How was antenna formula for uV/Meter Derived?
 

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

"Ian White, G3SEK" wrote in message ...
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).



This thread is about measuring the E field, but i've never heard
of measuring the H field as well, and then comparing the phase
difference.



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.



A degree of phase difference would be a boundary, just as the
-3dB cutoff point is a cut-off freq. for a filter.


Slick

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.

One could define far-field as the point where the E-field/H-field ratio
is and outward continues to be 377 ohm resistive.

The IEEE Dictionary defines "far-field region - The region of the field
of an antenna where the angular field distribution is essentially
independent of the distance from the antenna."
--
73, Cecil http://www.qsl.net/w5dxp



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On the Textron C band and S band antenna range[s] that I managed we had
a practical, that is PRACTICAL, limit of 1000 wavelengths to establish
the far field pattern. I don't know if it is/was technically justified
but it was contractually defined.

Deacon Dave, W1MCE

W5DXP wrote:

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.


One could define far-field as the point where the E-field/H-field ratio
is and outward continues to be 377 ohm resistive.

The IEEE Dictionary defines "far-field region - The region of the field
of an antenna where the angular field distribution is essentially
independent of the distance from the antenna."

I've been out of town since this was posted. There are a couple of
details I'd like to check in some texts before responding, so I'll
postpone my response until after I return home. In the meantime, I see
that Ian and Tom have made contributions, both to their usual high
standards. Both these gentlemen know what they're talking about, and I'm
glad to have them help out.

Roy Lewallen, W7EL

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

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.



I agree with you that the field is first converted by the antenna
before it can be measured.

But by definition, the E field is definitely related to voltage
potential.

Hugh Skillings' Fund. of Electric Waves: "Voltage from point 1 to
point 2 is the line integral of the electric field along any path from
point 1 to point 2. This is the amount by which point 1 is at a
higher potential than point 2."

Say PD = E^2/Z0 = H^2 * Z0. If you say the Power Density =
V^2/(R*m^2), and the R=Zo, then these will cancel, giving you E =
V/meter, which are the correct units. So here we are equating the
impedance of free space will a resistive impedance or load.

Roy, what do you think 1uV/meter really means in terms of how you
measure it? I mean, under what conditions must you have to measure
this 1uV/meter?

I'm starting to think that what this really means, is that an
exploring particle with a unit positive charge, when placed in a
electric field of 1uV/meter, will experience a change of voltage
potential of +1uV when it is moved directly towards an isotropic
radiator ("the potential of a point in space is the work required to
move to that point a unit positive charge, starting an infinite
distance away...potential increases as one positive charge is moved
closer to another positive charge" - Skilling).





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.)



But that's a static field, so we don't have to worry about
touching metallic objects that aren't grounded.




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



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.


Slick

Roy Lewallen wrote in message ...
I've been out of town since this was posted. There are a couple of
details I'd like to check in some texts before responding, so I'll
postpone my response until after I return home. In the meantime, I see
that Ian and Tom have made contributions, both to their usual high
standards. Both these gentlemen know what they're talking about, and I'm
glad to have them help out.

Roy Lewallen, W7EL



I look forward to your response Roy.


Slick

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

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.



I agree with you that the field is first converted by the antenna
before it can be measured.

But by definition, the E field is definitely related to voltage
potential.

Well, yes, speed (meters per second) is related to distance. Force
(Newtons) is related to work (Newton-meters). But speed isn't distance,
and force isn't work. The mass of the Earth is related to its orbital
velocity, and mass certainly isn't velocity. Worse yet, the impedance of
free space isn't a measure of the same thing as the characteristic
impedance of a transmission line. What I'm trying to illustrate is that
because two things are related doesn't make them the same thing, or
necessarily even close to the same thing.

Hugh Skillings' Fund. of Electric Waves: "Voltage from point 1 to
point 2 is the line integral of the electric field along any path from
point 1 to point 2. This is the amount by which point 1 is at a
higher potential than point 2."

Yes, that relates voltage and electric field. Don't overlook the bit
about the path, though.

Say PD = E^2/Z0 = H^2 * Z0. If you say the Power Density =
V^2/(R*m^2), and the R=Zo, then these will cancel, giving you E =
V/meter, which are the correct units. So here we are equating the
impedance of free space will a resistive impedance or load.

No, all you're doing is showing that they have the same dimensions. It
just doesn't seem to be sinking in that having the same dimensions
doesn't make two quantities the same thing. I've tried with the example
of torque and work, but that doesn't seem to be having any effect. Maybe
someone else can present some other examples, and maybe, just maybe,
with enough examples the concept will sink in.

