volts per meter... of what?
Clifton T. Sharp Jr.
the references raised another question.) Field strength is measured in
volts per meter, but "per meter" of *what*? Wire? Isotrope? 75-element
beam? All I'm sure of is that it's not per meter of distance from the
transmitter. I'm *reasonably* sure that I can't get a volt from a 75 uV/M
signal by using 13,333 meters of wire; however, that's hard to check. :-)
The new question: does this measurement apply only to the electric field?
(Not unreasonable, as a magnetic field through a conductor does induce a
voltage.)
--
+---------------------------------------------------------------------------+
| Cliff Sharp | Hate spam? Join The Great American Pink-Out! |
| WA9PDM | http://www.ybecker.net/pink/ |
+---------------------------------------------------------------------------+
kingpop
most cases)as measured in volts, millivolts, or microvolt per meter, is
inversely proportional to the wavelength and DISTANCE away from from the
source Hence in the expression uV/M, M= distance away from source in
meters.
Josh Nickel
or an insulated pair of conductors to measure the potential at two
separate points. If you know the potential difference and the physical
separation between the two points, then the E-field is just (pot.
diff)/(dist.). This '1 meter distance' simply says that an electric
field of x volts/m will induce a potential difference of x volts on two
conductors separated 1 m apart, or 2x volts on two conductors separted by
2 m. Sort of like a car cruising at 35 miles/hour will travel 35 miles
in 1 hour, or 70 miles in 2 hours. In both cases, the measurement is
assumed to be constant over the spatial or temporal variable; i.e. the car
travels at a constant 35 miles/hour (never slowing down or speeding up)
and the E-field incident on the antenna doesn't vary with distance
(assuming the source of the radiation is far away and we only vary in
distance locally).
For ionospheric static E-field measurements with sounding rockets or
balloons they usually use two conducting spheres held 1 m apart with an
insulating rod. These are typically deployed from the rocket at a given
altitude. A very high impedance voltmeter is then used to read the
voltage, and you obtain V/m. I believe lightning researchers and
triggerers use a similar setup on the ground to measure the static fields
produced during thunderstorms; when they know the E-field is above 12 kV/m
or so, they then try to trigger the lightning with small rockets.
I worked with a lightning detection researcher who measured quasi-static
electric fields with a calibrated 1-m antenna propped up with a tripod.
Incident electric fields simply produced a potential difference on the
antenna and he measured it with a high impedance voltmeter. I never
actually saw the antenna, we worked mostly on an ELF detection coil used
for measuring the magnetic signatures of sprites and elves
(lightning-related upper-atmospheric phenomena).
This leads to the second question - the answer is yes, if the magnetic
field is static. But for time-varying radio signals and propagating
electromagnetic radiation, the magnetic field can also be detected with a
conductor. Using a coil (like that in an AM radio), the change in magnetic
flux can be detected directly (magnetic field can then be calculated from
the magnetic flux). Also, since a time-varying magnetic field induces an
electric field (Faraday's law) the magnetic field could be calculated from
the measured electric field (provided that the intrinsic impedance of the
medium - usually air - is known). Just remember that magnetic field is
measured as Amperes/m, not Volts/m. And the magnetic field can only be
detected with a conductor if the field is time-varying or if the conductor
is physically moving in the magnetic field.
An example of this occured a few years ago on the space shuttle. They
performed an experiment where a 25 MILE long cable was towed by the
shuttle as it orbited the Earth. Since the Earth has a small but finite
magnetic field, an induced EMF (voltage caused by movement of a conductor
though a magnetic field) was created on the cable. Apparently this voltage
was underestimated - something like 20 kV was induced on the cable,
causing it to snap.
HTH,
Josh
Rich Griffiths
>
> The electromagnetic field strength from a radiating source (antenna in
> most cases)as measured in volts, millivolts, or microvolt per meter, is
> inversely proportional to the wavelength and DISTANCE away from from the
> source Hence in the expression uV/M, M= distance away from source in
> meters.
