equation: dBm to mW/cm^2

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

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

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 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.

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)

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

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).
 
                                       ...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:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

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)

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)

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