equation: dBm to mW/cm^2

Winston · May 4, 2012

I have a 1 GHz 1/4 W whip driving a detector that outputs
RF level data as a D.C. reading that is easily convertible to dBm.
However I need the reading presented in mW / cm^2

I understand that my antenna has a gain of 1 dBi.

In this chart http://www.geopathfinder.com/Conversion Chart.pdf

...I see that 0 dBm (for example) is equal to 0.0121 mW / cm^2

The chart refers to an equation source:
http://www.GeoPathfinder.com/Formulas For Unit Conversion Charts.pdf

...But I don't see how they derived mW / cm^2 using only the equations
presented in the chart and I would like to be able to calculate
mW / cm^2 inside a computer program rather than rely on a lookup table
or force my client to do the conversion. (Yikes!)

Google is my friend but this time I was not able to uncover the
equation I desire. May I have your thoughts on this please?

Thanks!

--Winston

Tim Wescott · May 4, 2012

On Thu, 03 May 2012 22:09:41 -0700, Winston wrote:

I have a 1 GHz 1/4 W whip driving a detector that outputs RF level data
as a D.C. reading that is easily convertible to dBm. However I need the
reading presented in mW / cm^2

I understand that my antenna has a gain of 1 dBi.

In this chart http://www.geopathfinder.com/Conversion Chart.pdf

..I see that 0 dBm (for example) is equal to 0.0121 mW / cm^2

The chart refers to an equation source:
http://www.GeoPathfinder.com/Formulas For Unit Conversion Charts.pdf

..But I don't see how they derived mW / cm^2 using only the equations
presented in the chart

You mean the site that has a page titled "Biologically Alien
Electromagnetic Energies and EMF Pollution (Electrosmog)" (http://
www.geopathfinder.com/9801.html)? For some reason, I don't want to trust
that site for technical correctness.

and I would like to be able to calculate mW /
cm^2 inside a computer program rather than rely on a lookup table or
force my client to do the conversion. (Yikes!)

Google is my friend but this time I was not able to uncover the equation
I desire. May I have your thoughts on this please?

What search terms are you using? Have you combed through the ARRL
Handbook?

I'm pretty sure that something like a quarter-wave dipole is only going
to be accurate at it's resonant frequency, and its sensitivity is going
to be in volts/wavelength, not watts/area. Of course there will be a
conversion, though.

If it's just field strength you want, an electrically short antenna
driving a correctly (and severely) mismatched load will probably be much
more accurate in broadband, but will still need calibration. And I
suspect it's not what you want.

Do you have the ARRL antenna book? Have you read it?

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

Winston · May 4, 2012

Tim Wescott wrote:

On Thu, 03 May 2012 22:09:41 -0700, Winston wrote:

I have a 1 GHz 1/4 W whip driving a detector that outputs RF level data
as a D.C. reading that is easily convertible to dBm. However I need the
reading presented in mW / cm^2

I understand that my antenna has a gain of 1 dBi.

In this chart http://www.geopathfinder.com/Conversion Chart.pdf

..I see that 0 dBm (for example) is equal to 0.0121 mW / cm^2

The chart refers to an equation source:
http://www.GeoPathfinder.com/Formulas For Unit Conversion Charts.pdf

..But I don't see how they derived mW / cm^2 using only the equations
presented in the chart

You mean the site that has a page titled "Biologically Alien
Electromagnetic Energies and EMF Pollution (Electrosmog)" (http://
www.geopathfinder.com/9801.html)? For some reason, I don't want to trust
that site for technical correctness.

Okay. Right now I'd happily settle for an equation published by
enthusiasts of trans-gendered seafood, if it were correct.

and I would like to be able to calculate mW /
cm^2 inside a computer program rather than rely on a lookup table or
force my client to do the conversion. (Yikes!)

Google is my friend but this time I was not able to uncover the equation
I desire. May I have your thoughts on this please?

What search terms are you using?

'dBm mW/cm^2 formula'
'dBm mW/cm^2 equation'

I've seen nomographs and charts using these terms
that show the relationship I'm seeking; just not the equation
that will reveal how many mW per square centimeter I can expect
from a given 1/4 - wave ground plane for a given electrical field
with a number of decibels normalized to one milliwatt.

Have you combed through the ARRL
Handbook?

That's a good suggestion. I didn't see
anything that looked useful within, however.
My copy has a tutorial on Electromagnetic
Compatibility but they manage to get through
it without revealing this equation.