>
> >
> > Checked my usual references and can't find an answer. (In fact, checking
> > the references raised another question.) Field strength is measured in
> > volts per meter, but "per meter" of *what*? Wire? Isotrope? 75-element
> > beam? All I'm sure of is that it's not per meter of distance from the
> > transmitter. I'm *reasonably* sure that I can't get a volt from a 75 uV/M
> > signal by using 13,333 meters of wire; however, that's hard to check. :-)
> >
> > The new question: does this measurement apply only to the electric field?
> > (Not unreasonable, as a magnetic field through a conductor does induce a
> > voltage.)
I believe Clifton has it about right; kingpop has got it wrong.
The figure in question (uV/m) is a gradient that is a measure of *local*
change. Except for the fact that the strength of a field (in most
practical cases) does diminish with distance from the source, this
figure has nothing to do with distance from the source.
It does indeed mean something like a 75 uV/m field will induce one Volt
in a 13,333-m wire. Of course, that's not too practical. If you tried
an experiment with a real source and a 13-km-long wire, the field
wouldn't be a constant value for that great a distance, and your
experiment would fail.
The figure is a measure of electric field strength. Magnetic field
strength is not measured in volts.
Rich W2RG
Gary Coffman
>Checked my usual references and can't find an answer. (In fact, checking
>the references raised another question.) Field strength is measured in
>volts per meter, but "per meter" of *what*? Wire? Isotrope? 75-element
>beam?
*Free space*. Remember that the wavefront is an expanding sphere
around the source. The field strength is volts per meter across *the
surface of that sphere*.
Free space has an impedance of 377 ohms. If we know the power density
on the surface of the sphere in watts per square meter at a certain distance,
and we can calulate that using the the ERP and the inverse square law, we
can then use V/m=sqrt(377*P/m^2) to give us the volts per meter field strength
at that location.
>All I'm sure of is that it's not per meter of distance from the
>transmitter.
Correct. That distance is a factor, since it determines the surface area
of the spherical wavefront at that distance, but it is not the distance
across which the voltage is measured.
>I'm *reasonably* sure that I can't get a volt from a 75 uV/M
>signal by using 13,333 meters of wire; however, that's hard to check. :-)
You can, but there is a caveat. The distance between the points on the
sphere have to be less than half a wavelength, otherwise you'll get phase
cancellation. So if the frequency is low enough that 13,333 meters is less
than half a wavelength, then you can get 1 volt from a field of 75 uV/M
using a 13,333 meter wire (more or less, it depends on the effective
aperture of the wire).
>The new question: does this measurement apply only to the electric field?
>(Not unreasonable, as a magnetic field through a conductor does induce a
>voltage.)
The measurement applies to signal *power density*, which involves both the
E-field and the H-field. When you make the measurement with a real antenna,
you use its effective aperture
A=(Gain*Lambda^2)/(4*pi)
in square meters and the received signal power to derive the watts/meter
power density.
P/m^2 = Pr/A
Then you use the impedance of free space to calculate the
volts/meter field strength
V/m =sqrt(377*P/m^2)
These equations are implemented by the meter scale calibration in the
field strength meter.
Gary
Gary Coffman KE4ZV | You make it |mail to ke...@bellsouth.net
534 Shannon Way | We break it |
Lawrenceville, GA | Guaranteed |
Gary Coffman
>The electromagnetic field strength from a radiating source (antenna in
>most cases)as measured in volts, millivolts, or microvolt per meter, is
>inversely proportional to the wavelength and DISTANCE away from from the
>source Hence in the expression uV/M, M= distance away from source in
>meters.
Incorrect. See my answer to Cliff.
Clifton T. Sharp Jr.
