Fred Abse wrote:
> On Sun, 06 May 2012 09:31:21 -0700, Winston wrote:
>
>> Fred Abse wrote:
>>> On Sat, 05 May 2012 15:14:59 -0700, Winston wrote:
>>>
>>>> Winston wrote:
>>>>
>>>> (...)
>>>>
>>>>> So, the on-axis 'near field' of an antenna with a gain of 10 is
>>>>> actually more like 50 wavelengths - rather than say 5 wavelengths -
>>>>> away?
>>>>>
>>>>> I conjecture that I *should* be calculating watts / cm to fall
>>>>> inversely with distance (rather than as the inverse cube...)
>>>>
>>>> Er. make that '(rather than as the inverse square...)'
>>
>> Yes, I meant W/cm^2 not W/cm.  My bad.
>>
>>> No. A lot of people get this wrong, in one way or the other.
>>>
>>> Power density (watts per unit area) falls inversely as the square of the
>>> distance.
>>
>> Yes, in the far field.  I've measured that to be the case. That ain't my
>> question.
>>
>> 1) The old tale about power density falling inversely (not as the
>>      inverse square) within the *near field* is true, yes? (I know. 'Just
>>      Measure It'.)  :)
>
> No, because the "near field" component *does not radiate power*.

OK.  Got it!  Thanks.

>> 2) The center lobe of a directional antenna with a gain of say 10
>>      would be expected to have a near field border about 10 x further from
>>      it than would an isotropic, all else being corrected, yes? (That
>>      sounds perfectly reasonable, but I don't want to assume.)
>>
>>
>
> A theoretical isotropic radiator, by definition, has no near field. It's
> a radiator only.
>
> Near field boundary is determined by physical dimensions, not theoretical
> gain.

Excellent news!

Thanks for helping me limit my ignorance.  :)

--Winston

Fred Abse wrote:
> On Sun, 06 May 2012 10:20:04 -0700, Winston wrote:

(...)

> There's no such thing as electric power density. Get these things straight
> in your mind, I think you're getting confused

I was referring to S, the power density.

(...)

> Kraus gives the near field boundary as R=2L^2/lambda (meters), where L is
> the antenna physical length.

Ah!  This puts that boundary at ~ 0.6 m for the system I'm
looking at. So that removes near field / far field from
'things I should be concerned about', no matter what.

That is extremely good news and I appreciate it.

The equation I cited earlier yields results that are very close
to the actual readings taken from a live system, so I am
happy to use it to check on the data I'll eventually gather.



Thanks again, Fred.


--Winston

On Sun, 06 May 2012 10:20:04 -0700, Winston wrote:

> 
> OK, but can we talk about electric power density as isolated from the
> magnetic density or voltage density?

Your terminology is confusing. Power has density (per unit area),
electromagnetic fields (voltage and magnetizing force) have intensity or
strength (per unit length). Consider a simple example: a capacitor has
two parallel plates, separated by one meter. The PD between the plates is
one volt, hence the field between the plates is one volt per meter
intensity.

>
> At what distance does electric power density stop falling as the inverse
> of distance and begin to fall as the square of distance? (This is the
> definition of 'near field' / 'far field', yes?)

Power density *always* follows an inverse square law.
Field intensity follows a first power law.

> 
> My current misunderstanding is that for an (admittedly theoretical)
> isotropic radiator, that distance is 'several wavelengths'.

See below.

> 
> If that is true, can we reasonably expect to see the shape of the near
> field be the same as the shape of the far field in the radiation pattern
> of an anisotropic transmitting antenna, though the *sizes* of those fields
> be quite different.
> 
> Is this correct?

No, see below


>> Power density (watts/square meter) falls off as the inverse square of
>> the distance.
> 
> Even within the near field? (Electric power density only)

There's no such thing as electric power density. Get these things straight
in your mind, I think you're getting confused, there is:

Electric field intensity - B Volts per meter.
Magnetic field intensity - H Amps (or amp turns if you like) per meter.
Power density - Watts per square meter.

All three are inseparable, and are linked by the intrinsic impedance of
free space = 377 ohms. It's just like Ohm's Law.

The near field has two components.

Firstly, an inductive component, which merely stores energy, returning it
to the antenna each half cycle. Analogous to a resonant circuit.
The inductive field *does not radiate power*, hence does not contribute to
power density. Its electric and magnetic components are antiphase.

Secondly, a component which radiates power as a field having an electric
(E) component, and a magnetic (H) component, mutually at right angles in
space, and in phase.

Energy flow (the Poynting vector) is at right angles to both E and H, and
flows away from the antenna, into space, and is equal to the vector
product of E and H.

In the far field, only the radiation component exists.

Kraus gives the near field boundary as R=2L^2/lambda (meters), where L is
the antenna physical length.

