Well, let's see. We can start with an isotropic antenna, which
distributes its power equally in all directions. I did that one three
days ago on this newsgroup, in the thread "Theoretical antenna
question". The result is that the power density from an isotropic source
at any distance r is

PD = P / (4 * pi * r^2)

where P is the total power radiated. Power density PD will be in
watts/square meter if P is in watts and r is in meters.

In the far field, the field strength E from any antenna is sqrt(PD *
Z0), where Z0 is the impedance of free space, very nearly 120 * pi ohms.
E is in volts/meter if PD is watts/meter^2 and Z0 is in ohms.
Substituting in the first equation gives

E = sqrt[(P * Z0) / (4 * pi * r^2)] ~ sqrt(30 * P) / r

So that's the field strength from an isotropic source. In free space,
the power density from a dipole in its most favorable direction is 1.64
times the power density from an isotropic source at the same distance.
This is the dipole's directivity which, if it's lossless, is the same as
the gain (2.15 dBi). You can find this derivation in nearly any antenna
text. So the field from a free-space dipole in its best direction is

E = sqrt(1.64 * 30 * P) / r = 7.01 * sqrt(P) / r

Roy Lewallen, W7EL

Dr. Slick wrote:
Hi Folks,

Someone once mentioned the following to me:


"You can calculate field strength from power and distance according to
this formula I found in an old broadcast engineering handbook...

e = 7 * sqrt(P) / d

where e = field strength in volts/meter, d = distance in meters, P =
power in watts. Antenna is assumed to be 1/2 wave dipole."


I'd like to know exactly how this simple formula was derived.

Any info greatly appreciated.


Thanks,

Slick

Roy Lewallen wrote in message ...
Well, let's see. We can start with an isotropic antenna, which
distributes its power equally in all directions. I did that one three
days ago on this newsgroup, in the thread "Theoretical antenna
question". The result is that the power density from an isotropic source
at any distance r is

PD = P / (4 * pi * r^2)

where P is the total power radiated. Power density PD will be in
watts/square meter if P is in watts and r is in meters.



That's just the power divided by the surface area of the outwardly
traveling EM wave that is a perfect sphere in the case of an isotropic
raditator.



In the far field, the field strength E from any antenna is sqrt(PD *
Z0), where Z0 is the impedance of free space, very nearly 120 * pi ohms.
E is in volts/meter if PD is watts/meter^2 and Z0 is in ohms.
Substituting in the first equation gives

E = sqrt[(P * Z0) / (4 * pi * r^2)] ~ sqrt(30 * P) / r



This is more proof that the "transformer" action between two
antennas is highly dependant on the impedance of the medium between
them.

Roy, i don't mean to be an overly inquisitive laid-off engineer
with too much time on my hands, but how was E = sqrt (PD * Zo)
derived exactly? This is really the key equation.

Here's something else I'm wondering about. If you get an answer
of 1 uV/meter, does this mean that a perfect conductor of 1 meter
length placed in this field (polarized with the E field) will measure
1uV RMS if you measure the AC voltage on the ends? In the real world,
what sort of receiving antenna do they use to measure E fields?
Obviously, the recieve antenna will affect the measurement...perhaps
you want something broadband, so as not to favor a particular
frequency (a resonance on the receive antenna will throw off the
reading)? Perhaps something as isotropic as possible, so orientation
is not as critical. How does the FCC measure it, what equipment do
they use?



Slick

Roy Lewallen wrote in message ...

In the far field, the field strength E from any antenna is sqrt(PD *
Z0), where Z0 is the impedance of free space, very nearly 120 * pi ohms.
E is in volts/meter if PD is watts/meter^2 and Z0 is in ohms.


Oh, i think it's something like this:

Power density is in Watts per square meter, or V**2/(R*m**2).
If you accept that R = impedance of free space = Z0, then these will
cancel, and you are left with E = sqrt (V**2/m**2) = V/m.