> The electromagnetic field strength from a radiating source (antenna in
> most cases)as measured in volts, millivolts, or microvolt per meter, is
> inversely proportional to the wavelength and DISTANCE away from from the
> source Hence in the expression uV/M, M= distance away from source in
> meters.
So if I get 75 uV/M standing one meter from a transmitter, then moving
13,333 meters away I'll get 1V/M?
Clifton T. Sharp Jr.
>
> On Sat, 26 Sep 1998 15:24:40 -0500, "Clifton T. Sharp Jr." <agen...@spambusters.ml.org> wrote:
>
> >Checked my usual references and can't find an answer. (In fact, checking
> >the references raised another question.) Field strength is measured in
> >volts per meter, but "per meter" of *what*? Wire? Isotrope? 75-element
> >beam?
>
> *Free space*. Remember that the wavefront is an expanding sphere
> around the source. The field strength is volts per meter across *the
> surface of that sphere*.
*ping* On goes the light bulb.
> Free space has an impedance of 377 ohms. If we know the power density
> on the surface of the sphere in watts per square meter at a certain distance,
> and we can calulate that using the the ERP and the inverse square law, we
> can then use V/m=sqrt(377*P/m^2) to give us the volts per meter field strength
> at that location.
*ping* Another light bulb.
> >I'm *reasonably* sure that I can't get a volt from a 75 uV/M
> >signal by using 13,333 meters of wire; however, that's hard to check. :-)
>
> You can, but there is a caveat. The distance between the points on the
> sphere have to be less than half a wavelength, otherwise you'll get phase
> cancellation. So if the frequency is low enough that 13,333 meters is less
> than half a wavelength, then you can get 1 volt from a field of 75 uV/M
> using a 13,333 meter wire (more or less, it depends on the effective
> aperture of the wire).
*ping*
> The measurement applies to signal *power density*, which involves both the
> E-field and the H-field. When you make the measurement with a real antenna,
> you use its effective aperture
>
> A=(Gain*Lambda^2)/(4*pi)
>
> in square meters and the received signal power to derive the watts/meter
> power density.
>
> P/m^2 = Pr/A
>
> Then you use the impedance of free space to calculate the
> volts/meter field strength
>
> V/m =sqrt(377*P/m^2)
>
> These equations are implemented by the meter scale calibration in the
> field strength meter.
*ping* Hey look, I'm a Christmas tree! Thanks, Gary.
Clifton T. Sharp Jr.
> HTH,
It certainly did. Thanks.
kingpop
and thought I had it right. But as thoughs more learned and eliquint
than I have pointed out I was off by a mile, I mean meter.
Clifton T. Sharp Jr. wrote:
>
> kingpop wrote:
> > The electromagnetic field strength from a radiating source (antenna in
> > most cases)as measured in volts, millivolts, or microvolt per meter, is
> > inversely proportional to the wavelength and DISTANCE away from from the
> > source Hence in the expression uV/M, M= distance away from source in
> > meters.
>
> So if I get 75 uV/M standing one meter from a transmitter, then moving
> 13,333 meters away I'll get 1V/M?
>
Richard Karlquist
Clifton T. Sharp Jr. <agen...@spambusters.ml.org> wrote:
>Checked my usual references and can't find an answer. (In fact, checking
>the references raised another question.) Field strength is measured in
>volts per meter, but "per meter" of *what*? Wire? Isotrope? 75-element
>transmitter. I'm *reasonably* sure that I can't get a volt from a 75 uV/M
>signal by using 13,333 meters of wire; however, that's hard to check. :-)
I think what you want to know is that the open circuit voltage
produced by a dipole with 1 meter long legs (i.e. total length =
2 meters) is 1 V when subjected to a field of 1 V/m, assuming that
the dipole length is small compared to a wavelength. For other lengths
and field strengths, everthing scales linearly. The reason for
the factor of 2 is that the average current in a short dipole is
half the current at the terminals. This factor approaches 1 for
dipoles with big end loading "hats" and is about 1.5 for a half wave
dipole. Hope that helps.
