equation: dBm to mW/cm^2

[Winston]

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

 

[Winston]

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.

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]

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]

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.

 

[Winston]

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.

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

 

[Fred Abse]

[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.

 

[Fred Abse]

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.

 

[Winston]

[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]

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 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]

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.

 

[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.

Aye aye, Sir.

--Winston

 

[Fred Abse]

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.

 

[Fred Abse]

Winston wrote:

(...)


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.

 

[Fred Abse]

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?

 

[Fred Abse]

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.



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.

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.

'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.

 

[Winston]

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

 

[Winston]

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

 

[Winston]

)


Yes.

Good. Thanks!

(...)

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?

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)

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.


(...)

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]

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


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.

 

[Fred Abse]

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.