Slick

Dr. Slick wrote:
Roy Lewallen wrote in message ...

Well, let's see. We can start with an isotropic antenna, which
distributes its power equally in all directions. I did that one three
days ago on this newsgroup, in the thread "Theoretical antenna
question". The result is that the power density from an isotropic source
at any distance r is

PD = P / (4 * pi * r^2)

where P is the total power radiated. Power density PD will be in
watts/square meter if P is in watts and r is in meters.




That's just the power divided by the surface area of the outwardly
traveling EM wave that is a perfect sphere in the case of an isotropic
raditator.

Right.


In the far field, the field strength E from any antenna is sqrt(PD *
Z0), where Z0 is the impedance of free space, very nearly 120 * pi ohms.
E is in volts/meter if PD is watts/meter^2 and Z0 is in ohms.
Substituting in the first equation gives

E = sqrt[(P * Z0) / (4 * pi * r^2)] ~ sqrt(30 * P) / r




This is more proof that the "transformer" action between two
antennas is highly dependant on the impedance of the medium between
them.

There's no doubt, and certainly no disagreement from me, that the mutual
coupling or "transformer action" between two antennas is strongly
affected by the medium between them.

Roy, i don't mean to be an overly inquisitive laid-off engineer
with too much time on my hands, but how was E = sqrt (PD * Zo)
derived exactly? This is really the key equation.

The power density is related to E and H fields by the Poynting vector, where

PD = E X H

I'm not going to derive this one -- you can find it in any
electromagnetics text.

In the far field in a lossless medium, E is in time phase with H.
Consequently, the magnitude of PD is simply |E| * |H|. The latter two
have to be RMS values for PD in watts; for peak values, you need an
addition factor of 1/2. Again in the far field, E/H = Z0, where Z0 is
the impedance of the medium. The definition of Z0 is generally defined
in terms of the permittivity and permeability of the medium, and the far
field E/H relationship follows from it. That's another one I won't
derive here, and that you can easily find in a text. In a lossless
medium, Z0 is purely real, making the math simple.

So, (dealing now only with magnitudes) given that PD = E * H and Z0 =
E/H, it follows that PD = E^2/Z0 = H^2 * Z0. Solving for E gives the
equation you're asking about.

Here's something else I'm wondering about. If you get an answer
of 1 uV/meter, does this mean that a perfect conductor of 1 meter
length placed in this field (polarized with the E field) will measure
1uV RMS if you measure the AC voltage on the ends?

No, it doesn't quite work out that way, because of the triangular
current distribution on the 1 meter wire (assuming that the wavelength
is 1 meter). The "effective height" of a wire that's short in terms
of wavelength turns out to be 1/2 the actual length.

In the real world,
what sort of receiving antenna do they use to measure E fields?

Near field E intensity is typically measured with a short probe. Far
field measurement is done with conventional antennas. In the far field,
once the E, H, or power density is known, the other two can be calculated.

Obviously, the recieve antenna will affect the measurement...perhaps
you want something broadband, so as not to favor a particular
frequency (a resonance on the receive antenna will throw off the
reading)? Perhaps something as isotropic as possible, so orientation
is not as critical. How does the FCC measure it, what equipment do
they use?

To my knowledge, the FCC doesn't do any measurements. Test labs doing
far field measurements typically use a conical dipole for the HF range,
and log periodic antenna for VHF and UHF. Although these are inherently
broadband, the dipole in particular varies a great deal with frequency.
So each antenna comes with a correction factor table. That's why EMI
measurement antennas, though simple, are expensive.

Some of the FCC Part 15 measurements I've been involved with are
actually done within the near field, but are done with standard antennas
nonetheless. Although the conversion from power density to field
strength isn't entirely valid, everybody plays by the same rules. I
think some of the FCC rules for safety now required for amateurs are
also in this category.

I haven't seen quantitative near field measurements being done, just
qualitative ones using a short probe.