Rick Karlquist, N6RK
ri...@area.com
www.karlquist.com
pmarkham
>I think what you want to know is that the open circuit voltage
>produced by a dipole with 1 meter long legs (i.e. total length =
>2 meters) is 1 V when subjected to a field of 1 V/m, assuming that
>the dipole length is small compared to a wavelength. For other lengths
>and field strengths, everthing scales linearly. The reason for
>the factor of 2 is that the average current in a short dipole is
>half the current at the terminals. This factor approaches 1 for
>dipoles with big end loading "hats" and is about 1.5 for a half wave
>dipole. Hope that helps.
This thread is destroying some of my ignorance. Thanks.
I am now prompted to ask the following questions:
What is the standard to which all measurements are scaled? Is there a simple
formula to express that relationship?
If the measuring dipole of of 2 meters in length is shorter (to what degree?)
than 1/2 wavelength at the frequency of measurement, and it measures "The
field strength (is) "in" volts per meter across *the surface of that sphere*."
then it appears, to me, reasonable to assume the 2 meter long measuring
dipole does not provide a directly comparable fs voltage at frequencies at
which the measuring dipole is longer than 1/2 wavelength due to "The distance
between the points on the sphere have to be less than half a wavelength,
otherwise you'll get phase cancellation."
How does one measure fs at hf frequencies or at vhf or uhf and compare the
measurements? If one designs and builds an antenna at vhf or uhf frequencies
to develop an antenna for hf, I suspect measurements can be made at those vhf
and uhf frequencies that will translate to equivalent meaningful values at hf.
TIA,
Pete/wa4hei
Clifton T. Sharp Jr.
> Clifton T. Sharp Jr. <agen...@spambusters.ml.org> wrote:
> >Checked my usual references and can't find an answer. (In fact, checking
> >the references raised another question.) Field strength is measured in
> >volts per meter, but "per meter" of *what*? Wire? Isotrope? 75-element
> >beam? All I'm sure of is that it's not per meter of distance from the
> >transmitter. I'm *reasonably* sure that I can't get a volt from a 75 uV/M
> >signal by using 13,333 meters of wire; however, that's hard to check. :-)
>
> produced by a dipole with 1 meter long legs (i.e. total length =
> 2 meters) is 1 V when subjected to a field of 1 V/m, assuming that
> the dipole length is small compared to a wavelength.
That just answered the only question I had remaining (and your timing is
fortuitous, because I just thought of it).
> For other lengths
> and field strengths, everthing scales linearly. The reason for
> the factor of 2 is that the average current in a short dipole is
> half the current at the terminals. This factor approaches 1 for
> dipoles with big end loading "hats" and is about 1.5 for a half wave
> dipole. Hope that helps.
Greatly. My brain seems to be working again. So, another question for the
merry group...
Are there cases where the performance of an isotrope or dipole, properly
tuned and matched for the test frequency, significantly differs from the
expected results as measured properly by the "short antenna" of the test
equipment? (Other than obstructions, propagation effects, etc., i.e. just
conditions as measured by the field strength tests.)
Clifton T. Sharp Jr.
> This thread is destroying some of my ignorance. Thanks.
You're not alone. :-)
> If the measuring dipole of of 2 meters in length is shorter (to what degree?)
> than 1/2 wavelength at the frequency of measurement, and it measures "The
> field strength (is) "in" volts per meter across *the surface of that sphere*."
> then it appears, to me, reasonable to assume the 2 meter long measuring
> dipole does not provide a directly comparable fs voltage at frequencies at
> which the measuring dipole is longer than 1/2 wavelength due to "The distance
> between the points on the sphere have to be less than half a wavelength,
> otherwise you'll get phase cancellation."
Let me take a stab at this question; that way if I stick my foot in my
mouth, I'll know I don't have it right yet.