I did see a couple things in the index that looked
hopeful, but they didn't pan out on the actual pages.

I'm pretty sure that something like a quarter-wave dipole is only going
to be accurate at it's resonant frequency, and its sensitivity is going
to be in volts/wavelength, not watts/area. Of course there will be a
conversion, though.

I suspect I am looking for the 'aperture' of my 1/4 wave ground plane
so that I can convert received power to received power/aperture.

I'm just parrotting what little info I *have* seen so
that is probably a misinterpretation.

I see a citation revealing that a 1/2 wave dipole has an aperture
measuring 5/16 of a wavelength for example.

If it's just field strength you want, an electrically short antenna
driving a correctly (and severely) mismatched load will probably be much
more accurate in broadband, but will still need calibration. And I
suspect it's not what you want.

I suspect that is true.

Do you have the ARRL antenna book? Have you read it?

I will have to slip out to my local library and paw their copy.
Thanks for your help.

--Winston

Winston · May 4, 2012

Winston wrote:

Tim Wescott wrote:

(...)

Okay. Right now I'd happily settle for an equation published by
enthusiasts of trans-gendered seafood, if it were correct.

I didn't have to go that far or even get into the jalopy.

From:
http://www.phys.hawaii.edu/~anita/new/papers/militaryHandbook/pwr-dens.pdf

Given a transmitting antenna with a gain of 10, a 100 W
transmitter will have an Effective Radiated Power of:

ERP = 10*log(100/.001) = 50 dBm

If our receiver is located 100 feet (or 3047.85 cm) away
from the transmitter, we will measure a power density of:

Pd =(100,000*10)/(4*PI()*3047.85^2) = 0.0086 mW/cm^2
^
(Where transmitted power is expressed in mW)

I'm so happy I could eat a tuna!

Thanks.

--Winston<--MMmmm Sushi.

Fred Abse · May 5, 2012

On Fri, 04 May 2012 11:08:20 -0500, Tim Wescott wrote:

I'm pretty sure that something like a quarter-wave dipole is only going to
be accurate at it's resonant frequency,

Correct, also, only in its direction of maximum radiation/sensitivity is
its gain figure correct.

and its sensitivity is going to be
in volts/wavelength, not watts/area. Of course there will be a
conversion, though.

They're the same thing (assuming you mean volts/meter, which is actually
not per wavelength, but per lineal meter):

Watts/area = volts/length squared divided by the intrinsic resistance of
free space (120*pi ~ 377 ohms).

W=V^2/R again!

The same applies to H field: watts/area = amp(turn)s/length squared times
377.

W=I^2R

SQRT(area) has dimensions of length.

SI units assumed.

The biggie here is that most simple formulas assume plane wave conditions.
ie. far-field. Near-field conditions are more complicated. I suspect that
the OP is trying to measure near fields.

Most amateur radio publications only concern themselves with far-field
conditions.

"Telecommunications Engineering", Duncan and Smith, (Van Nostrand) has
appropriate information, ISBN 0-442-30585.

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

Winston · May 5, 2012

Fred Abse wrote:

On Fri, 04 May 2012 11:08:20 -0500, Tim Wescott wrote:

I'm pretty sure that something like a quarter-wave dipole is only going to
be accurate at it's resonant frequency,

Correct, also, only in its direction of maximum radiation/sensitivity is
its gain figure correct.

This'll be a 1/4 wave ground plane so I assume
it'll receive equally well or poorly in all
directions.

and its sensitivity is going to be
in volts/wavelength, not watts/area. Of course there will be a
conversion, though.

They're the same thing (assuming you mean volts/meter, which is actually
not per wavelength, but per lineal meter):

Watts/area = volts/length squared divided by the intrinsic resistance of
free space (120*pi ~ 377 ohms).

W=V^2/R again!

The same applies to H field: watts/area = amp(turn)s/length squared times
377.

W=I^2R

SQRT(area) has dimensions of length.

SI units assumed.

The biggie here is that most simple formulas assume plane wave conditions.
ie. far-field. Near-field conditions are more complicated. I suspect that
the OP is trying to measure near fields.

Five wavelengths at 1000 MHz is about 1.5 m.
I'm interested in field intensity at that radius
and somewhat beyond, so the far-field answer will
work for me.

Most amateur radio publications only concern themselves with far-field
conditions.

That's reasonable.