Roy Lewallen, W7EL

Implicit in Roy's answer, but perhaps worth stating, is that the
result is for an antenna in freespace. If you are working with an
antenna over earth, things will be different because the energy is
radiated into a hemisphere instead of isotropically, and because of
interference and diffraction patterns, and absorption by the earth,
vegetation, etc. At some frequencies, absorption by things in the
atmosphere attenuates the signal, too.

Cheers,
Tom

Roy Lewallen wrote in message ...
Well, let's see. We can start with an isotropic antenna, which
distributes its power equally in all directions. I did that one three
days ago on this newsgroup, in the thread "Theoretical antenna
question". The result is that the power density from an isotropic source
at any distance r is

PD = P / (4 * pi * r^2)

where P is the total power radiated. Power density PD will be in
watts/square meter if P is in watts and r is in meters.

In the far field, the field strength E from any antenna is sqrt(PD *
Z0), where Z0 is the impedance of free space, very nearly 120 * pi ohms.
E is in volts/meter if PD is watts/meter^2 and Z0 is in ohms.
Substituting in the first equation gives

E = sqrt[(P * Z0) / (4 * pi * r^2)] ~ sqrt(30 * P) / r

So that's the field strength from an isotropic source. In free space,
the power density from a dipole in its most favorable direction is 1.64
times the power density from an isotropic source at the same distance.
This is the dipole's directivity which, if it's lossless, is the same as
the gain (2.15 dBi). You can find this derivation in nearly any antenna
text. So the field from a free-space dipole in its best direction is

E = sqrt(1.64 * 30 * P) / r = 7.01 * sqrt(P) / r

Roy Lewallen, W7EL

Dr. Slick wrote:
Hi Folks,

Someone once mentioned the following to me:


"You can calculate field strength from power and distance according to
this formula I found in an old broadcast engineering handbook...

e = 7 * sqrt(P) / d

where e = field strength in volts/meter, d = distance in meters, P =
power in watts. Antenna is assumed to be 1/2 wave dipole."


I'd like to know exactly how this simple formula was derived.

Any info greatly appreciated.


Thanks,

Slick

Thanks for making that clear, Tom. I should also have emphasized that
it's valid only in the far field, where the wave is essentially a plane
wave with E/H equal to the impedance of the medium.

Roy Lewallen, W7EL

Tom Bruhns wrote:
Implicit in Roy's answer, but perhaps worth stating, is that the
result is for an antenna in freespace. If you are working with an
antenna over earth, things will be different because the energy is
radiated into a hemisphere instead of isotropically, and because of
interference and diffraction patterns, and absorption by the earth,
vegetation, etc. At some frequencies, absorption by things in the
atmosphere attenuates the signal, too.

Cheers,
Tom

Roy Lewallen wrote in message ...

The power density is related to E and H fields by the Poynting vector, where

PD = E X H

I'm not going to derive this one -- you can find it in any
electromagnetics text.

In the far field in a lossless medium, E is in time phase with H.
Consequently, the magnitude of PD is simply |E| * |H|. The latter two
have to be RMS values for PD in watts; for peak values, you need an
addition factor of 1/2. Again in the far field, E/H = Z0, where Z0 is
the impedance of the medium. The definition of Z0 is generally defined
in terms of the permittivity and permeability of the medium, and the far
field E/H relationship follows from it. That's another one I won't
derive here, and that you can easily find in a text. In a lossless
medium, Z0 is purely real, making the math simple.

So, (dealing now only with magnitudes) given that PD = E * H and Z0 =
E/H, it follows that PD = E^2/Z0 = H^2 * Z0. Solving for E gives the
equation you're asking about.



Thanks Roy, i'm gonna look up these derivations.



Here's something else I'm wondering about. If you get an answer
of 1 uV/meter, does this mean that a perfect conductor of 1 meter
length placed in this field (polarized with the E field) will measure
1uV RMS if you measure the AC voltage on the ends?