The measuring antenna has to be *much* shorter than a wavelength, the
reason being that small changes in its length should not vary the quantity
you get by dividing the number of volts received by the length of the
antenna. If the antenna was a significant fraction of a wavelength, then
a small change in its size would make a nonproportional change in the
field strength measurement. In other words, if you make the antenna 10%
longer and that makes it 0.2% of a wavelength longer, you're definitely
avoiding resonance effects, while if you change the antenna 10% and it
makes it 30% of a wavelength longer, you're using too long an antenna.
It would follow that special techniques would have to be used on high
frequencies, because the measuring antenna will eventually get so small
as the frequency rises that the test gets skewed by capture area.
Reg Edwards
--
Reg G4FGQ
http://www.btinternet.com/~g4fgq.regp
Clifton T. Sharp Jr. <agen...@spambusters.ml.org> wrote in article
<360FF4...@spambusters.ml.org>...
Gary Coffman
>In article <6umu0t$1vj$1...@vax.area.com>, ri...@area.com (Richard Karlquist) wrote:
>
>>produced by a dipole with 1 meter long legs (i.e. total length =
>>2 meters) is 1 V when subjected to a field of 1 V/m, assuming that
>>and field strengths, everthing scales linearly. The reason for
>>the factor of 2 is that the average current in a short dipole is
>>half the current at the terminals. This factor approaches 1 for
>>dipoles with big end loading "hats" and is about 1.5 for a half wave
>>dipole. Hope that helps.
>
>
>
> What is the standard to which all measurements are scaled? Is there a simple
>formula to express that relationship?
>
>than 1/2 wavelength at the frequency of measurement, and it measures "The
>field strength (is) "in" volts per meter across *the surface of that sphere*."
>then it appears, to me, reasonable to assume the 2 meter long measuring
>dipole does not provide a directly comparable fs voltage at frequencies at
>which the measuring dipole is longer than 1/2 wavelength due to "The distance
>between the points on the sphere have to be less than half a wavelength,
>otherwise you'll get phase cancellation."
>
>measurements? If one designs and builds an antenna at vhf or uhf frequencies
>to develop an antenna for hf, I suspect measurements can be made at those vhf
>and uhf frequencies that will translate to equivalent meaningful values at hf.
Pete, one does *not* attempt to measure the volts/meter directly. That is sure
to be grossly in error because you can't make the measurement between two
separated points without introducing a conductor between them, and that changes
the nature of the measurement (changes aperture). I gave the procedure in my
previous post. Here it is again.
When you make the measurement with a real antenna, you use its effective
aperture
A=(Gain*Lambda^2)/(4*pi)
in square meters and the received signal power to derive the watts/meter
power density.
P/m^2 = Pr/A
Then you use the impedance of free space to calculate the
volts/meter field strength
V/m =sqrt(377*P/m^2)
These equations are implemented by the meter scale calibration in the
field strength meter.
In other words, you use a standard gain antenna and a specially calibrated
microwattmeter. You don't use probes and a voltmeter. That's Ok for measuring
static potential, but doesn't work for measuring RF potentials because you get
induction in the probe leads which throws the measurement off.
Gary Coffman
>Are there cases where the performance of an isotrope or dipole, properly
>tuned and matched for the test frequency, significantly differs from the
>expected results as measured properly by the "short antenna" of the test
>equipment? (Other than obstructions, propagation effects, etc., i.e. just
>conditions as measured by the field strength tests.)
I'm not sure what you're asking. In free space, you'll get the expected
field strength of the theoretically ideal pattern of the antenna. Near
Earth, the pattern will be altered by the reflection from Earth, and the
field strength measuring equipment will indicate that.
pmarkham
>In other words, you use a standard gain antenna and a specially calibrated
>microwattmeter.
Thanks for your patience, Gary. I believe we are now at the core of my
conceptual ignorance relative to the hardware used for qualitative fs
measurements.