Roy, what do you think 1uV/meter really means in terms of how you
measure it? I mean, under what conditions must you have to measure
this 1uV/meter?

As I mentioned before, it's usually measured with a short probe.

But electric field is actually defined in terms of the force on a
charge. You'll find an explanation in any basic physics text, as well as
many places on the Web. In Weidner and Sells, _Elementary Classical
Physics_, Vol. 2, the authors define electric field E as F/q, or the
force that would be exerted on a (sufficiently small) charge at the
point at which the field is being measured. They explain that the units
of electric field are newtons per coulomb which, it turns out, has the
same dimensions as volts per meter. So to your argument that electric
field is "related" to voltage, it's equally related to distance, force,
and charge. You can, in fact, find a bundle of other equivalent products
and quotients of units that are equivalent.

I'm starting to think that what this really means, is that an
exploring particle with a unit positive charge, when placed in a
electric field of 1uV/meter, will experience a change of voltage
potential of +1uV when it is moved directly towards an isotropic
radiator ("the potential of a point in space is the work required to
move to that point a unit positive charge, starting an infinite
distance away...potential increases as one positive charge is moved
closer to another positive charge" - Skilling).

Here we are again. Potential and voltage have the same dimensions, but
aren't necessarily equal. And as far as I can tell, "voltage potential"
is meaningless. To quote from Holt, _Electromagnetic Fields and Waves_,
"When the electromagnetic fields are static, as we shall see, the
voltage drop along a path equals the potential drop between the end
points of the path. Furthermore, these quantities [voltage and electric
potential] are also equal in *idealized* electric circuit diagrams, and
they are approximately equal in physical circuits, provided voltmeter
leads do not encircle appreciable time-changing magnetic flux." Pay
particular attention to the last qualification. When a time-changing
magnetic field is present, the voltage drop between two points depends
on the path taken, while the potential drop is simply the difference in
potential between the two points. So the voltage between two points in
an electromagnetic field can be just about anything you'd like it to be.

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.)



But that's a static field, so we don't have to worry about
touching metallic objects that aren't grounded.

Sorry, I don't see how that's relevant. If field and voltage were the
same, we'd be in trouble, static or not.


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).

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.

Actually, "far field" is often defined as a distance of 2L^2/lambda,
where L is the length of the antenna and lambda the wavelength, although
the authors generally admit to its being quite arbitrary (cf. Kraus,
_Antennas_). In Jordan and Balmain's _Electromagnetic Waves and
Radiating Systems_, they explain that the far field approximation is
valid at distances large compared to a wavelength and to the largest
dimension of the source. This is a somewhat conservative definition.
True far field wave characteristics occur only at an infinite distance
from a source, and how close you can come and still have the
characteristics be close enough to true far field depends on the
application as well as the antenna. So there's no strict definition. For
a lot of antennas and applications, field characteristics are close
enough to being far field at a distance of well under a wavelength. For
others, many wavelengths are required. Ian posted a good summary of
salient far field characteristics several days ago.

RF isn't any more nebulous than any other aspect of engineering.
Engineering is a practical discipline, so compromises and trade-offs are
universally necessary. Because we deal with real, physical objects and
are stuck with real measurements, the absolute precision of mathematics
and the pure sciences is never attainable. This is as true for using an
I-beam as it is for RF design. But the principles of RF are at least as
well known as the properties of I-beams. In fact, a good argument could
be made that RF is better known.

Roy Lewallen, W7EL

Roy Lewallen wrote in message ...
....
No, all you're doing is showing that they have the same dimensions. It
just doesn't seem to be sinking in that having the same dimensions
doesn't make two quantities the same thing. I've tried with the example
of torque and work, but that doesn't seem to be having any effect. Maybe
someone else can present some other examples, and maybe, just maybe,
with enough examples the concept will sink in.

I'd think the best examples come from dynamic systems, because the EM
wave is dynamic. In fact, you can define mechanical impedances, too.
You can have a long row of masses connected by springs, for example,
and spring force may be analogous to current, and mass displacement to
voltage. In a sound wave, the mass is the mass of the air (or other)
molecules, and the springs are the intermolecular forces (or elastic
collisions in air, if you will). Generally to propagate a wave,
you'll find you have two "state variables" whose states are
interrelated by differential equations, and the (a) solution to those
equations results in the description of the wave. The propagating
medium often has very low loss, and is propagating much higher power
levels than it's dissipating. So the (uV/meter)^2/377 has the units
of power, but that's not a power that's dissipated as it would be in a
377 ohm resistor...it's the power (at that frequency, or in that band)
per square meter passing that point in freespace.