"Telecommunications Engineering", Duncan and Smith, (Van Nostrand) has
appropriate information, ISBN 0-442-30585.

Thanks Fred!

--Winston

Jim Thompson · May 5, 2012

On Sat, 05 May 2012 11:42:57 -0700, Winston <Winston@Bigbrother.net> wrote:

[snip]

The biggie here is that most simple formulas assume plane wave conditions.
ie. far-field. Near-field conditions are more complicated. I suspect that
the OP is trying to measure near fields.

Five wavelengths at 1000 MHz is about 1.5 m.
I'm interested in field intensity at that radius
and somewhat beyond, so the far-field answer will
work for me.

[snip]

--Winston

Fred Abse · May 5, 2012

On Sat, 05 May 2012 12:05:01 -0700, Jim Thompson wrote:

On Sat, 05 May 2012 11:42:57 -0700, Winston <Winston@Bigbrother.net
wrote:

[snip]

The biggie here is that most simple formulas assume plane wave
conditions. ie. far-field. Near-field conditions are more complicated.
I suspect that the OP is trying to measure near fields.

Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field
intensity at that radius and somewhat beyond, so the far-field answer
will work for me.

[snip]

--Winston

So mW/cm^2 (power-density) is your absolute Watts divided by the surface
area of a sphere with radius equal to the distance of interest (uniform
radiator assumed).

Yup.

4 pi r^2

Modern practice is to use W/m^2, V/m, A/m.

That's where the apparent paradox regarding the inverse square law
originates. Watts per square meter does fall off as the square of the
distance. Volts per meter, which is what most people call "field
strength", falls of as the first power.

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

Fred Abse · May 5, 2012

On Sat, 05 May 2012 11:42:57 -0700, Winston wrote:

This'll be a 1/4 wave ground plane so I assume it'll receive equally well
or poorly in all directions.

Unfortunately no. There's no such thing in reality as the theoretical
isotropic radiator. The E-plane polar pattern will be similar to half a
half-wave dipole.

End-on, the gain/sensitivity will be zero, with maximum at right angles to
the antenna axis. Imagine, if you will, a half bagel, lying
cream-cheese-side-down on the ground plane

Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field
intensity at that radius and somewhat beyond, so the far-field answer
will work for me.

Five wavelengths is certainly getting there. The trick is to do the math
for a uniform isotropic, then adjust for the antenna gain in the direction
of interest.

It's easier with receiving antennas to use volts/meter, rather than
watts/square meter.

For example, the EMF induced in a half-wave dipole is usually taken to be:

e = E * lambda / pi, where e is the induced emf and E is the field
strength in volts per meter.

Using your antenna's stated isotropic gain of 1dB, and the theoretical
gain of a half wave dipole (2.15dB), that means that your antenna will be
1.15dB down on E lambda / pi.

This is (open circuit) EMF. For a perfectly matched antenna, the voltage
at the receiver terminals will be half that.

To obtain watts per square meter from volts per meter, square, and divide
by 377.

Forget centimeters, they're not used in engineering anymore.

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

Winston · May 5, 2012

Jim Thompson wrote:

On Sat, 05 May 2012 11:42:57 -0700, Winston<Winston@Bigbrother.net> wrote:

[snip]

The biggie here is that most simple formulas assume plane wave conditions.
ie. far-field. Near-field conditions are more complicated. I suspect that
the OP is trying to measure near fields.

Five wavelengths at 1000 MHz is about 1.5 m.
I'm interested in field intensity at that radius
and somewhat beyond, so the far-field answer will
work for me.

[snip]

--Winston

So mW/cm^2 (power-density) is your absolute Watts divided by the surface area
of a sphere with radius equal to the distance of interest (uniform radiator
assumed).

That sounds very reasonable isotropically speaking,
though I recall that the website example was for a
directional radiator with a gain of 10.

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.

I'm here to learn though, so I *am* willing to
listen to reason.

--Winston

Winston · May 5, 2012

Fred Abse wrote:

On Sat, 05 May 2012 11:42:57 -0700, Winston wrote:

This'll be a 1/4 wave ground plane so I assume it'll receive equally well
or poorly in all directions.

Unfortunately no. There's no such thing in reality as the theoretical
isotropic radiator. The E-plane polar pattern will be similar to half a
half-wave dipole.

End-on, the gain/sensitivity will be zero, with maximum at right angles to
the antenna axis. Imagine, if you will, a half bagel, lying
cream-cheese-side-down on the ground plane

Ah. I stand corrected. (Orthopedic sneakers).