The electric and magnetic components of the *inductive* field have
different spatial patterns. The electric component is concentrated around
the open end (where voltage is maximum), while the magnetic component is
concentrated around the point of maximum current.

The radiating field has both components spatially co-located.

Hence, the near and far field polar patterns are not the same.

> 
> 


Definitely get a copy of Kraus, and bone up on cylindrical and spherical
coordinate systems.

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

On Sun, 06 May 2012 09:52:13 -0700, Winston wrote:

> I'm just checking my basic understanding here.

At a quick glance, your quoted reference says not much more than I've been
saying. I suspect you're not understanding some of the basics, especially
the terminology.

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

On Sun, 06 May 2012 09:31:21 -0700, Winston wrote:

> Fred Abse wrote:
>> On Sat, 05 May 2012 15:14:59 -0700, Winston wrote:
>>
>>> Winston wrote:
>>>
>>> (...)
>>>
>>>> So, the on-axis 'near field' of an antenna with a gain of 10 is
>>>> actually more like 50 wavelengths - rather than say 5 wavelengths -
>>>> away?
>>>>
>>>> I conjecture that I *should* be calculating watts / cm to fall
>>>> inversely with distance (rather than as the inverse cube...)
>>>
>>> Er. make that '(rather than as the inverse square...)'
> 
> Yes, I meant W/cm^2 not W/cm.  My bad.
> 
>> No. A lot of people get this wrong, in one way or the other.
>>
>> Power density (watts per unit area) falls inversely as the square of the
>> distance.
> 
> Yes, in the far field.  I've measured that to be the case. That ain't my
> question.
> 
> 1) The old tale about power density falling inversely (not as the
>     inverse square) within the *near field* is true, yes? (I know. 'Just
>     Measure It'.)  :)

No, because the "near field" component *does not radiate power*. 

> 
> 2) The center lobe of a directional antenna with a gain of say 10
>     would be expected to have a near field border about 10 x further from
>     it than would an isotropic, all else being corrected, yes? (That
>     sounds perfectly reasonable, but I don't want to assume.)
> 
>

A theoretical isotropic radiator, by definition, has no near field. It's
a radiator only.

Near field boundary is determined by physical dimensions, not theoretical
gain.


-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Fred Abse wrote:
> On Sat, 05 May 2012 15:02:20 -0700, Winston wrote:

(...)

>> For my purpose, that 'half bagel' pattern will work just fine. (I assume
>> that the 'reception pattern' for a receiver
>>    exhibits the same geometry as does the 'radiation pattern' for a
>>    transmitter for a given antenna, yes?)
>
> Yes.

Good. Thanks!

(...)

>> So, the on-axis 'near field' of an antenna with a gain of 10 is actually
>> more like 50 wavelengths - rather than say 5 wavelengths - away?
>
> No, near field is dominated by induction, rather than radiation, which is
> not necessarily a function of antenna gain, From (now quite distant) memory, the
> inductive field is 3dB down at about 2/3 wavelength. I'll have to check
> this.

OK, but can we talk about electric power density
as isolated from the magnetic density or voltage density?

At what distance does electric power density stop falling as the
inverse of distance and begin to fall as the square of distance?
(This is the definition of 'near field' / 'far field', yes?)

My current misunderstanding is that for an (admittedly theoretical)
isotropic radiator, that distance is 'several wavelengths'.

If that is true, can we reasonably expect to see the shape of the
near field be the same as the shape of the far field in
the radiation pattern of an anisotropic transmitting antenna,
though the *sizes* of those fields be quite different.

Is this correct?

>> I conjecture that I *should* be calculating watts / cm to fall inversely
>> with distance (rather than as the inverse cube...) if my 1 GHz receiver
>> is within say 45 meters of the (gain of 10) antenna?  Holy Moley!
>
> There's no such thing as watts/cm. Power density is watts/area, ie.
> watts/square meter, or watts/square cm (if you must).

My bad. I meant to say W/cm^2 not W/cm.

> Power density (watts/square meter) falls off as the inverse square of the
> distance.

Even within the near field? (Electric power density only)

>> My application is EMC / agency compliance related, where W/cm^2 is the
>> lingua franca. So I want to be able to express field intensity using that
>> radix, even if it means doing the involved math to convert from W/m^2   :)
>
> What involved math?. For example 10mW/cm^2 is 100W/m^2. You can do that in
> your head, just multiply by 100^2 ;-)

You reveal the reason behind my original smiley.  :)

> Any regulatory standards in the last 30 years should be in SI units,
> anyway. Last thing I had to do in this area (which admittedly was for a
> European directive in about 1990), levels were quoted in W/m^2.

I'm equally happy with an answer expressed in W/m^2 or W/cm^2.