No, it doesn't quite work out that way, because of the triangular
current distribution on the 1 meter wire (assuming that the wavelength
is 1 meter). The "effective height" of a wire that's short in terms
of wavelength turns out to be 1/2 the actual length.



Well, that sounds reasonable and is why i asked what people used
on the receiving end for field-strength measurements. Would the above
be valid for wavelength than one meter? Perhaps i'm totally off
because the 1 uV/meter might mean that one would measure the 1uV RMS
across 1 meter of the medium in question (377 Ohms for free-space, no
wire involved at all?)?



In the real world,
what sort of receiving antenna do they use to measure E fields?

Near field E intensity is typically measured with a short probe. Far
field measurement is done with conventional antennas. In the far field,
once the E, H, or power density is known, the other two can be calculated.



Short probe that may be much shorter than one wavelength,
unfortunately?



Obviously, the recieve antenna will affect the measurement...perhaps
you want something broadband, so as not to favor a particular
frequency (a resonance on the receive antenna will throw off the
reading)? Perhaps something as isotropic as possible, so orientation
is not as critical. How does the FCC measure it, what equipment do
they use?

To my knowledge, the FCC doesn't do any measurements. Test labs doing
far field measurements typically use a conical dipole for the HF range,
and log periodic antenna for VHF and UHF. Although these are inherently
broadband, the dipole in particular varies a great deal with frequency.
So each antenna comes with a correction factor table. That's why EMI
measurement antennas, though simple, are expensive.



That's what i wanted to read, that they use a correction factor
table so that they can take out the natural resonance(s) of the
receive antenna. And it makes total sense that they would use a
different antenna for each frequency range.

The tricky part of making the correction factor table would be
that ideally, you would have an isotropic radiator that was perfectly
wide-band, and feed this with known input powers. This being
impossible, i would imagine that they might take advantage of the
reciprocity of antennas, and use the same antenna for both transmit
and receive, and then divide the correction factor by two, 1/2
correction for each antenna.



Some of the FCC Part 15 measurements I've been involved with are
actually done within the near field, but are done with standard antennas
nonetheless. Although the conversion from power density to field
strength isn't entirely valid, everybody plays by the same rules. I
think some of the FCC rules for safety now required for amateurs are
also in this category.

I haven't seen quantitative near field measurements being done, just
qualitative ones using a short probe.

Roy Lewallen, W7EL


Perhaps the far-field measurements would require too sensitive a
field-strength meter? Or maybe it's just more convenient to measure
up close.


Slick

Dr. Slick wrote:
Roy Lewallen wrote in message ...
. . .

No, it doesn't quite work out that way, because of the triangular
current distribution on the 1 meter wire (assuming that the wavelength
is 1 meter). The "effective height" of a wire that's short in terms
of wavelength turns out to be 1/2 the actual length.




Well, that sounds reasonable and is why i asked what people used
on the receiving end for field-strength measurements. Would the above
be valid for wavelength than one meter? Perhaps i'm totally off
because the 1 uV/meter might mean that one would measure the 1uV RMS
across 1 meter of the medium in question (377 Ohms for free-space, no
wire involved at all?)?

It's valid only as long as the length of the wire is a half
wavelength (actually, 1/4 wavelength for a single wire, half
wavelength for a dipole).

Here's the problem with that transformer concept again. A field is not a
voltage. So you can't measure it with a voltmeter. You can convert the
fields to voltages and currents by use of a transducer -- an antenna --
then you can measure the voltage and current from the antenna with
ordinary meters.

Good thing, too. Otherwise we'd all get electrocuted by the Earth's 100
volt/meter field. (And that's on a day with no storm nearby.)

The relationship between the fields, voltages, and currents is nicely
expressed by Maxwell's equations.





In the real world,

what sort of receiving antenna do they use to measure E fields?