The same is true, for example, of power going down a transmission
line, though the transmission line is much more dissipative than
freespace (unless it's made with superconductors). It's also true of
power transmitted by a driveshaft in a car: the torque you apply to
the input end twists the shaft, and that twist propagates down the
shaft. Yes, you can have reflections at the far end because of
impedance mismatches too.

I think a related concept is the difference between static (potential)
and dynamic energy. Both have units of energy, but they are rather
different things.

....


Here we are again. Potential and voltage have the same dimensions, but
aren't necessarily equal. And as far as I can tell, "voltage potential"
is meaningless. To quote from Holt, _Electromagnetic Fields and Waves_,
"When the electromagnetic fields are static, as we shall see, the
voltage drop along a path equals the potential drop between the end
points of the path. Furthermore, these quantities [voltage and electric
potential] are also equal in *idealized* electric circuit diagrams, and
they are approximately equal in physical circuits, provided voltmeter
leads do not encircle appreciable time-changing magnetic flux." Pay
particular attention to the last qualification. When a time-changing
magnetic field is present, the voltage drop between two points depends
on the path taken, while the potential drop is simply the difference in
potential between the two points. So the voltage between two points in
an electromagnetic field can be just about anything you'd like it to be.

It took me a long time to properly internalize that. It's not just
time-varying magnetic fields that cause trouble, either. A
temperature gradient along the voltmeter leads can cause an EMF also,
for example. A reminder: Kirchoff's voltage law is NOT that the sum
of voltage drops around a closed loop is zero, but rather that the sum
of voltage drops equals the sum of the EMFs in the loop. One such EMF
is because of any time-varying magnetic field enclosed by the loop
(and therefore may be different if you move the leads), but others are
chemical and thermal (actually two distinct thermal types). Both EMF
(electromotive force) and voltage drop have the units of "volts", but
they are not the same thing. I do like the statement in terms of a
closed loop much better than thinking about it between two points,
because Faraday's law says only that there is an EMF in any closed
loop enclosing a time-varying magnetic field, and that EMF is
proportional to the rate of change of the magnetic field. It does not
say that the EMF is uniformly distributed, nor that it is in any one
place, only that it exists. And Kirchoff's voltage law gives us a way
to apply that EMF (and any other EMFs which may also be in the same
loop because of other things happening) and understand why we see
voltage drops around the loop. Ohm's law tells us a lot about how the
drops will be distributed, according to the resistances in each
portion of the path. Faraday's law applies whether there is a
conductor in the loop or not. If there is a conductor the whole way
around, the EMF will drive a current in it--and that current will
create a magnetic field to oppose the one creating the EMF.

Cheers,
Tom

Tom Bruhns pointed out in an email that the statement quoted below isn't
a very good example. Because the mass of the Sun is so much greater than
that of the Earth, the Earth's orbital velocity isn't significantly
affected by the mass of the Earth.

So I'll restate that as follows:

The distance between the Earth and the Sun and the orbital velocity of
the Earth are related, but distance and velocity aren't the same thing.

Or,

The mass of the Sun and the orbital velocity of the Earth are related,
but the two aren't the same thing.

You get the drift.

Thanks, Tom, for pointing this out.

Roy Lewallen, W7EL

Roy Lewallen wrote:
. . .
The mass of the Earth is related to its orbital
velocity, and mass certainly isn't velocity. . .

I think I've done about as much as I can here, and the time spent is
getting out of proportion to the communication achieved. It's time for
someone else to take a crack at it, or for it to be taken to another
newsgroup.

Roy Lewallen, W7EL

Roy Lewallen wrote in message ...
I think I've done about as much as I can here, and the time spent is
getting out of proportion to the communication achieved. It's time for
someone else to take a crack at it, or for it to be taken to another
newsgroup.

Roy Lewallen, W7EL


Thanks for your input Roy.


Slick

Roy Lewallen wrote in message ...

But by definition, the E field is definitely related to voltage
potential.

Well, yes, speed (meters per second) is related to distance. Force
(Newtons) is related to work (Newton-meters). But speed isn't distance,
and force isn't work. The mass of the Earth is related to its orbital
velocity, and mass certainly isn't velocity. Worse yet, the impedance of
free space isn't a measure of the same thing as the characteristic
impedance of a transmission line. What I'm trying to illustrate is that
because two things are related doesn't make them the same thing, or
necessarily even close to the same thing.