Are the "standard gain antenna and a specially calibrated microwattmeter" of
unique design and used at "all" measured frequencies within a particular
range or are they any combination of antenna and meter that gives specific
results under specific conditions?
Phrased differently; if the 2 meter long dipole is the "standard antenna" for
hf fs measurements from 1.8-30 mhz, does the calibration of the
"microwattmeter" have compensation for the difference in antenna aperture or
does one use a different "standard antenna"? Can the "standard antenna" for fs
measurements in the 1.8 to 30 mhz range be described, loosely, as a 2 meter
long dipole? What is it for 144, 220, and 430 mHz?
TIA,
Pete/wa4hei
Richard Harrison
Near Earth, the pattern will be altered by reflection from Earth, and
the field strength measuring equipment will indicate that.
Hi Gary,
I`m not sure how Earth proximity skews field strength measurements. AM
BC band measurements were almost always made just a few feet above
ground level. They were usually as predicted and repeatable. The
calculations and charts of field strength at a distance from the
antennas must take into account that the measurements will be near
ground level, just as they take into account the characteristics of the
soil, etc. Do you have any further words of wisdom on this topic?
Best regards, Richard Harrison, KB5WZI
Clifton T. Sharp Jr.
I have one. I can't see how FS measurements at AM BCB frequencies, taken
several wavelengths above ground, would be very useful. :-)
--
+---------------------------------------------------------------------------+
Richard Karlquist
Gary Coffman <ke...@bellsouth.net> wrote:
>the nature of the measurement (changes aperture). I gave the procedure in my
>previous post. Here it is again.
>
...
>
>These equations are implemented by the meter scale calibration in the
>field strength meter.
>
>In other words, you use a standard gain antenna and a specially calibrated
>microwattmeter. You don't use probes and a voltmeter. That's Ok for measuring
>static potential, but doesn't work for measuring RF potentials because you get
>induction in the probe leads which throws the measurement off.
>
>Gary
>Gary Coffman KE4ZV | You make it |mail to ke...@bellsouth.net
>534 Shannon Way | We break it |
>Lawrenceville, GA | Guaranteed |
I agree that your procedure of using a standard gain antenna will work,
but probes and a voltmeter will also work, and is the only practical
way to make a primary standard at low frequencies where you can't
build a standard gain antenna. If you look at catalogs for manufacturers
of EMI testing equipment, you will see wideband measuring antennas
consisting of a 2 meter whip with a FET source follower having
a 50 ohm output. Volts out are equal to volts/meter field. They
use these to calibrate EMI test chambers. You never need to
worry about aperture area, gain or path loss, etc.
Rick Karlquist N6RK
ri...@area.com
www.karlquist.com
Gary Coffman
>Gary, KE4ZV wrote:
>Near Earth, the pattern will be altered by reflection from Earth, and
>the field strength measuring equipment will indicate that.
>
>Hi Gary,
>I`m not sure how Earth proximity skews field strength measurements. AM
>BC band measurements were almost always made just a few feet above
>ground level. They were usually as predicted and repeatable. The
>calculations and charts of field strength at a distance from the
>antennas must take into account that the measurements will be near
>ground level, just as they take into account the characteristics of the
>soil, etc. Do you have any further words of wisdom on this topic?
What I said is that the proxmity of Earth skews antenna *patterns*.
That's perfectly true, and any antenna modeling program will show
you the effect. Thus you won't get the pattern you'd expect from a
free space model of the antenna. In situ pattern *measurement*
will, of course, give you the actual pattern.
There is room for a surprise here if you aren't very careful to allow for
the effects of proximity to Earth when comparing your measurements
to calculated free space results. The classic example of this is the
dipole over Earth. At some mounting heights, it will show significant
gain over a free space dipole.
