source subtends an infinitesimal angle at
> > the receiving antenna. The mathematics becomes very complex otherwise.
> >
> > Jerry
> > --
> > Engineering is the art of making what you want from things you can get.
> > =AF=AF
>
> Hi Jerry,
>
> Thank you for your response. Let me ask you one follow up question to
> make sure I understand what you are saying.
> I am interested in antenna arrays in the far-field of point sources
> that emit plane wave radiation (like a GPS source for example).

These are contradictions in terms. The frame of refernce is the
reciever system. A "point source" is a source which is so small
that it resembles a point when viewed from the reciever. A simple
example is a star shining in the sky. It resembles a point.
Good examples of "distributed sources" are the sun and the
moon, that have clear angular extents.

A "point source" emits a spherical or cylindrical wave, which
energy dilutes as function of traveled distance. Also, the local
curvature of the wave front reduces as the distance to the
source increases. At some distance there is no point in dealing
with local curvature, and the plane wave approximation is used
instead.

> I am
> wondering if this is the same context to which you are referring. When
> you say that BOTH source and receiver "subtend an infinitesimal angle"
> with respect to each other, the only scenario I can envision that fits
> that criertion is perhaps when the source is on the horizon, so that
> the angle to which you are referring is essentially the complement of
> theta.

The source is never a point source, it has some physical extension.
However, for the far field to be valid, the angular extension of that
source, when viewed from the reciever, must be infitesimal.

Again, a star as example. The star Betelgeuze has a diameter
on the order of the diameter of the earth's orbit around the sun
http://en.wikipedia.org/wiki/Betelgeuze. Nevertheless, when viewed
from earth, it is seen as just a point.

Rune