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?)

Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field
intensity at that radius and somewhat beyond, so the far-field answer
will work for me.

Five wavelengths is certainly getting there. The trick is to do the math
for a uniform isotropic, then adjust for the antenna gain in the direction
of interest.

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...) if my 1 GHz receiver is
within say 45 meters of the (gain of 10) antenna? Holy Moley!

It's easier with receiving antennas to use volts/meter, rather than
watts/square meter.

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

For example, the EMF induced in a half-wave dipole is usually taken to be:

e = E * lambda / pi, where e is the induced emf and E is the field
strength in volts per meter.

Using your antenna's stated isotropic gain of 1dB, and the theoretical
gain of a half wave dipole (2.15dB), that means that your antenna will be
1.15dB down on E lambda / pi.

This is (open circuit) EMF. For a perfectly matched antenna, the voltage
at the receiver terminals will be half that.

'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).

To obtain watts per square meter from volts per meter, square, and divide
by 377.

Excellent! That is another good piece of info.

Forget centimeters, they're not used in engineering anymore.

Thanks for your help, Fred.

I sincerely appreciate it.

--Winston

Winston · May 5, 2012

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...)'

--Winston<--Magnets on the brain.

Jim Thompson · May 5, 2012

On Sat, 05 May 2012 14:21:58 -0700, Winston <Winston@Bigbrother.net> wrote:

Jim Thompson wrote:
On Sat, 05 May 2012 11:42:57 -0700, Winston<Winston@Bigbrother.net> wrote:

[snip]

The biggie here is that most simple formulas assume plane wave conditions.
ie. far-field. Near-field conditions are more complicated. I suspect that
the OP is trying to measure near fields.

Five wavelengths at 1000 MHz is about 1.5 m.
I'm interested in field intensity at that radius
and somewhat beyond, so the far-field answer will
work for me.

[snip]

--Winston

So mW/cm^2 (power-density) is your absolute Watts divided by the surface area
of a sphere with radius equal to the distance of interest (uniform radiator
assumed).

That sounds very reasonable isotropically speaking,
though I recall that the website example was for a
directional radiator with a gain of 10.

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.

I'm here to learn though, so I *am* willing to
listen to reason.

--Winston

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

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.

So measure it.

...Jim Thompson
--

| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice

Winston · May 6, 2012

Jim Thompson wrote:

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

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.

So measure it.

Aye aye, Sir.

--Winston

Fred Abse · May 6, 2012

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)

Fred Abse · May 6, 2012

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?

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

Fred Abse · May 6, 2012

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...)'

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.

Field intensity (volt per unit length) falls inversely as the distance.

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

Fred Abse · May 6, 2012

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

Fred Abse wrote:
On Sat, 05 May 2012 11:42:57 -0700, Winston wrote:

This'll be a 1/4 wave ground plane so I assume it'll receive equally
well or poorly in all directions.

Unfortunately no. There's no such thing in reality as the theoretical
isotropic radiator. The E-plane polar pattern will be similar to half a
half-wave dipole.

End-on, the gain/sensitivity will be zero, with maximum at right angles
to the antenna axis. Imagine, if you will, a half bagel, lying
cream-cheese-side-down on the ground plane

Ah. I stand corrected. (Orthopedic sneakers).

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.

Five wavelengths at 1000 MHz is about 1.5 m. I'm interested in field
intensity at that radius and somewhat beyond, so the far-field answer
will work for me.

Five wavelengths is certainly getting there. The trick is to do the math
for a uniform isotropic, then adjust for the antenna gain in the
direction of interest.

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.

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).

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

Field intensity (volts/meter) falls off as the inverse distance.

It's easier with receiving antennas to use volts/meter, rather than
watts/square meter.

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 ;-)

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.

For example, the EMF induced in a half-wave dipole is usually taken to
be:

e = E * lambda / pi, where e is the induced emf and E is the field
strength in volts per meter.

Using your antenna's stated isotropic gain of 1dB, and the theoretical
gain of a half wave dipole (2.15dB), that means that your antenna will
be 1.15dB down on E lambda / pi.

This is (open circuit) EMF. For a perfectly matched antenna, the
voltage at the receiver terminals will be half that.

'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.

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

Jim Thompson · May 6, 2012

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.

Winston · May 6, 2012

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

equation: dBm to mW/cm^2

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