Now here's where things can become tricky. The same thing can
happen to *your sampling antenna*. Thus its aperture won't be what
you calculated, and your field strength figures will be incorrect. You
have to be wary of this range effect and null it out by carefully calibrating
your antenna range to account for range reflections. This is tricky to do,
and very time consuming.
Richard Harrison
You have to be wary of this range effect and null it out by carefully
calibrating your antenna range to account for range reflections.
Thank you, Gary. Improvements have been made since I last ran pattern
surveys. The only thing I remember calibrating was the field strength
meter`s sensitivity, using it`s built-in calibration oscillator. But,
I`m old and forgetful.
Gary Coffman
>In article <3675e10b....@news.atl.bellsouth.net>, ke...@bellsouth.net (Gary Coffman) wrote:
>
>>microwattmeter.
>
>conceptual ignorance relative to the hardware used for qualitative fs
>measurements.
>
>Are the "standard gain antenna and a specially calibrated microwattmeter" of
>unique design and used at "all" measured frequencies within a particular
>range or are they any combination of antenna and meter that gives specific
>results under specific conditions?
>
>Phrased differently; if the 2 meter long dipole is the "standard antenna" for
>hf fs measurements from 1.8-30 mhz, does the calibration of the
>"microwattmeter" have compensation for the difference in antenna aperture or
>does one use a different "standard antenna"? Can the "standard antenna" for fs
>measurements in the 1.8 to 30 mhz range be described, loosely, as a 2 meter
>long dipole? What is it for 144, 220, and 430 mHz?
>
> TIA,
> Pete/wa4hei
For an absolute reference, there is an NBS standard antenna which can
be constructed for any particular frequency which will have a well defined
aperture. (This is a rather odd looking dipole above a metallic reference
plane.) This can be used with a standard calibrated microwattmeter.
For measurement at a different frequency, you have to scale the reference
antenna appropriately.
If you are satisfied with a secondary standard reference, you can use
any short dipole (short compared to a wavelength), and specially calibrate
the scale of the microwattmeter for the particular frequency range to be
measured. (But someone needs to have the absolute reference antenna
to do the initial calibration.)
Gary Coffman
>In article <3675e10b....@news.atl.bellsouth.net>,
>
>>the nature of the measurement (changes aperture). I gave the procedure in my
>>previous post. Here it is again.
>>
>...
>
>>
>>These equations are implemented by the meter scale calibration in the
>>field strength meter.
>>
>>static potential, but doesn't work for measuring RF potentials because you get
>>induction in the probe leads which throws the measurement off.
>
>but probes and a voltmeter will also work, and is the only practical
>way to make a primary standard at low frequencies where you can't
>build a standard gain antenna. If you look at catalogs for manufacturers
>of EMI testing equipment, you will see wideband measuring antennas
>consisting of a 2 meter whip with a FET source follower having
>a 50 ohm output. Volts out are equal to volts/meter field. They
>use these to calibrate EMI test chambers. You never need to
>worry about aperture area, gain or path loss, etc.
You have to use very special technique with this sort of sensing
arrangement in order to minimize errors due to multipath, mutual
coupling to the environment, etc. Repeatability is often poor, and
absolute accuracy even worse. The sense antenna has to be
*very* short compared to a wavelength to get any sort of reasonable
reading at all, which requires the measuring head to be very sensitive.
This technique isn't a good one at higher frequencies, though as
you note, you don't have much choice at low frequencies.
Gary Coffman
Field proofing an AM broadcast antenna doesn't require lab accuracy.
You can use fairly rough and ready techniques while still meeting FCC
requirements (if you don't get the predicted contour reading, you can
just wave the sense antenna around until you do, multipath and mutual
coupling to the local environment will assure a wide range of readings
will exist in any small area). That's not what I've been talking about.
PETER A BARRON
CODAN PTY LTD I suspect thats a down under firm
Broadband Antenna CODE 402
Is specs etc Chris VE3CBK the 19 set maniac would like info
Cheers
Peter
Ve...@iigs.net