Near field E intensity is typically measured with a short probe. Far
field measurement is done with conventional antennas. In the far field,
once the E, H, or power density is known, the other two can be calculated.




Short probe that may be much shorter than one wavelength,
unfortunately?

Yep.

. . .


Some of the FCC Part 15 measurements I've been involved with are
actually done within the near field, but are done with standard antennas
nonetheless. Although the conversion from power density to field
strength isn't entirely valid, everybody plays by the same rules. I
think some of the FCC rules for safety now required for amateurs are
also in this category.

I haven't seen quantitative near field measurements being done, just
qualitative ones using a short probe.

Roy Lewallen, W7EL



Perhaps the far-field measurements would require too sensitive a
field-strength meter? Or maybe it's just more convenient to measure
up close.

No, it's far field measurements that are more common. One problem with
making near field measurements is that the near field varies all over
the map with the type of antenna and the exact spot where you're making
the measurement. And it's of no importance at all to anything very far
away at all. I've only seen near field probing done to locate the source
of a problem emission. Compliance measurements are usually done with
far-field techniques, in or at least at the fringes of the far field.
The "within the near field" measurements I'm referring to are HF
measurements done at distances that aren't firmly in the far field. (The
far field boundary depends on the nature of the radiating structure, and
is nebulous anyway.) The FCC addresses this issue for Part 15 somewhat
in section 15.31(f).

Roy Lewallen, W7EL

Dr. Slick wrote:


I was going to ask you to define "far-field", and i thought maybe
people defined this as a number of wave-legnths away, but if it's
nebulous like a lot of RF topics, then i would certainly understand.


It's not nebulous at all. There is no boundary fence in the ether with a
sign saying "Here endeth the near field", but you *can* draw your own
lines.

That is not at all unusual in physics and engineering, nor is it limited
to RF problems.

In the idealized far field, the E and H fields are orthogonal, in phase
and have a ratio of 377 ohms. At any finite distance, you know that
you're not in the idealized far field, but until you come quite close to
the antenna you can't measure the difference in any way, so ideal
far-field conditions are a very good assumption. Coming closer to the
antenna, you enter the transition zone where you start to see measurable
and calculable deviations, but you can set an engineering criterion to
say you're still close enough to far-field conditions (a 1 degree E-H
phase difference is one example, but people can and do set different
criteria for different purposes).

Equally, you can begin in the near field and work your way outward.
You'll notice some strange behaviour of the E and H fields very close-in
(all of which turns out to be completely predictable if you try hard
enough) but farther out they get their act together and settle into
their correct far-field relationship.

The lack of ready-drawn boundaries doesn't make any of those ideas
"nebulous". They are exactly as clear - or exactly as nebulous - as the
way you choose to think about them. There's still the same rock-solid
physics underneath.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

"Ian White, G3SEK" wrote in message ...
....
In the idealized far field, the E and H fields are orthogonal, in phase
and have a ratio of 377 ohms. At any finite distance, you know that
you're not in the idealized far field, but until you come quite close to
the antenna you can't measure the difference in any way, so ideal
far-field conditions are a very good assumption.

(rest deleted...)

Again, just to make things clear, this is fine in freespace (or good
approximations of freespace), but when there are conductors and
dielectrics scattered about, it can be significantly disrupted. Of
course, all this is explained quite nicely by exactly the same
physical understanding that Ian mentioned for the fields in the
vicinity of the radiating structure. It's "only" a solution to a
bunch of differential equations and boundary conditions. Things are
just a lot simpler in freespace.

From a practical standpoint, it can be helpful to have an idea when
it's safe to say you're in the far field and when you're in the near
field. Somewhere in my files I have some articles about that
(awaiting better organization so I can actually find them! ;-). I
think the one I liked best was in "RF Design" magazine perhaps five
years ago. The estimate takes into account not only how far from the
antenna you are, but also how large the antenna is.

Cheers,
Tom