(...)

>> 'Sounds as if I need to build a prototype and calibrate
>>    it against a leveled 1 GHz source of known wattage (and known
>>    radiation pattern).
>
>
> That's how it's done. That's why we have expensive measuring antennas with
> NIST traceable calibration.
>
>
> I'd recommend that you get a copy of Kraus's "Antennas", if you really
> want to get into this stuff. I guess most university bookstores still have
> it, or can get it.

Cool!  Thanks!

--Winston

Fred Abse wrote:
> On Sat, 05 May 2012 14:21:58 -0700, Winston wrote:
>
>> The example implied that my receiver is *always* on-axis with the main
>> lobe on both planes, so I understand that the field strength would measure
>> the same as if the transmitter were 1000 W with an isotropic radiator, all
>> else being equal.
>
> Huh?

The equation I cited Friday afternoon from:
http://www.phys.hawaii.edu/~anita/new/papers/militaryHandbook/pwr-dens.pdf

..Provides an example of a 100 W transmitter radiating anisotropically
at a distance of 100 feet from the receiver, which is on axis with the
main radiated lobe of the transmitter's 10 x gain antenna.

I misremembered the example as being isotropic instead so let me
rephrase my question:

I tried different numbers for antenna gain in his equation.

In his example, his equation appeared to indicate that if this
transmitter were radiating isotropically instead of anisotropically,
we can reasonably expect to see the point of constant power density
(0.0086 mW/cm^2) to be 1/10 the distance (or only 10 feet) from the
transmitting antenna rather than 100 feet, given that the receiving
antenna is on-axis with the main lobe in the anisotropic condition.

Is this correct?

I'm just checking my basic understanding here.


--Winston

Fred Abse wrote:
> On Sat, 05 May 2012 15:14:59 -0700, Winston wrote:
>
>> Winston wrote:
>>
>> (...)
>>
>>> So, the on-axis 'near field' of an antenna with a gain of 10 is actually
>>> more like 50 wavelengths - rather than say 5 wavelengths - away?
>>>
>>> I conjecture that I *should* be calculating watts / cm to fall inversely
>>> with distance (rather than as the inverse cube...)
>>
>> Er. make that '(rather than as the inverse square...)'

Yes, I meant W/cm^2 not W/cm.  My bad.

> No. A lot of people get this wrong, in one way or the other.
>
> Power density (watts per unit area) falls inversely as the square of the
> distance.

Yes, in the far field.  I've measured that to be the case.
That ain't my question.

1) The old tale about power density falling inversely (not as the
    inverse square) within the *near field* is true, yes?
    (I know. 'Just Measure It'.)  :)

2) The center lobe of a directional antenna with a gain of say 10
    would be expected to have a near field border about 10 x further
    from it than would an isotropic, all else being corrected, yes?
    (That sounds perfectly reasonable, but I don't want to assume.)

Thanks for your help!  :)

--Winston

On Sun, 06 May 2012 08:17:54 -0700, Fred Abse <excretatauris@invalid.invalid>
wrote:

>On Sat, 05 May 2012 15:34:56 -0700, Jim Thompson wrote:
>
>> As soon as you have an anisotropic radiator, all "nice" equations are
>> meaningless.
>
>Some of them are.
>
>> 
>> Your roll-off with distance will be somewhere between 1/r and 1/r^2
>> depending on the lobe pattern, and how much you are off axis.
>
>I disagree. At any given angle, the radiated watts per steradian are
>constant, irrespective of distance. What changes with radial distance is
>the surface subtended by one steradian, which increases as r^2. Hence,
>along any radial line, power density always follows an inverse square law,
>and the field intensity, an inverse linear law.
>
>The lobe pattern merely defines the radiated power in a particular
>direction, ie watts per steradian along a *radial* line.
>
>> 
>> So measure it.
>
>Agreed.

Picky!  Picky!  Picky!  ;-)
 
                                       ...Jim Thompson
-- 

| James E.Thompson, CTO                            |    mens     |
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On Sat, 05 May 2012 15:34:56 -0700, Jim Thompson wrote:

> As soon as you have an anisotropic radiator, all "nice" equations are
> meaningless.

Some of them are.

> 
> Your roll-off with distance will be somewhere between 1/r and 1/r^2
> depending on the lobe pattern, and how much you are off axis.

I disagree. At any given angle, the radiated watts per steradian are
constant, irrespective of distance. What changes with radial distance is
the surface subtended by one steradian, which increases as r^2. Hence,
along any radial line, power density always follows an inverse square law,
and the field intensity, an inverse linear law.

The lobe pattern merely defines the radiated power in a particular
direction, ie watts per steradian along a *radial* line.

> 
> So measure it.

Agreed.

-- 
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)