I never said that the E field IS a voltage, only that they are
related, and that you can measure the strength of the E field by
measuring voltages, which is what you really do with a field strength
meter anyways.




Say PD = E^2/Z0 = H^2 * Z0. If you say the Power Density =
V^2/(R*m^2), and the R=Zo, then these will cancel, giving you E =
V/meter, which are the correct units. So here we are equating the
impedance of free space will a resistive impedance or load.

No, all you're doing is showing that they have the same dimensions. It
just doesn't seem to be sinking in that having the same dimensions
doesn't make two quantities the same thing. I've tried with the example
of torque and work, but that doesn't seem to be having any effect. Maybe
someone else can present some other examples, and maybe, just maybe,
with enough examples the concept will sink in.



Are you saying that Newton*Meters are different depending on the
situation? I don't think so. A Newton is a Newton. A meter should
always be a meter. And an Ohm should always be an Ohm.

I'm not saying that the impedance of free space is relating
voltage to current, because there is no current flow in free space.
But why do they use the same units (Ohms) to describe the relation of
the E field to the H field?

Hugh Skilling - "Intrinsic impedance is somewhat analogous to the
characteristic impedance of a transmission line. It has the
dimensions of Ohms, for the E is Volts/Meter and H is Amperes/meter."

Perhaps I'm asking the wrong NG.




But electric field is actually defined in terms of the force on a
charge. You'll find an explanation in any basic physics text, as well as
many places on the Web. In Weidner and Sells, _Elementary Classical
Physics_, Vol. 2, the authors define electric field E as F/q, or the
force that would be exerted on a (sufficiently small) charge at the
point at which the field is being measured. They explain that the units
of electric field are newtons per coulomb which, it turns out, has the
same dimensions as volts per meter. So to your argument that electric
field is "related" to voltage, it's equally related to distance, force,
and charge. You can, in fact, find a bundle of other equivalent products
and quotients of units that are equivalent.



The E field is certainly related to all these things. If you
could somehow accurately measure the repelling force on a unit
positive charge, that a static positive charge exerts upon it, then
you would know what the E field is in Newtons/Coulomb, which would
also be the Volts/meter.

If you could also measure the voltage potential of this unit
positive charge in the same positive field, at two different points
that are 1 meter apart (normal to the gradient of the E field), then
you should get the same answer as above.

This second situation is what i'm asking about when i say where
would the 1uV be measured. But in the real world, it would seem that
most field strength meters have amplifiers (LNAs) either before or
after rectification by Schottky diodes.

A field strength meter seems to relate the received power with
the field strength, but not actually measuring the uV/meter directly.
Certainly they must calibrate these instruments with a known amount of
power into an antenna of known characteristics, in an anechoic
chamber, at a fixed distance away, etc.




Here we are again. Potential and voltage have the same dimensions, but
aren't necessarily equal. And as far as I can tell, "voltage potential"
is meaningless. To quote from Holt, _Electromagnetic Fields and Waves_,
"When the electromagnetic fields are static, as we shall see, the
voltage drop along a path equals the potential drop between the end
points of the path. Furthermore, these quantities [voltage and electric
potential] are also equal in *idealized* electric circuit diagrams, and
they are approximately equal in physical circuits, provided voltmeter
leads do not encircle appreciable time-changing magnetic flux." Pay
particular attention to the last qualification. When a time-changing
magnetic field is present, the voltage drop between two points depends
on the path taken, while the potential drop is simply the difference in
potential between the two points. So the voltage between two points in
an electromagnetic field can be just about anything you'd like it to be.


I was simplifying the situation with the static E field case, but
the change of 1uV for every meter moved normal to the gradient of the
field can also apply to the AC situation, except that the +/- 1uV
would apply the RMS values.





RF isn't any more nebulous than any other aspect of engineering.
Engineering is a practical discipline, so compromises and trade-offs are
universally necessary. Because we deal with real, physical objects and
are stuck with real measurements, the absolute precision of mathematics
and the pure sciences is never attainable. This is as true for using an
I-beam as it is for RF design. But the principles of RF are at least as
well known as the properties of I-beams. In fact, a good argument could
be made that RF is better known.

Roy Lewallen, W7EL


You wrote: "(The far field boundary depends on the nature of the
radiating structure, and is nebulous anyway.)"

So i used the word "nebulous" after you did.

You're talking to a B.S. in Mechanical, so i know a bit about
I-beams, but i've done only RF since college, so I'm more of a EE now.

Among the Top Ten blunders of my life: not doing Electrical.


Slick